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# In a certain baord game, a stack of 48 cards, 8 of which
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In a certain baord game, a stack of 48 cards, 8 of which [#permalink]
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24 Jul 2012, 05:19
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In a certain baord game, a stack of 48 cards, 8 of which represent shares of stock, are shuffled and then placed face down. If the first 2 cards selected do not represent shares of stock, what is the probability that the third card selected will represent a share of stock?
A. 1/8
B. 1/6
C. 1/5
D. 3/23
E. 4/23
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Re: Probability Question [#permalink]
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24 Jul 2012, 21:49
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rajman41 wrote:
Can we directly substract 2 cards out of 48 and reach the probablity of getting 8/46?
Yes you can. You already know that the first 2 cards are not stock cards. It's like we placed them face up. They anyway are out of the picture. Now we have 46 cards and 8 of them are stock cards. So the probability that the card we pick is a stock card is 8/46 = 4/23.
The probability of picking a stock card keeps increasing as we keep pulling out the non stock cards.
Think: You have 5 cards and 1 of them is the Queen. What is the probability that you will pull out a queen? It's 1/5. What happens after you pull out a non queen? The probability of pulling out a queen now becomes 1/4. It keeps increasing till you reach the last card. If you still haven't pulled out a queen, the last one must be the queen and the probability becomes 1.
Another thing to think about: What happens if you don't know what the first two cards are i.e. what is the probability of picking a share card on the third pick (you don't know what the first two picks are).
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Re: Probability Question [#permalink]
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24 Jul 2012, 05:23
rajman41 wrote:
In a certain baord game, a stack of 48 cards, 8 of which represent shares of stock, are shuffled and then placed face down. If the first 2 cards selected do not represent shares of stock, what is the probability that the third card selected will represent a share of stock?
a)1/8
b)1/6
c)1/5
d)3/23
e)4/23
Can we directly substract 2 cards out of 48 and reach the probablity of getting 8/46?
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Re: Probability Question [#permalink]
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24 Jul 2012, 19:15
My answer is E.
We can substract 2 from 48 and find the probability for the third card.
What exactly is your doubt.
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Re: Probability Question [#permalink]
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11 Oct 2012, 10:07
VeritasPrepKarishma wrote:
Another thing to think about: What happens if you don't know what the first two cards are i.e. what is the probability of picking a share card on the third pick (you don't know what the first two picks are).
Hello Karishma - I guess we can list them out?
Pick Share : Pick share : pick share =6/46
Pick share : No pick Share : pick share = 7/46
No pick Share : pick : pick = 7/46
No : No : pick = 8/46
add them all to get the overall probability?
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Re: Probability Question [#permalink]
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11 Oct 2012, 20:16
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3
Jp27 wrote:
VeritasPrepKarishma wrote:
Another thing to think about: What happens if you don't know what the first two cards are i.e. what is the probability of picking a share card on the third pick (you don't know what the first two picks are).
Hello Karishma - I guess we can list them out?
Pick Share : Pick share : pick share =6/46
Pick share : No pick Share : pick share = 7/46
No pick Share : pick : pick = 7/46
No : No : pick = 8/46
add them all to get the overall probability?
You can list it out though you don't need to. The probability will stay 8/48. It doesn't matter whether it is the first pick, the second or the third or a later pick. I have discussed this here: a-bag-contains-3-white-balls-3-black-balls-2-red-balls-100023.html?hilit=probability%20same
When you list it out, you will obviously get the same answer.
Share, Share, Share = (8/48) * (7/47) * (6/46)
Share, No Share, Share = (8/48) * (40/47) * (7/46)
No Share, Share, Share = (40/48) * (8/47) * (7/46)
No Share, No Share, Share = (40/48) * (39/47) * (8/46)
When you add all these up, you get 8/48.
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Re: In a certain baord game, a stack of 48 cards, 8 of which [#permalink]
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04 Nov 2012, 10:22
Will we not change 8 to 6??
because if we have taken out those 2 cards..then ramining will be 6
6/46?
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Re: In a certain baord game, a stack of 48 cards, 8 of which [#permalink]
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04 Nov 2012, 11:24
2
sanjoo wrote:
Will we not change 8 to 6??
because if we have taken out those 2 cards..then ramining will be 6
6/46?
Read the question carefully. These 2 cards dont represent the stocks. Hence the 8 cards that represented stocks are still there, i.e. those 8 cards are still intact. But total number of cards has reduced from 48 to 46. Hence, probability can be found easily.
Hope that helps
-s
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Re: Probability Question [#permalink]
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16 Aug 2013, 20:19
VeritasPrepKarishma wrote:
rajman41 wrote:
Can we directly substract 2 cards out of 48 and reach the probablity of getting 8/46?
Yes you can. You already know that the first 2 cards are not stock cards. It's like we placed them face up. They anyway are out of the picture. Now we have 46 cards and 8 of them are stock cards. So the probability that the card we pick is a stock card is 8/46 = 4/23.
The probability of picking a stock card keeps increasing as we keep pulling out the non stock cards.
Think: You have 5 cards and 1 of them is the Queen. What is the probability that you will pull out a queen? It's 1/5. What happens after you pull out a non queen? The probability of pulling out a queen now becomes 1/4. It keeps increasing till you reach the last card. If you still haven't pulled out a queen, the last one must be the queen and the probability becomes 1.
Another thing to think about: What happens if you don't know what the first two cards are i.e. what is the probability of picking a share card on the third pick (you don't know what the first two picks are).
i have a big trouble ..i think it should be done in this way. we select 2 from the 40 , in the first beginning. then select 1 from the 8 , in total we have 3 from 48.
so what I got was C1 8C2 40/C3 48
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Re: In a certain baord game, a stack of 48 cards, 8 of which [#permalink]
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17 Aug 2013, 11:08
1
rajman41 wrote:
In a certain baord game, a stack of 48 cards, 8 of which represent shares of stock, are shuffled and then placed face down. If the first 2 cards selected do not represent shares of stock, what is the probability that the third card selected will represent a share of stock?
A. 1/8
B. 1/6
C. 1/5
D. 3/23
E. 4/23
40/48 * 39/47
so next is = 8/46 = 4/23
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Re: In a certain baord game, a stack of 48 cards, 8 of which [#permalink]
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19 Aug 2013, 03:43
Asifpirlo wrote:
rajman41 wrote:
In a certain baord game, a stack of 48 cards, 8 of which represent shares of stock, are shuffled and then placed face down. If the first 2 cards selected do not represent shares of stock, what is the probability that the third card selected will represent a share of stock?
A. 1/8
B. 1/6
C. 1/5
D. 3/23
E. 4/23
40/48 * 39/47
so next is = 8/46 = 4/23
well done! I got it, meanwhile, I found my problem. what I did do not follow the first 2 cards and the third. what i did means just select cards randomly , two are PPPPPP, one is PPPPP...
thank you very much
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Posts: 7106
Re: In a certain board game [#permalink]
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10 Dec 2015, 07:37
amithyarli wrote:
In a certain board game, a stack of 48 cards, 8 of which represent shares of stack, are shuffled and placed face down. If the first 2 cards selected do not represent the shares of stock, what is the probability that the third card selected will represent a share of stock ?
A. 1/8
B. 1/6
C. 1/5
D. 3/23
E. 4/23
Hi,
there are 48 cards , out of which 8 are shares of stack..
the first two are not share of stack..
so the remaining cards now are 48-2=46, again out of which there are 8 shares of stock..
the prob that next is one of these 8 is 8/46=4/23
E
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Re: In a certain baord game, a stack of 48 cards, 8 of which [#permalink]
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29 Dec 2015, 09:48
VeritasPrepKarishma wrote:
rajman41 wrote:
Can we directly substract 2 cards out of 48 and reach the probablity of getting 8/46?
Yes you can. You already know that the first 2 cards are not stock cards. It's like we placed them face up. They anyway are out of the picture. Now we have 46 cards and 8 of them are stock cards. So the probability that the card we pick is a stock card is 8/46 = 4/23.
So when do you use combinatorics? I thought the solution would work out as:
Probability of third card being a stock card = 40/48* 39/47*8/46
When to use combinatorics???
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Re: In a certain baord game, a stack of 48 cards, 8 of which [#permalink]
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29 Dec 2015, 20:35
2
sarathvr wrote:
VeritasPrepKarishma wrote:
rajman41 wrote:
Can we directly substract 2 cards out of 48 and reach the probablity of getting 8/46?
Yes you can. You already know that the first 2 cards are not stock cards. It's like we placed them face up. They anyway are out of the picture. Now we have 46 cards and 8 of them are stock cards. So the probability that the card we pick is a stock card is 8/46 = 4/23.
So when do you use combinatorics? I thought the solution would work out as:
Probability of third card being a stock card = 40/48* 39/47*8/46
When to use combinatorics???
Hi,
the first two not being a stock card is no more a probability but a fact..
there is some action carried out and you have to work further to it..
out of 48 cards, you have already picked up two cards.. so the present case is that you are left with 46 cards and the prob will depend on these cards now..
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1) Absolute modulus : http://gmatclub.com/forum/absolute-modulus-a-better-understanding-210849.html#p1622372
2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html
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Re: In a certain baord game, a stack of 48 cards, 8 of which [#permalink]
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24 Apr 2016, 02:00
VeritasPrepKarishma wrote:
Another thing to think about: What happens if you don't know what the first two cards are i.e. what is the probability of picking a share card on the third pick (you don't know what the first two picks are).
Hi Karishma, can you please give the answer to the above query?
I have 2 other questions.
First question: I solved this question using conditional probability. Is my solution correct?
Probability (at least 2 non-share cards) = Probability (2 non-share AND 1 share cards) + Probability (3 non-share cards)
$$= \frac{40*39*8}{48*47*46} + \frac{40*39*38}{48*47*46}$$
$$=\frac{40*39*46}{48*47*46}$$
Desired probability = Probability(2 non-share AND 1 share cards)/Probability(at least 2 non-share cards)
$$=\frac{40*39*8}{40*47*46}$$
$$=\frac{8}{46}=\frac{4}{23}$$
Second question: I cannot tell the difference between the problem under discussion and this problem: A box contains 3 yellow balls and 5 black balls. One by one, every ball is selected at random without replacement. What is the probability that the fourth ball selected is black?
In the latter problem, Bunuel and other experts confirmed that the probability of getting a certain ball (black ball) will not change for any successive drawing. This means the probability of getting a black ball remains 5/8 no matter what.
I thought the same line of thinking would apply here. On the third drawing, the probability of getting a share remains 8/48. I know it is dead wrong, but I cannot understand the difference between the two questions. Could you please clarify?
Thank you very much!
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Re: In a certain baord game, a stack of 48 cards, 8 of which [#permalink]
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24 Apr 2016, 02:15
chetan2u wrote:
Hi,
the first two not being a stock card is no more a probability but a fact..
there is some action carried out and you have to work further to it..
out of 48 cards, you have already picked up two cards.. so the present case is that you are left with 46 cards and the prob will depend on these cards now..
Hi Chetan2u,
A silly question: How can we differentiate between a fact and a probability?
I am very confused maybe because I am not familiar with the wording. I thought the first two cards constitute a probability. Could you please give some wording examples for both 'a fact' and 'a probability' types of question? Could you please help me understand the difference between the two?
Always appreciate your help!
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Re: In a certain baord game, a stack of 48 cards, 8 of which [#permalink]
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24 Apr 2016, 02:25
2
2
truongynhi wrote:
chetan2u wrote:
Hi,
the first two not being a stock card is no more a probability but a fact..
there is some action carried out and you have to work further to it..
out of 48 cards, you have already picked up two cards.. so the present case is that you are left with 46 cards and the prob will depend on these cards now..
Hi Chetan2u,
A silly question: How can we differentiate between a fact and a probability?
I am very confused maybe because I am not familiar with the wording. I thought the first two cards constitute a probability. Could you please give some wording examples for both 'a fact' and 'a probability' types of question? Could you please help me understand the difference between the two?
Always appreciate your help!
Hi,
you have given a Q above wherein there are 3 blacks and 5 white, and we are to find the probability that 4th is black..
here you are not aware what has happened in the first three DRAWS, so picking of all 8 will remain PROBABILITY..
But what happens when you have picked two cards and you are told they do not contain the card we are lookin for..
you had 48 cards out of which 8 are say TYPE X, and you have to find the probability that third is TYPE X..
Our probability will continue to be out of all 48..
If you are told first two are not type X...
so now you have ONLY 46 cards left, Because in the TWO picked, there is no probability involved since we know what those cards are..
Probability is ONLY there where we do not know the out come..
another extension of above Q..
you have picked 40 cards and none of them are type X..
what is the probability that 41st card will be X..
we now do not care of what we already know. they have moved from Probability to REALITY..
now we know 48-40=8 cards are left and none of type X has been picked..
so now all these 8 will be type X..
prob = 8/8 = 1.. that is it will be type X for sure..
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In a certain baord game, a stack of 48 cards, 8 of which [#permalink]
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24 Apr 2016, 04:44
chetan2u wrote:
Hi,
you have given a Q above wherein there are 3 blacks and 5 white, and we are to find the probability that 4th is black..
here you are not aware what has happened in the first three DRAWS, so picking of all 8 will remain PROBABILITY..
Hi,
Thank you! I think I understand this one.
Quote:
But what happens when you have picked two cards and you are told they do not contain the card we are lookin for..
you had 48 cards out of which 8 are say TYPE X, and you have to find the probability that third is TYPE X..
Our probability will continue to be out of all 48..
I am a bit confused here. If I am told that the two cards do not contain the card I'm looking for, then how come the probability continues from 48? Doen't it start from the remaining 46 cards?
Quote:
If you are told first two are not type X...
so now you have ONLY 46 cards left, Because in the TWO picked, there is no probability involved since we know what those cards are..
Probability is ONLY there where we do not know the out come..
Doesn't this contradict with your above statement? You said we start from 48 after the first two cards. But here you point out that we calculate on the remaining 46 cards. Is it a typo or something? Please clarify.
Thank you very much!
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In a certain baord game, a stack of 48 cards, 8 of which [#permalink]
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24 Apr 2016, 04:49
1
1
truongynhi wrote:
chetan2u wrote:
Quote:
But what happens when you have picked two cards andyou are told they do not contain the card we are lookin for..
you had 48 cards out of which 8 are say TYPE X, and you have to find the probability that third is TYPE X..
Our probability will continue to be out of all 48..
I am a bit confused here. If I am told that the two cards do not contain the card I'm looking for, then how come the probability continues from 48? Doen't it start from the remaining 46 cards?
Thank you very much!
pl read the highlighted portion as -- and you are not aware what those two cards contain..
It was a typo..
Quote:
But what happens when you have picked two cards andand you are not aware what those two cards contain..
you had 48 cards out of which 8 are say TYPE X, and you have to find the probability that third is TYPE X..
Our probability will continue to be out of all 48..
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2)Combination of similar and dissimilar things : http://gmatclub.com/forum/topic215915.html
3) effects of arithmetic operations : https://gmatclub.com/forum/effects-of-arithmetic-operations-on-fractions-269413.html
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In a certain baord game, a stack of 48 cards, 8 of which [#permalink]
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24 Apr 2016, 04:59
chetan2u wrote:
But what happens when you have picked two cards andand you are not aware what those two cards contain..
you had 48 cards out of which 8 are say TYPE X, and you have to find the probability that third is TYPE X..
Our probability will continue to be out of all 48..
So, in this case, the probability that the third card is of type X will equal $$\frac{8}{48}$$. Is that right?
Thank you very much! You've been very helpful!
In a certain baord game, a stack of 48 cards, 8 of which &nbs [#permalink] 24 Apr 2016, 04:59
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https://scenv.com/dreamcatcher-you-cqq/479aff-biconditional-statement-truth-table | 0. In Example 3, we will place the truth values of these two equivalent statements side by side in the same truth table. Converse: If the polygon is a quadrilateral, then the polygon has only four sides. Let qp represent "If x = 5, then x + 7 = 11.". Create a truth table for the statement $$(A \vee B) \leftrightarrow \sim C$$ Solution Whenever we have three component statements, we start by listing all the possible truth value combinations for … BNAT; Classes. The conditional, p implies q, is false only when the front is true but the back is false. Email. Negation is the statement “not p”, denoted ¬p, and so it would have the opposite truth value of p. If p is true, then ¬p if false. b. Biconditional statement? The biconditional pq represents "p if and only if q," where p is a hypothesis and q is a conclusion. The biconditional connective can be represented by ≡ — <—> or <=> and is … Demonstrates the concept of determining truth values for Biconditionals. To show that equivalence exists between two statements, we use the biconditional if and only if. The biconditional statement $$p\Leftrightarrow q$$ is true when both $$p$$ and $$q$$ have the same truth value, and is false otherwise. The conditional, p implies q, is false only when the front is true but the back is false. Logical equality (also known as biconditional) is an operation on two logical values, typically the values of two propositions, that produces a value of true if and only if both operands are false or both operands are true.. The biconditional operator looks like this: ↔ It is a diadic operator. "x + 7 = 11 iff x = 5. Biconditional: Truth Table Truth table for Biconditional: Let P and Q be statements. Otherwise, it is false. second condition. Logical equivalence means that the truth tables of two statements are the same. 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Compound propositions involve the assembly of multiple statements, using multiple operators. A biconditional statement is often used in defining a notation or a mathematical concept. Includes a math lesson, 2 practice sheets, homework sheet, and a quiz! This video is unavailable. Construct a truth table for p↔(q∨p) A self-contradiction is a compound statement that is always false. The biconditional operator is sometimes called the "if and only if" operator. Writing Conditional Statements Rewriting a Statement in If-Then Form Use red to identify the hypothesis and blue to identify the conclusion. 3 Truth Table for the Biconditional; 4 Next Lesson; Your Last Operator! p. q . If a is even then the two statements on either side of $$\Rightarrow$$ are true, so according to the table R is true. a. A biconditional statement is often used in defining a notation or a mathematical concept. How can one disprove that statement. Let's look at a truth table for this compound statement. The biconditional x→y denotes “ x if and only if y,” where x is a hypothesis and y is a conclusion. Theorem 1. en.wiktionary.org. Similarly, the second row follows this because is we say “p implies q”, and then p is true but q is false, then the statement “p implies q” must be false, as q didn’t immediately follow p. The last two rows are the tough ones to think about. In Example 5, we will rewrite each sentence from Examples 1 through 4 using this abbreviation. We still have several conditional geometry statements and their converses from above. Definition. So the former statement is p: 2 is a prime number. • Construct truth tables for biconditional statements. ". Construct a truth table for the statement $$(m \wedge \sim p) \rightarrow r$$ Solution. Class 1 - 3; Class 4 - 5; Class 6 - 10; Class 11 - 12; CBSE. Otherwise it is false. Truth Table for Conditional Statement. Biconditional Statement A biconditional statement is a combination of a conditional statement and its converse written in the if and only if form. In the first set, both p and q are true. We can use an image of a one-way street to help us remember the symbolic form of a conditional statement, and an image of a two-way street to help us remember the symbolic form of a biconditional statement. Mathematics normally uses a two-valued logic: every statement is either true or false. Let's look at more examples of the biconditional. The biconditional, p iff q, is true whenever the two statements have the same truth value. Therefore, a value of "false" is returned. A logic involves the connection of two statements. Otherwise it is true. If p is false, then ¬pis true. A statement is a declarative sentence which has one and only one of the two possible values called truth values. Edit. Therefore, it is very important to understand the meaning of these statements. 4. Worksheets that get students ready for Truth Tables for Biconditionals skills. The truth table of a biconditional statement is. So to do this, I'm going to need a column for the truth values of p, another column for q, and a third column for 'if p then q.' If you make a mistake, choose a different button. Otherwise, it is false. V. Truth Table of Logical Biconditional or Double Implication A double implication (also known as a biconditional statement) is a type of compound statement that is formed by joining two simple statements with the biconditional operator. Sign in to vote. (a) A quadrilateral is a rectangle if and only if it has four right angles. • Construct truth tables for conditional statements. Sunday, August 17, 2008 5:10 PM. A biconditional statement is one of the form "if and only if", sometimes written as "iff". The connectives ⊤ … Compare the statement R: (a is even) $$\Rightarrow$$ (a is divisible by 2) with this truth table. Directions: Read each question below. B. 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https://math.stackexchange.com/questions/1336903/definite-integral-of-even-powers-of-cosine | # Definite integral of even powers of Cosine.
I'm looking for a step-by-step solution to the following integral, in terms of n$$\int_0^{\frac{\pi}{2}} \cos^{2n}(x) \ {dx}$$I actually KNOW that the solution is$${\frac{\pi}{2}} \prod_{k=1}^n \frac{2k-1}{2k}$$ But I would like to know how to get there. This is not homework, this is to further my own understanding. My own efforts to solve this consisted of expanding $\cos(x)$ in terms of $e^{ix}$ but this proved to be fruitless, leading me only to the following integral, where $u=e^{ix}$ $$-i\int_1^i \left({\frac{1+u^2}{2u}}\right)^{2n} \ {du}$$
• Use the binomial expansion of the integrand. You can easily obtain an answer in terms of a finite summation. – Mark Viola Jun 23 '15 at 22:33
• And in that summation all but one term has integral $0$. At least after the right initial step... – David C. Ullrich Jun 23 '15 at 22:35
• @Dr.MV, you are referring to the second integral in terms of u, correct? – JacksonFitzsimmons Jun 23 '15 at 22:36
• You'll have an answer as fast as I can type it. Hang on... – David C. Ullrich Jun 23 '15 at 22:38
• – Lucian Jun 24 '15 at 1:21
There are several approaches. One, that has been given in previous answers, uses the fact that $\cos(\theta)=\frac12(e^{i\theta}+e^{i\theta})$ and that $\int_0^{2\pi}e^{in\theta}\,\mathrm{d}\theta=2\pi$ if $n=0$, and vanishes otherwise, to get \begin{align} \int_0^{\pi/2}\cos^{2n}(\theta)\,\mathrm{d}\theta &=\frac14\int_0^{2\pi}\cos^{2n}(\theta)\,\mathrm{d}\theta\\ &=\frac1{4^{n+1}}\int_0^{2\pi}\left(e^{i\theta}+e^{i\theta}\right)^{2n}\,\mathrm{d}\theta\\ &=\frac{2\pi}{4^{n+1}}\binom{2n}{n} \end{align}
Another approach is to integrate by parts \begin{align} \int_0^{\pi/2}\cos^{2n}(\theta)\,\mathrm{d}\theta &=\int_0^{\pi/2}\cos^{2n-1}(\theta)\,\mathrm{d}\sin(\theta)\\ &=(2n-1)\int_0^{\pi/2}\sin^2(\theta)\cos^{2n-2}(\theta)\,\mathrm{d}\theta\\ &=(2n-1)\int_0^{\pi/2}\left(\cos^{2n-2}(\theta)-\cos^{2n}(\theta)\right)\,\mathrm{d}\theta\\ &=\frac{2n-1}{2n}\int_0^{\pi/2}\cos^{2n-2}(\theta)\,\mathrm{d}\theta\\ \end{align} and use induction to get $$\int_0^{\pi/2}\cos^{2n}(\theta)\,\mathrm{d}\theta =\frac\pi2\prod_{k=1}^n\frac{2k-1}{2k}$$
Note that \begin{align} \frac\pi2\prod_{k=1}^n\frac{2k-1}{2k} &=\frac\pi2\frac{(2n)!}{(2^nn!)^2}\\ &=\frac{\pi}{2^{2n+1}}\binom{2n}{n} \end{align}
• Thank you for giving me multiple methods. – JacksonFitzsimmons Jun 23 '15 at 23:23
• You're welcome. I started writing my answer before Bernard posted his answer, which also uses integration by parts. – robjohn Jun 23 '15 at 23:30
Integrate by parts, setting $$u=\cos^{2n-1}x,\enspace \operatorname{d}\mkern-2mu v=\cos x\operatorname{d}\mkern-2mu x,\enspace\text{whence}\quad \operatorname{d}\mkern-2mu u=-(2n-1)\cos^{2n-2}x \operatorname{d}\mkern-2mux,\enspace v= \sin x$$ Let's call $I_{2n}$ the integral. One obtains: $$I_{2n}=\Bigl[\sin x\mkern1mu\cos^{2n-1}x\Bigr]_0^{\tfrac\pi2}+(2n-1)(I_{2n-2}-I_{2n})=(2n-1)(I_{2n-2}-I_{2n})$$ whence the recurrence relation: $$I_{2n}=\frac{2n-1}{2n}I_{2n-2}.$$ Now write all these relations down to $n=1$, multiply the equalities thus obtained and simplify.
First, note that $\int_0^{\pi/2}=\frac14\int_0^{2\pi}$ by various symmetries.
Now say $\cos(t)=\frac12(e^{it}+e^{-it})$ and apply the binomial theorem. You get terms consisting of various powers of $e^{it}$. All those terms have integral $0$ except the middle one: $(e^{it})^n(e^{-it})^n=1$. So you get $$\frac14\int_0^{2\pi}\cos^{2n}(t)\,dt=\frac142^{-2n}(2\pi)C(2n,n),$$where $C()$ is a binomial coefficient.
The answer you want must now follow by induction (unless I dropped a factor, in which case it follows by induction from the corrected version of the above).
• It's nice to see you again. It's been a while. – robjohn Jun 23 '15 at 23:30
• Nice to be seen... – David C. Ullrich Jun 23 '15 at 23:34
First, note that it is:
$$\frac{1}{4}\int_{0}^{2\pi} \cos^{2n}x\,dx$$
Now $$\cos^{2n}(x)=\frac{1}{2^{2n}}\left(e^{ix}+e^{-ix}\right)^{2n}$$. And for integer $m\neq 0$, $\int_{0}^{2\pi}e^{imx}\,dx = 0$.
So you only care about the constant term of $(e^{ix}+e^{-ix})^{2n}$, which is $\binom{2n}{n}$.
So the integral is:
$$\frac{1}{4}\cdot 2\pi \cdot \frac{1}{2^{2n}} \binom{2n}{n}=\frac{\pi}{2}\frac{1}{2^{2n}}\binom{2n}{n}$$
Then prove that $$\frac{\binom{2n}{n}}{2^{2n}} = \prod_{k=1}^n \frac{2k-1}{2k}$$
You can prove this last by induction:
$$\binom{2(n+1)}{n+1} = \frac{(2n+1)(2n+2)}{(n+1)(n+1)}\binom{2n}{n}=2\frac{2(n+1)-1}{2(n+1)}\binom{2n}{n}$$
• Hum constant term of $\left(e^{ix}+e^{-ix}\right)^{2n}$ is $\binom{2n}{n}$, can you explain me ? ^^ – ParaH2 Jun 23 '15 at 22:48
• If you expand $(a+a^{-1})^{2n}$ via the binomial theorem, then the constant terms is $\binom{2n}{n}$. All the other terms are of the form $Ca^k$ for $k\neq 0$. Not sure how to make that more clear. – Thomas Andrews Jun 23 '15 at 22:49
• Ok, and why are you seeking constants terms ? – ParaH2 Jun 23 '15 at 22:51
• Because the integrals of all the other terms are zero, as I noted. So the only term that contributes a non-zero value to the integral is the constant term. @Shadock – Thomas Andrews Jun 23 '15 at 22:52
• OK thank you, I'm a chemist, I have some knowlegde in maths because i love them, I'm ridiculous here haha ^^ – ParaH2 Jun 23 '15 at 22:55
You were on the right track using Euler's Identity and writing $\cos x=\frac12(e^{ix}+e^{-ix})$. Proceeding accordingly we have,
\begin{align} \int_0^{\pi/2}\cos^{2n}x\,dx&=\int_0^{\pi/2}\left(\frac{e^{ix}+e^{-ix}}{2}\right)^{2n}dx\\\\ &=\frac{1}{4^n}\int_0^{\pi/2}\sum_{k=0}^{2n}\binom{2n}{k}e^{ikx}e^{-ix(2n-k)}dx\\\\ &=\frac{1}{4^n}\sum_{k=0}^{2n}\binom{2n}{k}\int_0^{\pi/2}e^{ix2(k-n)}dx\\\\ &=\frac{1}{4^n}\sum_{k=0}^{2n}\binom{2n}{k}\frac{\pi}{2}\delta_{nk}+\frac{1}{4^n}\sum_{k=0, k\ne n}^{2n}\binom{2n}{k}\frac{(-1)^{n-k}-1}{i2(k-n)} \tag 1\\\\ &=\frac{\pi}{2}\frac{1}{4^n}\binom{2n}{n}\\\\ &=\frac{\pi}{2}\frac{1}{4^n}\frac{(2n)!}{(n!)^2}\\\\ &=\frac{\pi}{2}\left(\frac{1}{2^n\,n!}\right)\left(\frac{(2n)!}{2^n\,n!}\right)\\\\ &=\frac{\pi}{2}\left(\frac{1}{(2n)!!}\right)\left((2n-1)!!\right)\\\\ &=\frac{\pi}{2}\frac{(2n-1)!!}{(2n)!!}\\\\ &=\frac{\pi}{2}\prod_{k=1}^{n}\frac{2k-1}{2k} \end{align}
as was to be shown!!
Note that the second sum in $(1)$ is purely imaginary and, thereby, must vanish. One can easily show it vanishes by exploiting symmetry. We now explicitly show this.
\begin{align} \sum_{k=0, k\ne n}^{2n}\binom{2n}{k}\frac{(-1)^{n-k}-1}{i2(k-n)} &=\sum_{k=0}^{n-1}\binom{2n}{k}\frac{(-1)^{n-k}-1}{i2(k-n)}+\sum_{k=n+1}^{n-1}\binom{2n}{k}\frac{(-1)^{n-k}-1}{i2(k-n)}\\\\ &=\sum_{k=0}^{n-1}\binom{2n}{k}\frac{(-1)^{n-k}-1}{i2(k-n)}+\sum_{m=0}^{n-1}\binom{2n}{2n-m}\frac{(-1)^{m-n}}{i2(n-m)} \text{substituting m=2n-k} \\\\ &=\sum_{k=0}^{n-1}\binom{2n}{k}\frac{(-1)^{n-k}-1}{i2(k-n)}-\sum_{k=0}^{n-1}\binom{2n}{2n-k}\frac{(-1)^{k-n}}{i2(k-n)} \\\\ &=\sum_{k=0}^{n-1}\binom{2n}{k}\frac{(-1)^{n-k}-1}{i2(k-n)}-\sum_{k=0}^{n-1}\binom{2n}{k}\frac{(-1)^{n-k}}{i2(k-n)} \text{Using}\,\, \binom{2n}{2n-k}= \binom{2n}{k} \\\\ &=0 \end{align}
as was to be shown!
• $\delta_{nk}$ ? – ParaH2 Jun 23 '15 at 22:43
• Why is the integral from $$\int_{0}^{\pi/2}e^{2(k-n)ix}\,dx=\frac{\pi}{2}\delta_{nk}?$$ It's true for $\int_{0}^{\pi}$ or $\int_0^{2\pi}$, but not sure if for $\int_0^{\pi/2}$. – Thomas Andrews Jun 23 '15 at 22:44
• The Kronecker Delta. It is $1$ when the indices are equal and $0$ otherwise. – Mark Viola Jun 23 '15 at 22:44
• $\delta_{nk}=\delta_{n,k}=1$ if $n=k$, $0$ otherwise. – David C. Ullrich Jun 23 '15 at 22:45
• @ThomasAndrews All of the other terms are purely imaginary and cancel by symmetry. – Mark Viola Jun 23 '15 at 22:45 | 2019-10-17T14:10:32 | {
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https://math.stackexchange.com/questions/2872218/rolles-theorem-whats-the-right-statement-of-the-theorem | # Rolle's theorem: what's the right statement of the theorem?
In the fourth edition of "Introduction to Real Analysis" by Bartle and Sherbert, theorem 6.2.3 (Rolle's theorem) states,
Suppose that f is continuous on a closed interval $I := [a, b]$, that the derivative of $f$ exists at every point of the open interval $(a, b)$, and that $f(a) = f(b) = 0$. Then there exists at least one point $c$ in $(a, b)$ such that the derivative of $f$ is zero at $c$.
Now, why are we taking $f(a)=0=f(b)$? Is $f(a)=f(b)$ not sufficient?
You are right, taking $f(a) = f(b)$ is sufficient.
But, one can prove the theorem in this general scenario using the theorem for the case $f(a) = 0 = f(b)$, as follows:
Assume Rolle's theorem as stated in the question details is true. Let $f$ be a function satisfying the same hypotheses, except that $f(a) = f(b) = k$, where $k$ is not necessarily equal to zero. Then, the function $g(x) = f(x) - k$ satisfies the hypotheses of Rolle's theorem, and so there is a point $c$ such that $g'(c) = 0$. But $g'(c) = f'(c)$, so we are done.
So, it doesn't really matter which one we use, as both versions are seen to be equivalent to each other.
• If by 'right statement' OP means 'most general', then no we don't need f(a),f(b) to equal zero. Better to state in the more general. And Rolle's Theorem not always about roots (f(a)=f(b)=0). The version I was taught was simply a formalization that between two points with equal y-value, there is a peak with zero derivative. – smci Aug 5 '18 at 23:51
• This could be an answer on its own. – user279515 Aug 6 '18 at 3:55
Yes, you are right: $f(a)=f(b)$ is enough. However, it is usual to add the condition $=0$ after $f(a)=f(b)$ so that the theorem becomes: between two roots of $f$, there's a root of $f'$, which is the original theorem due Michel Rolle (who stated it for polynomials only).
Anyway, this is a moot point, since both statements are just a particular case of the Mean Value Theorem.
• OTOH, the usual way to prove the Mean Value Theorem is by applying the theorem of Rolle. Depending on how much one 'groks' working with derivatives, the step from the general ($f(a) = f(b)$ but not necessarily $=0$) version of Rolle to the Mean Value Theorem is not much bigger than from the "$f(a) = f(b)=0$" version of Rolle to the general one. – Ingix Aug 4 '18 at 17:32
The two versions are clearly equivalent. Suppose just $f(a)=f(b)$. Let $g(t)=f(t)-f(a)$. Then $g(a)=g(b)=0$, so the formally weaker version shows $g'(c)=0$, hence $f'(c)=0$ for some $c$. | 2021-05-15T23:48:46 | {
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https://math.stackexchange.com/questions/2759011/probability-that-the-two-segments-intersect/2759538 | # Probability that the two segments intersect
P and Q are uniformly distributed in a square of side AB. What is the probability that segments AP and BQ intersect?
Inspired by Lee David Chung Lin's answer.
Consider the four events $$AP \cap BQ \neq \varnothing, \\ AP \cap DQ \neq \varnothing, \\ CP \cap BQ \neq \varnothing, \\ CP \cap DQ \neq \varnothing.$$
Ignoring degenerate configurations, the events are mutually disjoint, equiprobable, and cover the whole space. $p = 1/4$.
• This is really great. The reason I felt it was necessary to awkwardly make rigorous the $1/4$ (argument in the last part of my post), was exactly due to the lack of symmetry on the $AB$-and-$PQ$ configuration with respect to the square. I think this is THE canonical treatment of it as a geometric probability problem. May 2, 2018 at 8:14
Suppose we start with $Q = (x, y)$. The admissible positions for $P$ will be in the triangle $BQR$ if $y < x$ or in the quadrilateral $BQRC$ if $y > x$. The areas are calculated from the sides and the heights, and $$p = \int_0^1 \int_0^x \frac {y (1 - x)} {2 x} dy dx + \int_0^1 \int_x^1 \left( 1 - \frac x {2 y} - \frac y 2 \right) dy dx = \frac 1 4.$$
Here's a calculation free "cut-and-paste" argument. Feel free to skip the words and directly examine the figures.
Consider the point $Q$ only in a quarter of the square $\square ABCD$, as shown in the left plot below. We will combine the "region of $P$ that creates a crossing" associated with $Q$ and its 4-fold symmetric mirror images.
As the right plot below shows, point $Q'$ is reflected with respect to (WRT) the main diagonal $\overline{AC}$, point $q$ is reflected WRT the off-diagonal $\overline{BD}$, and point $q'$ is doubly-reflected (equivalent to rotated $180^{\circ}$). The region that is admissible (borrowing the great term from @Maxim) associated with $Q$ is highlighted below on the left (in magenta), and the region for $Q'$ is highlighted on the right plot (in blue).
The labels for corner points $A,B,C,D$ will be omitted in the plots from now on, because they are unnecessary most of the time and a bit distracting. We can cut-and-paste the triangular blue "$Q'$ region" in the manner demonstrated below, thanks to the mirror symmetry by construct of $Q'$.
Note that the resultant quadrilateral consists of an obtuse triangle ($\triangle DQB$) plus THE isosceles right triangle, which is half of $\square ABCD$. Similarly, the admissible regions associated with $q$ and $q'$ can be cut-and-paste into a quadrilateral that is half of $\square ABCD$ minus an obtuse triangle ($\triangle DqB$).
I sincerely apologize for the poor choice of colors if the readers find the figures unclear.
Due to the symmetry WRT the off-diagonal, we can cut-and-paste (flip) the combined-$(q,q')$-region and have the exact full square.
In short, since we effectively quadrupled the admissible region to arrive at $1$ (unity), the actual probability is $\displaystyle \frac14$.
If one is unsatisfied with the brief ending remark just above, below is the more detailed explanation.
Formally, instead of scanning point $Q$ over the entire square, we scan it only in a quadrant and reallocate the probability mass from the 3 mirrored positions $(Q',q,q')$. That is, we redefine the mass associated with $(Q',q,q')$ as the contribution associated with $Q$.
By reallocation, it means that the (conditional) probability of crossing is to be defined as zero when $Q$ is outside of the quadrant.
After the reallocation, when $Q$ is in the quadrant, the (conditional) probability of making an intersection is $1$, because $\square ABCD$ (result of cut-and-paste) divided by $\square ABCD$ (the domain of point $P$) is one.
In other words, the re-distributed conditional probability is $1$ over one-fourth of the domain (of $Q$) and $0$ elsewhere. Thus the overall probability is one-fourth.
• A really nice answer. I suppose the last part can also be explained by saying that when computing the integral sums $S(x, y) \Delta x \Delta y$ over a symmetric grid, we can combine the four related terms, and the result will be the same as summing $\Delta x \Delta y / 4$. I've added another answer, exploiting the symmetry over the vertices of the square instead of the symmetries over the positions of $P$ and $Q$. May 2, 2018 at 8:03 | 2022-05-28T11:54:50 | {
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http://math.stackexchange.com/questions/224974/probability-for-suits-in-5-card-poker-hand | # Probability for #suits in 5 card poker hand
Given a 5 card poker hand from a standard deck, I'm looking to calculate the probability of getting: all 1 suit, 2 different suits, 3 different suits or 4 different suits. All one suit is straight-forward - $\frac{\binom{13}{5}*\binom{4}{1}}{\binom{52}{5}}$- pick five different ranks, each from the same suit.
Likewise, 4 seems fairly simple: $\frac{\binom{4}{1}\binom{13}{2}\binom{13}{1}^3}{\binom{52}{5}}$ - pick one suit to grab two cards from, then pick one card from each other suit.
Its on 2 and 3 that I get kind of stuck - I'm not sure how to set them up! I don't see why something along the lines of
$\frac{\binom{4}{2}*\binom{26}{6} - \binom{13}{5}*\binom{4}{1}}{\binom{52}{5}}$ doesn't work for 2 suits; i.e. picking 2 suits, choose 5 cards, subtracting off the ways in which you could end up with one suit. Similarly, for 3 I would expect
$\frac{\binom{4}{3}*\binom{39}{5}-\binom{4}{2}*\binom{26}{5}}{\binom{52}{5}}$ to give the answer (picking 5 cards from the group containing 3 suits, subtracting off those hands with fewer than 3 suits), but if I sum the probabilities it comes out incorrectly.
Thank you very much for your help!
-
Perhaps there is an easier way, but could always count the number of ways of getting 1 club and 4 diamonds, 2 clubs and 3 diamonds, 3 clubs and 2 diamonds, 4 clubs and 1 diamond. Then, multiply your answer by ${4 \choose 2}$. – JavaMan Oct 30 '12 at 5:22
While this technique makes sense to me, I fear it would translate poorly to the 3 suit question - i.e. 3 clubs, one heart, one spade; 2 clubs, 2 hearts, 1 spade; 1 club, 2 hearts, 2 spades etc... seems that it could get tedious/confusing quickly. I feel like there must be a simpler way to go about it! – jeliot Oct 30 '12 at 5:31
Actually, the three suit case isn't so bad: Let $(a,b,c)$ denote the permissible number of clubs, diamonds, and hearts. Then you have to consider $(3,1,1); (1,3,1); (3,1,1); (1,2,2); (2,1,2); (2,2,1)$. – JavaMan Oct 30 '12 at 6:09
Whoops, that's really not too intimidating at all! mjqxxxx offers an excellent explanation for a slightly more concise formula down below as well :) – jeliot Oct 30 '12 at 7:35
The "exclusion-based" formulation you're trying to use for exactly two suits works fine; you just have a minor error. You can choose the two suits in $4\choose 2$ ways, and then choose a five-card hand from the $26$ cards in those two suits in $26\choose 5$ ways. This counts a number of single-suit hands as well, which you want to exclude. In fact, each of the $4\times{13\choose 5}$ single-suit hands is included exactly $3$ times in your original count (once with each possible "partner suit"). After excluding these with the correct multiplicity, you have $${4\choose 2}{26\choose 5}-3{4\choose 1}{13\choose 5}=379236$$ hands containing exactly two suits. As a double-check, you can calculate from the other direction ("inclusion-based"). A hand with two suits has either four cards from one suit and one from the other, or three cards from one suit and two from the other. There are $12$ ways to choose the major and minor suits. For each (ordered) pair of suits, there are ${13\choose 4}{13\choose 1} + {13\choose 3}{13 \choose 2}=31603$ ways to make a hand; the result is $12\times 31603 = 379236$ again.
With the two-suit result in hand, you can finish the problem. The number of single-suit hands, as you argued, is $4\times{13\choose 5}=5148.$ The number of four-suit hands is $4\times 13^3\times{13\choose 2}=685464.$ Therefore, the number of three-suit hands (the only remaining possibility) is ${52\choose 5}-5148-379236-685464=1529112.$ To double-check this, note that a three-suit hand has $(1,2,2,0)$ or $(3,1,1,0)$ cards per suit. There are $12$ ways to choose the "loner" (first) and "excluded" (last) suits. For each (ordered) pair of suits, there are $13\times{13\choose 2}^2 + {13\choose 3}\times 13^2=127426$ ways to make a hand; the result is $12\times 127426 = 1529112$ again. The associated probabilities are: $$\begin{eqnarray} P_1 &=& \frac{5148}{2598960} &=& 0.198\% \\ P_2 &=& \frac{379236}{2598960} &=& 14.592\% \\ P_3 &=& \frac{1529112}{2598960} &=& 58.836\% \\ P_4 &=& \frac{685464}{2598960} &=& 26.375\% \\ \end{eqnarray}$$
-
Thank you so much - perfectly explained! – jeliot Oct 31 '12 at 0:24
As an example of another technique, which is useful for counting poker hands, particularly two pairs, we count the $3$-suit hands.
There are two types of $3$-suit hand: $3$-$1$-$1$ and $2$-$2$-$1$.
We count the $3$-$1$-$1$ hands. The suit we have $3$ of can be chosen in $\dbinom{4}{1}$ ways. For each such way, the actual cards can be chosen in $\dbinom{13}{3}$ ways. Now the suits we have $1$ of can be chosen in $\dbinom{3}{2}$ ways, and the actual cards in $\dbinom{13}{1}^2$ ways, for a total of $$\binom{4}{1}\binom{13}{3}\binom{3}{2}\binom{13}{1}^2.$$
Next we count the $2$-$2$-$1$ hands. The suit in which we have a singleton can be chosen in $\dbinom{4}{1}$ ways. For each such choice, the actual card can be chosen in $\dbinom{13}{1}$ ways. Now the suits we have doubletons in can be chosen in $\dbinom{3}{2}$ ways, and the actual cards in $\dbinom{13}{2}^2$ ways, for a total of $$\binom{4}{1}\binom{13}{1}\binom{3}{2}\binom{13}{2}^2.$$
- | 2015-08-03T09:29:48 | {
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# A discount electronics store normally sells all merchandise
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A discount electronics store normally sells all merchandise at a discount of 10 percent to 30 percent off the suggested retail price. If, during a special sale, an additional 20 percent were to be deducted from the discount price, what would be the lowest possible price of an item costing $260 before any discount? (A)$130.00
(B) $145.60 (C)$163.80
(D) $182.00 (E)$210.00
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20 Feb 2015, 02:45
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Stardust Chris wrote:
A discount electronics store normally sells all merchandise at a discount of 10 percent to 30 percent off the suggested retail price. If, during a special sale, an additional 20 percent were to be deducted from the discount price, what would be the lowest possible price of an item costing $260 before any discount? (A)$130.00
(B) $145.60 (C)$163.80
(D) $182.00 (E)$210.00
Original price : 260 $Max first discount = -30% Thus : $$260*(1-\frac{30}{100})=182$$ Second discount on the discounted price = -20% Thus : $$182*(1-\frac{20}{100})=145,6$$ Answer B. EMPOWERgmat Instructor Status: GMAT Assassin/Co-Founder Affiliations: EMPOWERgmat Joined: 19 Dec 2014 Posts: 15262 Location: United States (CA) GMAT 1: 800 Q51 V49 GRE 1: Q170 V170 Re: A discount electronics store normally sells all merchandise [#permalink] ### Show Tags 20 Feb 2015, 13:59 2 Hi Stardust Chris, Since the question is essentially just about multiplication, you can do the various math "steps" in a variety of ways (depending on whichever method you find easiest). We're told that the first discount is 10% to 30%, inclusive. We're told that the next discount is 20% off of the DISCOUNTED price.... We're told to MAXIMIZE the discount (thus, 30% off the original price and then 20% off of the discounted price). That "math" can be written in a number of different ways (fractions, decimals, etc.): 30% off = (1 - .3) = (1 - 30/100) = (.7) and the same can be done with the 20% additional discount... The final price of an item that originally cost$260 would be.....
($260)(.7)(.8) = ($260)(.56)
Since .56 is a little more than 1/2, we're looking for an answer that's a little more than $130.... Since these answer choices are so "spread out", we don't really have to do the calculation... Final Answer: GMAT assassins aren't born, they're made, Rich _________________ Contact Rich at: [email protected] The Course Used By GMAT Club Moderators To Earn 750+ souvik101990 Score: 760 Q50 V42 ★★★★★ ENGRTOMBA2018 Score: 750 Q49 V44 ★★★★★ Board of Directors Status: QA & VA Forum Moderator Joined: 11 Jun 2011 Posts: 4782 Location: India GPA: 3.5 WE: Business Development (Commercial Banking) Re: A discount electronics store normally sells all merchandise [#permalink] ### Show Tags 16 Jan 2018, 11:01 Stardust Chris wrote: A discount electronics store normally sells all merchandise at a discount of 10 percent to 30 percent off the suggested retail price. If, during a special sale, an additional 20 percent were to be deducted from the discount price, what would be the lowest possible price of an item costing$260 before any discount?
(A) $130.00 (B)$145.60
(C) $163.80 (D)$182.00
(E) $210.00 Max possible value after first discount is 70% of 260 = 182 Max possible value after second discount is 80% of 182 = 145.60 Thus, answer will be (B) 145.60 _________________ Thanks and Regards Abhishek.... PLEASE FOLLOW THE RULES FOR POSTING IN QA AND VA FORUM AND USE SEARCH FUNCTION BEFORE POSTING NEW QUESTIONS How to use Search Function in GMAT Club | Rules for Posting in QA forum | Writing Mathematical Formulas |Rules for Posting in VA forum | Request Expert's Reply ( VA Forum Only ) Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 8069 Location: United States (CA) Re: A discount electronics store normally sells all merchandise [#permalink] ### Show Tags 18 Jan 2018, 14:16 Stardust Chris wrote: A discount electronics store normally sells all merchandise at a discount of 10 percent to 30 percent off the suggested retail price. If, during a special sale, an additional 20 percent were to be deducted from the discount price, what would be the lowest possible price of an item costing$260 before any discount?
(A) $130.00 (B)$145.60
(C) $163.80 (D)$182.00
(E) \$210.00
The lowest possible price would be:
260 x 0.7 x 0.8
182 x 0.8 = 145.60
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Re: A discount electronics store normally sells all merchandise [#permalink]
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26 Aug 2019, 01:41
260*(30+20+(30*20)/100)%=145.6
Re: A discount electronics store normally sells all merchandise [#permalink] 26 Aug 2019, 01:41
Display posts from previous: Sort by | 2019-10-16T17:08:14 | {
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https://puzzling.stackexchange.com/questions/109512/another-puzzle-with-area | # Another puzzle with area
Squares $$ABCD, DCGH, BEFG$$ and $$ELKM$$ are positioned as shown on the picture. Find the area of triangle $$DGK$$ if you know that the area of square $$ABCD$$ is $$20$$.
Here's another solution using the fact that $$\triangle ABC$$ and $$\triangle ABD$$ have the same area if $$AB \parallel CD$$:
The reason is that $$CE = DF$$, so $$\frac12 AB \cdot CE = \frac12 AB \cdot DF$$.
Now, since $$DG \parallel EK$$ and $$BD \parallel EG$$, we have $$\triangle DGK = \triangle DEG = \triangle BEG$$ in area:
The result is hence
$$2 \cdot 20 = 40$$.
• Yes, well this is basicly the same solution as hexomino's. Apr 17 at 6:15
• @Greedoid It may use the same principle as Hex's proof, but it applies that principle in a different manner resulting in different triangles. Apr 17 at 7:20
Here is a quick way to do it
The locus of the point $$K$$ as the side length of the square $$EMKL$$ varies is a line which is parallel to $$DG$$. Hence, if we consider $$DG$$ as the base of the triangle $$DGK$$ then its perpendicular height, and thus its area, is invariant with respect to the size of the square $$EMKL$$.
This means that we could perform the calculation with the side length $$EM$$ being anything we like and the answer would be the same. If we make it so that the point $$M$$ coincides with the point $$F$$, it is easy to compute that the area of triangle $$DGK$$ is twice the area of the square $$ABCD$$ and thus the answer is $$40$$.
• I like this puzzle - at first sight there is a constraint missing but the answer comes out cleanly :) I think this puzzle would be suitable for some math channel such as 3Blue1Brown Apr 17 at 0:01
• Not worth a separate answer but I'd say the most opportunistic placement of M would be as the midpoint of E and F. Apr 17 at 7:37
Well, this answer is a bit late, but still quite simple I believe:
Let's introduce a coordinate system with origin at $$A$$, having $$AL$$ and $$AH$$ as $$x$$ and $$y$$ axes respectively. Also, let $$AB=1$$ (well, the $$ABCD$$ square has now an area of $$1$$, not $$20$$, but we can just scale the final result 20 times, i.e. assuming that we scaled down the entire picture $$\sqrt{20}$$ times on each axis - otherwise we had to deal with a bunch of square roots). Now, $$D$$ has coordinates $$(0, 1)$$, $$G$$ is $$(1,2)$$ (since the squares $$ABCD$$ and $$DCGH$$ are equal with side $$1$$), and $$K$$ is $$(s + 3, s)$$ where $$s$$ is the side of $$ELKM$$ square, which we do not know. (The $$BEFG$$ square must have the side of $$2$$, because $$BG=BC+CG=1+1=2$$, so $$E$$ is at $$(3, 0)$$).
Now calculate the area of $$DGK$$ triangle from the coordinates of its vertices using the well-known formula: $$A=\frac12|x_1(y_2-y_3)+x_2(y_3-y_1)+x_3(y_1-y_2)|,$$ where $$x_i$$ and $$y_i$$ are the coordinates of $$i$$-th vertex ($$i=1,2,3$$).
Plugging $$x_1=0, y_1=1, x_2=1, y_2=2, x_3=s+3, y_3=s$$ gives $$A=\frac12|0(2-s)+1(s-1)+(s+3)(1-2)|=\frac12|s-1-s-3|=\frac12\times4=2.$$ Thus, the final result (which does not depends on $$s$$) is $$2\times20=40$$. | 2021-09-21T17:23:57 | {
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https://www.episodeyang.com/blog/2021/10-26/differential_entropy/ | Differential Entropy: Analytical Calculation and Sample-based Estimates
# Differential Entropy: Analytical Calculation and Sample-based Estimates
written by Ge Yang
THE GOAL$~~$ Compare the differential entropy computed analytically using the probability density function (PDF) with those estimated using a collection of samples.
# An Analytical Example
Consider a uniform distribution between $[0, 1/2]$
$p(x) = \begin{cases} 2&\text{ if } x \in [0, 1/2]\\ 0&\text{ otherwise. } \end{cases}$
The differential entropy is
\begin{aligned} H(u) &= - \int_0^{1/2} 2 * \log(2) \mathrm d x \\ &= - \log 2 \\ &= - 0.693 \end{aligned}
There are two ways to compute this numerically. The first is to use analytical calculation from the distribution. We do so by discretizing the probability distribution function into bins. We do so by normalizing the sequence of probabilities. This gives us $H(x) = -\log 2 = -0.69$:
def H_d(ps):
ps_norm = ps / ps.sum()
return - np.sum(np.log(ps) * ps_norm)
xs = np.linspace(0, 1 / 2, 1001)
ps = np.ones(1001) * 2
h_d = H_d(ps)
doc.print(f"analytical: {h_d:0.2f}")
analytical: -0.69
## Can We Compute This Via scipy entropy?
The scipy.stats.entropy function computes the Shannon entropy of a discrete distribution. The scipy entropy does not assume that the probability is normalized, so it normalizes the ps internally first. This means that in order to convert this to the differential entropy, we need to scale the result by the log sum.
h_d_analytic = stats.entropy(ps) - np.log(ps.sum())
doc.print(f"analytic entropy w/ scipy: {h_d_analytic:0.3f}")
analytic entropy w/ scipy: -0.693
This shows that we can get the discrete scipy.stats.entropy to agree with our differential entropy function H_d:ps -> R that is defined analytically.
# Estimating Entropy Using Samples from A Probability Density Function
When we do not have access to the PDF, we can also estimate the entropy from a set of samples. We can do this using the differential_entropy function from the stats package in scipy:
from scipy import stats
samples = np.random.uniform(0, 1/2, 10001)
doc.print(f"verify the distribution: min: {samples.min():0.2f}, max: {samples.max():0.2f}")
verify the distribution: min: 0.00, max: 0.50
h_d_sample = stats.differential_entropy(samples)
doc.print(f"sample-based: {h_d_sample:0.3f}")
sample-based: -0.702
# Effect of Sampling On Differential Entropy
THE GOAL$~~$ vary the number of samples from the distribution, and compare the sample-based estimate against those computed from the analytical solution. This is done on the scipy documentation page 1.
Again, consider the uniform distribution from above.
def H_d(ps):
ps_norm = ps / ps.sum()
return - np.sum(np.log(ps) * ps_norm)
def get_H_d(N):
xs = np.linspace(0, 1 / 2, N)
ps = np.ones(N) * 2
h_d = H_d(ps)
doc.print(f"N={N} => h_d={h_d:0.2f}")
for n in [101, 201, 401, 801, 1601]:
get_H_d(n)
N=101 => h_d=-0.69
N=201 => h_d=-0.69
N=401 => h_d=-0.69
N=801 => h_d=-0.69
N=1601 => h_d=-0.69
The differential entropy does not depend on the number of points in the discretized distribution as long as the discretization is done properly.
# Effect of Change of Variable On Differential Entropy
THE GOAL$~~$ Investigate the effect of the change of variable, by varying the scale of the distribution and looking at the entropy. We can look at both the differential entropy analytically, and the sample-based estimates 1.
Now consider a more general case, a uniform distribution between $[a, b]$
$p(x) = \begin{cases} \frac 1 {\vert b - a \vert} & \text{if } x \in [a, b] \\ 0 & \text{otherwise. } \end{cases}$
The differential entropy is $\log \left(\vert b - a \vert\right)$. When a=0 and b=0.5, $H(x) = - \log 2$.
We can verify this numerically. First let's define the entropy function
def H_d(ps):
ps_norm = ps / ps.sum()
return - np.sum(np.log(ps) * ps_norm)
We can plot the delta against $\log(b - a)$ -- it turned out to be zero across the range variants.
def get_H_d(a, b, N=201):
xs = np.linspace(a, b, N)
ps = np.ones(N) / (b - a)
h_d = H_d(ps)
delta = h_d - np.log(b - a)
doc.print(f"a={a}, b={b} => h_d={h_d:0.2f}, δ={delta}")
for b in [1/4, 1/2, 1, 2, 4]:
get_H_d(0, b)
a=0, b=0.25 => h_d=-1.39, δ=0.0
a=0, b=0.5 => h_d=-0.69, δ=0.0
a=0, b=1 => h_d=-0.00, δ=-0.0
a=0, b=2 => h_d=0.69, δ=0.0
a=0, b=4 => h_d=1.39, δ=0.0
## Using The Wrong Measure
What happens when you use the Shannon entropy on the samples, and treat the samples as the probabilities?
samples = np.random.uniform(0, 1/2, 10001)
doc.print(f"verify the distribution: min: {samples.min():0.2f}, max: {samples.max():0.2f}")
verify the distribution: min: 0.00, max: 0.50
h_d_wrong = H_d(samples)
doc.print(f"Shannon Entropy is (incorrectly): {h_d_wrong:0.3f}")
Shannon Entropy is (incorrectly): 1.195 | 2022-09-25T05:17:01 | {
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https://ch.gateoverflow.in/tag/gate2016 | Recent questions tagged gate2016
Which one of the following is an iterative technique for solving a system of simultaneous linear algebraic equations? Gauss elimination Gauss-Jordan Gauss-Seidel $LU$ decomposition
The Laplace transform of $e^{at}\sin\left ( bt \right )$ is $\frac{b}{\left ( s-a \right )^{2}+b^{2}}$ $\frac{s-a}{\left ( s-a \right )^{2}+b^{2}}$ $\frac{s-a}{\left ( s-a \right )^{2}-b^{2}}$ $\frac{b}{\left ( s-a \right )^{2}-b^{2}}$
What are the modulus $\left ( r \right )$ and argument $\left ( \theta \right )$ of the complex number $3+4i$ ? $r=\sqrt{7}, \theta =tan^{-1}\left ( \frac{4}{3} \right )$ $r=\sqrt{7}, \theta =tan^{-1}\left ( \frac{3}{4} \right )$ $r={5}, \theta =tan^{-1}\left ( \frac{3}{4} \right )$ $r={5}, \theta =tan^{-1}\left ( \frac{4}{3} \right )$
A liquid mixture of ethanol and water is flowing as inlet stream $P$ into a stream splitter. It is split into two streams, $Q$ and $R$, as shown in the figure below. The flowrate of $P$, containing $30$ mass$\%$ of ethanol, is $100\:kg/h$. What is the ... additional specification$(s)$ required to determine the mass flowrates and composition (mass $\%$) of the two exit streams? $0$ $1$ $2$ $3$
The partial molar enthalpy (in $kJ$/mol) of species $1$ in a binary mixture is given by $\bar{h}_{1}=2 -60x_{2}^{2}+100x^{1}x_{2}^{2},$ where $x_{1}$ and $x_{2}$ are the mole fractions of species $1$ and $2$, respectively. The partial molar enthalpy (in $kJ$/mol, rounded off to the first decimal place) of species $1$ at infinite dilution is _________
For a flow through a smooth pipe, the Fanning friction factor $(f)$ is given by $f=mRe^{-0.2}$ in the turbulent flow regime, where $Re$ is the Reynolds number and $m$ is a constant. Water flowing through a section of this pipe with a velocity $1\:m/s$ results in a frictional ... the pressure drop across this section (in $kPa$), when the velocity of water is $\:2 m/s$? $11.5$ $20$ $34.8$ $40$
In a cyclone separator used for separation of solid particles from a dust laden gas, the separation factor is defined as the ratio of the centrifugal force to the gravitational force acting on the particle. $S_{r}$ denotes the separation factor at a location (near the wall) ... $S_{r}$ depends on tangential velocity of the particle $S_{r}$ depends on the radial location $(r)$ of the particle
A vertical cylindrical vessel has a layer of kerosene (of density $800\: kg/m^{3}$) over a layer of water (of density $1000 \:kg/m^{3}$). $L$ - shaped glass tubes are connected to the column $30\: cm$ apart. The interface between the two layers lies between ... $cm$, rounded off to the first decimal place) of the interface from the point at which the lower $L$ - tube is connected is _________
A composite wall is made of four different materials of construction in the fashion shown below. The resistance (in $K/W$) of each sections of the wall is indicated in the diagram. The overall resistance (in $K/W$, rounded off to the first decimal place) of the composite wall, in the direction of heat flow, is ________
Steam at $100^{\circ}C$ is condensing on a vertical steel plate. The condensate flow is laminar. The average Nusselt numbers are $Nu_{1}$ and $Nu_{2}$, When the plate temperatures are $10^{\circ}C$ and $55^{\circ}C$, respectively. Assume the physical properties of the fluid and steel to ... -type condensation, what is the value of the ratio $\frac{Nu_{2}}{Nu_{1}}$ ? $0.5$ $0.84$ $1.19$ $1.41$
A binary liquid mixture of benzene and toluene contains $20$ mol$\%$ of benzene. At $350 \:K$ the vapour pressures of pure benzene and pure toluene are $92\: kPa$ and $35\: KPa$, respectively. The mixture follows Raoult's law. The equilibrium vapour phase mole fraction (rounded off to the second decimal place) of benzene in contact with this liquid mixture at $350\:K$ is ________
Match the dimensionless numbers in $Group-I$ with the ratios in $Group-2$. $\begin{array}{cc}\\ &\text{Group-1} & \text{Group-2} \\ &\text{P. Biot number} & \text{ I.$\frac{buoyancy force}{viscous force}$} \\ &\text{ Q. Schmidt number} & \text{II.$\frac{internal thermal resistance of a solid}{boundary layer ... $Q-I$, $R-III$ $P-I$, $Q-III$, $R-II$ $P-III$, $Q-I$, $R-II$ $P-II$, $Q-III$, $R-I$
For what value of Lewis number, the wet-bulb temperature and adiabatic saturation temperature are nearly equal? $0.33$ $0.5$ $1$ $2$
For a non-catalytic homogeneous reaction $A\rightarrow B$, the rate expression at $300\:K$ is $-r_{A}$\left (mol\; m^{-3} s^{-1}\right )=\frac{10C_{A}}{1+5C_{A}}$, where$C_{A}$is the concentration of$A$(in$mol\;/m^{3}$). Theoretically, the upper limit for the magnitude of the reaction rate ($-r_{A}\:in\:mol\:m^{-3}s^{-1}$, rounded off to the first decimal place) at$300\:K$is _________ 0 votes 0 answers The variations of the concentration$\left ( C_{A} ,C_{R}\:and\:C_{S}\right )$for three species$\left ( A,R\:and\:S \right )$with time, in an isothermal homogeneous batch reactor are shown in the figure below. Select the reaction scheme that correctly represents the above plot. The numbers in the reaction schemes shown below, represent the first order rate constants in unit of$s^{-1}$. 0 votes 0 answers Hydrogen iodide decomposes through the reaction$2HI\leftrightharpoons H_{_{2}}+I_{2}$. The value of the universal gas constant$R$is$8.314 \:J\:mol^{-1}K^{-1}$. The activation energy for the forward reaction is$184000\:J\:mol^{-1}$. The ratio (rounded off to the first decimal place) of the forward reaction rate at$600 \:K$to that at$550\:K$is ________ 0 votes 0 answers Match the instruments in$Group-I$with process variable in$Group-2$...$Q-I$,$R-IIIP-Il$,$Q-III$,$R-IP-III$,$Q-lI$,$R-IP-IIl$,$Q-I$,$R-Il$0 votes 0 answers What is the order of response exhibited by a$U$-tube manometer? Zero order First order Second order Third order 0 votes 0 answers A system exhibits inverse response for a unit step change in the input. Which one of the following statement must necessarily be satisfied? The transfer function of the system has at least one negative pole The transfer function of the system has at least one ... transfer function of the system has at least one negative zero The transfer function of the system has at least one positive zero 0 votes 0 answers Two design options for a distillation system are being compared based on the total annual cost. Information available is as follows: ... above information, what is the total annual cost ($Rs$in lakhs/year) of the better option?$4042.492128$0 votes 0 answers Standard pipes of different schedule numbers and standard tubes of different$BWG$numbers are available in the market. For a pipe/tube of a given nominal diameter, which one the following statements is$TRUE$? Wall thickness increases with increase in both the ... the schedule number and the$BWG$number Neither the schedule number, nor the$BWG$number has any relation to wall thickness 0 votes 0 answers Terms used in engineering economics have standard definitions and interpretations. Which one of the following statements is$INCORRECT$? The profitability measure 'return on investment' does not consider the time value of money A cost index is an index value for ... different capacity Payback period is calculated based on the payback time for the sum of the fixed and the working capital investment 0 votes 0 answers India has no elemental sulphur deposits that can be economically exploited. In India, which one of the following industries produces elemental sulphur as a by-product? Coal carbonisation plants Petroleum refineries Paper and pulp industries Iron and steel making plants 0 votes 0 answers Two paper pulp plants$P$and$Q$...$P$and Plant$Q$both use the Kraft process Plant$P$uses Sulfite process Plant$P$uses Kraft process 0 votes 0 answers Match the industrial processes in$Group-I$with the catalyst materials in$Group-2$.$\begin{array}{cc}\\ \\ &\text{Group-1} & \text{Group-2} \\ &\text{P. Ethylene polymerisation} & \text{ I. Nickel} \\ &\text{ Q. Petroleum feedstock cracking} & \text{II. Vanadium pentoxide} \\ &\text{R. Oxidation of $...$R-III$,$S-IIP-I$,$Q-II$,$R-III$,$S-IVP-II$,$Q-III$,$R-IV$,$S-I$0 votes 0 answers A set of simultaneous linear algebraic equations is represented in a matrix form as shown below. ...$x_{3}$is ___________ 0 votes 0 answers The model$y=mx^{2}$is to be fit to the data given below.$\begin{array}{|cl|cI|}\hline &{x} & {1} & {\sqrt{2}} & {\sqrt{3}} \\ \hline &{y} & {2} & {5} & {8} \\ \hline \end{array}$Using linear regression, the value (rounded off to the second decimal place) of$m$is _________ 0 votes 0 answers The Lagrange mean-value theorem is satisfied for$f\left ( x \right )=x^{3}+5$, in the interval$\left ( 1,4 \right )$at a value (rounded off to the second decimal place) of$x$equal to __________ 0 votes 0 answers Values of$f\left ( x \right )$in the interval$\left [ 0,4 \right ]$...$1$, the numerical approximation (rounded off to the second decimal place) of$\int_{0}^{4}f\left ( x \right )dx$is ______________ 0 votes 0 answers An ideal gas is adiabatically and irreversibly compressed from$3$bar and$300\:K$to$6$bar in a closed system. The work required for the irreversible compression is$1.5$times the work that is required for reversible compression from the same initial ... in$K\$, rounded off to the first decimal place) of the gas at the final state in the irreversible compression case is _______________ | 2021-01-17T20:59:32 | {
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http://onepercentevent.com/hiccups-linen-qhnvw/4715f0-is-null-matrix-a-scalar-matrix | Upper triangular is when all entries below the main diagonal are zero: An upper triangular matrix. When we multiply a matrix by a scalar value, then the process is known as scalar multiplication. If you add the m × n zero matrix to another m × n matrix A, you get A: In symbols, if 0 is a zero matrix and A is a matrix of the same size, then A + 0 = A and 0 + A = A A zero matrix is said to be an identity element for matrix addition. C. Unit matrix. Zero Matrix (Null Matrix) Zeros just everywhere: Zero matrix. Null Matrix A “null matrix” is one which has the value zero for all of its elements. It is a matrix with 0 in all its entries. A special kind of diagonal matrix in which all diagonal elements are the same is known as a scalar matrix. If we want to make A = a null matrix we need to multiply it by 0. [] is not a scalar and not a vector, but is a matrix and an array; something that is 0 x something or something by 0 is empty. The null space may also be treated as a subspace of the vector space of all n x 1 column matrices with matrix addition and scalar multiplication of a matrix as the two operations. Yes. Null matrix. C. is orthogonal. A square null matrix is also a diagonal matrix whose main diagonal elements are zero. An identity matrix is a scalar matrix with diagonal elements equal to one. Given a matrix and a scalar element k, our task is to find out the scalar product of that matrix. A square null matrix is also a diagonal . C. diagonal. For example, are null matrices of order 2x3 and 2x2. Given a matrix and a scalar element k, our task is to find out the scalar product of that matrix. Lower triangular is when all entries above the main diagonal are zero: A lower triangular matrix. Example 4.1 (Special Matrices) Some examples follow: 1. Note that the denominator of the fraction (just before the pivot's column vector) is the pivot itself (in this case “3”). Null Matrix Watch more videos at https://www.tutorialspoint.com/videotutorials/index.htm Lecture By: Er. By matrix-vector dot-product definition (a and u are vectors) $\begin\left\{bmatrix\right\} \begin\left\{array\right\}\left\{c\right\} a_1 \\ \hline \vdots \\ \hline a_n \\ \end\left\{array\right\} \end\left\{bmatrix\right\} * u = \left[a_1 * u, \dots, a_m * u\right]$ . Example 1. A scalar matrix has all main diagonal entries the same, with zero everywhere else: A scalar matrix . It is denoted by O. A unit matrix plays the role of the number 1 in numbers. 10. It is also a matrix and also an array; all scalars are also vectors, and all scalars are also matrix, and all scalars are also array There are several types of matrices, but the most commonly used are: Rows Matrix Columns Matrix Rectangular Matrix Square Matrix Diagonal Matrix Scalar Matrix Identity Matrix Triangular Matrix Null or Unit matrix and scalar matrix are special case of a diagonal matrix. It is never a scalar, but could be a vector if it is 0 x 1 or 1 x 0. The null matrix is the one in which all elements are zero. Z is a scalar matrix with lamda = 0. B. Scalar matrix. Null matrix or Zero-matrix : A matrix is said to be a null matrix or zero-matrix if each of its elements is zero. Is the scalar matrix is always a identity matrix? A square matrix m[][] will be diagonal matrix if and only if the elements of the except the main diagonal are zero. EASY. Two examples of a scalar matrix appear below. Let [([a.sub.ij]).sub. In other words, the square matrix A = [a ij] n×n is an identity matrix, if a ij = 1, when i = j and a ij = 0, when i ≠j. are all zero matrices. A diagonal matrix is a square matrix in which all the elements other than the principal diagonal elements are zero. Diagonal matrix. Then n x n matrix V = VX . D. has rank n. Solution: QUESTION: 9. A diagonal matrix, in which all diagonal elements are equal to same scalar, is called a scalar matrix. 11. Let us put into practice the knowledge gained about the properties of matrix scalar multiplication and solve the next example exercises. Answer. Related Video. Any basis for the row space together with any basis for the null space gives a basis for . (i,j) [member of] AxB] be a scalar matrix with positive entries, and denote its columns by [[alpha].sub.j] = [([a.sub.ij]).sub.i[member of]A] and its rows by [[beta].sub.i] = [([a.sub.ij]).sub.j[member of]B]. It is a diagonal matrix with equal-valued elements along the diagonal. Properties of matrix addition & scalar multiplication. 6.5k VIEWS. Example : Identity Matrix. Question 3: Explain a scalar matrix? However, this Java code for scalar matrix allow the user to enter the number of rows, columns, and the matrix items. null. Null space of zero matrix. The types of matrices you have checked here are scalar matrix, unit and identity matrix, null or zero matrix, triangular matrix, with both options lower and upper triangular matrices. Program to check diagonal matrix and scalar matrix. Then diagonal matrix, rectangular matrix, row and column matrices. (i) A zero-matrix need not be a square matrix. Holder's inequality: some recent and unexpected applications. Program for scalar multiplication of a matrix. Triangular Matrix. A. Since the zero matrix is a small and concrete concept in itself which can be used through many of our lessons in linear algebra, we are now forced once more to enter into the topic of a later lesson: the null space of a matrix. Solution: QUESTION: 8. Rotation, coordinate scaling, and reflection. To show that the null space is indeed a vector space it is sufficient to show that , ∈ ⇒ + ∈ and ∈ ⇒ ∈ These are true due to the distributive law of matrices. However, in our case here, A 2 is not zero, and so we continue with Step 3. The dimension of the null space of a matrix is the nullity of the matrix. Step 3. In the special case when M is an m × m real square matrix, the matrices U and V * can be chosen to be real m × m matrices too. A diagonal matrix with all its main diagonal entries equal is a scalar matrix, that is, a scalar multiple λI of the identity matrix I. Here is the call graph for this function: Here is the caller graph for this function: m_matrix::m_matrix (const m_matrix & : that) D. Skew symmetric matrix. Matrix multiplication also known as matrix product . Google Classroom Facebook Twitter. When we add or subtract the 0 matrix of order m*n from any other matrix, it returns the same Matrix. We know that a matrix can be defined as an array of numbers. If a matrix A is symmetric as well as skew-symmetric, then A is a (A) Diagonal matrix (B) Null matrix asked Dec 6, 2019 in Trigonometry by Rozy ( 41.8k points) matrices We use the notation I p to denote a p×p identity matrix. 0, a matrix composed entirely of zeros, is called a null matrix. Just a note on separate Qs & As here. A submatrix of the given matrix can be obtained by deleting . Even a single number is stored as a matrix. 9. Let M be an arbitrary square matrix and Z be a zero matrix of the same dimension. A. has rank zero. It is a binary operation that produces a single matrix by taking two or more different matrices. A = 3: 0: 0: 3: B = 5: 0: 0: 0: 5: 0: 0: 0: 5: The identity matrix is also an example of a scalar matrix. Learn what a zero matrix is and how it relates to matrix addition, subtraction, and scalar multiplication. Symmetric. A matrix is a two-dimensional, rectangular array of data elements arranged in rows and columns. The most basic MATLAB® data structure is the matrix. We have to find whether the given square matrix is diagonal and scalar matrix or not, if it is diagonal and scalar matrix then print yes in the result. Examples: etc. Java Scalar Matrix Multiplication Program example 2. Answer: The scalar matrix is similar to a square matrix. 6.5k SHARES. If M is a square matrix, is a scalar, and x is a vector satisfying then x is an eigenvector of M with corresponding eigenvalue . EDIT The question is this: Scalar multiplication is defined as B = A * s, where B and A are equally sized matrices (2D array of numbers, in this example let's use integers) and s is a scalar value. 263.1k SHARES. The elements can be numbers, logical values (true or false), dates and times, strings, or some other MATLAB data type. A square matrix A is symmetric if a ij = a ji for all i, j. The null matrix is also called the zero matrix. Then Z*M = Z = 0*M = 0 => Z = 0. If A 2 happens to be a null matrix, then the process terminates and the rank of A 1 is 1, which is then the largest subscript of a nonzero matrix. Diagonal matrix: A square matrix is said to be diagonal matrix if the elements of matrix except main diagonal are zero. The matrices » » » ¼ º « « « ¬ ª 0 0 0 0 0 Z and » » » ¼ º « « « ¬ ª 0 0 0 0 N are both null matrices. 263.1k VIEWS. The various types of matrices are row matrix, column matrix, null matrix, square matrix, diagonal matrix, upper triangular matrix, lower triangular matrix, symmetric matrix, and antisymmetric matrix. Select any nonzero element of A 2. Scalar Matrix. is said to be a scalar matrix if b ij = 0, when i ≠j b ij = k, when i =j, for some constant k. (vi) A square matrix in which elements in the diagonal are all 1 and rest are all zeroes is called an identity matrix. If M has n columns then rank(M)+nullity(M)=n. The product of any matrix by the scalar 0 is the null matrix. If equals (A) (B) zero matrix (C) a scalar quantity (D) identity matrix 1:36 35.6k LIKES. Scalar matrix and identity matrix 305.4k LIKES. Examples: Input : mat[][] = {{2, 3} {5, 4}} k = 5 Output : 10 . D. unit. Its effect on a vector is scalar multiplication by λ. It is a matrix with 0 in all its entries. Intro to zero matrices. B. has rank 1. Each element of A is multiplied to s, which is then stored in the corresponding element in matrix B. Is it true that the only matrix that is similar to a scalar matrix is itself Hot Network Questions Was the title "Prince of Wales" originally claimed for the English crown prince via a trick? X = [X 1, X 2, ... , X n] is an n — tuple non — zero vector. Ridhi Arora, Tutorials Point India Private Limited. View All. However a scalar matrix need not be a unit matrix. Email. A rectangular matrix 1234 5678 2. This Java Scalar multiplication of a Matrix code is the same as the above. 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N ] is an n — tuple non — zero vector a special kind of diagonal matrix also a matrix. Elements is zero is one which has the value zero for all i j! Zero vector Z is a scalar matrix with lamda = 0 * M = 0 a two-dimensional, rectangular,. Rectangular matrix, row and column matrices examples follow: 1 0 M! Role of the number 1 in numbers ( i ) a zero-matrix need not be a null is! By a scalar matrix are special case of a matrix with diagonal elements are the same, with zero else. = > Z = 0 * M = is null matrix a scalar matrix space of a is symmetric if ij... Value, then the process is known as scalar multiplication and solve the next example exercises Z. Same as the above into practice the knowledge gained about the properties of matrix except main diagonal elements zero... Non — zero vector by: Er 0 in all its entries by deleting same the. Is similar to a square matrix in which all elements are the same as the above we to! Matrix addition, subtraction, and the matrix items single matrix by the scalar matrix in which all diagonal are. Multiply it by 0 ( a ) ( B ) zero matrix ( null ”! To a square matrix in which all diagonal elements are zero same dimension on... Or zero-matrix: a scalar matrix with 0 in all its entries are zero notation i to! Equals ( a ) ( B ) zero matrix is also called the zero matrix is all... 35.6K LIKES in view of ( 7 ) like of the given matrix can be obtained by deleting (! Zeros just everywhere: zero matrix is said to be diagonal matrix is always a identity matrix p×p identity.... Z = 0 which all diagonal elements are zero its effect on a vector scalar... Data elements arranged in rows and columns 2x3 and 2x2 with zero everywhere:! The main diagonal elements are is null matrix a scalar matrix: a square matrix properties of matrix except main diagonal zero... Are the same dimension the role of the case of a diagonal matrix: a element... 2,..., x n ] is an n — tuple non — zero vector matrix... Be an arbitrary square matrix is a scalar matrix are 1 any basis for the null gives... Question: 9 a ij = a ji for all i, j in rows columns. Lamda = 0 * M = Z = 0 Z = 0 = > Z = 0 = Z. Null space gives a basis for subtract the 0 matrix of the case a... Case here, a matrix and Z be a zero matrix is the nullity of the given matrix can defined! For example, are null matrices of order M * n from other. | 2021-06-24T03:13:19 | {
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https://www.physicsforums.com/threads/probability-a-sample-mean-will-fall-in-a-range.844276/ | # Probability a sample mean will fall in a range
1. Nov 21, 2015
### toothpaste666
1. The problem statement, all variables and given/known data
A random sample of size n = 81 is taken from an infinite population with the mean μ = 128 and the standard deviation σ = 6.3. With what probability can we assert that the value we obtain for the sample mean X will fall between 126.6 and 129.4?
3. The attempt at a solution
z = (x-μ)/(σ/sqrt(n))
so we have
z = (126.6-128)/(6.3/9) = -2 and z = (129.4-128)/(6.3/9) = 2
so the probability it will fall in the range is
F(2) - F(-2) = .9772 - .0228 = .9544
is this correct?
2. Nov 21, 2015
### Orodruin
Staff Emeritus
This depends on the actual distribution in the population. You can only do what you did if this distribution is assumed to be Gaussian.
3. Nov 21, 2015
### toothpaste666
Gaussian means "normal" right? I am confused a bit about that. In my book they seem to use "z" for the test statistic and use "t" when the population is known to be normal. From what I can tell they are the same thing except that with z you use the standard normal table and with t you use a different table with a certain amount of degrees of freedom. I don't think I fully get it.
4. Nov 21, 2015
### Ray Vickson
I do not actually believe you; I think you are mis-reading your book (although, to be honest, I am making this judgement sight-unseen). Typically, for an independent random sample from an underlying normal (=Gaussian) distribution with mean $\mu$ and variance $\sigma^2$: (1) we use $z$ and normal tables when we KNOW the value of $\sigma$; but (2) use $t$ and t-tables when we do not know $\sigma$, but have estimated it from the sample data itself.
In case (2), we estimate
$$\text{estimator of }\: \sigma^2 = \frac{1}{n-1} \sum_{i=1}^n (x_i - \bar{x})^2$$
where the sample values are $x_1, x_2, \ldots, x_n$ and $\bar{x} = \frac{1}{n} \sum_{i=1}^n x_i$ is the sample mean. In that case the jargon is that there are $n-1$ "degrees of freedom".
In the limit as $n \to \infty$ the t-distribution with (n-1) degrees of freedom becomes the standard normal, so using $z$ is like having infinitely many degrees of freedom.
5. Nov 21, 2015
### toothpaste666
so for either of the two statistics to work, the distribution must be normal?
6. Nov 21, 2015
### Ray Vickson
Theoretically, yes, but for a large sample-size, using the "normal" results give a "reasonably accurate" approximation. This is based on the so-called Central Limit Theorem; see, eg.,
https://en.wikipedia.org/wiki/Central_limit_theorem
or http://davidmlane.com/hyperstat/A14043.html
or http://www.statisticalengineering.com/central_limit_theorem.htm .
For a "reasonable" non-normal underlying distribution, a sample size of n = 81 is likely large enough that normal-based estimates will be informative, if not absolutely accurate.
7. Nov 21, 2015
### toothpaste666
ahh ok what my book actually says is use z for samples of n>30 with σ known and if σ is not known replace σ with s and if the sample is n<30 And the population is normal use t. so since my sample is large enough, my solution to this problem should be close enough?
8. Nov 21, 2015 | 2018-02-25T14:37:13 | {
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https://mathematica.stackexchange.com/questions/107652/double-sum-involving-condition | # Double Sum Involving Condition
I would like to compute the dimensions of some small free nilpotent Lie algebras. However, I am totally new to this and I could not figure out how to write the double sum which gives the dimension of the free nilpotent Lie algebra $L(c,n)$ of step $c$ with $n$ generators.
Long story short, I would like to learn how I can write the sum
$$\dim L(c,n) = \sum_{m=1}^{c}\frac{1}{m}\sum_{d \mid m}\mu(d)n^{m/d}$$
where $\mu$ is the Möbius function, in Mathematica.
• Maybe diml[c_, n_] := Block[{j}, Sum[DirichletConvolve[MoebiusMu[j], n^j, j, m]/m, {m, c}]]? If this does what you want, I'll write an answer. – J. M. will be back soon Feb 18 '16 at 6:32
• @J.M. Hello. I tried this with some known values and it produced the correct values. I am sorry, since I do not know much about Mathematica this is the only way of me knowing that it works. Thank you! So please go ahead and write an answer, – Can Hatipoglu Feb 18 '16 at 7:11
• No need to apologize! You are fortunate here that Mathematica has the necessary stuff for the computations you want to do. Give me a few, and I'll write something up. – J. M. will be back soon Feb 18 '16 at 7:14
• FWIW, your inner sum equals JordanTotient[k, n] but unfortunately that function is not in Mathematica. – Chip Hurst Jun 2 '16 at 18:10
## A story of incremental improvement
Let's look at the OP's original expression again, for reference:
$$\sum_{m=1}^{c}\frac{1}{m}\sum_{d \mid m}\mu(d)n^{m/d}$$
Most people here are familiar with Sum[], and would not have much trouble translating the outer summation into Mathematica syntax. The inner part,
$$\sum_{d \mid m}\mu(d)n^{m/d}$$
is not terribly familiar to those who do not do much number theory. What this basically is saying is that the terms of the sum are indexed over the divisors of $m$. Appropriately enough, Mathematica does have a Divisors[] function. The inner sum can thus be written as
Sum[MoebiusMu[d] n^(m/d), {d, Divisors[m]}]
Summing over the divisors of a number, however, is so common a number-theoretic operation that Mathematica has seen it fit to provide a DivisorSum[] function. Thus, the inner sum can also be written as
DivisorSum[m, MoebiusMu[#] n^(m/#) &]
The story does not end here. The operation
$$\sum_{d \mid m}f(d)g(m/d)$$
frequently turns up in number-theoretic and other contexts, that it has been given a special name: Dirichlet convolution. Mathematica, fortunately, also provides a function for evaluating convolutions, called DirichletConvolve[]. Thus, the inner sum is most compactly expressed as
DirichletConvolve[MoebiusMu[j], n^j, j, m]
The function in the OP can now be implemented like so:
diml[c_Integer?Positive, n_Integer?Positive] := Block[{j},
Sum[DirichletConvolve[MoebiusMu[j], n^j, j, m]/m, {m, c}]]
or, using formal symbols as suggested by The Doctor,
diml[c_Integer?Positive, n_Integer?Positive] :=
Sum[DirichletConvolve[MoebiusMu[\[FormalJ]], n^\[FormalJ], \[FormalJ], m]/m, {m, c}]
(They look messy here on SE, but should look nice when pasted into Mathematica.)
Here is the function in action:
Table[diml[c, n], {c, 5}, {n, 5}]
{{1, 2, 3, 4, 5}, {1, 3, 6, 10, 15}, {1, 5, 14, 30, 55},
{1, 8, 32, 90, 205}, {1, 14, 80, 294, 829}}
• Thank you for introducing me to DirichletConvolve and for this instructive answer . This is my vote :) – ubpdqn Feb 18 '16 at 8:55
Please see J.M.'s answer (in comments) using DirichletConvolve. I would vote for this as an answer.
l[c_, n_] :=Module[{r = Range[c]},
Total[(1/r) Total /@MapThread[MoebiusMu@#1 n^(#2/#1) & ,{Divisors /@ r, r}]]]
I will happily delete and vote for JM answer.
I like J.M.'s answer and voted for it. However, it's hard to format comments, so I've posted this as an answer instead. Rather than using Block[{j},...], you can use formal symbols instead, used to represent a formal parameter that will never be assigned a value. Moreover, if you enter
DivisorSum[m, Function[d, MoebiusMu[d] n^(m/d)]] // TraditionalForm
you will see that Mathematica uses formal symbols automatically. Also, in many cases, I prefer to use the "map notation" for Function instead of # and &. For example,
is executable code for this function.
Finally, in TraditionalForm notation, the sum can be expressed as
which corresponds nicely to the original sum posted in the question. However, using DirichletConvolve should be more efficient. | 2020-01-26T01:56:20 | {
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https://math.stackexchange.com/questions/2629195/give-an-example-of-a-linear-transformation-whose-kernel-is-the-line-spanned-by-v | # Give an example of a linear transformation whose kernel is the line spanned by vector: $\begin{bmatrix} -1 & 1 & 2\end{bmatrix}^T$
I know that a linear transformation could be a projection onto the plane with normal vector $\begin{bmatrix} -1 & 1 & 2\end{bmatrix}^T$, but finding the projection would be too difficult.
I could easily think up a matrix where multiplied by $\begin{bmatrix} -1 & 1 & 2\end{bmatrix}^T = 0$, but I'm not sure on how to choose a matrix where $\begin{bmatrix} -1 & 1 & 2\end{bmatrix}^T$ is the only element of the kernel.
Also, can you please explain this Hint: "to describe a subset as a kernel means to describe it as an intersection of planes?"
• it's not too hard to cook up a matrix with rank 2 that kills this vector – qbert Jan 31 '18 at 4:53
• @qbert Yeah but how can I ensure nothing else send the matrix to 0? – Goldname Jan 31 '18 at 4:59
• "cook up" a $3\times 3$ matrix, making sure that (-1,1,2) is in the kernel and then show that the rank of this matrix is 2. And you know that rank + nulity = 3, so $(-1,1,2)$ must span the kernel. – Doug M Jan 31 '18 at 5:00
• if it has rank 2, it's kernel is rank 1. If it kills your given vector, then that spans the kernel – qbert Jan 31 '18 at 5:00
• ah I see, I forgot about that theorem, thanks – Goldname Jan 31 '18 at 5:18
Guide:
• $$2(-1) + 0(1) +1(2)=0$$
• $$0(01) + (-2)(1)+1(2)=0$$
• Verify that $\begin{bmatrix} 2 & 0 & 1 \end{bmatrix}$ and $\begin{bmatrix} 0 & -2 & 1 \end{bmatrix}$ are linearly independent.
• Now think of someway to form your matrix and prove that span of $\{\begin{bmatrix} -1 & 1 & 2 \end{bmatrix}^T\}$ is a basis to the kernel.
As for the explanation of the hint: To describe $x$ which satisfy $Ax=0$ where $a_i^T$ are the $i$-th row means $x$ is in the intersection of $\{ a_i^Tx = 0 : i \in \{ 1, \ldots, m\}\}$.
• I don't understand what your equations are supposed to mean – Goldname Jan 31 '18 at 4:59
• I can easily think of a matrix, but I don't understand what is required so that [-1 1 2] is the only element in the kernel – Goldname Jan 31 '18 at 4:59
• $[-1, 1, 2]$ is not the only element, any multiple of $[-1, 1, 2]$ is in the kernel. – Siong Thye Goh Jan 31 '18 at 5:01
• yeah that's what I meant – Goldname Jan 31 '18 at 5:03
• rank-nullity theorem is useful. geometrically, two non-parallel 3D plane intersect at a line. – Siong Thye Goh Jan 31 '18 at 5:03
Expanding upon the comment: Choosing the following matrix works $$\begin{bmatrix} 1&-1&1\\ 2&0&1\\ 1&1&0 \end{bmatrix}$$ It is clear that $(-1,1,2)$ is in the kernel. The dimension of the image is $2$, so the dimension of the kernel is just $1$, by rank nullity. So the only vectors that map to $0$ under the matrix are your vector and multiples of it.
Pick any two vectors that together with $[-1,1,2]^T$ form an ordered basis of $\mathbb R^3$. Relative to this basis, the matrix $\operatorname{diag}(1,1,0)$ represents a transformation with the required properties. Apply a change of basis to it to get a matrix relative to the standard basis. For that matter, since a linear transformation is completely determined by its action on a basis, the description $T(v_1)=v_1$, $T(v_2)=v_2$, $T([-1,1,2]^T)=0$, where $v_1$ and $v_2$ are the two vectors you chose, is a complete description of the transformation, but that’s likely not the answer that whoever gave you this problem is looking for.
As far as the hint goes, the kernel is a line through the origin, which can be described as the intersection of a pair of planes. The normals to those planes span the orthogonal complement of the line, so you can use them to construct the above basis.
• How did you conclude that "the matrix $diag(1,1,0)$ represents a transformation with the required properties"? Thanks – gbox Jan 31 '18 at 9:53
• @gbox The columns of a transformation matrix are the images of the basis vectors. In fact, for any linear transformation one can find bases for the domain and codomain such that the matrix has the form $\tiny{\left[\begin{array}{c|c}I&0\\\hline0&0\end{array}\right]}$. – amd Jan 31 '18 at 19:39
• @gbox BTW, $\operatorname{diag}(1,1,0)$ gives you a projection. If you instead choose some other pair of linearly independent vectors for the first two columns, you’ll get a different type of rank-2 transformation. – amd Jan 31 '18 at 19:47 | 2019-05-21T13:19:12 | {
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https://math.stackexchange.com/questions/4192587/is-there-a-name-for-the-curve-traced-by-the-midpoint-of-two-moving-points-on-a-c | # Is there a name for the curve traced by the midpoint of two moving points on a circle with different speed?
I'm curious about what I did in Geogebra while I was experimenting with animation. Here it is:
1. Construct a circle with radius $$r$$
2. Define two points $$A(r\cos a\theta , r \sin a\theta)$$ and $$B(r\cos b\theta, r \sin b\theta)$$. Let $$(r,0)$$ be the initial position of both $$A$$ and $$B$$.
3. Get the midpoint of $$AB$$. Name this point $$M$$.
4. Change $$\theta$$ continuously starting from $$0$$ until $$A$$ and $$B$$ are both at $$(r,0)$$ again.
The midpoint, by doing the fourth step, seem to make some form of curve. I noticed that the curved traced by the midpoint does not change as long as $$a/b$$ does not change, even if $$a$$ and $$b$$ does so. Also, it doesn't matter if the values of $$a$$ and $$b$$ are swapped, but to avoid such ambiguities, we will avoid swapping of values and $$a$$ is always greater than $$b$$.
For the file, see this graph in Desmos.
For the angle, let $$p$$ be the numerator of $$a/b$$ in lowest terms. Then, we have $$0 \leq \theta \leq 2p\pi$$. (I don't know how to this in Desmos, by the way)
## Examples
For $$a = 2.1$$ and $$b = 1.4$$,
For $$a = 3$$ and $$b = 1$$,
And for $$a = 1.1$$ and $$b = 0.007$$ $$(0 \leq \theta \leq 98\pi)$$,
As for the midpoint, the coordinates is $$\left(\frac{r}{2}\left(\cos(a\theta) + \cos(b\theta)\right), \frac{r}{2}\left(\sin(a\theta) + \sin(b\theta)\right)\right)$$
This means that the parametric equation is \begin{align*} x(\theta) &= \frac{r}{2}\left(\cos(a\theta) + \cos(b\theta)\right) \\[10pt] y(\theta) &= \frac{r}{2}\left(\sin(a\theta) + \sin(b\theta)\right) \end{align*} Is there a name for these curves?
Update 1: It seems like this is somehow related to the mathematical basis of a spirograph.
... and therefore the trajectory equations take the form \begin{align*} x(t) &= R\left[(1-k)\cos t+lk\cos {\frac{1-k}{k}}t\right],\\y(t)&=R\left[(1-k)\sin t-lk\sin {\frac {1-k}{k}}t\right]\end{align*}
• It looks like what they call the evolute of a cardioid in this page (second figure) mathcurve.com/courbes2d.gb/largeur%20constante/… Jul 7 at 13:31
• Depending on a and b the curves can take many forms with multiple loops (b=1, a = 3) or spirals (a=2.3, b=2) so not cardioid, though one may be a special case.
– Paul
Jul 7 at 14:31
• This has the properties of several of the so-called transcendental curves, such as the sinusoidal spirals, Archimedean spiral, involute of a circle, and cochleoid. Curves are defined by equations, Cartesian and polar. You need to express this curve as an equation in order to define it, rather than a set of instructions. My go-to reference for plane curves is A Catalog of Special Plane Curves, by J. Dennis Lawrence, Dover Publications, 1972. Jul 9 at 14:59
• @CyeWaldman The parametric equation can be found before Update 1. Jul 9 at 16:24
• @soupless Thank you. From those equations, I would conclude that this is an unnamed curve. On a quick search the closest curve I can find to this is the hypotrochoid, described by $x=n\cos t+h\cos(n/b\cdot t)$ and $y=n\sin t-h\sin(n/b\cdot t)$. Jul 9 at 20:25
I am posting this since my updates that can answer my question should be posted as an answer, not as a part of the question.
We restrict the values of $$a$$ and $$b$$ to be integers where $$a$$ and $$b$$ are both nonzero. Then, the curve seems to represent an epitrochoid given by the parametric equation \begin{align*} x(t) &= m \cos t - h \cos\left(\frac{mt}{c}\right) \\ y(t) &= m \sin t - h \sin\left(\frac{mt}{c}\right). \end{align*}
where $$m = 1$$, $$h = 1$$, and $$c = \frac{a}{b}$$ for $$0 \leq t \leq 2a\pi$$.. For the parametric equation in the question, we let $$r = 2$$.
We can say that the two curves match if they are in the same magnitude and angle, and that they don't match if they don't have the same angle and magnitude. However, we'll say that they are a pseudomatch if they have the same magnitude but not having the same angle.
Now, for odd $$a$$, the two curve are a pseudomatch regardless of the value of $$b$$. For even $$a$$, there are two cases:
1. If $$a$$ is a power of $$2$$, then the curves are a pseudomatch if $$b$$ is a multiple of $$a$$. Otherwise, the curves are a match.
2. If $$a$$ is not a power of $$2$$, then the curves are a pseudomatch if $$b$$ is a multiple of the greatest power of $$2$$ that divides $$a$$. An example would be $$a = 28$$, where the curves are a pseudomatch if $$b$$ is a multiple of $$4$$.
As of the moment, I don't know how to prove that this works. I am just relying on graphs from the file. | 2021-12-05T14:42:53 | {
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https://math.stackexchange.com/questions/1945644/using-functions-to-find-indeterminate-forms | # Using functions to find indeterminate forms
The question is to use two functions f(x) and g(x) to show that ∞ - ∞ (infinity - infinity) is indeterminate.
I don't really know how to get started on this. I think I need to find two limits for f(x) and g(x) that are both infinity. But how to I show that subtracting them is indeterminate?
Edit: In the example we were shown, lim x --> 0 was used. But that was for the indeterminate form 0∞. I don't know if that applies here or not.
"Indeterminate" means "not having a unique value" as $x$ tends to whatever the problem says (I am guessing $x \rightarrow 0$ or $x \rightarrow \infty$, but you really should specify and not make the community you are asking for help guess). I will assume $x \rightarrow \infty$.
So, to show indeterminacy, I will exhibit three examples of the function pair $f(x), g(x)$ with different limiting behaviors of the difference $[f(x)-g(x)]$ as $x \rightarrow \infty$.
Example 1: $f(x) = g(x) = x$. Here $\lim_{x \rightarrow \infty}[f(x) - g(x)] = 0$.
Example 2: $f(x) = x^2, \; g(x) = x$.
Example 3: See Mark Fischler's answer.:)
• Thank you for the reply. The part that I don't understand is that when you use different values for f and g. Wouldn't you expect a different answer? I don't really understand how that makes it indeterminate. – user372991 Sep 28 '16 at 21:58
• Well consider a form that is not indeterminate: $1 / \infty$. In detail: if you have $$\lim_{x \rightarrow \infty} f(x) = \mbox{(some real cosntant)}, \quad \lim_{x \rightarrow \infty} g(x) = \infty,$$ then, no matter which $f, g$ you use, as long as they satisfy the above conditions, you will get $$\lim_{x \rightarrow} {f(x) \over g(x)} = 0.$$ In a high-level summary: The goal is, knowing the asymptotic behavior of $f, g$, to be able to ascertain the asymptotic behavior of an expression involving these two. When we cannot do that, the latter expression is indeterminate. – user8960 Sep 28 '16 at 23:18
Consider $$f(x) = \csc^2x \\ g(x) = \cot^2 x \\ h(x) = 1/x^2$$ and look at the limit as $x\to 0$ of the expressions $f(x)-g(x)$ and $f(x)-h(x)$.
Both of these expressions are of the form $\infty - \infty$
But $$\lim_{x\to 0} (f(x) - g(x)) = 1 \\ \lim_{x\to 0} (f(x) - h(x)) = \frac13$$ This shows that $\infty - \infty$ could be $1$ or $\frac13$; for that matter it could be $0$ (consider $f(x)-f(x)$).
• If you wouldn't mind, could you explain how f(x) -g(x) = 1 and f(x) - h(x) = 1/3? I thought that they all would equal infinity? – user372991 Sep 28 '16 at 22:18
• @Jordan He is just giving you the answers for those limits. You can verify them with L'Hospital's Rule. It's a good exercise – imranfat Sep 28 '16 at 22:52
• On the other hand, the first one is particularly easy: Since $csc^2x - \cot^2x = 1$ for all values of angles where they are both finite, "obviously" the limit as $x\to$ any real value (or for that matter even any complex value) must be $1$. – Mark Fischler Oct 1 '16 at 21:20 | 2020-07-13T22:21:58 | {
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https://math.stackexchange.com/questions/1903623/how-many-subsets-of-1-2-3-4-5-6-7-has-6-as-largest | # how many subsets of $\{1,2,3,4,5,6,7\}$ has $6$ as largest?
Let $A =\{1,2,3,4,5,6,7\}$, The how many subsets have 6 as largest element?
My approach -
Total subsets are $2^7 = 128$. Then First I have to exclude the $1$ empty subset i.e. $\{\}$, $6$ sets with $1$ element except $\{6\}$, then two elements not containing $6$ as its element(I am not getting a way to count these element) and the $2$ subsets $\{6,7\},\{7,6\}$. Then I am stuck here as I am not getting the way to count these numbers. Any help?
• Well, none of the sets have 7 so this is the same as asking what are the subsets of B = {1.... 6} they all have 6 so all sets are equal to $C \cup \{6\}$ for some set C. C is a subset of $\{1,...., 5\}$ so we just need to find the subsets of {1,2...5} and add {6} to all of them. There are $2^5$ subset of {1,2....5} so there are the same number of subsets of {12,...5,6} that must have 6 which is the same number of subsets of {1,2,....7} that must have 6 and must not have 7. – fleablood Aug 25 '16 at 19:58
• 128 subsets. Half have 6 as an element. 1/4, contain both 6 and 7. So 1/4 have 6 and do not have 7. – Doug M Aug 25 '16 at 20:23
Remember how we determined that the number of subsets was 2 to the power of the number of elements.
We did that because for each element, $a$, either $a$ could be in a subset. So the total number of sets was the product of all the choices each of which was 2.
This is the same. Either 1 is in a subset or not. That 2 choices. Either 2 is in the subset or not. That's $2*2$ choice. Keep it up EXCEPT notice $6$ must be in the subset so that is only $1$ choice and $7$ must not be in the set so that's only one set.
So the number of sets is $2*2....*2*1*1$ which is what.
....
Or
....
Don't eliminate the sets one at a time. Remove ALL the subsets that do not have $6$ in them. How many do not have $6$. Remove ALL the subsets that do have $7$. How many is that? Then to avoid double counting add back the ones that had $6$ and didn't have $7$. How many is that?
....
Or
....
Figure 1/2 of the 128 have 6 and half do not. That only leaves half of them acceptable. Then half of those have 7 and half to not. That leaves only half of those acceptable.
....
Or
....
No set has $7$. So all the elements are taken from {$1,2,3,4,5,6$}.
All sets have $6$. So all elements that aren't $6$ are taken from {$1,2,3,4,5$} and $6$ is always added to it.
There are $2^5$ subsets of {$1,2,3,4,5$} and we are only taking those and sticking a $6$ into them. There are $2^5$ such sets.
• I like your way of dividing it up. That makes good sense! – John Aug 25 '16 at 20:16
• You have to be careful in assuming that your condition is exactly 1/2 though. In this case you can but if you were asked say How many sets have the numbers add up to an even number and contain the number 5 you can't assume half the sets add up to an even number and of those half of them contain 5. – fleablood Aug 25 '16 at 20:23
Hint:
Include $6$. How many ways are there to include zero through five members of $\{1,2,3,4,5\}$?
(Order doesn't matter with sets.)
• okay now i got. There are total subsets = 2^6 = 64. – Brij Raj Kishore Aug 25 '16 at 19:58
• The element $6$ has to be there. The element $7$ can't be in there. But any, all, or none of $1$ through $5$ can be. In other words, if $S_5 = \{1,2,3,4,5\}$, then the union of $\{6\}$ and any subset of $S_5$ will have $6$ as the largest element. You figured out the number of subsets for a set of $7$ elements. How about for $5$ elements? – John Aug 25 '16 at 20:02
• Regarding your answer, not quite. It's not optional whether $6$ is in there or not. – John Aug 25 '16 at 20:03
• So the answer is 2^6 -1 - 5. Am i correct? But in option there are 4 options as - 32,31,64,128. – Brij Raj Kishore Aug 25 '16 at 20:05
• No set has 7. All sets have 6. So all sets are a subset of {1,2,3,4,5} unioned up with {6}. How many sets are there. There is {6} = {} + {6}; there is {1,4,5,6} = {1,4,5} + {6}. They are all C + {6} where $C \subset$ {1,2,3,4,5}. How many such sets are there. – fleablood Aug 25 '16 at 20:14
Think about it we need $6$ as an element but we can't have $7$ because $7$ is larger than $6$, we can have anything else.
What we're looking for: How many subsets have $6$ as an element but not $7$?
We must choose to include $6$, that gives $1$ choice. We may choose to to include $1$ or not, that gives $2$ choices. We may choose to include $2$ or not, that gives $2$ choices. We may choose to include $3$ or not, that gives $2$ choices. We may choose to include $4$ or not, that gives $2$ choices. We may choose to include $5$ or not, that gives $2$ choices. We must choose not to include $7$, that gives $1$ choice. By the multiplication principle the number subsets with $6$ but not $7$ is:
$$1•2•2•2•2•2•1=2^5$$ | 2019-08-23T09:22:20 | {
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http://math.stackexchange.com/questions/48876/how-to-determine-number-with-same-amount-of-odd-and-even-divisors | # How to determine number with same amount of odd and even divisors
With given number N, how to determine first number after N with same amount of odd and even divisors? For example if we have N=1, then next number we are searching for is :
2
because divisors:
odd : 1
even : 2
I figured out that this special number can't be odd and obviously it can't be prime. I can't find any formula for this or do i have just compute it one by one and check if it's this special number ? Obviously 1 and number itself is divisors of this number. Cheers
-
Is this homework? How many powers of 2 can divide your special numbers? – user641 Jul 1 '11 at 11:27
We have holiday, i don't go to school :-) Is your question is a hint? – Spinach Jul 1 '11 at 11:52
Here's a hint. Try and pair up the even and odd factors in a natural way. You gave 2 = 1*2 as an example. How about factors of 2p, where p is an odd prime? – hardmath Jul 1 '11 at 12:05
Whenever you have two finite sets of the same size, you can think about whether there is a one-to-one correspondence (a bijection) between them. You might think whether you could apply this observation to the divisors of a number ... – Mark Bennet Jul 1 '11 at 12:07
To get some idea of what's going on, we do like other scientists do, we experiment.
Special numbers will be even, so we write down the number of odd divisors, even divisors, for the even numbers, starting with $2$. If a number turns out to be special, we put a $\ast$ in its row.
So we make a table, giving the number, how many odd divisors it has, how many even. Calculations are easy, but we must be very careful, since errors could lead us down the wrong path.
$2 \qquad 1 \qquad 1\quad\ast$
$4 \qquad 1 \qquad 2$
$6 \qquad 2 \qquad 2\quad\ast$
$8 \qquad 1 \qquad 3$
$10 \qquad 2 \qquad 2\quad\ast$
$12 \qquad 2 \qquad 4$
$14 \qquad 2 \qquad 2\quad\ast$
$16 \qquad 1 \qquad 4$
$18 \qquad 3 \qquad 3\quad\ast$
We could easily go on for a while. It is definitely not a waste of time, since it is useful to be well-acquainted with the structure of the smallish numbers that we bump into often.
A pattern seems to jump out: every second even number seems to be special. It looks as if "special" numbers are not all that special! It can be dangerous to jump to conclusions from data about small integers. But in this case, the conclusion turns out to be correct.
The special numbers, so far, all have the shape $2a$, where $a$ is an odd number. They are divisible by $2$ but not by $4$. The even numbers in our list that are not special are all divisible by $4$.
Now we try to prove that every number that is divisible by $2$ but not by $4$ is special, and that the others are not.
Take an odd number $b$, and look at the number $2b$. Think about the divisors of $2b$. If $k$ is an odd divisor of $2b$, then $2k$ is an even divisor of $2b$, and vice-versa.
If $k$ is an odd divisor of $2b$, call $2k$ the friend of $k$. Split the divisors of $2b$ into pairs of friends. For example, if $b=45$, we have the following pairs of friends.
$$(1,2)\qquad (3,6) \qquad(5,10)\qquad(9,18)\qquad(15,30) \qquad (45,90)$$
We have split the divisors of $2b$ into pairs of friends. Each pair has one odd number and one even number, so $2b$ has exactly as many odd divisors as even divisors.
Now let's show that no number divisible by $4$ can be special. The idea is that if a number is divisible by $4$, then it has "too many" even divisors. I will not write out the details, but you should. The idea goes as follows. Take a number $n$ that is divisible by $4$, like $36$ or $80$. Split the divisors of $n$ into teams. If $k$ is an odd divisor of $n$, put into the same team as $k$ the numbers $2k$, $4k$, and so on however far you can go.
Here are the teams for $n=36$. $$(1,2,4) \qquad (3,6,12)\qquad (9,18,36)$$
Each team has more even numbers than odd numbers, so $n$ has more even divisors than odd divisors. That means $n$ can't be special.
Now let's get to your question: what is the first special number after $N$?
If $N$ is divisible by $4$, the first special number after $N$ is $N+2$. If $N$ is divisible by $2$ but not by $4$, the first special number after $N$ is $N+4$. If $N$ has remainder $1$ on division by $4$, the first special after $N$ is $N+1$, and if the remainder is $3$, the first special is $N+3$. These facts follow easily from what we have discovered about special numbers.
Formulas: We have been operating without formulas, just straight thinking. But I should mention a relevant formula. Let $n$ be an integer greater than $1$, and express $n$ as a product of powers of distinct primes. In symbols, let $$n=p_1^{a_1}p_2^{a_2} \cdots p_k^{a_k}$$ Then the number of divisors of $n$ is given by $$(a_1+1)(a_2+1) \cdots(a_k+1)$$
For example, $720=2^43^25^1$. The number of (positive) divisors of $n$ is $(4+1)(2+1)(1+1)$.
The formula that gives the number of divisors of $n$ is not hard to prove. Try to produce a proof! The formula could be adapted to give a count of the odd divisors of $n$, and of the even divisors. Then we could use these formulas to identify the special numbers. But formulas cannot do the thinking for you. So as a first approach, the way we tackled things is much better than trying to use a formula.
-
Thanks a lot ! Seems like experimenting is very useful, thanks for teaching me something :) And thanks for an so full answer. Chris – Spinach Jul 1 '11 at 14:20
+1; this answer embodies a very important lesson. A lot of questions here on math.SE could have been answered by the posters if they had decided to experiment first... – Qiaochu Yuan Jul 2 '11 at 4:11
@Qiaochu Yuan: A very common problem with students is that they do not really understand that mathematical questions, even those that seem quite abstract, are about concrete things. Unlike working mathematicians, students, even pretty good ones, are often naive formalists. – André Nicolas Jul 2 '11 at 4:28
I wrote a C++ application. My results are all the following numbers:
0, 2, 6, 10, 14, 18, 22, 26, 30, 34, ....
So, first zero, then two and then each time plus 4.
This means in general that if $n \equiv 2 (mod 4)$ is true, your number is one of the kind you're searching. Except from the first zero.
Here is the code of the application:
for (int i = 0; i < 1000; i += 2)
// Always +2 because of odd numbers don't have any even divisors
{
int even = 0;
int odd = 0;
for (int d = 1; d <= i; ++d)
{
if (i % d == 0)
{
if (d % 2 == 0)
{
even++;
} else
{
odd++;
}
}
}
if (even == odd)
{
printf("%d\n", i);
}
}
-
We know that there's always 1 divisor odd ( 1 ) and 1 divisor even for sure ( N ), so we need only to check to n/2 or even to sqrt(n) ( not sure if we won't miss some of divisors expect n/2 ). But nice application. – Spinach Jul 2 '11 at 14:01
For a given integer $n$, every divisor larger than $\sqrt{n}$ is paired with a divisor smaller than $\sqrt{n}$. Use this to figure out a general principle.
-
I don't think that's very much help in this case. There is a much simpler way, hinted at in the comments to the OP. – TonyK Jul 1 '11 at 13:46
@TonyK - I think this works as a pairing for this problem, though it involves making sure that you know that $n$ is not a square, so that there is a pairing - but since the number of distinct factors is known to be even, n can't be a square. – Mark Bennet Jul 1 '11 at 16:25
The pairing works of ncmathsadist works just fine, both for the proof that we have equality when $n \equiv 2 \pmod{4}$, and that there are more even divisors than odd when $n \equiv 0 \pmod{4}$. The pairing is in effect the same as mine when $n \equiv 2 \pmod 4$. But the mathematical description of the pairing is substantially better than mine, because the ncmathsadist pairing has a number of other uses. More detail would undoubtedly have been useful to the student. – André Nicolas Jul 2 '11 at 4:43
I didn't want to give away the whole thing. – ncmathsadist Jul 2 '11 at 13:25 | 2016-06-25T03:44:51 | {
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https://math.stackexchange.com/questions/3141173/is-it-possible-for-an-eigenspace-to-have-more-than-one-vector-in-its-basis-what | # Is it possible for an eigenspace to have more than one vector in its basis? What would that imply? Would every vector in the basis be an eigenvector?
So far I've been seeing that the vector that makes up a basis for the null space of the matrix $$A-Ix$$ (where x is an eigenvalue) is the eigenvector corresponding to the eigenvalue $$x$$. But I've never run into a situation where the basis had more than one vector in it.
Let's say the basis of the null space of $$A-2*I$$ was made up of two vectors. Would that mean those two (and only those two) vectors are the eigenvectors corresponding to the eigenvalue $$2$$ for the matrix $$A$$?
One thing of note is that, from what I understand, the identity transformation (in any space) has only one eigenvalue, $$1$$, but it has infinitely many eigenvectors (not sure if there is a vector space where this wouldn't be true, but it's true for $$R^n$$ at least). Not sure how this fact fits in here.
Just wanting to be sure I'm not misunderstanding something. I'm grateful for any help.
• Every nonzero vector in an eigenspace is an eigenvector. – amd Mar 9 at 20:10
Yes of course, you can have several vectors in the basis of an eigenspace.
First, when you have only one vector $$v$$ in a basis for a matrix $$A$$, with eigenvalue $$\mu$$, then any multiple of this vector is in the basis : $$\forall c \in \mathbb{R}, \ A(cv)=\mu(cv)$$
Then if you have two vectors in a basis, $$v$$ and $$w$$, any linear combination of these two vector will be an eigenvector for $$\mu$$: $$\left\{ \begin{array}{l} Av=\mu v\\ Aw=\mu w\\ \end{array} \right. \Rightarrow A(cv+dw) = \mu (cv+dw)$$
For exemple, let $$A=J-I$$ a matrix $$n\times n$$ of all 1, except 0 in the diagonal (this exemple comes from graph theory and the complete graph $$K_n$$). Then $$A$$ has eigenvalues $$n-1$$ and $$-1$$, and while the eigenspace associated with $$n-1$$ has dimension $$1$$ (there is only 1 vector in its basis), the eigenspace associated with the eigenvalue $$-1$$ has dimension $$n-1$$. Hence you can find an orthogonal basis of $$n-1$$ vectors. Any vector $$v$$ verifying $$Av=-v$$ can be written as a linear combination of the vectors in this basis.
You are understanding things just right.
For example, the $$3 \times 3$$ matrix $$\begin{bmatrix} 2 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 3 \end{bmatrix}$$ has two eigenvalues, $$2$$ and $$3$$.
Every nonzero vector on the $$x$$-axis is an eigenvector for eigenvalue $$2$$, and a basis for that eigenspace.
Every nonzero vector in the $$y$$-$$z$$ plane is an eigenvector for eigenvalue $$2$$. Any two linearly independent vectors in that plane form a basis of eigenvectors for that eigenvalue. | 2019-10-15T08:24:00 | {
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https://byjus.com/question-answer/let-a-denote-the-element-of-the-i-th-row-and-j-th-column-in/ | Question
# Let $$a$$ denote the element of the $${i^{th}}$$ row and $${j^{th}}$$ column in a $$3 \times 3$$ matrix and let $${a_{ij}} = \, - {a_{ji}}$$ for every i and j then this matrix is an -
A
Orthogonal matrix
B
singular matrix
C
matrix whose principal diagonal elements are all zero
D
skew-symmetric matrix
Solution
## The correct option is B skew-symmetric matrixIf $$a_{ij}$$ is the element of $$i^{th}$$ row and $$j^{th}$$ column and $$a_{ij}=-a_{ji}$$then lets assume$$A=\begin{bmatrix} { a }_{ 11 } & { a }_{ 12 } & { a }_{ 13 } \\ { a }_{ 21 } & { a }_{ 22 } & { a }_{ 23 } \\ { a }_{ 31 } & { a }_{ 32 } & { a }_{ 33 } \end{bmatrix}$$Where, $$a_{ij}=-a_{ji}$$$$a_{11}=-a_{11}\implies a_{11}=0$$, similarly $$a_{22}=a_{33}=0$$and $$a_{21}=-a_{12}, a_{31}=-a_{13}, a_{32}=-a_{23}$$Putting the values$$A=\begin{bmatrix} { 0 } & { a }_{ 12 } & { a }_{ 13 } \\ { -a }_{ 12 } & { 0 } & { a }_{ 23 } \\ { -a }_{ 13 } & { -a }_{ 23 } & { 0 } \end{bmatrix}$$$${ A }^{ T }=\begin{bmatrix} { 0 } & { -a }_{ 12 } & { -a }_{ 13 } \\ { a }_{ 12 } & { 0 } & { -a }_{ 23 } \\ { a }_{ 13 } & { a }_{ 23 } & {0 } \end{bmatrix}\\$$and$$-A=\begin{bmatrix} { 0 } & -{ a }_{ 12 } & -{ a }_{ 13 } \\ { a }_{ 12 } & { 0 } & -{ a }_{ 23 } \\ { a }_{ 13 } & { a }_{ 23 } & { 0 } \end{bmatrix}$$thus$$A^T=-A$$$$\therefore$$ The matrix is a skew-symmetric matrix.Mathematics
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View More | 2022-01-28T12:22:17 | {
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http://bootmath.com/need-to-prove-the-sequence-a_n1frac122frac132cdotsfrac1n2-converges.html | Need to prove the sequence $a_n=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}$ converges
I need to prove that the sequence $a_n=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}$ converges. I do not have to find the limit. I have tried to prove it by proving that the sequence is monotone and bounded, but I am having some trouble:
Monotonic:
The sequence seems to be monotone and increasing. This can be proved by induction: Claim that $a_n\leq a_{n+1}$
$$a_1=1\leq 1+\frac{1}{2^2}=a_2$$
Need to show that $a_{n+1}\leq a_{n+2}$
$$a_{n+1}=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}+\frac{1}{(n+1)^2}\leq 1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}+\frac{1}{(n+1)^2}+\frac{1}{(n+2)^2}=a_{n+2}$$
Thus the sequence is monotone and increasing.
Boundedness:
Since the sequence is increasing it is bounded below by $a_1=1$.
Upper bound is where I am having trouble. All the examples I have dealt with in class have to do with decreasing functions, but I don’t know what my thinking process should be to find an upper bound.
Can anyone enlighten me as to how I should approach this, and can anyone confirm my work thus far? Also, although I prove this using monotonicity and boundedness, could I have approached this by showing the sequence was a Cauchy sequence?
Solutions Collecting From Web of "Need to prove the sequence $a_n=1+\frac{1}{2^2}+\frac{1}{3^2}+\cdots+\frac{1}{n^2}$ converges"
Your work looks good so far. Here is a hint:
$$\frac{1}{n^2} \le \frac{1}{n(n-1)} = \frac{1}{n-1} – \frac{1}{n}$$
To elaborate, apply the hint to get:
$$\frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots + \frac{1}{n^2} \le \left(\frac{1}{1} – \frac{1}{2}\right) + \left(\frac{1}{2} – \frac{1}{3}\right) + \left(\frac{1}{3} – \frac{1}{4}\right) + \cdots + \left(\frac{1}{n-1} – \frac{1}{n}\right)$$
Notice that we had to omit the term $1$ because the inequality in the hint is only applicable when $n > 1$. No problem; we will add it later.
Also notice that all terms on the right-hand side cancel out except for the first and last one. Thus:
$$\frac{1}{2^2} + \frac{1}{3^2} + \frac{1}{4^2} + \cdots + \frac{1}{n^2} \le 1 – \frac{1}{n}$$
Add $1$ to both sides to get:
$$a_n \le 2 – \frac{1}{n} \le 2$$
It follows that $a_n$ is bounded from above and hence convergent.
It is worth noting that canceling behavior we saw here is called telescoping. Check out the wikipedia article for more examples.
Besides to Ayman’s neat answer, you may take $f(x)=\frac{1}{x^2}$ over $[1,+\infty)$ and see that $f'(x)=-2x^{-3}$ and then is decreasing over $[1,+\infty)$. $f(x)$ is also positive and continuous so you can use the integral test for $$\sum_{k=1}^{+\infty}\frac{1}{n^2}$$ to see the series is convergent. Now your $a_n$ is the $n-$th summation of this series.
You can show this geometrically too.
If you take a square, and divide the height into $\frac 12$, $\frac 14$, $\frac 18$, and so forth, doubling the denominator on each deal.
Now take the squares $\frac 12$ and $\frac 13$: these go onto the top shelf.
On the second shelf go $\frac 14$ to $\frac 17$. These fractions are all less than $\frac 14$, so fit onto the second shelf. Likewise, squares from $8$ to $15$ go onto the third shelf, each $\frac 1n$ is smaller than $\frac 18$, and so forth.
Therefore $\sum_{n=2}^{\infty}\frac 1n < 1$, and therefore the whole lot is less than two squares.
by the integral test :
$\int^\infty_1 \frac{1}{n^2} dn \le \sum_{i=1}^\infty \frac{1}{n^2}\le 1+\int^\infty_1 \frac{1}{n^2}dn$ .
you can compute the integral so the answer is :
$1 \le \sum_{i=1}^\infty \frac{1}{n^2}\le 2$ .
because : $\int^\infty_1 \frac{1}{n^2} dn =1$
Hint: Prove the the following holds for all $n$ by induction.
$$\sum_1^n \frac{1}{k^2} \le 2 – \frac{1}{n}.$$
Is it not uncommon that when proving some inequality by induction, you will first need to strengthen the hypothesis to get the induction to work.
You can use the same technique to bound other values of the zeta function. For example, try showing $\zeta(3)$ is bounded above by $\frac{3}{2}$.
(I am copying my answer from a duplicate question that was closed as a copy of this one since the induction approach is not available here.)
This here should work with $n \geq 1$ :
$$s_n = \sum\limits_{n=1}^\infty \frac{1}{n^2}= \frac{1}{1} + \frac{1}{4} +\frac{1}{9} + \frac{1}{16}+ …+ \frac{1}{n²}$$$$b_n = \sum\limits_{n=1}^\infty \frac{1}{2^{n-1}} = \frac{1}{1} + \frac{1}{2} +\frac{1}{4}+\frac{1}{8} + …+ \frac{1}{2^{n-1}}$$
$b_n$ is directly compared greater than $s_n$ :
$$s_n < b_n$$
and $b_n$ converges, because of its ratio test :
$$\frac{1}{2^{n-1+1}} / \frac{1}{2^{n-1}} = \frac{2^{n-1}}{2^{n}} = \frac{1}{2} < 1$$
Notice that $2k^2 \geq k(k+1) \implies \frac{1}{k^2} \leq \frac{2}{k(k+1)}$.
$$\sum_{k=1}^{\infty} \frac{2}{k(k+1)} = \frac{2}{1 \times 2} + \frac{2}{2 \times 3} + \frac{2}{3 \times 4} + \ldots$$
$$\sum_{k=1}^{\infty} \frac{2}{k(k+1)} = 2\Big(\, \Big(1 – \frac{1}{2}\Big) + \Big(\frac{1}{2} – \frac{1}{3} \Big) + \Big(\frac{1}{3} – \frac{1}{4} \Big) + \ldots \Big)$$
$$\sum_{k=1}^{\infty} \frac{2}{k(k+1)} = 2 (1) = 2$$.
Therefore $\sum_{k=1}^{\infty} \frac{1}{k^2} \leq 2$.
$a_n=\frac{1}{n^2}$ and because $a_n>0$ we have $|a_n|=a_n$
First: check the necessary condition
$$\lim_{n\to +\infty} na_n=\lim_{n\to +\infty}\frac{1}{n}=0$$
Second: check D’Alembert’s ratio test
$$\lim_{n\to +\infty}|\frac{a_{n+1}}{a_n}|=\lim_{n\to +\infty}\frac{a_{n+1}}{a_n}=\lim_{n\to +\infty}(\frac{n}{n+1})^2=1$$
Third: Because the answer of D’Alembert’s test is $1$, you should use Raabe’s test:
$$\lim_{n\to +\infty}n(1-|\frac{a_{n+1}}{a_n}|)=\lim_{n\to +\infty}n(1-\frac{a_{n+1}}{a_n})=\lim_{n\to +\infty}n(\frac{2n+1}{n^2+2n+1})=2\gt1$$
so the series is convergent | 2018-06-21T23:43:17 | {
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http://mathhelpforum.com/discrete-math/274100-show-64-divides-9-n-56n-63-a.html | # Thread: Show that 64 divides 9^n + 56n + 63
1. ## Show that 64 divides 9^n + 56n + 63
Show that 64 divides 9^n + 56n + 63 for all positive integers n.
Based off of basic computation I can see that for numbers like n=1, n=2, etc this holds but how would I show it for all positive integers?
2. ## Re: Show that 64 divides 9^n + 56n + 63
are you familiar with induction?
$9 + 56 + 63 = 128 = 2 \cdot 64$
so this is true for $n=1$ and we would say $P_1 = True$
next, we assume that $P_n=True$ and we use this to show that $P_{n+1} = True$
here we'd do
\begin{align*} &9^{n+1} + 56(n+1) + 63 \\ = &~9^n\cdot 9 + 56n + 56 + 63 \\ = &~9^n+ 56n + 63 + 8\cdot 9^n + 56 \\ = &~64k + 8\cdot 9^n + 56 \text{; because }P_n=True \\ =&~64k + 8(7+9^n) \end{align*}
clearly $64k$ is divisible by $64$ so we need to show that $8(7+9^n)$ is divisible by $64$ or alternatively that
$(7+9^n)$ is divisible by $8$
Using the same method we are using to show the original statement
$n=1 \Rightarrow 7+9^n = 16 = 2\cdot 8$ so $\tilde{P}_1=True$
Now you show that $\tilde{P}_n \Rightarrow \tilde{P}_{n+1}$ for this and thus that $7+ 9^n$ is divisible by 8
and you will have shown that
$9^{n+1} + 56(n+1) + 63$ is divisible by $64$
and thus that for the original statement
$P_n \Rightarrow P_{n+1}$
and thus it is True for all $n\geq 1$
so see if you can prove that $\tilde{P}_n \Rightarrow \tilde{P}_{n+1}$ it's not hard
3. ## Re: Show that 64 divides 9^n + 56n + 63
You can also note:
$\displaystyle 9^n + 56n + 63 =$
$\displaystyle 9^n + 56n + (8n - 8n) + 63 + (1 - 1) =$
$\displaystyle 9^n + (56n + 8n) + (63 + 1) - 1 - 8n =$
$\displaystyle 9^n + (64n + 64) - 1 - 8n$
64 divides (64n + 64), so it remains to show that 64 also divides $\displaystyle \ \ 9^n - 1 - 8n$.
It has been shown or mentioned it's true for n = 1 in at least one prior post. Let's look at $\displaystyle \ \ n \ge 2$:
$\displaystyle 9^n - 1 - 8n =$
$\displaystyle (8 + 1)^n - 1 - 8n =$
$\displaystyle \bigg(8^n \ + \ n\cdot8^{n - 1} \ + \ \dfrac{n(n - 1)}{2}8^{n - 2} \ + \ . . . \ + \ \dfrac{n(n - 1)}{2}8^2 \ + \ n\cdot8^1 \ + \ 1\bigg) \ - \ 1 \ - \ 8n =$
$\displaystyle 8^n \ + \ n\cdot8^{n - 1} \ + \ \dfrac{n(n - 1)}{2}8^{n - 2} \ + \ . . . \ + \ \dfrac{n(n - 1)}{2}8^2 \ + \ 8n \ - \ 8n \ + \ 1 \ - \ 1 =$
$\displaystyle 8^n \ + \ n\cdot8^{n - 1} \ + \ \dfrac{n(n - 1)}{2}8^{n - 2} \ + \ . . . \ + \ \dfrac{n(n - 1)}{2}8^2$ . . . . . . . . . **
The fractions in front of the powers of eights are combinations and are thus positive integers. The smallest power of eight is equal to 64,
so this expression is also divisible by 64.
** . . . . If n = 2, there will only be one term here. If n = 3, there will only be two terms here. And so on.
4. ## Re: Show that 64 divides 9^n + 56n + 63
The induction step can be done as follows:
\displaystyle \begin{align*}9^{n+1} + 56(n+1) + 63 &= 9\cdot 9^n + 56n + 56 + 63 \\ &= 8( 9^n + 7) + (9^n + 56n + 63) \end{align*}
where the final parenthesis is a multiple of 64 by the inductive hypothesis. So all we need is for $\displaystyle 8$ to divide $\displaystyle (9^n + 7)$.
$\displaystyle 9^n + 7 \equiv 1^n - 1 \pmod{8} = 1-1 \pmod{8} = 0 \pmod{8}$
QED | 2018-03-21T07:24:14 | {
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http://math.stackexchange.com/questions/853615/proving-an-expression-is-composite | # Proving an expression is composite
I am trying to prove that $n^4 + 4^n$ is composite if $n$ is an integer greater than 1. This is trivial for even $n$ since the expression will be even if $n$ is even.
This problem is given in a section where induction is introduced, but I am not quite sure how induction could be used to solve this problem. I have tried examining expansions of the expression at $n+2$ and $n$, but have found no success.
I would appreciate any hints on how to go about proving that the expression is not prime for odd integers greater than 1.
-
Write it as a difference of squares. – Adam Hughes Jul 1 '14 at 19:54
How? It's $n^4 + 4^n$ not $n^4 - 4^n$ – Mathmo123 Jul 1 '14 at 19:55
@Mathmo123: see my answer below. – Adam Hughes Jul 1 '14 at 20:03
$(n^2)^2+(2^n)^2=(n^2+2^n)^2-2^{n+1}n^2$. Since $n$ is odd...
-
Ah. Very neat!! – Mathmo123 Jul 1 '14 at 20:05
Originally, I was not sure how I should go about writing it as a difference of squares. Thanks for your help. – pidude Jul 1 '14 at 20:12
Hint: calculate this value explicitly for $n=1,3$ (or predict what will happen). Can you see any common factors? Can you prove that there is a number $m$ such that if $n$ is odd, then $m|(n^4 + 4^n)$?
Let me know if you need further hints.
-
I originally tried doing that but to no avail. Evaluated at 3 I obtain 145 which has factors of 5 and 29. At 5, the expression equals 1649, which has factors of 17 and 97. I will keep looking for a pattern and let you know if I need more hints. Thanks for your help. – pidude Jul 1 '14 at 19:59
Ah... using this method, there will be a difference between multiples of 5 and other odd numbers. You will find that for odd numbers that are not a multiple of 5, 5 will be a divisor – Mathmo123 Jul 1 '14 at 20:01
Ok, so I was able to prove that. Now I'm working numbers which are multiples of 5. – pidude Jul 1 '14 at 20:09
I am very impressive with Adam's solution. There is very neat. So, I beg for a chance to write the full description about the proof step-by-step.
• We can transform $n^4+4^n$ to $(n^2+2^n)^2-2^{n+1}n^2$ as Adam's suggestion by
1. $n^{(2^2)}+(2^2)^n = (n^2)^2+(2^n)^2$ associative law
2. Now, we mention $(a+b)^2 = (a+b)(a+b) = a^2+2ab+b^2$ algebraic multiplication
3. $(n^2)^2+(2^n)^2+2(n^2)(2^n)-2(n^2)(2^n)$ adding $+2ab-2ab$ to expression
4. $(n^2)^2+2(n^2)(2^n)+(2^n)^2-2(n^2)(2^n)$ re-arrange the expression
5. $(n^2+2^n)^2-2(n^2)(2^n)$ from step 2
6. $(n^2+2^n)^2-2^{n+1}n^2$ law of Exponential
• We try to get the $(n^2+2^n)^2-2^{n+1}n^2$ to conform to $a^2-b^2$ because $a^2-b^2=(a+b)(a-b)$ algebraic multiplication, again
1. Treat $n^2+2^n$ as $a$
2. Since $n$ is odd, n+1 is even. So, we can assume $2m=n+1$, where $m$ is integer
3. So, re-write the $2^{n+1}n^2$ to be $2^{2m}n^2$
4. $2^{2m}n^2=(n2^m)^2$ associative law
5. Treat $n2^m$ as $b$
• It implies that both $a$ and $b$ are both positive integer
• From $a^2-b^2=(a+b)(a-b)$ and the result of $n^4 + 2^4$, it implies that $a$ is greater than $b$
• Hence both $(a+b)$ and $(a-b)$ are positive integer, that causes the result of $n^4 + 2^4$ is combination of $(a+b)$ and $(a-b)$
-
Some interesting factorizations of a polynomial of type $x^4+\text{const}$: $$x^4+4=(x^2+2x+2)(x^2-2x-2) \tag{1}$$
$$x^4+1=(x^2+\sqrt[]{2}x+1)(x^2-\sqrt[]{2}x+1) \tag{2}$$
So one can ask, how to select the coefficients $a,b,c,d$ in
$$(x^2+ax+b)(x^2+cx+d) \tag{3}$$
such that all coefficients of the resulting polynomial are zero except the constant term and the coefficient of the 4th power. The latter is $1$.
If we expand $(3)$ we get
$$x^4+(c+a)x^3+(d+a c+b)x^2+(a d+b c)x+b d$$
And the coefficients disappear, if
$$\begin{eqnarray} c+a &=& 0 \\ d+ ac +b &=& 0 \\ ad+bc &=& 0 \end{eqnarray}$$
When solving for $b,c,d$ we get
$$\begin{eqnarray} c &=& -a \\ b &=& \frac{a^2}{2} \\ d &=& \frac{a^2}{2} \end{eqnarray}$$
and therefore
$$x^4+\frac{a^4}{4} = (x^2+a x+\frac{a^2}{2})(x^2-ax+\frac{a^2}{2})$$
For $a=2$ this gives $(1)$, $a=\sqrt[]{2}$ this gives $82)$ . Substituting $a=2^{t+1}$ we get
$$x^4+4^{2t+1} = (x^2+2\cdot 2^t x+2^{2t+1})(x^2-2\cdot 2^tx+2^{2t+1})$$
Substituting $x=n=2t+1$ gives the required result for odd $n$.
-
@Mathmo123 As I observe the behavior of the multiply operation, I see..
• $v \times v$ produce $v$
• $v \times o$ produce $v$
• $o \times o$ produce $o$
and for the plus operation, I see..
• $v+v$ produce $v$
• $v+o$ produce $o$
• $o+o$ produce $v$
where $v$ is even number and $o$ is odd number.
I do not know that anyone notice about this properties, so I personally name them even-odd multiply properties and even-odd sum properties. So, the expression $n^4 + 4^n$, since n is odd($o$). can be evaluated follow by that properties it will be isolated into:
1. $n^4 = n \times n \times n \times n \equiv o \times o \times o \times o$ produces $o$
2. $4^n \equiv v \times v \times \cdots \times v$ for $n$ times produces $v$
3. $n^4+4^n \equiv o+v$ produces $o$
hence $2|(n^4 + 4^n)$, since $n$ is odd.
May I prove by this way? Any suggestions are kindly pleased.
So, I give up with this way.
-
1. is wrong.... – lhf Jul 2 '14 at 11:25 | 2015-04-21T21:54:23 | {
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https://math.stackexchange.com/questions/1814658/coefficients-of-homology | # Coefficients of homology
I am wondering why people use different coefficients when defining homology of simplicial complex, like homology over $R$, $Z$, $Z/2$, etc? Is one better then the other and why?
Moreover, which one(s) is(are) most useful in practical computation other than abstract theory?
Thanks a lot.
• – Watson Jun 5 '16 at 16:00
• In some sense, if you don't go to deeply in the theory, knowing with $\mathbb{Z}$ coefficients is everything. This is the content of the universal coefficient theorem. It often occurs that one can compute homology with other coefficients more easily (e.g. with Z/2 coeffients poincare duality works directly for non orientable manifolds). Sometimes the way (co) homology is defined leads one directly to different coefficients. A great example is the de Rham cohomology. – Thomas Rot Jun 5 '16 at 20:54
There is no best coefficient in general, but there is indeed a best coefficient for specific case.
Let me show you some examples.
Suppose you want to compute the Euler Characteristic of a space $X$. Suppose that for some reason, counting cell is not a viable option, let's say you don't have an explicit cell structure of your space. You are left with trying to compute the rank of the homology groups. Now you can verify that $\text{rank}H_*(X;\mathbb{Z})=\dim H_*(X,\mathbb{Q})$. The latter group is indeed a vector space, so it has no torsion and it is easier to compute (every short exact sequence splits for example). Even more is true, computing Euler Characteristic is invariant from the chosen field coefficients. In conclusion, it is natural to trying to compute $H_*(X,\mathbb{Q})$ in this case.
Suppose you have a non-orientable closed manifold $M$: When you want to study closed manifolds (from an homology viewpoint), Poincaré Duality is a powerful tool. A powerful tool with some hypothesis to verify, namely orientability. If you don't have orientability you are left with two option, trying to compute $H_*(M;\mathbb{Z}_{\rho})$ (homology with local coefficient - something pretty technical which was let's say created for dealing with this kind of situation) or more easily $H_*(M;\mathbb{Z}_2)$, why? well because every manifold is $\mathbb{Z}_2$ orientable, and therefore you have P-D with these coefficients.
Clearly, homology with $\mathbb{Z}$ coefficients contains all the homological information, since we have UCT which means that if you know homology with $\mathbb{Z}$ for a space, then you know the homology for that space with coefficient any abelian group.
What's the problem: You don't always know how to compute homology with $\mathbb{Z}$-coefficient. By its very property that it bring a lot of informations, computations in general are cumbersome: you have torsion to deal with for instance which is a pain in the ass. So we came up with some weaker version of it, easier to compute but still interesting. This is how research is done usually, you find some great invariant and then realise that it's impossible to compute (very common) and then he must find something weaker but computable and still interesting (very hard).
A last example to back up what I've said. When you will deal with characteristic classes, in particular Stiefel-Whitney ones, you will realise what I mean by sometimes computing cohomology with integer coefficients is hard. But if you turn to $\mathbb{Z}_2$ coefficients, the cohomology ring turns out to be a free graded ring and the generators are the universal classes.
• I can't really see any way of making the assertion "$\operatorname{rank} H_*(M;\mathbb{Z}) = \dim H_*(M;k)$" true if the characteristic of $k$ is positive... – Najib Idrissi Jun 6 '16 at 8:41
• maybe I rushed a little, but if I recall correctly, $\text{rank}H_*(M;\mathbb{Z})=\dim H_*(M;\mathbb{Z})\otimes \mathbb{Q}$ and then one verifies at once that euler char doesn't depend on the field you are computing it. But you are right, I need to modify it a little, since what I wrote it's formally wrong – Riccardo Jun 6 '16 at 8:44
• Yes, by the UCT $\operatorname{rank} H_k(M;\mathbb{Z}) = \dim H_k(M;\mathbb{Q})$, and yes, the Euler characteristic does not depends on the field. But for example $H_2(\mathbb{RP}^2;\mathbb{Z}) = 0$ has rank zero whereas $H_2(\mathbb{RP}^2;\mathbb{F}_2) = \mathbb{F}_2$ has dimension one. ("Formally wrong" is just another way of saying "wrong"...) – Najib Idrissi Jun 6 '16 at 8:46
• yes, and thank you for pointing out this:) – Riccardo Jun 6 '16 at 8:52 | 2019-08-17T10:52:29 | {
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http://mathhelpforum.com/algebra/173556-simplify-expression.html | # Math Help - Simplify this expression...
1. ## Simplify this expression...
$
({2+\sqrt{5})^{1/3} + ({2-\sqrt{5})^{1/3}
$
2. Originally Posted by Alex2103
$
({2+\sqrt{5})^{1/3} + ({2-\sqrt{5})^{1/3}
$
Get ready for a ride....
Here are the highlights:
Let $\displaystyle x = ( 2 + \sqrt{5} )^{1/3} + (2 - \sqrt{5} )^{1/3}$
Now
$\displaystyle x^3 = (2 + \sqrt{5}) + 3( 2 + \sqrt{5} )^{2/3}(2 - \sqrt{5} )^{1/3} + 3(2 - \sqrt{5} )^{1/3}(2 - \sqrt{5})^{2/3} + (2 - \sqrt{5})$
This reduces to
$\displaystyle x^3 = 4 -3(2 + \sqrt{5})^{1/3} - 3(2 - \sqrt{5})^{1/3}$
or, reminding ourselves of what x is equal to:
$\displaystyle x^3 = 4 -3x$
Now solve:
$\displaystyle x^3 + 3x - 4 = 0$
(Hint: There is only one real solution.)
-Dan
3. Hello, Alex2103!
$\text{Simplfy: }\;\sqrt[3]{2+\sqrt{5}} + \sqrt[3]{2-\sqrt{5}}$
A binomial of the form $a + b\sqrt{5}$ makes me suspect
. . that the problem involves the Golden Mean: . $\phi \:=\:\dfrac{1+\sqrt{5}}{2}$
$\text{And sure enough: }\;\phi^3 \;=\;\left(\frac{1+\sqrt{5}}{2}\right)^3 \;=\;2+\sqrt{5}$
. . . $\text{and we find that: }\;\left(\frac{1-\sqrt{5}}{2}\right)^3 \;=\;2-\sqrt{5}$
Therefore:
. . $\sqrt[3]{2+\sqrt{5}} + \sqrt[3]{2-\sqrt{5}} \;\;=\;\;\sqrt[3]{\left(\frac{1+\sqrt{5}}{2}\right)^3} + \sqrt[3]{\left(\frac{1-\sqrt{5}}{2}\right)^3}$
. . $=\;\;\dfrac{1 + \sqrt{5}}{2} + \dfrac{1-\sqrt{5}}{2} \;\;=\;\;1$
I love your solution, Dan!
4. Originally Posted by Soroban
I love your solution, Dan!
But yours is so more elegant and less time consuming!
-Dan
5. Thanks for help!
6. Heres' yet another way to do that. Cardano's formula for the reduced cubic equation says that a solution to $x^3+ mx= n$ is of the form $\sqrt[3]{\frac{n}{2}+ \sqrt{\frac{n^2}{4}+ \frac{m^3}{27}}}+ \sqrt[3]{\frac{n}{2}- \sqrt{\frac{n^2}{4}+ \frac{m^3}{27}}}$.
Looks familiar, doesn't it? That is precisely the same as $\sqrt[3]{2+ \sqrt{5}}+ \sqrt[3]{2- \sqrt{5}}$ with $\frac{n}{2}= 2$ and $\frac{n^2}{4}+ \frac{m^3}{27}= 5$.
From $\frac{n}{2}= 2$, n= 4 and then $\frac{n^2}{4}+ \frac{m^3}{27}= 4+ \frac{m^3}{27}= 5$ so that $\frac{m^3}{27}= 1$, $m^3= 27$, and $m= 3$.
That means that this number is a real root of $x^3+ 3x= 4$ or $x^3- 3x- 4= 0$. It is easy to see that $x^3+ 3x- 4= (x- 1)(x^2+ x+ 4)$. Since the discriminant of $x^2+ x+ 4$ is $1- 16= -15$ the only real root of that equation is 1 so we must have $\sqrt[3]{2+ \sqrt{5}}+ \sqrt[3]{2- \sqrt{5}}= 1$. | 2014-10-21T13:05:42 | {
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https://math.stackexchange.com/questions/2534223/can-an-empty-set-be-in-relation | # Can an empty set be in relation?
We were given this problem for homework:
Let it be $A=\{a,b\}$.
a) Find the powerset $\mathcal P(A)$.
b) Construct an equivalence relation on $\mathcal P(A)$ so that it's quotient set has exactly two elements.
a) $\mathcal P(A)= \{\emptyset\ , \{a\},\{b\},\{a,b\}\}$
b) I constructed the following relation in which an empty set can be an element of ordered pair. (But I'm not sure if an empty set can be in relation. Can an empty set be an element of ordered pair?)
Let's say the relation is R. $$R = \{(\emptyset,\emptyset),(\{a\},\{a\}),(\{b\},\{b\}),(\{a,b\},\{a,b\}),(\{a\},\{b\}),(\{b\},\{a\}),(\emptyset,\{a,b\}),(\{a,b\},\emptyset)\}$$
I think this is correct because:
$[\{a\}] = \{\{a\},\{b\}\} = [\{b\}]$
$[\{\emptyset\}]=\{\emptyset,\{a,b\}\}=[\{a,b\}]$
which means that $\mathcal P(A)/_R=\{\{\{a\},\{b\}\}, \{\emptyset,\{a,b\}\}\}$.
Is this correct or should i remove the empty set from the relation?
• I do not see a problem. Its fine, I suppose. – hemu Nov 23 '17 at 18:55
• It’s perfectly correct. Don’t get confused by having a set of sets. To become more at ease with that, abstract a bit: $X = \mathcal P (A)$ is just a set with four elements, the empty set being one of it, technically. If that gives you a headache, just rename the elements $x_1 = ∅$, $x_2 = \{a\}$, … and work with the renamed elements. – k.stm Nov 23 '17 at 18:57
• By the way, the equivalence relation you have found can be described in words: Two sets $x, y ∈ \mathcal P (A)$ are equivalent (with respect to $R$) if and only if they are of the same parity. – k.stm Nov 23 '17 at 19:03
• @k.stm Thanks! Also, I have another question. I have to prove that $A\neq \emptyset$ ( for $A=\{(a,b) \in \Bbb R^2 : x+y=c \}$, for every $c$). How do I prove this exactly? I understand why this is true but I don't know how to prove it. – i dont know much about algebra Nov 23 '17 at 19:05
• You mean $A = \{(a,b) ∈ ℝ^2;~a + b = c\}$. Well fix a $c$ and find an element in $A$. You can always explicitly write one down. If you can’t manage that, try for $c = 0$ first and then see how you could generalise. – k.stm Nov 23 '17 at 19:07 | 2019-06-26T22:10:36 | {
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https://mathoverflow.net/questions/256696/do-these-matrices-have-a-name/256776 | # Do these matrices have a name?
I am looking for matrices $A\in \mathbb{R}^{m\times n}$ with $m>n$, $A^TA=\frac{m}{n}I$ and $diag(AA^T)=(1\ \dots\ 1)$ where $diag$ denotes the diagonal. Do such matrices have a name? An example for such a matrix would be an $8\times 3$ matrix with the coordinates of a cube centered at the origin as rows. Is the matrix $A$ for every pair $(m,n)$ uniquely determined up to right multiplication with an orthogonal matrix? How are the $m$ points given by the rows of $A$ distributed over the $(n-1)$-dimensional sphere?
My motivation is that I have an unknown $x\in \mathbb{R}^n$. I assume that I can measure the scalar product $a^Tx$ with any unit vector $a\in S^{n-1}$. The measurements are assumed to be i.i.d. Now we assume we have $m$ measurements and write the corresponding vectors as the rows of a matrix $A\in \mathbb{R}^{m\times n}$ and the vectors as a matrix $b\in \mathbb{R}^m$. The best possible reconstruction (see Gauss-Markov) of $x$ is given by the least square solution $(A^TA)^{-1}A^Tb$ where $b\in \mathbb{R}^n$ denotes the measurements. The covariance matrix for the reconstruction is $(A^TA)^{-1}$. This means that if $A$ satisfies the properties above then the components of the reconstructed $x$ are uncorrelated. Also I think that $A$ gives us the best possible way to measure $x$, i.e. minimizing the standard deviation.
• Is $m$ a power of $2$? It would be very nice if it were. – Rodrigo de Azevedo Dec 8 '16 at 16:28
• In general no. But if you have some insights for that case it would definitely be interesting. – user35593 Dec 8 '16 at 23:08
• In general if A is such a matrix then e can build new matrices by concatenating copies of A. Also we can permute the rows of A to get a new solution. – user35593 Dec 9 '16 at 7:29
• If we have solutions with the same n we can build new solutions by concatenating. One could introduce the notion of a 'prime' matrix with this property. I.e. a matrix which does not contain a submatrix with n columns with the same property – user35593 Dec 9 '16 at 7:44
• The rows of this matrix form a tight frame of unit vectors. See e.g. ams.org/notices/201306/rnoti-p748.pdf – Terry Tao Dec 9 '16 at 17:11
Assuming that the Hadamard Conjecture is true, if $m$ is a multiple of $4$, then a thin $m \times n$ matrix that satisfies the given constraints is given by
$$\boxed{\mathrm A := \frac{1}{\sqrt n} \mathrm H_m^{\top} \mathrm S_n}$$
where
• $\mathrm H_m \in \{\pm 1\}^{m \times m}$ is a Hadamard matrix. Thus, the $m$ rows of $\mathrm H_m$ are orthogonal, i.e., $$\mathrm H_m \mathrm H_m^{\top} = m \mathrm I_m$$
• $\mathrm S_n$ is a thin $m \times n$ matrix whose $n$ columns are chosen from the $m$ columns of the $m \times m$ identity matrix. Thus, the $n$ columns of $\mathrm S_n$ are orthonormal, i.e.,
$$\mathrm S_n^{\top} \mathrm S_n = \mathrm I_n$$
Hence,
$$\mathrm A^{\top} \mathrm A = \frac{1}{n} \mathrm S_n^{\top} \mathrm H_m \mathrm H_m^{\top} \mathrm S_n = \frac{m}{n} \mathrm S_n^{\top} \mathrm S_n = \frac{m}{n} \mathrm I_n$$
as desired. Let $\mathrm e_k$ and $\mathrm h_k$ denote the $k$-th columns of $\mathrm I_m$ and $\mathrm H_m$, respectively. Hence,
$$\mathrm e_k^{\top} \mathrm A \mathrm A^{\top} \mathrm e_k = \| \mathrm A^{\top} \mathrm e_k \|_2^2 = \frac 1n \| \mathrm S_n^{\top} \mathrm H_m \mathrm e_k \|_2^2 = \frac 1n \| \mathrm S_n^{\top} \mathrm h_k \|_2^2 = \frac 1n \sum_{k=1}^n (\pm 1)^2 = \frac nn = 1$$
for all $k \in \{1,2,\dots,m\}$, as desired. Note that we used the fact that the entries of $\mathrm h_k$ are $\pm 1$.
If $m$ is a power of $2$, then $\mathrm H_m$ can be built recursively using the Sylvester construction
$$\mathrm H_{2k} = \begin{bmatrix} \mathrm H_k & \mathrm H_k\\ \mathrm H_k & -\mathrm H_k\end{bmatrix} \qquad \qquad \qquad \mathrm H_1 = 1$$
which builds (symmetric) Walsh matrices. If $m$ is not a power of $2$, we can use the Paley construction instead.
Example
Let $m = 8$ and $n = 3$. Since $8$ is a power of $2$, we can use the Sylvester construction to build $\mathrm H_8$.
Using MATLAB,
>> H1 = 1;
>> H2 = [H1,H1;H1,-H1];
>> H4 = [H2,H2;H2,-H2];
>> H8 = [H4,H4;H4,-H4]
H8 =
1 1 1 1 1 1 1 1
1 -1 1 -1 1 -1 1 -1
1 1 -1 -1 1 1 -1 -1
1 -1 -1 1 1 -1 -1 1
1 1 1 1 -1 -1 -1 -1
1 -1 1 -1 -1 1 -1 1
1 1 -1 -1 -1 -1 1 1
1 -1 -1 1 -1 1 1 -1
Let the $3$ columns of $\mathrm S_3$ be the first $3$ columns of $\mathrm I_8$
>> I8 = eye(8);
>> H8 * I8(:,[1,2,3])
ans =
1 1 1
1 -1 1
1 1 -1
1 -1 -1
1 1 1
1 -1 1
1 1 -1
1 -1 -1
Note that the last four rows are copies of the first four rows. Hence, let the $3$ columns of $\mathrm S_3$ be the 2nd, 3rd and 5th columns of $\mathrm I_8$
>> H8 * I8(:,[2,3,5])
ans =
1 1 1
-1 1 1
1 -1 1
-1 -1 1
1 1 -1
-1 1 -1
1 -1 -1
-1 -1 -1
Note that the $8$ rows are now the $8$ vertices of the cube $[-1,1]^3$.
We build matrix $\mathrm A$ by normalizing the rows
>> A = inv(sqrt(3)) * H8 * I8(:,[2,3,5])
A =
0.5774 0.5774 0.5774
-0.5774 0.5774 0.5774
0.5774 -0.5774 0.5774
-0.5774 -0.5774 0.5774
0.5774 0.5774 -0.5774
-0.5774 0.5774 -0.5774
0.5774 -0.5774 -0.5774
-0.5774 -0.5774 -0.5774
Is the constraint $\mathrm A^{\top} \mathrm A = \frac 83 \mathrm I_3$ satisfied?
>> A' * A
ans =
2.6667 0 0
0 2.6667 0
0 0 2.6667
It is. Are the diagonal entries of $\mathrm A \mathrm A^{\top}$ equal to $1$?
>> A * A'
ans =
1.0000 0.3333 0.3333 -0.3333 0.3333 -0.3333 -0.3333 -1.0000
0.3333 1.0000 -0.3333 0.3333 -0.3333 0.3333 -1.0000 -0.3333
0.3333 -0.3333 1.0000 0.3333 -0.3333 -1.0000 0.3333 -0.3333
-0.3333 0.3333 0.3333 1.0000 -1.0000 -0.3333 -0.3333 0.3333
0.3333 -0.3333 -0.3333 -1.0000 1.0000 0.3333 0.3333 -0.3333
-0.3333 0.3333 -1.0000 -0.3333 0.3333 1.0000 -0.3333 0.3333
-0.3333 -1.0000 0.3333 -0.3333 0.3333 -0.3333 1.0000 0.3333
-1.0000 -0.3333 -0.3333 0.3333 -0.3333 0.3333 0.3333 1.0000
They are. | 2021-04-14T14:34:31 | {
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https://math.stackexchange.com/questions/1398696/why-does-lh%C3%B4pitals-rule-work-for-sequences?noredirect=1 | Why does L'Hôpital's rule work for sequences?
Say, for the classic example, $\frac{\log(n)}{n}$, this sequence converges to zero, from applying L'Hôpital's rule. Why does it work in the discrete setting, when the rule is about differentiable functions?
Is it because at infinity, it doesn't matter that we relabel the discrete variable, $n$, with a continuous variable, $x$, and instead look at the limit of $\frac{\log(x)}{x}$?
But then what about the quotients of sequences that go to the indeterminate form $\frac{0}{0}$? Why is it OK to use L'Hôpital's rule, as $n$ goes to zero?
I haven't found anything on Wikipedia or Wolfram about the discrete setting.
Thanks.
• The counterpart of L'Hospital's rule for sequences is the so-called Stolz Theorem. – Zhanxiong Aug 16 '15 at 0:28
• Thanks for the link, @Zhanxiong. – User001 Aug 16 '15 at 0:52
• possible duplicate of Is there a discrete version of de l'Hôpital's rule? – J. M. isn't a mathematician Aug 16 '15 at 21:26
• @Guesswhoitis.I don't think that's the same question. He asks "why do people differentiate sequences?". – user223391 Aug 16 '15 at 22:07
• I disagree. Some dots would need to be connected in order to be able to say this question is a duplicate of the other one. – Robert Soupe Aug 17 '15 at 2:03
There IS a L'Hospital's rule for sequences called Stolz-Cesàro theorem. If you have an indeterminate form, then:
$$\lim\limits_{n\to\infty} \frac{s_n}{t_n}=\lim\limits_{n\to\infty} \frac{s_n-s_{n-1}}{t_n-t_{n-1}}$$
$$\lim\limits_{n\to\infty}\frac{\ln(n)}{n}=\lim\limits_{n\to\infty}\frac{\ln\left(\frac{n}{n-1}\right)}{n-n+1}=\lim\limits_{n\to\infty}\ln\left(\frac{n}{n-1}\right)=0$$
But that isn't your question. Your question is, why do people "differentiate"? Basically because the real case covers the discrete case.
Recall the definition of limits for real and discrete cases.
Definition. A sequence, $s_n\colon \Bbb{N}\to \Bbb{R},$ converges to $L$ as $n\to\infty$, written $\lim\limits_{n\to\infty} s_n=L$ iff for all $\epsilon>0$ there is some $N$ such that for all $n\in \Bbb{N}$ with $n>N$, $|s_n-L|<\epsilon$.
Definition. A function, $f(x) \colon \Bbb{R}\to \Bbb{R}$ converges to $L$ as $x\to\infty$, written $\lim\limits_{x\to\infty} f(x)=L$ iff for all $\epsilon>0$ there is some $X$ such that for all $x\in \Bbb{R}$ with $x>X$, $|f(x)-L|<\epsilon$.
So if $f(x)$ is a real valued function that agrees with a sequence, $s_n$ on integer values, then $\lim\limits_{x\to\infty} f(x)=L$ implies $\lim\limits_{n\to\infty} s_n=L$.
• Thanks so much @avid19 - this was an awesome explanation. – User001 Aug 16 '15 at 0:51
• My +1. I want to add here that there is a catch. There are cases when $f(n) \to L$ as $n \to \infty$ but $f(x)$ has no limit as $x \to \infty$. – Paramanand Singh Aug 16 '15 at 5:10
• @ParamanandSingh: The most obvious example would be $f(x) = \sin(2\pi x)$ – R.. GitHub STOP HELPING ICE Aug 16 '15 at 14:19
Your explanation is not really precise, it does matter whether you use the discrete or continuous variable. However, there is a theorem in mathematical analysis that states that the following is equivalent:
• $\lim_{x \rightarrow c} f(x) = A$
• For every sequence $\{x_n\}$ such that $\forall n \in \mathbb{N} : x_n \in D(f), x_n \neq c$ and that $\lim_{n \rightarrow \infty} x_n = c$ it is true that $\lim_{n \rightarrow \infty} f(x_n) = A$.
In simpler words, once you know the limit of a function in continuous variable, like $\lim_{x \rightarrow \infty} \frac{\log x}{x} = 0$, you also know the limit of any sequence you get by "picking out" points of this function's domain, in your case specifically you take the sequence $\{x_n\} = \{n\}$. Notice that the conditions of the second statement are met, since $\frac{\log x}{x}$ is defined for every $n$, $\lim_{n \rightarrow \infty} n = \infty$ and $n \neq \infty$ for every $n$ as well.
This also solves the problem for the limits of indeterminate form "$\frac{0}{0}$", or any other.
Hope that helps :),
Epsiloney
Fact: If a function $f:\Bbb R\to\Bbb R$ is continuous, $x_n\to x$ then $f(x_n)\to f(x)$.
You can see here Stolz–Cesàro theorem for the "l'Hôpital's rule" for sequences.
This uses the fact that if $\displaystyle\lim_{x\to\infty}f(x)=L, \text{ then } \lim_{n\to\infty}f(n)=L$ since
$\displaystyle\lim_{x\to\infty}f(x)=L\iff$ for every $\epsilon>0,$ there is an M such that if $x\ge M,$ then $\big|f(x)-L\big|<\epsilon$ and
$\displaystyle\lim_{n\to\infty}f(n)=L\iff$ for every $\epsilon>0,$ there is an N such that if $n\ge N,$ then $\big|f(n)-L\big|<\epsilon$ $\;\;$(with $n\in\mathbb{N}$) | 2020-06-02T13:42:44 | {
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https://math.stackexchange.com/questions/2808027/finding-the-probability-density-function-of-z-x-y | # Finding the probability density function of $Z = X + Y$
### The problem
Let there be two independent random variables given. Let's name them $X$ and $Y$. Suppose the variables have both normal distribution $N(\mu, \lambda^2).$
Let's define third random variable $Z = X + Y$. The task is to find the density function of $Z$.
### My attempt
I know that the density function can be easily calculated using the distribution function (I would need to differentiate it). That's why I started with finding the distribution. $$F_Z(t) = Pr(Z \le t) = Pr(Y + X \le t) = Pr(Y \le t - X) = \int \limits_{-\infty}^{\infty}Pr(Y \le t-x)f_X(x) dx$$ The equation above leads us to the following one: $$\int \limits_{-\infty}^{\infty} \int \limits_{-\infty}^{t-x} f_Y(y)f_X(x) \mbox{d} y\mbox{d}x \tag{1}.$$ Using the knowledge about $f_X$ and $f_Y$ we can write $(1)$ in this form: $$\theta\int \limits_{-\infty}^{\infty} \int \limits_{-\infty}^{t-x}e^{-\frac{(y- \mu)^2}{\lambda^2}} e^{-\frac{(x- \mu)^2}{\lambda^2}} \mbox{d} y\mbox{d}x,$$ where $\theta = \frac{1}{\sqrt{2 \lambda^2}}.$
I wonder if my attempt was correct because the integral is quite horrible. Is it possible to calculate it in a "nicer" way?
Is $(1)$ equivalent to this expression $(f_X*f_Y)(x)$?
If yes I would get the following integral: $$\int \limits_{\mathbb{R}} e^{-\frac{(y- \mu)^2}{\lambda^2}} e^{-\frac{(x- y - \mu)^2}{\lambda^2}} \mbox{d}y.$$
Is there a method of calculating such things?
• en.wikipedia.org/wiki/… – saulspatz Jun 4 '18 at 18:50
• @Hendrra In general, sum of random variables $X$ and $Y$ with densities $f$ and $g$ has density being convolution of $f$ and $g$, you are correct – Jakobian Jun 4 '18 at 19:04
• @Adam Thank you. What about my first approach? – Hendrra Jun 4 '18 at 19:33
From your expression, $F_Z(t) = \int_\mathbb{R}P(Y \leq t-x)f_X(x)dx$, differentiate to obtain the density (you may need to justify why you can differentiate under the integral),
$$f_Z(t) = \int_\mathbb{R}f_Y(t-x)f_X(x)dx = \int_\mathbb{R}\frac{1}{\sqrt{2\pi}}e^{-(t-x-\mu)^2/2\lambda^2}\frac{1}{\sqrt{2\pi}}e^{-(x-\mu)^2/2\lambda^2}dx$$
which we recognize as $(f_Y*f_X) (x)$. To actually evaluate the integral, expand the exponent terms inside the integrand,
$$-\frac{1}{2\lambda^2}\left[(t-x-\mu)^2 + (x-\mu)^2\right] = -\frac{1}{2\lambda^2}(t^2-2\mu t + 2\mu^2 + \underbrace{2x^2-2tx}_{(*)})$$
Complete the square on the $(*)$ term, $2x^2 - 2tx = 2\left(x-\frac{t}{2}\right)^2-\frac{t^2}{2}$. Putting things together,
$$-\frac{1}{2\lambda^2}(t^2-2\mu t + 2\mu^2 + 2x^2-2tx) = -\frac{1}{4\lambda^2}(t^2-4\mu t + 4\mu^2)-\frac{1}{\lambda^2}(x-t/2)^2$$
So, we can pull out the $t$ terms from the integrand to be left with,
$$f_Z(t) = \frac{1}{\sqrt{4\pi\lambda^2}}e^{-(t-2\mu)^2/4\lambda^2}\underbrace{\int_\mathbb{R}\frac{1}{\sqrt{\pi\lambda^2}}e^{-(x-t/2)^2/\lambda^2}dx}_{=1} = \frac{1}{\sqrt{4\pi\lambda^2}}e^{-(t-2\mu)^2/4\lambda^2}$$
The integral above equals $1$ because the integrand is the density function of a $N(t/2,\lambda^2/2)$ distribution.
• At the end, did you intend $N(2 \mu,2 \lambda^2)$ for the mean and variance rather than $t/2, \lambda^2 /{2}$? – Henry Jun 5 '18 at 16:31
• Sorry, I was referring to the density function inside the final integral (the one that equals 1). The distribution of $Z$ is definitely $N(2\mu,2\lambda^2)$. – Flowsnake Jun 5 '18 at 16:33
• So was I. I think a density $f_Z(t)= \frac{1}{\sqrt{4\pi\lambda^2}}e^{-(t-2\mu)^2/4\lambda^2}$ suggests a normal distribution for $Z$ with a mean of $2 \mu$ and a variance of $2 \lambda^2$ – Henry Jun 5 '18 at 16:37
• The density function I was referring to was, $\frac{1}{\sqrt{\pi\lambda^2}}e^{-(x-t/2)^2/\lambda^2}$, which is the density of a $N(t/2,\lambda^2/2)$. – Flowsnake Jun 5 '18 at 16:49
The problem can also be solved by using the following property: If $X\sim N(\mu_1,\sigma_1^2)$ and $Y \sim N(\mu_2,\sigma_2^2)$ and are independent, then $aX + bY \sim N(a\mu_1 + b\mu_2,a^2\sigma_1^2 +b^2\sigma_2^2)$ and thus in your case $Z \sim N(2\mu,2\lambda^2)$.
Edit: I just noticed the comment by @saulspatz above who already pointed this out. | 2019-08-24T06:53:13 | {
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https://math.stackexchange.com/questions/4455265/the-coupon-collectors-most-collected-coupon | # The coupon collector's most collected coupon
Suppose a coupon collector is collecting a set of $$n$$ coupons that he receives one-by-one uniformly at random.
If the collector stops exactly when the collection is complete, we know the expected number of coupons in his collection is $$n*H[n]$$. What is the expected number of copies, $$M$$, of his most collected coupon?
When $$n = 2$$, then $$M = 2$$ because after the first coupon he will collect the other coupon with probability p ~ Geom(1/2). So he will collect the same coupon one additional time on average, and that coupon is certain to be his most collected coupon.
I don't know the exact value for any $$n$$ larger than 2, but found some approximate values by simulation:
n M
3 2.8415
4 3.4992
5 4.0259
6 4.4633
7 4.8377
8 5.1649
9 5.4560
EDIT 1:
For small n enumerating the small possibilities converges faster than simulation, and from this approach I hypothesize that $$M[3] = (15 + 6 \sqrt{5})/10$$ but don't have any real argument to support that claim.
EDIT 2:
With some more thought, I find that M has an explicit sum formula. The probability the collection terminates with a given distribution of coupons $$v = \{c_1, ..., c_{n-1}\}$$ other than the final collected coupon is just the multinomial coefficient $$(c_1; ...; c_{n-1})$$ over $$n^\text{total # of coupons in the collection}$$. This is an infinite sum over n-1 variables. Here is Mathematica code the computes the sum for terms where the collection has no more than k copies of any coupon:
M[n_, k_] := Sum[Max[v]*(Multinomial @@ v)/n^Total[v], {v, Tuples[Range[1, k], n-1]}]
Mathematica can't actually evaluate this though for $$k \rightarrow \infty$$ though.
• You might be interested in Exercise 2.20 of Boucheron, Lugosi, and Massart's ""Concentration Inequalities: A Nonasymptotic Theory of Independence", which asks to prove an upper bound of $$e\log n \cdot ce^{W((1-c)/(ce))}$$for this expectation, when collecting $m = cn\log n$ coupons. ($W$ being the Lambert function.) May 21 at 6:52
Value of $$M_3$$. I will denote $$M$$ as $$M_n$$ to emphasize the dependence on $$n$$. Then by using A230137, we get
\begin{align*} M_3 &= \sum_{n=2}^{\infty} \frac{\sum_{k=1}^{n-1} \binom{n}{k} \max\{k,n-k\}}{3^n} \\ &= \sum_{n=1}^{\infty} \frac{\left( 4^n + \binom{2n}{n} \right)n - 2(2n)}{3^{2n}} + \sum_{n=1}^{\infty} \frac{\left( 4^n + \binom{2n}{n} \right)(2n+1) - 2(2n+1)}{3^{2n+1}} \\ &= 3\left(\frac{1}{2} + \frac{1}{\sqrt{5}}\right) \\ &\approx 2.84164, \end{align*}
which is the same as what OP conjectured.
Asymptotic Formula of $$M_n$$. A heuristic seems suggesting that
$$M_n \sim e \log n \qquad \text{as} \quad n \to \infty,$$
The figure below compares simulated values of $$M_n$$ and the values of the asymptotic formula.
Heuristic argument. First, we consider the Poissonized version of the problem:
1. Let $$N^{(1)}, \ldots, N^{(n)}$$ be independent Poisson processes with rate $$\frac{1}{n}$$. Then the arrivals in each $$N^{(i)}$$ model the times when the collector receives coupons of type $$i$$.
2. Let $$T$$ be the first time a complete set of coupons is collected. Then we know that $$T$$ has the same distribution as the sum of $$n$$ independent exponential RVs with respective rates $$1$$, $$\frac{n-1}{n}$$, $$\ldots$$, $$\frac{1}{n}$$. (This is an example of the exponential race problem.) For large $$n$$, this is asymptotically normal with mean $$\sum_{i=1}^{n}\frac{n}{i} \sim n \log n$$ and variance $$\sum_{i=1}^{n}\frac{n^2}{i^2} \sim \zeta(2)n^2$$. This tells that $$T \sim n \log n$$ in probability, and so, it is not unreasonable to expect that
$$M_n = \mathbf{E}\Bigl[ \max_{1\leq i\leq n} N^{(i)}_T \Bigr] \sim \mathbf{E}\Bigl[ \max_{1\leq i\leq n} N^{(i)}_{n\log n} \Bigr]. \tag{1}$$
holds as well.
3. Note that each $$N^{(i)}_{n\log n}$$ has a Poisson distribution with rate $$\log n$$. Now, we fix $$\theta > 1$$ and choose an integer sequence $$(a_n)$$ such that $$a_n > \log n$$ and $$a_n/\log n \to \theta$$ as $$n \to \infty$$. Then
\begin{align*} \mathbf{P}\Bigl( N^{(i)}_{n\log n} > a_n \Bigr) &\asymp \frac{(\log n)^{a_n}}{a_n!}e^{-\log n} \\ &\asymp \frac{(\log n)^{a_n}}{(a_n)^{a_n} e^{-a_n+\log n} \sqrt{a_n}} \\ &\asymp \frac{1}{n^{\psi(\theta) + 1 + o(1)}}, \qquad \psi(\theta) = \theta \log \theta - \theta, \end{align*}
where $$f(n) \asymp g(n)$$ means that $$f(n)/g(n)$$ is bounded away from both $$0$$ and $$\infty$$ as $$n\to\infty$$. This shows that
$$\mathbf{P}\Bigl( \max_{1\leq i\leq n} N^{(i)}_{n\log n} \leq a_n \Bigr) = \biggl[ 1 - \frac{1}{n^{\psi(\theta) + 1 + o(1)}} \biggr]^n \xrightarrow[n\to\infty]{} \begin{cases} 0, & \psi(\theta) < 0, \\[0.5em] 1, & \psi(\theta) > 0. \end{cases}$$
So, by noting that $$\psi(e) = 0$$, we get
$$\max_{1\leq i\leq n} N^{(i)}_{n\log n} \sim e \log n \qquad \text{in probability}. \tag{2}$$
This suggests that $$M_n \sim e \log n$$ as well.
4. I believe that this heuristic argument can be turned into an actual proof. To this, we need the following:
1. We have to justify the relation $$\text{(1)}$$ in an appropriate sense. We may perhaps study the inequality
$$\max_{1\leq i\leq n} N^{(i)}_{(1-\varepsilon)n\log n} \leq \max_{1\leq i\leq n} N^{(i)}_T \leq \max_{1\leq i\leq n} N^{(i)}_{(1+\varepsilon)n\log n} \qquad \text{w.h.p.}$$
and show that the probability of the exceptional event is small enough to bound the difference of both sides of $$\text{(1)}$$ by a vanishing qantity.
2. Note that $$\text{(2)}$$ alone is not enough to show this. So, we need to elaborate the argument in step 3 so that it can actually prove the relation $$\mathbf{E}\bigl[ \max_{1\leq i\leq n} N^{(i)}_{n\log n} \bigr] \sim e \log n$$. | 2022-06-26T11:57:52 | {
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https://mathematica.stackexchange.com/questions/92776/constructing-tri-diagonal-matrices | # Constructing Tri-Diagonal Matrices
Fractional Calculus Course, we are instructed to create an $n \times n$ Tri-Diagonal matrix in the form of:
\begin{array} a A &= \begin{bmatrix} 2 & -1 & 0 & 0 & ... & 0 & 0 & 0\\ -1 & 2 & -1 & 0 & 0 & 0&... & 0 \\ 0 & -1 & 2 & -1 & 0& 0... & 0 & 0\\ 0 & 0 &-1 & 2 & -1 & ... & 0 & 0\\ &&&\vdots&&& \\ 0 & 0 & 0&... & 0 & -1 & 2 & -1\\ 0 & 0 & 0 & 0&... & 0 & -1 & 2 \\ \end{bmatrix} \end{array}
This is where my dilemma begins. I am not sure how to create this Tri-Diagonal Matrix. I came across the "SparseArray" command upon my research and it help me create a Tri-Diagonal Matrix, but I am having a hard time manipulating it to get the $-1, 2, -1$ pattern I am looking for.
mat = SparseArray[ {i_, j_} /; Abs[i - j] <= 1 :> 1, {10, 10}];
mat // MatrixForm
Above is the command I used for a $10 \times 10$ matrix. But the tri-diagonal entries were all 1's.
Thus my question is, how would I create the tri-diagonal matrix $n \times n$ I desire? Is there a way to create a function so I can simply manipulate the value of $n$ to get a new matrix without typing (or copy-pasting) the entire code again?
• Look up Band in the help files. – march Sep 1 '15 at 17:51
• Just curious ...What is a Fractional Calculus Course ? – Dr. belisarius Sep 1 '15 at 18:19
• Fractional Calculus is a course that focuses on the applications of fraction derivatives and fraction integrals. So imagine taking half the derivative of f(x) or the two-third integral of f(x). I'm still new to the material so I can't give you a better example that this, sorry. – Kevin_H Sep 1 '15 at 19:09
• Proposed duplicate: (13004). Related: (13796) – Mr.Wizard Sep 2 '15 at 0:07
SparseArray[{{i_, i_} :> 2, {i_, j_} :> -1 /; Abs[i - j] == 1}, {10, 10}] // MatrixForm
or
SparseArray[{Band[{1, 1}] -> 2, Band[{2, 1}] -> -1, Band[{1, 2}] -> -1}, {10, 10}] //
MatrixForm
The following is sometimes useful:
m = {{{a, b}, {b, a}}};
d = {10, 10};
SparseArray[{Band[{2, 2}, d] -> m, Band[{1, 1}, d] -> m}, d] // MatrixForm
Without SparseArray:
n = 10;
Total[
{DiagonalMatrix[Array[-1 &, n - 1], -1],
DiagonalMatrix[Array[2 &, n]],
DiagonalMatrix[Array[-1 &, n - 1], 1]}
]
Or strictly using Array:
Array[Which[#1 == #2, 2, Abs[#1 - #2] == 1, -1, True, 0] &, {10, 10}]
AND, what the heck? One more for more flexible applications:
a = {2, -1, -2, 3};
n = 10;
With[{a1 = PadRight[a, n]}, (Array[a1[[Abs[#1 - #2] + 1]] &, {n, n}])]//MatrixForm
• Your second snippet is what was usually done before SparseArray[] came along. – J. M. is away Sep 2 '15 at 0:57
This should be faster (where n is your square dimension, e.g. 1000 for 1kX1k), easily extends to n-diagonal symmetric with no performance impact:
ToeplitzMatrix[PadRight[{2, -1}, n]] | 2019-06-17T05:48:39 | {
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http://www.kanunpiyano.com/i-m-uaq/biconditional-statement-truth-table-4204cd | # biconditional statement truth table
We will then examine the biconditional of these statements. Sign up using Google Sign up using Facebook Sign up using Email and Password Submit. Compare the statement R: (a is even) $$\Rightarrow$$ (a is divisible by 2) with this truth table. We can use the properties of logical equivalence to show that this compound statement is logically equivalent to $$T$$. Give a real-life example of two statements or events P and Q such that P<=>Q is always true. Venn diagram of ↔ (true part in red) In logic and mathematics, the logical biconditional, sometimes known as the material biconditional, is the logical connective used to conjoin two statements and to form the statement "if and only if", where is known as the antecedent, and the consequent. But would you need to convert the biconditional to an equivalence statement first? V. Truth Table of Logical Biconditional or Double Implication. I've studied them in Mathematical Language subject and Introduction to Mathematical Thinking. Otherwise it is true. For example, the propositional formula p ∧ q → ¬r could be written as p /\ q -> ~r, as p and q => not r, or as p && q -> !r. You can enter logical operators in several different formats. So to do this, I'm going to need a column for the truth values of p, another column for q, and a third column for 'if p then q.' We can use an image of a one-way street to help us remember the symbolic form of a conditional statement, and an image of a two-way street to help us remember the symbolic form of a biconditional statement. The biconditional statement $$p\Leftrightarrow q$$ is true when both $$p$$ and $$q$$ have the same truth value, and is false otherwise. (true) 3. Two formulas A 1 and A 2 are said to be duals of each other if either one can be obtained from the other by replacing ∧ (AND) by ∨ (OR) by ∧ (AND). We still have several conditional geometry statements and their converses from above. Whenever the two statements have the same truth value, the biconditional is true. A tautology is a compound statement that is always true. second condition. Also, when one is false, the other must also be false. Therefore, the sentence "A triangle is isosceles if and only if it has two congruent (equal) sides" is biconditional. The truth tables above show that ~q p is logically equivalent to p q, since these statements have the same exact truth values. Learn the different types of unary and binary operations along with their truth-tables at BYJU'S. Email. If a is odd then the two statements on either side of $$\Rightarrow$$ are false, and again according to the table R is true. When we combine two conditional statements this way, we have a biconditional. Copyright 2010- 2017 MathBootCamps | Privacy Policy, Click to share on Twitter (Opens in new window), Click to share on Facebook (Opens in new window), Click to share on Google+ (Opens in new window), Truth tables for “not”, “and”, “or” (negation, conjunction, disjunction), Analyzing compound propositions with truth tables. You passed the exam if and only if you scored 65% or higher. If no one shows you the notes and you do not see them, a value of true is returned. So the former statement is p: 2 is a prime number. B. A→B. The conditional, p implies q, is false only when the front is true but the back is false. Post as a guest. Compound propositions involve the assembly of multiple statements, using multiple operators. Hence Proved. Logical equality (also known as biconditional) is an operation on two logical values, typically the values of two propositions, that produces a value of true if and only if both operands are false or both operands are true. a. Therefore, it is very important to understand the meaning of these statements. 3. A biconditional is true except when both components are true or both are false. Converse: If the polygon is a quadrilateral, then the polygon has only four sides. Ask Question Asked 9 years, 4 months ago. The following is truth table for ↔ (also written as ≡, =, or P EQ Q): The statement sr is also true. So we can state the truth table for the truth functional connective which is the biconditional as follows. Two line segments are congruent if and only if they are of equal length. Therefore the order of the rows doesn’t matter – its the rows themselves that must be correct. first condition. If I get money, then I will purchase a computer. A biconditional statement is one of the form "if and only if", sometimes written as "iff". The biconditional connective can be represented by ≡ — <—> or <=> and is … When P is true and Q is true, then the biconditional, P if and only if Q is going to be true. Writing this out is the first step of any truth table. Theorem 1. The truth table for the biconditional is . The truth table for ⇔ is shown below. biconditional A logical statement combining two statements, truth values, or formulas P and Q in such a way that the outcome is true only if P and Q are both true or both false, as indicated in the table. When we combine two conditional statements this way, we have a biconditional. Otherwise, it is false. To learn more, see our tips on writing great answers. For each truth table below, we have two propositions: p and q. We start by constructing a truth table with 8 rows to cover all possible scenarios. Otherwise it is false. A biconditional statement is really a combination of a conditional statement and its converse. Biconditional Statements (If-and-only-If Statements) The truth table for P ↔ Q is shown below. In this section we will analyze the other two types If-Then and If and only if. According to when p is false, the conditional p → q is true regardless of the truth value of q. In this post, we’ll be going over how a table setup can help you figure out the truth of conditional statements. Watch Queue Queue. 13. s: A triangle has two congruent (equal) sides. The biconditional operator is sometimes called the "if and only if" operator. Otherwise it is true. You are in Texas if you are in Houston. A biconditional is true only when p and q have the same truth value. The biconditional x→y denotes “ x if and only if y,” where x is a hypothesis and y is a conclusion. Construct a truth table for ~p ↔ q Construct a truth table for (q↔p)→q Construct a truth table for p↔(q∨p) A self-contradiction is a compound statement that is always false. A biconditional statement will be considered as truth when both the parts will have a similar truth value. It is helpful to think of the biconditional as a conditional statement that is true in both directions. The statement qp is also false by the same definition. • Identify logically equivalent forms of a conditional. 4. T. T. T. T. F. F. F. T. T. F. F. T. Example: We have a conditional statement If it is raining, we will not play. Logical equality (also known as biconditional) is an operation on two logical values, typically the values of two propositions, that produces a value of true if and only if both operands are false or both operands are true.. The conditional statement is saying that if p is true, then q will immediately follow and thus be true. Is there XNOR (Logical biconditional) operator in C#? Is this statement biconditional? b. You passed the exam iff you scored 65% or higher. Class 1 - 3; Class 4 - 5; Class 6 - 10; Class 11 - 12; CBSE. The statement pq is false by the definition of a conditional. Name. All Rights Reserved. P Q P Q T T T T F F F T F F F T 50 Examples: 51 I get wet it is raining x 2 = 1 ( x = 1 x = -1) False (ii) True (i) Write down the truth value of the following statements. To show that equivalence exists between two statements, we use the biconditional if and only if. The truth table for the biconditional is Note that is equivalent to Biconditional statements occur frequently in mathematics. text/html 8/17/2008 5:10:46 PM bigamee 0. Next, we can focus on the antecedent, $$m \wedge \sim p$$. Now you will be introduced to the concepts of logical equivalence and compound propositions. It is a combination of two conditional statements, “if two line segments are congruent then they are of equal length” and “if two line segments are of equal length then they are congruent”. en.wiktionary.org. (true) 4. The conditional operator is represented by a double-headed arrow ↔. All birds have feathers. We will then examine the biconditional of these statements. Construct a truth table for the statement $$(m \wedge \sim p) \rightarrow r$$ Solution. Definition: A biconditional statement is defined to be true whenever both parts have the same truth value. The conditional, p implies q, is false only when the front is true but the back is false. Mathematicians abbreviate "if and only if" with "iff." In the truth table above, when p and q have the same truth values, the compound statement (p q) (q p) is true. If and only if statements, which math people like to shorthand with “iff”, are very powerful as they are essentially saying that p and q are interchangeable statements. A biconditional statement is often used in defining a notation or a mathematical concept. NCERT Books. This is reflected in the truth table. The biconditional statement $$p\Leftrightarrow q$$ is true when both $$p$$ and $$q$$ have the same truth value, and is false otherwise. Sign up to get occasional emails (once every couple or three weeks) letting you know what's new! T. T. T. T. F. F. F. T. F. F. F. T. Note that is equivalent to Biconditional statements occur frequently in mathematics. And the latter statement is q: 2 is an even number. Symbolically, it is equivalent to: $$\left(p \Rightarrow q\right) \wedge \left(q \Rightarrow p\right)$$. • Construct truth tables for biconditional statements. To help you remember the truth tables for these statements, you can think of the following: Previous: Truth tables for “not”, “and”, “or” (negation, conjunction, disjunction), Next: Analyzing compound propositions with truth tables. Construct a truth table for (p↔q)∧(p↔~q), is this a self-contradiction. If given a biconditional logic statement. Let qp represent "If x = 5, then x + 7 = 11.". In a biconditional statement, p if q is true whenever the two statements have the same truth value. Thus R is true no matter what value a has. A biconditional statement is really a combination of a conditional statement and its converse. The biconditional statement $$p\Leftrightarrow q$$ is true when both $$p$$ and $$q$$ have the same truth value, and is false otherwise. The biconditional operator looks like this: ↔ It is a diadic operator. A biconditional statement is one of the form "if and only if", sometimes written as "iff". If a is even then the two statements on either side of $$\Rightarrow$$ are true, so according to the table R is true. Notice that the truth table shows all of these possibilities. Similarly, the second row follows this because is we say “p implies q”, and then p is true but q is false, then the statement “p implies q” must be false, as q didn’t immediately follow p. The last two rows are the tough ones to think about. The statement rs is true by definition of a conditional. The conditional operator is represented by a double-headed arrow ↔. A discussion of conditional (or 'if') statements and biconditional statements. Principle of Duality. Unit 3 - Truth Tables for Conditional & Biconditional and Equivalent Statements & De Morgan's Laws. Make truth tables. Let's put in the possible values for p and q. A polygon is a triangle iff it has exactly 3 sides. The biconditional connects, any two propositions, let's call them P and Q, it doesn't matter what they are. BOOK FREE CLASS; COMPETITIVE EXAMS. As we analyze the truth tables, remember that the idea is to show the truth value for the statement, given every possible combination of truth values for p and q. Biconditional Statement A biconditional statement is a combination of a conditional statement and its converse written in the if and only if form. When x 5, both a and b are false. Continuing with the sunglasses example just a little more, the only time you would question the validity of my statement is if you saw me on a sunny day without my sunglasses (p true, q false). Sign up or log in. Conditional: If the quadrilateral has four congruent sides and angles, then the quadrilateral is a square. Biconditional statement? Implication In natural language we often hear expressions or statements like this one: If Athletic Bilbao wins, I'll… Since, the truth tables are the same, hence they are logically equivalent. Let's look at a truth table for this compound statement. They can either both be true (first row), both be false (last row), or have one true and the other false (middle two rows). Writing Conditional Statements Rewriting a Statement in If-Then Form Use red to identify the hypothesis and blue to identify the conclusion. Based on the truth table of Question 1, we can conclude that P if and only Q is true when both P and Q are _____, or if both P and Q are _____. Compound Propositions and Logical Equivalence Edit. Edit. Truth table is used for boolean algebra, which involves only True or False values. Negation is the statement “not p”, denoted ¬p, and so it would have the opposite truth value of p. If p is true, then ¬p if false. biconditional statement = biconditionality; biconditionally; biconditionals; bicondylar; bicondylar diameter; biconditional in English translation and definition "biconditional", Dictionary English-English online. When one is true, you automatically know the other is true as well. In this implication, p is called the hypothesis (or antecedent) and q is called the conclusion (or consequent). Having two conditions. ", Solution: rs represents, "You passed the exam if and only if you scored 65% or higher.". This video is unavailable. How to find the truth value of a biconditional statement: definition, truth value, 4 examples, and their solutions. Definition: A biconditional statement is defined to be true whenever both parts have the same truth value. A statement is a declarative sentence which has one and only one of the two possible values called truth values. ". It's a biconditional statement. In the first conditional, p is the hypothesis and q is the conclusion; in the second conditional, q is the hypothesis and p is the conclusion. Includes a math lesson, 2 practice sheets, homework sheet, and a quiz! Hope someone can help with this. Truth Table Generator This tool generates truth tables for propositional logic formulas. If no one shows you the notes and you see them, the biconditional statement is violated. Examples. Truth Table for Conditional Statement. The biconditional pq represents "p if and only if q," where p is a hypothesis and q is a conclusion. Directions: Read each question below. Select your answer by clicking on its button. The biconditional, p iff q, is true whenever the two statements have the same truth value. More examples of the biconditional if and only if... ] in other words, logical statement p q! Analyze the other two types If-Then and if and only if I get money, then polygon... Iff : rewrite each of the following is a compound statement p q, since these statements the... That this compound statement that is equivalent to p q, is false only when the front is regardless! Rows to cover all possible scenarios ( If-Then statements ) the truth tables the. Language and code, conditional, p is true whenever the two possible values called truth values of the is! Math lesson, 2 practice sheets, homework sheet, and their converses from above a counter-example statement rs true... ) operator in c # ] [ 3 ] this is often abbreviated as iff... Logic formulas not see them, a value of true is returned shows you the and! You the notes and you do not see them, a: it is a and! Sides, then the polygon is a conclusion iff you scored 65 or... 2 practice sheets, homework sheet, and q are logically biconditional statement truth table also how to do it without a... Be correct exam Question: know how to do a truth table also be false the! Conjunction of two conditional statements am alive them in mathematical language subject and Introduction to mathematical Thinking tables for logic., calculator guides, and q table setup can help you figure out the way it.... Or false ~q p is logically equivalent to p q, is this a self-contradiction is a hypothesis q! Double implication on the antecedent, \ ( ( m \wedge \sim p ) \Rightarrow r\ ).! 10 ; Class 11 - 12 ; CBSE \sim p\ ) be.! - 3 ; Class 4 - 5 ; Class 6 - 10 ; Class 11 - 12 CBSE! Problem packs components are true and biconditional statements frequently in mathematics therefore the order of the following is rectangle. Q will immediately follow and thus be true whenever both parts have the truth. Start by constructing a truth table for any two propositions, let 's look a... We can focus on the truth or falsity of its components, calculator guides and...: \ ( ( m \wedge \sim p\ ) - truth tables of two statements, agree! The statement rs is true by definition of a complicated statement depends on truth! Writing this out is the biconditional x→y denotes “ x if and only if '' ... Whenever both parts have the same, hence they are logically equivalent for boolean algebra, which involves only or! A triangle iff it has two congruent ( equal ) sides..., Solution: xy represents the sentence: x + 7 = 11 iff x = 5. represents... 'S put in the table below, we have two propositions, 's... In Example 3, we will rewrite each of the two statements have the same truth,. And the latter statement is saying that if p is false only the... Falsity of its components ll be going over how a table setup can help you figure the. = b and b = c, then x = 5, then the quadrilateral four. Be false an equivalence statement first ^ q ) and q, is false ; otherwise, it is important... New free lessons and adding more study guides, calculator guides, calculator guides, and their from! Unit 3 - truth tables for conditional & biconditional and equivalent statements side by side in the truth... Is equivalent to \ ( m \wedge \sim p\ ) q are true p and q ''! Occur frequently in mathematics logical equivalencies to determine how the truth table for p↔ ( q∨p a... N'T matter what they are the possible truth values of this statement: ( p. a is. Feedback to your answer is provided in the first row naturally follows definition. Guides, calculator guides, and a biconditional use a truth table logical. Form if x + 2 = 7 if and only if q ' I 'll try! Is false adding more study guides, calculator guides, calculator guides, calculator guides, calculator guides, contrapositive. I am breathing if and only if q ' '' instead of if and only q! Step of any truth table for biconditional pq represents p if and only if they are of length! Sentence: x + 2 = 7 if and only if y, ” x... 11 iff x = 5 '' is not biconditional, and a biconditional state the truth for. My attempt to explain these topics: implication, conditional, p if and only one of statement! Its inverse, converse, and q raining and b are false Example. Using Email and Password Submit false by the symbol ↔ or ⇔ one shows you the notes you! B is given by ; a are confident about your work. learn the types. ( If-Then statements ) the truth or falsity of its components polygon is a sentence. Practice sheets, homework sheet, and contrapositive out is the first row naturally this... You agree to receive useful information and to our privacy policy you the notes and you do not see,... The conclusion ( or consequent ) a triangle has two congruent ( equal ) sides. p.. Are always posting new free biconditional statement truth table and adding more study guides, and problem.! Following sentences using iff I 've studied them in mathematical language subject and Introduction to Thinking... True in both directions to omit such columns if you are in Houston defined we. Prime number as a conditional statement and its converse conditional statements sides, then +!, equivalence and compound propositions involve the assembly of multiple statements, you may choose omit! Agree to receive useful information and to our privacy policy you automatically know the other must also false... Byju 's of its components logically equivalent exactly 3 sides. topics: implication, conditional, equivalence compound... Congruent sides and angles, then I will purchase a computer construct a truth table for this compound statement -. Same definition which has one and only one of the form if and if... And y is a hypothesis and blue to identify the conclusion know the two... Logical equivalence and biconditional, its inverse, converse, and a biconditional is true whenever both have... When the front is true whenever both parts have the same truth value, the first of! ~ ( ~P ^ q ) and q are logically equivalent to \ ( T\ ) of the following,! You 'll learn about what it does n't matter what value a has are listed in the below... The latter statement is logically equivalent to: \ ( ( m \sim... Examples 1 through 4 using this abbreviation: x + 7 = 11 iff =! Through 4 using this abbreviation two statements are the same truth value, calculator guides, calculator guides, q. For biconditional statement truth table and why it comes out the truth or falsity of components... Can look at the truth values of the given statement is often used in defining a notation or mathematical... Can disprove via a counter-example logical equivalence to show that this compound (... Has one and only if... ] x→y denotes “ x if and only if q ' of! New free lessons and adding more study guides, calculator guides, calculator guides, guides... Lesson, 2 practice sheets, homework sheet, and q are true = 5 ''. Of q to discuss examples both in natural language and code you passed exam... Also “ biconditional statement truth table implies that p < = > q, its,! Both components are true a triangle iff it has two congruent ( equal ).., since these statements table is used for boolean algebra, which is a conclusion [ ]. Is helpful to think of the form ' p if and only if y, where. Is provided in the RESULTS BOX answer is provided in the table below, will... You know what 's new 10 ; Class 6 - 10 ; Class -. Its components regardless of the biconditional uses a two-valued logic: every statement is used. More examples of the form if x = 5 '' is not.! For conditional & biconditional and equivalent statements side by side in the same value! Operator is sometimes called the hypothesis and q, is true by definition of a conditional statement has a arrow! Of any truth table for any two inputs, say a and b we. Iff it has exactly 3 sides. form biconditional statement truth table red to identify the (... Statement, p if and only if q, since these statements pq represents p and. Final exam Question: know how to do it without using a Truth-Table consequent.... Similar truth value choose a different button propositions involve the assembly of multiple statements you... ( p↔~q ), is true, then I will purchase a computer a statement... Definition: a biconditional statement will be introduced to the concepts of logical equivalence and biconditional statements rectangle and... Functional connective which is the first step of any truth biconditional statement truth table for this statement. A conditional q is false, the truth table which involves only true or values. Such columns if you are confident about your work. b is given ;! | 2021-02-27T00:51:39 | {
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https://codereview.stackexchange.com/questions/128810/calculate-the-lattice-paths-in-an-nn-lattice-project-euler-15 | # Calculate the lattice paths in an n*n lattice (Project Euler #15)
Project Euler #15 is to find out all the distinct possible ways there are to move from the first point (1,1) to point (n,n) in an n*n lattice.
def lattice_paths_of_n(n):
list2 = []
my_list = []
for i in range(1, n+2):
list2.append(i)
for i in range(1, n+2):
my_list.append(list2)
for i in range(0,n+1):
for f in range(0,n+1):
if f == 0 or i == 0:
my_list[i][f] = 1
else:
my_list[i][f] = my_list[i-1][f]+my_list[i][f-1]
return my_list[n][n]
print(lattice_paths_of_n(20))
However, this function is extremely inefficient and I would appreciate it if you could give me suggestions to optimize the same. I tried to find a more efficient solution to this problem on this site, but couldn't find one in Python 3.
• What leads you to believe it's extremely inefficient? (I'm not arguing it isn't, just asking) – Insane May 20 '16 at 3:10
• Well, I basically used brute force; whereas I can use a more efficient intelligent way to calculate the answer. – Aradhye Agarwal May 20 '16 at 3:11
• Won't spoil the fun for you, but there is a closed form solution that involves permutations: you just need to compute a couple of factorials to get to the answer. – Jaime May 20 '16 at 4:51
• @Jaime Well this is what Code Review is! Feel free to write it up as an answer with an explanation :) – Insane May 20 '16 at 10:03
• @Jaime if you do answer, I would appreciate it if you also provided a logical explanation behind your formula and it's derivation. – Aradhye Agarwal May 20 '16 at 10:31
To move from the top left corner of an $n\times n$ grid to the opposite one you have to move $n$ times to the right and $n$ times down. So in total you will do $2n$ moves. If you could make those in any order there would be $(2 n)!$ ways of doing them. But you don't have that freedom, because the order within the movements to the right and within the movements down is fixed, e.g. you have to move from row 4 to row 5 before you can move from row 5 to row 6. So of the $n!$ ways the movements to the right can be ordered, only one is valid, and similarly with the movements down.
Summing it all up, the closed form answer to that problem is:
$$\frac{(2n)!}{n!n!}$$
Unsurprisingly this is the same as $C_{2n, n}$, the combinations of $2n$ items taken $n$ at a time. You could think of the same problem as having $2n$ movements and having to choose $n$ of those to make e.g. the movements down, leaving the others for the movements right.
With Python's arbitrary size integers you could simply calculate that as:
import math
def pe15(n):
n_fact = math.factorial(n)
return math.factorial(2 * n) / n_fact / n_fact
To give this answer a little more meat, in most other languages you don't have the luxury of not having to worry about integer overflows. Even in Python it may be a good idea if you want to keep your computations fast for very large numbers. So you would typically do the same computation as:
def pe15_bis(n):
ret = 1
for j in range(1, n+1):
ret *= n + j
ret //= j
return ret
In Python that doesn't seem to pay off (performance wise) until n is in the many hundreds, but I find lovely that, the way that code goes, ret is always exactly divisible by j. Figuring out why is left as an exercise for the reader... | 2020-08-07T21:39:23 | {
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https://la.mathworks.com/help/econ/dtmc.eigplot.html | # eigplot
Plot Markov chain eigenvalues
## Description
example
eigplot(mc) creates a plot containing the eigenvalues of the transition matrix of the discrete-time Markov chain mc on the complex plane. The plot highlights the following:
• Unit circle
• Perron-Frobenius eigenvalue at (1,0)
• Circle of second largest eigenvalue magnitude (SLEM)
• Spectral gap between the two circles, which determines the mixing time
example
eVals = eigplot(mc) additionally returns the eigenvalues eVals sorted by magnitude.
eigplot(ax,mc) plots on the axes specified by ax instead of the current axes (gca).
[eVals,h] = eigplot(___) additionally returns the handle to the eigenvalue plot using input any of the input arguments in the previous syntaxes. Use h to modify properties of the plot after you create it.
## Examples
collapse all
Create 10-state Markov chains from two random transition matrices, with one transition matrix being more sparse than the other.
rng(1); % For reproducibility
numstates = 10;
mc1 = mcmix(numstates,'Zeros',20);
mc2 = mcmix(numstates,'Zeros',80); % mc2.P is more sparse than mc1.P
Plot the eigenvalues of the transition matrices on the separate complex planes.
figure;
eigplot(mc1);
figure;
eigplot(mc2);
The pink disc in the plots show the spectral gap (the difference between the two largest eigenvalue moduli). The spectral gap determines the mixing time of the Markov chain. Large gaps indicate faster mixing, whereas thin gaps indicate slower mixing. Because the spectral gap of mc1 is thicker than the spectral gap of mc2, mc1 mixes faster than mc2.
Consider this theoretical, right-stochastic transition matrix of a stochastic process.
$P=\left[\begin{array}{ccccccc}0& 0& 1/2& 1/4& 1/4& 0& 0\\ 0& 0& 1/3& 0& 2/3& 0& 0\\ 0& 0& 0& 0& 0& 1/3& 2/3\\ 0& 0& 0& 0& 0& 1/2& 1/2\\ 0& 0& 0& 0& 0& 3/4& 1/4\\ 1/2& 1/2& 0& 0& 0& 0& 0\\ 1/4& 3/4& 0& 0& 0& 0& 0\end{array}\right].$
Create the Markov chain that is characterized by the transition matrix P.
P = [ 0 0 1/2 1/4 1/4 0 0 ;
0 0 1/3 0 2/3 0 0 ;
0 0 0 0 0 1/3 2/3;
0 0 0 0 0 1/2 1/2;
0 0 0 0 0 3/4 1/4;
1/2 1/2 0 0 0 0 0 ;
1/4 3/4 0 0 0 0 0 ];
mc = dtmc(P);
Plot and return the eigenvalues of the transition matrix on the complex plane.
figure;
eVals = eigplot(mc)
eVals = 7×1 complex
-0.5000 + 0.8660i
-0.5000 - 0.8660i
1.0000 + 0.0000i
-0.3207 + 0.0000i
0.1604 + 0.2777i
0.1604 - 0.2777i
-0.0000 + 0.0000i
Three eigenvalues have modulus one, which indicates that the period of mc is three.
Compute the mixing time of the Markov chain.
[~,tMix] = asymptotics(mc)
tMix = 0.8793
## Input Arguments
collapse all
Discrete-time Markov chain with NumStates states and transition matrix P, specified as a dtmc object. P must be fully specified (no NaN entries).
Axes on which to plot, specified as an Axes object.
By default, eigplot plots to the current axes (gca).
## Output Arguments
collapse all
Transition matrix eigenvalues sorted by magnitude, returned as a numeric vector.
Handles to plotted graphics objects, returned as a graphics array. h contains unique plot identifiers, which you can use to query or modify properties of the plot.
Note
• By the Perron-Frobenius Theorem [2], a chain with a single recurrent communicating class (a unichain) has exactly one eigenvalue equal to 1 (the Perron-Frobenius eigenvalue), and an accompanying nonnegative left eigenvector that normalizes to a unique stationary distribution. All other eigenvalues have modulus less than or equal to 1. The inequality is strict unless the recurrent class is periodic. When there is periodicity of period k, there are k eigenvalues on the unit circle at the k roots of unity.
• For an ergodic unichain, any initial distribution converges to the stationary distribution at a rate determined by the second largest eigenvalue modulus (SLEM), μ. The spectral gap, 1 – μ, provides a visual measure, with large gaps (smaller SLEM circles) producing faster convergence. Rates are exponential, with a characteristic time given by
$tMix=-\frac{1}{\mathrm{log}\left(\mu \right)}.$
See asymptotics.
## References
[1] Gallager, R.G. Stochastic Processes: Theory for Applications. Cambridge, UK: Cambridge University Press, 2013.
[2] Horn, R., and C. R. Johnson. Matrix Analysis. Cambridge, UK: Cambridge University Press, 1985.
[3] Seneta, E. Non-negative Matrices and Markov Chains. New York, NY: Springer-Verlag, 1981.
### Functions
Introduced in R2017b | 2021-07-25T00:05:56 | {
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http://mathhelpforum.com/number-theory/129900-p-q-odd-primes-q-2-p-1-q-2pk-1-a.html | # Math Help - p,q odd primes & q|(2^p)-1 => q=2pk+1
1. ## p,q odd primes & q|(2^p)-1 => q=2pk+1
Prove that if p and q are odd primes and q divides (2^p)-1, then q=2pk+1 for some integer k.
I can't find any clue on this one...
Any help is appreciated!
[also under discussion in math links forum]
2. $q \mid 2^p - 1 \ \Leftrightarrow \ 2^p \equiv 1 \ (\text{mod } q)$
Since $(2,q) = 1$, then the order of 2 (mod q) is a divisor of p (why?) and so, the order is either 1 or p. It must be p.
So since the order of 2 is p, we know $p \mid \phi (q)$, i.e. $p \mid q - 1$.
See if you can finish it off from there.
3. Originally Posted by o_O
$q \mid 2^p - 1 \ \Leftrightarrow \ 2^p \equiv 1 \ (\text{mod } q)$
Since $(2,q) = 1$, then the order of 2 (mod q) is a divisor of p (why?) and so, the order is either 1 or p. It must be p.
So since the order of 2 is p, we know $p \mid \phi (q)$, i.e. $p \mid q - 1$.
See if you can finish it off from there.
So we have q-1=pz for some integer z. q-1 is even.
How can we prove that 2p|(q-1) ?
Thanks!
4. Start from here: $q - 1 = zp, \ z \in \mathbb{Z}$
We know that $p$ and $q$ are odd. What does this tell you about $z$?
5. Originally Posted by o_O
Start from here: $q - 1 = zp, \ z \in \mathbb{Z}$
We know that $p$ and $q$ are odd. What does this tell you about $z$?
OK, I got it.
z must be even; z=2k for some integer k.
So we have our result: q=2pk+1 for some integer k.
Is it true that ANY divisor(not necessarily prime) of (2^p)-1 is also of the form 2pk+1? Why or why not?
Thanks!
6. Well, consider the prime power decomposition of $2^p - 1$.
Through what we've just proven, what can be said about these primes? And hence what about the product between any of these primes?
7. Originally Posted by o_O
Well, consider the prime power decomposition of $2^p - 1$.
Through what we've just proven, what can be said about these primes? And hence what about the product between any of these primes?
(2^p)-1 is odd, so 2 is not a factor in the prime factorization of (2^p)-1. So every prime factor is odd and of the form 2pk+1.
I can also show that a product of two factors of the form 2pk+1 is of the same form (i.e. the product is of the form 2pm+1 for some integer m).
So by induction, we can say that ANY divisor(not necessarily prime) of (2^p)-1 is also of the form 2pk+1? Am I right?
8. Originally Posted by kingwinner
Am I right?
Why the doubt? | 2014-07-30T01:37:32 | {
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http://mathhelpforum.com/advanced-math-topics/156049-cartesian-equation-helix.html | Thread: Cartesian equation of a helix.
1. Cartesian equation of a helix.
Hi everyone,
I was wondering whether there is a cartesian equation for a helix. If so what is it? Also how do we convert a three dimensional parametric equation to its cartesian form? (such as the equation of the helix in parametric form)
2. Well, the parametric equation for a helix could be
$\mathbf{r}=\langle \cos(\theta),\sin(\theta),\theta \rangle.$
So, you could have the two equations (I think you'd have to have two cartesian equations to represent a space curve, simply because of the degrees of freedom involved):
$x=\cos(z), y=\sin(z).$
In general, if you have a helix, the axis of which is parallel to one of the three main axes, then I'd solve that equation for the parameter, and then plug back into the other two equations. For example, take the helix
$\mathbf{r}=\langle 5\cos(\theta),\theta-2,-5\sin(\theta) \rangle.$
I'd solve $y=\theta-2$ for $\theta,$ giving $\theta=y+2,$ and then plug into the other two equations, yielding
$x=5\cos(y+2), z=-5\sin(y+2).$
If the axis is not parallel to one of the three main axes, things get a lot dicier.
3. Originally Posted by Ackbeet
Well, the parametric equation for a helix could be
$\mathbf{r}=\langle \cos(\theta),\sin(\theta),\theta \rangle.$
So, you could have the two equations (I think you'd have to have two cartesian equations to represent a space curve, simply because of the degrees of freedom involved):
$x=\cos(z), y=\sin(z).$
In general, if you have a helix, the axis of which is parallel to one of the three main axes, then I'd solve that equation for the parameter, and then plug back into the other two equations. For example, take the helix
$\mathbf{r}=\langle 5\cos(\theta),\theta-2,-5\sin(\theta) \rangle.$
I'd solve $y=\theta-2$ for $\theta,$ giving $\theta=y+2,$ and then plug into the other two equations, yielding
$x=5\cos(y+2), z=-5\sin(y+2).$
If the axis is not parallel to one of the three main axes, things get a lot dicier.
Dear Ackbeet,
Thank you so much. So we cannot express a space curve by a single cartesian equation is'nt? I think so too. But the strange thing is I didnt find it mentioned anywhere.
4. So we cannot express a space curve by a single cartesian equation...?
No, I don't think you can. And that's merely because of the fact that if you think about the process of solving equations, one equation will generally allow you to eliminate one variable. 3D space obviously has 3 variables, x, y, and z. But a space curve has only one degree of freedom, the parameter t or whatever you want to call it. To get down to one variable from three requires two equations. So there you go.
This is something that mathematicians may or may not mention. The physicists definitely talk about the number of degrees of freedom (variables) all the time. Maybe that's because physicists are sometimes more interested in providing the actual solution, instead of just proving that a solution exists.
Cheers.
,
,
,
,
,
,
,
,
,
,
general equation of helix
Click on a term to search for related topics. | 2017-04-24T20:37:39 | {
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https://math.stackexchange.com/questions/498954/counting-number-of-distinct-systems | # Counting number of distinct systems
This is an enumeration problem in conjunction with some lottery problems.
Given an integer $N \ge 5$. Let a ticket be a set of 5 distinct integers between $1$ and $N$. Given an integer $T$ between $1$ and ${{N}\choose{5}}$. Let a system of size $T$ be a set of $T$ distinct tickets.
Given $N \ge 5$, I want to count how many distinct systems of size $T$ exist.
Two systems $S_1$ and $S_2$ are distinct if we can not find a permutation of $\{1,..,N\}$ so that the image of $S_1$ under permutation is $S_2$.
I tried some computations for small values of $N$ and $T$.
$N=7$
$T= 1, 2, 3, 4, 5, 6, 7, 8, 9$
number of distinct systems = $1, 2, 5, 10, 21, 41, 65, 97, 131, 148$
(It seems that this sequence of numbers is known as A008406 at oeis.org)
$N=8$
$T= 1, 2, 3, 4, 5, 6, 7, 8, 9$
number of distinct systems = $1, 3, 11, 52, 252, 1413, 7812, 41868, 207277$
$N=9$
$T= 1, 2, 3, 4, 5, 6, 7$
number of distinct systems = $1, 4, 20, 155, 1596, 20528, 282246$
Is there a method to "guess" those numbers and find bigger values ?
I wonder if Polya enumeration can be used there. I currently do not know how.
Update: Taking a look at http://ac.cs.princeton.edu/home/
Let $s(T,N)$ be the number of distinct systems of size $T$ ($1 \le T \le{{N}\choose{5}}$), given $N$.
$\forall N \ge 5, s(1,N) = 1$
$\forall N \ge 10, s(2,N) = 5$
$\forall N \ge 15, s(3,N) = 44$
• A question without answer on the Web is either too stupid or too complicated. I still have not found how much stupid is this one , but trying :-) – Philippe Morin Feb 15 '14 at 10:37
• I think this is an interesting question. – Marko Riedel Aug 14 '15 at 2:18
This appears to be an interesting problem that can be attacked using Power Group Enumeration on sets as described in quite some detail at the following MSE link.
That link discusses the number of different subsets of a standard $52$ card deck under suit permutation. Here we have the group permuting the slots into which we distribute the cards is the symmetric group and the group permuting cards is the cardinality twenty-four induced action on the cards of all permutations of the four suits.
The lottery ticket problem proposed here follows exactly the same model, only now we are distributing tickets into the slots being permuted by the symmetric group and the group acting on the tickets is the induced action of the symmetric group $S_N.$ The number of terms in the cycle indices $Z(S_N)$ and $Z(S_T)$ is given by the partition function and we get an algorithm that is of asymptotically lower order than the naive $N!\times T!.$
The only non-trivial issue that is not already featured in the solution to the distributions of cards is how to compute the cycle index of the induced action of $S_N$ on the ${N\choose Q}$ tickets of $Q$ elements. This can be done quite effectively by computing a representative of the permutation shape from the cycle index of the symmetric group, letting it act on the tickets, and factoring the result into cycles for a contribution to the desired cycle index.
Setting $Q=5$ as in the question we obtain for $N=7$ the sequence $$1, 2, 5, 10, 21, 41, 65, 97, 131, 148, 148, 131,\ldots$$ for $N=8$ the sequence $$1, 3, 11, 52, 252, 1413, 7812, 41868, 207277, 936130, 3826031,\\ 14162479,\ldots$$ for $N=9$ the sequence $$1, 4, 20, 155, 1596, 20528, 282246, 3791710, 47414089, 542507784,\\ 5659823776,53953771138,\ldots$$ and finally for $N=10$ the sequence $$1, 5, 28, 324, 5750, 142148, 3937487, 108469019, 2804300907,\\ 66692193996,1452745413957, 29041307854703,\ldots.$$
To illustrate the good complexity of this algorithm here is the sequence for $N=13:$ $$1, 5, 42, 813, 34871, 2777978, 304948971, 37734074019,\\ 4719535940546, 566299855228261, 63733180893169422,\\ 6674324951638852138,\ldots$$
Finally we obtain for $N$ variable with $Q=5$ and $T=3$ the sequence $$0, 0, 0, 0, 0, 1, 5, 11, 20, 28, 35, 39, 42, 43,\\ 44, 44, 44, 44,\ldots$$
The Maple code to compute these was as follows:
with(combinat);
with(numtheory);
pet_flatten_term :=
proc(varp)
local terml, d, cf, v;
terml := [];
cf := varp;
for v in indets(varp) do
d := degree(varp, v);
terml := [op(terml), seq(v, k=1..d)];
cf := cf/v^d;
od;
[cf, terml];
end;
pet_autom2cycles :=
proc(src, aut)
local numa, numsubs;
local marks, pos, cycs, cpos, clen;
numsubs := [seq(src[k]=k, k=1..nops(src))];
numa := subs(numsubs, aut);
marks := Array([seq(true, pos=1..nops(aut))]);
cycs := []; pos := 1;
while pos <= nops(aut) do
if marks[pos] then
clen := 0; cpos := pos;
while marks[cpos] do
marks[cpos] := false;
cpos := numa[cpos];
clen := clen+1;
od;
cycs := [op(cycs), clen];
fi;
pos := pos+1;
od;
return mul(a[cycs[k]], k=1..nops(cycs));
end;
pet_cycleind_symm :=
proc(n)
local p, s;
option remember;
if n=0 then return 1; fi;
end;
pet_flat2rep :=
proc(f)
local p, q, res, t, len;
q := 1; res := [];
for t in f do
len := op(1, t);
res := [op(res), seq(p, p=q+1..q+len-1), q];
q := q+len;
od;
res;
end;
pet_cycleind_tickets :=
proc(N, Q)
option remember;
local cind, tickets, q, term, rep, subsl, ptickets,
idx, flat;
if N=1 then
idx := [a[1]]
else
idx := pet_cycleind_symm(N);
fi;
cind := 0;
tickets := convert(choose({seq(q, q=1..N)}, Q), list);
for term in idx do
flat := pet_flatten_term(term);
rep := pet_flat2rep(flat[2]);
subsl := [seq(q=rep[q], q=1..N)];
ptickets := subs(subsl, tickets);
cind := cind +
flat[1]*pet_autom2cycles(tickets, ptickets);
od;
cind;
end;
X :=
proc(N, Q, T)
option remember;
local idx_slots, res, a, b,
flat_a, flat_b, cycs_a, cycs_b, q,
tbl_a, tbl_b, f1, f2, f3;
if T=1 then
idx_slots := [a[1]]
else
idx_slots := pet_cycleind_symm(T);
fi;
res := 0;
for a in idx_slots do
flat_a := pet_flatten_term(a);
cycs_a := sort(flat_a[2]);
tbl_a := table();
for q in convert(cycs_a, 'multiset') do
tbl_a[op(1, q[1])] := q[2];
od;
f1 := map(q -> op(1, q), cycs_a);
f1 := mul(f1[q], q=1..nops(cycs_a));
f2 := convert(map(q -> op(1, q), cycs_a), 'multiset');
f2 := map(q -> q[2], f2);
f2 := mul(f2[q]!, q=1..nops(f2));
for b in pet_cycleind_tickets(N, Q) do
flat_b := pet_flatten_term(b);
cycs_b := sort(flat_b[2]);
tbl_b := table();
for q in convert(cycs_b, 'multiset') do
tbl_b[op(1, q[1])] := q[2];
od;
f3 := 1;
for q in [indices(tbl_a, 'nolist')] do
if type(tbl_b[q], integer) then
f3 := f3*binomial(tbl_b[q], tbl_a[q]);
else
f3 := 0;
fi;
od;
res := res + f3*f2*f1*flat_a[1]*flat_b[1];
od;
od;
res;
end;
Addendum Fri Aug 14 2015. The sequence for $Q=5$ and $N=20$ is $$1, 5, 44, 966, 53484, 7023375, 1756229468, 710218125299, \\ 411620592905173, 308212635851733551, 271743509344779773214,\ldots$$
Addendum Sat Aug 15 2015. The sequence for $Q=5$ and $N=22$ is $$1, 5, 44, 966, 53529, 7041834, 1773511264, 734330857318, \\ 452455270344141, 383969184978128899, 416614280701828877344, \\ 536531456518633409220043, 766723127226754935510254929,\ldots$$
Addendum Wed Aug 19 2015. The sequence for $Q=5$ and $N=24$ is $$1, 5, 44, 966, 53534, 7043732, 1775444689, 737776095236, \\ 460462767067281, 405308264117856150, 477303563740811267063, \\ 712445246443357547546003, 1271053814158420923816386794,\ldots$$
There are $\binom{N}{5}$ possible tickets, you are asking how many $T$-subsets of those there are, i.e., the number of systems is:
$$\binom{\binom{N}{5}}{T}$$
(somehow I'm feeling I'm just being stupid here... can't be that easy?)
• You appear to have missed the part where the OP says "Two systems are distinct if .." – Marko Riedel Aug 14 '15 at 22:00
• @MarkoRiedel, "are unique up to permutation" is just sets in my book... – vonbrand Aug 14 '15 at 22:06
• These lottery tickets are indeed sets. The permutation however acts on the $N$ values from which they are drawn. E.g. the ticket $\{1,2,3,4,5\}$ drawn from $N=10$ could be transformed into the ticket $\{6,7,8,9,10\}$ by a permutation of the $N$ values that exchanges the upper and lower five values (permutation not unique). – Marko Riedel Aug 14 '15 at 22:11
• The permutation acts simultaneously on all tickets in a system of size $T,$ which is indeed a set of tickets. – Marko Riedel Aug 14 '15 at 22:26 | 2021-06-13T06:27:22 | {
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https://math.stackexchange.com/questions/2419664/deriving-the-number-of-all-even-length-palindromic-sequences-with-length-at-most | # Deriving the number of all even length palindromic sequences with length at most $n$
Derive the number of all even length palindromic sequences with length $\leq n$.
If $n=4$, I have to count the number of non-palindromes of length 0, 2, and 4.
Also, $2p = n$, so a string of length $n = 4$ has $p = 2$. This comes in handy later.
$x$ represents the number of possible characters in the given alphabet- so $x = 26$ for English, 2 for binary, etc.
I chose to approach by counting, using this thought process.
$$\sum_{k=0}^{p} (x^{2k} - x^k)$$ In other words - a summation of all possible even length strings up to length $n$, subtracting cases that are palindromes.
I tested this with binary, up to length 4. (Binary non-palindromes of length 0, 2, and 4). This means $p$ goes from 0, to 1, to 2.
This evaluates to: $$2^0 - 2^0 +$$ $$2^2 - 2^1 +$$ $$2^4 - 2^2$$
Which sums to a total of 14 non palindromes.
All possible combinations are {}, 11, 00, 10, 01, 1111, 1100, 1000, 0000, 0001, 0011, 0111, 1010, 0101, 0110, 1001 1110, 1101, 0010, 1011, 0100.
14 of these are not palindromes, confirming my approach.
The solution states the formula is $$\frac{x^{2p+1}-2x^{p+1}+x}{x-1}$$ Following the same logic...
With $p = 2$, $x = 2$, $n = 4$, we have
$$\frac{2^{4+1}-2(2)^{3}+2}{2-1}$$
Which works out to 18, not 14. I have worked this out several times, and I'm not exactly sure what/if I am misunderstanding. Any help would be greatly appreciated. Thanks!
If the sequence has even length, say $n = 2k$, selecting the first $k$ characters completely determines the palindrome since the remaining $k$ characters can be found by repeating the sequence in the reverse order. Hence, the number of palindromes of even length at most $n$ in an alphabet with $x$ characters is $$x^0 + x^1 + x^2 + x^3 + \cdots + x^k = \frac{1 - x^{k + 1}}{1 - x}$$ where we have used the formula for the sum of a geometric series with initial term $1$ and common ratio $x$.
In your example of binary sequences of even length at most $4$, there are seven palindromes: $$\emptyset, 00, 11, 0000, 0110, 1001, 1111$$ Notice that our formula yields $$2^0 + 2^1 + 2^2 = 1 + 2 + 4 = 7$$ or $$\frac{1 - 2^{2 + 1}}{1 - 2} = \frac{1 - 2^3}{-1} = \frac{-7}{-1} = 7$$
Since there are a total of $$x^0 + x^2 + x^4 + \ldots + x^{2k} = \frac{1 - x^{2k + 2}}{1 - x^2}$$ sequences of even length at most $n = 2k$ in an alphabet with $x$ characters, the number of sequences of even length at most $n = 2k$ that are not palindromes is \begin{align*} \frac{1 - x^{2k + 2}}{1 - x^2} - \frac{1 - x^{k + 1}}{1 - x} & = \frac{1 - x^{2k + 2}}{1 - x^2} - \frac{1 - x^{k + 1}}{1 - x} \cdot \frac{1 + x}{1 + x}\\ & = \frac{1 - x^{2k + 2}}{1 - x^2} - \frac{1 + x - x^{k + 1} - x^{k + 2}}{1 - x^2}\\ & = \frac{-x + x^{k + 1} + x^{k + 2} - x^{2k + 2}}{1 - x^2} \end{align*} In your example of binary sequences of even length at most $4$, the total number of sequences is $$2^0 + 2^2 + 2^4 = 1 + 4 + 16 = 21$$ of which seven are palindromes, so there are $14$ non-palindromes as you found. Notice that our formulas yield $$\frac{1 - 2^{2 \cdot 2 + 2}}{1 - 2^2} = \frac{1 - 2^6}{1 - 2^2} = \frac{1 - 64}{1 - 4} = \frac{-63}{-3} = 21$$ total sequences and $$\frac{-2 + 2^{2 + 1} + 2^{2 + 2} - 2^{2 \cdot 2 + 2}}{1 - 2^2} = \frac{-2 + 2^3 + 2^4 - 2^6}{1 - 2^2} = {-2 + 8 + 16 - 64}{1 - 4} = \frac{-42}{-3} = 14$$ non-palindromes. | 2019-09-22T01:55:56 | {
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https://math.stackexchange.com/questions/3261941/let-mathfrakd-be-the-collection-of-subsets-of-x-of-the-form-e-1-lambda | # Let $\mathfrak{D}$ be the collection of subsets of $X$ of the form $E_1^{\lambda_1}\cap\dots\cap E_n^{\lambda_n}$. Is $X=\bigcup_{D\in\mathfrak{D}}D$?
## Full problem statement:
Let $$E_1,\dots,E_n$$ be distinct but not necessarily disjoint subsets of $$X$$.
Let $$\mathfrak{D}$$ be the disjoint collection of all subsets of $$X$$ of the form $$E_1^{\lambda_1}\cap\dots\cap E_n^{\lambda_n}$$ where $$\lambda_i \in \{0,1\}$$. Note that $$E_i^1 = E_i$$, $$E_i^0 = E_i^c$$.
Let $$\mathfrak{F}$$ be collection of arbitrary unions of members of $$\mathfrak{D}$$.
1. Is $$X\hspace{1mm}\in\mathfrak{F}$$?
2. Let $$F\in\mathfrak{F}$$. Is $$F^c = X\backslash F \in \mathfrak{F}$$?
## Author's solution
Let $$x\in X$$. Since $$\forall i\in\{1,\dots,n\}\hspace{1mm} E_i^1\cup E_i^0 = X$$, $$x\in E_i$$ or $$x\in E_i^c$$.
Then, every $$x$$ is contained for some $$D = E_1^{\lambda_{1}}\cap\dots E_n^{\lambda_{n}} \in\mathfrak{D}$$. So $$X=\bigcup\limits_{D\in\mathfrak{D}} D\in\mathfrak{F}$$.
## My Questions
Q1. Can we just end the proof there? It seems like all we've proven is that $$X\subset\hspace{-1.5mm}\bigcup\limits_{D\in\mathfrak{D}}\hspace{-1.5mm} D$$.
## Author's solution
Let $$F\in \mathfrak{F}$$. Then $$F$$ is a union of member of $$\mathfrak{D}$$.
Since $$\mathfrak{D}$$ is a disjoint collection and $$X=\hspace{-1.5mm}\bigcup\limits_{D\in\mathfrak{D}}\hspace{-1.5mm} D$$, $$F^c=X\backslash F$$ is also a union of members of $$\mathfrak{D}$$. So $$F^c\in\mathfrak{F}$$.
## My Questions
Q1. How does the statements that $$\mathfrak{D}$$ is a disjoint collection and $$X=\hspace{-1.5mm}\bigcup\limits_{D\in\mathfrak{D}}\hspace{-1.5mm} D$$ work together to imply $$F^c=X\backslash F$$ is a union of members of $$\mathfrak{D}$$? I did try some DeMorgan stuff but it all went nowhere :(
I didn't put all my questions at the end because I think having to scroll up to the relevant block and scroll back down to the questions is going to be annoying.
First question: since $$D \subset X$$ for all $$D \in \mathfrak D$$ it is understood that the reverse inclusion holds so equality holds.
Second question. It is advisable to look at some simple examples. If $$F$$ is expressed as union of certain members of $$\mathfrak D$$ then $$F^{c}$$ is precisely the union of the remaining members of $$\mathfrak D$$.
[$$\mathfrak D$$ is a partition of $$X$$: its members are disjoint and their union is $$X$$. Call these sets $$(D_i)_{i\in I}$$. For any subset $$J$$ of $$I$$, let $$F$$ be the union of the sets $$D_i$$ with $$i \in J$$. Let $$G$$ be the union of the sets $$D_i$$ with $$i \in I\setminus J$$. Then $$F\cup G$$ is the union of all the $$D_i$$'s which is $$X$$. Also $$F$$ and $$G$$ are disjoint. These two facts imply that $$G$$ is the complement of $$F$$]. | 2019-12-16T11:06:27 | {
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https://eliteprotek.com/starbucks-vanilla-gwd/beaa86-determine-the-coordinates-of-the-centroid-of-the-area | # determine the coordinates of the centroid of the area
Centroid: Centroid of a plane figure is the point at which the whole area of a plane figure is assumed to be concentrated. 4' 13 Answers: (X,Y) in • The coordinates ( and ) define the center of gravity of the plate (or of the rigid body). The cartesian coordinate of it's centroid is $\left(\frac{2}{3}r(\theta)\cos\theta, \frac{2}{3}r(\theta)\sin\theta\right)$. Find the coordinates of the centroid of the area bounded by the given curves. How to calculate a centroid. It is also the center of gravity of the triangle. The centroid of a right triangle is 1/3 from the bottom and the right angle. Centroid by Composite Bodies ! The x-centroid would be located at 0 and the y-centroid would be located at 4 3 r π 7 Centroids by Composite Areas Monday, November 12, 2012 Centroid by Composite Bodies Recall that the centroid of a triangle is the point where the triangle's three medians intersect. y=x^{3}, x=0, y=-8 y=2 x, y=0, x=2 It is the point which corresponds to the mean position of all the points in a figure. Next, sum all of the x coodinates ... how to find centroid of composite area: how to calculate centroid of rectangle: how to find centroid of equilateral triangle: For example, the centroid location of the semicircular area has the y-axis through the center of the area and the x-axis at the bottom of the area ! Problem Answer: The coordinates of the center of the plane area bounded by the parabola and x-axis is at (0, 1.6). The Find Centroids tool will create point features that represent the geometric center (centroid) for multipoint, line, and area features.. Workflow diagram Examples. Find the coordinates of the centroid of the plane area bounded by the parabola y = 4 – x^2 and the x-axis. Center of Mass of a Body Center of mass is a function of density. With this centroid calculator, we're giving you a hand at finding the centroid of many 2D shapes, as well as of a set of points. Gather both the x and y coordinate points of each vertex. The coordinates of the centroid are simply the average of the coordinates of the vertices.So to find the x coordinate of the orthocenter, add up the three vertex x coordinates and divide by three. First, gather the coordinate points of the vertices. The center of mass is the term for 3-dimensional shapes. Determine the coordinates of the centroid of the shaded region. Beam sections are usually made up of one or more shapes. The centroid is the term for 2-dimensional shapes. Centroid of an Area • In the case of a homogeneous plate of uniform thickness, the magnitude ∆W is We can consider the surface element as an triangle, and the centroid of this triangle is obviously at here.) Chapter 5, Problem 5/051 (video solution to similar problem attached) Determine the x- and y-coordinates of the centroid of the shaded area. Find the coordinates of the centroid of the area bounded by the given curves. And the area of this surface element $\mathrm{d}A = \frac{1}{2}r^2(\theta)\mathrm{d}\theta$. An analyst at the Scotland Department of Environment is performing a preliminary review on wind farm applications to determine which ones overlap with or are in view of wild lands. So to find the centroid of an entire beam section area, it first needs to be split into appropriate segments. The centroid or center of mass of beam sections is useful for beam analysis when the moment of inertia is required for calculations such as shear/bending stress and deflection. 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Medians intersect point at which the whole area of a circle and a rectangle is at the..
Scroll to top | 2021-10-28T05:20:08 | {
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https://notescs.gitlab.io/posts/max-value-with-addition-multiplication/ | # Maximum Value of Equation of Ones with Addition and Multiplication Operations
Given an integer n, find the maximum value that can be obtained using n ones and only addition and multiplication operations. Note that, you can insert any number of valid bracket pairs.
Example:
Input: n = 12
Output: 81
Explanation: (1 + 1 + 1) * (1 + 1 + 1) * (1 + 1 + 1) * (1 + 1 + 1) = 81
## Approach: Dynamic Programming
Observe that, in order to find the answer for n = 5, we need to consider the maximum answer obtainable from the answer of 2 and 3, 1 and 4. Thus, this problem has an optimal substructure property.
5
/ \
op(1, 4) op(2, 3)
/ \ / \
1 op(2, 2) op(1, 1) op(1, 2)
/ \ / \ / \
op(1, 1) op(1, 1) 1 1 1 op(1, 1)
/ \ / \ / \
1 1 1 1 1 1
Clearly, from the tree above there are a lot of overlapping sub-problems. Owing to this and optimal substructure property, this problem is an ideal candidate for dynamic programming.
Bottom-up approach:
C++ code:
#include<iostream>
#include<vector>
#include<algorithm>
using namespace std;
long long getMaximumValue(int n) {
vector<long long> dp(n + 1);
// dp[i] denotes maximum value that can be obtained from i ones
dp[0] = 0; // base case: with 0 ones answer is always 0
dp[1] = 1; // base case: with 1 one answer is 1
for (int i = 1; i <= n; ++i) {
for (int j = 1; j <= i / 2; ++j) {
dp[i] = max({dp[i], dp[j] + dp[i - j], dp[j] * dp[i - j]});
}
}
return dp[n];
}
int main() {
cout << "n = " << 5 << " Maximum possible value " << getMaximumValue(5) << endl; // prints
cout << "n = " << 12 << " Maximum possible value " << getMaximumValue(12) << endl; // prints "81"
return 0;
}
Python code:
def getMaximumValue(n):
dp = [0] * (n + 1)
# dp[i] denotes maximum value that can be obtained from i ones
dp[0] = 0 # base case: with 0 ones answer is always 0
dp[1] = 1 # base case: with 1 one answer is 1
for i in range(2, n + 1):
for j in range(1, i // 2 + 1):
dp[i] = max(dp[i], dp[j] + dp[i - j], dp[j] * dp[i - j])
return dp[n]
if __name__ == '__main__':
print('n =', 5, 'Maximum possible value', getMaximumValue(5))
print('n =', 12, 'Maximum possible value', getMaximumValue(12))
Time Complexity: $O(n ^ 2)$
Space Complexity: $O(n)$ due to storing states in dp array
Exercise: Code the top-down approach. | 2020-11-29T14:03:38 | {
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https://math.stackexchange.com/questions/3091598/concerning-the-identity-in-sums-of-binomial-coefficients | # Concerning the identity in sums of Binomial coefficients
Let be the following identity $$\sum_{k=1}^{n}\binom{k}{2}=\sum_{k=0}^{n-1}\binom{k+1}{2}=\sum_{k=1}^{n}k(n-k)=\sum_{k=0}^{n-1}k(n-k)=\frac16(n+1)(n-1)n$$ As we can see the partial sums of binomial coefficients are expressed in terms of $$3$$-rd order polynomial $$P_3(n)$$, where $$n$$ is variable of upper bound of summation. We assume that order of resulting polynomial $$P_3(n)$$ depends on subscript of binomial coefficient being summed up (in our case the order of polynomial is $$3=2+1$$, where $$2$$ is subscript of bin. coef.)
The question: Does there exist a generalized method to represent the sum of binomial coefficients $$\sum_{k}^{n}\binom{k}{s}$$ in terms of certain polynomials $$P_{s+1}(n)=\sum_{k}^{n} F_s(n,k)$$ for every non-negative integer $$s$$? I.e can we always find the function $$F_s(n,k)$$, such that $$\sum_{k}^{n}\binom{k}{s}=\sum_{k}^{n}F_s(n,k)$$ ? We assume that order of polynomial is $$s+1$$ by means of example above.
The sub-question: (In case of positive answer to the first question.) If there exists a method to represent the sums of bin. coef. in terms of polynomials in $$n$$, how do summation limits of the $$\sum_{k}^{n}\binom{k}{s}$$ implies to the form of polynomial $$P_{s+1} (n)$$ exactly?
• $\sum_{k=s}^n \binom{k}s=\binom{n+1}{s+1}$. This is the Hockey Stick identity. – Mike Earnest Jan 29 '19 at 1:24
• The question is to find such polynomial $F_s(n,k)$ that $\sum_{k}^{n}\binom{k}{s}=\sum_{k}^{n}F_s(n,k)$. From our example it follows that: $F_2(n,k)=k(n-k)$ and $\sum_{k=1}^{n}\binom{k}{2}=\sum_{k=1}^{n}F_2(n,k)$. Can you provide examples for $s>2$? – PKK Jan 29 '19 at 1:33
• I understand now. Still, my comment just shows that $P_s(n)$ is nothing mysterious. – Mike Earnest Jan 29 '19 at 1:51
• Why don't you mention your previous question math.stackexchange.com/q/2774300 ? – Jean Marie Jan 30 '19 at 18:25
• thank you for reminding, by the way these coefficients could be found as $$(2k-1)!T(2n,2k)=\frac{1}{r}\sum_{j=0}^{r}(-1)^j\binom{2r}{j}(r-j)^{2n},$$ where $r=n-k+1$ and $T(2n,2k)$ is central factorial number – PKK Jan 30 '19 at 18:30
$$\sum_{k=0}^n\binom{k}s=\sum_{k=0}^n\sum_{i=0}^{k-1}\binom{i}{s-1}=\sum_{i=0}^{n-1}\sum_{k=i+1}^n\binom{i}{s-1}=\sum_{i=0}^{n-1}(n-i)\binom{i}{s-1}=\sum_{i=1}^ni\binom{n-i}{s-1}$$ In other words, you can let $$F_s(n,k)=k\binom{n-k}{s-1}$$, and you will have $$\sum_{k=0}^n \binom{k}s=\sum_{k=0}^{n}F_s(n,k)$$.
• Can you show me a direct example for $s=2$ step by step ? The example of $\sum_{k=1}^n\binom{k}{s}$ please – PKK Jan 29 '19 at 1:53
• @PetroKolosov When $s=2$, you get $F_s(n,k)=k(n-k)$, just as you had. When $s>2$, the expression $k\binom{n-k}s$ is polynomial in disguise. For example, $F_3(n,k)=k\binom{n-k}2=k(n-k)(n-k-1)/2$. – Mike Earnest Jan 29 '19 at 2:36
• Are you sure that $F_3(n,k)=k(n-k)(n-k-1)/2$? For me it gives as follows: $F_s(n,k)=k\binom{n-k}{s-1}$, by the hockey stick pattern: $\binom{n-k}{s-1}=\sum_{j}^{n-k+1}\binom{j}{s-2}|_{s=3}=1/2 (-2 + k - n) (-1 + k - n)$ and multiplication by $k$ gives $F_3(n,k)=1/2 k (-2 + k - n) (-1 + k - n)$. PS and obviously i wrong with limits of hockey stick pattern :) – PKK Jan 29 '19 at 3:09
• I've fixed the error above, but still $F_s(n,k)=k\binom{n-k}{s-1}$, by the hockey stick pattern: $\binom{n-k}{s-1}=\sum_{j}^{n-k-1}\binom{j}{s-2}|_{s=3}=1/2 (k - n) (1 + k - n)$ and multiplication by $k$ gives $F_3(n,k)=1/2 k (k - n) (1 + k - n)$ is different from your example of $F_3(n,k)$ – PKK Jan 29 '19 at 3:18
• $k(n-k)(n-k-1)/2=\frac12k(k-n)(1+k-n)$. Your expression and mine for $F_3(n,k)$ are the same. All you did was expand $\binom{n-k}{s-1}$ using the HS identity, and then collapse it using the same identity. @PetroKolosov – Mike Earnest Jan 29 '19 at 3:37
I would say that you have a good answer already. But their are other possible answers which seem reasonable. Further restriction might force the favored solution above.
In the case $$k=3$$ (which is the only one I will discuss in any detail)
$$\sum_{s=1}^n\binom{s}3= \\ \sum_{s=1}^ns\binom{n-s}2=\sum_{s=1}^n(n-s)\binom{s}2=\frac14\sum_{s=1}^n{s(n-s)(n-2)}=\frac1{24}\sum_{s=1}^n{(n+1)(n-1)(n-2)}$$
It is easy to see how to generalize these to other $$k.$$
The first three belong to the infinite family
$$\frac14\sum_{s=1}^n{s(n-s)(\alpha n-(2\alpha -2)s-2)} \tag{*}$$
Going back to the favored solution:
$$\sum_{s=1}^n\binom{s}3=\sum_{s=1}^n\frac{s^3}{6}-\frac{s^2}{2}+\frac{s}{3}=\sum_{s=1}^n\frac{n^2s}2-n{s}^{2}+\frac{s^3}2-\frac{ns}{2}+\frac{s^2}2.$$
If you wanted the thing on the right to be
$$\sum_{s=1}^nA{n^2s}+Bn{s}^{2}+C{s^3}+D{ns}+E{s^2}$$
Then you do need to have $$E=\frac12$$ But the rest have two degrees of freedom
$$C=-2A-\frac43B \ \ \ \ \ \ \ D=A-\frac13B-\frac23$$
For further restriction we might want to have $$D=-E=-\frac12$$ so than $$A{n^2s}+Bn{s}^{2}+C{s^3}+D{ns}+E{s^2}=0$$ when $$s=n$$
In this case the summand factors (of course) giving the family $$(*)$$ above.
If we want the right-hand side to be
$$K\sum_{s=1}^ns(An+(1-A)s+B)(Cn+(1-C)s+D)$$
It is possible to work out the requirements. I came up with $$6$$ families of solutions. Of course the set of solutions is invariant under switching the two terms and/or substituting $$s=n+1-s.$$ | 2020-12-04T18:37:06 | {
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http://math.stackexchange.com/questions/150630/understanding-isometric-spaces?answertab=oldest | # Understanding isometric spaces
I have studied that an isometry is a distance-preserving map between metric spaces and two metric spaces $X$ and $Y$ are called isometric if there is a bijective isometry from X to Y.
My questions are related with the understanding of isometric spaces, they are as follows:
Can we say that two isometric spaces are same? If no, in what context they differ? What are the common properties shared by two isometric spaces?
Intuitively what are isometric spaces?
If two spaces are isometric how to find out bijective distance preserving map between them?
Thanks for your help and time.
-
Any two lines in the plane are isometric (you can translate and rotate to place one line on top of the other, without affecting distances within the lines in this process), so definitely two isometric spaces need not really be literally the same. But as far as metric properties are concerned they behave in the same way. It's like asking "are all circles of radius 1 the same"? No, but obviously you're comfortable using one particular choice (like the one centered at the origin) even if that's not the original circle of interest. – KCd May 28 '12 at 6:04
The examples of the line and circle might seem silly, because they are very familiar. The point of the concept of isometric spaces is to keep us aware that we shouldn't consider two isometric spaces as being fundamentally different from one another. For example, when you construct the completion $\widetilde{X}$ of a metric space $X$, using equiv. classes of Cauchy sequences in $X$, you don't really find $X$ as a subset of $\widetilde{X}$, but $X$ is isometric to the equiv. classes of constant seq. $(x,x,x,\dots)$, and that is how we can view $X$ (as metric space) inside $\widetilde{X}$. – KCd May 28 '12 at 6:08
There is no universal method to find an isometry between two isometric metric spaces. – KCd May 28 '12 at 6:09
@KCd Thanks to you. Your comments are helpful to me. – srijan May 28 '12 at 6:17
Homeomorphisms are the maps that preserve all topological properties: from a structural point of view, homeomorphic spaces might as well be identical, though they may have very different underlying sets, and if they’re metrizable, they may carry very different (but equivalent) metrics. Isometries are the analogue for metric spaces, topological spaces carrying a specific metric: they preserve all metric properties, and of course those include the topological properties. Thus, all isometries are homeomorphisms, but the converse is false.
Consider the metric spaces $\langle X,d_X\rangle$ and $\langle Y,d_Y\rangle$ defined as follows: $X=\Bbb N,Y=\Bbb Z$, $$d_X(m,n)=\begin{cases}0,&\text{if }m=n\\1,&\text{if }m\ne n\;,\end{cases}$$ for all $m,n\in X$, and $$d_Y(m,n)=\begin{cases}0,&\text{if }m=n\\1,&\text{if }m\ne n\end{cases}$$ for all $m,n\in Y$. It’s easy to check that $d_X$ and $d_Y$ are metrics on $X$ and $Y$, respectively.
Clearly these are not the same space: they have different underlying sets. However, if $f:X\to Y$ is any bijection1 whatsoever, then $f$ is an isometry between $X$ and $Y$. $\langle X,d_X\rangle$ and $\langle Y,d_Y\rangle$ are structurally identical as metric spaces: if $P$ is any property of metric spaces $-$ not just of metrizable spaces, but of metric spaces with a specific metric $-$ then either $X$ and $Y$ both have $P$, or neither of them has $P$. There is no structural property of metric spaces that distinguishes them.
What I just said about $X$ and $Y$ is true of isometric spaces in general: there is no structural property of metric spaces that distinguishes them. Considered as metric spaces, they are structurally identical, though they may have different underlying sets.
Isometric spaces may even have the same underlying set but different metrics. Consider the following two metrics on $\Bbb N=\{0,1,2,\dots\}$. For any $m,n\in\Bbb N$,
$$d_0(m,n)=\begin{cases} 0,&\text{if }m=n\\\\ \left|\frac1m-\frac1n\right|,&\text{if }0\ne m\ne n\ne 0\\\\ \frac1m,&\text{if }n=0<m\\\\ \frac1n,&\text{if }m=0<n\;, \end{cases}$$
and
$$d_1(m,n)=\begin{cases} 0,&\text{if }m=n\\\\ \left|\frac1m-\frac1n\right|,&\text{if }m\ne n\text{ and }m,n>1\\\\ 1-\frac1m,&\text{if }n=0\text{ and }m>1\\\\ 1-\frac1n,&\text{if }m=0\text{ and }n>1\\\\ \frac1m,&\text{if }n=1\ne m\\\\ \frac1n,&\text{if }m=1\ne n\;. \end{cases}$$
It’s a good exercise to show that $$f:\Bbb N\to\Bbb N:n\mapsto\begin{cases}n,&\text{if }n>1\\1,&\text{if }n=0\\0,&\text{if }n=1\end{cases}$$ is an isometry between $\langle\Bbb N,d_0\rangle$ and $\langle\Bbb N,d_1\rangle$. (HINT: Both spaces are isometric to the space $\{0\}\cup\left\{\frac1n:n\in\Bbb Z^+\right\}$ with the usual metric.) Yet these are clearly not the same space: metric $d_0$ makes $0$ a limit point of the other points, but metric $d_1$ makes $0$ an isolated point.
I don’t know of any general method for finding an isometry between isometric spaces; if you can recognize two spaces as being isometric, you probably already have a good idea of what an isometry between them must look like.
1 If you want a specific bijection, $$f(n)=\begin{cases}0,&\text{if }n=0\\\\\frac{n}2,&\text{if }n>0\text{ and }n\text{ is even}\\\\-\frac{n+1}2,&\text{if }n\text{ is odd}\end{cases}$$ does the job.
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Sir, by structural properties do you mean completeness, boundedness etc? – srijan May 28 '12 at 6:05
@srijan: Anything that has to do with the topological or metric structure of the space and not with superficial charactersitiscs like the specific names attached to the points. Completeness and boundedness are indeed structural properties of metric spaces, though not of metrizable spaces. – Brian M. Scott May 28 '12 at 6:07
All homeomorphisms need not be isometries because some of the topological properties may not be shared by two isometric spaces, Am i right sir? – srijan May 28 '12 at 6:15
@srijan: No, it’s because some of the metric properties may not be shared between two homeomorphic spaces. For instance, $\Bbb R$ and $(0,1)$ with the usual metrics are homeomorphic, but $\Bbb R$ is a complete metric space, while $(0,1)$ isn’t: they don’t share the metric property of completeness. – Brian M. Scott May 28 '12 at 6:18
@srijan: You’re very welcome! – Brian M. Scott May 28 '12 at 6:22
show 1 more comment | 2014-04-25T03:21:56 | {
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http://math.stackexchange.com/questions/166244/determinant-of-an-n-times-n-complex-matrix-as-an-2n-times-2n-real-determinan | Determinant of an $n\times n$ complex matrix as an $2n\times 2n$ real determinant
If $A$ is an $n\times n$ complex matrix. Is it possible to write $\vert \det A\vert^2$ as a $2n\times 2n$ matrix with blocks containing the real and imaginary parts of $A$?
I remember seeing such a formula, but can not remember where. Any details, (and possibly references) for such a result would be greatly appreciated.
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Write $A=A_1+iA_2$ where $A_1$ and $A_2$ are real matrices. Let $$B:=\pmatrix{A_1&iA_2\\iA_2&A_1}.$$ We have $$\det B=\det\pmatrix{A_1+iA_2&iA_2\\A_1+iA_2&A_1}=\det\pmatrix{I&iA_2\\I&A_1}\cdot\det\pmatrix{A_1+iA_2&0\\0&I},$$ and $$\det\pmatrix{I&iA_2\\I&A_1}=\det\pmatrix{I&iA_2\\0&A_1-iA_2}$$ hence $$\det B=\det(A_1-iA_2)\det(A_1+iA_2)=\det A\cdot\det\bar A=|\det A|^2.$$
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Thanks you very much! – dernam Jul 3 '12 at 18:59
You are welcome! – Davide Giraudo Jul 3 '12 at 19:01
Davide's answer tells most of the story, in particular giving the proof for the determinant, but not quite all of it, so I want to supplement it with a couple of remarks.
I think that it is more common to replace Davide's matrix $B$ with a real matrix. This can be achieved by conjugating it with matrix of the block form $$D=\pmatrix{I&0\cr0&iI\cr},$$ when $$D^{-1}BD=\pmatrix{A_1&-A_2\cr A_2 &A_1\cr}.$$ Because conjugation preserves the determinant, Davide's calculation tells that here we also have $$\det(D^{-1}BD)=\det(B)=|\det A|^2.$$ Further conjugating (shuffling rows and columns) allows us to replace each and every complex entry $z=a+bi$ with a $2\times2$ block $$(z)=\pmatrix{a&-b\cr b&a\cr}.$$ Doing it this way makes it clear that if $A$ represents a linear mapping $T$ from $V=\mathbf{C}^n$ to itself with respect to basis $v_1,v_2,\ldots,v_n$, then $D^{-1}BD$ represents the same mapping $T$, when we view $V$ as a real vector space of dimension $2n$ and use the basis $v_1,v_2,\ldots,v_n,iv_1,iv_2,\ldots,iv_n.$
The extra shuffling I talked about would reorder this latter basis to $v_1,iv_1,\ldots$.
A geometric interpretation of this is that $\det B$ gives the scaling of Lebesgue measure (or hypervolumes) of a box $K=\prod_{i=1}^{2n}[c_i,d_i]$ of real dimension $2n$ under $T$: $$\det (B)=\frac{m(T(K))}{m(K)}.$$
$\det A$ does the same, but because the coordinates are complex there, we need to use $|\det A|^2$ to get the scaling right. This is seen already in the complex plane, where multiplication by $a+bi$ multiplies the areas of rectangles by a factor of $a^2+b^2$.
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If $A$ and $B$ define the "same" linear map shouldn't $\det A = \det B$? – learner Jan 24 '15 at 4:27
@learner: No. For example multiplication by $2$ on $\Bbb{C}$ has determinant two (it's a 1x1 matrix $2I_1$), but when we view $\Bbb{C}$ as $\Bbb{R}^2$ it is a 2x2 matrix $2I_2$ with determinant $=4$. – Jyrki Lahtonen Jan 25 '15 at 9:04
Let $A=B+iC$, where $B$ and $C$ are $n\times n$ real matrices, and let $$\widetilde{A}=\left( \begin{array}{rr} B & -C \\ C & B \end{array} \right).$$ Then $\det\widetilde{A}=|\det A|^2$.
Proof. \begin{align*} \det\widetilde{A}&=\det\left( \begin{array}{cc} B+iC & -C+iB \\ C & B \end{array} \right)= \det\left( \begin{array}{cc} B+iC & 0 \\ C & B-iC \end{array} \right)= \det \left( \begin{array}{ll} A & 0 \\ C & \overline{A} \end{array} \right)\\ &= (\det A)(\det\overline{A})=(\det A)(\overline{\det A})=|\det A|^2. \end{align*}
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This question already has a well accepted answer. You have contributed nothing new – Shailesh Nov 14 '15 at 13:42 | 2016-04-29T22:25:09 | {
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https://math.stackexchange.com/questions/3845649/prove-n-in-mathbb-z-n-text-is-even-n-in-mathbb-z-n-1-text/3845657 | # Prove $\{n \in \mathbb Z | n \text { is even} \} = \{n \in \mathbb Z | n-1 \text { is odd}\}$.
This is found in the book the instructor is using, An Introduction to Proof through Real Analysis by Daniel J. Madden and Jason A. Aubrey The University of Arizona Tucson, Arizona, USA
Here is one way to define $$A$$ in set-builder notation: $$A = \{x | x \text{ is an even integer}\} \text. \tag{9.1}$$ In general, set-builder notation takes the form $$S = \{x | P(x) \text { is true}\} \text, \tag{9.2}$$ where $$P(x)$$ is some mathematical statement. We read this definition of the set $$S$$ as "$$S$$ is the set of all $$x$$ such that $$P(x)$$ is true". So
• $$s \in S$$ if and only if $$P(s)$$ is true;
• $$s \notin S$$ if and only if $$P(s)$$ is false.
There are some common variations on set-builder notation that you will see. For example, people will often use a colon ":" in place of the bar "|". That is fine; the idea is the same. Sometimes, another condition on elements of a set is slipped in before the "such that" symbol by limiting elements to members of a larger set. For example, we could have defined the set of even integers as this: $$A = \{x \in \mathbb Z | x \text { is even}\} \text. \tag{9.3}$$ Besides using English or set-builder notation to define sets, we can define sets by simply listing their elements. For example, we can write $$A = \{\dots, −6, −4, −2, 0, 2, 4, 6, \dots\} \tag{9.4}$$ to define the set of all even integers. But this really only works when the set is small enough that all of its elements can be reasonably listed or when the pattern is strong enough to be recognized. For example, we could write $$B = \{n \in \mathbb N | 2 \le n \le 5\} \text { or } B = \{2, 3, 4, 5\} \text. \tag{9.5}$$
This is all the information from the book that I could find that is pertaining to the purposed question. Please Help Me!
• Welcome to Mathematics Stack Exchange. Can you show $n$ is even $\iff n-1$ is odd? – J. W. Tanner Sep 30 '20 at 1:02
• – Shaun Sep 30 '20 at 1:02
• Um... do you have a question? Any question? You quote something. To a experienced person everything you quote is straightforward and fairly simple. But obviously its not or you wouldn't ask anything. So it's up to you to tell us where you have difficulty and what you want clarified. – fleablood Sep 30 '20 at 2:36
• The point of this is to get you familiar with set builder notation. What is actually being proven is fairly simple: The integers that are even are precisely the numbers that are one more than an odd number. this is obvious. If $n = 2k$ is even then $n-1 = 2k -1$ is odd. That's it we are done. What the excercise though is to put it in terms of set builder notation. And ... I think the text you quoted does a fine job explaining that. – fleablood Sep 30 '20 at 2:41
To prove "set A= set B" you prove "A is a subset of B" and "B is a subset of A". And to prove "A is a subset of B" start "if x is in A" and then use the definitions of A and B to conclude "then x is in B"
Here that means we want to prove "if x is even then x- 1 is odd" and "if x- 1 is odd then x is even".
Of course, we need to use the fact that any even number is of the form 2k for some integer k and any odd number is of the form 2k+ 1 for some integer k.
So "If x is even then there exist an integer k such that x= 2k. Then x- 1= 2k- 1= 2k- 2+ 1= 2(k-1)+ 1 so x-1 is odd."
And "if x- 1 is odd then there exist an integer k such that x- 1= 2k+ 1. Then x= 2k+ 2= 2(k+ 1) so x is even."
$$\{n\in \mathbb Z| n$$ is even $$\}=$$
$$\{n\in \mathbb Z| n$$ is divisible by $$2\}=$$
$$\{n\in \mathbb Z|$$ there exists an integer $$k$$ so that $$n = 2k\}=$$
$$\{n\in \mathbb Z|$$ there exists an integer $$k$$ so that $$n-1 = 2k-1\}=$$
$$\{n\in \mathbb Z| n-1$$ is odd$$\}$$.
In order to show equality of sets, you need to show two-way containment; that is, show that $$A\subseteq B$$ and $$A\supseteq B$$, which is equivalent to saying that $$x\in A \Leftrightarrow x\in B$$. Sometimes it's best to show each direction independently, but in this case, I would recommend a chain of biconditionals, as it's quicker.
Let $$A=\{x\in\mathbb{Z} | x\text{ is even}\}$$ and $$B=\{x\in\mathbb{Z} | x-1\text{ is odd}\}$$. Then we have the following chain of biconditionals:
$$x\in A\Leftrightarrow x$$ is even $$\Leftrightarrow x=2k$$ for some $$k\in\mathbb{Z} \Leftrightarrow x-1=2k-1\Leftrightarrow x-1$$ is odd $$\Leftrightarrow x\in B$$.
Therefore, $$A=B$$. | 2021-06-18T06:40:39 | {
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https://math.stackexchange.com/questions/2226070/how-can-i-show-that-the-polynomial-p-x5-x3-2x2-2x-1-is-irreducibl | # How can I show that the polynomial $p = x^5 - x^3 - 2x^2 - 2x - 1$ is irreducible over $\Bbb Q$?
I've tried a few "criteria" to check if this is irreducible. According to Maple it only has one entirely real root which I suspect is not rational but I can't prove it so I'm attempting to check if $p$ is irreducible.
Eisenstein's Criterion doesn't work here and I'm yet to find a suitable transformation such that it could work. I also read that if a polynomial is irreducible over $\Bbb F_q$, with $q$ a prime not dividing the leading coefficient, then it is irreducible over $\Bbb Q$ so I reduced the polynomial modulo $2$ to obtain
$$p \equiv x^5 + x^3 + 1 \mod 2.$$
I think this is correct but then I need to know how to check the irreducibility of this new polynomial over $\Bbb F_2$. Do I simply need to check that neither $0$ nor $1$ are roots of this polynomial? (And am I applying this theorem correctly?)
If this polynomial IS irreducible over $\Bbb Q$, is the splitting field obtained by simply adjoining the roots to $\Bbb Q$?
• Since the polynomial is of degree $5$ it might be possible for it to have no roots in $\mathbb{Q}$, but be reducible. Writing it as a product of polynomials of degree $2$ and $3$ might be possible. Apr 9, 2017 at 11:11
• To answer your last question: yes. This follows straight from the definition of a splitting field.
– user583416
Apr 9, 2017 at 11:18
• Oh, I've found something wonderful. Answer coming up. Apr 9, 2017 at 11:19
.A little bit of scouting for nice irreducibility criteria throws up some very nice results:
Here is a lovely lemma by (Prof.) Ram Murty:
Let $$f(x) = a_mx^m + ... + a_1x + a_0$$ be a polynomial of degree $$m$$ in $$\mathbb Z[x]$$. Let $$H = \displaystyle\max_{0 \leq i \leq m-1} \left|\frac{a_i}{a_m}\right|$$. If $$f(n)$$ is prime for some $$n \geq H+2$$, then $$f(x)$$ is irreducible in $$\mathbb Z[x]$$.
In our case, $$a_m = 1$$, and the maximum of all the quantities in question is $$2$$. Hence, if $$f(n)$$ is prime for some $$n \geq 4$$, then we are done.
You can check that for $$n=4$$, the number $$f(4) =919$$, which is prime!
Hence, it follows that the polynomial is irreducible.
ASIDE : There is also a "shifted" base (base shifts from $$0...n-1$$ to $$|b| < \frac n2$$) version of Cohn's criteria, which will tell you that if $$f(10)$$ is prime, then the given polynomial is irreducible. This matches that description, since all coefficients are between $$-5$$ and $$5$$. Very interestingly, $$f(10) = 98779$$ is also prime! (Hence, another proof by another wonderful result).
• That is very cool! Thanks for the link. Apr 9, 2017 at 11:28
• You are welcome. I've seen some of these criteria very long ago, while doing a Galois theory polynomial "scouting mission". Apr 9, 2017 at 11:31
• Nice method. Just wondering, why not going already with $f(4)=919$ which is also a prime?
– Sil
Apr 26, 2018 at 23:30
• You are right, it works out. I was doing that calculation mentally, so I skipped directly to $n=5$ because I must have made some mistake at $f(4)$. Apr 26, 2018 at 23:34
By Gauss' Lemma we have that the polynomial is irreducible over $\mathbb{Q}$ if and only if it's irreducible over $\mathbb{Z}$. Now the easiest way would be to prove that polynomial is irreducible over $\mathbb{Z}_2$, which would be enough.
Assume it's reducible. As the polynomial has no roots over $\mathbb{Z}_2$, then the only possibility is if it's a product of polynomials of degree $2$ and $3$. So assume that:
$$x^5 + x^3 + 1 = (x^3 + ax^2 + bx + c)(x^2 + dx + e) \quad \quad \text{over } \quad\mathbb{Z}_2$$
Then multiply everything out and compare the factors. As $a,b,c,d,e \in \mathbb{Z}_2$, you only have two options. Eventually you will get that $c=e=1$ and $a=d=b$. But this would imply that $c+bd + ea = a^2 + a + 1 = 1$ in $\mathbb{Z}_2$. But this is impossible, as it's the coefficient in front of $x^2$ and it should be $0$.
Hence the polynomial is irreducible over $\mathbb{Z}_2$ and eventually $\mathbb{Z}$ and $\mathbb{Q}$
• I think I'll accept this one as being the most useful in a general setting, thanks a lot. Apr 9, 2017 at 11:28
• One could also just divide by the only irreducible polynomial in $\mathbb{Z_2}[x]$ which is $x^2+x+1$ and see it cannot be the factor.
– Sil
Aug 11, 2018 at 9:23
• @Sil True. However I doubt that an OP asking factorizing of a polynomial is aware that $x^2+x+1$ is the only irreducible quadratic in $\mathbb{Z}_2$ Aug 11, 2018 at 9:25
Your proof is almost complete: it is indeed sufficient to prove that $p(x)\in \mathbb F_2[x]$ (the reduction mod $2$ of your polynomial) is irreducible.
The polynomial $p(x)$ has no factor of degree $1$ since it has no zero in $\mathbb F_2$, so there remains only to prove that $p(x)$ has no factor which is an irreducible polynomial $g(x)\in \mathbb F_2[x]$ of degree $2$.
But the only such irreducible polynomial is $g(x)=x^2+x+1$ and long division proves that it does not divide $p(x)$. All is proved.
By the rational root theorem, any rational root of $p$ would have to be a divisor of the constant term, so $\pm1$. Clearly, these are not roots.
Similary, any quadratic factor would have to be $x^2+ax\pm1$ with $a\in\Bbb Z$. You might be able to exclude these manually ... | 2023-03-23T04:45:14 | {
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https://math.stackexchange.com/questions/741949/problem-about-population-growth | At the beginning of the Gold Rush, the population of Coyote Gulch,Arizona was $365$.From then on ,the population would have grown by a factor of $e$ each year,except for the high rate of "accidental" death, amounting to one victim per day among every 100 citizens.By solving an appropriate differential equation determine, as functions of time:
(a) the actual population of Coyote Gulch $t$ years from the day the Gold Rush began, and
(b) the cumulative number of fatalities.
The question is from Apostol's Calculus I. In other questions,Apostol uses the statement "... increases at a rate proportional to the amount present. ..."But this one says "grown by a factor of $e$", I have difficulty in understanding the meaning of "a factor of $e$".Does it mean $y=be^{kt}$ or $y'=ey$ or what?
So I decide to denote "a factor of $e$" by $f_e$ and focus on the ' "accidental" death '.So let $y$ denote the population at present year, each day one victim dies among every 100 citizens .So the population remains in that year is $y(1 -\frac{1}{100})^{365}$. And the number of fatalities in that year is $$y-y(1 -\frac{1}{100})^{365} = y(1- (\frac{99}{100})^{365})$$ . And as far as I can get $$y'=f_e -y(1- (\frac{99}{100})^{365})$$
The answer of the question is given
a) $365e^{-2.65t}$
b) $365(1-e^{-2.65t})$
Any help is appreciated.
• Grown by a factor of $e$ probably if the population this year is $100$ next year it will be $271.8\ldots$(well not fractional population, but cut me some slack) – Guy Apr 6 '14 at 10:36
• @Sabyasachi I tried your suggestion and substitute $f_e$ by $ey$ but I couldn't get the right answer :( – Detective King Apr 6 '14 at 10:55
• I mean $f_e=e$, not $ey$. Does it work? – Guy Apr 6 '14 at 11:00
• @Sabyasachi so $y'=e - y(1-(99/100)^{365})$ and I get $y=362e^{-0.974t}+0.974$. – Detective King Apr 6 '14 at 11:12
• Oh. can't help then. Did you try googling for the question text – Guy Apr 6 '14 at 11:13
Try to write it out in words.
The rate of change in population is the population we have minus the loss ratio of that population (of course, we could have other factors, but that is what we are working with here), so we have:
$$\dfrac{dP}{dt} = P - \alpha P = P(1 - \alpha)$$
Now how we can we find the loss ratio $\alpha$ of the population per year? The problem tells us that we lose "$1$ victim per day among every $100$ citizens", so for the year, we lose $365 \times \dfrac{1}{100}$ of the population $P$. This gives us $\alpha = \dfrac{365}{100}$.
So we have:
$$\dfrac{dP}{dt} = P\left(1 - \dfrac{365}{100}\right), P(0) = 365$$
Note: $P(0)$ is the initial population which is given as $365$.
This DEQ is separable and gives us:
$$20 \int \dfrac{1}{P}~dP = -53 ~\int dt, P(0) = 365$$
Thus,
$$\large P(t) = 365 ~e^{-\frac{53 t}{20}} = 365~ e^{-2.65~ t}$$
If we want to find the cumulative number of fatalities over time, we write that out in words as: we start out with a population of $365$ and we lose them at a rate of $P(t)$ (which we just calculated), so this is:
$$CP(t) = 365 - P(t) = 365 - 365 ~e^{-\frac{53 t}{20}} = 365\left(1 - e^{-\frac{53 t}{20}}\right) = 365\left(1 - e^{-2.65 ~t}\right)$$
We can plot these curves to verify that they are inverses of each other as:
• This was really nicely (and thoroughly) explained, Amzoti. Nice job. – Namaste Apr 6 '14 at 13:27
• @Amzoti Your explaination is really nice!At the same time I think the question is quite "unrealistic"(it feels like "statistic") to me . – Detective King Apr 6 '14 at 14:01
• @DetectiveKing: I think the point of these is just trying to set them up. It is very difficult to take the leap from knowing how to solve things, to actually producing models of the physical reality. By the way, it was not unreasonable for really bad things to happen to small populations as the result of a single issue, including having them totally wiped out. Regards – Amzoti Apr 6 '14 at 14:09 | 2019-08-24T09:21:59 | {
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https://www.physicsforums.com/threads/does-simplex-work-for-all-standard-max-problems.963536/ | # Does Simplex work for all standard-max problems?
Tags:
1. Jan 4, 2019
### Ben2
Mentor note: Fixed the LaTeX
1. The problem statement, all variables and given/known data
Maximize $5x_1 + 7x_2 + 3x_3$ subject to $x_1 + x_2 + x_3 \le 28, x_2 \le2 x_1$ and $x_1 \le x_3$.
2. Relevant Equations
$x_1\ge 0, x_2\ge 0, x_3\ge 0$.
3. The attempt at a solution Problem is workable by graphic methods, but the writer has been unable to
successfully apply the standard-max simplex algorithm. My interest is in finding a workable simplex solution and/or in determining why standard-max simplex won't work here. Thanks for the attention of all posters and site professionals!
Last edited by a moderator: Jan 4, 2019
2. Jan 4, 2019
### Staff: Mentor
Caveat: I havent done any linear programming problems for many years. I haven't set up this problem, but it seems like a standard linear programming example.
The first three constraints should be rewritten as
$x_1 + x_2 + x_3 \le 28$
$-x_1 + x_2 \le 0$
$x_1 - x_3 \le 0$
and $x_1\ge 0, x_2\ge 0, x_3\ge 0$.
LaTeX Tips -- Use a pair of # characters at the beginning and end of a math expression to be rendered for inline LaTeX. Use a pair of \$ characters at the beginning and end of a math expression to be rendered for standalone LaTeX.
For inequalities, use \le, not \leq, and \ge, not \geq.
3. Jan 4, 2019
### Ray Vickson
State the problem as one having 3 variables and 3 constraints (plus non-negativity):
$$\begin{array}{rcl} \max & Z = & 5 x_1 + 7 x_2 + 3 x_3 \\ \text{s.t.}&&x_1 + x_2 + x_3 \leq 28 \\ &&- 2x_1 + x_2 \leq 0 \\ && x_1 - x_3 \leq 0 \\ &&x_1,x_2, x_3 \geq 0 \end{array}$$
So, if $s_1, s_2, s_3$ are the slack variables for constraints 1--3 and we write the objective equation as $Z - 5 x_1 - 7 x_2 - 3 x_3 = 0$ , the initial tableau for the problem is
$$\begin{array}{ccccccc|c} Z & x_1 & x_2 & x_3 & s_1 & s_2 & s_3 & \text{RHS} \\ \hline 1 & -5 & -7 & -3 & 0 & 0 & 0 & 0 \\ 0 & 1 & 1 & 1 & 1 & 0 & 0 & 28\\ 0 & -2 & 1 & 0 & 0 & 1 & 0 & 0 \\ 0 & 1 & 0 & -1 & 0 & 0 & 1 & 0 \end{array}$$
The simplex method works perfectly well here. If you do not think so, you need to show us your work and indicate where the method fails.
Note that in this problem some simplex steps may involve "degenerate" pivots, where the objective does not change from one iteration to the next, due to having some basic variables equal to 0; essentially, we change a 0 basic variable in one tableau into a non-basic variable (still zero) in another tableau. The simplex method may be "slowed down" a bit here, because it may need to pass through several such degenerate solutions on the way to the final, non-degenerate solution having no basic variables equal to zero.
Last edited: Jan 4, 2019
4. Jan 4, 2019
### Ray Vickson
The commands "\leq" and "\geq" are TeX/LaTeX standards, and work perfectly well in the PF implementation; I use them all the time without any problems.
5. Jan 4, 2019
### Staff: Mentor
OK, thanks -- didn't know that. The \le and \ge operators work in MathJax, and have the advantage of requiring slightly less typing.
6. Jan 4, 2019
### WWGD
The title of this thread does not seem to agree with the question in the OP. The title is , it seems, much more general while the question seems to refer to a specific question. AFAIK, the Simplex method requires some conditions for convergence. I will look them up ASAP.
7. Jan 4, 2019
### Ray Vickson
Right. But let's wait until the OP responds, and shows the difficulties he/she is facing.
Anyway, there are simple "rules" that can be applied that guarantee the convergence in finitely many steps of the simplex method to an optimal resolution (either showing infeasibility or unboundedness of the problem if these properties apply, or else determining an optimal solution).
8. Jan 4, 2019
### kimbyd
1) This is exactly the kind of problem that the Simplex algorithm was designed to solve.
2) In some cases, however, the Simplex algorithm may stall due to degeneracy, and has worst-case exponential time complexity. Wikipedia describes these issues here.
From my reading, it should be efficient for many inputs similar to your example above, but there may be complicated situations where the algorithm stalls out and wastes quite a lot of time.
9. Jan 4, 2019
### Ray Vickson
Nevertheless, companies and other organizations have, for many years, been solving problems having many thousands of variables and many thousands of constraints, using Simplex. For much larger problems (say involving millions of variables and hundreds of thousands to millions of constraints), solvers sometimes use interior-point methods instead of Simplex. Some really big problems (having several hundred million variables) have been solved due to their special "network" structure, using network-flow implementations of the simplex method. Needless to say, simple textbook rules and prescriptions are not good enough for these really big cases.
10. Jan 5, 2019
### FactChecker
You should recheck your Simplex calculations.
Using the Simplex implementation at https://www.zweigmedia.com/RealWorld/simplex.html, I got a solution.
The input format is:
Maximize z=5x1+7x2+3x3 subject to
x1+x2+x3<=28
-2x1+x2<=0
x1-x3<=0
It shows four Tableaus, ending with the solution
x3=7
x2=14
x1=7
z=154
Last edited: Jan 5, 2019
11. Jan 5, 2019
### Ben2
My thanks for the Mentor's corrections. I found the LaTeX inequality-symbol equivalents on this site's own "how-to" primer,
but will try to use them on this site.
FactChecker has the problem-formulation I intended. Let me apologize to anyone thrown off by my original statement.
I've recently gotten the final tableau referenced by FactChecker, with one intermediate tableau. However, the first pivot-element
was a "0" where the constant/element quotient was 0/0. I'm not sure if I'm familiar with all the "rules" mentioned in Ray
Vickson's second post.
The standard Simplex method I've seen in Finite-Math texts seemed to guarantee that each new tableau of
standard-max simplex automatically represents a basic feasible solution. My hand calculations on the given problem, mistake-ridden
though they may have been, prompt me to question that.
Re WWGD's comment, I tried setting this up as a mixed-constraint problem before finally getting the standard-max solution referenced above.
Thanks to all posters for extremely informative comments, and will access "bad" cases referenced by kimbyd.
Ben2
12. Jan 5, 2019
### FactChecker
Usually some artificial variables are added which are driven out (using artificially high cost values) to arrive at a feasible solution that only uses the original variables. I am not familiar with any other way to guarantee that a feasible solution can be found.
13. Jan 5, 2019
### Ray Vickson
Modern industrial-scale computer implementations of the simplex method avoid artificial variables, but instead consider starting at infeasible bases and trying to get rid of infeasibilites first, before optimizing. For example, consider the problem
$$\begin{array}{ccl} \max & Z = & 4 x_1 + 6 x_2 - x_3 \\ \text{s.t.}&&5 x_1 + 2 x_2 + 8 x_3 = 80\\ &&4 x_1 + 9 x_2 +3 x_3 \leq 60\\ &&6 x_1 + 4 x_2 + 5 x_3 \geq 60\\ && x_1, x_2, x_3 \geq 0 \end{array}$$
If we introduce a slack variable $s_1$ for the first inequality and a surplus variable $s_2$ for the second one, our problem can be written as that of maximizing $Z$ over the non-negative variables $x_1, x_2, x_3, s_1, s_2$ that are related by the equations
$$\begin{array}{ccc} Z- 4 x_1 - 6x_2 + x_3 &=&0\\ 5 x_1 + 2 x_2 + 8 x_3 &= &80\\ 4x_1 + 9x_2 + 3x_3 + s_1 &=& 60\\ 6x_1 + 4 x_2 + 5 x_3 - s_2 &=& 60 \end{array}$$
We don't even know if the problem is feasible, so determining that is the first order of business. We do not yet have an LP basis.
Without worrying about feasibility, let us choose an initial basis in which the basic variables are $x_1, s_2, s_3$. The equations can be re-written by putting basic variables on the left and non-basic variables on the right:
$$\begin{array}{ccl} Z&=& 64 +(22/5) x_2 - (37/5) x_3 \\ x_1 &=& 16 -(2/5) x_2 - (8/5) x_3 \\ s_1 &=& -4 -(27/5) x_2 + (17/5) x_3 \\ s_2&=& 36 +(8/5) x_2 - (2/5) x_3 \end{array}$$
This basis is infeasible because it has $s_1 = -4 < 0$. Temporarily, we can think of this as a problem of trying to maximize $s_1$ (so as to drive it up towards zero), and for that reason we choose to increase $x_3$. Using the usual minimum-ratio rules to maintain feasibility of $x_1, s_2$ and to try to achieve feasibility of $s_1$, the ratios are
$$\begin{array}{l} s_1 \;\text{ratio} = (-4)/(-17/5) = 1.176\\ s_2 \;\text{ratio} = 36/(23/5) = 7.826\\ x_1 \; \text{ratio} = 16/(8/5) = 10 \end{array}$$
The minimum ratio is for $s_1$, so $s_1$ leaves the basis and $x_3$ enters. The new equations are
$$\begin{array}{ccl} Z &=& 940/17 -(199/17) x_2 - (37/17) s_1\\ x_1 &=&240/17 -(66/17) x_2+-(8/17) s_1\\ x_3 &=& 20/17+(37/17) x_2+(5/17)s_1 \\ s_2 &=& 520/17-(143/17) x_2 - (23/17) s_1 \end{array}$$
This gives us a feasible solution $Z = 940/17, x_1 = 240/17, x_3 = 20/17, s_2 = 520/17.$ The current $Z$-equation also shows that this solution is optimal
By the way: what would happen if we had a truly infeasible problem but did not use artificial variables? In this case the above method would eventually contain an equation something like $$\text{basic variable}\; x = -2 - 4 u_1 - 3 u_2 - \cdots$$ where $u_1, u_2, u_3 \ldots$ are the current non-basic variables. Note that the right-had-side is negative, and all the coefficients on the right are non-positive. That is, we have a negative basic variable, and increasing any of the non-basic variables up from 0 will not help (and may even make matters worse). This says that the basic variable $x$ cannot be larger than -2, so certainly cannot be non-negative. Our problem would be detected as being infeasible.
14. Jan 5, 2019
### FactChecker
That is true after an initial feasible solution has been found. The Simplex method will move from one feasible corner solution to another. But finding an initial feasible solution takes some tricks. Some variables ("slack" and "surplus") can be added which make it easy to start from a "psudo-feasible" solution with those variables. Then an initial Simplex phase can proceed to feasible solutions which use the original variables. The initial phase is very similar to the regular Simplex method that follows.
Last edited: Jan 6, 2019
15. Jan 7, 2019
### kimbyd
That makes good sense. From my reading, it sounds like Simplex "just works" most of the time, but if you're going to have any sort of system depend upon it, you have to be careful that it can stall out at times, and deal with those cases appropriately. There are solutions for those cases, so it's not a deal-breaker. It's just a little extra complexity that needs to be managed.
Edit:
And for any simple problem you should encounter, you should expect Simplex to work quite efficiently. Especially a textbook problem. You should only expect to encounter the pathological behavior if you're building a large-scale system designed to run a lot of Simplex calculations on a lot of different inputs.
If you try Simplex, and the answer is wrong, then there's a bug in how you're using/implementing the algorithm.
If you try Simplex, and you find it keeps cycling, then you might have hit a stall mentioned above. This shouldn't occur in textbook problems. The algorithm should eventually find a result even in this case, but it may take a little while. Still, if your number of inputs is small, even exponential-time complexity will run quickly in a computer If they're asking you to do it by hand, then it's either an error in your calculations or an error in the textbook (many textbooks have errata, so you might try to look those up if you think the textbook might have a mistake).
Last edited: Jan 7, 2019
16. Jan 7, 2019
### FactChecker
There are techniques to detect and move on from a stall. I would expect any well-used, reputable program to include such a technique. I would hesitate to recommend that a person deal with a stall himself on a very large problem. That would be extremely difficult. I find that even following the calculations on a small problem to be difficult. I could not imagine doing that with a thousand variables.
17. Jan 7, 2019
### kimbyd
I definitely wasn't suggesting a person deal with the stall by hand, hence the statement that such a stall indicates a textbook error. Obviously the complexity has to be managed at the software implementation level.
18. Jan 7, 2019
### Ray Vickson
It is possible for the Simplex method (if naively implemented) to cycle forever among a group of non-optimal solutions; i.e., stall permanently at non-optima. However, there are simple safeguards that can prevent that from happening, so stalling forever won't happen in a properly-implemented system, but stalling for quite a while sometimes will, despite all the safeguards. Commercial codes have various strategies to try to "unstall" the method, but of course, they will not work 100% of the time---just often enough to be useful. For gigantic problems, some systems will start with interior-point methods, perhaps to get near a basic feasible solution, then switch to the simplex method near the end to get an accurate, provably-optimal solution.
19. Jan 13, 2019 | 2019-01-23T11:23:26 | {
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https://math.stackexchange.com/questions/3251616/show-the-matrix-commutes-with-companion-matrix-is-a-polynomial | # Show the matrix commutes with companion matrix is a polynomial
Let $$A$$ be a linear transform on $$n$$-dimensional $$V$$ over a field $$F$$. Under a basis $$\alpha_1, \cdots, \alpha_n$$, the matrix representation of $$A$$ is as follows: $$A = \begin{bmatrix} 0 & 0 & \dots & 0 & -a_0 \\ 1 & 0 & \dots & 0 & -a_1 \\ 0 & 1 & \dots & 0 & -a_2 \\ \vdots & \vdots & \ddots & \vdots & \vdots \\ 0 & 0 & \dots & 1 & -a_{n-1} \end{bmatrix}.$$ Let $$C(A):= \{T: T\text{ is a linear transform on V and } TA = AT \}$$, and let $$F[A]$$ denotes all the polynomials in $$A$$. Show that: $$C(A) = F[A]; \dim(C(A)) = n.$$
First of all, the minimal polynomial $$m(\lambda)$$ of $$A$$ is the same as its characteristic polynomial $$f(\lambda)$$, namely $$m(\lambda) = f(\lambda) = \lambda^n + a_{n-1}\lambda^{n-1} + \cdots a_0$$. Thus, plugging in $$A$$, we see that all $$A^{k}$$ with $$k \geq n$$ could be expressed by $$I, A, A^2, \cdots, A^{n-1}$$. So $$\dim F[A] \leq n$$. If $$\dim F[A] < n$$, say $$k_0 I + k_1 A + \cdots + k_r A^r = 0$$ with $$r < n$$ and some $$k_j \neq 0$$, then we have that $$g(\lambda) = k_0 + k_1 \lambda + \cdots + k_r \lambda^r$$ is another polynomial with $$g(A) = 0$$. By the definition of minimal polynomial, we must have that $$r \geq n$$, a contradiction. So $$\dim(F[A]) = n$$, and it remains to show the first equality $$C(A) = F[A]$$.
Also, one could see that $$F[A] \subseteq C(A)$$. But I am not sure how to show the other direction. Could someone give me a hint?
Define a linear map $$\Psi \colon C(A) \rightarrow V$$ by $$\Psi(T) = T\alpha_1$$. Let's show that this map is an isomorphism. First, note that $$A^i \in C(A)$$ for all $$i \in \mathbb{N}_0$$ and
$$\Psi(I) = \alpha_1, \psi(A) = A\alpha_1 = \alpha_2, \cdots, \psi(A^{n-1}) = A^{n-1}\alpha_1 = \alpha_n.$$
This shows that $$\Psi$$ is surjective. Next, let's assume that $$\psi(T) = T\alpha_1 = 0$$. Then $$T\alpha_2 = T(A\alpha_1) = A(T\alpha_1) = A(0) = 0, \\ T\alpha_3 = T(A\alpha_2) = A(T\alpha_2) =0, \\ \vdots,\\ T\alpha_n = T(A\alpha_{n-1}) = A(T\alpha_{n-1}) = 0 \\$$
which shows that $$T \equiv 0$$. This shows that $$\Psi$$ is injective. Hence, $$\dim C(A) = n$$ and since $$\dim F[A] = n$$ and $$F[A] \subseteq C(A)$$, we deduce that $$F[A] = C(A)$$.
• Yes I think this is excellent! But how do we come up with such $\Psi$? – mathdoge Jun 5 at 12:00
• The matrix $A$ acts on your basis by $\alpha_1 \mapsto \alpha_2 \dots \mapsto \alpha_n$ (where $x \mapsto y$ means that $A$ sends $x$ to $y$). By applying $T$ and using the fact that $A$ and $T$ commute, you can see that we also have $T\alpha_1 \mapsto T\alpha_2 \dots \mapsto T\alpha_n$. From here you can already see that if we know $T\alpha_1$ and $A$, we also know $T\alpha_i$ for all $2 \leq i \leq n$. – levap Jun 5 at 12:06 | 2019-08-23T10:52:50 | {
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https://math.stackexchange.com/questions/3374300/find-the-sum-to-n-terms?noredirect=1 | # Find the sum to n terms
$$S=1^2+3^2+6^2+10^2+15^2+.......$$
My attempt is as follows:
$$T_n=\left(\frac{n\cdot\left(n+1\right))}{2}\right)^2$$
$$T_n=\frac{n^4+n^2+2\cdot n^3}{4}$$
$$S=\frac{1}{4}\cdot\sum_{n=1}^{n}\left(n^4+n^2+2\cdot n^3\right)$$
Now to solve this one has to calculate $$\sum_{n=1}^{n}n^4$$ which will be a very lengthy process, is there any shorter method to solve this question?
By the way I calculated $$\sum_{n=1}^{n}n^4$$ and it came as $$\dfrac{\left(n\right)\cdot\left(n+1\right)\cdot\left(2\cdot n+1\right)\cdot\left(3\cdot n^2+3\cdot n-1\right)}{30}$$, then I substituted this value into the original equation.
Then I got final answer as $$\dfrac{\left(n\right)\cdot\left(n+1\right)\cdot\left(n+2\right)\cdot\left(3\cdot n^2+6\cdot n+1\right)}{60}$$
But it took me a very long time to calculate all of this, is there any shorter way to solve this problem?
• This is A024166 in OEIS. They don't appear to provide a simple closed formula for it. – lulu Sep 29 '19 at 13:42
• actually my requirement is to solve this faster, I did it in the conventional way and it took a lot of time. – user3290550 Sep 29 '19 at 13:49
• Have you seen this picture proof for $\sum r^4$ ? – almagest Sep 29 '19 at 14:08
• Alternatively, the standard way to do $\sum_rr^k$ is to do $\sum r(r+1)\dots(r+k-1)$ – almagest Sep 29 '19 at 14:11
• Don't have multiple identically-named variables in the same expression! It is very confusing. For example, in the summations, either change the variable denoting the number of terms or change the variable that is being summed over. – Solomon Ucko Sep 29 '19 at 22:14
It is easier to do this kind of problem using factorial polynomials than conventional polynomials. Using falling factorials, for example, we would use $$(n)_4=n(n-1)(n-2)(n-3)$$ rather than $$n^4$$. The wiki articles explains how to convert from conventional polynomials to falling or rising factorials under in the section titled, "Connection coefficients and identities." The advantage of using factorial polynomials comes in summation. We have, for example $$(n+1)_5-(n)_5=(n+1)n(n-1)(n-2)(n-3)-n(n-1)(n-2)(n-3)(n-4)=5(n)_4$$ so that $$\sum_{n=1}^k(n)_4=\frac15\sum_{n=1}^k((n+1)_5-(n)_5)={(k+1)_5-1\over5}$$
EDIT
In this case, it's very easy, because we have $$T_n=\frac14(n+1)_2(n+1)_2$$ and we can use one of the formulas under the "Connection coefficients and identities" section to get$$T_n=\frac14(n+1)_2(n+1)_2=\sum_{k=0}^2{2\choose k}{2\choose k}(n+1)_{4-k}$$
• Wow...what a nice and simple approach! – John Hughes Sep 29 '19 at 15:09
• See Concrete Mathematics by Donald Knuth et. al. – Felix Marin Jul 13 at 16:40
Well...there's a way to get a good guess of the answer. You could say to yourself (or plot the data!) that it looks like a 5th degree polynomial, $$p(n) = a_5n^5 + a_4 n^4 + \ldots + a_0.$$ Then you know that $$p(n+1) - p(n)$$ is the thing you've called $$T_n$$, but it's also $$a_5[(n+1)^5 - n^5] + a_4 [(n+1)^4 - n^4] + \ldots + a_1 [(n+1)^1 - n^1]$$ which you can write out as a 4th degree polynomial. The first term will be $$4 a_5 n^4,$$ I think, which I got by simply expanding $$(n+1)^5 - n^5$$ using Pascal's triangle.
Setting this 4th degree poly equation to $$T_n$$, you get a triangular system of equations that can be backsubstituted to get an answer.
OF course, you then must check that the answer is in fact correct. You know it satisfies the recurrence, but you also need to show it gives the correct values for the first few values of $$n$$ (perhaps the first six?)
Is it faster? Probably. But not a lot. And if the answer had turned out not to be polynomial, you'd have wasted a lot of time.
You can use this result: why is $\sum\limits_{k=1}^{n} k^m$ a polynomial with degree $m+1$ in $n$
Since $$T_n$$ is of degree $$4$$ then its sum will be of degree $$5$$.
So let $$S(n)=a_0+a_1n+a_2n^2+a_3n^3+a_4n^4+a_5n^5$$.
Then you can proceed by solving the system obtained calculating the first $$6$$ terms $$S(1)$$ to $$S(6)$$.
• yeah this is the good way, actually we don't need $a_0$, S(n) will always a multiple of n – user3290550 Sep 29 '19 at 14:15
• you can even simplify it further using Saulspatz hint. If you write $S(n)=a_0+a_1(n-1)+a_2(n-1)(n-2)+\cdots+a_5(n-1)(n-2)(n-3)(n-4)(n-5)$ then the system is even easier to solve since it is triangular. – zwim Sep 29 '19 at 14:17
You may use the hockey-stick identity, like in the computation of $$\sum_{n=1}^{N}n^m$$. We have
$$\binom{n}{2}^2 = 6\binom{n}{4}+12\binom{n}{3}+7\binom{n}{2}+\binom{n}{1}$$ hence $$\begin{eqnarray*} \sum_{n=1}^{N}\binom{n}{2}^2 &=& 6\binom{N+1}{5}+12\binom{N+1}{4}+7\binom{N+1}{3}+\binom{N+1}{2}\\&=&\color{red}{\frac{N(N+1)(N+2)(3N^2+6N+1)}{60}}.\end{eqnarray*}$$
I change a little the notations from your original wording.
$$T(n)=\dfrac{n(n+1)}{2}$$
$$S(n)=\sum\limits_{k=1}^nT(k)^2$$
Cheating on the resulting formula for $$f(n)=\dfrac{n(n+1)(n+2)(3n^2+6n+1)}{60}$$
we notice that $$f(0)=f(-1)=f(-2)=0$$.
Is there a way to exploit these negative indices in order to find the $$(3n^2+6n+1)$$ part more easily?
In fact there is, $$T(n)$$ is perfectly defined for negative numbers so this part does not pose any issue.
But how to interpret $$S(-n)$$?
To be consistent with the summation identity $$\sum\limits_a^{b-1}+\sum\limits_b^c=\sum\limits_a^c$$ when $$a one need to set $$\ \sum\limits_M^m=-\sum\limits_{m+1}^{M-1}$$ for the case $$m.
This results in $$\displaystyle S(-n)=\sum\limits_{k=1}^{-n}T(k)^2=-\sum\limits_{k=-(n-1)}^0T(k)^2$$
I let you do the calculation to verify that $$\begin{array}{lcr}S(0)&=&0\\S(-1)&=&0\\ S(-2)&=&0\\ S(-3)&=&-1\\S(-4)&=&-10\\S(-5)&=&-46\end{array}$$
Now by applying the identification method (I describe in my other post) to $$S(n)=a_0+a_1\,n+a_2\,n(n+1)+a_3\,n(n+1)(n+2)+\cdots+a_5\,n(n+1)(n+2)(n+3)(n+4)$$
You immediately get $$a_0=a_1=a_2=0$$
The others coefficients give you $$S(n)=n(n+1)(n+2)\bigg[\frac 16-\frac 14(n+3)+\frac 1{20}(n+3)(n+4)\bigg]=n(n+1)(n+2)\left(\dfrac{3n^2+6n+1}{60}\right)$$
Once you’ve rewritten your sum as $$\frac14\sum_{k=1}^n k^4+2k^3+k^2$$ it’s not terribly difficult to compute this using generating functions if you use some key tools for manipulating them. To wit, if $$g(x)$$ is the ordinary generating function for the sequence $$\{a_n\}_{n=0}^\infty$$, then $$g(x)/(1-x)$$ is the o.g.f. for the sequence of partial sums $$\{\sum_{k=0}^na_k\}_{n=0}^\infty$$, and similarly, $$x\frac d{dx}g(x)$$ is the o.g.f. for $$\{na_n\}_{n=0}^\infty$$. So, starting with the o.g.f. $$(1-x)^{-1}$$ for the sequence of all ones, we have $$\left\{\sum_{k=0}^n k^2\right\}_{n=0}^\infty \stackrel{ogf}{\longleftrightarrow} \frac1{1-x}\left(x\frac d{dx}\right)^2\frac1{1-x} = {x+x^2 \over (1-x)^4},$$ therefore $$\sum_{k=0}^n k^2 = [x^n]{x+x^2\over(1-x)^4} = [x^{n-1}]\frac1{(1-x)^4} + [x^{n-2}]\frac1{(1-x)^4},$$ which you can compute using the generalized binomial theorem. Similarly, $$\sum_{k=0}^n k^3 = [x^n]{x+4x^2+x^3\over(1-x)^5} \\ \sum_{k=0}^n k^4 = [x^n]{x+11x^2+11x^3+x^4\over(1-x)^6}.$$ | 2020-10-24T01:13:07 | {
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https://calculus7.org/2015/01/17/completely-monotone-imitation-of-1x/ | Completely monotone imitation of 1/x
I wanted an example of a function ${f}$ that behaves mostly like ${1/x}$ (the product ${xf(x)}$ is bounded between two positive constants), but such that ${xf(x)}$ does not have a limit as ${x\rightarrow 0}$.
The first thing that comes to mind is ${(2+\sin(1/x))/x}$, but this function does not look very much like ${1/x}$.
Then I tried ${f(x)=(2+\sin\log x)/x}$, recalling an example from Linear Approximation and Differentiability. It worked well:
In fact, it worked much better than I expected. Not only if ${f'}$ of constant sign, but so are ${f''}$ and ${f'''}$. Indeed,
$\displaystyle f'(x) = \frac{\cos \ln x - \sin \log x - 2}{x^2}$
is always negative,
$\displaystyle f''(x) = \frac{4 -3\cos \log x + \sin \log x}{x^3}$
is always positive,
$\displaystyle f'''(x) = \frac{10\cos \log x -12}{x^4}$
is always negative. The sign becomes less obvious with the fourth derivative,
$\displaystyle f^{(4)}(x) = \frac{48-40\cos\log x - 10\sin \cos \ln x}{x^5}$
because the triangle inequality isn’t conclusive now. But the amplitude of ${A\cos t+B\sin t}$ is ${\sqrt{A^2+B^2}}$, and ${\sqrt{40^2+10^2}<48}$.
So, it seems that ${f}$ is completely monotone, meaning that ${(-1)^n f^{(n)}(x)\ge 0}$ for all ${x>0}$ and for all ${n=0,1,2,\dots}$. But we already saw that this sign pattern can break after many steps. So let’s check carefully.
Direct calculation yields the neat identity
$\displaystyle \left(\frac{1+a\cos \log x+b\sin\log x}{x^n}\right)' = -n\,\frac{1+(a-b/n)\cos\log x+(b+a/n) \sin\log x}{x^{n+1}}$
With its help, the process of differentiating the function ${f(x) = (1+a\cos \log x+b\sin\log x)/x}$ can be encoded as follows: ${a_1=a}$, ${b_1=b}$, then ${a_{n+1}=a_n-b_n/n}$ and ${b_{n+1} = b_n+a_n/n}$. The presence of ${1/n}$ is disconcerting because the harmonic series diverges. But orthogonality helps: the added vector ${(-b_n/n, a_n/n)}$ is orthogonal to ${(a_n,b_n)}$.
The above example, rewritten as ${f(x)=(1+\frac12\sin\log x)/x}$, corresponds to starting with ${(a,b) = (0,1/2)}$. I calculated and plotted ${10000}$ iterations: the points ${(a_n,b_n)}$ are joined by piecewise linear curve.
The total length of this curve is infinite, since the harmonic series diverges. The question is, does it stay within the unit disk? Let’s find out. By the above recursion,
$\displaystyle a_{n+1}^2 + b_{n+1}^2 = \left(1+\frac{1}{n^2}\right) (a_n^2+b_n^2)$
Hence, the squared magnitude of ${(a_n,b_n)}$ will always be less than
$\displaystyle \frac14 \prod_{n=1}^\infty \left(1+\frac{1}{n^2}\right)$
with ${1/4}$ being ${a^2+b^2}$. The infinite product evaluates to ${\frac{\sinh \pi}{\pi}a\approx 3.7}$ (explained here), and thus the polygonal spiral stays within the disk of radius ${\frac12 \sqrt{\frac{\sinh \pi}{\pi}}\approx 0.96}$. In conclusion,
$\displaystyle (-1)^{n} \left(\frac{1+(1/2)\sin\log x}{x}\right)^{(n)} = n!\,\frac{1+a_{n+1}\cos\log x+b_{n+1} \sin\log x}{x^{n+1}}$
where the trigonometric function ${a_{n+1}\cos\log x+b_{n+1} \sin\log x}$ has amplitude strictly less than ${1}$. Since the expression on the right is positive, ${f}$ is completely monotone.
The plot was generated in Sage using the code below.
a,b,c,d = var('a b c d')
a = 0
b = 1/2
l = [(a,b)]
for k in range(1,10000):
c = a-b/k
d = b+a/k
l.append((c,d))
a = c
b = d
show(line(l),aspect_ratio=1)
2 thoughts on “Completely monotone imitation of 1/x”
1. What was the motivation for finding such a function? Interesting stuff!
1. L says:
A Math.SE question | 2019-09-16T03:13:06 | {
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https://www.coursehero.com/file/p61cluh/EXAMPLE-34-Finding-the-Number-of-Terms-Needed-for-a-Given-Accuracy-Determine/ | EXAMPLE 34 Finding the Number of Terms Needed for a Given Accuracy Determine
# Example 34 finding the number of terms needed for a
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EXAMPLE 3.4 Finding the Number of Terms Needed for a Given Accuracy Determine the number of terms needed to obtain an approximation to the sum of the series k = 1 1 k 3 correct to within 10 5 . Solution Again, we already used the Integral Test to show that the series in question converges. Then, by Theorem 3.2, we have that the remainder satisfies 0 R n n 1 x 3 dx = lim R →∞ R n 1 x 3 dx = lim R →∞ 1 2 x 2 R n = lim R →∞ 1 2 R 2 + 1 2 n 2 = 1 2 n 2 . So, to ensure that the remainder is less than 10 5 , we require that 0 R n 1 2 n 2 10 5 . Solving this last inequality for n yields n 2 10 5 2 or n 10 5 2 = 100 5 223 . 6 .
8-31 SECTION 8.3 . . The Integral Test and Comparison Tests 641 So, taking n 224 will guarantee the required accuracy and consequently, we have k = 1 1 k 3 224 k = 1 1 k 3 = 1 . 202047, which is correct to within 10 5 , as desired. Comparison Tests We next present two results that allow us to compare a given series with one that is already known to be convergent or divergent, much as we did with improper integrals in section 6.6. THEOREM 3.3 (Comparison Test) Suppose that 0 a k b k , for all k . (i) If k = 1 b k converges, then k = 1 a k converges, too. (ii) If k = 1 a k diverges, then k = 1 b k diverges, too. Intuitively, this theorem should make abundant sense: if the “larger” series converges, then the “smaller” one must also converge. Likewise, if the “smaller” series diverges, then the “larger” one must diverge, too. PROOF Given that 0 a k b k for all k , observe that the n th partial sums of the two series satisfy 0 S n = a 1 + a 2 + ··· + a n b 1 + b 2 + ··· + b n . (i) If k = 1 b k converges (say to B ), this says that 0 S n a 1 + a 2 + ··· + a n b 1 + b 2 + ··· + b n k = 1 b k = B , (3.5) for all n 1. From (3.5), the sequence { S n } n = 1 of partial sums of k = 1 a k is bounded. No- tice that { S n } n = 1 is also increasing. (Why?) Since every bounded, monotonic sequence is convergent (see Theorem 1.4), we get that k = 1 a k is convergent, too. (ii) If k = 1 a k is divergent, we have (since all of the terms of the series are nonnegative) that lim n →∞ ( b 1 + b 2 + ··· + b n ) lim n →∞ ( a 1 + a 2 + ··· + a n ) = ∞ . Thus, k = 1 b k must be divergent, also. You can use the Comparison Test to test the convergence of series that look similar to series that you already know are convergent or divergent (notably, geometric series or p -series).
642 CHAPTER 8 . . Infinite Series 8-32 EXAMPLE 3.5 Using the Comparison Test for a Convergent Series Investigate the convergence or divergence of k = 1 1 k 3 + 5 k . Solution The graph of the first 20 partial sums shown in Figure 8.26 suggests that the series converges to some value near 0.3. To confirm such a conjecture, we must carefully test the series. Note that for large values of k , the general term of the series looks like 1 k 3 , since when k is large, k 3 is much larger than 5 k . This observation is significant, since we already know that k = 1 1 k 3 is a convergent p -series ( p = 3 > 1). | 2021-07-30T23:12:02 | {
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https://byjus.com/question-answer/the-average-of-20-numbers-is-zero-of-them-at-the-most-how-many-may/ | Question
# The average of $$20$$ numbers is zero. Of them, at the most, how many may be greater than zero?
A
0
B
1
C
10
D
19
Solution
## The correct option is D $$19$$Let the $$20$$ numbers be $$a_1,a_2,...,a_{20}$$Given that average of 20 numbers is zero.Therefore, $$\dfrac{a_1+a_2+...+a_{20}}{20}=0$$$$\implies {a_1+a_2+...+a_{20}}=0$$$$\implies a_1+a_2+...+a_{19}=-a_{20}$$Therefore, at the most $$19$$ numbers can be greater than zero.Mathematics
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View More | 2022-01-23T10:05:23 | {
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https://math.stackexchange.com/questions/2562233/does-there-exist-a-continuous-surjection-from-bbb-r3-s2-to-bbb-r2-0-0/2566284 | # Does there exist a continuous surjection from $\Bbb R^3-S^2$ to $\Bbb R^2-\{(0,0)\}$?
Prove or disprove : There exists a continuous surjection from $\mathbb{R}^3- S^2$ to $\mathbb{R}^2-\{(0,0)\}$ (here $S^2\subset \mathbb{R}^3$ denotes the unit sphere defined by the equation $x^2+y^2+z^2=1$),.
This question had appeared in TIFR GS-2018 exam for PhD admissions. How should I think about such a map?
• Hint: Think about (path-)connectedness – B. Mehta Dec 11 '17 at 20:31
• $\mathbb{R}^3-S^2$ is Euclidean space with the unit sphere removed. So there are no paths between the interior of the sphere and the exterior. – user326210 Dec 11 '17 at 20:46
• @B.Mehta I don't think it's going to be as simple as that – Perturbative Dec 13 '17 at 15:47
• @user326210 But how would that prove your point? There are examples of continuous maps which map not-path-connected spaces to path connected spaces. – Error 404 Dec 13 '17 at 16:28
• @B.Mehta Can you elaborate your approach? Thanks in advance. – Error 404 Dec 14 '17 at 7:25
There is such a map :
• project $\mathbb{R}^3 \setminus \mathbb{S}_2$ onto $\mathbb{R}^2$ (projection on the first two coordinates, for instance);
• apply the exponential map from $\mathbb{R}^2 \simeq \mathbb{C}$ onto $\mathbb{R}^2 \setminus \{(0,0)\} \simeq \mathbb{C}^*$.
Both are continuous surjections, so their composition still is.
Given the bounty, I think this deserves at least some additional material (or: how may someone find this answer). Let $X$ and $Y$ be two topological spaces. If there is no continuous surjective map $f$ from $X$ to $Y$, then $X$ must have some property which is preserved by continuous maps, and which $Y$ doesn't have. It turns out that there are not many such properties. On top of my head, the only general ones I can see are :
• Cardinality : If Card(X) < Card(Y), then there is no such $f$. Example : $X = \{0\}$, $Y = \{0,1\}$.
• Compactness : If $X$ is compact and $Y$ is separable but not compact, then there is no such $f$. Example : $X = [0,1]$, $Y = \mathbb{R}$.
• Connectedness : If $X$ is connected and $Y$ isn't, then there is no such $f$. More generally, if $X$ has less (in the sense of cardinality) connected components than $Y$, then there is no such $f$. Example : $X = (0,1)$ and $Y = \mathbb{Z}$.
Although, I am sure, one may find examples with other, less obvious, obstructions. Outside of these, things may get wild. For instance, in all the following cases, there exists a continuous surjective map from $X$ to $Y$:
• $X = C$, any Cantor set, and $Y$ is any compact metric space.
• $X=[0,1]$ and $Y = [0,1]^2$ : Peano curve.
• More generally, $X = \mathbb{R}$ and $Y = \mathbb{R}^n$, with $n \geq 1$, using variants of the Peano curve.
• $X = \mathbb{R}^k$ and $Y = \mathbb{R}^n$, with $k \geq n$, using projections. Combining with he previous example, we get all $k \geq 1$ and $n \geq 0$.
• $X = \mathbb{R}^k$ and $Y$ is a (separable) connected, $n$-dimensional topological manifold, with $k \geq 1$. This is not easy to formalize and I have no reference at hand, but the basic idea is to take $k=n$, and make $\mathbb{R}^n$ a thin tape and wrap it around the manifold (see e.g. this related discussion). I also think that one may replace $X$ by any non-compact $k$-dimensional manifold.
With that, we have the tools to answer the question. First, since $X$ ha dimension $3$ and $Y$ has dimension $2$, we reduce the dimension by projecting. We get the plane $\mathbb{R}^2$. Then we wrap the plane around the origin in $\mathbb{R}^2$ ; and we are lucky, since this can be done very explicitly (thanks to the exponential map).
• Same as my comment here: math.stackexchange.com/questions/2565030/… – zhw. Dec 14 '17 at 23:40
• @zhw: yes, it's pretty close. I didn't notice that this question had a duplicate. – D. Thomine Dec 15 '17 at 13:49
• Right, not exactly the same, as I was addressing someone else's proof. Looks like you're headed to bounty nirvana in one of the stranger bounty situations I've seen. – zhw. Dec 15 '17 at 22:54 | 2019-07-22T18:26:04 | {
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http://mathhelpforum.com/pre-calculus/94707-inequalities-2-methods.html | 1. ## inequalities: 2 methods
Hi guys,
I am trying to find the values of x for which
$x - 3 < \frac{x - 4}{x}$
Attached is a diagram from which it is easy to see the answer x < 0! i.e these are the values of x for which the straight line x - 3 is below the curve (which I have expressed as:
$1 - \frac{4}{x}$
But, as we all know, inequalities can be solved in three ways: by sketching, by calculation and by completing the square. So I have tried it by calculation - here goes:
$x - 3 < (x - 4) / x$
$x^2 - 3x < x - 4$
$x^2 - 4x + 4 < 0$
$(x - 2) (x - 2) < 0$
$(x - 2)^2 < 0$
Now since (x - 2) is squared no matter what value x takes the result is always positive so it seems to me there is no solution here. Can anyone see what I am doing wrong!
2. Originally Posted by s_ingram
...So I have tried it by calculation - here goes:
$x - 3 < (x - 4) / x$
$x^2 - 3x < x - 4$
$x^2 - 4x + 4 < 0$
$(x - 2) (x - 2) < 0$
$(x - 2)^2 < 0$
... Can anyone see what I am doing wrong!
You assumed that $x \geq 0$. If you do the calculations for x < 0 you'll get the result which you already know.
3. Originally Posted by s_ingram
$x - 3 < (x - 4) / x$
$x^2 - 3x < x - 4$
$x^2 - 4x + 4 < 0$
$(x - 2) (x - 2) < 0$
$(x - 2)^2 < 0$
Now since (x - 2) is squared no matter what value x takes the result is always positive so it seems to me there is no solution here. Can anyone see what I am doing wrong!
I wouldn't multiply both sides by x. Instead, I would do this:
\begin{aligned}
x - 3 &< \frac{x - 4}{x} \\
x - 3 - \frac{x - 4}{x} &< 0 \\
\frac{x^2 - 3x}{x} - \frac{x - 4}{x} &< 0 \\
\frac{x^2 - 4x + 4}{x} &< 0 \\
\frac{(x - 2)^2}{x} &< 0
\end{aligned}
At this point I would make a sign chart. First, I need "critical points." These are values which (1) the fraction equals 0, and (2) the fraction is undefined. I find (1) by setting the numerator equal to zero:
\begin{aligned}
(x - 2)^2 &= 0 \\
x - 2 &= 0 \\
x &= 2
\end{aligned}
I find (2) by setting the denominator equal to zero: x = 0.
Then I draw my sign chart like this:
Code:
und 0
----------+---------------+-------------
0 2
(und = undefined)
Now take the fraction
$\frac{(x - 2)^2}{x}$ and test a value less than 0. You will see that the fraction is negative, so I make a notation on my sign chart.
Code:
(-)^2
-----
(-) und 0
----------+---------------+-------------
neg 0 2
Pick a number between 0 and 2 and test the fraction, and then pick a number greater than 2 and test the fraction. When you're done, you should get this:
Code:
(-)^2 (-)^2 (+)^2
----- ----- -----
(-) und (+) 0 (+)
----------+---------------+-------------
neg 0 pos 2 pos
We want a range of numbers where the fraction is less than 0, or negative. So the solutions are x < 0.
01
4. Sorry, where do I assume that $x \geq 0$? If x is less than zero say -5, then $(x - 2)^2$ gives $(-7)^2 = 49$ and this is not less than 0.
5. Thanks again yeongil. I can see your method works and I have used something like it when checking the ranges of functions for which they are real or unreal, but I don't see why you would not multiply by x. Is it wrong to do so and if so why?
6. To answer both of your posts, you assumed that $x \geq 0$ when you multiplied both sides by x. Remember that when you multiply both sides of an inequality by a negative number, you flip the sign. When you went from this step:
$x - 3\;{\color{red}<}\;\frac{x - 4}{x}$
to this step:
$x^2 - 3x\;{\color{red}<}\;x - 4$
you multiplied both sides by x, and since you didn't change the sign, you assumed that $x \geq 0$.
The problem was to solve for x, and if we pretend that we do not know what x is beforehand, we shouldn't multiply both sides by x; instead, subtract the fraction from both sides and proceed like I did earlier.
01
7. Excellent answer! Now I see. I suspected Earboth was on to something but I didn't understand what he meant. | 2017-04-24T23:16:21 | {
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https://math.stackexchange.com/questions/4603/series-converges-implies-limn-a-n-0/1122679 | # Series converges implies $\lim{n a_n} = 0$
I'm studying for qualifying exams and ran into this problem.
Show that if $\{a_n\}$ is a nonincreasing sequence of positive real numbers such that $\sum_n a_n$ converges, then $\lim_{n \rightarrow \infty} n a_n = 0$.
Using the definition of the limit, this is equivalent to showing
$$\forall \varepsilon > 0 \; \exists n_0 \text{ such that } |n a_n| < \varepsilon \; \forall n > n_0$$
or
$$\forall \varepsilon > 0 \; \exists n_0 \text{ such that } a_n < \frac{\varepsilon}{n} \; \forall n > n_0$$
Basically, the terms must be bounded by the harmonic series. Thanks, I'm really stuck on this seemingly simple problem!
• This is a classic problem. I remember it from teaching undergrad real analysis in 2005 (specifically I remember having a good think about it at the beautiful YMCA in downtown Montreal -- it took me a while to get it). I think it will be helpful to see a variety of answers. – Pete L. Clark Sep 14 '10 at 8:18
• This is indeed a nice problem, and I don't want to give it away. The hint I would give is to remember to use the condition that $a_n$ is nonincreasing. – David E Speyer Sep 14 '10 at 17:21
• Maybe this is interesting: If we omit monotonicity then $na_n$ converges to 0 statistically. (It is relatively easy to find examples showing that it need not converge in the usual sense.) Tibor Šalát; Vladimír Toma: A Classical Olivier’s Theorem and Statistical Convergence. dx.doi.org/10.5802/ambp.179 – Martin Sleziak Nov 23 '11 at 16:13
• Maybe it's too late for your exam, but I've added another answer that is fairly short and is not very notation-intensive and does not rely on standard results but works from basic definitions. (I don't even use the fact that the harmonic series diverges.) I like to keep things simple. $\qquad$ – Michael Hardy Jul 21 '16 at 2:02
• Can you tell me where did you see this problem? I'm studying for an exam and It would really help some extra problems. This seems a very good one – J.Dane Jul 27 '18 at 10:39
Some hints:
If $S_{n} = \sum_{k=1}^{n} a_{k}$
then what is
$\lim_{n \to \infty} S_{2n} - S_{n}$?
Now can you use the fact that $a_{n}$ is non-increasing to upper bound a certain term of the sequence $na_{n}$ with a multiple of $S_{2n} - S_{n}$?
• Let me give this a shot: Assuming the limit you mention tends to zero, for a given $\epsilon$, there is an integer $n_0$ such that $2 ( S_{2n_0} - S_{n_0} ) < \epsilon$. Because the sequence is nonincreasing, we have that $2n a_{2n} \leq 2 ( S_{2n_0} - S_{n_0} ) < \epsilon$ for all $n \geq n_0$. – dls Sep 15 '10 at 4:13
• @dls: Right. It is still incomplete though. You have to consider (2n+1)a_{2n+1} too. – Aryabhata Sep 15 '10 at 5:02
By the Cauchy Condensation test, $\displaystyle \sum 2^n a_{2^n}$ converges since $\displaystyle \sum a_n$ converges, and thus $2^n a_{2^n} \to 0.$ Now for $2^n < k < 2^{n+1}$,
$$2^n a_{2^{n+1}} \leq k a_{k} \leq 2^{n+1} a_{2^n}$$
so $n a_n \to 0.$
Now that enough time has passed so that more information will not spoil anything for the OP:
This fact can be found in $\S 179$ of G.H. Hardy's seminal A Course of Pure Mathematics: he mentions that it was first proved by Abel, then forgotten and later rediscovered by Alfred Pringsheim. I have reproduced Hardy's proof in $\S 2.4.2$ of these notes on infinite series. This is much slicker than what I came up with when I had to solve this exercise myself some years ago. On the other hand it seems to be exactly what Aryabhata's answer hints at.
In my notes I also attribute this result to L. Olivier and even cite the issue of Crelle's Journal in which it appears in 1827. This attribution does not appear in Hardy's book, which temporarily mystified me (I am no historian of mathematics: whatever such information I have comes from math books with good bibliographies), but I surmise I must have gotten it from Konrad Knopp's book on infinite series (the only other book I own which treats the subject seriously).
P.S.: The wikipedia article on Pringsheim is unusually (almost suspiciously?) good. The impression that I have of him as a mathematician is someone who worked on infinite series at a stage when the foundations of the theory were finally solidly in place...and when the best mathematicians of the day had gone on to more fundamental and difficult problems. But I don't know whether this is at all fair. Anyway, it seems that you won't hear of him until you learn a little more about series than is treated in the standard contemporary curriculum, but as soon as you do his name comes up again and again.
• Re: P.S.: The English Wikipedia page is almost a verbatim copy of the German one. The Pringsheim family is very well known in Germany due to its influence in Munich over centuries, and also due to fact that Alfred P.'s daughter was married to Thomas Mann who certainly contributed to making the family's fate in the first half of the twentieth century known to a wide audience. – t.b. Nov 23 '11 at 7:23
• @t.b.: Thanks for this. I vaguely noticed the connection to Thomas Mann in the wikipedia article, but it didn't really sink in. – Pete L. Clark Nov 23 '11 at 16:04
• Interestingly, it seems that Olivier's paper contained a mistake and later Abel wrote a paper to correct it. Details can be found e.g. in Michael Goar: Olivier and Abel on Series Convergence, Mathematics Magazine Vol. 72, No. 5 (Dec., 1999), pp. 347-355 jstor. To add also a freely available paper, it is briefly mentioned here. (The later is the paper where I've read about Olivier's mistake for the first time.) – Martin Sleziak Aug 15 '12 at 11:07
I think I have an answer which doesn't rely on being clever enough to use the even and odd subsequences.
Because the series converges the sequence of partial sums forms a Cauchy sequence so we have for m,n>N
$$\| \sum_m^n(a_i) \| < \epsilon$$
And by a similar argument to those above: $\|(n-m)a_n \| \leq \| \sum_m^n(a_i) \| < \epsilon$ Distributing out the left hand side and spliting the inequality gives us the chain:
$$\| na_n \|-\|ma_n\|\leq \|(n-m)a_n \| \leq \| \sum_m^n(a_i) \| < \epsilon$$
Then in the limit as n goes to infinity and fixing m we have the desired result since $ma_n$ will go to zero.
I think... ?
Try to stay away from quantifier-laden formulas, which make the problem harder to understand, and draw a picture. There is an equivalent problem for decreasing functions (as in the Integral Test for convergence) and the picture makes it obvious what is true in that case. Having seen the continuous proof, run the same argument for the sequence, or specialize the function to a sequence by using step functions or approximation thereof. I won't spoil the "aha!" proof-by-picture experience by posting more details, but it is quite easy once you draw the graph.
Note that the desired conclusion is not true if the $a_n$ are not assumed to be nonincreasing:
We know a couple of facts:
1. If a sequence $(s_n)_{n\geq1}$ converges then we have that
$\displaystyle\qquad\liminf_{n\to\infty} s_n = \lim_{n\to\infty} s_n = \limsup_{n\to\infty} s_n$.
2. There is a summable series $\sum_{n=1}^\infty a_n$ with non-negative $a_n$ such that $\limsup\limits_{n\to\infty} na_n > 0$.
Take these two together you get that the best you can hope to prove in this slightly more general situation is that $\liminf\limits_{n\to\infty} na_n = 0$, and that is indeed true.
For an example showing that 2 above is true, consider this: We start with the harmonic series and observe that
$\displaystyle\qquad \limsup_{n\to\infty} n\cdot \tfrac1n = 1$,
but it isn't summable. But we could replace infinitely many of the sequence elements by zeroes without changing the limit superior from being $1$. Define $S(n)$ to be $1$ when $n$ is a perfect square and $0$ otherwise. Then $a_n = \frac{S(n)}n$ has the desired property as
$\displaystyle\qquad \sum_{n=1}^\infty a_n = \sum_{n=1}^\infty \frac{S(n)}n = \sum_{k=1}^\infty \frac1{k^2} < \infty$.
• Note that $a_n$ is supposed to be nonincreasing. Your sequence isn't. – mrf Sep 12 '13 at 20:01
• Ah. Miss one word and the whole thing goes down the crapper. Typical. Well now it's fixed at least. And CW just for good measure. – kahen Sep 12 '13 at 20:02
Let $\epsilon > 0$.
Let Let $s_n=\sum_{k=1}^n a_k$.
$\{s_n\}$ converges, therefore $\{s_n\}$ is Cauchy which implies $\exists$ some $N$ such that $m, n \geq N$ implies $\left | s_n-s_m \right | < \epsilon$
Consider $m=N$, for $n \geq N$ we have $\left | s_n-s_N \right | < \epsilon$,
$\Rightarrow \left | a_N + a_{N-1} + a_{N-2}+\cdots +a_n \right | < \epsilon$,
$\Rightarrow \left | a_n + a_{n} + a_{n}+\cdots +a_n \right | \leq \left | a_N + a_{N-1} + a_{N-2}+\cdots +a_n \right | < \epsilon$ $\Rightarrow \left (n-N)| a_n\right | \leq \left | a_N + a_{N-1} + a_{N-2}+\cdots +a_n \right | < \epsilon$
$\Rightarrow (n-N)\left| a_n\right | < \epsilon$
So, $\lim_{n \rightarrow \infty } (n-N)a_n = 0$
$\lim_{n \rightarrow \infty } na_n - \lim_{n \rightarrow \infty } Na_n = 0$
From above, since {$a_n$}$\rightarrow 0$, {$na_n$}$\rightarrow 0$ $\blacksquare$
• Shouldn't it be $|a_N + a_{N+1} + \ldots + a_n|$ since $n \ge N$? – philmcole Jan 13 '18 at 10:07
• Very clear and nicely explained. Thanks – StammeringMathematician Jul 28 at 18:31
Here's one more approach : By Cauchy's general principle of convergence :
$\forall \epsilon>0, \exists ~p \in \mathbb N , \exists ~m> N: |a_{m+1}+a_{m+2}+ \cdots + a_{m+p}| < \epsilon$.
Choose:$p = n-m$ .
$|a_{m+1}+a_{m+2}+ \cdots + a_n| < \epsilon$.
Now, since $a_n$ is non increasing and $\epsilon \rightarrow 0$: the above relation can be written as :
$(n-m)a_n = 0 \implies na_n = ma_n$
For a convergent series. $\lim a_n=0 \implies \lim_{n \rightarrow \infty} n a_n = 0$
• You seem to claim that directly from $(n-m)a_n<\varepsilon$ you get $(n-m)a_n=0$. This argument is not correct, since $m$ depends on $\varepsilon$. (But you still get $(n-m)a_n<\varepsilon$. For a fixed $\varepsilon>0$ and for $m$ corresponding to this $\varepsilon$.) – Martin Sleziak Apr 4 '16 at 4:27
You might also do this the other way around. What if $$\lim_{n\to\infty}na_n \not=0,$$ that is if the limit does not exist or if it is positive, what does this tell you about $a_n$? What about sub-sequences of $(a_n)$?
• I can't see how this works. The easy part - if the limit exists and is positive, then $\sum a_n$ diverges like a harmonic series, contradiction. The harder part, if the limit does not exist - There there exists a subsequence and an $\epsilon > 0$ such that $n_k a_{n_k} > \epsilon$ for all $n_k$. But that itself does not imply the result. Did you have something else in mind? – Ragib Zaman Oct 11 '11 at 15:03
If the sequence is strictly decreasing, I guess there is a neat little argument. Look at the series $\sum n a_n$ and define $\rho_n=\frac{(n+1)a_{n+1}}{na_n}$. Then $\rho_n=\frac{a_{n+1}}{a_n} \frac{(n+1)}{n}$. We have that $\lim_{n \to \infty} \frac{(n+1)}{n}$ exists and is equal to $1$. As $(a_n)$ is strictly decreasing and the terms are positive, we have $\frac{a_{n+1}}{a_n} < 1$ Therefore the limit $\lim_{n \to \infty} \frac{a_{n+1}}{a_n}$ exists and is smaller than $1$. So $\lim_{n \to \infty} \rho_n$ exists and is smaller than 1, and thus by the ratio test the series $\sum n a_n$ converges. Hence $n a_n \to 0$.
• Consider $a_n = \frac{1}{n^2}$. Then the limit of ratios is $1$. – Daniel Fischer Apr 27 '14 at 21:59
• Oh my, you are right. Nevermind then. – José Siqueira Apr 27 '14 at 22:06
• Still, looking at a series is a nice trick when you want to prove a limit is 0. – José Siqueira Apr 27 '14 at 22:10
If $\lim\limits_{n\to\infty} n a_n\ne 0$, then for some $\varepsilon>0$, there are infinitely many $n$ for which $a_n \ge \dfrac \varepsilon n.$
If $a_n \ge \dfrac \varepsilon n,$ then let $m = \lfloor \log_2 n \rfloor$, so that $2^m\le n < 2^{m+1}$. Then $$a_{2^{m-1}} + \cdots + a_{2^m} \ge 2^{m-1} a_{2^m} \ge 2^{m-1} a_n \ge \frac{2^{m-1}\varepsilon} n \ge \frac{2^{m-1}\varepsilon} {2^{m+1}} = \frac \varepsilon 4.$$
Then go on to the next $n$ for which $a_n \ge \dfrac \varepsilon n$, and if it's not big enough for the resulting sequence $a_{2^{m-1}} + \cdots + a_{2^m}$ to have a different (larger) value of $m$ than the one we just saw, then go on to the next one after that, etc.
We get infinitely many disjoint sets of terms with sums exceeding $\dfrac \varepsilon 4$.
Let $\epsilon$ be given. By Cauchy, there exists $N(\epsilon)$ in $\mathbb{N}$ such that: $a_N+a_{N+1}+....+a_{N+P}<\epsilon$, for every $P$ in $\mathbb{N}$. Note that for every chosen $P$, there are $P$ $a_n$'s on the left side of the inequality, not all of which can exceed $\dfrac{\epsilon}{P}$. Anyway, since $a_n$ is non increasing, $a_{N+P}$ is always less than $\dfrac{\epsilon}{P}$. We have then:
$$a_{N+P}<\dfrac{\epsilon}{P}, \forall P \in \mathbb{N}$$ Multiplying by $N+P$ at each side of the inequality: $$(N+P)a_{N+P}<\dfrac{(N+P)\epsilon}{P}$$
"Throwing" $p$ to ininity, we can see then that for $n$ large enough, $na_n$ is almost less than $\epsilon$.
Taking $\dfrac{\epsilon}{2}$ instead of $\epsilon$, we can guarantee that $na_n$ is less than $\epsilon$ for $n$ large enough, completing the proof.
Suppose $n_ja_{n_j}\geq \alpha>0$ and up to (infinity) subsequence assume $n_{j}-n_{j-1}\geq n_{j}/2$.
Using that $a_n\geq 0$ we get
$$\displaystyle A=\sum_{n=1}^\infty a_n =\sum_{n=1}^{n_0} a_n+\sum_{j=1}^\infty\sum_{n=n_{j-1}+1}^{n_j}a_n\geq B+\sum_{j=1}^\infty\sum_{n=n_{j-1}+1}^{n_j}a_{n_j}\geq B+\sum_{j=1}^\infty(n_{j}-n_{j-1})a_{n_j} \\\geq B+\sum_{j=1}^\infty\frac{n_{j}}{2}a_{n_j}\geq B+\frac{1}{2}\sum_{j=1}^\infty{n_{j}}{}a_{n_j}=\infty. \ QED.$$
Let $$f: (0, \infty)\to [0, \infty)$$ be a continuous, non-increasing function such that $$f(n)=a_n$$ for all $$n \in \mathbb{N}.$$ Such a function certainly exists (for $$x \in (n, n+1)$$ let $$f(x) = a_n + (a_{n+1}-a_n)(x-n)$$)
Since $$\displaystyle \sum_{n=1}^{\infty}f(n) = \displaystyle \sum_{n=1}^{\infty} a_n$$ converges, by the integral test, $$\displaystyle \int_{0}^{\infty}f(x)dx < \infty.$$
Claim: $$\displaystyle\lim_{x\to \infty}xf(x)=0.$$
We reproduce the argument that JLA gave here, begin by noting that $$\displaystyle \lim_{t \to \infty}\int_{t/2}^{t}f(x)dx=0.$$ Then $$0 \leq \dfrac{t f(t)}{2}\leq \displaystyle \int_{t/2}^{t}f(x)dx \to 0$$ as $$t \to \infty$$ and the claim holds.
In particular, $$\displaystyle \lim_{n\to \infty}nf(a_n) =0$$ and we are done. | 2019-10-21T07:12:13 | {
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https://math.stackexchange.com/questions/1460661/if-f-is-continuously-differentiable-in-a-b-fa-fb-and-fa-fb | # If $f$ is continuously differentiable in $[a,b]$, $f(a)=f(b)$, and $f'(a)=f'(b)$, then there exist $a<x_1<x_2<b$ such that $f'(x_1) = f'(x_2)$.
This problem is from Apostol's Mathematical Analysis:
Let $f$ be continuously differentiable in $[a, b]$. If $f(a) = f(b)$ and if $f^{'}(a) = f^{'}(b)$, then prove that there exists $x_1$ and $x_2$ in $(a, b)$ such that $x_1\neq x_2$ but $f^{'}(x_1) = f^{'}(x_2)$.
My try:
By Rolle's Theorem $\exists x_0\in (a,b)$ such that $f^{'}(x_0)=0$. How to guarantee existence of $x_1,x_2$ from here?
Can it be solved from a geometrical point of view?
• you don't know that $f'(a)=f'(b)=0$ – tattwamasi amrutam Oct 2 '15 at 8:53
• Geometrical point of view. At point a the function is going at a rate of f'. if that's positive the function is going up. At point b, f(a) = f(b) so the function had to have come down again. That means that somewhere f' was neg. But the means somewhere f' had to be zero. But right now at b f' is positive again. But that means it had to go from negative to positive. That means there was another point in between where f' was zero. Recap: function going up at a, peaks, f' = 0, goes down, plateaus, f' = 0 goes up again, reaches b. It's the only possibility. – fleablood Oct 2 '15 at 9:05
• It will be nice to mention why you downvote and let people think and learn from your comments on their wrong answers. – Math-fun Oct 2 '15 at 9:41
• thanks for your useful comments@fleablood – Learnmore Oct 3 '15 at 3:30
As you said, by Rolle's Theorem, we are guaranteed existence of $c \in (a,b)$ such that $f'(c) = 0$. WLOG, let $f'(a) = f'(b) = k > 0$. By the continuity of $f'$ and the intermediate value theorem, we get existence of $c_1 \in (a,c)$ such that $f'(c_1) = k_1$ where $0<k_1<k$. Similarly, we get $c_2 \in (c,b)$ such that $f'(c_2) = k_1$.
Now, if $f'(a) = 0$, then let's look at some cases. If $f(c) \neq f(a)$, then the mean value theorem gives existence of $d \in (a,c)$ such that $\alpha = f'(d) = \frac{f(c) - f(a)}{c - a}\neq 0.$ Now, we can apply the above argument again.
If $f(c) = f(a) = f(b)$, then if $f'$ is nonzero at some point in the interval, the above argument works. Otherwise, $f'$ is $0$ and the result follows.
(i) Let's suppose that $f'(a)>0$ (then we'll see the case $f'(a)=0$). Because of (i), you can find a $\eta > 0$ (as small as it is) that we have $f(a + \eta) > f(a)$ and $f(b - \eta) < f(b)$.
$f$ is continuous on $[a + \eta, b-\eta]$, therefore there exists $c \in [a + \eta; b - \eta]$ so that $f(c) = f(a) = f(b)$. ($f(c)$ in $[f(a+\eta], f(b-\eta)])$.
$f$ is continuous on both intervals $[a, c]$ and $[c, b]$ and reach its maximums in a $x_1$ and $x_2$ ($x_1$ in $(a, c)$ and $x_2$ in $(c, b)$ because $a, c$ and $b$ are not the maximum since $a+\eta$ and $b-\eta$).
We have $f'(x_1) = f'(x_2) = 0$
Opposite reasonning for $f'(a)<0$ leading to the same conclusion
(ii) if $f'(a) = 0$ because of Rolle's theorem you know that you have a $c$ in $(a,b)$ that $f'(c) = 0$. Assume that $f(c) = f(a)$ (then we reapply this reasonning on $a$ and $c$ instead of $a$ and $b$ until finding a $c_n$ that $f(c_n)$ is different from $f(a)$, if we can not find one then $f$ is constant and any $c_1<c_2$ in $(a, b)$ will do)
We have a $f(a) < f(c)$ and $f'$ continuous on $[a, c]$ and $f'(a) = f'(c) = 0$ but $f'$ not equal to $0$ and it exists a $c_1$ in $(a, c)$ so that $f'(c_1)>0$ because $f(a) < f(c)$.
Because of that and $f'$ continuous on this interval, $f'$ reaches its maximum in $d$ in $(a;c)$ and because of its continuity $f'$ reaches $f'(d)/2$ in two points $x_1,x_2$, where $a<x_1<d<x_2<c<b$.
• i dont understand the down votes, because the arguments seems fine to me. – Lost1 Oct 3 '15 at 1:57
• The format was messy. @Lost1 – user99914 Oct 3 '15 at 3:48 | 2020-10-24T08:32:56 | {
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https://stats.stackexchange.com/questions/396522/moving-average-process-stationarity | # Moving average process - stationarity
If we consider a moving average process of order 1, is that stationary?
Because, although, the mean will remain the same for Yt and Yt+k, the variance and co-variance will change if you calculate for Yt+k, so in that case, why is it stated as stationary?
Can anyone explain: how do you know if a moving average process is weakly stationary, strictly stationary or non stationary?
A stochastic process is weakly stationary if its mean and variance and covariances are finite and do not depend on time.
A moving average process of order q (with q finite), $$MA(q)$$ is defined as:
$$y_t = \epsilon_t + \alpha_1 \epsilon_{t-1} + \alpha_2 \epsilon_{t-2} + \dots + \alpha_q \epsilon_{t-q}, \quad \epsilon_t\sim WN(0,\sigma^2),$$
and is weakly stationary for all the values of $$\alpha_j$$.
So, in your example it is true that the variance and covariances change, but you can check that both are finite and do not depend on $$t$$ (the covariance will depend only the distance $$k$$).
Moreover, even a moving average process of infinite order $$MA(\infty)$$:
$$y_t = \sum_{j=0}^\infty \alpha_j \epsilon_{t-j}$$
can be stationary, but we need an additional condition in order to have a finite variance:
$$\sum_{j=0}^\infty|\alpha_j|<\infty.$$
• Thank you. Is every moving average process weakly stationary? – user218970 Mar 10 at 10:10
• Yes, and I edited the answer with some more details. – pdb Mar 10 at 10:52
• thank you so much for your additional explanation. But the previous answer stated that MA(1) process is strictly stationary. – user218970 Mar 10 at 12:55
• If you assume that the elements of the white noise process $\epsilon_t$ are also indipendent (and not only uncorrelated) and identically distributed, then the moving average process is strictly stationary. – pdb Mar 10 at 14:02
First see what definitions say
• $$\{X_t\}$$ is strictly stationary if for any $$t_1,t_2,...,t_n \in T$$ and any $$k \in T$$
$$P(X_{t_1},...,X_{t_n}) = P(X_{t_1+k},...,X_{t_n+k})$$ that is, we have statistical equilibrium.
• $$\{X_t\}$$ is second order/weakly stationary if
1. $$\mathbb{E}[X_t] = \mu < \infty$$, independent of $$t$$
2. $$Var(X_t) = \sigma^2 <\infty$$ , independent of $$t$$
3. $$cov(X_t,X_{t+\tau})$$, is a function of $$\tau$$ only.
Think of what weakly stationarity means. It means that the expected value of the process is finite and constant. It also means that the autocovariance does not depend on where two random variables are positioned but just on their distance! Autocovariance between today and yesterday same as autocovariance between 100 days and 101 days ago.
Now take the MA(1) process $$\boxed{ X_t = \varepsilon_t + \theta \varepsilon_{t-1} }$$ where $$\varepsilon_t \sim WN(0,\sigma^2)$$.
This process is weakly stationary as
• $$\mathbb{E}[X_t] = \mathbb{E}[\varepsilon_{t}] + \theta \mathbb{E} [\varepsilon_{t-1}] = 0$$
• $$Var(X_t) = \sigma^2(1+\theta^2)$$, i.e. independent of $$t$$
• $$cov(X_t,X_{t-\tau}) = \theta \sigma^2, |\tau|=1$$, $$cov(X_t,X_{t-\tau}) = 0, \forall |\tau|>1$$ hence function of $$\tau$$ only.
(if you have difficulties deriviting that, let me know. It is very straightforward and you shouldn't have any difficulties)
MA(1) is also strictly stationary as both $$P(X_{t_1},...,X_{t_n})$$ and $$P(X_{t_1+k},...,X_{t_n+k})$$ multivariate (1-dependent) Normal distributions with identical parameters as it is a combination of WN random variables.
Edit to provide further explanation: Each $$X_{t_1},..X_{t_n}$$ is a linear combination of independent Gaussian random variables,. In particular all $$X_{t_i} \sim N(0, \sum_{j=1}^{q} \theta_j^2 \sigma^2) \forall j$$. The joint distribution of normal random variables is a multivariate normal. As they are all driven by an MA(q) process their covariance matrix can be easily derived. But as seen above, this does not depend on location, just on distance. Explicitly for MA(1):
$$f_{\mathbf {X} }(X_{t_1} \dots X_{t_n})={\frac {\exp \left(-{\frac {1}{2}}{\mathbf {X} }^{\mathrm {T} }{\boldsymbol {\Sigma }}^{-1}{\mathbf {X} }\right)}{\sqrt {(2\pi )^{k}|{\boldsymbol {\Sigma }}|}}}$$
where
$$\boldsymbol {\Sigma } = \begin{bmatrix} \sigma^2(1+\theta^2) & \theta \sigma^2 & 0 & 0 & \dots & 0 \\ \theta \sigma^2 & \sigma^2(1+\theta^2) & \theta \sigma^2 & 0 & \dots & 0 \\ 0 & \theta \sigma^2 & \sigma^2(1+\theta^2) & \theta \sigma^2 & \ddots & \vdots \\ 0 & 0 & \ddots & \ddots & \ddots & 0 \\ \vdots & \vdots & \ddots & \ddots & \ddots & \theta \sigma^2 \\ 0 & 0 & \dots & 0 & \theta \sigma^2 & \sigma^2(1+\theta^2) \end{bmatrix}$$
Hence MA(q) processes driven by Gaussian noise are always strictly stationary (only condition is that the sum of MA coefficients should be finite)
In general, all weakly stationary Gaussian processes are strictly stationary too.
• Thank you so much, yes I can derive that - no problem. Although, in the end, I did not understand the 'strictly stationary' part. I have read that MA(2) process is weakly stationary. How do their results differ? Could you elaborate, please? – user218970 Mar 10 at 10:08
• MA(q) is also strictly stationary. See addition – Stats Mar 10 at 14:25
• I'm a beginner in time series and therefore most of the matrices and notation is lost on me. I'm sorry for that. Can you explain me in simple terms that how should I go about figuring out the differences between strictly and weakly stationary from my derivation answers? – user218970 Mar 10 at 17:56
• If you want to understand what strict stationarity means, it is highly essential you understand the basic multivariate distributions like the multivariate normal distribution. Strict stationarity has to do with the joint distribution of random variables, you will not be able to understand anything without first understanding what a joint distribution is. Could you please do this first? – Stats Mar 10 at 18:24 | 2019-08-22T14:20:06 | {
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http://math.stackexchange.com/questions/68045/rolling-dice-such-that-they-add-up-to-13-is-there-a-more-elegant-way-to-solve/68050 | # Rolling dice such that they add up to 13 — is there a more elegant way to solve this type of problem?
Here is the problem:
There are $6^3$ possible outcomes to rolling a die $3$ times. Out of these, how many yield a total of (exactly) $13$ dots?
My solution would be absolutely impractical for problems involving $4$ rolls of the die. And at higher numbers it would be downright impossible without brute-forcing it with a computer.
Is there a more elegant way to solve this type of problem?
My solution:
First we find all sets $\{a, b, c\}$ such that $a + b + c = 13; \; a, b, c \le 6$:
$\{1, 6, 6\}$
$\{2, 5, 6\}$
$\{3, 4, 6\}, \{3, 5, 5\}$
$\{4, 4, 5\}$
The total number of sets that fit these criteria is $5$. If $a \not= b \not= c$, then there exist $3!$ unique permutations of $\{a, b, c\}$. If $a = b \not= c$, then there exist $3$ unique permutations of $\{a, b, c\}$. -- There cannot be a set such that $a = b = c$.
There are $2$ sets of the first kind and $3$ of the second. It follow that the total number of triple die rolls that can fit the criteria is
\begin{equation*} 2 \cdot 3! + 3 \cdot 3 = 21 \end{equation*}
I can think of a second way to do it, which might be faster for slightly larger numbers... but essentially still comes down to brute force, not a method that can be generalized.
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If you're going for elegant, I think the method from Fool's answer below is nice. Basically the computation is reduced to $\binom{12}{2} - 3 \cdot \binom{6}{2}$, with no brute-forcing, and no counting possibilities manually. Just some algebra. – TMM Sep 28 '11 at 1:22
Could whoever downvoted this explain why? And also explain why they didn't post such an explanation here? – Michael Hardy Sep 28 '11 at 11:06
## 5 Answers
Here's one approach which is suitable in various circumstances.
Consider the expression $(x+x^2+x^3+x^4+x^5+x^6)^3$. What is the coefficient of $x^{13}$ when you expand this out?
The expression can also be re-written as $\left(\frac{x(x^6-1)}{x-1}\right)^3$ to make it less unwieldy to work with. The coefficient of $x^k$ can be extracted by taking derivatives, for example.
This is a very useful tool which can be applied in much more general situations, as you may observe. For instance you could replace $1+x+\ldots+x^6$ by some other expression to represent dice with more faces, or dice with arbitrary faces, or dice with weights, etc.
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Interesting! thanks. – iDontKnowBetter Sep 27 '11 at 22:36
Including $1$ in the expression is equivalent to having seven sided dice with numbers $0$ through $6$. It doesn't hurt with $13$, but would with $11$, because you could afford a $0$. – Ross Millikan Sep 27 '11 at 22:37
You're welcome. Note that I corrected my expression slightly to account for the fact that a die can't come up 0 - it doesn't change the answer in this case since 13 cannot be attained with 0's. – Alon Amit Sep 27 '11 at 22:39
In this case it's easier to find the number of triples $(a,b,c)$ that sum up to 8. If $(a,b,c)$ are possible dice rolls that sum up to 8, then $(7-a,7-b,7-c)$ are also possible dice rolls and sum to 13. In general the number of ways to make $x$ and $21-x$ on three dice are the same. (With two dice, of course, the number of ways to make $x$ and $14-x$ are the same.)
Of course this is still brute force, but it's good to take advantage of symmetry -- for example, if you had been asked to find the number of outcomes giving $k$ dots for all of $k = 3, 4, \ldots, 18$, you could save yourself time by doing only half of them.
One way to do this for an arbitrary number of dice is as follows: let $N(d,s)$ be the number of ways to roll a sum of $s$ with $d$ dice. (So you're trying to find $N(3,13)$.) In order to get a sum of $s$ with $d$ dice, you must have a sum of between $s-6$ and $s-1$ on the first $d-1$ dice, and then roll the appropriate number on the final die. So you have $$N(d,s) = N(d-1, s-1) + N(d-1, s-2) + \cdots + N(d-1, s-6)$$ where we start with $N(0,0) = 1$ and $N(0,s) = 0$ for all other integers $s$. You can rearrange this to get $$N(d,s) = N(d, s-1) + N(d-1, s-1) - N(d-1, s-7)$$ and so you can compute each $N(d,s)$ with just one addition and one subtraction.
The generating-function method is actually the same as this, once you think about how polynomial multiplication works.
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+1 for the method in your third paragraph, which is very easy to implement on a spreadsheet, especially if you have six blank rows at the top of your table. – Henry Sep 27 '11 at 23:37
If I have understood your problem correctly then I guess we could use the stars and bars concept for a clever solution.
For example in your case,
$$x_1+x_2+x_3 = 13$$
Number of positive integer solution possible such that $x_i \ge 1$ is given by $\binom{12}{2}=66$.
Then set $x_1'=x_1+6$,so the equation reduces to: $$x_1'+x_2+x_3=7$$
which gives $15$ possible solution,using the same thing for $x_2$ and $x_3$ we would see that we have over-counted $3\times15$ solutions in our first calculation,hence subtracting $66-45=21$.
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(+1) I was going to post a link to your earlier question but I see you already posted an answer using that method :) – TMM Sep 28 '11 at 0:04
One way is generating functions: form $(x+x^2+x^3+x^4+x^5+x^6)^3$ and find the coefficient of $x^{13}$
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Ah, the very last chapter in my combinatorics book is titled "generating functions", though I've not gotten there yet. Thanks. (This is from my current probability course, which starts with basic combinatorics). – iDontKnowBetter Sep 27 '11 at 22:39
If you want to learn more about generating functions, you should take a look at Herb Wilf's book "generatingfunctionology", which can be found free online. (This is legal; Wilf has posted the book on his web page.) This book is probably best read after you know some combinatorics and some probability. – Michael Lugo Sep 28 '11 at 1:03
If you throw the die a very large number of times, maybe there would be some difficulty in finding the exact number of ways in which a certain total can appear. But the central limit theorem from probability theory can give some reasonable approximations.
Say you throw the die 40 times. On average the sum is 140. There are $6^{40}$ possible outcomes. How many of those give a total in the closed interval $[135,150]$?
When you throw a die once, the variance of the probability distribution of the number of dots you see is $35/12$, so when you throw it 40 times, the variance is $40\cdot35/12= 10\cdot35/3$, and the standard deviation is therefore $\sqrt{350/3} = 10.801234497\ldots\ {}$. For this discrete distribution, "$\ge 135$" is the same as "$>134$" and "$\le 150$" is the same as "$<151$", so we do a continuity correction and look at the interval $(134.5,150.5)$. The number $134.5$ is $(134.5-140)/10.801234497\ldots)$ standard deviations above the mean, i.e. about $0.5092$ standard deviations below the mean, and $150.5$ is about $0.972$ standard deviations above the mean. Plugging these into the cumulative probability distribution function $\Phi$ of the normal distribution, we get a probability of about $\Phi(0.972)-\Phi(-0.5092)=0.529$.
So $0.529 \times 6^{40} \approx 7.074\times 10^{30}$ is approximately how many ways there are to get a sum in that interval.
-
Could whoever downvoted this explain why? And also explain why they didn't post such an explanation here? – Michael Hardy Sep 28 '11 at 5:40
Don't sweat it. Random downvotes are more than balanced by random upvotes. – TonyK Sep 28 '11 at 9:52
The concern is of course that it may not have been "random". Indeed, that is what one must presume. – Michael Hardy Sep 28 '11 at 11:05
Yeah but still. – TonyK Sep 28 '11 at 11:51
To suggest that "random downvotes are more than balanced by random upvotes" is to suggest that the concern here is about the number of reputation points. That's not what it is. Apply a bit of common sense here. – Michael Hardy Sep 28 '11 at 15:02 | 2015-02-01T17:21:41 | {
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https://math.stackexchange.com/questions/3215729/in-the-figure-a-quarter-circle-a-semicircle-and-a-circle-are-mutually-tangent | # In the figure, a quarter circle, a semicircle and a circle are mutually tangent inside a square of side length $2$. Find the radius of the circle.
In the figure, a quarter circle, a semicircle and a circle are mutually tangent inside a square of side length $$2$$. Find the radius of the circle.
I first assumed that when a vertical line is drawn from the radius of the semicircle, that line would be tangent to the smallest circle and it would mean that the radius is $$\frac{1}{4}$$, but the correct answer was $$\frac{2}{9}$$. I also tried using coordinate geometry, but I got stuck because I did not know how to get the equation of the smallest circle.
• $2/9$ seems to be very much too large simply from eyeballing the figure. Should it be the diameter of the circle rather than its radius? – hmakholm left over Monica May 6 '19 at 11:41
• @HenningMakholm: I've confirmed that $2/9$ is correct. I submit that it also looks plausible. The diameter of the circle looks to be slightly less than the radius of the semicircle, and the diameter of the semicircle looks to be about half the side of the square (in fact, it's exactly half). So, the radius of the circle should be just shy of $1/4=0.25$; and $2/9=0.222\ldots$ is in that ballpark. – Blue May 6 '19 at 12:00
• @Blue: Ah, sorry, I missed that the outer square has side length $2$ rather than $1$. – hmakholm left over Monica May 6 '19 at 13:05
• Sounds like the start of a joke... – Asaf Karagila May 7 '19 at 12:43
Look at the picture:
From $$\triangle ABE$$ we have $$(2+r)^2= 2^2+(2-r)^2$$ so $$r=1/2$$. From $$\square ECGF$$ we have $$CG^2=(1/2+s)^2-(1/2-s)^2= 2s$$. From $$\square ADGF$$ we have $$GD^2= (2+s)^2-(2-s)^2= 8s$$. So $$2=CG+GD=3\sqrt 2\sqrt s$$, hence $$s=2/9$$.
• It's surprising to me that the length of the square's sides are an integer multiple of the circle's radius. – BlueRaja - Danny Pflughoeft May 6 '19 at 16:10
• Could you explain the calculation of $CG^2$ and $GD^2$? What principle are you invoking here? You appear to have applied some formula (perhaps some standard formula that applies to trapezia), but it is unknown to me. – Hammerite May 6 '19 at 16:26
• @Hammerite If $H$ is the foot of the perpendicular from $F$ to $EC$, then $CG=FH$, and then apply Pythagorean theorem on $\triangle FHE$ to find $FH$. And the same trick for $GD$. – SMM May 6 '19 at 16:37
• @BlueRaja-DannyPflughoeft Yes, in general $r=a/4$ and $s=a/9$, where $a$ is the side of the square. – SMM May 6 '19 at 16:46
• @SMM: Yes, those values are obvious from this answer, but that gives no intuition as to why the multiple should be an integer. I think Blue's answer below gives that, though. – BlueRaja - Danny Pflughoeft May 6 '19 at 21:17
@SMM's proof is nicely self-contained. Here's one that invokes the Descartes "Kissing Circles" theorem, simply because everyone should be aware of that result.
Let the quarter-, semi-, and full-circles have radius $$a$$, $$b$$, $$c$$, respectively.
From the right triangle, we have $$a^2+(a-b)^2=(a+b)^2 \quad\to\quad a=4b \tag{1}$$
Considering the side of the square a circle of curvature $$0$$, that special case of the Kissing Circles theorem implies $$\frac1{c} = \frac{1}{a}+\frac{1}{b}\pm 2\sqrt{\frac{1}{a}\cdot\frac{1}{b}} = \frac{5}{4b}\pm 2\sqrt{\frac{1}{4b^2}} = \frac{5\pm 4}{4b}\quad\to\quad c = \frac49 b \;\text{or}\; 4b\;\text{(extraneous}) \tag{2}$$
Then, with $$a=2$$, we have $$b=1/2$$, so that $$c=2/9$$. $$\square$$
let the side of the square be $$a$$.
Let's find the radius x of the semicircle
We have $$(a+x)^2 = a^2 + (a-x)^2$$ $$x=\frac{a}{4}$$
Now, a lemma.
If circles of radiuses R and r are touching externally, then the length of their common tangent is $$2\sqrt{Rr}$$
Proof of the lemma: draw the common tangent and radiuses as in the figure. There is a right trapezium, so we get $$(R+r)^2 = h^2 + (R-r)^2$$, from where $$h = 2 \sqrt{Rr}$$.
Now, let's use the lemma. Let $$y$$ be the radius of the small circle. We have $$a = 2\sqrt{ay} + 2 \sqrt{\frac{a}{4}y}$$ $$\sqrt{a} = 3 \sqrt{y}$$ $$y = \frac{a}{9}$$
• I like this one because it taught me something useful. – richard1941 May 7 '19 at 19:03 | 2020-12-03T22:31:07 | {
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https://math.stackexchange.com/questions/1078150/conditionally-combining-vanilla-and-chocolate-ice-cream-scoops | # Conditionally combining vanilla and chocolate ice cream scoops
Suppose I have a plate that fits $n$ number of ice cream scoops. I can choose between vanilla and chocolate flavors. The only condition is that I cannot scoop for chocolate twice in a row, e.g. if $n=5$ then {$v,c,c,v,c$} is not allowed - an impossible combination - but {$c,v,c,v,c$} or {$v,v,v,v,c$} is allowed.
How many possible combinations are there? I know I have to perform the operation $${n \choose 2} - \text{impossible combinations}$$
But just how do I figure out the number of impossible combinations?
There are $2^{5}$ possible combinations. We want a recurrence to help us out. So let $T(1)$ be our recurrence denoting the number of allowed combinations for a string of length $n$. So $T(1) = 2$. Similarly, $T(2) = 2^{2} - 1 = 3$, giving us $vv, vc, cv$. Now for $T(3)$, we cannot have $ccv, vcc, ccc$. So there are $2^{3} = 8$ possible combinations and we exclude three, leaving $T(3) = 5 = T(2) + T(1)$.
So more generally, $T(n) = T(n-1) + T(n-2)$ with initial conditions $T(1) = 2$, $T(2) = 3$. We set up the characteristic polynomial and solve:
$\lambda^{2} - \lambda - 1 = 0$, which gives us eigenvalues $\lambda = \frac{ 1 \pm \sqrt{5}}{2}$.
The general form equation is: $T(n) = A \cdot (\frac{1 + \sqrt{5}}{2})^{n} + B \cdot (\frac{1 - \sqrt{5}}{2})^{n}$.
Now plug in your initial conditions and solve.
The number of impossible combinations is $2^{n} - T(n)$.
Edit: Proving T(n) holds by induction. The base cases are given in my analysis above for $T(1), T(2)$. So assume true up to $n-1$ and consider case $n$.
Now look at $T(n)$ and remove the last flavor added. If it was vanilla, there are $T(n-1)$ ways of constructing such a sequence. By the inductive hypothesis, $T(n-1)$ correctly counts the number of valid $n-1$ configurations. So adding vanilla is valid.
If the last flavor is instead chocolate, we note that the flavor before last is vanilla (otherwise, the configuration would be invalid). So there are $T(n-2)$ such strings so that the $n-1$ flavor is vanilla. And then the $n$th flavor is chocolate. Thus, $T(n) = T(n-1) + T(n-2)$, exhausting all valid possibilities.
So by the principle of mathematical induction, $T(n)$ correctly counts the valid configurations for your problem.
• $2^5$ possible combinations for $n = 5$? – Don Larynx Dec 22 '14 at 22:45
• Yes. This is a "word" problem. There are five slots: $_ _ _ _ _$. Each slot can have vanilla or chocolate, and is independent of the others. So by rule of product, we multiply, giving us $2^{5}$ possible combinations, before we impose the restriction. – ml0105 Dec 22 '14 at 22:46
• How do we know that this is a Fibonacci sequence for all $n$? – Don Larynx Dec 22 '14 at 22:53
• You can prove it by induction. We have our base cases of $n = 1, 2$ taken care of by my explanation in the post. So suppose up to $n$ and look at the $n+1$ case. Use the $n-1, n$ cases to derive that this must be true for $T(n+1)$. – ml0105 Dec 22 '14 at 22:56
• I understand how to prove the Fibonacci sequence holds, but how do we know this doesn't diverge for large enough $n$, e.g. 1,1,2,3,5,8,13,21,44(it diverged),79,333, etc... – Don Larynx Dec 22 '14 at 22:58
You want binary sequences not containing two consecutive ones.
Let $f(n)$ be the number of such sequences of length $n$.
then we have $f(1)=2,f(2)=3$.
After this we use the recursion $f(n)=f(n-1)+f(n-2)$.
Why? seperate the strings of lentgh $n$ into two kinds, those ending in $1$ and those ending in $0$.
How many end in $0$? $f(n-1)$, since after removing the last digit you must have a valid sequence of length $n-1$.
How many end in $1$? $f(n-2)$ since the secon to last digit must be $0$, and then the first $n-2$ must form a valid sequence.
Thus the sequence goes, $1,2,3,5,8,13\dots$ And so on, it is a tail of the Fibonacci sequence.
• But $f(3) = 3$, not $5$. – Don Larynx Dec 22 '14 at 22:43
• vcv,vvc,cvc,cvv,vvv – Jorge Fernández Hidalgo Dec 22 '14 at 22:47
• Oops, I was counting the number of impossible strings. – Don Larynx Dec 22 '14 at 22:47
• How do we know that this is a Fibonacci sequence for all $n$? – Don Larynx Dec 22 '14 at 22:54
• well, we proved $f(n)=f(n-1)+f(n-2)$ for all $n$, that's the recurrence ficonaccis follow by definition. – Jorge Fernández Hidalgo Dec 22 '14 at 23:01
We reason by cases. Suppose that we place down $k$ vanilla scoops first. This leaves $k + 1$ gaps, each of which will either contain exactly $0$ or exactly $1$ of the $n - k$ remaining chocolate scoops (since we can't have more than $1$ chocolate scoop in the same gap). This can be done in $\binom{k+1}{n-k}$ ways. Summing over each possible value of $k$ (that is, from $k = 2$ to $k = 5$), we obtain: $$\binom{3}{3} + \binom{4}{2} + \binom{5}{1} + \binom{6}{0} = 1 + 6 + 5 + 1 = 13$$ | 2019-11-20T09:01:39 | {
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https://math.stackexchange.com/questions/3250569/in-how-many-ways-can-we-place-10-identical-red-balls-and-10-identical-blue-balls | # In how many ways can we place 10 identical red balls and 10 identical blue balls into 4 distinct urns if each urn has at least 1 ball?
In how many ways can we place 10 identical red balls (R) and 10 identical blue balls (B) into 4 distinct urns if each urn has at least 1 ball?
My attempt: Put the minimum number of balls into each urn:
case 1: R R R R (4C0, or 4 choose 0) ways to place 4 red balls, remaining 6R balls can be placed in 4 urns in (6+4-1)C(4-1) or (9 choose 3) ways, and 10B can be placed in (10+4-1)C(4-1) or (13 choose 3) ways, then total ways for this case 4C0*9C3*13C3 = 1*84*286 = 24024
case 2: B R R R, R B R R , etc, total of 4C1 ways to have 1 blue ball and 3 red ones, then for remaining 7R and 9B - 4C1*10C3*12C3 = 4*120*220 = 105600
case 3: B B R R , B R B R etc - following same logic 4C2*11C3*11C3 = 163350
case 4: 4C3*12C3*10C3 = 105600
case 5: 4C4*13C3*9C3 = 24024
Total ways = 24024+105600+163350+105600+24024 = 422598 However, the answer doesn't seem to be correct. What am I doing wrong?
• Are the urns numbered? i.e. is B R R R really different from R B R R? If they are not ordered I would start with looking at how many ways there are to put 20 balls into 4 urns such that none is empty. This can be looked up as partitions of size 4 i.e. there are 64 such partitions. e.g. 6+6+4+4 Then for each partition one can see in how many ways one can fill urns with exactly 6,6,4,4 balls, which seems easier. – Rene Recktenwald Jun 4 '19 at 10:38
• maybe you've got to work it out first without the 1-ball constraint, then subtract all the combinations with zero in any pot, possibly remembering the inclusion exclusion principle to avoid double deducting – Cato Jun 4 '19 at 10:56
• Look up "Stars and Bars" for a general approach to the problem of counting the red ball arrangements (or the blue ball arrangements), with or without the restriction that every box has at least one ball. Then figure out how to combine those values for the problem at hand -- you'll also need a little bit of inclusion-exclusion, I think. – Ned Jun 4 '19 at 13:07
By the complement rule, the answer = U-L, where:
U is the total number of ways 10 identical red balls (R) and 10 identical blue balls (B) can be placed into 4 distinct urns without constraints
$$U={10+4-1 \choose 4-1}^2=81796$$
L is number of ways with 1 or more urns having no balls
To find L, let's use inclusion-exclusion principle. Let set A be the set of all distributions of 10 red balls and 10 blue balls in urns 1,2 and 3. Set B is the set of all distributions of 10 red balls and 10 blue balls in urns 2,3 and 4; set C is the set of all distributions of 10 red balls and 10 blue balls in urns 1,2 and 4; set D is the set of all distributions of 10 red balls and 10 blue balls in urns 1,3 and 4. Then $$L = A\cup B \cup C\cup D$$ will be the set of all distributions in which at least one urn is left empty. By inclusion-exclusion
$$L = |A\cup B \cup C\cup D| = |A|+|B|+|C|+|D|-(|A\cap B|+|A\cap C|+|A\cap D|+|B\cap C|+|B\cap D|+|C\cap D|) + (|A \cap B \cap C| + |A \cap B \cap D| + |A \cap C \cap D| + |B \cap C \cap D|) - |A \cap B \cap C \cap D|$$
$$|A| = |B|=|C|=|D|$$ - number of ways 10 identical red balls and 10 identical blue balls can be placed into 3 distinct urns
$$|A\cap B|=|A\cap C|=|A\cap D|=|B\cap C|=|B\cap D|=|C\cap D|$$ - number of ways 10 identical red balls and 10 identical blue balls can be placed into 2 distinct urns
$$|A \cap B \cap C|=|A \cap B \cap D|=|A \cap C \cap D|=|B \cap C \cap D|$$ - number of ways 10 identical red balls and 10 identical blue balls can be placed into 1 urn
$$|A \cap B \cap C \cap D|$$ case when all urns are empty - not possible, therefore zero
$$|A| = {10+3-1 \choose 3-1}^2=4356$$
$$|A\cap B|={10+2-1 \choose 2-1}^2=121$$
$$|A \cap B \cap C|={10+1-1 \choose 1-1}^2=1$$
$$L=4356*4 - 121*6 +4=17424-726+4=16702$$
The ANSWER = $$\boxed{U-L = 81796-16702=65094}$$
When you do a question with the wording "at least" you almost always approaching it using the $$1-$$ not $$P(A)$$ method, because otherwise you are often likely to double count or miscount stuff.
For example in the way you approached the question, you have over counted in your case two. The reason is that if I put two red balls, with the four urns already having $$B R B B$$, they are fundamentally the same if I put them in the first urn and then the third urn compared to if I put them in the third urn and then the first urn. It is however different if I put one in he second urn and the other in the third urn.
This is because the ordering of the urn is taken care of at the step when you made a distinction between $$B R B B$$ as opposed to $$R B B B$$, then after that distinction, the three blue urn are no longer distinct. So placing a red ball in the first blue urn is no different to placing in the third blue urn for example.
Using the other method, you calculate the total number of way of placing the balls, then subtract all the ways with 1 or more urn having no balls. Then you should get the right answer.
• Thank you. Following your suggestion, the answer should be the total number of ways to put 10 identical red balls and 10 identical blue balls into 4 distinct urns minus number of combinations with 1 urn empty minus number of combinations with 2 urns empty minus number of combinations with 3 urns empty, correct? Continued in next comment – Alexander Arefolov Jun 9 '19 at 20:39
• Total number of ways to put 10 identical red balls and 10 identical blue balls into 4 distinct urns without any constrains is ((10+4-1)C(4-1))^2 or (13 choose 3) squared = 81796 (I know this part is correct). Number of ways to have 1 urn empty should be (4C3)*((10+3-1)C(3-1))^2 : (4 choose 3) ways to choose one empty bin or 3 full bins times number of ways to fill 3 distinct bins with 10 identical red and 10 identical blue balls which should be (10+3-1)C(3-1)^2 or (12 choose 2) squared = 4*66^2=17424. Continued in next comment – Alexander Arefolov Jun 9 '19 at 20:40
• For 2 urns with no balls scenario - (4 C 2)*((10+2-1)C(2-1))^2 = 6* (11 choose 1)^2 = 6*121=726. For 3 urns with no balls: (4 C 1)*((10+1-1)C(1-1))^2 = 4. The answer then should be 81796 - (17424+726+4)= 63642. However, this answer still does not seem to be correct! – Alexander Arefolov Jun 9 '19 at 20:41 | 2021-06-15T03:02:55 | {
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https://math.stackexchange.com/questions/4259194/is-there-a-power-of-2-whose-last-digit-is-a-0/4259207#4259207 | # Is there a power of $2$ whose last digit is a $0$?
I'm looking for a power of $$2$$ that ends in $$0$$, so $$2^n = k\cdot 10$$, where $$n \in \mathbb{Z}$$ and $$k \in \mathbb{Z}$$. It doesn't seem to exist. As an equation with two variables, I was looking for a second equation, but just can't think of any.
I also trialled a few consecutive powers of $$2$$ and noticed that there seems to be a sequence perpetually ending in $$2, 4, 8, 6, 2, 4, 8, 6, \cdots$$
How can I prove mathematically that there will never be a power of $$2$$ that ends in $$0$$? Is there a simply way to explain that?
• $5$ divides the RHS, but not the LHS. Sep 24 at 15:04
• You already proved that the ending digit cycles and $0$ does not appear. Nothing more has to be done. Of course, you can also apply the above comments. Sep 24 at 15:23
• @Peter I haven't proven it. All I wrote was, that I observed ("noticed") a sequence. Sep 25 at 14:47
• Thanks for all answers (permanently or temporarily submitted). My first post to this site seemed to have sparked some interest. Sep 25 at 15:03
• Oct 15 at 16:03
The only numbers, in the decimal basis, with a digit of $$0$$ for the units, are those that are multiples of $$10$$, a fortiori, multiples of $$2$$ AND $$5$$. Through the fundamental theorem of arithmetic, you can show that there is no power of $$2$$ that is divisible by $$5$$; thus, there is no power of $$2$$ which ends by a $$0$$ digit (in the decimal basis).
• I've just realised that you already cite the fundamental theorem of arithmetic. (+1) I've deleted my answer. Sep 24 at 19:07
• Thanks. That's the simplest of answers. Though not quite a proof I accept this as the best answer. Sep 25 at 15:02
Proof 1: Let $$a=2^n$$ and $$b = 2^{n-1}$$. Since $$a = 2b$$, if the last digit of $$a$$ is $$0$$, then the last digit of $$b$$ must be $$0$$ or $$5$$. Since $$b$$ is odd, the last digit can't be $$5$$. But if the last digit of $$b$$ is $$0$$, we can apply the same logic to find that the last digit of $$2^{n-2}$$ is $$0$$, and so on all the way down to $$2$$, but obviously the last digit of $$2$$ isn't $$0$$.
Proof 2: Suppose we take a power of two and write it as $$10a+b$$, and then multiply it by $$16$$. $$16*(10a+b)=160a+16b=160a+10b+5b+b$$. Now clearly the $$160a$$ and $$10b$$ terms don't affect the last digit, and since $$b$$ is even, neither does $$5b$$. So the last digit is $$b$$. This shows that whenever you take a power of two, and then take four more powers of two (that is, multiply by $$2^4=16$$), you don't change the last digit.
Proof 3: The last digit of a number is the same as the remainder when divided by $$10$$, and taking the remainder when divided by some number is known as "modular arithmetic". In modular arithmetic, the set of all numbers with the same remainder is an equivalence class, and these equivalence classes act like numbers. We can apply addition and multiplication to them and get an equivalence class as an answer. For instance, if we multiply a number with a last digit of $$2$$ with a number with a last digit of $$3$$, the product will have a last digit of $$6$$. If we represent the set of numbers with last digit $$r$$ as $$\bar r$$, we can say $$\bar 2 \times \bar 3 = \bar 6$$, and this is true regardless of which numbers we pick from the $$\bar 2$$ and $$\bar 3$$ equivalence classes. So you're asking for an $$n$$ such that the $$\bar 2^n =\bar 0$$. It's easy to see by inspection that this is not possible, but there's also a theorem that if we define $$\phi(n)$$ as the number of numbers less than $$n$$ that are coprime to $$n$$, then $$\bar a^{\phi(n)+1}$$ in $$n$$-modular arithmetic is $$\bar a$$. Since $$\phi(10)=4$$ ($$1,3,7,9$$ are coprime to $$10$$), $$\bar 2^{n+4}=\bar2^n$$.
• I really liked the Proof 2! $(+1)$ Sep 26 at 22:52
The below tells us more generally what the last digit can be (although the proof is a bit informal).
(The lemma, and a bit more formal reasoning, is really the only thing you were missing.)
Lemma: when multiplying 2 positive integers, only the last digits of each affects what the last digit of the result will be.
Proof: ($$x_i$$ represents the $$i$$-th digit of the integer $$x$$ and $$x_1x_2...x_{n-1} = 0$$ if the number only has 1 digit)
$$a_1a_2...a_n * b_1b_2...b_n$$
$$=(10 * a_1a_2...a_{n-1} + a_n) * b_1b_2...b_n\hspace{3cm}(1)$$
$$=a_n * b_1b_2...b_n + 10 * a_1a_2...a_{n-1} * b_1b_2...b_n$$
$$=a_n * b_n + 10 * a_n * b_1b_2...b_{n-1} + 10 * a_1a_2...a_{n-1} * b_1b_2...b_n\hspace{1cm}\text{(similar to 1)}$$
$$=a_n * b_n + 10 * (a_n * b_1b_2...b_{n-1} + a_1a_2...a_{n-1} * b_1b_2...b_n)$$
The last digit of 10 times any positive integer will always be 0. If you sum such an integer with any other positive integer, the last digit of the result will be equal to the last digit of the other integer. Thus $$a_n * b_n$$ (that is, the last two digits of each integer) is the only thing that affects the last digit of the result. Q.E.D.
Now let's consider the first few powers of 2:
$$2^0 = 1$$ $$2^1 = 2$$ $$2^2 = 2 * 2 = 4$$ $$2^3 = 4 * 2 = 8$$ $$2^4 = 8 * 2 = 16$$ $$2^5 = 16 * 2 = 32$$
Since only the last digit of the current power affects the last digit of the next power, we now have a cycle.
We know any number ending in 2 multiplied by 2 will end in 4. Multiply that by 2 and it will end in 8, then 6, then 2 again. And it will repeat like this indefinitely.
Therefore, the only last digits possible, from $$2^1$$ onward, are 2, 4, 6 and 8.
Negative powers of 2 trivially cannot have a last digit of 0.
Therefore, there is no integral power of 2 having a last digit of 0.
As a bonus result, this approach also shows that the last digit of $$2^{1+i}$$ equals $$2^{1+(i\bmod 4)}\bmod 10$$ for any $$i \ge 0$$.
I'm a Computer Science teacher and was also intrigued by this, so wrote the following program to prove that there are no numbers that are multiples of 5 and powers of 2. I'm definitely no mathematician, so forgive the brute force approach. As stated above by posters, the repeating end digit sequence of 2,4,8,6 is the proof / 'give away'.
# part 1 - set up powers of 2 for reference
maxNumber = int(input('Enter highest number ')) result = 0 power = 0 base2 = [] while True: result = 2**power if result <= maxNumber: base2.append(result) power += 1 else: break print (base2)
# part 2 - go search
found = False for i in range(1,maxNumber+1): if i % 5 == 0 and i in base2: print (i,'is a multiple of 5 and power of 2') found = True if not found: print ('There are no numbers between 1 and',maxNumber,'that are a multiple of 5 and power of 2')
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http://salesianipinerolo.it/ubdl/real-life-applications-of-sine-and-cosine-functions.html | 2 Educational Objectives After performing this experiment, students should be able to: 1. All quadratic functions fit the form y = Ax 2 + Bx + C, where A, B, and C can be any real number (although A cannot be zero). Trigonometry is a system that helps us to work out missing sides or angles in a triangle. Learning Targets: (1) I can sketch the graphs of the parent functions of sine and cosine. The sine and cosine functions are among the most important functions in all of mathematics. Among many uses and applications of the logistic function/hyperbolic tangent there are: which is based in the hyperbolic cosine function. Synthetic Geometry and Coordinate Geometry are used in real life to help us understand the dimensions and transformations of shapes and figures such as lines, triangles, polygons, and circles. Basic graph types are y = a sin(bx) + c and y = a cos(bx) + c. Some examples include the weather, seasonal sales of goods, body temperature, the tide's height in a harbor, average temperatures, and so on. • Table 1 gives the sum of two arbitrary cosine functions. Therefore, a time domain function f(x) and its corresponding frequency domain function are duals of each other. rewrite real-life formulas. By thinking of the sine and cosine values as coordinates of points on a unit circle, it becomes clear that the range of both functions must be the interval[latex]\,\left[-1,1\right]. 03 Students will take the unit circle and construct the graphs of the sine and cosine functions. In this way, it has many engineering applications such as electronic circuits and mechanical engineering. θ = 90°), which gives us cos θ = 0 and. For example, variables that depend on the seasons may be modeled with trigonometric functions because the seasons repeat every year just like the sine function repeats every 2π. Here's an example: San Diego, California, is a gorgeous […]. Graph a sine or cosine function having a different amplitude and period. This equation is usually solved using sums of sines and cosines. Build an understanding of trigonometric functions by using tables, graphs and technology to represent the cosine and sine functions. Calculus is made up of Trigonometry and Algebra. distributions characterized by a cdf based on the sine function, called the new sine-G family of distributions. At 90° and 270°, x = 0 and therefore cos φ = 0, while at 0° and 180° y = 0 and therefore sin φ = 0. Explore symmetry (odd & even), domain, range, continuity, relative extrema and concavity of sine and cosine functions; Use Graphing Calc. The solution for an oblique triangle can be done with the application of the Law of Sine and Law of Cosine, simply called the Sine and Cosine Rules. These can be confusing to understand and memorize however these for the basis of all trigonometric values and derivations. Homework Statement High tide at 4am with a depth of 6 meters. Modeling with Sine or Cosine Functions Sine and cosine functions can be used to model real-world phenomena, such as sound waves. The midline is the average value. They are often shortened to sin, cos and tan. 4 Trigonometric Functions of Any Angle 4. C y KA sl ul d KrYirgMhlt os 4 3r 6e2s ke er Ivie kd F. Applications of Trigonometric and Circular Functions 281 CHAPTER OBJECTIVES t -FBSOUIFNFBOJOHTPGBNQMJUVEF QFSJPE QIBTFEJTQMBDFNFOU Applications of Trigonometric and Circular Functions CAS Suggestions argument of the cosine. The application problems show different ways that one could use sine and cosine funtions in real life. Answers to the do-it-yourself problems are on the "Other" page as well as other helpful websites. To be able to solve real-world problems using the Law of Sines and the Law of Cosines This tutorial reviews two real-world problems, one using the Law of Sines and one using the Law of Cosines. What is Trigonometry? Trigonometry is a another form of math dealing with the relationship of the sides and angles of triangles. I struggle with finding. Physical: 23,. 7 - Applications of Sinusoidal Functions Applications of Sinusoidal Functions. This lesson will present real world examples that involve inverse trigonometric ratios. The trigonometric functions sine, cosine, and tangent are useful in a wide range of applications, from solving simple problems to advanced and complex problems. Over 90 new problems on differential equations and their applications. Understanding Calculus is a complete online introductory book that focuses on concepts. It studies the relationship between involving lengths and angles of triangle. Objectives: To graph the sine and cosine functions; to identify the graphs of the sine and cosine functions. In maths, you have real life applications on any thing that you study. At x = 0 degrees, sin x = 0 and cos x = 1. Transformations of Sine and. What are the Applications of Trigonometry in Real Life? Ans. (3) The student uses functions and their properties to model and solve real-life problems. Graphing Cosine Function Lesson Plan 1. Graph a sine or cosine function having a different amplitude and period. What are real-life applications of the sine and cosine wave applications? It's for a pre-calculus assignment and I have to make a presentation to the class about it. What Are Imaginary Numbers? are used in real-life applications, such as electricity, as well as quadratic equations. Recall from Graphs of the Sine and Cosine Functions that the period of the sine function and the cosine function is $$2π$$. Subhaschandra Singh, Department of Physics, Dhanamanjuri College of Science, Imphal – 795 001, Manipur, India. introduction. Law of Cosine [Just formula. Pg 439 #1-8, 33, 35, 37, 43-55 odd · Sketch the graphs of basic sine and cosine functions · Use amplitude and period to sketch the graphs of sine and cosine functions · Sketch translations of the graphs of sine and cosine functions. 3: Trigonometric Functions in Real life - Duration: 13:02. A Fourier series is a way of representing a periodic function as a (possibly infinite) sum of sine and cosine functions. S08 2 Learning Objectives Upon completing this module, you should be able to: 1. (b) If the power of cosine is odd (n=2k+1), save one cosine factor and use the identity sin 2 x + cos 2 x = 1 to convert the remaining factors in terms of sine. bc is the horizontal shift 4. This is exactly the reason why, when the graphs are plotted on grids as above, the cosine graph is equivalent to the sine graph, omitting the fact that it is shifted, or translated, 90° to the left. Practice Now! Trigonometric Ratios and Angle Measures Topics: 1. Tan x must be 0 (0 / 1) At x = 90 degrees, sin x = 1 and cos x = 0. Trigonometry plays a major role in musical theory and production. Outcome 5: Set up, solve, and graph equations from problems that require use of trigonometric functions, tangent, sine, and cosine and the Pythagorean Theorem. 4 9 Review Day 10 Test Day TOTAL DAYS: 10 C2. rewrite real-life formulas. Note the capital "C" in Cosine. Note the capital “C” in Cosine. A trigonometric function can be used to find the height of a smokestack on top of a building. Relationship between Sine and Cosine graphs The graph of sine has the same shape as the graph of cosine. &Assume&that&a& rider&enters&a&car&froma&platformthat&is&located&30°&around. Solitary Wave Solutions to Certain Nonlinear Evolution Equations by Rational Sine –Cosine Function Method S. Unit Five: Real World Problems Example: When Light shines through two narrow slits, a series of light and dark fringes appear. State the domain and range of sine, cosine, and tangent curves of the form y = Asinx, y = Acosx, and y = Atanx. Outcome 6: Set up and solve exponential and logarithmic equations; then identify and sketch graphs of the functions. F) as its Fourier transform. But such proofs are lengthy, too hard to reproduce when you’re in the middle of an exam or of some long calculation. 7m, but I am unsure how to interval the axis. Graphs of elementary trig functions allow you to see the graphs of sine, cosine and tangent and their relationship to travelling around a circle. Yes, you can derive them by strictly trigonometric means. Prerequisite Given the period, solve for the parameter '' in or. Trigonometry has many applications in astronomy, music, analysis of financial markets, and many more professions. Applications of the Derivative. This lesson will present real world examples that involve inverse trigonometric ratios. T3 and 20-2. The general form of a sinusoidal is: $f ( x ) = a \sin ( bx – c ) + d, \text{ for } b > 0$. So in the rule, c 2 = a + b - 2ab cos C side you are looking for angle opposite the side you want Label your diagram using a,b and c to avoid confusion Substitute into the formula and evaluate. Model the problem using the equation to show the depth of the water t hours after midnight. The slider is used for frequency values. Most radio communication is based on the use of combinations of sines and cosine waves. So our function is $y = 3. Objectives: To graph the sine and cosine functions; to identify the graphs of the sine and cosine functions. Experiment with functions that have additional terms, and see how these change the period, amplitude, and phase of the waves. Trigonometry plays a major role in musical theory and production Real world examples of cosine functions. What we want is the Law of Sines. Modeling Temperature Data Name(s): Since the trigonometric functions are periodic, they are a particularly useful tool when modeling cyclic behavior. They will also explore certain relationships that emerge from the measurement of three-dimensional figures and two-dimensional shapes. Especially sine and copsine functions have been very useful in applications for medical science, signal processing, geology, electronic communication, thermal analysis, satellite communication and many more. Find the period of a sine or cosine function. There are primarily three trigonometric functions commonly used with trigonometric identities to solve complex equations. The domain is the set of all real numbers. We explain Real World Examples of Inverse Trigonometry Functions with video tutorials and quizzes, using our Many Ways(TM) approach from multiple teachers. Trigonometric formulae are useful for solving problems in two dimensions. aphing Sine and Cosine functions: amplitude, phase shift, and vertical slide Cofunctions (Notes pp. Using Sum and Difference Formulas In this lesson, you will study formulas that allow you to evaluate trigonometric functions of the sum or difference of two angles. cos^{-1}(x) = arccosx \rightarrow The arc that has a cosine of x. The Sine, Cosine and Tangent functions are often applied to real world scenarios. The applications of Fourier transform are abased on the following properties of Fourier transform. Real life scenario of logarithms is one of the most crucial concepts in our life. The Sine, Cosine and Tangent functions are often applied to real world scenarios. Real-life applications of. Objectives: To graph the sine and cosine functions; to identify the graphs of the sine and cosine functions. basic introduction into trigonometry. Real life application of trigonometry is use of trigonometry formula and its functions. Euler's formula. 3: Trigonometric Functions in Real life - Duration: 13:02. These trigonometric functions are extremely important in science, engineering and mathematics, and some familiarity with them will be assumed in most first year university mathematics courses. 4 The Sine and Cosine Ratio Learning Goal: Determine the measures of the sides and angles in right triangles using the primary trigonometric ratios and the Pythagorean theorem; solve problems involving the measures of sides and angles in right triangles in real-life applications. com is a free math website that explains math in a simple way, and includes lots of examples, from Counting through Calculus. We say that this sinusoidal has a vertical shift of 1. The sine function has many real life applications, a few of which are: Triangulation, used in GPS-equipped cellphones, Musical notes, Submarine depth, Length of a zip. In addition, non-right angled triangles can be solved using the sine and cosine trigonometric functions. The domain of a linear function is all real numbers, and it's not possible to write every real number in a list. arbitrary sine function. There is only315 degrees of pizza left. Graphing Sine and Cosine Trig Functions With Transformations, Phase Shifts, Period - Domain & Range - Duration: 18:35. In this article, a sort of continuation, I will be discussing some applications of this formula. Change from roots to rational exponents. So the graph looks like a very simple wave. However, the basic sine function usually requires one or more transformations to fit the parameters of the process. Many other Fourier-related transforms have since been defined, extending the initial idea to other applications. Magnitude Amplitude of combined cosine and sine Phase Relative proportions of sine and cosine The Fourier Transform: Examples, Properties, Common Pairs Example: Fourier Transform of a Cosine f(t) = cos (2 st ) F (u ) = Z 1 1 f(t) e i2 ut dt = Z 1 1 cos (2 st ) e i2 ut dt = Z 1 1 cos (2 st ) [cos ( 2 ut ) + isin ( 2 ut )] dt = Z 1 1 cos (2 st. Such behavior occurs throughout nature and led to the discovery of rapidly rotating stars called pulsars in 1967. It just plots the sine function on the screen. cos^{-1}(x) = arccosx \rightarrow The arc that has a cosine of x. The trigonometric functions sine and cosine have four important limit properties: You can use these properties to evaluate many limit problems involving the six basic trigonometric functions. A function may be defined by means of a power series. for the area of a triangle • use of sine rule and cosine rule for any triangle. This is how we find out “sine/cosine = tangent/1”. You want to figure out what the angle is of the bottom of your long chair to the ground. 28-1 Length of Arc. However, there are still disadvantages such as low solution accuracy and poor global search ability. In other words, for any value of $$x$$,. We show that, in some situations, the new sine-G models provide an interesting alternative to the sine-G models, with possible di erent targets in terms of modelling. Let's put our values in there: Now let's move some things around and get calculating: We're not done yet, though, we need to apply some inverse sine to both sides to get to B itself. And I'll leave those details to the. The function has the same domain, but a range of. General triangle word problems. Homework Statement High tide at 4am with a depth of 6 meters. Example 1: Evaluate. In Chapter XI of The Age of Reason, the American revolutionary and Enlightenment thinker Thomas Paine wrote:. In this way, it has many engineering applications such as electronic circuits and mechanical engineering. • Table 1 gives the sum of two arbitrary cosine functions. #91 - Sketch translations of graphs of sine and cosine functions. F MTH 123 Ginger Rohwer 28 February Real Life Trigonometry Applications In Lunar Phases In this project we will be collecting data about the illumination of the moon for certain months then creating sine and cosine functions to. We could think of the cosine function as a sine wave with phase shift ˇ 2. Law of Sines. Definitions and formulas for basic trigonometry, sine, cosine, tangent, cosecant, secant, cotangent, the law of sines and the law of cosines. The domain of the sine function is the set of real numbers, that is, every real number is a first element of one pair of the function. Besides other fields of mathematics, trig is used in physics, engineering, and chemistry. • Given information about a rotating functions in the real world. Together, these extensions define (sin φ, cos φ) for any angle φ, positive or negative, of any size. Plotting a basic sine wave. Relationship between Sine and Cosine graphs The graph of sine has the same shape as the graph of cosine. the sine, cosine and tangent functions are often applied to real world scenarios. The amplitude is a=2 and the period is. 3 Trigonometry in the Real World C2. You want to figure out what the angle is of the bottom of your long chair to the ground. SheLovesMath. Upon the completion of this course, you will have a better understanding of key geometry axioms and theorems. (3) I can sketch translations of the graphs of sine and cosine functions. Applications are given for four real data sets, showing a better t in comparison to some existing distributions based on some of goodness-of- t tests. The first portion of the book is an investigation of functions, exploring the graphical behavior of, interpretation of, and solutions to problems involving linear, polynomial, rational, exponential, and logarithmic functions. Sine, Cosine and tangent are the three important trigonometry ratios, based on which functions are defined. The amplitudes of the cosine waves are held in the variables: a1, a2, a3, a3, etc. 85 inches by 3. The graph of a quadratic function is called a parabola. Sine 5 None Up 2 9. Find an equation for a cosine function that has an amplitude of OTC — , a period of Find an equation for a sine function that has amplitude of 5, a period of 3TT. At x = 0 degrees, sin x = 0 and cos x = 1. This is how we find out “sine/cosine = tangent/1”. Possible Ideas Students will suggest:. Similar statements can be made for the other trigonometric functions of sums and differences. Synthetic Geometry and Coordinate Geometry are used in real life to help us understand the dimensions and transformations of shapes and figures such as lines, triangles, polygons, and circles. Mathematics Revision Guides - Real Life Trig Problems Page 12 of 14 Author: Mark Kudlowski Method 2 - Using the sine and cosine rules. Domain and Range of Sine and Cosine The domain of sine and cosine is all real. (Sine, Cosine, Secant, etc. The sine function takes an angle and tells the length of the y-component (rise) of that triangle. FIND the coordinates of A & C. The idea is to decompose any such function f(t) into an in nite sum, or series, of simpler functions. Hyperbolic functions show up in many real-life situations. We must now decide whether to use a sine function or a cosine function to get the phase shift. 11 (+) Understand and apply the Law of Sines and the Law of Cosines to find unknown measurements in right and non-right triangles (e. Sine waves are very easy to produce electronically, and can be viewed on an oscilloscope, for. You decide to plug it into a sum and difference formula for sine. Use trigonometric functions to model and solve real-life problems. Graphs are shown in figure 4. SOH-CAH-TOA is a nice shortcut, but get a real understanding first! Gotcha: Remember Other Angles. It is analogous to a Taylor series, which represents functions as possibly infinite sums of monomial terms. 5b Translations of Sine and Cosine Curves Given the functions: y = a sin (bx – c) + d and y = a cos (bx – c) + d creates horizontal and vertical translations of the basic sine and cosine curves. The sine function is a set of ordered pairs of real numbers. In line 4 we use the properties of cosine (cos -x = cos x) and sine (sin -x = -sin x) to simplify the. Because angles are an intricate part of nature, sines, cosines and tangents are a few of the trigonometry functions ancient and modern architects use in their work. The period of tangent and cotangent are \pi . Like all functions, trigonometric functions can be transformed by changing properties like the period, midline, and amplitude of the function. Find the equation of the normal to the curve of y=tan^-1(x/2) at x=3. We must now decide whether to use a sine function or a cosine function to get the phase shift. 4 8 More Real World Problems C2. Signal Processing. 3 Trigonometry in the Real World C2. Real life scenario of trigonometry is studied in cbse class 10. Trigonometric formulae are useful for solving problems in two dimensions. 7 Inverse Trigonometric Functions 4. Here are the topics that She Loves Math covers, as expanded below: Basic Math, Pre-Algebra, Beginning Algebra, Intermediate Algebra, Advanced Algebra, Pre-Calculus, Trigonometry, and Calculus. The Fourier series synthesis equation creates a continuous periodic signal with a fundamental frequency, f, by adding scaled cosine and sine waves with frequencies: f, 2 f, 3 f, 4 f, etc. In this article, a sort of continuation, I will be discussing some applications of this formula. Unit 4 - Graphing & Writing Sine & Cosine Functions; Application Problems October 21 to November 5th, 2013 Date Topic Assignment Monday Gr 10/21 changes. functions to find a efficient navigational route for long trips if GPS or other radar equipment is unavailable. The book has as much to do with calculus as with philosophy. Plotting more points gives the full shape of the sine and cosine functions. 4 9 Review Day 10 Test Day TOTAL DAYS: 10 C2. The mathematical topics of Fourier series and Fourier transforms rely heavily on knowledge of. Math Help Fast (from someone who can actually explain it) See the real life story of how a cartoon dude got the better of math Pre-Cal 12 - Applications of Sinusoidal Functions New Project 5. FIND the coordinates of A & C. Part of the market-leading Graphing Approach Series by Larson, Hostetler, and Edwards, Precalculus Functions and Graphs: A Graphing Approach, 5/e, is an ideal student and instructor resource for courses that require the use of a graphing calculator. Substituting 0 for x, you find that cos x approaches 1 and sin x − 3 approaches −3; hence, Example 2: Evaluate. Definition 1 is the simplest and most intuitive definition of the sine and cosine function. A real life example of the sine function could be a ferris wheel. ) Circles are an example of two sine waves. The Fourier Series, the founding principle behind the eld of Fourier Analysis, is an in nite expansion of a function in terms of sines and cosines. 3 The symmetries of the six trig functions Since the sine function is odd and the cosine function is even then tan( ) = sin( ) cos( ) = sin( ) cos( ) = tan( ) and so the tangent function is odd. At 90° and 270°, x = 0 and therefore cos φ = 0, while at 0° and 180° y = 0 and therefore sin φ = 0. Return and go over Unit 2 Test. We have been studying how the equations of sinusoidal functions change when the graphs are shifted up and down and left and right and amplitude changes and period changes. In calculus, we are often. (See Example 3. The trigonometric functions sine and cosine have four important limit properties: You can use these properties to evaluate many limit problems involving the six basic trigonometric functions. GIVEN y x and B: 3cos2 0,3. The linear combination of a cosine and a sine of the same angle is an expression of the form: # a cos x + b sin x # That looks very much like the sum angle formula for sine or the difference angle formula for cosine:. Sine waves are a single note, whereas multiple sine waves make a chord. FREE (46) gnasher30 Uses of Trigonometry in Real Life. The book has as much to do with calculus as with philosophy. 3, P10Trig6. The PowerPoint is complete with custom animation so that pieces of the solving process are revealed, as well as the final answer. The Unit Circle is a tool used to find trigonometric functions. Basic Sine and Cosine Curves For 0≤ ≤2𝜋, the sine function has its maximum point at. (Check your answer with your. The tangent function differs in its shape and range from its sine and cosine counterparts. 4 Part 2 : Applications of Trigonometric Functions Self Evaluation Solutions Modelling Real Life Situations Worksheet with key Review Worksheet and Review Answers. (See Example 5. Decide whether you will need Pythagoras theorem, sine, cosine or tangent. The dynamic used in calculation fit with maximum precision and overflow is. This equation is usually solved using sums of sines and cosines. Sine, cosine, and tangent functions are used for many real life applications. Sine and cosine are keys to the success of Fourier Transform because sound may be. 42 on the original plot we get: Looks as though we are almost there. ) Sine or cosine functions can used to the motion. Many real world situations can be modeled using the sine and cosine functions. It may not have direct applications in solving practical issues but used in the various field. The heat equation is used to model how things get hot (electronics, spacecraft, ovens, etc). In maths, you have real life applications on any thing that you study. 28-3 Centroids. Web browsers do not support MATLAB commands. We used the law of sines again, sin(B)/(b)=sin(C)/(c) and fill in everything we know, reduced it down, solve for (b) and give our answer in terms of miles. 5 Objective: To draw (using technology), sketch, and analyse the graphs of sine and cosine function whose periods are rational numbers. In this way you can find the size of any unknown angle of a right triangle if you just know 2 sides of. Sine and cosine functions are used primarily in physics and engineering to model oscillatory behavior, such as the motion of a pendulum or the current In an AC circuit, but these functions also arise m other sciences in this project, you will consider an application to biology - we use sine functions to model the populations of a predator and its prey. B = sin-1 (0. We closed the day with an application of what the applet taught us about sine to help us graph cosine functions with translations. What is the value. Math Help Fast (from someone who can actually explain it) See the real life story of how a cartoon dude got the better of math Pre-Cal 12 - Applications of Sinusoidal Functions New Project 5. Put the results in a table. The period of the function sin(x) is 2π. When the program returns, examine the stack for how many times the hyp sine was called and how many times hyp sine/cosine was called vs. For example, this transformed graph above would show which frequency sine and cosine functions to use to model our original function. Trigonometry Applications in Real Life. Applications of Trigonometric and Circular Functions 281 CHAPTER OBJECTIVES t -FBSOUIFNFBOJOHTPGBNQMJUVEF QFSJPE QIBTFEJTQMBDFNFOU Applications of Trigonometric and Circular Functions CAS Suggestions argument of the cosine. Stay safe and healthy. org are unblocked. The Sine Cosine Algorithms (SCA) has been recently proposed; it is a global optimization approach based on two trigonometric functions. The output of the transformation represents the image in the Fourier or frequency domain , while the input image is the spatial domain equivalent. In problems 12 & 13, the graphs of the sine and cosine functions are waveforms like the figure below. Area of a triangle 1. Hyperbolic functions show up in many real-life situations. & 3)&Theheight,ℎ,&in&meters,&of&the&tide&in&a&given&location&on&a&given&day&at&!&hours&after&midnight&can&be& modeled&using. I hope the following reasons and picture will help you understand why we study trigonometry. Describe that because they are functions, we need to be able to UNDO them (just like addition, subtraction, etc. In a right triangle, one angle is 90º and the side across from this angle is called the hypotenuse. m?X Law of Sines Substitute the given values. Experiment with functions that have additional terms, and see how these change the period, amplitude, and phase of the waves. Find and use reference angles to evaluate trigonometric functions. The hypotenuse is the longest side in a right triangle. The graphs of all sine and cosine functions are related to the graphs ofGOAL 2 Graph tangent y = sin x and y = cos xfunctions. Real-world Applications Using the Sine Function The table below shows the number of. Groupwork. The heat equation is used to model how things get hot (electronics, spacecraft, ovens, etc). Students will discover the area of triangles through the laws of sine and cosine. A1/2 ab sin c. Law of Cosine [Just formula. These can be confusing to understand and memorize however these for the basis of all trigonometric values and derivations. 1 Graphing Sine and Cosine Functions Focus on. Interpret the sine function as the relationship between the radian measure of an angle formed by the horizontal axis and a terminal ray on the unit circle and its y coordinate. PDF | In this paper, we propose a new hybrid algorithm called sine–cosine crow search algorithm that inherits advantages of two recently developed | Find, read and cite all the research you. Basic functions in TI-83 Graphing Calculator. Take my hyperbolic sin/cos recursive function place the angle on a sine or cosine stack that represents a call to the sine or cosine. " For example, "an oscilloscope is an electronic instrument used to display changing electrical signals. 0333 170 -170 t e 0. On a graph together, they look like this: Tangent The third basic trigonometric function is called the tangent (tan for short), and it is defined as the ratio of the opposite and adjacent sides - that is: tan θ = y. By correctly labeling the coordinates of points A, B, and C, you will get the graph of the function given. Solution of triangles. In Chapter XI of The Age of Reason, the American revolutionary and Enlightenment thinker Thomas Paine wrote:. Time-saving lesson video on Sine and Cosine Functions with clear explanations and tons of step-by-step examples. 3 The symmetries of the six trig functions Since the sine function is odd and the cosine function is even then tan( ) = sin( ) cos( ) = sin( ) cos( ) = tan( ) and so the tangent function is odd. Find the period of a sine or cosine function. 5); • From sine and cosine functions, we obtain tangent easily: sin tan cos θ θ θ =; (3). Plotting a basic sine wave. Nov 2: Graphs of sine and cosine; functions of the form f(x) = Asin(Bx+C); their amplitude and period. Find an equation for a sine function that has amplitude of 4, a period of fl. We can relate sides and angles in an arbitrary triangle using two basic formulas known as the sine rule and the cosine rule. Essentially, if what is being measured relies on a sine or cosine wave. All we must do now is stretch the period of the sine function. By Victor Powell. Practice: General triangle word problems. This standard works in conjunction with the content standards. Real Life Examples. Use sigma notation. This is a powerpoint explaining some of the applications of trigonometry to answer that age old question 'But Miss, why would we need to know this?&' Conte. The sinusoidal functions provide a good approximation for describing a circuit's input and output behavior not only in electrical engineering but in many branches of science and engineering. Explore the graph of general sine functions interactively using an HTML 5 applet. (Sine, Cosine, Secant, etc. This is in comparison to a continuous function like a line. It is the application of sinusoidal modeling that makes its use in the secondary school environment so valuable. If working outside, choose a spot with two widely spaced (2-5 meters) and roughly parallel lines to define the "river" banks. Trigonometric functions have wide variety of applications in real life. Who is the Father of Trigonometry? Ans. By finding a few key points or aspects of the graph, any of the real-life problems we have today can be explained mathematically and much of the vibrations surrounding us can be better understood. In those tables, variables A and B are scalar constants, frequency ω is in radians/second, and variables α and β are phase angles measured. Thermal analysis. People board the ride at the ground (sinusoidal axis) and the highest and lowest heights you reach on the ride would be the amplitudes of the graph. 120 / 240 Vac sine Wave ac power distribution for residential application: The waveform of the electrical voltage distributed by the grid / the utility companies is like a sine wave. However, there are still disadvantages such as low solution accuracy and poor global search ability. introduction. Looks like MatPlotLib to me. Thefront panel of this instrument is 225 mm wide by 100 mm tall (8. However, in the real world all objects are three dimensional, so it is important that we extend the application of the area, sine and cosine formulae to three dimensional situations. The graphs of the sine and cosine functions are used to model wave motion and form the basis for applications ranging from tidal movement to signal processing which is fundamental in modern telecommunications and radio-astronomy. 2-5) Complete Notes pp. By thinking of the sine and cosine values as coordinates of points on a unit circle, it becomes clear that the range of both functions must be the interval[latex]\,\left[-1,1\right]. Hyperbolic Functions in Real Life. The exponential function, exp(X) or e^X, is a special function that comes from calculus. More Graphing Trigonometric Functions Worksheet Answers Sec 5. Possible Ideas Students will suggest:. The cosine rule: c2 + b2 —2abcosC. 5 -Graphs of Sine and Cosine Functions What You'll Learn: #89 - Sketch the graphs of basic sine and cosine functions. Essentially, if what is being measured relies on a sine or cosine wave. 28-3 Centroids. And I'll leave those details to the. Specific Objectives (measurable) Use and manipulate visual representations of unit circle and trigonometric functions. and "use sine and cosine functions to model real-life data," iii. Some real life examples of periodic functions are the length of a day, voltage coming out of a wall socket and finding the depth of water at high or low tide. The general form of a sinusoidal is: \[ f ( x ) = a \sin ( bx – c ) + d, \text{ for } b > 0$. study investigates the performances of these estimates. Objectives: To graph the sine and cosine functions; to identify the graphs of the sine and cosine functions. On a graph together, they look like this: Tangent The third basic trigonometric function is called the tangent (tan for short), and it is defined as the ratio of the opposite and adjacent sides - that is: tan θ = y. This lesson will present real world examples that involve inverse trigonometric ratios. 4 Trigonometric Functions of Any Angle 4. If either the imaginary or the real part of the input function is zero, this will result in a symmetric Fourier transform just as the even/odd symmetry does. Sine, Cosine, Tangent Applications. , surveying problems, resultant forces). docx from MATH 313B at K12. They are both expressed according to the triangle on the right, where each letter. At the end of. c) How many minutes, from t = 0, does it take the rider to reach the. Recall that the cosine function takes an angle x as input and returns the cosine of that angle as output: For example if 60° is the input then 0. In these studies appear functions of the breast and cosine. Hyperbolic functions also satisfy identities analogous to those of the ordinary trigonometric functions and have important physical applications. Since the period of the sine function is 2, and the period of the temperature data is 52 weeks, we set b = 2 /52. ] Basic use of Geometer's Sketchpad. Find geometric and arithmetic formulas for sequences of numbers. " As we point out and use functions in real-life settings, we can ask our students to keep alert for other input-output situations in the real world. More specifically, trigonometry is about right-angled triangles, where one of the internal angles is 90°. PPT TRIGONOMETRY in REAL LIFE - authorSTREAM Presentation. Trigonometry is not the work of any one person or nation. The function is assumed to be a function of time and the function values must be between -1 and 1. Crankshaft design (optimization of a function on a closed interval) Math 141 Projects, Spring 1998 1. When sound waves are produced by a musical instrument, they move in a reprising pattern and trigonometry functions such as cosine and sine can be used to represent them. In these trigonometry graphs, X-axis values of the angles are in radians, and on the y-axis its f (a), the value of the function at each given angle. The amplitudes of the cosine waves are held in the variables: a1, a2, a3, a3, etc. These include sine, cosine and tangent functions. &Assume&that&a& rider&enters&a&car&froma&platformthat&is&located&30°&around. The function has the same domain, range, and form as the sine function, but is offset by π/2 radians (90°). Fourier originally defined the Fourier series for real-valued functions of real arguments, and using the sine and cosine functions as the basis set for the decomposition. So our cosine curve will be shifted to the right by approximately `0. REAL LIFE APPLICATIONS AND REFLECTION; SUMMARY OF CONCEPT AND PROCESS - UNIT CIRCLE. There are many uses of sin,cos,tan in real life. REAL LIFE APPLICATIONS AND REFLECTION; SUMMARY OF CONCEPT AND PROCESS - UNIT CIRCLE. GIVEN y x and B: 3cos2 0,3. Identify sine, cosine, and tangent as trigonometric functions as well as relationships between side lengths in right triangles. #92 - Use sine and cosine functions to model real-life data. Learning resources you may be interested in. Our teacher asked up to come up with a "real" Example of Either a Sine Wave, or Cosine Wave. The student will evaluate trigonometric equations using trigonometric identities and special formulas. With the sine law. Pg 439 #1-8, 33, 35, 37, 43-55 odd · Sketch the graphs of basic sine and cosine functions · Use amplitude and period to sketch the graphs of sine and cosine functions · Sketch translations of the graphs of sine and cosine functions. Many other Fourier-related transforms have since been defined, extending the initial idea to other applications. Real-life applications of. The next day, we continued by using the second applet (with some of the parameters changed) to walk us through the process of graphing sine functions with translations *and* changes to period/amplitude. Derived, but not applied. Applications of this branch of mathematics in real life are many and varied. Graphs are shown in figure 4. In other words, B measures the time it takes (how fast or slow) for our function to complete a full wave/cycle, as nicely stated by Khan Academy. Of the six functions in basic trigonometry, the sine, cosine and tangent are the most important to architecture because they allow the architect to easily find the opposite and adjacent values related to an angle or hypotenuse, translating a diagonal vector into horizontal and vertical vectors. The period of the function sin(x) is 2π. The dynamic used in calculation fit with maximum precision and overflow is. I need this last bit. In most encoders, this waveform is "squared off" inter-. For a low number of counts per revo-lution the waveform approaches a triangle while at higher counts the waveform becomes sinusoidal. This is how I like to introduce sine and cosine graphs this unit (after spending time with the unit circle and rotations it is a great way to see how we get the sinusoidal graph from a circle, see my blog post here for details ). In line 4 we use the properties of cosine (cos -x = cos x) and sine (sin -x = -sin x) to simplify the. The slope of a function Powers of sine and cosine; 3. Then ask learners to. Thermal analysis. basic sine and cosine functions Section 4. More specifically, trigonometry is about right-angled triangles, where one of the internal angles is 90°. 2a: Interpret the sine function as the relationship between the radian measure of an angle formed by the horizontal axis and a terminal ray on the unit circle and its y coordinate. Unit Five: Real World Problems Example: When Light shines through two narrow slits, a series of light and dark fringes appear. They are both expressed according to the triangle on the right, where each letter. Modeling Temperature Data Name(s): Since the trigonometric functions are periodic, they are a particularly useful tool when modeling cyclic behavior. Trigonometric Functions in Real Life There are many actions that complete some sort of regular cycle periodically and can be modeled by trigonometric functions. FREE (9) Popular paid. Objectives: To graph the sine and cosine functions; to identify the graphs of the sine and cosine functions. five bio-medical science dataset and one sine dataset problems. When I consider how to address the Precalculus objectives “to solve real-life problems involving harmonic motion”. 11 (+) Understand and apply the Law of Sines and the Law of Cosines to find unknown measurements in right and non-right triangles (e. Domain, Range, and Period of the Sine Function. In this section, we explore transformations of the sine and cosine functions and use them to model real life situations. We will describe the numbered controls and their functions. You will see updates in your activity feed. As we know, in our maths book of 9th-10th class, there is a chapter named LOGARITHM is a very interesting chapter and its questions are some types that are required techniques to solve. Reply Pingback: Law of Cosines (Cont. circular functions. You and your friend had planned to walk to the movies together, but now it's five minutes to show time and no sign of him. Hyperbolic Functions in Real Life. Precalculus: An Investigation of Functions is a free, open textbook covering a two-quarter pre-calculus sequence including trigonometry. (See Example 3. Collapse menu 1 Analytic Geometry. This sheet describes the range, domain and period for each of the trig functions. My teacher gave me an example where in the sine curve was used in "biorhythms". At x = 0 degrees, sin x = 0 and cos x = 1. Introducing circular functions with. The graphs of all sine and cosine functions are related to the graphs of y =sinx and y=cos x which are shown below. 1 4 Sine Law C2. b: Apply the six trigonometric functions in relation to a right triangle to solve real-world applications and problems in mathematical settings. Real world uses of hyperbolic trigonometric functions. 1 Laboratory (Homework) Objective The objective of this laboratory is to learn basic trigonometric functions, conversion from rectangular to polar form, and vice-versa. Experiment with functions that have additional terms, and see how these change the period, amplitude, and phase of the waves. #92 - Use sine and cosine functions to model real-life data. We have already derived the derivatives of sine and. The relationship between trigonometric ratios; the circular functions sinx, cosx, and tanx; amplitude, their periodic nature, and their graphs. Line 1 just restates Euler’s formula. Find area of a sector of a circle. 9 - 5 Applications to Navigation and Surveying Try the quiz at the bottom of the page! go to quiz We can use trigonometry to work with navigation problems as well as surveying problems. θ = 90°), which gives us cos θ = 0 and. The applications of Fourier transform are abased on the following properties of Fourier transform. 13 Check It Out! Example 2c Find the measure. Trigonometry plays a major role in musical theory and production. 2-5) Complete Notes pp. Angles: Real Life Applications of. Change from roots to rational exponents. 1 – solve problems, including those that arise from real-world applications (e. Understanding Calculus is a complete online introductory book that focuses on concepts. Title: Applications of Sine and Cosine Graphs Standard(s): MA3A3. 1 Laboratory (Homework) Objective The objective of this laboratory is to learn basic trigonometric functions, conversion from rectangular to polar form, and vice-versa. Similar statements can be made for the other trigonometric functions of sums and differences. An equation that can be used to model these data is of the form: y = A cos B(x - C) + D,. 5: Graphs of Sine & Cosine Functions KHöJLO (L Gra hs of the "Parent" Functions. A discrete function is a function where both the domain and range can be listed as distinct elements in a set. This provides a breathtaking example of how a simple idea involving geometry and ratio was abstracted and developed. Besides other fields of mathematics, trig is used in physics, engineering, and chemistry. Plotting a basic sine wave. Law of sines and cosines In most of the practical applications, related to trigonometry, we need to calculate the angles and sides of a scalene triangle and not a right triangle. This would be a Fourier series with only one term, and would return the desired function with the magnitude changed. Be able to split the limits in order to correctly find the area between a function and the x axis. If you can remember the graphs of the sine and cosine functions, you can use the identity above (that you need to learn anyway!) to make sure you get your asymptotes and x-intercepts in the right places when graphing the tangent function. The linear combination of a cosine and a sine of the same angle is an expression of the form: # a cos x + b sin x # That looks very much like the sum angle formula for sine or the difference angle formula for cosine:. Sin-1 is called "inverse sine" because it does the opposite of the sine function. sine and cosine Although sin(x) and cos(x) will create an n-petaled roses inscribed in the unit circle, what is the difference between them? The graph with the sine appears tangent to the positive x axis, while the cosine version has a petal centered at the positive x axis. Sine 5 None Up 2 9. In line 3 we plug in -x into Euler’s formula. bc is the horizontal shift 4. real world examples of trigonometry Applications of Trigonometry in Real life. 15: Graphs of Sine and Cosine Functions. Many new applications from business, medicine, life and social sciences—based on current real-world data. – Hipparchus was a Greek astronomer who lived between 190-120 B. As you know, our basic trig functions of cosine, sine, and tangent can be. Recall from Graphs of the Sine and Cosine Functions that the period of the sine function and the cosine function is $$2π$$. OVERVIEW The students will learn how to interpret and graph an inverse trig. This is exactly the reason why, when the graphs are plotted on grids as above, the cosine graph is equivalent to the sine graph, omitting the fact that it is shifted, or translated, 90° to the left. Use special triangles to determine geometrically the values of sine, cosine, tangent for π/3, π/4 and π/6, and use the unit circle to express the values of sine, cosines, and tangent for x, π + x, and 2π – x in terms of their values for x, where x is any real number. We should now understand that any variable that is cyclical, harmonic, oscillating, or periodic in nature can be modeled graphically by a sine or cosine wave. You and your friends order a pizza. At the end of. Learn how to graph trigonometric functions and how to interpret those graphs. Angle in standard position. If you can remember the graphs of the sine and cosine functions, you can use the identity above (that you need to learn anyway!) to make sure you get your asymptotes and x-intercepts in the right places when graphing the tangent function. c) How many minutes, from t = 0, does it take the rider to reach the. I know for sure that pilots use the trig. These functions are most conveniently defined in terms of the exponential function, with sinh z = 1 / 2 (e z − e −z) and cosh z = 1 / 2 (e z + e. Scroll down the page for examples and solutions. Sinusoidal functions graph wave forms. Date: 12/21/98 at 13:04:53 From: Doctor Santu Subject: Re: Trigonometry and music Dear Elizabeth, Certainly the functions sine and cosine have a connection to music. 2 to 5 Tuesday 10/22 Graphing Sine and Cosine functions cont'd. Applications of sinusoidal functions Description. Plotting a basic sine wave. Find the period of a sine or cosine function. View Notes - NOTESTrigonometry 3. Let's take a look at navigation. In addition, how do I know if this the graph of sine or cosine?. The midline is the average value. The applica-tions listed below represent a small sample of the applications. 1 4 Sine Law C2. 4 The Sine and Cosine Ratio Learning Goal: Determine the measures of the sides and angles in right triangles using the primary trigonometric ratios and the Pythagorean theorem; solve problems involving the measures of sides and angles in right triangles in real-life applications. (2) I can use amplitude and period to help sketch the graphs of sine and cosine functions. 3 sin 50 Use the inverse sine function to find m?X. Trigonometric Functions Arbitrary angles and the unit circle We’ve used the unit circle to define the trigonometric functions for acute angles so far. Due to its exploration ability it has been applied to solve many real-life applications. Answers to the do-it-yourself problems are on the "Other" page as well as other helpful websites. In these trigonometry graphs, X-axis values of the angles are in radians, and on the y-axis its f (a), the value of the function at each given angle. The sine function just bounces back and forth (since you're basically tracing the outline of a circle). There is of course no point in simply copying this info here. The sine and cosine functions are among the most important functions in all of mathematics. "The sine and cosine functions are defined for all real numbers, and these functions have many real-world applications. y= cos 2x SOLUTION a. So I differentiate the sine function twice and I get, I claim minus the sine function. By thinking of sine and cosine as coordinates of points on a unit circle, it becomes clear that the range of both functions must be the interval [−1,1]. We say that this sinusoidal has a vertical shift of 1. We then use the sine rule to find the side labelled a: sin12 40 sin16 a sin12. PPT TRIGONOMETRY in REAL LIFE - authorSTREAM Presentation. 0028 Open image in a new page Graph of e = 170 cos(120πt - π/3). The hypotenuse is the longest side in a right triangle. 2 to 5 Tuesday 10/22 Graphing Sine and Cosine functions cont'd. So that part, I hope, is clear. Just as with the sine function: • the domain for the cosine function is all real numbers; • the range of the cosine function is -1 ≤ y ≤ 1; • the cosine function is periodic and will repeat this pattern over intervals of 2π. The Sine Ratio Passy's World of Mathematics. • Apply addition or subtraction identities for sine, cosine, and tangent. I know for sure that pilots use the trig. The main goal is to illustrate how this theorem can be used to evaluate various types of integrals of real valued functions of real variable. In this article, a sort of continuation, I will be discussing some applications of this formula. Analysis of beams in mechanics (polynomial integration and optimization of a function on a closed interval) 2. Law of Sine. As you see, $$y = 1 + \sin x$$ merely raises the graph of sine one unit. The law of sines is a formula that helps you to find the measurement of a side or angle of any triangle. Double angle identities for sine and cosine. If it helps, you can draw a rough sketch to view this triangle, but this is optional. When we first learn the sine and cosine functions, we learn how to use them to find missing side-lengths & angles in right-angled triangles. The student is expected to: (A) use functions such as logarithmic, exponential, trigonometric, polynomial, etc. This would be a Fourier series with only one term, and would return the desired function with the magnitude changed. 2 to 5 Tuesday 10/22 Graphing Sine and Cosine functions cont'd. The sine and cosine graphs. In mathematics, the trigonometric functions (also called circular functions, angle functions or goniometric functions) are real functions which relate an angle of a right-angled triangle to ratios of two side lengths. We check out triangle ABD and work out that angle ADB = 49°. and "use sine and cosine functions to model real-life data," iii. Applications of this branch of mathematics in real life are many and varied. A Fourier series is a way of representing a periodic function as a (possibly infinite) sum of sine and cosine functions. 4)AFerris&wheel&has&a&diameter&of&20&mand&is&4&mabove&ground&level&at&its&lowest&point. graphs of sine and cosine functions. 42 on the original plot we get: Looks as though we are almost there. 03 Real-world Applications of Trigonometric Functions. He is considered the father of trigonometry. Let's take a look at navigation. Peggy and Raymond then present groups with several real-world situations to work with. 5 –Graphs of Sine and Cosine Functions What You’ll Learn: #89 - Sketch the graphs of basic sine and cosine functions. (sine or cosine) inputs. 6ep7yrwj7rao, o7e5upezd1u3u, gw9joobvhirz2sm, w5yw1hcyy3, hgc3hy119dy, x0bluc6ksm4p, nztz99cmdv1t, ihnhyyxs5u9, f6mhu9u964bou, z52phaeu284nuz, ue99gp4bhf, 3x0oekbpvdy, jm2sczb2mnj1, lr41lj4kntz, bknyyhwsu4v, xlgusjsmamqhbf, olijtkg53n, n0oq7e9mjj, lordvnua01, xat1ajnfbhtu1j, ct63zp7jwogh, odkeljziwabr, 2otnx4k3rxnfck8, b6d986skqf, sbxpdkzn2rrj, 1hpc8sts6g2up, k4lj0jtibfrei, 19ltt17gps | 2020-05-25T20:54:06 | {
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https://math.stackexchange.com/questions/2137223/solving-the-congruence-x2-equiv-4-mod-105-is-there-an-alternative-to-using/2137308 | # Solving the congruence $x^2 \equiv 4 \mod 105$. Is there an alternative to using Chinese Remainder Theorem multiple times?
I'm trying to solve $$x^2 \equiv 4 \mod 105.$$
This is of course equivalent to $$(x+2)(x-2) \equiv 0 \mod 105$$
which is also equivalent to the system of congruences
$$(x+2)(x-2) \equiv 0 \mod 3$$ $$(x+2)(x-2) \equiv 0 \mod 5$$ $$(x+2)(x-2) \equiv 0 \mod 7$$
which have solutions of $$x \equiv 1\ \text{or}\ 2$$ $$x \equiv 2\ \text{or}\ 3,\ \text{and}$$ $$x \equiv 2\ \text{or}\ 5$$ respectively.
Now, I could in principle take each combination of {$1,2$}, {$2,3$}, and {$2,5$} and use the Chinese Remainder Theorem to solve each system, but that seems incredibly tedious.
Is there a simpler way?
I'll note that this is a homework question, so it seems likely that there's a trick. Unfortunately I can't spot one.
• no trick....................................At the same time, some of your writing suggests you don't know what happens. What are the values $\pmod {15},$ just combine results for $3,5$ before going on – Will Jagy Feb 9 '17 at 21:14
• I do understand that I can just use the values that I obtain $\mod 15$ and then solve the system involving those values and the values $\mod 7$. Is that what you're getting at? – Alex Feb 9 '17 at 21:19
• sure. How many values $\pmod {15},$ and what are they? – Will Jagy Feb 9 '17 at 21:21
There are a couple ways to optimize. First you need only compute half of the $8$ combinations since if $\,x\equiv (a,b,c)\bmod (7,5,3)\,$ then $\,-x\equiv (-a,-b,-c)\bmod (7,5,3).\$ Then use CRT to solve it for general $\,a,b,c\,$ to get $\,x\equiv 15a+21b-35c\,$ Use that to compute those $4$ values. It's very easy
e.g. note $\ x\equiv (2,\color{#0a0}{-2},\color{#c00}2)\equiv 15(2)+21(\color{#0a0}{-2})-35(\color{#c00}2)\equiv 23\pmod{105}$
Negating it $\ (-2,\color{#0a0}2,\color{#c00}{-2})\equiv -x\equiv -23\equiv 82\pmod{105}.\$ Of course $\pm(2,2,2)\equiv \pm2.\$
Do as above for $\,(-2,2,2),\ (2,2,-2)\$ to get the other$\,4\,$ solutions (a couple minutes work).
Remark There are also other ways we can exploit the negation symmetry on the solution space, i.e. if $x$ is a root so too is $-x$ since $\,x^2\equiv 2\,\Rightarrow\, (-x)^2\equiv x^2 \equiv 2\pmod{\!105}.\,$ Below is one such method, selected primarily because it reveals how to view Rob's answer in CRT language.
By CCRT = Constant case CRT: $\,x\equiv (2,2,2)\pmod{\!7,5,3}\iff x\equiv 2\pmod{\!105}.$ Its negation is $\,(-2,-2,-2)\,$ corresponding to $\,-2\pmod{\!105}.\,$ For other "nontrivial" solutions, either $x$ or $-x$ has one entry $\equiv -2$ and both others $\equiv 2,\,$ say $\, x\equiv (-2,2,2)\pmod{p,q,r}.\,$ Again by CCRT $\,x\equiv (2,2)\pmod{q,r}\iff x\equiv 2\pmod{qr}$ by $\,q,r\,$ coprime. So we reduce from $3$ to the $2$ congruences below. Solving them by Easy CRT, using $p$ coprime to $qr,\,$ we get
\quad\ \ \begin{align} x&\equiv -2\!\pmod p\\ x&\equiv\ \ \,2\!\pmod{qr}\end{align}\!\iff x\equiv 2\ +\ qr\left[\,\dfrac{-4}{qr}\ \bmod\ p\,\right]\pmod{pqr}\,
\qquad \qquad\qquad\qquad\! \begin{align} p=7\,\ \Rightarrow\,\ x &\equiv 2 + 3\cdot 5(-4/(\color{#c00}{3\cdot 5)})\bmod 7)\equiv 2+3\cdot 5(3)\equiv 47\\[.2em] p=5\,\ \Rightarrow\,\ x&\equiv 2 + 3\cdot 7(-4/(\color{#c00}{3\cdot 7}))\bmod 5)\equiv 2+3\cdot 7(1)\equiv 23\\[.2em] p=3\,\ \Rightarrow\,\ x&\equiv 2 + 5\cdot 7{(-4/\!\!\!\underbrace{(\color{#0a0}{5\cdot 7})}_{\large \equiv\ \color{#c00}{1}\ {\rm or}\ \color{#0a0}{-1}}}\!\!\!)\bmod 3)\equiv 2\color{}{ +}5\cdot 7(1)\equiv 37\\ \end{align}
We arranged the above to exploit easy inverses (of $\,\color{#c00}{1}$ or $\color{#0a0}{-1})\,$ just as in the first solution (cf. my comment below). So $\,47,23,37\,$ and their negatives $\,58,82,68\,$ are all the nontrivial solutions.
The method in Rob's answer is essentially equivalent to the above (without the CRT language), except it doesn't take advantage of the easy inverses, instead solving the congruences by brute force (sometimes this may be quicker than general methods when the numbers are small enough).
There is also another CRT optimization used implicitly in Rob's answer. Namely a change of variables $\ y = x\!-\!2\,$ is performed to shift one of the congruences into the form $\,y\equiv 0,\,$ which makes it easy to eliminate explicit use of CRT. We show how this works for the prior congruences, using the mod Distributive Law $\,ca\bmod cn =\, c(a\bmod n)\quad\qquad$
$\qquad qr\mid x\!-\!2\,\ \Rightarrow\,\ x\!-\!2\bmod{pqr}\, =\, qr\!\!\!\!\!\!\overbrace{\left[\dfrac{x\!-\!2}{qr}\bmod p\right]}^{\large\quad\ \ x\ \equiv\ -2\pmod{p}\ \ \Rightarrow}\!\!\!\!\!\!\! =\, qr\left[\dfrac{-4}{qr}\bmod p\right]$
That's the same solution for $\,x\!-\!2\,$ that Easy CRT gave above. So the mod Distributive Law provides a "shifty" way to apply CRT in operational form - one that often proves handy.
• Thanks. This seems like a bit more reasonable of an approach than just explicitly using CRT $8$ times. It's not entirely clear to me why we have $x \equiv 15a+21b-35c$. I recognize that those are the three different ways we can multiply {$3,5,7$}, but why? – Alex Feb 10 '17 at 14:51
• @Alex It follows by applying the general CRT formula, namely \begin{align} x\ &\equiv\ (a,b,c)\!\!\pmod{7,5,3}\\[.3em] \overset{\rm CRT}\iff x\ &\equiv\ a(3\cdot 5)\left[\dfrac{1}{3\cdot 5}\right]_7 + b(3\cdot 7)\left[\dfrac{1}{3\cdot 7}\right]_5 + c(5\cdot 7) \left[\dfrac{1}{5\cdot 7}\right]_3 \pmod{105}\\[.3em] &\equiv \quad a\,(15)\,(1/1)_7\ \ +\ \ b\,(21)\,(1/1)_5\ \ +\ \ c\,(35)\,(1/(-1))_3\\[.3em] &\equiv\ \ 15\,a\ +\ 21\,b\ -\ 35\,c \end{align} Note that the problem was designed to make this very easy, i.e. all the inverses are inverses of $\pm1.$ $\quad$ – Bill Dubuque Feb 10 '17 at 15:42
• What you describe as a computation by "brute force" in my answer amounts to a quick pass over a list of $12$ natural numbers $n$ less than $100$ to decide whether $n\pm4$ is a multiple of one of $3$, $5$, $7$. You seem to be doing much more calculation than I did. – Rob Arthan Feb 10 '17 at 23:05
• @Rob First, "brute force" is a very common name for such exhaustive searches. It's not meant to denigrate such methods (which are often useful). Second, if you show all your work (as I do) then it will be of comparable length (if not more). Third, the above is efficient for any size numbers, but brute force searching is not. Finally, above I do not aim to optimize brevity, Rather, I aim to highlight the essence of the matter, from a more general viewpoint. – Bill Dubuque Feb 10 '17 at 23:14
• @BillDubuque: first: the essence of this problem to me is "how can a quadratic equation have more than two roots in a commutative ring?". Second: tabulating the quick pass over a list of 12 small natural numbers is much quicker than any application of the CRT formula. – Rob Arthan Feb 10 '17 at 23:26
Here's a method for avoiding CRT altogether (which comes from thinking about why there can be more than two roots to a quadratic in a commutative ring). If $x$ is any solution, we have $$(x - 2)(x + 2) \equiv 0 \bmod 105.$$ So in addition to the easy solutions $x = 2$ and $x = 103$, we get a solution $x = y + 2$ for any $y$ such that $$y(y + 4) \equiv 0 \bmod 105.$$ I.e., for any $y$ such that $y$ and $y + 4$ are a "complementary" pair of zero divisors in the ring $\Bbb{Z}_{105}$. Any such pair has one of the following forms (possibly with the order of the factors reversed). \begin{align*} 3m &\cdot 35n & \quad &\mbox{with 1 \le n \le 2}\\ 15m &\cdot 7n & \quad &\mbox{with 1 \le m \le 6}\\ 5m &\cdot 21n & \quad &\mbox{with 1 \le n \le 4} \end{align*} It doesn't take long to work through the $2+6+4$ possibilities for $y$ or $y + 4$ to get that $y(y+4)$ must be one of:
$$21 \cdot 25\\ 35 \cdot 39 \\ 45 \cdot 49\\ 56 \cdot 60\\ 66 \cdot 70\\ 80 \cdot 84$$
Giving 23, 37, 47, 58, 68 and 82 as the not-so-easy solutions.
• As someone currently taking a ring theory class, I think this solution is really neat. Could you clarify why we have bounds on $n$ and $m$ in the third chunk of centred math? I don't exactly follow why a "complementary pair of zero divisors" must have one of those forms. It makes sense to me that you can split up the prime factors $3,5,7$ but I can't figure out why the $n$ and $m$ are necessary, and how they're chosen. – Alex Feb 9 '17 at 22:56
• @Alex: I think you understand that to get a non-trivial solution to $ab=0$ in $\Bbb{Z}_{105}$, you need to split the prime factors of $105$ between $a$ and $b$. Then you can multiply either $a$ or $b$ by any number you like, but as the equivalence class mod $105$ is all that matters, you only need to look at multiples between $0$ and $104$. My case analysis just picks the bigger of $a$ and $b$ to give the smallest number of cases. Then you just test whether the resulting multiple of $a$ (or $b$) $\pm4$ is a multiple of the "other" factor. – Rob Arthan Feb 9 '17 at 23:09
• @Alex Though CRT is not explicitly used above, in fact the method is a special case of general CRT optimizations, as I describe in the Remark in my answer. If one does many manual CRT calculations it proves handy to be familiar with these and related ideas. – Bill Dubuque Feb 10 '17 at 22:32 | 2019-12-13T10:50:07 | {
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https://math.stackexchange.com/questions/3116473/selecting-2-cards-from-a-full-deck-of-cards | # Selecting $2$ cards from a full deck of cards.
I thought of a problem earlier and I am quite clueless on how to solve it, or begin solving it, because I cannot find a way to easily compute the amount of combinations of $$2$$ cards that sum to a certain value $$x$$. Anyway, the first part of the problem is as follows
We have a full deck of $$52$$ cards, and randomly select $$2$$ cards from this deck. We look at the cards and compute the sum of the values of the cards, ace being $$1$$ and K, Q and J being $$13$$, $$12$$ and $$11$$ respectively. We then shuffle the cards back into the deck and randomly select $$2$$ cards once more. What is the probability that the sum of the value of these $$2$$ cards, is the same as the sum of the values of the first $$2$$ selected cards?
This brought me to think of another problem, which is comparable. It is as follows
Let's say we have a full deck of $$52$$ cards, we randomly select $$2$$ cards, and we do this twice, yielding $$2$$ sets of $$2$$ cards. What is the probability that the sum of the values of the cards in the first set, equals the sum of the values of the cards in the second set?
Again, I'm quite confused about this problem, because I cannot think of an easy way to compute the amount of possible configurations of two cards, that sum to some value $$x$$.
Any help on solving these problems is appreciated. Furthermore, what would be a good guesstimate for these probabilities that could be given without any computations?
• Have you tried to work out the number of combinations for any specific sums? How many combinations add up to $2?$ What about $3?$ Try a few examples and look for pattern. – saulspatz Feb 17 at 17:34
• I'm thinking a decent guesstimate is in the area of $1/338 = 1/(2 \cdot 13^2)$. – Robert Shore Feb 17 at 17:35
• @robjohn As a guesstimate or as a result of a computation? – S. Crim Feb 17 at 18:29
• @S.Crim: It was a computation, but I had a replacement in there that shouldn't have been, so I need to compute again. – robjohn Feb 17 at 18:35
• @saulspatz I've managed to solve the first problem, but am now stuck on the second problem because it is confusing to me what happens when the sum is unknown. – S. Crim Feb 17 at 18:36
For an interview, I'd try to compare the second scenario to the first.
Say that you draw $$6$$ and $$3$$ the first time. If you drew from a separate deck, you'd have $$16$$ ways to make $$9,$$ but if you draw from the same deck, you only have $$9$$ ways. In the worst case you draw two aces, or two Kings, and the number of ways to succeed goes down from $$6$$ to $$1.$$ In the best case, you draw two $$7$$s and the number of ways goes down from 96 to 91. $$102$$ to $$97$$.
Of course, the better cases, near the middle will occur more frequently than the worse cases, near the ends, so I would guess that the probability of success in the second case might be about $$\frac34$$ of the probability of success in the first case.
Drat! Now I'm going to have to work them out and see how close I came.
EDIT
In response to the OP's that we don't have an explicit answer for the first problem either, in an interview, I would do the same kind of estimation. There are $${52\choose2}={52\cdot51\over2}\approx 1300$$ ways to chose two cards. For a sum of $$2$$ we must choose two Aces, and there are only $$6$$ ways to do that, so about half a percent chance of success. If the sum is $$14,$$ we have $$102$$ chances of success, so around $$8\%.$$ Of course, the sums near $$14$$ will arise more frequently than the sums near $$2$$ or $$26$$ so they should be weighted more heavily. I'd guess around $$5$$ percent.
With one deck, the probability of success is $${328888\over6497400}\approx .05062$$ and with two decks, the probability of success is $${365392\over7033104}\approx.05195$$ I computed the exact probabilities with a script, and confirmed them through simulation.
• Hah! That sounds good. However, for the first scenario, we cannot give an explicit probability, e.g. 60%, right? Because our probability depends on the outcome of the first sum, as you stated. So then how exactly would we compare the two situations in terms of probability of getting equal sums? – S. Crim Feb 17 at 20:30
• @S.Crim I was assuming that assuming that you'd already somehow come up with answer for the first scenario. In one of your comments, you said "I've managed to solve the first problem." I'll add some more to my answer. – saulspatz Feb 17 at 21:38
• Yeah I solved the first problem to the point where, knowing the outcome of the sum, I can compute the probability of drawing this sum again after reshuffling the drawn pair back in. So I cannot compute some explicit formula or something for the probability yet. – S. Crim Feb 17 at 21:56
Hint:
$$\begin{array}{|l|l|l|l|l|l|l|l|l|l|l|l|l|l|} \hline + & \color{red}{1} & \color{red}{2} & \color{red}{3} & \color{red}{4} & \color{red}{5} & \color{red}{6} & \color{red}{7} & \color{red}{8} & \color{red}{9} & \color{red}{10} & \color{red}{11} & \color{red}{12} & \color{red}{13} \\ \hline \color{red}{1} & 2 & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 \\ \hline \color{red}{2} & 3 & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 \\ \hline \color{red}{3} & 4 & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 \\ \hline \color{red}{4} & 5 & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 \\ \hline \color{red}{5} & 6 & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 \\ \hline \color{red}{6} & 7 & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 \\ \hline \color{red}{7} & 8 & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 \\ \hline \color{red}{8} & 9 & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 \\ \hline \color{red}{9} & 10 & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 \\ \hline \color{red}{10} & 11 & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 & 23 \\ \hline \color{red}{11} & 12 & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 & 23 & 24 \\ \hline \color{red}{12} & 13 & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 & 23 & 24 & 25 \\ \hline \color{red}{13} & 14 & 15 & 16 & 17 & 18 & 19 & 20 & 21 & 22 & 23 & 24 & 25 & 26 \\ \hline \end{array}$$
The row of red numbers is the possible values for the 1st card, the column of red numbers is the possible values for the 2nd card. How many ways are there to get, say, a total sum of $$9$$?
• I see, very helpful. This is probably a good tool for me to try and find the solution. Thanks alot! – S. Crim Feb 17 at 18:14
• This allowed me to find the solution to first scenario, but I'm still stuck on the second one. Because if we do not know the value of the sum yet, then how can we compute the probability that the two sums are equal? I imagine we again have to look at possibilities or amounts of configurations but I'm not sure. – S. Crim Feb 17 at 18:20
• Use the Law of Total Probability. – Graham Kemp Feb 18 at 22:46
On the problem you thought about
We select cards $$c_1, c_2, c_3, c_4$$ from the same deck; there are $$U:=52\cdot 51\cdot 50\cdot 49= 6497400$$ possible cases, and we wish to count the cases for which $$N(c_1)+N(c_2)=N(c_3)+N(c_4) \label{eq1}\tag{1}.$$
Let $$V$$ denote the count of cases that satisfy $$(\ref{eq1})$$, and let us start by figuring out how many cases there are that satisfy $$N(c_1)+N(c_2) =n= N(c_3)+N(c_4),\label{eq2}\tag{2}$$ for some fixed $$n$$:
As robjohn says, there are $$W(n)=208-16\,|n-14|-4\,[\,2\mid n\,]$$ ways we can select $$c_1, c_2$$ so that they sum to $$n$$. (The formula for $$W(n)$$ should be clear from the square provided by cansomeonehelpmeout.) Fix such a choice of $$c_1, c_2$$; if we want to continue selecting $$c_3, c_4$$ such that they sum to $$n$$ too, then they must be chosen from the remaining ways $$W(n)-Q,$$ where $$Q$$ is the number of ways to select two cards that sum to $$n$$ such that at least one of the two cards is either $$c_1$$ or $$c_2$$. When $$N(c_1)\neq N(c_2)$$ (which is always true if $$n$$ is odd), contemplating on some specific cases, it does not take much effort to realize that $$Q$$ is always $$14$$. On the other hand, if $$n$$ is even, of all the choices we could have made for $$c_1, c_2$$, exactly $$12$$ of them have $$N(c_1)=N(c_2)$$, and for each one of these we have $$Q=10$$ and not $$14$$. Therefore, we let $$Q=14$$ and we add $$(14-10)\cdot12$$ to the total when $$n$$ is even to compensate for the extras we have subtracted, and we get the number of cases satisfying $$(\ref{eq2})$$ as $$W(n)(W(n)-14)+4\cdot 12\,[\,2\mid n\,]. \label{eq3}\tag{3}$$
All we need to do now is to sum $$(\ref{eq3})$$ over all possible $$n$$: $$V=\sum_{n=2}^{26}\left(W(n)(W(n)-14)+4\cdot 12\,[\,2\mid n\,]\right).$$ The probability that we get $$(\ref{eq1})$$ will then be: $$\frac{V}{U}.$$
Note: The computation by hand is time consuming. I would use a program to calculate the sum and give a final answer, but the software I use is a very old version of Maple, and it is very cumbersome.
Ways to draw $$n$$ with $$2$$ cards: $$W(n)=\overbrace{208-16\,|n-14|}^{16\,(13-|n-14|)}-4\,[\,2\mid n\,]$$ That is, there are $$13-|n-14|$$ ways to choose $$1\le a,b\le13$$ so that $$a+b=n$$. There are $$4\cdot4=16$$ card choices for $$a$$ and $$b$$, except if $$2\mid n$$ and $$a=b$$, when there are only $$4\cdot3=12$$, so we subtract $$4$$ if $$2\mid n$$. $$\sum_{n=2}^{26}W(n)=2652$$ The probability of drawing $$n$$ is then $$\frac{W(n)}{2652}$$ The probability of drawing the same thing twice in a row, replacing between draws, is then $$\frac1{2652^2}\sum_{n=2}^{26}W(n)^2=\frac{22837}{439569}\doteq0.051953$$
• Thanks alot for the answer. Could you elaborate a little more on the formule you gave for $W(n)$? What does the $208$ mean, and what exactly does it mean that something only occurs when $2|n$? Perhaps I am just unfamiliar with notations. – S. Crim Feb 18 at 15:08
• @S.Crim: I have hopefully made the explanation better. – robjohn Feb 19 at 8:05
• Thanks alot, that indeed clarifies some things. – S. Crim Feb 19 at 14:45
• @S.Crim: Is something still unclear? $[\dots]$ are Iverson Brackets and $a\mid b$ means that $a$ divides $b$; e.g. $[\,2\mid n\,]$ is $1$ when $n$ is even and $0$ when $n$ is odd. – robjohn Feb 19 at 15:05
• I think I get it now, yeah. Thanks for the clarifications on the notation. I was unfamiliar with the Iverson Brackets before this, and its been a while since I've worked with the $a|b$ notation. Thanks alot! – S. Crim Feb 19 at 17:45 | 2019-04-24T06:32:25 | {
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http://shey.madzuli.it/regular-polygon-questions.html | # Regular Polygon Questions
24, 25 Homework Help Interior Angles of Regular Polygons Find the measure of an interior angle of the regular polygon. Polygon Questions and Answers - Math Discussion. The Area of a Trapezoid [1/27/1996] My daughter has been assigned to derive the formula for the area. Use your knowledge of the sums of the interior and exterior angles of a polygon to answer the following questions. 180(x-3) + 180 C. This is of course made far more obnoxious by the fact that IsPolygon is a valid rule token, so your rule will seem to be correct, and even pass the rule-checker, but silently fail when you try to actually repour the. Practice questions. 21-23 Example 4: Exs. Ask the class if they agree with where everyone placed their shapes or if they disagree to explain why they think a shape should be moved. Read each question carefully before you begin answering it. Use whatever is shorter in your language. Each interior angle of a regular polygon is the sum of the interior angles of the polygon divided by the number of sides. Is this a inductive reasoning or deductive re. Polygons are multi-sided shapes with different properties. Also, l == [1] * n (1 repeated n times) case is just the regular polygon untouched, which is a valid input. Poly- means "many" and -gon means "angle". Example: How many sides does a regular polygon have if one interior angle. Use Microsoft Paint to create a polygon picture. Name the quadrilaterals shown: rhombus, parallelogram, square, rectangle, or trapezoid. Polygons Questions and Answers - Math Discussion. You can only use the formula to find a single interior angle if the polygon is regular!. Buuuut then you had some questions about the quant section—specifically question 13 of the second Quantitative section of Practice Test 1. Age 11 to 14 Short. Example of how to draw regular polygons using a \foreach loop inside a path. Polygons are two-dimensional, closed, plane shapes composed of a finite number of straight sides that meet at points called vertices. Each student will receive an Area of Irregular Polygons Mini Lesson Examples. 360/x -10x Kudos for the right answer and explanation. Have students come up to the T-chart and place their shapes on the correct side. Enter the number of sides on the polygon. A polygon with the fewest number of sides is a triangle. The moral of this story- While you can use our formula to find the sum of. If you attempt to use geoshape or geotrace questions in forms authored prior to 3. Regular polygon. The problem concerns a polygon with twelve sides, so we will let n = 12. Return from the Polygon Game page to the Math Play homepage. 9 degrees are in the measure of an interior angle of a regular seven sided polygon. Regular Polygons Printable regular polygon sheet which includes picture and name of each of the 10 polygons from triangle, through to dodecagon. Read each question carefully before you begin answering it. Videos, worksheets, 5-a-day and much more. In this video, we'll talk about a few things you need to know about polygons for the GMAT, some of their basic names of regular polygons, and the sum of the interior angles of a polygon. Polygons can be regular or. Polygon diagonals. Find out which polygons are regular and which are not. Polygons problems with solutions for CAT exam. (The other option is to use shape=regular polygon from \usetikzlibrary{decorations. 652 Responses. Download the set (5 Worksheets). The definition of a regular polygons is a polygon that is both equiangular and equilateral. Central angle = 20° Supposing that the polygon is regular, that's all sides are of equal length and all central angles are equal. Formula to find the sum of interior angles of a n-sided polygon is = (n - 2) ⋅ 180 ° By using the formula, sum of the interior angles of the above polygon is = (9 - 2) ⋅ 180 ° = 7 ⋅ 180 ° = 126 0 ° Formula to find the measure of each interior angle of a n-sided regular polygon is. It is the point which is equidistant from each vertex. Unanswered Questions. Angles of Polygons Interior and Exterior Angles of a Polygon After all, there are only so many concepts that are "fair game" on the GMAT, so some questions will be designed to be more difficult by putting these concepts together in unexpected ways. Find the area of a regular hexagon with side length of 10 cm and apothem length of 8. You will have to read all the given answers and click over the correct answer. The number of these congruent triangles is the same as that of its sides. A regular polygon is a polygon that is both equiangular and equilateral. Quiz questions focus on the types of polygons and the. Vocabulary. A regular polygon is a closed, 2-dimensional figure consisting exclusively of line segments of equal length. 6-1 Polygons & Review 29 TEST Friday, 1/25/13 6-1: Properties and Attributes of Polygons I can name polygons with up to ten sides. Consider a random irregular convex polygon, for example, the 6-side polygon I want to define a function that, given a certain parameter r (roundness), rounds each. A regular polygon is a bounded, not self-intersecting polygon in which all edges have the same length and all interior angles are identically, say, $\theta$. 3 of the tikz manual. Regular Polygons A regular polygon has lines that are all the same length and it also has all the same angles. This is part of our collection of Short Problems. Each interior angle of a regular polygon is the sum of the interior angles of the polygon divided by the number of sides. in a kite not all the sides are the same then its angles are not all equal then it is not a regular polygon. A regular polygon has all sides and angles equal (e. Area= (2)(apothem)(perimeter) is the formula for finding the area of a regular polygon. Try to get 100%. My goal is to build a map as this one, where the color represents the number of overlapping polygons, here from 0 to 6:. Level 1 Level 2 Level 3 Exam-Style Description Reminders More Angle Activities Fill in the table with the names and angle sizes of the first eight regular polygons. Practice questions. We can make "pencilogons" by aligning multiple, identical pencils end-of-tip to start-of-tip together without leaving any gaps, as shown above, so that the enclosed area forms a regular polygon (the example above left is an 8-pencilogon). THE NUMBER OF INTERSECTION POINTS MADE BY THE DIAGONALS OF A REGULAR POLYGON BJORN POONEN AND MICHAEL RUBINSTEIN Abstract. I can classify a polygon as concave or convex and regular or irregular. Matching Polygons Matching game for kindergarten, preschool and 1st grade. 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For example if a polygon has 41 sides, it would be called a 41-gon. …And how about the pentagon…which kinda looks like a home plate. I also make them available for a student who wants to do focused independent study on a topic. Examples of regular polygon: In the adjoining figure of an equilateral triangle ABC there are three sides i. Last week we looked at questions on polygons inscribed in a circle. Convex or Concave Polygons A polygon is either convex or concave. the polygon is the same length and all of its angles have the same degree both equilateral and e Design a class named RegularPolygon that contains: A private int data field named n that defines the number of sides in the polygon. Both equilateral and equiangular conditions are required. Add ScratchPad. † Can any quadrilateral tile the plane? † Can any irregular polygon tile the plane?. Preview and details. Videos, worksheets, 5-a-day and much more. A polygon is called a REGULAR polygon when all of its sides are of the same length and all of its angles are of the same measure. Prepare a Presentation. Grade 3 geometry worksheet: Properties of polygons Author: K5 Learning Subject: Grade 3 Geometry Worksheet Keywords: Grade 3 geometry worksheet 2D shapes, lines, angles, parallel, area, perimiter Created Date: 4/28/2017 12:38:52 PM. The Overflow Blog How to develop a defensive plan for your open-source software project. Find the interior angle sum for each polygon. Now that we’ve been through all of our polygon rules and formulas, let’s look at a few different types of polygon questions you’ll see on the SAT. Coloring Polygons Worksheets. Exam Style Questions Ensure you have: Pencil, pen, ruler, protractor, pair of compasses and eraser 10. If you're seeing this message, it means we're having trouble loading external resources on our website. Find out which polygons are regular and which are not. Definition of Polygon explained with real life illustrated examples. org are unblocked. Read each question carefully before you begin answering it. A hexagon (six-sided polygon) can be divided into four triangles. Concave polygon. What is the value So, you were trying to be a good test taker and practice for the GRE with PowerPrep online. The dodecagon would have sides of alternating lengths, with ratio $4:\sqrt{12}$; as a specific size for the 30-60-90 triangle is not given. Being equilateral is not enough. Buuuut then you had some questions about the quant section—specifically question 13 of the second Quantitative section of Practice Test 1. 30x+180/x D. Each Interior Angle of a Regular Polygon - MathHelp. Grade 5 Other Polygons CCSS: 5. A regular polygon is one that has equal sides. † Can any quadrilateral tile the plane? † Can any irregular polygon tile the plane?. A "Polygon" object contains the polygon outline. Examples of regular polygons include the equilateral triangle and square. Name the quadrilaterals shown: rhombus, parallelogram, square, rectangle, or trapezoid. User: Approximately how many degrees are in the measure of an interior angle of a regular seven sided polygon? 102. 👍If you like this resource, then please rate it and/or leave a comment💬. Videos, worksheets, 5-a-day and much more. of sides = 360°/20° = 18. We show a quadrilateral and a pentagon below. Regular polygon D. The vertices of an N-vertex polygon are located at the angles (2*Math. a square is a regular quadrilateral). Theorems include: opposite sides are congruent, opposite angles are congruent, the diagonals of a parallelogram bisect each other, and conversely, rectangles are parallelograms with congruent diagonals. How to use polygon in a sentence. What is the probability that the center of the circle lies inside the. Instructions. Chapter 12. The Corbettmaths Practice Questions on Frequency polygons. The measure of an exterior angle of a convex regular polygon is 45. It is easy to show [39] that there are just five possibilities (vertex figure = triangle, square, tetrahedron, octahedron and cube) and each leads to a unique net. 30x+180/x D. A polygon is a shape that has any number of straight sides, such as a triangle, square or hexagon. Regular polygons have congruent edges and congruent. Central angle = 20° Supposing that the polygon is regular, that's all sides are of equal length and all central angles are equal. I can classify a polygon as concave or convex and regular or irregular. provided a robust regular polygon detection method [4] [5], which derived the α posteriori probability for a mixture of regular polygons and thus the probability density function. A Polygon is a closed figure made up of lines segments (not curves) in two-dimensions. So, if we know all the interior angles other than x, then we can find x. A hexagon has 6 sides. Return from the Polygon Game page to the Math Play homepage. There are 9 polygon maths worksheets. Length = Students who took this test also took : Parallelograms (8. (Must be 18 years old to sign up. Print full size. Area of Polygons Choose the best answer. A polygon in which all the sides are the same length and all the measures have the same measure is regular polygon. Decagons may be irregular or regular. An octagon is any eight-sided polygon, and the sum of its angles is 1080°, as we saw above. If you are not sure about the answer then you can check the answer using Show Answer button. PRACTICE: Polygons – Assignment Worksheet Monday, 1/28/13 6-1: Properties and Attributes of Polygons. A minimum of three line segments are required for making a closed figure, thus a polygon with a minimum of three sides is known as Triangle. The problem concerns a polygon with twelve sides, so we will let n = 12. Selecting Objects Inside Polygon Is there a way within a source drawing to select all objects within a predetermined boundary/polygon? The regular select routine is limited to defining a polygon but not selecting an existing one. The Area of a Trapezoid [1/27/1996] My daughter has been assigned to derive the formula for the area. Similar polygons worksheet answers. Award Information. The simplest polygons are triangles (three sides), quadrilaterals (four sides), and pentagons (five sides). For this lesson, students will find the area of irregular polygons by separating it into smaller identifiable regular polygons. A regular polygon has equal length sides with equal angles between each side. You can now earn points by answering the unanswered questions listed. Instructions. Calculators may be used. Typical Polygon Questions. Irregular polygons are polygons that have unequal angles and unequal sides, as opposed to regular polygons which are polygons that have equal sides and equal angles. 18 sides 8. [1] 6) One interior angle of a regular polygon is 135°. An irregular polygons have sides of differing length and angles of differing measure. So, the above regular polygon has 9 sides. A "Polygon" object contains the polygon outline. Polygons are multi-sided shapes with different properties. In addition to the examples given by Karen: octagons (eight sides), nonagons (nine sides), decagons (ten sides), hendecagons (eleven sides), dodecagons (twelve sides), tridecagons (thirteen sides), tetradecagons (fourteen sides), pendecagons (fift. Do you really know everything about these 2-D polygons and. In these applets, a pentagon and a triangle will be used to illustrate this definition. Learn regular+polygon geometry with free interactive flashcards. How many diagonals does an octagon have?. Question: regular polygon Tags are words are used to describe and categorize your content. Identifying shape attributes, Understanding polygons. The word 'apothem' can also refer to the length of that line segment. Expressions are given for two side lengths. A regular polygon is always convex. Categories of polygons based on the number of sides, and whether they are concave or convex. Show Answer Example 3. Each page is designed with the latest design principles. So a rotation of 72, 144, 216, 288, 360,. Higher Tier. next solving the apothem of octagon. Sum of the interior angles of a polygon = (N - 2) x 180° The number of diagonals in a polygon = 1/2 N(N-3) The number of triangles (when you draw all the diagonals from one vertex) in a polygon = (N - 2) Polygon Parts. Khan Academy is a 501(c)(3) nonprofit organization. Draw the lines of symmetry for each regular polygon, fill in the table including an expression for the number of lines of symmetry in a n-sided polygon. Learn vocabulary, terms, and more with flashcards, games, and other study tools. A Lesson for Kindergarten, First, and Second Grade. In this worksheet, decide whether a polygon is regular or irregular. com, a math practice program for schools and individual families. Polygons are 2-dimensional shapes. All regular star polygons are non-convex polygons. If a regular polygon has x angles each measuring q degrees, then what is the value of q ? A. The Corbettmaths Practice Questions on Frequency polygons. For example, a triangle has 3 sides and 3 angles. Examples of regular polygon: In the adjoining figure of an equilateral triangle ABC there are three sides i. Like 0 Dislike 0 Reply Quote Follow. General Questions: 1. Common Core Standards: Grade 3 Geometry, Grade 4 Geometry. Polygon features One Click Demo Install, so demo content can be installed in minutes. Do you really know everything about these 2-D polygons and. MathScore EduFighter is one of the best math games on the Internet today. Videos, worksheets, 5-a-day and much more. First, take a look at the question itself: The hexagon above has interior angles whose measures are all equal. PRACTICE: Polygons – Assignment Worksheet Monday, 1/28/13 6-1: Properties and Attributes of Polygons. A triangle with all sides and angles equal is known as an equilateral triangle. The following diagram gives the formula to find the area of a regular polygon using the perimeter and the apothem. We also compute the number of regions formed by. I passed my intro programming class with a B because my prof gave us 20 min. 3 A polygon has 2 or more sides and angles. Examples: Regular: Not regular: More Geometry Subjects Circle Polygons Quadrilaterals Triangles Pythagorean Theorem Perimeter Slope Surface Area Volume of a Box or Cube Volume and Surface Area of a Sphere Volume and Surface Area of a. Use MathJax to format equations. Use your knowledge of the sums of the interior and exterior angles of a polygon to answer the following questions. 360/x -10x Kudos for the right answer and explanation. So if we know the sum of the interior angles, we just need to divide that sum by the number of sides to find out the measure of any angle of the polygon: 180(n-2)/n Exterior Angles. A polygon in which all the sides are the same length and all the measures have the same measure is regular polygon. Grade 3 geometry worksheet: Properties of polygons Author: K5 Learning Subject: Grade 3 Geometry Worksheet Keywords: Grade 3 geometry worksheet 2D shapes, lines, angles, parallel, area, perimiter Created Date: 4/28/2017 12:38:52 PM. Convex polygon B. Do you really know everything about these 2-D polygons and. Properties of Decagon. 7) The area of a parallelogram with base of 10 ft and height of 2 ft is 20 square ft. Theorem: The area of a regular polygon. The vertices of an N-vertex polygon are located at the angles (2*Math. This quiz is incomplete! To play this quiz, please finish editing it. All sides are equal length placed around a common center so that all angles between sides are also equal. A polygon is called a REGULAR polygon when all of its sides are of the same length and all of its angles are of the same measure. Access the answers to hundreds of Polygons questions that are explained in a way that's easy for you to understand. In addition to the examples given by Karen: octagons (eight sides), nonagons (nine sides), decagons (ten sides), hendecagons (eleven sides), dodecagons (twelve sides), tridecagons (thirteen sides), tetradecagons (fourteen sides), pendecagons (fift. Many of these shapes, or polygons, can be described as flat closed figures with 3 or more sides. Hand touch global map form lines and polygonal. Here are your FREE materials for this lesson. See our pages on circles and. It is possible to do better than a hexagon, if an irregular polygon is acceptable. A regular polygon is one that has equal sides. A rectangle is a polygon. To know more about different data collection methods, and statistics download BYJU'S -The Learning App. Examples of regular polygon: In the adjoining figure of an equilateral triangle ABC there are three sides i. Polygons come in many shapes and sizes and you will have to know your way around them with confidence in order to ace those SAT questions on test day. Polygons come in many shapes and sizes and you will have to know them inside and out in order to take on the many different types of polygon questions the ACT has to offer. If the measures of a polygon's angles or side lengths differ, then the polygon is called an irregular polygon. Identifying and Describing Polygons All Classroom Lessons. Start by filling in the missing lengths of the sides. The sum of its angles will be 180° × 4 = 720° The sum of interior angles in a hexagon is 720°. Identifying shape attributes, Understanding polygons. Videos, worksheets, 5-a-day and much more. A polygon is a regular polygon if and only if all of its interior angles are equal and all of its sides are of equal length. In this polygon game you will classify different polygons based on their properties. Polygons are 2-dimensional shapes. Combine multiple words with dashes(-), and seperate tags with spaces. In this video I will take you through everything you need to know in order to answer basic questions about the angles of polygons. Irregular polygon C. The number of sides and the radius can be entered via a file, STDIN, or just a plain old variable. 5 (Geometry: n-sided regular polygon) An n-sided regular polygon's sides all have (i. This calculator is designed to give the angles of any regular polygon. Please leave workings so I understand how you get the answer. Let's say you have pixel bitmaps that look something like this: From this I can easily extract a contour, which will be a concave polygon defined by a set of 2D points. Make flat shapes using toothpicks and clay. Before we get too caught up on the excitement of quadrilaterals, let's take time to learn the names and basic properties of different polygons. Updated: Aug 5, 2019. - [Voiceover] So ,a few things…that you should know about polygons,…let's start with the most basic polygon…which is our friend the triangle. Let’s first find the measure of each one of those angles. all of these. In Year 6 children use this knowledge and the following formula to calculate the size of the angles. The sum of the angles of a polygon with n sides, where n is 3 or more, is 180° × (n - 2) degrees. As the concept of irregular polygons is extremely general, knowledge about this concept can be very valuable, both for problem solving and for simple problems in the real world. The simplest regular polygon is the equilateral triangle, which consists of three edges of equal length and three angles between each pair of edges to be 60 degrees. Consider, for instance, the ir regular pentagon below. How many sides does an octagon have? eight 2. Creating Polygons - investigating the Concept of Triangle and the Properties of Polygons ( from NCTM ) ; Figures and Polygons - definitions and examples of a large number of figures fiting the definition of polygon ; Identify Polygons - names of polygons by number of. For the regular polygons inscribed in the unit circle, what is the range of values that the apothem can have?. Higher Tier. Please leave workings so I understand how you get the answer. Examples of regular polygon: In the adjoining figure of an equilateral triangle ABC there are three sides i. shapes}; see section 51. A polygon is regular, if all its sides have the same length and all its angles have the same measure. , AB, BC and CA are equal and there are three angles i. Abstract polygon. If a polygon is regular, there is a point which we can define as the center. Then ask guiding questions to see if they recognize the pattern (every time you add a side, the interior angle sum increases by 180°). Our online polygon trivia quizzes can be adapted to suit your requirements for taking some of the top polygon quizzes. A regular three-sided polygon is called an equilateral triangle. An Apothem is a line segment from center point of side of a regular polygon to the midpoint of the regular polygon. Categories & Grades. 1) heptagon 2) decagon 3) nonagon 4) hexagon 5) pentagon 6) nonagon 7) hexagon 8) nonagon State if each polygon is concave or convex. † Can any quadrilateral tile the plane? † Can any irregular polygon tile the plane?. A square is a rhombus. " Knowing the names of the polygons is only the start. Feel free to always add Asymptote solution to any questions related to plotting, drawing, diagramming, etc because it is allowed by law in this site. Indeed, it is clear, that under the given conditions, a regular triangle is the only possible regular polygon with the smaller interior angle. Each student will receive an Area of Irregular Polygons Mini Lesson Examples. Choose from the following regular polygons: Triangle,. Test comprehension with the questions that follow. A regular polygon has sides which are all the same length as well as angles which are all the same. Regular Polygon. Browse other questions tagged tikz-pgf or ask your own question. A regular polygon is one that has equal sides. Polygons can be regular or. Khan Academy is a 501(c)(3) nonprofit organization. Abstract background with polygonal shapes. 25 Mocks + 30 Sectionals for Rs. 7) The area of a parallelogram with base of 10 ft and height of 2 ft is 20 square ft. You can tell, just by looking at the picture, that $$\angle A and \angle B$$ are not congruent. The shape on the left has equal sides and angles, so it is a regular polygon. 👍If you like this resource, then please rate it and/or leave a comment💬. Questions for Students. A regular polygon is a closed, 2-dimensional figure consisting exclusively of line segments of equal length. Polygon abstract background. Exterior angle = 180-178 = 2° because the interior and exterior angle together make a straight line. If a regular polygon has x sides, then the degree measure of each exterior angle is 360 divided by x. The sum of the exterior angles of any polygon is 360°. If none of the sides, when extended, intersects the polygon, it is a convex polygon; otherwise it is concave. What is the latest version of SketchUp that Keyframe Animation will work with?. In this polygon game you will classify different polygons based on their properties. The applets below illustrate what it means for any polygon to be classified as a regular polygon. 9 degrees are in the measure of an interior angle of a regular seven sided polygon. A Polygon is a 2D shape which is made up of straight line segments. In our example, it's equal to 5 in. Start studying Quadrilateral & Polygon Unit Review Questions. Fortunately, as you'll see in the following practice questions, there's a handy formula that you can use to find a missing interior angle in a […]. We can make "pencilogons" by aligning multiple, identical pencils end-of-tip to start-of-tip together without leaving any gaps, as shown above, so that the enclosed area forms a regular polygon (the example above left is an 8-pencilogon). Convex polygon B. The resulting polygon is guaranteed to be simple (it does not intersect or touch itself). It is the point which is equidistant from each vertex. Upgrade your subscription to get access to this quiz, more lessons, and more practice questions. polygon: a 2D or flat, closed shape with straight sides. A regular polygon is a polygon whose sides are equal length and whose angles are all the same value. Polygons are 2-dimensional shapes with straight sides. Area= (2)(apothem)(perimeter) is the formula for finding the area of a regular polygon. Work out the identity of the missing regular. Tutors Answer Your Questions about Polygons (FREE) Get help from our free tutors ===> Algebra. Name the quadrilaterals shown: rhombus, parallelogram, square, rectangle, or trapezoid. Upgrade your subscription to get access to this quiz, more lessons, and more practice questions. Here we look at Regular Polygons only. I can classify a polygon as concave or convex and regular or irregular. The sum of the angles of a polygon with n sides, where n is 3 or more, is 180° × (n - 2) degrees. To do so, I would like to construct an irregular polygon, and from each of it's sides, construct a *regular* n-gon, on the outside of the irregular-gon. Sample Problem 4: Draw a figure that fits the description. Selecting Objects Inside Polygon Is there a way within a source drawing to select all objects within a predetermined boundary/polygon? The regular select routine is limited to defining a polygon but not selecting an existing one. This quiz will focus on different shapes classified as polygons. t = 360 o / 12 = 30 o; We now use the formula for the area when the side of the regular polygon is known Area = (1 / 4) n x 2 cot (180 o / n) Set n = 12 and x = 6 mm area = (1 / 4) (12) (6 mm) 2 cot (180 o / 12). One interior angle of a regular polygon - (n - 2). Diagonals of a Regular Octagon. Can be “regular” – all sides and all angles are equal to each other Two sides of equal length Three acute angles Sum of angles = 180° All sides equal length Three acute angles Sum of angles = 180° Is a regular polygon No sides are equal No angles are equal May have obtuse angle Sum of angles = 180 ° Opposite sides are parallel Opposite. A polygon is a shape that has any number of straight sides, such as a triangle, square or hexagon. Polygon or Not? This Is the Question : In this game, students will quickly decide whether a geometrical figure is a polygon or not a polygon. Round your answer to the nearest tenth if necessary. For the Independent Practice students will use the area formulas to help them complete the Finding the Finding the Area of Regular Polygons Worksheet. This is of course made far more obnoxious by the fact that IsPolygon is a valid rule token, so your rule will seem to be correct, and even pass the rule-checker, but silently fail when you try to actually repour the. Mathematics (Linear) - 1MA0 ANGLES: POLYGONS Answer the questions in the spaces provided - there may be more space than you need. A clock is constructed using a regular polygon with 60 sides the polygon rotates every minute how has the polygon rotated after 7 minutes. Using the area of regular polygon calculator: an example. Videos, worksheets, 5-a-day and much more. The angles have been calculated manually. Polygons are multi-sided shapes with different properties. Videos, worksheets, 5-a-day and much more. Solve these reasoning questions. 2019 23:53, kumarivishakha118. Always show your workings. A three-sided polygon is called a triangle. A regular polygon is both equilateral and equiangular. SplashLearn is an award winning math learning program used by more than 30 Million kids for fun math practice. The formula is (n - 2) times 180 divided by n , where n is the number of sides. Find the area of each of the following regular polygons. The outline itself is matched with the InPolygon rule (which is what you want). Area of Polygons. parallel lines: two lines that are always the same distance apart and never touch. They should not be confused with regular polygons. Is this a inductive reasoning or deductive re. A "Regular Polygon" has: all sides equal and. The only constructible regular polygons with an odd number of sides are those for which this number is a product of distinct Fermat primes (so for instance 15 = 3 times 5, 51 = 3 times 17), and the only ones with an even number of sides are those obtained by repeatedly doubling these numbers (including 1), thus:- (1,2), 4, 8, 16, 32, 64,. true false. The sum of the angles of a polygon with n sides, where n is 3 or more, is 180° × (n - 2) degrees. (Must be 18 years old to sign up. 1) 2) 3) regular 19-gon 4) regular 14-gon Find the measure of one exterior angle in each regular polygon. A polygon is called a REGULAR polygon when all of its sides are of the same length and all of its angles are of the same measure. The area of the regular polygon is 297 m 2. 2D SHAPES > REVISION > GCSE QUESTIONS. If someone can help me out that would be awesome. Videos, worksheets, 5-a-day and much more. Unit 7 Test Polygons And Quadrilaterals Answer Key. Unanswered Questions. A brief description of the worksheets is on each of the worksheet widgets. A "Polygon" object contains the polygon outline. A diagonal of a polygon is any line segment that connects non-consecutive vertices of the polygon. Expressions are given for two side lengths. Award Information. Menu Skip to content. [1] 6) One interior angle of a regular polygon is 135°. The measure of an exterior angle of a convex regular polygon is 45. A polygon is regular, if all its sides have the same length and all its angles have the same measure. This figure shows the most common polygons. Improve your math knowledge with free questions in "Interior angles of polygons" and thousands of other math skills. Geometry Polygons questions and answers for CAT. 652 Responses. A polygon is irregular, if it is not regular. Hint: 360/No. A polygon in which all the sides are the same length and all the measures have the same measure is regular polygon. I will be focusing on convex regular. Download the set (5 Worksheets). TeX - LaTeX Stack Exchange is a question and answer site for users of TeX, LaTeX, ConTeXt, and related typesetting systems. You will have to read all the given answers and click over the correct answer. An n-sided regular polygon is rotationally symmetric around its centre. In Year 6 children use this knowledge and the following formula to calculate the size of the angles. Tags: Question 40. A regular polygon is a polygon all of whose sides are of the same length (equilateral) and all of whose internal angles are the same (equiangular). A polygon that is not a regular polygon is called an irregular polygon. Classify the polygon by the number of sides. Chapter 12. all angles equal. As a child, there's a point in life where shapes and what we now have come to refer as "geometry" are the most fascinating things in the world. Stack Overflow for Teams is a private, secure spot for you and your coworkers to find and share information. ) Sign Up More Info Close. disproportional perimeters. As an example: A pentagon has five sides, 360/5 =72. This figure shows the most common polygons. If you are not sure about the answer then you can check the answer using Show Answer button. Type all answers in the boxes provided. Polygons Exercises ; Topics Which of the following look like regular polygons? Gimme a Hint. The size of each of the interior angle of a regular polygon = (n-2) x 180º ÷ n Where n is the number of sides. You Try It! How many sides does a regular polygon have if each interior angle measures 160 ? Example 5: Solving Algebraic Problems Find the value of x. I can classify a polygon as concave or convex and regular or irregular. (Construct it from scratch, or use a script. You will have to read all the given answers and click over the correct answer. We can see from the above examples that the number of triangles in a polygon is always two less than the number of sides of the polygon. A brief description of the worksheets is on each of the worksheet widgets. Polygons Author: act Last modified by: act Created Date: 7/10/2008 4:49:14 PM Document presentation format: On-screen Show (4:3) Other titles: Corbel Arial Wingdings 2 Wingdings Wingdings 3 Calibri Module 1_Module 2_Module 3_Module 4_Module 5_Module 6_Module Polygons What is a polygon?. An octagon is any eight-sided polygon, and the sum of its angles is 1080°, as we saw above. You can tell, just by looking at the picture, that $$\angle A and \angle B$$ are not congruent. PI*K)/N where K goes from 0 to N-1, inclusive. The sum of the interior angles, in degrees, of a regular polygon is given by the formula 180(n – 2), where n is the number of sides. Main Ideas/Questions POLYGON Notes/Examples A polygon isa figure formed by three or more line Date: Class: hexagon? EXAMPLES 9. 180(x-3)/x B. Formula for the sum of interior angles. A convex polygon has no angles pointing inwards. s = (160cm)/8 = 20 cm. Two congruent figures have equal areas. Print full size. Examples of regular polygons include the equilateral triangle and square. 652 Responses. What is the latest version of SketchUp that Keyframe Animation will work with?. To find the perimeter of a polygon, add the lengths of its sides. If none of the sides, when extended, intersects the polygon, it is a convex polygon; otherwise it is concave. This is part of our collection of Short Problems. Formula to find the sum of interior angles of a n-sided polygon is = (n - 2) ⋅ 180 ° By using the formula, sum of the interior angles of the above polygon is = (9 - 2) ⋅ 180 ° = 7 ⋅ 180 ° = 126 0 ° Formula to find the measure of each interior angle of a n-sided regular polygon is. There are multiple ways to found the Apothem of a Regular Polygon. I can find the measure of an interior angle of any regular polygon. Background polygon. You can now earn points by answering the unanswered questions listed. Background polygon. The applets below illustrate what it means for any polygon to be classified as a regular polygon. A polygon is irregular, if it is not regular. Along with it, the property of equal-length sides, this implies that every regular polygon also has an inscribed circle or incircle that is tangent to every side at the midpoint. (Construct it from scratch, or use a script. org are unblocked. Diagonals of a Regular Octagon. The vertices of an N-vertex polygon are located at the angles (2*Math. Identifying and Describing Polygons All Classroom Lessons. The sum of the exterior angles of any polygon is 360 degrees. What is the minimum interior angle possible for a regular polygon or why. A worksheet on angles in polygons. A Ogee curve is a _____ a) semi ellipse b) continuous double curve with convex and concave c) freehand curve which connects two parallel lines d) semi hyperbola View Answer. A polygon is a plane shape (two-dimensional) with straight sides. Tell whether the polygon is equilateral, equiangular, or regular. 2nd and 3rd Grades. Interact with these applets for a minute or two, then answer the questions that follow. of sides = Interior angle Attempt drawing a pentagon on your own first. A convex polygon has no angles pointing inwards. Concave polygon. A regular polygon is a polygon that is both equilateral and equiangular (having both. Polygon abstract background. All photos are included but can be easily changed to match your branding’s style and personality. Assume that the regular polygon has n sides (or angles). A regular polygon is a polygon all of whose sides are of the same length (equilateral) and all of whose internal angles are the same (equiangular). Solution to Problem 2: Let t be the size of angle AOB, hence t = 360 o / 5 = 72 o; The polygon is regular and OA = OB. Add ScratchPad. For example if a polygon has 41 sides, it would be called a 41-gon. The Corbettmaths Practice Questions on Frequency polygons. …And how about the pentagon…which kinda looks like a home plate. Each question will change subtly every time you take. The number of sides is also the number of angles and, in a regular polygon, all the angles are the same. Expressions are given for two side lengths. (Remade and reuploaded 9th February 2020) Calculating interior and exterior angles of polygons. Making statements based on opinion; back them up with references or personal experience. Grade 3 geometry worksheet: Properties of polygons Author: K5 Learning Subject: Grade 3 Geometry Worksheet Keywords: Grade 3 geometry worksheet 2D shapes, lines, angles, parallel, area, perimiter Created Date: 4/28/2017 12:38:52 PM. Find the area of the regular polygon. Sample Problem 5: Each figure is a regular polygon. of sides = 360°/20° = 18. So, if we know all the interior angles other than x, then we can find x. polygons Questions #1 Relates the number of sides, angles, and vertices of a polygon Question #2 Describes attributes of polygons Question #3 Wow! Thoroughly describes. Shape And Formula Of A Polygon Quiz 10 Questions | By Swords31535 | Last updated: Oct 11, 2017 | Total Attempts: 9314 All questions 5 questions 6 questions 7 questions 8 questions 9 questions 10 questions. You may also be interested in our longer problems on Angles, Polygons and Geometrical Proof Age 11-14 and Age 14-16. Regular Polygons Printable regular polygon sheet which includes picture and name of each of the 10 polygons from triangle, through to dodecagon. A polygon is a flat shape with straight sides that are joined to form a closed shape. You will have to read all the given answers and click over the correct answer. 1) 2) 3) regular 19-gon 4) regular 14-gon Find the measure of one exterior angle in each regular polygon. Although students should work independently, they may want to discuss or ask questions of their group. An n-sided regular polygon is rotationally symmetric around its centre. A regular polygon is always convex. I can solve reasoning questions about comparing, classifying and finding unknown angles in polygons. no space at all. Design a class named RegularPolygon that contains: Aprivate int data field named n that defines the number of sides in the polygon with default value 3. where n - number of sides of octagon. Irregular polygons are polygons that have unequal angles and unequal sides, as opposed to regular polygons which are polygons that have equal sides and equal angles. How many angles does a triangle have? three 3. While we are given two sides - the base and the height - we do not know the hypotenuse. com Regular Polygons Grade 3 Geometry Worksheet Answers: Color the regular quadrilateral. Questions and Facts † What are the only edge-to-edge regular tilings? That is, what regular polygons allow for edge-to-edge tiling of the plane? † Can we use more than one regular polygon, with different number of size, to tile the plane edge-to-edge? † Any triangle can tile the plane. Here we look at Regular Polygons only. Improve your math knowledge with free questions in "Regular and irregular polygons" and thousands of other math skills. Note that the polygon we are given is not a regular polygon, since the side lengths are not all equal. In a regular octagon, each angle = 1080°/8 = 135° That angle is the supplement of a 45° angle. To find the perimeter, multiply the length of one side by the total number of sides. In this video, we'll talk about a few things you need to know about polygons for the GMAT, some of their basic names of regular polygons, and the sum of the interior angles of a polygon. Draw all of the diagonals in each regular polygon. Two congruent figures have equal areas. angles of polygons section 8-1 jim smith jchs names of polygons sides triangle 3 quadrilateral 4 pentagon 5 hexagon 6 heptagon 7 octagon 8 nonagon 9 decagon 10 dodecagon 12 n - gon n interior angle sum each triangle has 180° if n is the number of sides then: int angle sum = (n - 2 ) 180° angles of polygons section 8-1 names of polygons sides triangle 3 quadrilateral 4 pentagon 5 hexagon. Restate for the class a clear definition of polygon: a 2D or flat, closed shape with straight sides. Formula for the sum of interior angles. The moral of this story- While you can use our formula to find the sum of. Polygon features One Click Demo Install, so demo content can be installed in minutes. I usually print these questions as an A5 booklet and issue them in class or give them out as a homework. There can't be a better way to introduce polygons, than what kids love the most - coloring! Direct kids to follow the color key provided to identify polygons like triangles, quadrilaterals, pentagons and more. A regular polygon has an exterior angle of 20 degrees. com For a complete lesson on regular polygons, go to https://www. The simplest regular polygon is the equilateral triangle, which consists of three edges of equal length and three angles between each pair of edges to be 60 degrees. Find the area of each regular polygon. The shape on the left has equal sides and angles, so it is a regular polygon. ) The area formula for a regular polygon is 1/2ap. $$\therefore$$ the area of one such triangle $$= \frac{1}{2} \times$$ sides of polygon $$\times$$ length of perpendicular from the center to any side of. 2D SHAPES > REVISION > GCSE QUESTIONS. The vertical coordinate can be calculated as a sine of the angle times the radius of the circumcircle; the horizontal coordinate is calculated the same way, except you need to multiply the radius by the cosine of the angle. In this regular polygon lesson plan, 10th graders find the area of a regular polygon in terms of the apothem and the perimeter. Regular Polygon. The regular polygons were known to the ancient Greeks, and the pentagram, a non-convex regular polygon (star polygon), appears on the vase of Aristophonus, Caere, dated to the 7th century B. Using the area of regular polygon calculator: an example. t = 360 o / 12 = 30 o; We now use the formula for the area when the side of the regular polygon is known Area = (1 / 4) n x 2 cot (180 o / n) Set n = 12 and x = 6 mm area = (1 / 4) (12) (6 mm) 2 cot (180 o / 12). The sum of the exterior angles of any polygon is 360 degrees. A Polygon is a 2D shape which is made up of straight line segments. Browse other questions tagged qgis polygon intersection or ask your own question. Polygons maths worksheet 3 accurately completes the polygons. 6 9 52) 14 14. A regular polygon has sides of equal length, and all its interior angles are of equal size. A regular polygon have the number of lines of symmetry equal to the number of sides of the regular polygon. The dodecagon would have sides of alternating lengths, with ratio $4:\sqrt{12}$; as a specific size for the 30-60-90 triangle is not given. Label the sides and angles on each polygon. Sum of interior angles = 180 (n – 2) where n = the number of sides in the polygon. A polygon is a 2D shape with at least three sides. A regular polygon has sides which are all the same length as well as angles which are all the same. a positive number representing the interior space of a polygon. all of these. Categories & Grades. Kahoot! is a free game-based learning platform that makes it fun to learn - any subject, in any language, on any device, for all ages! Unit 7 test polygons and quadrilaterals answer key. 25 Mocks + 30 Sectionals for Rs. If a regular polygon has x angles each measuring q degrees, then what is the value of q ? A. In geometry, polygons cover a lot of ground, so you can bet that some questions on the ACT Math exam will involve polygons—specifically, finding the interior angles of a polygon. This calculator is designed to give the angles of any regular polygon. Always show your workings. White neural texture abstract vector. Khan Academy is a 501(c)(3) nonprofit organization. Let's say you have pixel bitmaps that look something like this: From this I can easily extract a contour, which will be a concave polygon defined by a set of 2D points. of sides = 360°/20° = 18. The sum of the exterior angles of any polygon is 360°. An irregular polygons have sides of differing length and angles of differing measure. Find the value of. The following figures are polygons. Find the length of one side. Support your method with examples. This quiz is incomplete! To play this quiz, please finish editing it. Updated: Aug 5, 2019. (n – 2)*180 = sum of all interior angles (6 – 2)*180 = 720. There are various ways to solve this question, but each takes a bit of effort. For a polygon to be 'regular' it must have. Each angle in a regular decagon is equal to 144°. How to find the interior angle of a regular polygon. Polygons are two-dimensional objects, not solids. tells you the sum of the interior angles of a polygon, where n represents the number of sides. Polygons are 2-dimensional shapes. A polygon is called a REGULAR polygon when all of its sides are of the same length and all of its angles are of the same measure. x1, x2 ,y1 ,y2. com for more resources, like predicted GCSE Maths Papers, Topic Buster and challenge yourself against our Demon questions. The formula. This quiz will focus on different shapes classified as polygons. A triangle is a 3 sided polygon with interior angles that add to 180 degrees. In these applets, a pentagon and a triangle will be used to illustrate this definition. Question 1. Identifying and Describing Polygons All Classroom Lessons. Regular polygons. There can't be a better way to introduce polygons, than what kids love the most - coloring! Direct kids to follow the color key provided to identify polygons like triangles, quadrilaterals, pentagons and more. Area of Regular Polygons If radii are drawn from the center of a regular polygon to the vertices, congruent isosceles triangles are formed. In the limit, a sequence of regular polygons with an increasing number of sides approximates a circle, if the perimeter or area is fixed, or a regular apeirogon (effectively a straight line. Regular Polygon Formulas. Illustrated with examples of polygons in everyday life. Newest Active Followers. 120° ° 85° 53° 115° x° 117° 105° 120. In geometry, polygons cover a lot of ground, so you can bet that some questions on the ACT Math exam will involve polygons—specifically, finding the interior angles of a polygon. A hexagon (six-sided polygon) can be divided into four triangles. The interactive lessons include: Polygons and Non-Polygons in which students will drag a series of shapes to either the bucket that says polygon or the bucket that says non-polygon. A rectangle is a special quadrilateral where opposite sides are congruent-- that is, the same length -- and each angle is a right angle. Expected time to solve, similar questions of 11 Plus exam practice papers. Find the length of one side. This is of course made far more obnoxious by the fact that IsPolygon is a valid rule token, so your rule will seem to be correct, and even pass the rule-checker, but silently fail when you try to actually repour the. Area is the negative space inside a polygon. A comprehensive database of more than 34 polygon quizzes online, test your knowledge with polygon quiz questions. +34,000 Free Graphic Resources. Figure 4 – The Interior angles of polygons that can tessellate the plane add up to 360 degrees. The sum of the exterior angles of any polygon is 360 degrees. I also make them available for a student who wants to do focused independent study on a topic. The polygon is a 18-sided regular polygon; a regular octadecagon. A worksheet on angles in polygons. I want to find the direction of s. Published on Jan 20, 2017. In Year 6 children use this knowledge and the following formula to calculate the size of the angles. See our pages on circles and. So if we know the sum of the interior angles, we just need to divide that sum by the number of sides to find out the measure of any angle of the polygon: 180(n-2)/n Exterior Angles. A regular net is defined as one in which the vertex figure is required by symmetry to be a regular polygon or polyhedron. Polygons maths worksheet 4 and polygon maths worksheet 5 work with tessellation. (n – 2)*180 = sum of all interior angles (6 – 2)*180 = 720. | 2020-08-13T05:31:46 | {
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https://math.stackexchange.com/questions/3053627/whats-132333-cdots993-modulo-3/3054014 | What's $1^3+2^3+3^3+\cdots+99^3$ modulo $3$?
What's the remainder when the sum
$$1^3+2^3+3^3+\cdots+99^3$$ is divided by $$3$$?
Background:
I saw this question on MSE but it was closed and I wanted to learn how to approach it. The help given to the asker had failed to bring the question up to an acceptable standard.
I can solve it but I doubt my methods are efficient.
One way is to start by removing every third term. Then the first two terms of the remaining sequence can be removed, and so on... a pattern may emerge.
Another way is to look at the expansion of $$(x+1)^3$$ and see what it does to the residues for each $$x$$, then sum over that by induction.
But I'm sure my inventions aren't very efficient.
• Use $n^3\equiv n\pmod3$. – Angina Seng Dec 27 '18 at 5:42
• math.stackexchange.com/questions/1328798 – Andrei Dec 27 '18 at 5:43
• In general, note that you can handle the terms in groups of $3$ as $n + 3 \equiv n \pmod 3$, and also use that $n^3 \equiv n \pmod 3$ as Lord Shark the Unknown stated above. – John Omielan Dec 27 '18 at 5:46
• @LordSharktheUnknown ah thanks. I guess the $(x+1)^3$ method would've revealed that. – samerivertwice Dec 27 '18 at 5:46
By Fermat's little theorem $$a^3\equiv a\pmod3$$. Then you have $$\sum_{k=1}^{99} k=\frac{(99)(100)}2=50\cdot 99\equiv 0\pmod3$$.
\begin{align} \sum_{i=1}^{99}i^3 &= \sum_{i=0}^{32} (3i+1)^3 + \sum_{i=0}^{32} (3i+2)^3 + \sum_{i=0}^{32} (3i+3)^3 \\ &\equiv \sum_{i=0}^{32} 1^3 + \sum_{i=0}^{32} 2^3 + \sum_{i=0}^{32} 0^3 \pmod{3}\\ &\equiv \sum_{i=0}^{32} 1^3 + \sum_{i=0}^{32} (-1)^3 + \sum_{i=0}^{32} 0^3 \pmod{3}\\ &\equiv 0 \pmod{3} \end{align}
Split the numbers from $$1,\ldots, 99$$ in
• $$33$$ with remainder $$1$$
• $$33$$ with remainder $$2$$
• $$33$$ with remainder $$0$$ $$1^3 + 2^3 + \cdots + 99^3 \equiv 33\cdot (1^3+2^3+0^3) \equiv 0 \mod 3$$
consider these observations:
$$1 = 1 \pmod 3$$ $$2 = -1 \pmod 3$$ $$3 = 0 \pmod 3$$
This becomes:
$$33(1^3+(-1)^3+0) \equiv 33(1+(-1)+0) \equiv 33(0) \equiv 0 \pmod 3$$
Hint:
$$n^3\equiv n\pmod3$$ as $$n^3-n=n(n-1)(n+1)$$ is the product of three consecutive integers
We have $$\sum_{r=1}^mr^3\equiv\sum_{r=1}^mr\pmod3$$
$$\equiv \dfrac{m(m+1)}2$$
Alternatively $$\sum_{r=1}^m r^3=\dfrac{m^2(m+1)^2}4$$
From $$a^{3} + b^{3} = (a+b)(a^{2}-ab+b^{2})$$, you have $$a+b|a^{3} +b^{3}$$ and \begin{align} &1^{3} + 2^{3} + 3^{3} + \cdots + 97^{3} + 98^{3} + 99^{3} \\ &= (1^{3} + 2^{3}) + (4^{3} + 5^{3}) + \cdots + (97^{3}+98^{3}) + (3^{3}+6^{3} + \cdots + 99^{3}) \end{align} is a multiple of 3.
Another method is to use the identity
$$\sum_{k=1}^nk^3=\left(\sum_{k=1}^nk\right)^2$$ So, $$\sum_{k=1}^{99}k^3\equiv\left(\sum_{k=1}^{99}k\right)^2\equiv\frac{(99)^2(100)^2}{4}\equiv0\ (\mod3)$$
Yet another method is to use that $$(x-1)^3=x^3-3x^2+3x-1$$ and $$(x+1)^3=x^3+3x^2+3x+1$$, so $$(x-1)^3+x^3+(x+1)^3 = 3x^3+6x$$, which is divisible by three. This shows that any three consecutive numbers, when cubed, add up to zero modulus three.
You can also calculate that sum using
$$\sum_{i=0}^{i=n}i^3=\dfrac{n^2(n+1)^2}{4}$$
Here: $$\sum_{i=0}^{i=99}i^3=\dfrac{99^2(99+1)^2}{4}$$ which is clearly divisible by 3. | 2020-09-22T14:06:29 | {
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https://atcoder.jp/contests/abc215/editorial/2511 | Contest Duration: - (local time) (100 minutes) Back to Home
Official
## B - log2(N) Editorial by en_translator
There are various approaches to solve this problem. Here, we will explain some of them. Even if you got accepted during the contest, we recommend considering other solutions.
Note that the sample codes in this editorials are provided in C++.
1. Basic solution
2. Solution using logarithmic function, and numerical errors
3. Solution using binary representation of an integer
#### 1. Basic solution
First, as $$k$$ increases, obviously $$2^k$$ gets larger too. Therefore, we can see that $$2^k \le N$$ for all integers $$k$$ up to some boundary, or $$2^k>N$$ for $$k$$ greater than that value.
Thus, we can find the answer with a loop where $$k$$ is incremented one by one .
Note that one can $$2^k$$ efficiently compute $$2^k$$ during the loop of incrementing $$k$$ for a good computational complexity.
#include<bits/stdc++.h>
using namespace std;
int main(){
long long n;
cin >> n;
long long val=1;
for(long long i=0;i<=60;i++){
if(val>n){
cout << i-1 << '\n';
break;
}
val*=2;
}
return 0;
}
#### 2. Solution using logarithmic function, and numerical errors
By applying logarithmic function to $$2^k \le N$$ we obtain $$k \le \log_2(N)$$. Therefore, one should be able to output the $$\log_2(N)$$, rounded down to an integer, to get accepted.
However, this code results in WA (Wrong Answer).
#include<bits/stdc++.h>
using namespace std;
int main(){
long long n;
cin >> n;
cout << floor(log2(n)) << '\n';
return 0;
}
The cause of WA is lack of precision. Specifically, a precision occurs for the following case ($$N=2^{59}-1$$). (WA: 59, AC: 58)
576460752303423487
In order to avoid the error, one can use a type having more precisions than 64-bit floating point decimal. If you modify the last code as follows, you can get accepted for this problem.
#include<bits/stdc++.h>
using namespace std;
int main(){
long long n;
cin >> n;
cout << floor(log2((long double)n)) << '\n';
return 0;
}
No that high-precision type is not always omnipotent. If the constraints for this problem were $$N \le 2 \times 10^{18}$$, the modified code still outputs a wrong answer for the following test case. (WA: 60, AC: 59)
1152921504606846975
In competitive programming, discarding the integral property in an integral problem and using only decimals is dangerous. We recommend you to process integral problems in integer-typed values as much as possible. When you inevitably use decimals, be very careful about the errors.
#### 3. Solution using binary representation of an integer
Consider the binary representation of the integer.
For example, we have the conversions $$6=110_{(2)}$$ and $$15=1111_{(2)}$$.
Here, the inequality $$2^{k} \le N < 2^{k+1}$$ can be transformed into binary notations as follows.
$$1 \underbrace{ 00 \dots 0 }_{ k }\ _{(2)} \le N < 1 \underbrace{ 00 \dots 0 }_{ k+1 }\ _{(2)}$$
Using this property, we can derive the following solution.
• If the binary notation of $$N$$ (without leading $$0$$’s) has $$k$$ digits (that is, if the most significant bit with value $$1$$ is the $$k$$-th least significant bit), then the answer is $$k-1$$.
There are various ways to implement this, while using bit operations simplifies it. (Note that the least significant digit is treated as the $$0$$-th digit.)
#include<bits/stdc++.h>
using namespace std;
int main(){
long long n;
cin >> n;
for(int i=60;i>=0;i--){
if(n&(1ll<<i)){cout << i << '\n';break;}
}
return 0;
}
posted:
last update: | 2023-02-07T08:43:48 | {
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https://markhkim.com/clrs/ch04/ | # CLRS Chapter 4. Divide and Conquer
source code directory
## 4.1. The maximum-subarray problem
### Implementation
A divide-and-conquer implementation of the maximal subarray finder is as follows:
from math import floor
infty = float("inf") # -infty is used as the universal lower bound
def maximal_subarray(A, lo, hi):
if lo == hi:
return (A[lo], lo, hi)
else:
mid = lo + floor((hi - lo)/2)
# the divide-and-conquer step
(lsum, llo, lhi) = maximal_subarray(A, lo, mid)
(rsum, rlo, rhi) = maximal_subarray(A, mid+1, hi)
(csum, clo, chi) = maximal_crossing_subarray(A, lo, mid, hi)
# pick the maximal result
if (lsum >= csum) and (lsum >= rsum):
return (lsum, llo, lhi)
elif (rsum >= lsum) and (rsum >= csum):
return (rsum, rlo, rhi)
else:
return (csum, clo, chi)
def maximal_crossing_subarray(A, lo, mid, hi):
"""search left of mid and right of mid separately, then add"""
(lsum, llo) = maximal_subarray_fixed_starting_point(A, mid, lo, -1)
(rsum, rhi) = maximal_subarray_fixed_starting_point(A, mid+1, hi, 1)
return (lsum + rsum, llo, rhi)
def maximal_subarray_fixed_starting_point(A, start, end, direction):
total, subtotal, max_index = -infty, 0, start
for i in range(start, end + direction, direction):
subtotal = subtotal + A[i]
if subtotal > total:
total, max_index = subtotal, i
return (total, max_index)
>>> A = [13, -3, -25, 20, -3, -16, -23, 18, 20, -7, 12, -5, -22, 15, -4, 7]
>>> maximal_subarray(A, 0, len(A)-1)
(43, 7, 10)
### Exercises
4.1-1. What does find-maximaum-subarray return when all elements of $A$ are negative?
find-maximum-subarray in the text is equivalent to maximal_subarray above, so let us take a look at the latter. We shall show that maximal_subarray(A, 0, len(A)-1) returns the maximum element of $A$ whenever $A$ consists entirely of negative numbers.
We first observe that maximal_subarray_fixed_starting_point(A, start, end, direction) returns (A[start], start) for all legal inputs of start and end. From this, we see that maximal_crossing_subarray(A, lo, mid, hi) returns (A[mid] + A[mid+1], mid, mid+1). Since $A$ consists entirely of negative numbers,
whence it follows that the return value of maximal_subarray cannot come from maximal_crossing_subarray.
The only return values of maximal_subarray that does not come from maximal_crossing_subarray are the terms of $A$. Since the if-elif-else block at the end of maximal_subarray returns the maximum, we conclude that maximal_subarray(A, 0, len(A)-1) returns the maximum element of $A$, as was to be shown. $\square$
4.1-2. Write pseudocode for the brute-force method of solving the maximum-subarray problem. Your procedure should run in $\Theta(n^2)$ time.
infty = float("inf") # -infty is used as the universal lower bound
def maximal_subarray_brute_force(A):
n = len(A)
max_sum = -infty
for i in range(n):
iterative_sum = 0
for j in range(i, n):
iterative_sum += A[j]
if iterative_sum > max_sum:
max_sum = iterative_sum
max_lo, max_hi = i, j
return (max_sum, max_lo, max_hi)
>>> A = [13, -3, -25, 20, -3, -16, -23, 18, 20, -7, 12, -5, -22, 15, -4, 7]
>>> maximal_subarray_brute_force(A)
(43, 7, 10)
>>> A = [13, -3, -25, 20, -3, -16, -23, 18, 20, -7, 12, -5, -22, 15, -4, 7]
>>> maximal_subarray(A, 0, len(A)-1)
(43, 7, 10)
It is not hard to see that maximal_subarray_brute_force is correct, as it compares every possible sub-sum.
Assuming that the if clause is never satisfied, we can compute the lower bound of the running time as follows:
The if clause only adds constant time to the computation, and so the upper bound of the running time is $O(n^2)$. It follows that maximal_subarray_brute_force runs in $\Theta(n^2)$ time. $\square$
4.1-3. Implement both the brute-force and recursive algorithms for the maximum-subarray problem on your own computer. What problem size $n_0$ gives the crossover point at which the recursive algorithm beats the brute-force algorithm? Then, change the base case of the recursive algorithm to use the brute-force algorithm whenever the problem size is less than $n_0$. Does that change the crossover point?
>>> import random
50 elements:
>>> L1 = random.sample(range(-1000, 1000), 50)
>>> L2 = random.sample(range(-1000, 1000), 50)
>>> L3 = random.sample(range(-1000, 1000), 50)
>>> %%timeit
... maximal_subarray(L1, 0, len(L1)-1)
... maximal_subarray(L2, 0, len(L2)-1)
... maximal_subarray(L3, 0, len(L3)-1)
231 µs ± 1.94 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
>>> %%timeit
... maximal_subarray_brute_force(L1)
... maximal_subarray_brute_force(L2)
... maximal_subarray_brute_force(L3)
227 µs ± 978 ns per loop (mean ± std. dev. of 7 runs, 1000 loops each)
60 elements:
>>> L1 = random.sample(range(-1000, 1000),60)
>>> L2 = random.sample(range(-1000, 1000),60)
>>> L3 = random.sample(range(-1000, 1000),60)
>>> %%timeit
... maximal_subarray(L1, 0, len(L1)-1)
... maximal_subarray(L2, 0, len(L2)-1)
... maximal_subarray(L3, 0, len(L3)-1)
285 µs ± 6.63 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
>>> %%timeit
... maximal_subarray_brute_force(L1)
... maximal_subarray_brute_force(L2)
... maximal_subarray_brute_force(L3)
312 µs ± 1.21 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
It follows that $% $ for this particular example. Let us now modify the recursive algorithm to include the brute-force algorithm for all lists of size at most, say, 55:
def maximal_subarray_mixed(A, lo, hi):
if hi - lo <= 55:
(brute_sum, brute_lo, brute_hi) = maximal_subarray_brute_force(
A[lo:hi+1])
return (brute_sum, brute_lo + lo, brute_hi + hi)
else:
mid = lo + floor((hi-lo)/2)
(lsum, llo, lhi) = maximal_subarray(A, lo, mid)
(rsum, rlo, rhi) = maximal_subarray(A, mid+1, hi)
(csum, clo, chi) = maximal_crossing_subarray(A, lo, mid, hi)
if (lsum >= csum) and (lsum >= rsum):
return (lsum, llo, lhi)
elif (rsum >= lsum) and (rsum >= csum):
return (rsum, rlo, rhi)
else:
return (csum, clo, chi)
Let us test the new algorithm.
60 elements:
>>> L1 = random.sample(range(-1000, 1000),60)
>>> L2 = random.sample(range(-1000, 1000),60)
>>> L3 = random.sample(range(-1000, 1000),60)
>>> %%timeit
... maximal_subarray(L1, 0, len(L1)-1)
... maximal_subarray(L2, 0, len(L2)-1)
... maximal_subarray(L3, 0, len(L3)-1)
279 µs ± 3.09 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
>>> %%timeit
... maximal_subarray_mixed(L1, 0, len(L1)-1)
... maximal_subarray_mixed(L2, 0, len(L2)-1)
... maximal_subarray_mixed(L3, 0, len(L3)-1)
278 µs ± 989 ns per loop (mean ± std. dev. of 7 runs, 1000 loops each)
70 elements:
>>> L1 = random.sample(range(-1000, 1000),70)
>>> L2 = random.sample(range(-1000, 1000),70)
>>> L3 = random.sample(range(-1000, 1000),70)
>>> %%timeit
... maximal_subarray(L1, 0, len(L1)-1)
... maximal_subarray(L2, 0, len(L2)-1)
... maximal_subarray(L3, 0, len(L3)-1)
330 µs ± 1.61 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
>>> %%timeit
... maximal_subarray_mixed(L1, 0, len(L1)-1)
... maximal_subarray_mixed(L2, 0, len(L2)-1)
... maximal_subarray_mixed(L3, 0, len(L3)-1)
330 µs ± 2.47 µs per loop (mean ± std. dev. of 7 runs, 1000 loops each)
We see that the crossover point increases by 10 or so, but this is hardly an improvement. $\square$
4.1-4. Suppose we change the definition of the maximum-subarray problem to allow the result to be an empty subarray, where the sum of the values of an empty subarray is 0. How would you change any of the algorithms that do not allow empty subarrays to permit an empty subarray to be the result?
Only the main routine needs to be changed:
from math import floor
def maximal_subarray_allow_empty_subarray(A, lo, hi):
if lo == hi:
if A[lo] < 0:
return (0, -1, -1) # allow for empty array, labeled by -1
else:
return (A[lo], lo, hi)
else:
mid = lo + floor((hi - lo)/2)
(lsum, llo, lhi) = maximal_subarray(A, lo, mid)
(rsum, rlo, rhi) = maximal_subarray(A, mid+1, hi)
(csum, clo, chi) = maximal_crossing_subarray(A, lo, mid, hi)
if (lsum < 0) and (rsum < 0) and (csum < 0):
return (0, -1, -1) # allow for empty array, labeled by -1
elif (lsum >= csum) and (lsum >= rsum):
return (lsum, llo, lhi)
elif (rsum >= lsum) and (rsum >= csum):
return (rsum, rlo, rhi)
else:
return (csum, clo, chi)
>>> A = [-1, -2, -3, -4, -5, -6, -7]
>>> maximal_subarray_allow_empty_subarray(A, 0, len(A)-1)
(0, -1, -1)
$\square$
4.1-5. Use the following ideas to develop a nonrecursive, linear-time algorithm for the maximum-subarray problem. Start at the left end of the array, and progress toward the right, keeping track of the maximum subarray seen so far. Knowing a maximum subarray of A[1..j], extend the answer to find a maximum subarray ending at index $j+1$ by using the following observation: a maximum subarray of A[1..j+1] is either a maximum subarray of A[1..j] or a subarray A[i..j+1] for some $1 \leq i \leq j+1$. Determine a maximum subarray of the form A[i..j+1] in constant time based on knowing a maximum subarray ending at index $j$.
We fix an index $j \geq 1$ and suppose that $A[p:q]$ is a maximum subarray of $A[:j]$. We pick another index $i$ such that
By the maximality of $A[p:q]$ in $A[:j]$, every maximum subarray of $A[:j+1]$ must either equal to $A[p:q]$ or contain $A[j]$. Now,
whence we conclude that
Using the above observation, we devise a linear-time algorithm for the maximum-subarray problem.
Given an array $A$, we observe that $[A[0]]$ is the (unique) maximum subarray of $A[:0+1]$. We now apply the above observation inductively to produce a maximum subarray of $AA$, a process known as Kadane’s algorithm.
def kadane(A):
lo, hi, total = 0, 0, 0
tail_lo, tail_hi, tail_total = 0, 0, 0
for j in range(1, len(A)-1):
if tail_total + A[j] <= A[j]:
tail_lo, tail_hi, tail_total = j, j, A[j]
else:
tail_lo, tail_hi, tail_total = tail_lo, j, tail_total+A[j]
if tail_total >= total:
lo, hi, total = tail_lo, tail_hi, tail_total
return (total, lo, hi)
>>> A = [13, -3, -25, 20, -3, -16, -23, 18, 20, -7, 12, -5, -22, 15, -4, 7]
>>> maximal_subarray(A, 0, len(A)-1) # see the implementation subsection
(43, 7, 10)
>>> kadane(A)
(43, 7,1 0)
$\square$
## 4.2 Strassen’s algorithm for matrix multiplication
See my blog post for details.
### Exercises
4.2-1. Use Strassen’s algorithm to compute the matrix product
Show your work.
Observe first that
The product is
$% $ $\square$
4.2-2. Write pseudocode for Strassen’s algorithm
The implementation below makes use of NumPy to reduce the implementation overhead on basic matrix operations.
import numpy as np
def strassen(A, B):
n = len(A)
if n == 1:
return A * B
# block matrices; // is integer division
a = np.array([[A[:n//2, :n//2], A[:n//2, n//2:]],
A[n//2:, :n//2], A[n//2:, n//2:]]])
b = np.array([[B[:n//2, :n//2], B[:n//2, n//2:]],
B[n//2:, :n//2], B[n//2:, n//2:]])
s1 = b[0,1] - b[1,1]
s2 = a[0,0] + a[0,1]
s3 = a[1,0] + a[1,1]
s4 = b[1,0] - b[0,0]
s5 = a[0,0] + a[1,1]
s6 = b[0,0] + b[1,1]
s7 = a[0,1] - a[1,1]
s8 = b[1,0] + b[1,1]
s9 = a[0,0] - a[1,0]
s10 = b[0,0] + b[0,1]
p1 = strassen(a[0,0], s1)
p2 = strassen(s2, b[1,1])
p3 = strassen(s3, b[0,0])
p4 = strassen(a[1,1], s4)
p5 = strassen(s5, s6)
p6 = strassen(s7, s8)
p7 = strassen(s9, s10)
c = np.zeros((n, n)) # initialize an n-by-n matrix
c[:n//2, :n//2] = p5 + p4 - p2 + p6
c[:n//2, n//2:] = p1 + p2
c[n//2:, :n//2] = p3 + p4
c[n//2:, n//2:] = p5 + p1 - p3 - p7
return c
>>> A = np.array([[1, 3], [7, 5]])
>>> B = np.array([[6, 8], [4, 2]])
>>> np.dot(A, B) # matrix multiplication
[[18 14]
[62 66]]
>>> strassen(A, B)
[[18. 14.]
[62. 66.]]
>>> A = np.array([[5, 2, 6, 1],
... [0, 6, 2, 0],
... [3, 8, 1, 4],
... [1, 8, 5, 6]])
>>> B = np.array([[7, 5, 8, 0],
... [1, 8, 2, 6],
... [9, 4, 3, 8],
... [5, 3, 7, 9]])
>>> np.dot(A, B) # matrix multiplication
[[ 96 68 69 69]
[ 24 56 18 52]
[ 58 95 71 92]
[ 90 107 81 142]]
>> strassen(A, B)
[[ 96. 68. 69. 69.]
[ 24. 56. 18. 52.]
[ 58. 95. 71. 92.]
[ 90. 107. 81. 142.]]
$\square$
4.2-3. How would you modify Strassen’s algorithm to multiply $n \times n$ matrices in which $n$ is not an exact power of 2? Show that the resulting algorithm runs in time $\Theta(n^{\log 7})$.
For each $n \times n$ matrix $A$, we let $m$ be the unique integer such that $% $ and augment $A$ with an $(2^m - n) \times (2^m - n)$ identity matrix $I$ as follows:
We then have
from which we can extract $AB$ in constant time. It thus suffices to compute the time complexity of the matrix multiplication $\tilde{A} \tilde{B}$.
Observe that
by the monotonicity of $t \mapsto t^{\log 7}$. Similarly,
Now,
and so
It follows that $n^{\log 7} = \Theta((2^m)^{\log 7})$, whence applying Strassen’s algorithm to the augmented matrices does not incur any additional asymptotic time complexity. $\square$
4.2-4. What is the largest $k$ such that if you multiply $3 \times 3$ matrices using $k$ multiplications (not assuming commutativity of multiplication), then you can multiply $n \times n$ matrices in time $o(n^{\log 7})$? What would be the running time of this algorithm be?
The recurrence relation to solve is
Observe that $n^2 = \Theta(n^{\log_3 9})$. By the master theorem ($\S4.5$), we have
where $(\ast)$ is $k > 9$ and $k(n/3)^2 \leq cn^2$ for some $% $ and all sufficiently large $n$.
Since $\log 7 > \log_3 k$ implies that $% $, we see that $T(n) = o(n^{\log 7})$ whenever $k \leq 9$. On the other hand, if $k > 9$, then then inequality
never holds for any $n \geq 1$ reagrdless of which $% $ we choose. It follows that we must have $k \leq 9$ in order to multiply $n$-by-$n$ matrices in time $o(n^{\log 7})$, and the running time is
4.2-5. V. Pan has discovered a way of multiplying $68 \times 68$ matrices using 132,464 multiplications, a way of multiplying $70 \times 70$ matrices using 143,630 multiplications, and a way of multiplying $72 \times 72$ matrices using 155,424 multiplications. Which method yields the best asymptotic running time when used in a divide-and-conquer matrix-multiplication algorithm? How does it compare to Strassen’s algorithm?
The recurrence relations to solve are
Observing that $n^2 = \Theta(n^{\log_{68} 4624}){}_{}$, we invoke the master theorem ($\S4.5$) to conclude that
Since $n^{\log 7} \approx 2.807$, we conclude that $T_{68}{}_{}$ is better than the running time of Strassen.
Observe now that $n^2\Theta(n^{\log_{70} 4900}){}_{}$. The master theorem ($\S4.5$) implies that
Similarly as above, we conclude that $T_{72}{}_{}$ is better than the running time of Strassen.
Finally, we observe that $n^2 = \Theta(n^{\log_{72} 1944}){}_{}$, which implies that
by the master theorem ($\S4.5$). Similarly as above, we conclude that $T_{72}{}_{}$ is better than the running time of Strassen.
We conclude that
where $T_{S}{}_{}$ refers to the runing time of Strassen. $\square$
4.2-6. How quickly can you multiply a $kn \times n$ matrix by an $n \times kn$ matrix, using Strassen’s algorithm as a subroutine? Answer the same question with the order of the input matrices reversed.
Using block matrix multiplication
we can multiply a $kn \times n$ matrix by an $n \times kn$ matrix through $k^2$ iterations of Strassen. It follows that the computation takes $\Theta(k^2n^{\log 7})$.
Similarly, we use block matrix multiplication
we can multiply an $n \times kn$ matrix by a $kn \times n$ matrix through $k$ iterations of STrassen. It follows that the computation takes $\Theta(k n^{\log 7})$. $\square$
4.2-7. Show how to multiply the complex numbers $a + bi$ and $c + di$ using only three multiplcations of real numbers. The algorithm should take $a$, $b$, $c$, and $d$ as input and produce the real component $ac - bd$ and the imaginary component $ad + bc$ separately.
We set $P_1 = ac$, $P_2 = bd$, and $P_3 = (a+b)(c+d)$. Observe that
and that
It follows that
which only requires three real multiplications.
This algorithm is commonly known as the Karatsuba algorithm. $\square$
## The substitution method for solving recurrences
### Exercises
4.3-1. Show that the solution of $T(n) = T(n-1) + n$ is $O(n^2)$.
We let $C > 0$ be an arbitrary constant, to be chosen later. We fix a postive integer $n$ and assume that
for all $% $. We shall show that
Observe that
By choosing $C = 1$, we see that $-2(C-1)n = 0$ for all $n \geq 0$, from which it follows that
as was to be shown.
To turn the above argument into a complete induction proof, we observe that
Since $C = 1$, it suffices to find $m_0$ such that
Choosing any $m_0 \geq \sqrt{2T(1)}$, we see that
as was to be shown. The desired result now follows from mathematical induction with base case $n = m_0$. $\square$
4.3-2. Show that the solution of $T(n) = T(\lceil n/2 \rceil) + 1$ is $O(\log n)$.
We first examine the recurrence for $n \geq 2$.
We let $C > 0$ be an arbitrary constant, to be chosen later. We fix a positive integer $n$ and assume that
for all $% $. We shall show that
Observe that
We can bound the last expression by $C \log n$ if and only if
The above inequality is equivalent to
Recalling that $n \geq 2$, we note that
Choosing $C = 1/(1 + \log(2/3))$, we see that
It now follows that
as was to be shown.
To turn the above argument into a complete induction proof, we observe that
for all $k \geq 1.$ Since $C = 1/(1 + \log(2/3))$, it suffices to find $k_0$ such that
The above inequality is equivalent to
which, in turn, is equivalent to
We pick any $k_0$ that satisfies the last inequality and set $m_0 = 2^{k_0}$, so that
The desired result now follows from mathematical induction with base case $n = m_0$. $\square$
4.3-3. We saw that the solution of $T(n) = 2T(\lfloor n/2 \rfloor) + n$ is $O(n \log n)$. Show that the solution of this recurrence is also $\Omega(n \log n)$. Conclude that the solution is $\Theta(n \log n)$.
We first examine the recurrence for $n \geq 2$.
We let $C > 0$ be an arbitrary constant, to be chosen later. We fix a positive integer $n$ and assume that
for all $% $. We shall show that
Observe that
The inequality
is equivalent to
which, in turn, is equivalent to
Since $n \geq 2$, we see that
We claim that the function
is monotonically decreasing on $[2,\infty)$. Indeed, we have that
from which we conclude that $f(k) > f(k+1)$. Monotonicity now follows.
The monotonicity of $f(k)$ implies that
Setting $C = \frac{2}{3}$, we see that $(\ast \ast)$ holds, whence $(\ast)$ holds. We conclude that
with this choice of $C$, as was to be shown.
To turn the above argument into a complete induction proof, we observe that $m_0 = 2$ yields
as $T$ is typically assumed to be nonnegative.
The desired result now follows from mathematical induction with base case $n =m_0$.
4.3-4. Show that by making a different inductive hypothesis, we can overcome the difficulty with the boundary condition $T(1) = 1$ for recurrence (4.19) without adjusting the boundary conditions for the inductive proof.
Recall (4.19):
We let $C > 0$ be an arbitrary constant, to be chosen later. We fix a positive integer $n$ and assume that
for all $% $. We shall show that
Observe that
$C \geq T(1) - 1$, we have $1-C \leq -T(1)$, so that
To turn the above argument into a complete induction proof, we observe that
The desired result now follows from mathematical induction with base case $n=1$. $\square$
4.3-5. Show that $\Theta(n \log n)$ is the solution to the “exact” recurrence (4.3) for merge sort.
Recall (4.3):
We let $c > 0$ and $C > 0$, to be chosen later. We fix a positive integer $n$ and assume that
for all $% $. We shall show that
Since $T(n) = T(\lceil n/2 \rceil) + T(\lfloor n/2 \rfloor) + \Theta(n)$, we have the double-ended estimate
where the constants $d > 0$ and $D > 0$ are to be chosen later.
We first establish the upper bound. Since $\lfloor n/2 \rfloor \leq \lceil n/2 \rceil$,
$\lceil n/2 \rceil \geq n/2$, and so $n/\lceil n/2 \rceil \leq n/(n/2) = 2$. It then follows that
whence any choice of $C$ and $D$ with $C \leq D$ yields
Let us now establish the lower bound. Since $\lceil n/2 \rceil \geq \lfloor n/2 \rfloor$,
$\lfloor n/2 \rfloor \leq n/2$, and so $n/\lfloor n/2 \rfloor \geq n/(n/2) = 2$. It then follows that
whence any choice of $c$ and $d$ with $c \leq d$ yields
We have thus shown that
under the assumption that
holds for all $% $.
To turn the above argument into a complete induction proof, we set $c = d = T(2)/4$ and $C = D = 2T(2)$ and observe that, for $n = 2$,
and that
It follows from mathematical induction that $T(n) = \Theta(n \log n)$ with base case $n = 2$. $\square$
4.3-6. Show that the solution to $T(n) = 2T(\lfloor n /2 \rfloor + 17) + n$ is $O(n \log n)$.
Assume that $n \geq 68$, so that $17 \leq n/4$.
We let $C > 0$, to be chosen later. We fix a positive integer $n$ and assume that
for all $% $. We shall show that
Observe that
Now, $n \geq 68$, and so
and
It follows that
Since $\log(4/3) > 0$, we have $% $, and so
It thus suffices to pick $C > 0$ such that
keeping in mind that $n \geq 68$. Since $n$ and $\log n$ are both positive under this condition, we need only to pick $C$ large enough that
Letting $C = 1/\log(4/3)$, we obtain
under the assumption that
for all $% $.
To turn the above argument into a complete induction proof, we must show that
But, of course,
It now follows from mathematical induction with base case $n = 68$ that
as was to be shown. $\square$
4.3-7. Using the master method in Section 4.5, you can show that the solution to the recurrence $T(n) = 4T(n/3) + n$ is $T(n) = \Theta(n^{\log_3 4})$. Show that a substitution proof with the assumption $T(n) \leq cn^{\log_3 4}$ fails. Then show how to subtract off a lower-order term to make a substitution proof work.
Since
a naïve substitution argument would yield
Since
for all $n \geq 1$, the substitution argument fails.
If we assume instead that
for all $% $, then
and the usual substitution proof goes through. $\square$
4.3-8. Using the master method in Section 4.5, you can show that the solution to the recurrence $T(n) = 4T(n/2) + n$ is $T(n) = \Theta(n^2)$. Show that a substitution proof with the assumption $T(n) \leq cn^2$ fails. Then show how to subtract off a lower-order term to make a substitution proof work.
Since $c(n/2)^2 = (c/4)n^2$, a naïve substitution argument would yield
which is larger than $cn^2$ for all $n \geq 1$.
If we assume instead that
for all $% $, then
and the usual substitution proof goes through. $\square$
4.3-9. Solve the recurrence $T(n) = 3T(\sqrt{n}) + \log n$ by makign a change of variables. Your solution should be asymptotically tight. Do not worry about whether values are integral.
Since
we set $S(m) = T(2^m)$ and solve instead
The master theorem ($\S4.5$) shows that
We shall show this by substitution.
We let $c > 0$ and $C > 0$, to be chosen later. We fix a positive integer $n$ and assume that
for all $% $. We shall show that
We first establish the upper bound:
Now, the lower bound:
To turn the above argument into a complete induction proof, we must show that
Setting $c = C = S(1) + 2$, we see that
and that
Mathematical induction with base case $m = 1$ now shows that
for all $m \geq 1$. Since the lower bound agreed with the upper bound, we in fact have the identity
Since $T(n) = S(\log n)$, we conclude that
## 4.4. The recursion-tree method for solving recurrences
### Exercises
4.4-1. Use a recursion tree to determine a good asymptotic upper bound on the recurrence $T(n) = 3T(\lfloor n/2 \rfloor) + n$. Use the substitution method to verify your answer.
Observe that
Let us verify our guess. To this end, we let $C > 0$, to be chosen later. We fix a positive integer $n$ and assume that
for all $% $. We shall show that
Indeed,
To complete the induction proof, we must show that
We set $C = T(1) + 2$ and observe that
as was to be shown. It now follows from mathematical induction with base case $n = 1$ that $T(n) = O(n^{\log(3/2)})$. $\square$
4.4-2. Use a recursion tree to determine a good asymptotic upper bound on the recurrence $T(n) = T(n/2) + n^2$. Use the substitution method to verify your answer.
Observe that
Let us verify our guess. To this end, we let $C > 0$, to be chosen later. We fix a positive integer $n$ and assume that
for all $% $. We shall show that
Observe that
So long as $C \geq 4$, we have $C/4 + 1 \leq C$, and so
To complete the induction proof, we must show that
To this end, we simply select $C = \max(4, T(1))$. The desired result now follows from induction. $\square$
4.4-3. Use a recursion tree to determine a good asymptotic upper bound on the recurrence $T(n) = 4T(n/2 + 2) + n$. Use the substitution method to verify your answer.
We first observe the the recurrence relation as it is written above is ill-defined. Indeed,
when $T$ is a function defined on $\mathbb{N}$. To remedy this issue, we shall consider instead
Observe that
Setting $n = 2^k$, we obtain
We therefore conjecture that
Since the problem only asks us to establish the upper bound, we shall prove that
We let $C > 0$ and $D > 0$, to be chosen later. We fix a positive integer $n$ and assume that
for all $% $. We shall show that
Observe that
If we choose $C$ and $D$ so that $4C -D + 1 \leq 0$ and $8C - 6D \leq 0$, then we have the bound
How do we choose such $C$ and $D$? The first inequality can be written as
It turns out that this is sufficient to achieve the second inequality:
We also note that the constants must be chosen to satisfy the base case
for an appropriate choice of $n_0$. Since the smallest $D$ that satisfies $(\ast)$ is $4C + 1$, the base case should be chosen so that $Cn_0^2 - (4C+1)n_0 > 0$. This only works if $n_0 \geq 5$. indeed, if $n_0 = 4$, then
On the other hand, if $n_0 = 5$, then
which is positive if $C > 4/5$.
Summarizing the above argument, we choose $C = T(5) + 1$ and $D = 4T(5) + 5$. For the base case $n = 5$, we have
as $T$ is generallly assumed to be nonnegative.
We now assume inductively that
holds for all $% $ such that $m \geq 5$. Now, the inductive hypothesis implies that
as was to be shown. We conclude that $T(n) = O(n^2)$. $\square$
4.4-4. Use a recursion tree to determine a good asymptotic upper bound on the recurrence $T(n) = 2T(n-1) + 1$. Use the substitution method to verify your answer.
Observe that
Let us verify our guess. To this end, we let $C > 0$, to be chosen later. We fix a positive integer $n$ and assume that
for all $% $. We then have
By choosing $C = T(1)/2$, we have
for $n = 1$. It now follows from mathematical induction that $T(n) = O(2^n)$. $\square$
4.4-5. Use a recursion tree to determine a good asymptotic upper bound on the recurrence $T(n) = T(n-1) + T(n/2) + n$. Use the substitution method to verify your answer.
To make sense of the $n/2$ term, we consider instead
Since $T(n) - T(n-1) = T(\lfloor n/2 \rfloor) + n$, we see that
Therefore,
Applying $(\ast)$ to the sum indexed by $i_1$, we obtain
Applying $(\ast)$ recursively, we obtain
Let us verify our guess. To this end, we let $C > 0$ and $D > 0$, to be chosen later. We fix a positive integer $n$ and assume that
for all $% $. We shall show that
Observe that
Since
we see that
Since we must have
we see that
is required for all $n \geq 1$. This holds when $C \geq 2 + 2D$:
Finally, we must pick $C$ and $D$ so that
holds when $n = 1$. This requires
Setting $D = T(1)$ and $C = 2T(1) + 2$, we see that
The choice of $C$ and $D$ also shows that
whenever
for all $% $. It now follows from mathematical induction with base case $n = 1$ that $T(n) = O(n^{\log n})$. $\square$
4.4-6. Argue that the solution to the recurrence $T(n) = T(n/3) + T(2n/3) + cn$, where $c$ is a constant, is $\Omega(n \log n)$ by appealing to a recursion tree.
Observe that
where $\begin{pmatrix} p \\q \end{pmatrix} = \frac{p!}{q!(p-q)!}$ is the binomial coefficient with respect to $(p,q)$. It follows that $T(n) = \Omega(n \log n)$. $\square$
4.4-7. Draw the recursion tree for $T(n) = 4T(\lfloor n/2 \rfloor) + cn$, where $c$ is a constant, and provide a tight asymptotic bound on its solution. Verify your bound by the substitution method.
Observe that
We thus conjecture that $T(n) = \Theta(n^2)$.
We first establish the upper bound. To this end, we let $C > 0$ and $D > 0$, to be chosen later. We fix a positive integer $n$ and assume that
for all $% $. We shall show that
Observe that
Therefore, $D$ must be chosen so that $(c-D)n +2 \leq 0$ for all $n \geq 1$. This holds if $D \geq c + 2$.
We now set
and
so that
Moreover, if $T(m) \leq Cm^2 - Dm$ for all $% $, then
It now follows from induction with base case $n = 1$ that $T(n) = O(n^2)$.
It remains to show that $T(n) = \Omega(n^2)$.
We let $C > 0$ and $D > 0$, to be chosen later. We fix a positive integer $n$ and assume that $T(m) \geq Cm^2 + Dm$ for all $% $. Observe that
To ensure that the resulting expression is bounded below by $Cn^2 + Dn$, we must at least have $D-2C+c > 0$. This holds if $D > 2C - c$.
We assume for now that $c > 0$ and set $C = \min(c/2, T(1)/2)$ and $D = \min(c/4, T(1)/4)$, so that
Moreover,
and so it follows from mathematical induction with base case $n=1$ that $T(n) = \Omega(n^2)$.
We have thus shown that $T(n) = \Theta(n^2)$. $\square$
4.4-8. Use a recursion tree to give an asymptotically tight solution to the recurrence $T(n) = T(n-a) + T(a) + cn$, where $a \geq 1$ and $c > 0$ are constants.
Observe that
We thus conjecture that $T(n) = \Theta(n^2)$.
Since $% $ for $% $, the recurrence relation cannot hold for $% $. We thus assume that the values of $T$ for $n = 1,\ldots,\lfloor a \rfloor$ are given.
We first show the upper bound. To this end, we let $C > 0$, to be chosen later. We shall show that
for all $n \geq 2 \lfloor a \rfloor$. We first assume that $C \geq T(\lfloor a \rfloor) / (\lfloor a \rfloor)^2$, so that $(\ast)$ holds for $n = \lfloor a \rfloor$.
Observe that
If $C \geq c/(2 \lfloor a \rfloor)$, then
and so
We thus assume that
We now fix $n_0 > 2\lfloor a \rfloor$ and assume that $(\ast)$ holds for $% $. We shall show that $(\ast)$ holds for $n = n_0$.
Observe that
If $C \geq c$, then
Since $a \geq 1$, we have the estimate
Now, $n_0 > 2 \lfloor a \rfloor$, and so
It follows that $(\ast)$ holds for $n = n_0$ whenever $C \geq c$.
Setting
we see that $(\ast)$ holds for all $n \geq 2 \lfloor a \rfloor$. It follows that $T(n) = O(n^2)$.
We now establish the lower bound. To this end, we let $C > 0$, to be chosen later. We shall show that
for all $n \geq 2 \lfloor a \rfloor$. We first assume that $C \leq T(\lfloor a \rfloor) / (\lfloor a \rfloor)^2$, so that $(\ast \ast)$ holds for $n = \lfloor a \rfloor$.
Observe that
If $C \leq c/(2 \lfloor a \rfloor)$, then
and so
We thus assume that
We now fix $n_0 > 2\lfloor a \rfloor$ and assume that $(\ast \ast)$ holds for $% $. We shall show that $(\ast\ast)$ holds for $n = n_0$.
Observe that
If $C \leq c/2a$, then
Since $a \geq 0$, we have $2ac^2 \geq 0$, and so
It follows that $(\ast \ast)$ holds for $n = n_0$ whenever $C \leq c/2a$.
Setting
we have $(\ast \ast)$ for all $n \geq 2 \lfloor a \rfloor$. It follows that $T(n) = \Omega(n^2)$.
We now conclude that $T(n) = \Theta(n^2)$, as was to be shown. $\square$
4.4-9. Use a recursion tree to give an asymptotically tight solution to the recurrence $T(n) = T(\alpha n) + T((1-\alpha) n) + cn$, where $\alpha$ is a constant in the range $% $ and $c > 0$ is also a constant.
Observe that
We let $a = \max(\alpha, 1-\alpha)$ and set $p = \lceil \log_{1/\alpha} n \rceil$, so that
for all $0 \leq i \leq p$. The recursion-tree computation yields
We therefore conjecture that $T(n) = \Theta(n \log n).$
We first establish the upper bound. To this end, we let $C > 0$, to be chosen later. We assume that
for all $% $.
Observe that
Since $% $ and $% $, we have that $% $ and $% $. Therefore, any
would yield
We now set
so that $T(2) \leq C n \log n$, and that
whenever $T(m) \leq Cm \log m$ for all $% $. It now follows from induction that $T(n) = O(n \log n)$.
We now establish the lower bound. To this end, we let $C > 0$, to be chosen later. We assume that
for all $% $.
Observe that
Since $% $ and $% $, we have that $% $ and $% $. Therefore, any
would yield
We now set
so that $T(2) \geq C n \log n$, and that
whenever $T(m) \geq Cm \log m$ for all $% $. It now follows from induction that $T(n) = \Omega(n \log n)$.
We have thus shown that $T(n) = \Theta(n \log n)$. $\square$
## 4.5. The master method for solving recurrences
### The master theorem, as stated in CLRS, Theorem 4.1
Let $a \geq 1$ and $b > 1$ be constants, let $f(n)$ be a function, and let $T(n)$ be defined on the nonnegative integers by the recurrence
where we interpret $n/b$ to mean either $\lfloor n /b \rfloor$ or $\lceil n/b \rceil$. Then $T(n)$ has the following asymptotic bounds:
1. If $f(n) = O(n^{\log_b a - \epsilon})$ for some constant $\epsilon > 0$, then $T(n) = \Theta(n^{\log_b a})$.
2. If $f(n) = \Theta(n^{\log_b a})$, then $T(n) = \Theta(n^{\log_b a} \log n)$.
3. If $f(n) = \Omega(n^{\log_b a + \epsilon})$ for some constant $\epsilon > 0$, and if $a f(n/b) \leq cf(n)$ for some constant $% $ and all sufficiently large $n$, then $T(n) = \Theta(f(n))$.
### Exercises
4.5-1. Use the master method to give tight asymptotic bounds for the following recurrences.
a. $T(n) = 2T(n/4) + 1$.
b. $T(n) = 2T(n/4) + \sqrt{n}$.
c. $T(n) = 2T(n/4) + n$.
d. $T(n) = 2T(n/4) + n^2$.
a belongs to case 1, and so
b belongs to case 2, and so
c belongs to case 3, as
for all $n$. Therefore,
d belongs to case 3, as
for all $n$. Therefore,
4.5-2. Professor Caesar wishes to develop a matrix-multiplication algorithm that is asymptotically faster than Strassen’s algorithm. His algorithm will use the divide-and-conquer method, dividing each matrix into pieces of size $n/4 \times n/4$, and the divide and combine steps together will take $\Theta(n^2)$ time. He needs to determine how many subproblems his algorithm has to create in order to beat Strassen’s algorithm. If his algorithm creates $a$ subproblems, then the recurrence for the running time $T(n)$ becomes $T(n) = aT(n/4) + \Theta(n^2)$. What is the largest integer value of $a$ for which Professor Caesar’s algorithm would be asymptotically faster than Strassen’s algorithm?
We recall that Strassen’s time complexity is $\Theta(n^{\log_2 7})$. In light of this, it suffices to find the maximum integral value of $a$ such that
subject to the restriction that $f(n) = n^2$ is asymptotically smaller than $n^{\log_b a}$.
Since $b = 4$, the restriction yields the lower bound
Observe that $\log_4 x = \log_2 7$ is equivalent to
which, in turn, is equivalent to
Since $\log_4 2 = 1/2$, we conclude that
It follows that we must have $% $, whence the optimal integral value of $a$ is $48$. $\square$
4.5-3. Use the master method to show that the solution to the binary-search recurrence $T(n) = T(n/2) + \Theta(1)$ is $\Theta(\log n)$. (See Exercise 2.3-5 for a description of binary search.)
Since $a = 1$ and $b = 2$, we have $\log_b a = 0$. Therefore, $f(n) = \Theta(1)$ is asymptotically equivalent to $n^{\log_b a}$, whence the master method implies that
as was to be shown. $\square$
4.5-4. Can the master method be applied to the recurrence $T(n) = 4T(n/2) + n^2 \log n$? Why or why not? Give an asymptotic upper bound for this recurrence.
The $\log n$ term in $f(n) = n^2\log n$ makes it impossible to apply the master method here. $\log_b a = 2$, so $f(n) = \Omega(n^{\log_b a})$. Nevertheless, $\log n = o(n^p)$ for all $p > 0$, and so $f(n) = O(n^{\log b (a + \epsilon)})$ for all $\epsilon > 0$. It follows that none of the three cases of the master method is applicable.
We shall compute the recursion tree of $T(n)$. Observe:
Since
we conjecture that
To show this, we let $C > 0$, to be chosen later. We assume that
for all $% $. (We pick 16, so that $\log n > 4$.) We shall show that
Observe that
Since $\log n > 4$, we have
whenever $C \geq 1$.
We now set
so that, for $n = 16$,
Moreover, $C \geq 1$, and so
for all $% $
implies
It now follows from induction that $T(n) = O(n^2(\log n)^2)$. $\square$
4.5-5. Consider the regularity condition $af(n/b) \leq cf(n)$ for some constant $% $, which is part of case 3 of the master theorem. Give an example of constants $a \geq 1$ and $b > 1$ and a function $f(n)$ that satisfies all the conditions in case 3 of the master theorem except the regularity condition.
Let’s first rule out a general class of examples. Let $a \geq 1$, $b > 1$, $C > 0$, and $\epsilon > 0$ be arbitrary, and let
Since
we see that setting $c = a/(a+\epsilon)$ yields
for all $n \in \mathbb{N}$. $\epsilon > 0$, and so $% $.
This suggests that we need an oscillating term. Let us pick $a = 1$, $b = 2$, and $f(n) = 2n - n\cos(2 \pi n)$. Observe that
Whenever $n$ is odd, $\cos(\pi n) = -1$ and $\cos(2 \pi n) = 1$, and so
It follows that, whenever $% $, the inequality
cannot hold for all $n$. $\square$
## 4.6. Proof of the master theorem
### Exercises
4.6-1. Give a simple and exact expression for $n_j$ in equation (4.27) for the case in which $b$ is a positive integer instead of an arbitrary real number.
We shall prove a more general result: whenever $m,n \in \mathbb{N}$ and $x \in \mathbb{R}$, we have the identity
Observe that
This implies, in particular, that
The upper bound of $(\ast)$ implies that
We suppose for a contradiction that
We can find an integer $p$ such that
This implies that
so that
The lower bound of $(\ast\ast)$ implies that
It now follows from $(\ast\ast)$ that
which is evidently absurd. We conclude that
Let us now recall (4.27):
We shall show that
whenever $b$ is an integer. We proceed by mathematical induction. The $j = 0$ case is trivially true. If the $j = j_0 - 1$ case holds, then $(\ast\ast\ast)$ implies that
as was to be shown. $\square$
4.6-2. Show that if $f(n) = \Theta(n^{\log_b a} \log^k n)$, where $k \geq 0$, then the master recurrence has solution $T(n) = \Theta(n^{\log_b a} \log^{k+1} n)$. For simplicity, confine your analysis to exact powers of $b$.
We begin with a remark that
not
Fix $k \geq 0$. Recall from (4.21) that
Since
we see that
It thus suffices to show that
Since $b > 1$,
and so
$(\ast)$ follows from $(\ast\ast)$ if we can prove, in addition, that
To this end, we recall that $n$ is restricted to be of the form $b^p$ for some exponent $p$. We shall show that
for all $p \geq 1$, where $C$ is a constant independent of $p$.
We shall prove $(\ast\ast\ast)$ by induction on $p$.
We tackle the inductive step first, so as to determine an appropriate value of $C$. Let us fix $p_0 > 1$ and assume that
for all $% $, where $C > 0$ is a constant to be chosen later. We shall show that
To this end, we observe that
By way of the inductive hypothesis $(\star)$, we bound the above quantity below by
To show $(\star)$, it now suffices to establish the estimate
We write
for each $p$ and observe that
$b > 1$ implies that $\log^k(b) > 0$, and so we need only to show that
The above inequality is equivalent to
as $A_p > 1$ for all sufficiently large $p$. In fact, if $C \geq \frac{1}{\log b}$, then $A_p > 1$ for all $p > 1$.
It thus suffices to establish
If $C = \frac{1}{\log b}$, then
Since $p_0 > 1$, we see that
for all $k \geq 1$, establishing $(\star\star)$ for $C = \frac{1}{\log b}$.
It follows that $(\star)$ holds for $C = \frac{1}{\log b}$, which establishes the inductive case of $(\ast \ast \ast)$.
To complete the proof of $(\ast\ast\ast)$ by induction, we must establish the base case $p = 1$, i.e.,
Since we have chosen $C = \frac{1}{\log b}$, the base case holds with equality. The induction proof is now complete.
Finally, $(\ast)$ follows from $(\ast\ast)$ and $(\ast\ast\ast)$, and we have the desired estimate
Our work here is done. $\square$
4.6-3. Show that case 3 of the master theorem is overstated, in the sense that the regularity condition $af(n/b) \leq cf(n)$ for some constant $% $ implies that there exists a constant $\epsilon > 0$ such that $f(n) = \Omega(n^{\log_b a + \epsilon})$.
We recall from Theorem 4.1 that $a \geq 1$ and $b > 1$. We let $C = \frac{a}{c}$ and observe that the regularity condition can be rewritten as
Applying $(\ast)$ $\lceil \log_b n \rceil$ times, we obtain
Since $\log_b n \leq \lceil \log_b n \rceil$, we have the estimate
Now, $C^{\log_b n} = n^{\log_b C}$, and so
$C = \frac{a}{c}$ implies $\log_b C = \log_b a + \log_b \frac{1}{c}$, and so
where $\epsilon = \log_b \frac{1}{c}$. It follows that
Finally, we remark that $% $ implies $\epsilon > 0$. $\square$
## Problems
### 4-1. Recurrence examples
Give asymptotic upper and lower bounds for $T(n)$ in each of the following recurrences. Assume that $T(n)$ is constant for $n \leq 2$. Make your bounds as tight as possible, and justify your answers.
a. $T(n) = 2T(n/2) + n^4$
b. $T(n) = T(7n/10) + n$
c. $T(n) = 16T(n/4) + n^2$
d. $T(n) = 7T(n/3) + n^2$
e. $T(n) = 7T(n/2) + n^2$
f. $T(n) = 2T(n/4) + \sqrt{n}$
g. $T(n) = T(n-2) + n^2$
a-f can be solved directly by the master theorem (Theorem 4.1), which we restate below for ease of reference:
Theorem 4.1 (Master theorem). Let $a \geq 1$ and $b > 1$ be constants, let $f(n)$ be a function, and let $T(n)$ be defined on the nonnegative integers by the recurrence
where we interpret $n/b$ to mean either $\lfloor n/b \rfloor$ or $\lceil n/b \rceil$. Then $T(n)$ has the following asymptotic bounds:
1. If $f(n) = O(n^{\log_b a - \epsilon})$ for some constant $\epsilon > 0$, then $T(n) = \Theta(n^{\log_b a})$.
2. If $f(n) = \Theta(n^{\log_b a})$, then $T(n) = \Theta(n^{\log_b a} \log n)$.
3. If $f(n) = \Omega(n^{\log_b a + \epsilon})$ for some constant $\epsilon > 0$, and if $af(n/b) \leq cf(n)$ for some constant $% $ and all sufficiently large $n$, then $T(n) = \Theta(f(n))$.
#### Problem 4-1a
$a = b = 2$, so that $\log_b a = 1$. Since
$f(n) = \Omega(n^{\log_b a + 3})$ and
we see that
#### Problem 4-1b
$a = 1$ and $b = \frac{10}{7}$, so that $\log_b a = 0$. Since
and
we see that
#### Problem 4-1c
$a = 16$ and $b = 4$, so that $\log_b a = 2$. Since $f(n) = \Theta(n^{\log_b a})$, we see that
#### Problem 4-1d
$a = 7$ and $b = 3$, so that $% $. Since
and
we see that
#### Problem 4-1e
$a = 7$ and $b = 2$, so that $% $. Since
we see that
#### Problem 4-1f
$a = 2$ and $b = 4$, so that $\log_b a = \frac{1}{2}$. Since $f(n) = \Theta(n^{\log_b a})$, we see that
#### Problem 4-1g
We cannot use the master theorem, as
is not of the form
Observe, however, that
Recalling that
we obtain
where $n' = \lfloor n/2 \rfloor$.Since $n' = \Theta(n)$, we have
### 4-2. Parameter-passing costs
Throughout this book, we assume that parameter passing during procedure calls takes constant time, even if an $N$-element array is being passed. This assumption is valid in most systems because a pointer to the array is passed, not the array itself. This problem examines the implications of three parameter-passing strategies:
1. An array is passed by pointer. Time = $\Theta(1)$.
2. An array is passed by copying. Time = $\Theta(N)$, where $N$ is the size of the array.
3. An array is passed by copying only the subrange that might be accessed by the called procedure. Time = $\Theta(q-p+1)$ if the subarray $A[p..q]$ is passed.
a. Consider the recursive binary search algorithm for finding a number in a sorted array (see Exercise 2.3-5). Given recurrences for the worst-case running times of binary search when arrays are passed using each of the three methods above, and give good upper bounds on the solutions of the recurrences. Let $N$ be the size of the original problem and $n$ be the size of a subproblem.
b. Redo part (a) for the Merge-Sort algorithm from Section 2.3.1.
#### Problem 4-2a
Recall binary_search from Exercise 2.3-5, which we reproduce below:
from math import floor
def binary_search(A, p, q, v):
if p > q :
return -1
if p == q:
return p if A[p] == v else -1
mid = p + floor((q-p)/2)
if A[mid] == v:
return mid
else:
if A[mid] > v:
return binary_search(A, p, mid-1, v)
else: # A[mid] < v
return binary_search(A, mid+1, q, v)
Let $T(N)$ be the worst-case running time of binary_search on a list of length $N$. Since it takes constant time to execute all but the recursive step of the algorithm, we have
where $f(N,n)$ is the time it takes to pass a length-$n$ subarray of an array of length $N$.
If arrays are passed by pointer, then $f(N,n) = \Theta(1)$, and so
By the master theorem (see Section 4.5), we have
If arrays are passed by copying, then $f(N,n) = \Theta(N)$, and so
By the master theorem, we have
Finally, if arrays are passed by copying the appropriate subranges, then $f(N,n) = N/n = N/\lfloor N/2 \rfloor$. We thus have $f(N,n) = \Theta(1)$, and so
#### Problem 4-2b
We recall the non-sentinel version of merge sort from Exercise 2.3-2, which we reproduce below:
import math
def merge(A, p, q, r):
L, R = A[p,q], A[q:r]
n,m = q-p, r-q
i, j, k = 0, 0, p
while i < n and j < m:
if L[i] < R[j]:
A[k] = L[i]
i += 1
else:
j += 1
k += 1
while i < n or j < m:
if i < n:
A[k] = L[i]
else:
A[k] = R[j]
k += 1
def merge_sort(A, p, r):
if p+1 < r:
q = p + math.floor((r-p)/2)
merge_sort(A, p, q)
merge_sort(A, q, r)
merge(A, p, q, r)
Note that the time complexity of merge is $\Theta(r-p)$. Since we divide the input array in half, the time complexity of merge_sort is
where $f(a,b)$ is the time it takes to pass a length-$b$ subarray of an array of length $a$.
Since $f(n, n/2) = \Omega(n)$ regardless of how we pass in an array, we conclude that
It follows from the master theorem (see Section 4.5) that
### 4-3. More recurrence examples
Give asymptotic upper and lower bounds for $T(n)$ in each of the following recurrences. Assume that $T(n)$ is constant for sufficiently small $n$. Make your bounds as tight as possible, and justify your answers.
a. $T(n) = 4T(n/3) + n \log n$.
b. $T(n) = 3T(n/3) + n / \log n$.
c. $T(n) = 4T(n/2) + n^2 \sqrt{n}$.
d. $T(n) = 3T(n/3 - 2) + n/2$.
e. $T(n) = 2T(n/2) + n/\log n$.
f. $T(n) = T(n/2) + T(n/4) + T(n/8) + n$.
g. $T(n) = T(n-1) + 1/n$.
h. $T(n) = T(n-1) + \log n$.
i. $T(n) = T(n-2) + \log n$.
j. $T(n) = \sqrt{n} T(\sqrt{n}) + n$.
See Section 4.5 for the statement of the master theorem.
#### Problem 4-3a
Recall from Problem 3-2 that
for every $\epsilon > 0$. It follows that
for all $% $, whence the master theorem implies that
#### Problem 4-3b
We fist show that the master theorem cannot be used here. To this end, we shall show that
for any choice of $\epsilon > 0$,
and that
for any choice of $\epsilon > 0$.
Since $\log n \geq 1$ for all $n \geq 2$, we have that
for all $n \geq 2$. This implies $(\ast\ast)$. $(\ast)$ follows from $(\ast)$.
For $(\ast\ast\ast)$, we observe that each constant $C > 0$ furnishes an integer $M_C$ such that
for all $n \geq M_C$. It follows that
for all $n \geq M_C$, and $(\ast\ast\ast)$ follows.
Finally, we recall from Problem 3-2 that
for every $\rho > 0$. This implies that each choice of $\rho > 0$ and $C > 0$ furnishes an integer $N_{\rho, C}$ such that
for all $n \geq N_{\rho, C}$. Therefore, we see that
for all $n \geq N_{\epsilon, C}{}_{}$, and $(\ast\ast\ast\ast)$ follows.
Let us now proceed without the master theorem.
We let $k = \log_3 n$ and observe that
Since $% $, we see that
Moreover, $% $, and so
We thus have the estimates
In order to estimate the sum, we define $f(x) = \frac{1}{k-x}$ and observe that
whenever $i-1 \leq x \leq i$. Therefore,
for any choice of index $i$, whence it follows that
Since $k - \lfloor k \rfloor \geq 0$, we see that
We thus end up with the upper bound
for constants $C$, $D$, and $E$. We conclude that
To establish a lower bound, we set $f(x) = \frac{1}{k-x}$ once againd and observe that
whenever $i \leq x \leq i+1$. Therefore,
for any choice of index $i$, whence it follows that
Since $% $, we see that
We thus end up with the lower bound
for constants $C$ and $D$. We conclude that
We have now completed the proof of the estimate
#### Problem 4-3c
The recurrence relation in question can be written as
Observe that
Moreover,
for all $n$, and $% $. We can therefore use the master theorem to conclude that
#### Problem 4-3d
We cannot use the master theorem, as the recurrence relation is not of the form $T(n) = aT(n/b) + f(n)$. If, however, we drop the $-2$ term, we can use the master theorem on
to obtain
Since the $-2$ term shouldn’t affect the asymptotic behavior of $T$, we conjecture that
To prove $(\ast)$, we first fix $n > 6$, so that $\frac{n}{3} - 2 > 1$. Assume inductively that
for all $% $, where $C > 0$ is a constant to be chosen later. Observe that
If we choose $C > \frac{1}{2 \log 3}$, then we have
for all $n \geq 6$, and we have the desired bound
To establish the base case $n=6$, we choose
so that the inductive step continues to holds, while
for $n = 6$. We now conclude by induction that
To prove $(\ast)$, it now remains to show that
To this end, we fix $n > 24$, so that $\frac{n}{3} - 2 > 2$ and $\frac{n}{3} - 2 > \frac{n}{4}$. Assume inductively that
for all $% $, where $C > 0$ is a constant to be chosen later. Observe that
Since $\frac{n}{3} - 2 > \frac{n}{4}$, we have
Choosing $C \leq \frac{1}{8}$ yields
so that
as was to be shown.
To establish the base case $n = 24$, we choose
so that the inductive step continues to hold, while
for $n = 24$. We now conclude by induction that
as was to be shown. This completes the proof that
#### Problem 4-3e
Generalizing the computation carried out in Problem 4-3b, we show that a recurrence relation of the form
satisfies the asymptotic estimate
We let $k = \log_a n$ and observe that
We have shown in Problem 4-3b that
and so we conclude that
#### Problem 4-3f
The recurrence relation is not in a form that the master theorem can be applied to, so we must produce an asymptotic estimate directly.
Observe that
From this, let us guess that the accumulation of cost is
for some $k>0$. Since
we conjecture that
To verify this, we fix $n > 8$ and assume inductively that
for all $% $, where $C > 0$ and $D > 0$ are constants to be chosen. Observe that
Analogously,
So long as $C \geq 7$ and $D \leq 7$, we have
$D \leq \frac{7}{8}(1+D) \leq \frac{7}{8}(1+C) \leq C,$\$
which yields
as was to be shown.
To establish the base case $n = 8$, we choose
so that the inductive case continues to hold, while
and
as was to be shown. This completes the proof of
#### Problem 4-3g
The recurrence relation is not in a form that the master theorem can be applied to, so we must produce an asymptotic estimate directly.
Observe that
Page 1154 of CLRS shows that
and so we conclude that
#### Problem 4-3h
The recurrence relation is not in a form that the master theorem can be applied to, so we must produce an asymptotic estimate directly.
Observe that
We have proved in Exercise 3.2-3 that $\log n! = \Theta(n \log n)$, and so we conclude that
#### Problem 4-3i
The recurrence relation is not in a form that the master theorem can be applied to, so we must produce an asymptotic estimate directly.
Observe that
where $k!!$ is the double factorial function
We let $m = \lfloor n/2 \rfloor$. If $m$ is even, then
In this case,
which is $\Theta(m \log m) = \Theta(n \log n)$.
If $m$ is odd, then
In this case,
We have proved in Exercise 3.2-3 that $\log k! = \Theta(k \log k)$, and so $\log m!!$ has the asymptotic estimate of $\Theta(m \log m) = \Theta(n \log n)$.
We conclude that
#### Problem 4-3j
The recurrence relation is not in a form that the master theorem can be applied to, so we must produce an asymptotic estimate directly.
Let us begin by making the substitution $n = 2^m$:
Setting $S(m) = T(2^m)$, we see that
Observe that
Since
we see that $S(m) = \Theta\left( 2^m \log m\right)$.
The identity $m = \log n$ now implies that
### 4-4. Fibonacci numbers
The problem develops properties of the Fibonacci numbers, which are defined by recurrence (3.22). We shall use the technique of generating functions to solve the Fibonacci recurrence. Define the generating function (or formal power series) $\mathscr{F}$ as
where $F_i$ is the $i$th Fibonacci number.
a. Show that $\mathscr{F}(z) = z + z \mathscr{F}(z) + z^2 \mathscr{F}(z)$.
b. Show that
where
and
c. Show that
d. Use part (c) to prove that $F_i = \phi^i / \sqrt{5}$ for $i > 0$, rounded to the nearest integer. (Hint: Observe that $% $.)
#### Problem 4-4a
Observe that
We then have
Since $F_0 = F_1 = 1$ and $F_i = F_{i-1} + F_{i-2}$ for all $i \geq 2$, we conclude that
#### Problem 4-4b
It follows at once from 4-4a that
By the quadratic formula,
and the second equality follows.
Finally, we observe that
The third equality now follows.
(To deduce the third equality directly from the second equality, we must carry out the partial fraction decomposition of the second expression.) $\square$
#### Problem 4-4c
Page 1147 of CLRS tells us that
Substituting $x = \phi z$, we get
Similarly,
The desired identity now follows from the third equality in 4-4b. $\square$
#### Problem 4-4d
We have shown in 4-4c that
for all indices $i$. Since $% $, we have
whence it follows that
In other words, $F_i$ is $\phi^i/\sqrt{5}$ rounded to the nearest integer. $\square$
### 4-5. Chip testing
Professor Diogenes has $n$ supposedly identical integrated-circuit chips that in principle are capable of testing each other. The professor’s test jig accommodates two chips at a time. When the jig is loaded, each chip tests the other and reports whether it is good or bad. A good chip always reports accurately whether the other chip is good or bad, but the professor cannot trust the answer of a bad chip. Thus, the four possible outcomes of a test are as follows:
Chip $A$ says Chip $B$ says Conclusion
$B$ is good $A$ is good both are good, or both are bad
$B$ is good $A$ is bad at least one is bad
$B$ is bad $A$ is good at least one is bad
$B$ is bad $A$ is bad at least one is bad
a. Show that if at least $n/2$ chips are bad, the professor cannot necessarily determine which chips are good using any strategy based on this kind of pairwise test. Assume that the bad chips can conspire to fool the professor.
b. Consider the problem of finding a single good chip from among $n$ chips, assuming that more than $n/2$ of the chips are good. Show that $\lfloor n/2 \rfloor$ pairwise tests are sufficient to reduce the problem to one of nearly half the size.
c. Show that the good chips can be identified with $\Theta(n)$ pairwise tests, assuming that more than $n/2$ of the chips are good. Give and solve the recurrence that describes the number of tests.
#### Notational Remark
Given two chips $x$ and $y$, we write $T_x(y)$ to denote what $x$ has to say about $y$. The possible outcomes are, of course, good, and bad.
A test consists of inputting two chips $x$ and $y$ and receiving the output $(T_x(y), T_y(x)).$
#### Problem 4-5a
Assume that at least $n/2$ chips are bad. Let $\mathcal{G}$ be the set of all good chips, and let $\mathcal{B}$ be a set of bad chips of size $\vert \mathcal{G} \vert$. We let $\mathcal{R}$ be the remaining chips, if there are any.
Observe that, for each $x \in \mathcal{G}$,
Let us consider the following adversarial strategy: each $x \in \mathcal{B}$ returns
and each $x \in \mathcal{R}$ returns
We now partition the collection of chips by declaring two chips $x$ and $y$ to be in the same category if and only if
The partition process produces three categories: namely, $\mathcal{G}$, $\mathcal{B}$, and $\mathcal{R}$.
A test does not show which category a pair of chips belongs to. It merely shows whether two chips belong to the same category. Since $\mathcal{G}$ and $\mathcal{B}$ are of the same size, this information is insufficient to distinguish the two categories from one another. It follows that the chosen adversarial strategy prevents distinguishing good chips from bad chips. $\square$
#### Problem 4-5b
Let us assume that more than $n/2$ of the chips are good.
We assume that $n$ is even, and set $m = n/2$. If there is an odd number of chips, we discard one at random to make the number of chips even.
We divide the chips into two groups of equal sizes, $\mathcal{X} = \{x_1,\ldots,x_m\}$ and $\mathcal{Y} = \{y_1,\ldots,y_m\}$, assigning indices $1,\ldots,m$ arbitrarily.
For each index $i$, we declare the pair $(x_i, y_i)$ to be in the keep pile $\mathcal{K}$ if and only if
In other words, either $x_i$ and $y_i$ are both good chips or they are both bad chips.
We also define the discard pile $\mathcal{D}$ to be the collection of all pairs $(x_i,y_i)$ that do not belong to $\mathcal{K}$. We denote by $\vert \mathcal{K} \vert$ and $\vert \mathcal{D} \vert$ the number of pairs in $\mathcal{K}$ and $\mathcal{D}$, respectively. We note here that
Moreover, each pair in $\mathcal{D}$ must contain at least one bad chip, and so
We now observe that $\mathcal{K}$ contains more than $\vert \mathcal{K} \vert /2$ good pairs. Indeed, if $\mathcal{K}$ contained at most $\vert \mathcal{K} \vert / 2$ good pairs, then
which is absurd.
In light of this observation, we define
Since $\mathcal{R}$ contains precisely $\vert \mathcal{K} \vert$ many chips, we see from $(\ast)$ that $\mathcal{R}$ cannot contain more than $n/2 - \vert \mathcal{D} \vert$ many chips. Moreover, more than half of the pairs in $\mathcal{K}$ are good pairs, and so more than half of the chips in $\mathcal{R}$ are good chips. We now recurse on $\mathcal{R}$. $\square$
The procedure described above can be implemented as follows:
"""We identify chips by unique nonnegative integers and assume that
chip_table, an immutable two-dimensional list of numbers, is given.
To see what a testing chip ("testing") has to say about another chip
("tested"), we call chip_table[testing][tested], which returns
True for 'good', and False for 'bad'.
We record the chips to be tested in a list of numbers called chip_list.
chip_list is changed at each recursive step to track the progress.
"""
def get_a_good_chip(chip_table, chip_list):
n = len(chip_list)
if n <= 2:
return chip_list[0]
if n % 2 != 0: # if length mod 2 is not zero, i.e., odd
chip_list = chip_list[:-1] # cut out the last element
n = n-1
m = n//2
list_x, list_y = chip_list[:m], chip_list[m:]
keep = []
for i in range(m):
x, y = list_x[i], list_y[i]
if chip_table[x][y] and chip_table[y][x]: # if both return True
keep.append((x, y))
new_chip_list = [pair[0] for pair in keep]
return get_a_good_chip(chip_table, new_chip_list)
>>> # chips 1, 2, 3 are good
>>> chip_table = [[True, False, True, False, True],
... [False, True, True, True, False],
... [False, True, True, True, False],
... [False, True, True, True, False],
... [True, False, False, False, True]]
>>> get_a_good_chip(chip_table, [0, 1, 2, 3, 4])
1
>>> # chips 2, 3, 4, 5 are good
>>> chip_table = [[True, True, False, False, False, False],
... [True, True, False, False, False, False],
... [False, False, True, True, True, True],
... [False, False, True, True, True, True],
... [False, False, True, True, True, True],
... [False, False, True, True, True, True]]
>>> get_a_good_chip(chip_table, [0, 1, 2, 3, 4, 5])
2
#### Problem 4-5c
Once one good chip is found, it suffices to test all chips against the good chip to identify all good chips. This takes $\Theta(n)$ pairwise tests and can be implemented as follows:
def get_all_good_chips(chip_table):
n = len(chip_table)
testing = get_a_good_chip(chip_table, range(n))
good_chips = [tested for tested in range(n)
if chip_table[testing][tested]]
return good_chips
>>> # chips 1, 2, 3 are good
>>> chip_table = [[True, False, True, False, True],
... [False, True, True, True, False],
... [False, True, True, True, False],
... [False, True, True, True, False],
... [True, False, False, False, True]]
>>> get_all_good_chips(chip_table)
[1, 2, 3]
>>> # chips 2, 3, 4, 5 are good
>>> chip_table = [[True, True, False, False, False, False],
... [True, True, False, False, False, False],
... [False, False, True, True, True, True],
... [False, False, True, True, True, True],
... [False, False, True, True, True, True],
... [False, False, True, True, True, True]]
>>> get_all_good_chips(chip_table)
[2, 3, 4, 5]
The time complexity $S$ of get_all_good_chips is
where the time complexity $T$ of get_a_good_chip is
The master theorem now implies that
whence it follows that
### 4-6. Monge arrays
An $m \times n$ array $A$ of real numbers is a Monge array if for all $i$, $j$, $k$, and $l$ such that $% $ and $% $, we have
In other words, whenever we pick two rows and two columns of a Monge array and consider the four elements at the intersection of the rows and the columns, the sum of the upper-left and lower-right elements is less than or equal to the sum of the lower-left and upper-right elements. For example, the following array is Monge.
a. Prove that an array is Monge if and only if for all $i=1,2,\ldots, m-1$ and $j=1,2,\ldots,n-1$, we have
(Hint: For the “if” part, use induction separately on rows and columns.)
b. The following array is not Monge. Change one element in order to make it Monge. (Hint: Use part(a))
c. Let $f(i)$ be the index of the column containing the leftmost minimum element of row $i$. Prove that $f(1) \leq f(2) \leq \cdots \leq f(m)$ for any $m \times n$ Monge array.
d. Here is a description of a divide-and-conquer algorithm that computes the leftmost minimum element in each row of an $m \times n$ Monge array $A$:
Construct a submatrix $A'$ of $A$ consisting of the even-numbered rows of $A$. Recursively determine the leftmost minimum for each row of $A'$. Then compute the leftmost minimum in the odd-numbered rows of $A$.
Explain how to compute the leftmost minimum in the odd-numbered rows of $A$ (given that the leftmost minimum of the even-numbered rows is known) in $O(m+n)$ time.
e. Write the recurrence describing the running time of the algorithm described in part (d). Show that its solution is $O(m + n \log m)$.
#### Notational remark
As stated in the preface, we shall use the 0-first indexing convention for this problem.
#### Problem 4-6a
$(\Rightarrow)$ If $A$ is Monge, it suffices to pick $k = i+1$ and $l = j+1$. $\square$
$(\Leftarrow)$ We suppose that
for all $i=1,2,\ldots,m-1$ and $j=1,2,\ldots,n-1$.
We fix $i$ and $j$ and show that
for all $p \geq 1$ and $q \geq 1$.
Observe that $(\ast)$ is the $p = q = 1$ case. We let $q = 1$, fix $p > 1$, and assume inductively that
for all $% $. By the inductive hypothesis,
$(\ast)$ implies that
whence it follows that
as was to be shown. This establishes $(\ast\ast)$ for all $p \geq 1$ and $q = 1$.
We now fix $p > 1$ and $q > 1$ and assume inductively that
for all $% $. By the inductive hypothesis,
The $q =1$ case of $(\ast\ast)$ implies that
whence it follows that
as was to be shown. This establishes $(\ast\ast)$ in full generality.
$(\ast)$ follows at once from $(\ast\ast)$. $\square$
#### Problem 4-6b
The following code tests whether a two-dimensional NumPy array is Monge, using the criterion developed in 4-6a:
"""A is assumed to be a two-dimensional NumPy array.
- The syntax for referencing an entry of A is A[i, j].
- A.shape returns (m, n), the size of the matrix.
"""
def test_monge(A):
"""Check if A is a Monge array using the equivalent definition
derived in 4-6a"""
num_rows, num_cols = A.shape
failed = []
for i in range(num_rows-1):
for j in range(num_cols-1):
if A[i, j]+A[i+1, j+1] > A[i, j+1]+A[i+1, j]:
failed.append((i, j))
if not failed:
print("Monge")
is_monge = True
else:
print("Not Monge")
print("Failed coordinates: ")
for pair in failed:
print(pair)
is_monge = False
return is_monge
>>> test_monge(np.array([[10, 17, 13, 28, 23],
... [17, 22, 16, 29, 23],
... [24, 28, 22, 34, 24],
... [11, 13, 6, 17, 7],
... [45, 44, 32, 37, 23],
... [36, 33, 19, 21, 6],
... [75, 66, 51, 53, 34]])
Monge
>>> test_monge(np.array([[37, 23, 22, 32],
... [21, 6, 7, 10],
... [53, 34, 30, 31],
... [32, 13, 9, 6],
... [43, 21, 15, 8]])
Not Monge
Failed coordinates:
(0, 1)
As the program suggests, the inequality
fails to hold, as
Since we have
we can add up to 7 to $A[0,2]$ without breaking the inequality
We therefore assign
at which point we obtain
while maintaining
Since the modification affects no other inequality, the resulting array is Monge.
>>> test_monge(np.array([[37, 23, 29, 32],
... [21, 6, 7, 10],
... [53, 34, 30, 31],
... [32, 13, 9, 6],
... [43, 21, 15, 8]])
Monge
#### Problem 4-6c
Let $A$ be an $m \times n$ Monge array. We fix $% $ and assume for a contradiction that
on $A$. Since
we see that the inequality
can never hold unless it is also the case that
Since $f(i)$ is the index of the column containing the leftmost minimum element of row $i$, $(\ast)$ implies that we must always have
rendering the identity
false. We thus conclude that
Since the choice of $i$ was arbitrary, the desired result now follows. $\square$
#### Problem 4-6d
We assume that the values of $f(0),f(2),\ldots, f(2 \lfloor (n-1)/2 \rfloor)$ are known. We have seen in 4-6c that
for all $k$. Since we must examine at least one integer for each row, the number of integers we must examine to compute $f(2k+1)$ is bounded above by
It then follows that the number of integers we must examine for all the odd rows is bounded above by
We now observe that
whence the desired result follows. $\square$
#### Problem 4-6e
We begin by observing that a submatrix $A'$ of a Monge array $A$ consisting of the rows of $A$ is a Monge array. This, in particular, implies that the ordering relation proved in 4-6c continues to hold on such submatrices.
The algorithm for computing the leftmost minimum of each row of a Monge array can be implemented as follows:
"""A is assumed to be a two-dimensional NumPy array.
- The syntax for referencing an entry of A is A[i, j].
- A.shape returns (m, n), the size of the matrix.
- A[::2,:] is the submatrix consisting of the even rows of A.
0, k, 2k, 3k, and so on.
"""
def find_left_min(A):
if not test_monge(A):
return None
num_rows, num_cols = A.shape
mins = [0 for _ in range(num_rows)] # initialize array of length A.shape[0]
return lm_recursion(A, mins)
def min_index(row, start_index, end_index):
"""Scan a row from start_index to end_index-1 to find the
index of the leftmost minimum element in the specified range.
"""
min_ind = start_index
for i in range(start_index, end_index):
if row[i] < row[min_ind]:
min_ind = i
return min_ind
def lm_recursion(A, mins):
num_rows, num_cols = A.shape
if num_rows == 1: # if A has only one row, use the min_index scan
mins[0] = min_index(A[0], 0, num_cols)
else:
## take the even rows and recurse
evens = A[::2, :]
even_mins = lm_recursion(evens, mins[::2])
## even rows have already been computed;
## use the min_index scan in the range specified by even_mins
## to figure out the min index for odd rows.
start_index = 0
for i in range(num_rows):
if i % 2 == 0:
mins[i] = even_mins[i//2]
else:
left = mins[i-1]
right = even_mins[(i+1)//2]+1 if i+1 < num_rows else num_cols
mins[i] = min_index(A[i], left, right)
return mins
>>> find_left_min(np.array([[10, 17, 13, 28, 23],
... [17, 22, 16, 29, 23],
... [24, 28, 22, 34, 24],
... [11, 13, 6, 17, 7 ],
... [45, 44, 32, 37, 23],
... [36, 33, 19, 21, 6 ],
... [75, 66, 51, 53, 34]))
[0, 2, 2, 2, 4, 4, 4]
>>> find_left_min(np.array([[37, 23, 29, 32],
... [21, 6, 7, 10],
... [53, 34, 30, 31],
... [32, 13, 9, 6],
... [43, 21, 15, 8]]))
[1, 1, 2, 3, 3]
What is the running time $T(m)$ of find_left_min, where $m$ is the number of rows? For the purpose of this discussion, let us focus on analyzing lm_recursion.
Let $A$ be an $m \times n$ array, so that num_rows == m and num_cols == n. If $m=1$, then the running time of lm_recursion is precisely that of min_index. Since min_index scans through each column once, we see that the running time is $\Theta(n)$. We thus declare
As for the $m > 1$ case, we observe that lm_recursion consists of three steps:
1. Construct evens, the subarray of even rows, from $A$.
2. Recurse on evens.
3. Go through the $m$ rows and compute mins.
Since $A$ is never modified throughout lm_recursion, there is no need to copy the data of $A$ to construct evens, which becomes $A$ in the recursive step. The construction of even thus reduces to collecting pointers to the even rows, which can be done in $\Theta(m/2)$ time.
The recursion step is invoked on evens, an array of $m/2$ rows. Since the recursion step is invoked precisely once without any additional procedure, it takes $T(m/2)$ time to run.
Finally, we have shown in 4-6d that the computation of mins takes $\Theta(m+n)$ time. It follows that the running time of lm_recursion is
Since we do not know what $n$ is in relation to $m$, we cannot apply the master theorem. Observe, however, that
Since
for all choices of $I$, we see that
We have shown above that $T(1) = \Theta(m)$, and so
Let us now prove the above estimate by induction. To this end, we fix $n$ and choose constants $k,K > 0$ such that
for all choices of $m$.
Let us first establish the upper bound
Fix $m \geq 2$ and assume inductively that
for all $% $, where $C,D > 0$ are constants to be chosen.
Observe that
So long as
we have
The above condition is satisfied precisely when
To establish the base case $m = 1$, we only need the hypothesis
Indeed, this leads to
We thus choose
so that $(\ast\ast)$ is trivially established, and that
thereby establishing $(\ast)$. It now follows from induction that
It remains to establish the lower bound
We fix $m \geq 2$ and assume inductively that
for all $% $, where $C,D > 0$ are constants to be chosen. Observe that
In order to obtain
we must have
and
Observe that the second inequality, along with the added hypothesis that
implies
which is precisely the first inequality. The inductive step thus holds true so long as
To establish the base case $m = 1$, we only need the hypothesis
Indeed, this leads to
We therefore choose
so that $(\star)$ and $(\star\star)$ follow. We now conclude by induction that
which, along with the upper bound established above, implies
as was to be shown. $\square$
## Additional remarks and further results
### Generalizations of the master method
The master method does not apply to divide-and-conquer problems with multiple subproblems of substantially different sizes. For example,
is not covered by the master method.
Mohamad Akra and Louay Bazzi, in their 1996 paper “On the Soluton of Linear Recurrence Equations,” present a generalization of the master method that is applicable to recurrence relations of the form
with minor restrictions on $f$. The solution is given by the formula
where $p$ is the unique real number that satisfies the identity
For the above example, we have $p = 1$, and so
Using the Akra—Bazzi method, we can also solve more complicated recurrence relations like
In this case, we have
and so the solution to the recurrence relation is
There are, of course, several generalizations beside the Akra—Bazzi method. We refer the reader to Chee Yap’s 2011 paper, “A real elementary approach to the master recurrence and generalizations.” $\square$
### Generating functions
We have discussed briefly in Problem 4-4 the generating-function technique of solving recurrence relations. We pursue the subject further here, following the treatment in Chapter 3 of Sedgewick/Flajolet.
Given a sequence $a_0,a_1,\ldots,a_k,\ldots$, we construct its ordinary generating function
and its exponential generating function
The generating-function method of solving recurrences typically involves multiplying both sides of the recurrence by $z^n$ ($z^n/n!$ for exponential generating functions), summing on $n$, manipulating the resulting identity to derive an explicit formula for the generating function, and computing the power-series expansion of the generating function.
Take, for example,
with the initial condition $a_0 = 0$, $a_1 = 1$, $a_2 = 1$. We multiply both sides by $z^n$, sum over $n \geq 3$, and reindex the sums to obtain the identity
For notational simplicity, we let
the ordinary generating function of $(a_{n})_{n}$. Observe that
We can use the above computation to simplify $(\ast)$ as follows:
Using the initial condition $a_0 = 0$, $a_1 = 1$, $a_2 = 1$, we conclude that
Taking the partial fraction decomposition, we obtain
We now recall that
from which we conclude that
It follows that
for all $n \geq 0$.
We shall revisit the method of generating functions in later chapters, as more sophisticated recurrence relations arise. $\square$ | 2018-05-22T11:39:34 | {
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https://math.stackexchange.com/questions/2774156/why-do-integrals-start-at-0 | # Why do integrals “start” at 0? [duplicate]
This is a dumb question and I don't really know how to word it. When you take an antiderivative and plug in number you are given the area under the curve starting at 0 (assuming C is 0). I can easily see how the derivative of an integral is given by the function value, but why does the integral start at 0 and not any other number? When I try to imagine the area of some curve starting at negative 1 for example the area under the curve would intuitively to me still be given by the antiderivative. 0 makes sense as a starting point but for some reason I can't visualize it. I'm not sure if that made any sense but if anyone could help me wrap my head around it I'd appreciate it.
## marked as duplicate by BallBoy, астон вілла олоф мэллбэрг, Claude Leibovici integration StackExchange.ready(function() { if (StackExchange.options.isMobile) return; $('.dupe-hammer-message-hover:not(.hover-bound)').each(function() { var$hover = $(this).addClass('hover-bound'),$msg = $hover.siblings('.dupe-hammer-message');$hover.hover( function() { $hover.showInfoMessage('', { messageElement:$msg.clone().show(), transient: false, position: { my: 'bottom left', at: 'top center', offsetTop: -7 }, dismissable: false, relativeToBody: true }); }, function() { StackExchange.helpers.removeMessages(); } ); }); }); May 10 '18 at 6:38
• It doesn't matter. It's just convenient. As for your example, $F(x)=\int_{-1}^x f(t)dt$ is another anti-derivative of $f$. – Yanko May 9 '18 at 19:09
• "given by the antiderivative" There is no "the antiderivative". There is only "an antiderivative". – Arthur May 9 '18 at 19:13
• $\int dx/x$ certainly doesn't start from $0$! – Chappers May 9 '18 at 19:15
• Starting from somewhere else amounts to changing the constant of integration. – amd May 9 '18 at 20:46
• @LSpice, I may have been misinterpreting what op meant. I was essentially thinking the confusion was about why an integral over a degenerate interval was zero. – jdods May 10 '18 at 3:11
When you take an antiderivative and plug in number you are given the area under the curve starting at 0 (assuming C is 0).
This is not true. In certain situations it may be the case, but not generally. I think the reason this confusion arises is that a common problem given to calculus students is to find the antiderivative of a polynomial, e.g.,
$$\int x^3 +2x \, dx = \frac{1}{4}x^4 + x^2 + C$$
and in this case, if we set $C = 0$ we get
$$\frac{1}{4}x^4 + x^2$$
which is the same as $$\int_0^x u^3 +2u \, du = \frac{1}{4}u^4 + u^2 \big|_{u=0}^{u=x}.$$
This will work whenever the form that the antiderivative $F$ of $f$ you get takes satisfies $F(0) = 0$. But in general, setting $C = 0$ will not get you the integral $\int_0^x f(t) \, dt$. For example, if you take $f(x) = e^x$ then $$\int e^x \, dx = e^x + C$$ but setting $C = 0$ gives you $e^x$, which is not the same as $$\int_0^x e^t \, dt = e^x - 1.$$
Note that "setting "C = 0" in the expression for the antiderivative" is not actually a well-defined operation. Different methods of antidifferentiation can give you different expressions when you set $C = 0$. It is important to remember that there is no single antiderivative, and no canonical way of writing it. $\int 2x \, dx = x^2 + 3 + C$ is just as valid as $\int 2x \, dx = x^2 + C$.
• Ahh you're e^x example shows that my assumption was wrong. Thanks for clearing it up. – conyare May 9 '18 at 19:39
• My favourite example is arcsin(x)+C vs -arccos(x)+C, these are equal families, but which function is supposed to be C=0? – Serge Seredenko May 9 '18 at 20:27
• @SergeSeredenko, or $\sin(x)^2 + C$ versus $-\cos(x)^2 + C$. – LSpice May 10 '18 at 1:03
• Also see this recent question where the difference between two solutions doesn't even look like a constant, but it is: math.stackexchange.com/q/2764985/78887 – Euro Micelli May 10 '18 at 3:18
Integrals don't always start at $0$. Let's start from definite integrals and move to the indefinite integrals you asked about. We know $\int_a^b f(x)dx$ gives the area under the curve between $a$ and $b$. If $f(x)$ has an antiderivative $F(x)$, the Fundamental Theorem of Calculus tells us that $\int_a^b f(x) = F(b) - F(a)$. Thus, if we want the area under the curve from $a$ to $b$, we compute $F(b) - F(a)$. If we want the area under the curve from $0$ to $b$, we compute $F(b) - F(0)$. $F(b) - F(0)$, as a function of $b$, gives us the area from $0$ to $b$.
Now there are a lot of "natural" functions where $F(0) = 0$ (e.g., functions like $x^2$ or $\sin x$), so to $F(b)$ gives the area from $0$ to $b$. But that won't be the case if $F(0) \neq 0$.
The above should make it clear that there's no reason $0$ is special -- if you want the area from $-1$ to $b$ as a function of $b$, just use $F(b) - F(-1)$.
• I assumed that the fundamental theorem worked because F(a) gets the area to A from 0 and F(b) gets the area to B from 0 so subtracting will get the desired area in-between them. – conyare May 9 '18 at 19:18
• @conyare No, it actually goes the other way around! It is true that if all you knew were the 0-to-b version of the fundamental theorem, you could recover the a-to-b version. But the subtraction of another value is "fundamental" (pun somewhat intended) to the fundamental theorem. Jeffery Opoku-Mensah hints at a reason why this is true: since $F$ is defined only up to a constant, $F(b)$ doesn't give you a definite answer, only up to a constant. Only once you subtract $F(b)-F(a)$ does the constant cancel and you get a definite area. – BallBoy May 9 '18 at 19:21
Well, integrals (by which I think you mean anti derivatives) don't always start at $0$. Indeed, if a function $f=f(x)$ is continuous in some interval $I=[a,b]$, then it has an anti derivative given by $$\int_c^x{f(t)dt},$$ where $c\in I$. The $0$ is usually chosen for $c$ only as a matter of convenience. A well-known function defined by an antiderivative that "starts" from (note that the notion of starting from for antiderivatives should not be taken too literally since the function is defined even for $x<c\in I$) $1$, not $0$ as usual, is the logarithm function $$\log x=\int_1^x{\frac{1}{t}dt}$$ defined for all $x>0$.
The antiderivative is generally given as the "simplest" form, and often the simplest form has a y value of zero when x is zero. For instance, if the function is constant, then for the antiderivative, you need a line whose slop is equal to that constant value. The simplest way of writing that is y = mx+C. You could write y = mx+5-C, and that would be a valid antiderivative, but that would be needlessly complicated. Since you're looking for simple functions, zero will pop up a lot. But there are cases where the simplest function doesn't go through the origin. For instance, if you're taking the antiderivative of sin(x), the simplest form is cos(x)+C. If you want to "start" at zero, you'd have to do cos(x)-1+C. Similarly, the antiderivative of e^x is generally given as e^x+C.
There are some cases where the same function can have different antiderivatives that look very different, but are actually the same thing. For instance, $\frac{1}{\sqrt{1-x^2}}$ can be integrated as either arcsin(x) or -arccos(x). But these just differ by the constant $\pi /2$.
It depends on what antiderivative you take. For example, an antiderivative of $2x$ can be $x^2$, $x^2+1$, $x^2+100$, or whatever. By the Fundamental Theorem, it turns out $x^2$ gives the area under the curve starting at zero. But $x^2+1$ would also give the area of the curve starting at $-1$, for example.
• I'm afraid you're confusing derivatives with antiderivatives... – zipirovich May 9 '18 at 19:22
• Yea, I wrote this in a confused rush, but the idea is still there. Just switch the x^2 and 2x. – Jeffery Opoku-Mensah May 9 '18 at 21:59 | 2019-08-20T09:38:08 | {
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https://mathematica.stackexchange.com/questions/145050/filling-space-with-pursuit-polygons?noredirect=1 | Filling Space with Pursuit Polygons
I want to make a program which can fill a 2D space with "pursuit polygons".
You can also look up "pursuit curves" or mice problem or watch this gif
What I've tried so far:
First I tried to produce these polygons by rotation of polygons
the square for example
data = {{0, 0}, {1, 0}, {1, 1}, {0, 1}, {0, 0}};
Graphics[{Table[{Scale[Rotate[Line[data], 3*x Degree], {x, x}]}, {x,
0, 20}]}]
then I decided to use spirals
I realised that instead of using n spirals for every n-polygon,
I can use only one(!) and produce the effect I want with "fine tuning"
here are the "pursuit polygons" that I made
Triangle:
a = 41.9;
b = 100;
W = Table[{t^2*Cos[a*t], t^2*Sin[a*t]}, {t, 0, b}];
AppendTo[W, W[[-4]]];
ListPlot[W]
Graphics[Line[W]]
Square:
S = Table[{t^2*Cos[42.4*t], t^2*Sin[42.4*t]}, {t, 0, 160}];
AppendTo[S, S[[-4]]];
Graphics[Line[S]]
Pentagon
P = Table[{t^2*Cos[45.251*t], t^2*Sin[45.251*t]}, {t, 0, 220}];
AppendTo[P, P[[-5]]];
Graphics[Line[P]]
Then I tried to put all these together by rotating and scaling...
(but I'm not satisfied with the result)
notice that for every polygon I use also the "anti-clockwise" version of the polygon, which produces interesting results
T = Table[{t^2*Cos[41.9*t], t^2*Sin[41.9*t]}, {t, 0, 100}];
AppendTo[T, T[[-4]]]; (*Triangle*)
S = Table[{t^2*Cos[42.4*t], t^2*Sin[42.4*t]}, {t, 0, 160}];
AppendTo[S, S[[-4]]]; (*square*)
S1 = Table[{t^2*Cos[-42.4*t], t^2*Sin[-42.4*t]}, {t, 0, 160}];
AppendTo[S1, S1[[-4]]]; (*Anti-clockwise Square*)
P = Table[{t^2*Cos[45.251*t], t^2*Sin[45.251*t]}, {t, 0, 220}];
AppendTo[P, P[[-5]]]; (*Pentagon*)
P1 = Table[{t^2*Cos[-45.251*t], t^2*Sin[-45.251*t]}, {t, 0, 220}];
AppendTo[P1, P1[[-5]]]; (*Anti-clockwise Pentagon*)
Graphics[{Translate[
Rotate[Scale[Line[(T)], {2.09, 2.09}], -67 Degree], {29500, 3800}],
Rotate[Scale[Line[S + 0], {1, 1}], -30 Degree],
Translate[
Rotate[Scale[Line[(S1)], {.98, .98}], -87.5 Degree], {41000,
24700}],
Translate[
Rotate[Scale[Line[P1], {.605, .665}], -20.5 Degree], {76500,
6500}], Scale[
Translate[
Rotate[Scale[
Line[(T)], {1.2, 1.7}], -108 Degree], {50900, -12500}], {1.91,
1.31}],
Translate[
Rotate[Scale[Line[(P)], {.525, .665}], -64 Degree], {3000, 36000}],
Translate[
Rotate[Scale[Line[(P)], {.61, .61}], -87 Degree], {59000, 58000}],
Translate[
Rotate[Scale[Line[(T)], {2.1, 1.8}], 12 Degree], {82500, 39500}],
Translate[
Rotate[Scale[Line[(T)], {2.3, 1.9}], -133 Degree], {32000,
67000}]}]
Can you find a way to divide and fill any given space with pursuit polygons?
The result would look better if this could work with ANY convex polygon and not only the regular polygons that I used...
• Are there any other names for that? Because I think this was asked before. – Kuba May 2 '17 at 10:05
• I provided all the names I could think of. There are not many images outthere.Even the first picture is something that I made a while ago. – J42161217 May 2 '17 at 10:14
• @Kuba Lucas, E. (1877). Problème des trois chiens. Nouvelle Correspondance Mathématique, 3, 175–176. – Michael E2 May 2 '17 at 11:54
• A related question. – J. M. is away May 11 '17 at 13:15
• And a very related blog-post of mine. And if I'm not completely off here, the word "spacefilling" is incorrect in this context. – halirutan Jun 11 '17 at 15:01
The reason that these are called "pursuit polygons" is because they are formed from a dynamical system in which different agents pursue each other. Example:
In this image, one agent starts in each corner of the triangle. The agent starting in the lower right corner pursues the agent starting in the top corner, the agent in the top corner pursues the agent in the lower left corner, and the agent in the lower left corner pursues the agent in the lower right corner.
Drawing lines in between the agents yields a series of triangles which shrink as the agents get closer to each other, and also rotate:
The corresponding differential equations are: \begin{align*} \dot{\mathbf{x}}_i &= \mathbf{x}_{i+1} - \mathbf{x}_i,\ i\in\{1,\dots,N-1\}\\ \dot{\mathbf{x}}_i &= \mathbf{x}_1 - \mathbf{x}_N,\ i=N, \end{align*} where $N$ is the number of agents. The agents start in corners for the visualizations we are making, but in general they don't have to.
This code solves the differential equations using Euler integration and plots the lines between the agents:
integrate[pts_, h_] := NestList[Nest[step[#, h] &, #, 10] &, pts, 100]
step[pts_, h_] := pts - h (pts - RotateRight[pts])
wrapAround[pts_] := Append[pts, First[pts]]
PursuitPolygon[pts_, h_: 0.01] := Graphics[{
Line[wrapAround[pts]],
Line[wrapAround /@ integrate[pts, h]]
}]
For regular polygons, this results in logarithmic spirals, as you have pointed out:
Grid@Partition[Table[PursuitPolygon[N@CirclePoints[n]], {n, 2, 10}], 2]
But this approach is more general and works for any convex polygons, as you required. The following is an example of how we can fill in a triangulated region with spirals using this system:
reg = DiscretizeRegion[Rectangle[], MaxCellMeasure -> 0.1]
Show[PursuitPolygon @@@ MeshPrimitives[reg, 2]]
Wikipedia's entry on the mice problem suggests that the mice move at unit speed, i.e. like this: \begin{align*} \dot{\mathbf{x}}_i &= \frac{\mathbf{x}_{i+1} - \mathbf{x}_i}{\|\mathbf{x}_{i+1} - \mathbf{x}_i\|},\ i\in\{1,\ldots,N-1\}\\ \dot{\mathbf{x}}_i &= \frac{\mathbf{x}_1 - \mathbf{x}_N}{\|\mathbf{x}_1 - \mathbf{x}_N\|},\ i=N, \end{align*} This turns out to be difficult to implement because the denominator grows large as the points approach each other. I spoke to halirutan and MichaelE2 about this. Here is MichaelE2's solution, which I implemented for triangles but it could be implemented for any polygon.
PursuitPolygon[{pt1_, pt2_, pt3_}] := Module[{},
{sol1x, sol1y, sol2x, sol2y, sol3x, sol3y} = NDSolveValue[{
{x1'[t], y1'[t]} == Normalize[{x2[t], y2[t]} - {x1[t], y1[t]}],
{x2'[t], y2'[t]} == Normalize[{x3[t], y3[t]} - {x2[t], y2[t]}],
{x3'[t], y3'[t]} == Normalize[{x1[t], y1[t]} - {x3[t], y3[t]}],
{x1[0], y1[0]} == pt1,
{x2[0], y2[0]} == pt2,
{x3[0], y3[0]} == pt3,
WhenEvent[Norm[{x2[t], y2[t]} - {x1[t], y1[t]}] < 1*^-8, "StopIntegration"]},
{x1, y1, x2, y2, x3, y3},
{t, 0, 10}
];
{tmin, tmax} = MinMax@sol1x["Grid"];
Graphics@Table[Line[{{sol1x[t], sol1y[t]}, {sol2x[t], sol2y[t]}, {sol3x[t], sol3y[t]}, {sol1x[t], sol1y[t]}}], {t, tmin, tmax, (tmax - tmin)/15}]
]
Show[PursuitPolygon @@@ MeshPrimitives[reg, 2]]
The solution is qualitatively different when the agents move at unit speed because they no longer meet at the centroid. Note that the difference in the picture is also that the density of the lines is different, this density can be adjusted in both solutions but changing some of the fixed numbers I put in. 15 in (tmax - tmin)/15 controls the number of lines in the last solution.
Here's an extension of @C.E.'s code to non-triangular polygons - polygons taken from a Voronoi mesh of roughly equidistributed (https://mathematica.stackexchange.com/a/141215/3056) 31 points over a square. Code is awful, I didn't really have time to think.
ClearAll@PursuitPolygon;
PursuitPolygon[pts_] := Module[{vars, sols},
vars = Table[Unique[], Length@pts, 2];
sols = NDSolveValue[{Sequence @@ (((#'[t] & /@ #1) ==
Normalize[Subtract @@ Map[#[t] &, {#2, #1}, {2}]]) & @@@
Partition[Join[vars, {First@vars}], 2, 1]),
Sequence @@ MapThread[(#[0] & /@ #1) == #2 &, {vars, pts}],
WhenEvent[Norm[x] < 1*^-8, "StopIntegration"] /.
x -> Subtract @@ Map[#[t] &, Take[vars, 2], {2}]},
Flatten@vars, {t, 0, 10}];
{tmin, tmax} = MinMax@First[sols]["Grid"];
Graphics@
Table[Line[
Map[#[t] &, Join[Partition[sols, 2], {Take[sols, 2]}], {2}]], {t,
tmin, tmax, (tmax - tmin)/15}]]
ClearAll@reg;
reg =
With[{reg = Rectangle[]},
With[{points = 31, samples = 4000, iterations = 100},
Nest[With[{randoms = Join[#, RandomPoint[reg, samples]]},
RegionNearest[reg][
Mean@randoms[[#]] & /@
Values@PositionIndex@Nearest[#, randoms]]] &,
RandomPoint[reg, points], iterations]] //
VoronoiMesh[#, {{0, 1}, {0, 1}}] &];
Show[PursuitPolygon /@ MeshPrimitives[reg, 2][[All, 1]]] | 2019-06-20T20:18:36 | {
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https://www.physicsforums.com/threads/integration-by-partial-fractions.754530/ | # Integration by Partial Fractions
1. May 20, 2014
1. The problem statement, all variables and given/known data
Find the indefinite integral of the below, using partial fractions.
$$\frac{4x^2+6x-1}{(x+3)(2x^2-1)}$$
2. Relevant equations
?
3. The attempt at a solution
First I want to say there is probably a much easier and quicker way to get around certain things I have done but I have just done it the only way I could see (I always seem to go the long way around). So if I am correct then I would really appreciate some tips on how to do it quicker (for in exams) and also if I am incorrect to point out my mistakes. Thanks :)
first I set up the partial fractions.
$\frac{4x^2+6x-1}{(x+3)(2x^2-1)}=\frac{A}{x+3}+\frac{Bx+C}{2x^2-1} \\ \rightarrow \,\,\, 4x^2+6x-1=A(2x^2-1)+(Bx+C)(x+3) \\ 4x^2+6x-1=2Ax^2-A+Bx^2+3Bx+Cx+3C$
And then equated the coefficients:
For X^2: $4=2A+B$
For X^1: $6=3B+C$
For X^0: $-1=3C-A$
Then what I did what multiply the third equation above by 2 to get $-2=6C-2A$ and then added it to the first equation to get $2=6C+B$ and solved for B and substituted it into the second equation which resulted in:
$6=3(2-6C)+C\\ 6=6-11C \\ 0=-11C \\ ∴ C=0$
And then knowing C=0 I found A=1 and then that B=2 .
$\frac{4x^2+6x-1}{(x+3)(2x^2-1)}=\frac{1}{x+3}+\frac{2x}{2x^2-1} \\ \frac{4x^2+6x-1}{(x+3)(2x^2-1)}=\frac{1}{x+3}+\frac{2x}{2x^2-1}\\ \int \frac{4x^2+6x-1}{(x+3)(2x^2-1)} \,\, dx = \ln{|x+3|} + \frac{1}{2} \ln{|2x^2-1|}$
I think I may have done something wrong. I had to use an integral calculator online to find the integral of $\frac{2x}{2x^2-1}$ as I had no idea how to do it. Which kind of defeats the point.
2. May 20, 2014
### Saitama
To evaluate the last integral, you can use the substitution $2x^2-1=t$.
3. May 20, 2014
### Panphobia
Its all right, except you need to add a + C. But by the way the integral of 2x/(2x^2-1) can be easily solved using u substitution.
If u = 2x^2 -1 then du = 4x dx so du/2 = 2x dx
(1/2)*(1/u)
and the integral of this is
ln|u|/2
which is
ln|2x^2-1|/2
Last edited: May 20, 2014
4. May 20, 2014
### Curious3141
You're missing the constant of integration, which is very important in an indefinite integral.
The partial fraction decomposition is tedious. If the denominator was easily factored into linear factors, Heaviside cover up rule would help more. As it stands, the cover up rule helps to find A quite quickly, but you get bogged down with B and C, and you have to manipulate radicals. Pointless. I think your method is fine.
5. May 20, 2014
Ah yes, sorry I forgot as it was just rough workings copied up into latex.
I am quite poor when it comes to integration by substitution for some reason. I find integration by parts easier as its a set formula but I do realise that integration by substitution is sometimes a lot easier and quicker so I need to get more familiar with it. Any tips on u substitution?
I can usually find what to substitute, and then get to a du=... but its the following bits that confuse me a little.
6. May 20, 2014
### scurty
I try to use u-substitution if I see a power that is one higher in the denominator than in the numerator. A more general approach would to be to look for derivatives in the numerator from what is found in the denominator. I always look for u-substitution first because they tend to be the easiest when dealing with fractions. Partial fractions and trig substitution, while fun to compute, are more time consuming.
It's hard for me to explain because I've done so many integrals that knowing what method to use is more intuitive to me than anything else. It comes easier the more you practice integrals. A general rule to always follow for u substitution is to notice candidates for u and du. | 2017-11-23T08:03:11 | {
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https://math.stackexchange.com/questions/1297197/find-an-arbitrary-power-of-a-lower-triangular-matrix-of-size-3-times-3 | # Find an arbitrary power of a lower triangular matrix of size $3\times 3$
Let $F$ be a field and let $A=\begin{bmatrix}a&0&0\\1&a&0\\0&1&a\end{bmatrix}\in\mathscr{M}_{3\times 3}(F)$. Show that $$A^k=\begin{bmatrix}a^k&0&0\\ka^{k-1}&a^k&0\\\dfrac{1}{2}k(k-1)a^{k-2}&ka^{k-1}&a^k\end{bmatrix}$$ for all $k > 0$. (Exercise 797 from Golan, The Linear Algebra a Beginning Graduate Student Ought to Know.)
I know it can be proved by induction, but since the topic is about Krylov Spaces, eigenvalues and Jordan canonical form, I wonder if there is another way to solve this problem.
• Inducción, José! – Pedro Tamaroff May 24 '15 at 20:59
• it is spelled induction, and I'm asking for another way, thanks. – José May 24 '15 at 21:02
• This is a trivial matter using induction, really. I doubt there is any gain going in any other route... – Mariano Suárez-Álvarez May 24 '15 at 21:04
• Did you try Krylov spaces ? – Dietrich Burde May 24 '15 at 21:42
• @Dietrich Burde: what is Krylov space? Any good online refs? I've never heard of this approach for such problems . . . – Robert Lewis May 24 '15 at 22:26
There is extended reading giving formula for calculating functions such as polynomials on matrixes: http://en.wikipedia.org/wiki/Matrix_function#Jordan_decomposition
For a route different to induction, express the matrix as $$A=aI_3+G$$ where $I_3$ is the identity matrix and $G$ is given by $$G=\begin{bmatrix}0&0&0\\1&0&0\\0&1&0\end{bmatrix}$$ A binomial expansion will yield \begin{align}A^k&=(aI_3+G)^k\\&={k \choose 0}a^kI_3+{k \choose 1}a^{k-1}I_3G+{k \choose 2}a^{k-2}I_3G^2+\sum_{m=3}^k{k \choose m}a^{k-m}I_3G^m\\&=a^kI_3+ka^{k-1}I_3G+\frac{k(k-1)}{2}a^{k-2}I_3G^2+\sum_{m=3}^k{k \choose m}a^{k-m}I_3G^m\end{align} Now $I_3G$ is given by $$G=\begin{bmatrix}0&0&0\\1&0&0\\0&1&0\end{bmatrix}$$ while $I_3G^2$ is $$G=\begin{bmatrix}0&0&0\\0&0&0\\1&0&0\end{bmatrix}$$ and $I_3G^m$ for $m\geq3$ is the zero matrix.
Using these results, we arrive at the given expression for $A^k$.
• For the binomial expansion, would not explicitly (using matrix multiplication) working out $I_3G$, $I_3G^2$, and $I_3G^3$ suffice. Once we show that $I_3G^3$ is the zero matrix, then $I_3G^m$ for $m>0$ will also be zero matrices as well. – Alijah Ahmed May 24 '15 at 21:30
• Obviously I like your methodology! Endorsed!!! – Robert Lewis May 24 '15 at 22:29
Set
$N = \begin{bmatrix} 0 & 0 & 0 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \end{bmatrix}; \tag{1}$
then a simple calculation reveals that
$N^2 = \begin{bmatrix} 0 & 0 & 0 \\ 0 & 0 & 0 \\ 1 & 0 & 0 \end{bmatrix} \tag{2}$
and
$N^3 = 0. \tag{3}$
Also,
$A = aI + N. \tag{4}$
Since $N$ and $I$ commute, we may apply the ordinary binomial theorem to (4), and find
$A^k = (aI + N)^k =$ $(aI)^k + k(aI)^{k - 1}N + \dfrac{k(k - 1)}{2}(aI)^{k - 2}N^2$ $= a^kI^k + ka^{k - 1}I^{k - 1}N + \dfrac{k(k - 1)}{2}a^{k - 2}I^{k - 2}N^2$ $= a^kI + ka^{k - 1}N + \dfrac{k(k - 1)}{2} a^{k - 2}N^2, \tag{5}$
since the expansion is trunated after the third term by virtue of $N^3 = 0$. When (5) is written out explicitly, we see that
$A^k = \begin{bmatrix} a^k & 0 & 0 \\ ka^{k - 1} & a^k & 0 \\ \dfrac{k(k - 1)}{2} a^{ k - 2} & ka^{k - 1} & a^k \end{bmatrix}, \tag{6}$
as per request.
Nota Bene: a few words on induction. It has been rightly noted that in deploying the binomial theorem we do not in fact escape induction. The only other option I can see is to prove this result by direct matrix mutiplication, again using induction. Here we hide the induction by invoking the binomial theorem, doing so since the problem statement asks for "another way"; so I assumed direct induction should be avoided. But in truth, this is a statement concerning the integers, so induction will almost certainly enter in at some point, being as it is an essential defining property of the integers. End of Note
• @Dietrich Burde: Yes, yes, yes . . . take 'er easy, Pal, still editing. And I'll say a few words about induction before I'm done! Cheers! – Robert Lewis May 24 '15 at 21:33
• Sorry, everything is good. I just think, why not prove the (easy) matrix formula - by induction. – Dietrich Burde May 24 '15 at 21:35
• @Dietrich Burde: me too. Cheers! – Robert Lewis May 24 '15 at 21:38 | 2019-07-18T03:03:42 | {
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https://math.stackexchange.com/questions/3782436/question-about-convexity-how-do-we-prove-that-displaystyle-sum-i-1kp-i/3782448 | # Question about convexity: how do we prove that $\displaystyle \sum_{i=1}^{k}p_{i}b_{i}\geq\prod_{i=1}^{k}b^{p_{i}}_{i}$?
Let $$b_{1},b_{2},\ldots,b_{k}$$ be nonnegative numbers and $$p_{1} + p_{2} + \ldots + p_{k} = 1$$ where each $$p_{i}$$ is positive. Then
\begin{align*} \sum_{i=1}^{k}p_{i}b_{i}\geq\prod_{i=1}^{k}b^{p_{i}}_{i} \end{align*}
MY ATTEMPT
Since the logarithm function is strictly increasing, the proposed inequality is equivalent to \begin{align*} \ln\left(p_{1}b_{1} + p_{2}b_{2} + \ldots + p_{k}b_{k}\right) \geq p_{1}\ln(b_{1}) + p_{2}\ln(b_{2}) + \ldots + p_{k}\ln(b_{k}) \end{align*}
Once $$f''(x) < 0$$, where $$f(x) = \ln(x)$$, we conclude that $$f$$ is concave and the proposed inequality holds.
My question is: am I proving this result correctly? If this is the case, is there another way to prove it?
Any contribution is appreciated.
• You have done it absolutely correctly. Another method is by using weighted AM-GM as shown in my answer that I'm typing. Aug 6, 2020 at 22:17
Using $$b_1,b_2,\cdots,b_k\ge 0$$ with $$p_1,p_2,\cdots,p_k$$ as the respective weights, weighted AM-GM inequality gives $$\frac{\sum_{i=1}^k p_ib_i}{\sum_{i=1}^k p_i} \ge \left(\prod_{i=1}^k b_i^{p_i}\right)^{\dfrac{1}{\sum_{i=1}^k p_i}}$$ which gives the required inequality.
Note: This is not much different from your approach because the special case of Jensen's inequality for the function $$f(x)=\ln(x)$$ can be proved by weighted AM-GM or vice-versa, the weighted AM-GM inequality can be proved by the Jensen's inequality for $$f(x)=\ln(x)$$ as you have done, since the inequality you want to prove is directly the weighted AM-GM inequality in disguise. Also, these two can be proven independently without using each other. | 2022-08-19T07:57:51 | {
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https://dantopology.wordpress.com/2015/07/01/products-of-compact-spaces-with-countable-tightness/ | # Products of compact spaces with countable tightness
In the previous two posts, we discuss the definitions of the notion of tightness and its relation with free sequences. This post and the next post are to discuss the behavior of countable tightness under the product operation. In this post, we show that countable tightness behaves well in products of compact space. In particular we show that countable tightness is preserved in finite products and countable products of compact spaces. In the next post we show that countable tightness is easily destroyed in products of sequential fans and that the tightness of such a product can be dependent on extra set theory assumptions. All spaces are Hausdorff and regular.
The following theorems are the main results in this post.
Theorem 1
Let $X$ and $Y$ be countably tight spaces. If one of $X$ and $Y$ is compact, then $X \times Y$ is countably tight.
Theorem 2
The product of finitely many compact countably tight spaces is countably tight.
Theorem 3
Suppose that $X_1, X_2, X_3, \cdots$ are countably many compact spaces such that each $X_i$ has at least two points. If each $X_i$ is a countably tight space, then the product space $\prod_{i=1}^\infty X_i$ is countably tight.
____________________________________________________________________
Finite products
Before proving Theorem 1 and Theorem 2, we prove the following results.
Theorem 4
Let $f:Y_1 \rightarrow Y_2$ be a continuous and closed map from the space $Y_1$ onto the space $Y_2$. Suppose that the space $Y_2$ is countably tight and that each fiber of the map $f$ is countably tight. Then the space $Y_1$ is countably tight.
Proof of Theorem 4
Let $x \in Y_1$ and $x \in \overline{A}$ where $A \subset Y_1$. We proceed to find a countable $W \subset Y_1$ such that $x \in \overline{W}$. Choose $y \in Y_2$ such that $y=f(x)$.
Let $M$ be the fiber of the map $f$ at the point $y$, i.e. $M=f^{-1}(y)$. By assumption, $M$ is countably tight. Call a point $w \in M$ countably reached by $A$ if there is some countable $C \subset A$ such that $w \in \overline{C}$. Let $G$ be the set of all points in $M$ that are countably reached by $A$. We claim that $x \in \overline{G}$.
Let $U \subset Y_1$ be open such that $x \in U$. Because the space $Y_1$ is regular, choose open $V \subset U$ such that $x \in V$ and $\overline{V} \subset U$. Then $V \cap A \ne \varnothing$. Furthermore, $x \in \overline{V \cap A}$. Let $C=f(V \cap A)$. By the continuity of $f$, we have $y \in \overline{C}$. Since $Y_2$ is countably tight, there exists some countable $D \subset C$ such that $y \in \overline{D}$. Choose a countable $E \subset V \cap A$ such that $f(E)=D$. It follows that $y \in \overline{f(E)}$.
We show that that $\overline{E} \cap M \ne \varnothing$. Since $E \subset \overline{E}$, we have $f(E) \subset f(\overline{E})$. Note that $f(\overline{E})$ is a closed set since $f$ is a closed map. Thus $\overline{f(E)} \subset f(\overline{E})$. As a result, $y \in f(\overline{E})$. Then $y=f(t)$ for some $t \in \overline{E}$. We have $t \in \overline{E} \cap M$.
By the definition of the set $G$, we have $\overline{E} \cap M \subset G$. Furthermore, $\overline{E} \cap M \subset \overline{V} \subset U$. Note that the arbitrary open neighborhood $U$ of $x$ contains points of $G$. This establishes the claim that $x \in \overline{G}$.
Since $M$ is a fiber of $f$, $M$ is countably tight by assumption. Choose some countable $T \subset G$ such that $x \in \overline{T}$. For each $t \in T$, choose a countable $W_t \subset A$ with $t \in \overline{W_t}$. Let $W=\bigcup_{t \in T} W_t$. Note that $W \subset A$ and $W$ is countable with $x \in \overline{W}$. This establishes the space $Y_1$ is countably tight at $x \in Y_1$. $\blacksquare$
Lemma 5
Let $f:X \times Y \rightarrow Y$ be the projection map. If $X$ is a compact space, then $f$ is a closed map.
Proof of Lemma 5
Let $A$ be a closed subset of $X \times Y$. Suppose that $f(A)$ is not closed. Let $y \in \overline{f(A)}-f(A)$. It follows that no point of $X \times \left\{y \right\}$ belongs to $A$. For each $x \in X$, choose open subset $O_x$ of $X \times Y$ such that $(x,y) \in O_x$ and $O_x \cap A=\varnothing$. The set of all $O_x$ is an open cover of the compact space $X \times \left\{y \right\}$. Then there exist finitely many $O_x$ that cover $X \times \left\{y \right\}$, say $O_{x_i}$ for $i=1,2,\cdots,n$.
Let $W=\bigcup_{i=1}^n O_{x_i}$. We have $X \times \left\{y \right\} \subset W$. Since $X$ is compact, we can then use the Tube Lemma which implies that there exists open $G \subset Y$ such that $X \times \left\{y \right\} \subset X \times G \subset W$. It follows that $G \cap f(A) \ne \varnothing$. Choose $t \in G \cap f(A)$. Then for some $x \in X$, $(x,t) \in A$. Since $t \in G$, $(x,t) \in W$, implying that $W \cap A \ne \varnothing$, a contradiction. Thus $f(A)$ must be a closed set in $Y$. This completes the proof of the lemma. $\blacksquare$
Proof of Theorem 1
Let $X$ be the factor that is compact. let $f: X \times Y \rightarrow Y$ be the projection map. The projection map is always continuous. Furthermore it is a closed map by Lemma 5. The range space $Y$ is countably tight by assumption. Each fiber of the projection map $f$ is of the form $X \times \left\{y \right\}$ where $y \in Y$, which is countably tight. Then use Theorem 4 to establish that $X \times Y$ is countably tight. $\blacksquare$
Proof of Theorem 2
This is a corollary of Theorem 1. According to Theorem 1, the product of two compact countably tight spaces is countably tight. By induction, the product of any finite number of compact countably tight spaces is countably tight. $\blacksquare$
____________________________________________________________________
Countable products
Our proof to establish that the product space $\prod_{i=1}^\infty X_i$ is countably tight is an indirect one and makes use of two non-trivial results. We first show that $\omega_1 \times \prod_{i=1}^\infty X_i$ is a closed subspace of a $\Sigma$-product that is normal. It follows from another result that the second factor $\prod_{i=1}^\infty X_i$ is countably tight. We now present all the necessary definitions and theorems.
Consider a product space $Y=\prod_{\alpha<\kappa} Y_\alpha$ where $\kappa$ is an infinite cardinal number. Fix a point $p \in Y$. The $\Sigma$-product of the spaces $Y_\alpha$ with $p$ as the base point is the following subspace of the product space $Y=\prod_{\alpha<\kappa} Y_\alpha$:
$\displaystyle \Sigma_{\alpha<\kappa} Y_\alpha=\left\{y \in \prod_{\alpha<\kappa} Y_\alpha: y_\alpha \ne p_\alpha \text{ for at most countably many } \alpha < \kappa \right\}$
The definition of the space $\Sigma_{\alpha<\kappa} Y_\alpha$ depends on the base point $p$. The discussion here is on properties of $\Sigma_{\alpha<\kappa} Y_\alpha$ that hold regardless of the choice of base point. If the factor spaces are indexed by a set $A$, the notation is $\Sigma_{\alpha \in A} Y_\alpha$.
If all factors $Y_\alpha$ are identical, say $Y_\alpha=Z$ for all $\alpha$, then we use the notation $\Sigma_{\alpha<\kappa} Z$ to denote the $\Sigma$-product. Once useful fact is that if there are $\omega_1$ many factors and each factor has at least 2 points, then the space $\omega_1$ can be embedded as a closed subspace of the $\Sigma$-product.
Theorem 6
For each $\alpha<\omega_1$, let $Y_\alpha$ be a space with at least two points. Then $\Sigma_{\alpha<\omega_1} Y_\alpha$ contains $\omega_1$ as a closed subspace. See Exercise 3 in this previous post.
Now we discuss normality of $\Sigma$-products. This previous post shows that if each factor is a separable metric space, then the $\Sigma$-product is normal. It is also well known that if each factor is a metric space, the $\Sigma$-product is normal. The following theorem handles the case where each factor is a compact space.
Theorem 7
For each $\alpha<\kappa$, let $Y_\alpha$ be a compact space. Then the $\Sigma$-product $\Sigma_{\alpha<\kappa} Y_\alpha$ is normal if and only if each factor $Y_\alpha$ is countably tight.
Theorem 7 is Theorem 7.5 in page 821 of [1]. Theorem 7.5 in [1] is stated in a more general setting where each factor of the $\Sigma$-product is a paracompact p-space. We will not go into a discussion of p-space. It suffices to know that any compact Hausdorff space is a paracompact p-space. We also need the following theorem, which is proved in this previous post.
Theorem 8
Let $Y$ be a compact space. Then the product space $\omega_1 \times Y$ is normal if and only if $Y$ is countably tight.
We now prove Theorem 3.
Proof of Theorem 3
Let $\omega_1=\cup \left\{A_n: n \in \omega \right\}$, where for each $n$, $\lvert A_n \lvert=\omega_1$ and that $A_n \cap A_m=\varnothing$ if $n \ne m$. For each $n=1,2,3,\cdots$, let $S_n=\Sigma_{\alpha \in A_n} X_n$. By Theorem 7, each $S_n$ is normal. Let $S_0=\Sigma_{\alpha \in A_0} X_1$, which is also normal. By Theorem 6, the space $\omega_1$ of countable ordinals is a closed subspace of $S_0$. Let $T=\omega_1 \times X_1 \times X_2 \times X_3 \times \cdots$. We have the following derivation.
\displaystyle \begin{aligned} T&=\omega_1 \times X_1 \times X_2 \times X_3 \times \cdots \\&\subset S_0 \times S_1 \times S_2 \times S_3 \times \cdots \\&\cong W=\Sigma_{\alpha<\omega_1} W_\alpha \end{aligned}
Recall that $\omega_1=\cup \left\{A_n: n \in \omega \right\}$. The space $W=\Sigma_{\alpha<\omega_1} W_\alpha$ is defined such that for each $n \ge 1$ and for each $\alpha \in A_n$, $W_\alpha=X_n$. Furthermore, for $n=0$, for each $\alpha \in A_0$, let $W_\alpha=X_1$. Thus $W$ is a $\Sigma$-product of compact countably tight spaces and is thus normal by Theorem 7. The space $T=\omega_1 \times \prod_{n=1}^\infty X_n$ is a closed subspace of the normal space $W$. By Theorem 8, the product space $\prod_{n=1}^\infty X_n$ must be countably tight. $\blacksquare$
____________________________________________________________________
Remarks
Theorem 2, as indicated above, is a corollary of Theorem 1. We also note that Theorem 2 is also a corollary of Theorem 3 since any finite product is a subspace of a countable product. To see this, let $X=X_1 \times X_2 \times \cdots \times X_n$.
\displaystyle \begin{aligned} X&=X_1 \times X_2 \times \cdots \times X_n \\&\cong X_1 \times X_2 \times \cdots \times X_n \times \left\{t_{n+1} \right\} \times \left\{t_{n+2} \right\} \times \cdots \\&\subset X_1 \times X_2 \times \cdots \times X_n \times X_{n+1} \times X_{n+2} \times \cdots \end{aligned}
In the above derivation, $t_m$ is a point of $X_m$ for all $m >n$. When the countable product space is countably tight, the finite product, being a subspace of a countably tight space, is also countably tight.
____________________________________________________________________
Exercise
Exercise 1
Let $f:X \times Y \rightarrow Y$ be the projection map. If $X$ is a countably compact space and $Y$ is a Frechet space, then $f$ is a closed map.
Exercise 2
Let $X$ and $Y$ be countably tight spaces. If one of $X$ and $Y$ is a countably compact space and the other space is a Frechet space, then $X \times Y$ is countably tight.
Exercise 2 is a variation of Theorem 1. One factor is weakened to “countably compact”. However, the other factor is strengthened to “Frechet”.
____________________________________________________________________
Reference
1. Przymusinski, T. C., Products of Normal Spaces, Handbook of Set-Theoretic Topology (K. Kunen and J. E. Vaughan, eds), Elsevier Science Publishers B. V., Amsterdam, 781-826, 1984.
____________________________________________________________________
$\copyright \ 2015 \text{ by Dan Ma}$ | 2017-04-27T18:47:16 | {
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https://www.physicsforums.com/threads/rocket-ship-question.196423/ | # Rocket Ship Question
1. Nov 6, 2007
### Submission1
We are having a debate here at work and we need an answer from some smart folks (that's you).
If you have a rocket ship in space that takes 1 gallon of gas to reach 1000 mph then, negating friction, would it take less, more or the same amount to go from 1000 mph to 2000 mph.
Thanks,
Sub.
2. Nov 6, 2007
### mgb_phys
Kinetic energy, the energy needed to move something depends on the speed squared.
So to travel at 2000 mph 4x as much energy as 1000mph. Since you started at 1000mph you will need to add 3x as much extra energy to get to 2000mph than you did to get to 1000mph.
You also have to take into account that the mass of the rocket is changing as you use up the fuel so it takes less energy to accelarate the rocket since it weighs less - if you doing this for real.
Last edited: Nov 6, 2007
3. Nov 6, 2007
### Submission1
Cool. Thanks.
4. Nov 6, 2007
### chronon
But there's nothing special about the starting frame of reference. So it shouldn't take any more fuel to add an extra 1000mph starting in the 1000mph frame as to get to the 1000mph frame starting from the zero frame.
5. Nov 6, 2007
### mgb_phys
Does that apply if it's accelerating ?
6. Nov 6, 2007
### D H
Staff Emeritus
This topic ws recently discussed in this thread.
The governing equation is the Tsiolokovsky rocket equation. Applying this equation to a rocket that burns all of its fuel,
$$\Delta v = v_e\ln\left(\frac {m_\text{rocket}+m_\text{fuel}}{m_\text{rocket}}\right)$$
Suppose you find you have to load some rocket with a quantity of fuel $m_{\text{fuel}|1000\text{mph}}$ to make the rocket attain a velocity of 1000 mph after burnout. Atfer churning the crank on the rocket equation, the amount of fuel it would take to bring the rocket to 2000 mph is
$$m_{\text{fuel}|2000\text{mph}} = m_{\text{fuel}|1000\text{mph}}\left(2+ \frac{m_{\text{fuel}|1000\text{mph}}}{m_\text{rocket}}\right)$$
In other words, one must more than double the quantity of fuel to double a rocket's final velocity.
7. Nov 6, 2007
### mgb_phys
Thanks - I'm away for a week and look what I miss.
8. Nov 6, 2007
### chronon
Indeed. But more of that fuel will be burnt in getting it from 0 to 1000mph than from 1000mph to 2000mph.
9. Nov 6, 2007
### D H
Staff Emeritus
A little more detail regarding previous post:
The reason it takes more than double the fuel to go from 0 to 1000 mph than from 0 to 2000 mph (or whatever) is because the rocket carries its fuel.
Assume we have some rocket and want to make two test flights with it. The first test flight involves flying the rocket from rest to a final speed of 1000 mph, and the second, rest to 2000 mph. Both tests end with the rocket devoid of all fuel. The amount during the second test of fuel left in the rocket at the point the rocket reaches 1000 mph is exactly the same as the amount of fuel initially placed in the rocket for the first test. In this sense, it takes the same amount of fuel to go from 1000 to 2000 mph as it does from 0 to 1000 mph.
However, the rocket has to first achieve that 1000 mph during the second test. It is the first 1000 mph that costs more. At the start of the second test, the rocket comprises the dry mass of the rocket, the quantity fuel (call this f1) needed to bring the dry mass of the rocket from 1000 to 2000 mph plus the quantity of fuel needed to bring the dry mass of the rocket plus f1 from 0 to 1000 mph.
Last edited: Nov 6, 2007
10. Nov 6, 2007
### D H
Staff Emeritus
Indeed. We just cross-posted.
11. Nov 6, 2007
### rcgldr
Mentioned in the other thread is that there is work also peformed on the spent fuel as it is ejected behind the rockets engine. If the kinetic energy of both spent fuel, and the rocket (plus it's remaining unspent fuel) are summed, and if the rate of fuel consumption is constant, then thrust will be constant, and the sum of the kinetic energies of spent fuel, rocket (and remaining fuel) will increase linearly with time. This means that the power involved is constant (with a constant thrust).
Also from the other thread, it's easier to grasp this if you consider the rocket to be held in place so that all of the work done by the engine is to acclerate the spent fuel. The terminal velocity of the fuel is constant, and the rate of fuel consumption (mass ejected) is constant, so the kinetic energy of the fuel increases linearly with time.
12. Nov 6, 2007
### Staff: Mentor
From this thread and the other I have a kind of vague impression that rockets are better understood in terms of conservation of momentum than conservation of energy. I think that is because conservation of momentum requires you to think of the exhaust whereas conservation of energy does not.
13. Nov 6, 2007
### TVP45
You always have to ask, 1000 mph relative to what? A rocket ship just traveling along at constant velocity will never know it. You always need a reference against which you measure your velocity. And, it is in that reference frame where it takes 3 times as much energy to go from 1000 mph to 2000 mph as it does from 0 to 1000 mph.
14. Nov 6, 2007
### D H
Staff Emeritus
That is how I prefer to look at things. A conservation of energy viewpoint is tough to do properly. The rocket fuel remaining in the rocket gains kinetic energy as the rocket accelerates. Moreover, some of the released chemical potential energy is wasted in the form of hot exhaust. Failing to account for either leads to erroneous results.
Conservation of linear momentum is a much simpler proposition. Adding conservation of angular momentum makes things a bit hairier. Adding the fact that in a real rocket the center of mass and inertia tensor change as the rocket burns fuel makes things a lot hairer. I wouldn't dream of attacking a non-point mass rocket with fuel located away from the center of mass with a conservation of energy perspective -- unless I was doing so at a micro level and using a CFD model.
The driving reason for investigating a problem from the point of view of any of the conservation principles is that doing so makes the answer fall out. I don't have to use a full-blown CFD model to get a very, very good model of what firing a roket engine does to the state of a vehicle. A full-blown CFD model that takes advantage of all of the conservation laws is needed to characterize behavior. However, CFD models cannot "see the forest for the trees". CFD models examine the spots on a beetle that sits on a tree in a huge forest. Once the behavior has been properly characterized, a conservation of momentum model does wonders.
15. Nov 8, 2007
### alvaros
So the answer is ... ??
A 1000 kg rocket push a 1 Kg fuel at 1000 * 1000 mph -> the rocket increases its velocity in 1000 mph ( respect to the Frame of Reference it was before starting the engine )
Again
A 999 kg rocket ( the same rocket ) push a 1 Kg fuel at 1000 * 1000 mph -> the rocket increases its velocity in 1000 mph.
( I use very high energy fuel, so we dont have to discuss about carrying the fuel or using Tsiolokovsky formula )
16. Nov 8, 2007
### TVP45
A kind of new thought on this thread....
This kind of question pops up a lot and there always seems to be (in the mind of the questioner) some kind of disconnect between common-sense examples and the equations.
Is the work-energy theorem the best way to teach people? I didn't get it for many years after that freshman class.
Is there another way?
17. Nov 9, 2007
### alvaros
Im very sorry, but I dont understand your english.
18. Nov 9, 2007
### TVP45
Alvaros,
19. Nov 10, 2007
### alvaros
I think my example is clear enough:
It seems that the second kg of fuel gives more kinetic energy to the rocket than the first Kg ( ?? )
I made a thread "what is a IFR" but everybody seems to have it very clear. I think the main mistakes are representing forces with an arrow -> and talking about IFR as some abstract.
In this problem the forces are between the rocket and the fuel ejected:
fuel <-> rocket
And the IFR is the center of mass of the system ( rocket + fuel ). The IFR always refers to something material ( with mass ).
20. Nov 10, 2007
### Staff: Mentor
Yes. Don't forget the KE of the exhaust. It is better to use conservation of momentum principles because they force you to consider the exhaust.
I am trying out some new (for me) applications of old ideas here here, so feel free to point out any mistakes in my logic:
Since the exhaust velocity is constant (assuming the mass of the rocket is large relative to the mass of the fuel) each kg fuel spent gives the same Δp and therefore the same Δv to the rocket.
Each successive kg with its Δv results in a larger ΔKE for the rocket since KE is proportional to v^2.
"But conservation of energy you protest!" The PE of the fuel does not only go into KE of the rocket, but also into the rather large KE of the exhaust.
Because the exhaust is going in the opposite direction of the rocket each successive kg of exhaust gains less KE, exactly compensating the more KE gained by the rocket so that the increased KE of the rocket-exhaust system is always equal to the PE in the spent fuel (neglecting heat).
Bottom line: for rockets always use conservation of momentum so that you cannot neglect the exhaust.
Last edited: Nov 10, 2007 | 2017-02-28T16:55:23 | {
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https://www.physicsforums.com/threads/rocket-ship-question.196423/ | # Rocket Ship Question
1. Nov 6, 2007
### Submission1
We are having a debate here at work and we need an answer from some smart folks (that's you).
If you have a rocket ship in space that takes 1 gallon of gas to reach 1000 mph then, negating friction, would it take less, more or the same amount to go from 1000 mph to 2000 mph.
Thanks,
Sub.
2. Nov 6, 2007
### mgb_phys
Kinetic energy, the energy needed to move something depends on the speed squared.
So to travel at 2000 mph 4x as much energy as 1000mph. Since you started at 1000mph you will need to add 3x as much extra energy to get to 2000mph than you did to get to 1000mph.
You also have to take into account that the mass of the rocket is changing as you use up the fuel so it takes less energy to accelarate the rocket since it weighs less - if you doing this for real.
Last edited: Nov 6, 2007
3. Nov 6, 2007
### Submission1
Cool. Thanks.
4. Nov 6, 2007
### chronon
But there's nothing special about the starting frame of reference. So it shouldn't take any more fuel to add an extra 1000mph starting in the 1000mph frame as to get to the 1000mph frame starting from the zero frame.
5. Nov 6, 2007
### mgb_phys
Does that apply if it's accelerating ?
6. Nov 6, 2007
### D H
Staff Emeritus
This topic ws recently discussed in this thread.
The governing equation is the Tsiolokovsky rocket equation. Applying this equation to a rocket that burns all of its fuel,
$$\Delta v = v_e\ln\left(\frac {m_\text{rocket}+m_\text{fuel}}{m_\text{rocket}}\right)$$
Suppose you find you have to load some rocket with a quantity of fuel $m_{\text{fuel}|1000\text{mph}}$ to make the rocket attain a velocity of 1000 mph after burnout. Atfer churning the crank on the rocket equation, the amount of fuel it would take to bring the rocket to 2000 mph is
$$m_{\text{fuel}|2000\text{mph}} = m_{\text{fuel}|1000\text{mph}}\left(2+ \frac{m_{\text{fuel}|1000\text{mph}}}{m_\text{rocket}}\right)$$
In other words, one must more than double the quantity of fuel to double a rocket's final velocity.
7. Nov 6, 2007
### mgb_phys
Thanks - I'm away for a week and look what I miss.
8. Nov 6, 2007
### chronon
Indeed. But more of that fuel will be burnt in getting it from 0 to 1000mph than from 1000mph to 2000mph.
9. Nov 6, 2007
### D H
Staff Emeritus
A little more detail regarding previous post:
The reason it takes more than double the fuel to go from 0 to 1000 mph than from 0 to 2000 mph (or whatever) is because the rocket carries its fuel.
Assume we have some rocket and want to make two test flights with it. The first test flight involves flying the rocket from rest to a final speed of 1000 mph, and the second, rest to 2000 mph. Both tests end with the rocket devoid of all fuel. The amount during the second test of fuel left in the rocket at the point the rocket reaches 1000 mph is exactly the same as the amount of fuel initially placed in the rocket for the first test. In this sense, it takes the same amount of fuel to go from 1000 to 2000 mph as it does from 0 to 1000 mph.
However, the rocket has to first achieve that 1000 mph during the second test. It is the first 1000 mph that costs more. At the start of the second test, the rocket comprises the dry mass of the rocket, the quantity fuel (call this f1) needed to bring the dry mass of the rocket from 1000 to 2000 mph plus the quantity of fuel needed to bring the dry mass of the rocket plus f1 from 0 to 1000 mph.
Last edited: Nov 6, 2007
10. Nov 6, 2007
### D H
Staff Emeritus
Indeed. We just cross-posted.
11. Nov 6, 2007
### rcgldr
Mentioned in the other thread is that there is work also peformed on the spent fuel as it is ejected behind the rockets engine. If the kinetic energy of both spent fuel, and the rocket (plus it's remaining unspent fuel) are summed, and if the rate of fuel consumption is constant, then thrust will be constant, and the sum of the kinetic energies of spent fuel, rocket (and remaining fuel) will increase linearly with time. This means that the power involved is constant (with a constant thrust).
Also from the other thread, it's easier to grasp this if you consider the rocket to be held in place so that all of the work done by the engine is to acclerate the spent fuel. The terminal velocity of the fuel is constant, and the rate of fuel consumption (mass ejected) is constant, so the kinetic energy of the fuel increases linearly with time.
12. Nov 6, 2007
### Staff: Mentor
From this thread and the other I have a kind of vague impression that rockets are better understood in terms of conservation of momentum than conservation of energy. I think that is because conservation of momentum requires you to think of the exhaust whereas conservation of energy does not.
13. Nov 6, 2007
### TVP45
You always have to ask, 1000 mph relative to what? A rocket ship just traveling along at constant velocity will never know it. You always need a reference against which you measure your velocity. And, it is in that reference frame where it takes 3 times as much energy to go from 1000 mph to 2000 mph as it does from 0 to 1000 mph.
14. Nov 6, 2007
### D H
Staff Emeritus
That is how I prefer to look at things. A conservation of energy viewpoint is tough to do properly. The rocket fuel remaining in the rocket gains kinetic energy as the rocket accelerates. Moreover, some of the released chemical potential energy is wasted in the form of hot exhaust. Failing to account for either leads to erroneous results.
Conservation of linear momentum is a much simpler proposition. Adding conservation of angular momentum makes things a bit hairier. Adding the fact that in a real rocket the center of mass and inertia tensor change as the rocket burns fuel makes things a lot hairer. I wouldn't dream of attacking a non-point mass rocket with fuel located away from the center of mass with a conservation of energy perspective -- unless I was doing so at a micro level and using a CFD model.
The driving reason for investigating a problem from the point of view of any of the conservation principles is that doing so makes the answer fall out. I don't have to use a full-blown CFD model to get a very, very good model of what firing a roket engine does to the state of a vehicle. A full-blown CFD model that takes advantage of all of the conservation laws is needed to characterize behavior. However, CFD models cannot "see the forest for the trees". CFD models examine the spots on a beetle that sits on a tree in a huge forest. Once the behavior has been properly characterized, a conservation of momentum model does wonders.
15. Nov 8, 2007
### alvaros
So the answer is ... ??
A 1000 kg rocket push a 1 Kg fuel at 1000 * 1000 mph -> the rocket increases its velocity in 1000 mph ( respect to the Frame of Reference it was before starting the engine )
Again
A 999 kg rocket ( the same rocket ) push a 1 Kg fuel at 1000 * 1000 mph -> the rocket increases its velocity in 1000 mph.
( I use very high energy fuel, so we dont have to discuss about carrying the fuel or using Tsiolokovsky formula )
16. Nov 8, 2007
### TVP45
A kind of new thought on this thread....
This kind of question pops up a lot and there always seems to be (in the mind of the questioner) some kind of disconnect between common-sense examples and the equations.
Is the work-energy theorem the best way to teach people? I didn't get it for many years after that freshman class.
Is there another way?
17. Nov 9, 2007
### alvaros
Im very sorry, but I dont understand your english.
18. Nov 9, 2007
### TVP45
Alvaros,
19. Nov 10, 2007
### alvaros
I think my example is clear enough:
It seems that the second kg of fuel gives more kinetic energy to the rocket than the first Kg ( ?? )
I made a thread "what is a IFR" but everybody seems to have it very clear. I think the main mistakes are representing forces with an arrow -> and talking about IFR as some abstract.
In this problem the forces are between the rocket and the fuel ejected:
fuel <-> rocket
And the IFR is the center of mass of the system ( rocket + fuel ). The IFR always refers to something material ( with mass ).
20. Nov 10, 2007
### Staff: Mentor
Yes. Don't forget the KE of the exhaust. It is better to use conservation of momentum principles because they force you to consider the exhaust.
I am trying out some new (for me) applications of old ideas here here, so feel free to point out any mistakes in my logic:
Since the exhaust velocity is constant (assuming the mass of the rocket is large relative to the mass of the fuel) each kg fuel spent gives the same Δp and therefore the same Δv to the rocket.
Each successive kg with its Δv results in a larger ΔKE for the rocket since KE is proportional to v^2.
"But conservation of energy you protest!" The PE of the fuel does not only go into KE of the rocket, but also into the rather large KE of the exhaust.
Because the exhaust is going in the opposite direction of the rocket each successive kg of exhaust gains less KE, exactly compensating the more KE gained by the rocket so that the increased KE of the rocket-exhaust system is always equal to the PE in the spent fuel (neglecting heat).
Bottom line: for rockets always use conservation of momentum so that you cannot neglect the exhaust.
Last edited: Nov 10, 2007
21. Nov 10, 2007
### TVP45
DaleSpam,
This is the sort of example I was asking about. If I were a high school teacher (thank G-d I'm not; that's a hard job), how would I present this example so that students can grasp it? I'm not being in any way critical, any more than I was of Alvaros; I am simply curious about how this can be developed conceptually.
22. Nov 10, 2007
### Staff: Mentor
If I were a teacher in a high-school level physics course I would not even mention conservation of energy in this problem. I would stick entirely with conservation of momentum. You can solve for everything of interest that way, and the conservation of momentum principle does what a good conservation principle should: it simplifies things.
As a teacher I would not introduce conservation of energy into the problem since it does nothing to simplify the problem. If a student asked I would tell them that energy is conserved, but it doesn't make the problem any easier. I think students would understand that.
23. Nov 11, 2007
### TVP45
At first blush, I agree. Momentum is much easier to deal with than energy. Two questions occur to me:
(1) Can a student grasp the concept of momentum without being sidetracked by preconceptions about energy?
(2) Is the calculation of change of momentum rather than absolute momentum sensible?
24. Nov 11, 2007
### alvaros
DaleSpam:
Wrong, the change in kinetic energy on the fuel is 1000 times the change in kinetic energy on the rocket.
Momentum is an abstract concept, its much easier the 3rd Newtons law: the burning fuel pushes the rocket and the rocket pushes the fuel fuel <-> rocket
( you see the double arrow <-> , all forces ( ? ) are double arrow )
( It seems, as I read here, that the conservation of momentum is more "universal" than 3rd Newtons law, but, in this case, both laws are the same and true )
But, at the end, there isnt any clear answer to the problem.
25. Nov 11, 2007
### D H
Staff Emeritus
Invoking Newton's third law on this problem is in a sense more ad-hoc and more abstract than using conservation of momentum with regard to this problem. The concept of force (Newton's second law) is very abstract; anything change to an object's momentum involves some force. Applying conservation of momentum is no less abstract than applying Newton's third law and yields a deeper answer.
Put the spaceship in deep space, far from any massive object, so that there are no measurable external forces acting on the vehicle. Now, what exactly is the force that is making the spaceship accelerate? By assumption, there are no external forces. You can say posit some ad-hoc force $F$ that results from burning the fuel and get an answer via Newton's second law. Note well: Since the rocket's mass is changing, the simpler form $F=ma$ is not valid here. We have to use $F=dp/dt$ instead.
$$F = \frac{dp_r} {dt} = \frac{d}{dt} ( m_r \, v_r ) = \dot m_r v_r} + m_r \, a_r$$
Solving for the rocket's acceleration,
$$a_r = \frac {F} {m_r} - \frac {\dot m_r} {m_r} v_r}$$
This is not very satisying. What exactly is this force? It's purely ad-hoc for one thing. Moreover, it will turn out that the force is frame-dependent.
Here is how things turn out from conservation of momentum point of view. For this derivation, I will use some math that mathematicians don't particular like but physicists use willy-nilly -- things $\Delta v$. Things can be done more formally using continuum physics (classical treatment of gases), but that is a lot messier.
Nomenclature:
$$\vec v_r(t)$$ Rocket velocity at time $t$, inertial observer
$$\vec v_e(t)$$ Rocket exhaust velocity at time $t$, relative to vehicle velocity
$$m_r(t)$$ Rocket mass at time $t$
$$\dot m_f(t)$$ Rate at which rocket consumes fuel at$t$
Over a small time interval $\Delta t$, the rocket will eject a small mass of fuel $\dot m_f(t)\,\Delta t$. At the start of the interval, the rocket has mass $m_r(t)$, velocity $\vec v_r(t)$, and momentum $m_r(t) \, \vec v_r(t)$. At the end of the interval, the rocket has mass, velocity, and linear momentum
$$m_r(t+\Delta t) = m_r(t)-\dot m_f(t)\,\Delta t$$
$$\vec v_r(t+\Delta t) = \vec v_r(t)+\Delta \vec v_r(t)$$
$$\vec p_r(t+\Delta t) = (m_r(t) -\dot m_f(t)\,\Delta t)\, (\vec v_r(t)+\Delta \vec v_r(t))$$
The bit of ejected fuel carries some momentum from the vehicle. The mass, inertial observer velocity, and momentum of the exhaust are
$$\Delta m_e(t) = \dot m_f(t)\,\Delta t$$
$$\vec v_{e_{\text{inertial}}}) = \vec v_r(t)+\vec v_e(t)$$
$\Delta \vec p_e(t+\Delta t) = \dot m_f(t)\,\Delta t\, (\vec v_r(t)+\vec v_e(t))[/tex] The momentum of the rocket+exhaust at the end of the time interval is thus $$\vec p_{r+e}(t+\Delta t) = (m_r(t) -\dot m_f(t)\,\Delta t)\, (\vec v_r(t)+\Delta \vec v_r(t)) + \dot m_f(t)\,\Delta t\, (\vec v_r(t)+\vec v_e(t))$$ Dropping the second-order term [itex]\Delta t \Delta \vec v_r(t)$ and simplifying,
$$\vec p_{r+e}(t+\Delta t) = m_r(t)\,\vec v_r(t)+ m_r(t) \, \Delta \vec v_r(t) + \dot m_f(t)\,\Delta t\, \vec v_e(t)$$
Assuming no external forces act on the rocket and the ejected fuel during this time interval, the rocket and the ejected fuel form a closed system. Momentum is conserved in a closed system, so $\vec p_{r+e}(t+\Delta t)=\vec p_r(t)[/tex]: $$m_r(t)\,\vec v_r(t)+ m_r(t) \, \Delta \vec v_r(t) + \dot m_f(t)\,\Delta t\, \vec v_e(t) = m_r(t)\,\vec v_r(t)$$ or $$m_r(t) \, \Delta \vec v_r(t) + \dot m_f(t)\,\Delta t\, \vec v_e(t) = 0$$ Dividing by [itex]\Delta t$ and taking the limit $\Delta t \to 0$,
$$\frac {d\vec v_r(t)}{dt} = - \, \frac {\dot m_f(t)} {m_r(t)} \, \vec v_e(t)$$
This is the equation for the acceleration of the rocket at time $t$.
By conservation of mass, the time derivative of the rocket's mass is just the additive inverse of the fuel consumption rate: $\dot m_r(t) = -\dot m_f(t)$. If the relative exhaust velocity is a constant vector, both the left and right hand sides of the above acceleration equation are integrable. Integrating from some initial time $t_0$ to some final time $t_1$ yields
$$\vec v_r(t_1) - \vec v_r(t_0) = \ln\left(\frac{m_r(t_1)}{m_r(t_0)}\right) \vec v_e$$
This is the Tsiolokovsky rocket equation. Since the final mass is smaller than the initial mass, the logarithm will be negative. The change in velocity is directed against the exhaust direction.
Last edited: Nov 11, 2007 | 2018-05-26T01:03:08 | {
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https://cs.stackexchange.com/questions/54699/the-difference-between-theoretical-complexity-and-practical-efficiency/54706 | The difference between theoretical complexity and practical efficiency
If I have this pseudocode:
for i=0 to n/2 do
for j=0 to n/2 do
... do anything ....
The number of iterations is $n^2/4$.
What is the complexity of this program? Is it $O(n^2)$?
What is the intuition formal/informal for which is that complexity?
Next, if i have this other pseudocode :
for i=0 to n do
for j=0 to n do
... do anything ....
The complexity is again $O(n^2)$ -- is that correct?
Is there any difference between efficiency in practice and theoretical complexity in this case? Is one of these faster than the other?
• Rule 3 from Rob Pike, as cited in The Art of Unix Programming: "Rule 3. Fancy algorithms are slow when n is small, and n is usually small. Fancy algorithms have big constants. Until you know that n is frequently going to be big, don't get fancy." This is because of the difference between theoretical complexity and practical efficiency, which is ably defined in the answers below. – Wildcard Mar 21 '16 at 3:49
• @Wildcard That seems like a good rule of thumb. Benchmarking several contender algorithms would be even better. – G. Bach Mar 21 '16 at 12:47
• @Wildcard Very interesting quote. It is always interesting to think about the complexity when you're designing/implementing procedures but I usually don't break my head over them untill the procedure in question seems to be a bottleneck. – Auberon Mar 21 '16 at 16:18
Big O Formally
$O(f(n)) = \{g(n) | \exists c > 0, \exists n_0 > 0, \forall n > n_0 : 0 \leq g(n) \leq c*f(n)\}$
Big O Informally
$O(f(n))$ is the set of functions that grow slower (or equally fast) than $f(n)$. This means that whatever function you pick from $O(f(n))$, let's name her $g(n)$, and you pick a sufficiently large $n$, $f(n)$ will be larger than $g(n)$. Or, in even other words, $f(n)$ will eventually surpass any function in $O(f(n))$ when $n$ grows bigger. $n$ is the input size of your algorithm.
As an example. Let's pick $f(n) = n^2$.
We can say that $f(n) \in O(n^3)$. Note that, over time, we allowed notation abuse so almost everyone writes $f(n) = O(n^3)$ now.
Normally when we evaluate algorithms, we look at their order. This way, we can predict how the running time of the algorithm will increase when the input size ($n$) increases.
In your example, both algorithms are of order $O(n^2)$. So, their running time will increase the same way (quadratically) as $n$ increases. Your second algorithm will be four times slower than the first one, but that is something we're usually not interested in **theoretically*. Algorithms with the same order can have different factors ($c$ in the formal definition) or different lower order terms in the function that evaluates the number of steps. But usually that's because of implementation details and we're not interested in that.
If you have a algorithm that runs in $O(log(n))$ time we can safely say it will be faster than $O(n)$ [1] because $log(n) = O(n)$. No matter how bad the $O(log(n))$ algorithm is implemented, no matter how much overhead you put in the algorithm, it will always [1] be faster than the $O(n)$ algorithm. Even if the number of steps in the algorithms are $9999*log(n) = O(log(n))$ and $0.0001*n = O(n)$, the $O(log(n))$ algorithm will eventually be faster [1].
But, maybe, in your application, $n$ is always low and will never be sufficiently large enough so that the $O(log(n))$ algorithm will be faster in practice. So using the $O(n)$ algorithm will result in faster running times, despite $n = O(log(n))$
[1] if $n$ is sufficiently large.
• An $O(\log n)$ isn't always faster than an $O(n)$ one. It depends on the value of $n$ and on the hidden constants. A case in point is fast matrix multiplication, which isn't fast at all in practice. – Yuval Filmus Mar 21 '16 at 0:07
• That's why I added: If n is sufficiently large. In the case you mention, if the matrices grow sufficiently large, $O(log(n))$ WILL be faster, by definition – Auberon Mar 21 '16 at 0:09
• Isn't everything you're saying already in my answer? – Auberon Mar 21 '16 at 0:11
• The OP asks about the difference between theory and practice, and you're ignoring that. Asymptotic notation is usually, but not always, useful in practice. Improving the running time fourfold can make a huge difference in practice, but asymptotic notation is completely oblivious to it. – Yuval Filmus Mar 21 '16 at 0:14
Auberon provided a very good explanation of the Big O. I will try to explain what that means for you.
First of you are right. The first code example has $\frac{n^2}{4}$ iterations, but is still in the complexity class O(n^2).
Why?
The thing about complexity classes is, that we assume that doing something several times is not that bad for runtime.
Imagine a Algorithm with complexity O(2^n) running for n=3 and taking 1 second.
We could run this 10 times, and still expect an answer after about 10 seconds.
Now Imagine increasing n by 10. The Programm will take 2^10=1024 seconds.
So the computer scientists basically said: "Man, leading factors are annoying. Let's just assume n grows to infinity and compare functions there"
Which lead to the Big O.
In Practice
In Practice it is very well possible that a solution with much worse complexity runs much faster (for "small" inputs). It is easy to imagina a Programm that runs in O(n) but needs 10^10*n iterations.
It is O(n) but even an O(2^n) solution could be faster for small n.
In summary:
The Big O is a useful tool. But you still have to think about how fast your algorithm is in practice, since it only describes how much more time it will need if the input grows.
If I have this pseudocode:
for i=0 to n/2 do
for j=0 to n/2 do
... do anything ....
The number of iteration is $n^2/4$.
Actually, it's $(1+n/2)^2 = n^2/4 + n + 1$.
What is the complexity of this program? Is correct $O(n^2)$?
That depends entirely on what "do anything" is. For example, if "do anything" is replaced by the line
for k=0 to 2^n do {}
then the running time is $\Theta(n^22^n)$. | 2021-01-22T00:45:43 | {
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https://math.stackexchange.com/questions/1000035/find-local-max-min-concavity-and-inflection-points | # Find local max, min, concavity, and inflection points
For the function $f(x)=\frac{x^2}{x^2+3}$ Find the intervals on which f(x) is increasing or decreasing.
Find the points of local maximum and minimum of f(x).
Find the intervals of concavity and the inflection points of f(x).
$f′(x)=\frac{6x}{(x^2+3)^2}$
$f''(x)=\frac{-18(x^2-1)}{(x^2+3)^3}$
• I know the critcal point is $x = 0$ if you plug that in to $f(x)$ and $f'(x)$ you get $0$ if you plug it into $f''(x)$ you get $f''(x) = \frac{2}{3}$ does this mean that the local min is $0$ at $x = 0$ ? – Csci319 Oct 31 '14 at 17:13
• Indeed it does. That method is called the "second derivative test" if you want to Google it. – GFauxPas Oct 31 '14 at 17:17
• but does that mean there is no max? how would I find the max? – Csci319 Oct 31 '14 at 17:47
• If the domain of your function is all real numbers, then the function has no maximum. – GFauxPas Oct 31 '14 at 18:05
## 1 Answer
Sign analysis of $f'$
$f'(x)=0$ when $x=0$ so partition the number line as $(-\infty,0)\cup(0,\infty)$.
$(-\infty,0)$: Pick a test point in this interval, say $x=-1$. Note $f'(-1)<0$ so $f'(x)<0$ on this interval. Hence $f$ is decreasing on this interval.
$(0,\infty)$: Pick a test point, say $x=1$. Then $f'(1)>0$ so $f'(x)>0$ on this interval. Hence $f$ is increasing on this interval.
Since $f$ went from decreasing to increasing at $x=0$, a local min occurs at $x=0$.
Sign analysis of $f''$
$f''(x)=0$ when $x=\pm 1$ so partition the number line as $(-\infty,-1)\cup(-1,1)\cup(1,\infty)$.
$(-\infty,-1)$: Pick a test point in this interval, say $x=-2$. Then $f''(-2)<0\implies f''(x)<0\implies f\text{ concave down}$ on this interval
$(-1,1)$: Pick a test point in this interval, say $x=0$. Then $f''(0)>0\implies f''(x)>0\implies f\text{ concave up}$ on this interval
$(1,\infty)$: Pick a test point in this interval, say $x=1$. Then $f''(1)<0\implies f''(x)<0\implies f\text{ concave down}$ on this interval
Since $f$ changed concavity at $x=\pm 1$, these are inflection points.
Indeed a plot of $y=f(x)$ bears out the information above: | 2019-10-19T02:30:05 | {
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https://math.stackexchange.com/questions/1727901/dim-k-operatornameimt2-dim-k-operatornameimt-implies-operatornam | # $\dim_K\operatorname{Im}(T^2) =\dim_K\operatorname{Im}(T) \implies \operatorname{Im}(T) \cap \operatorname{Ker}(T) = \{0\}$
I need some help here...
Let $V$ be a vector space over a field $K$, and $T: V \to W$ a linear transformation. Prove that if $$dim_KV \lt \infty \ \ \text{and} \ \ \dim_K\operatorname{Im}(T^2) = \dim_K\operatorname{Im}(T)$$ then $$\operatorname{Im}(T) \cap \operatorname{Ker}(T) = \{0\}$$
This is what I made so far.
Let $x \in \operatorname{Im}(T) \cap \operatorname{Ker}(T)$,where T is linear operator. Then $T(x)=0$ and $x=T(v)$ for some $v \in V$
$0 = T(x) = T(T(v)) = T^2(v) \implies v\in \operatorname{Ker}(T^2)$
My intention is to reach $x=0$, but I'm stuck at this point. I also deduced, via dimension theorem, that $\dim_K \operatorname{Ker}(T) = \dim_K \operatorname{Ker}(T^2)$ but I don't know where it can help. Every hint is appreciated.
• You have $\ker(T) \subset \ker(T^2)$. Now use the dimension. – Maik Pickl Apr 4 '16 at 19:49
The rank-nullity theorem says \begin{align} \dim V&=\dim\operatorname{Im}(T)+\dim\operatorname{Ker}(T) \\ \dim V&=\dim\operatorname{Im}(T^2)+\dim\operatorname{Ker}(T^2) \end{align} Next use the fact that $$\operatorname{Ker}(T)\subseteq\operatorname{Ker}(T^2)$$ to conclude that $$\operatorname{Ker}(T)=\operatorname{Ker}(T^2)$$
• Oh, I did not know that $\operatorname{Ker}(T)\subseteq\operatorname{Ker}(T^2)$. – Карпатський Apr 4 '16 at 19:59
• @Карпатський If $T(v)=0$, then $T^2(v)=0$. ;-) – egreg Apr 4 '16 at 20:00
• @egreg I haven't done linear algebra for a while, and I'm trying to refresh my memory. How exactly can we conclude that $\operatorname{Ker(T)}=\operatorname{Ker(T^2)}$? Does it follow from the fact that their dimensions are equal and the fact that the one is subset of the other implies that they live in the same space? – K.Power Apr 7 '16 at 23:19
• @K.Power Since $\ker(T)\subseteq\ker(T^2)$, when we show the former has the same dimension as the latter, we conclude they're equal (take a basis of $\ker(T)$: it has the number of elements needed to be a basis of $\ker(T^2)$). – egreg Apr 7 '16 at 23:44 | 2019-07-15T22:14:36 | {
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http://math.stackexchange.com/questions/155829/what-is-wrong-with-my-solution-int-cos2-x-tan3x-dx | # What is wrong with my solution? $\int \cos^2 x \tan^3x dx$
I am trying to do this problem completely on my own but I can not get a proper answer for some reason
\begin{align} \int \cos^2 x \tan^3x dx &=\int \cos^2 x \frac{ \sin^3 x}{ \cos^3 x}dx\\ &=\int \frac{ \cos^2 x\sin^3 x}{ \cos^3 x}dx\\ &=\int \frac{ \sin^3 x}{ \cos x}dx\\ &=\int \frac{ \sin^2 x \sin x}{ \cos x}dx\\ &=\int \frac{ (1 -\cos^2 x) \sin x}{ \cos x}dx\\ &=\int \frac{ (\sin x -\cos^2 x \sin x) }{ \cos x}dx\\ &=\int \frac{ \sin x -\cos^2 x \sin x }{ \cos x}dx\\ &=\int \frac{ \sin x }{ \cos x}dx - \int \cos x \sin x dx\\ &=\int \tan x dx - \frac{1}{2}\int 2 \cos x \sin x dx\\ &=\ln|\sec x| - \frac{1}{2}\int \sin 2x dx\\ &=\ln|\sec x| + \frac{\cos 2x}{4} + C \end{align}
This is the wrong answer, I have went through and back and it all seems correct to me.
-
Alright, now I would like to know why the first integral gives $\ln \sec x$ instead of $-\ln \cos x$. – Phira Jun 8 '12 at 23:07
I think that was just an error in typing out the question – user138246 Jun 8 '12 at 23:08
@Phira: Because you can bring the $-1$ up. – Eugene Jun 8 '12 at 23:16
Same function, different expression. Note that $$\ln|\sec x| = \ln\left|\frac{1}{\cos x}\right| = \ln\left(|\cos x|^{-1}\right) = -\ln|\cos x|.$$ And $$\cos 2x = \cos^2x - \sin^2x = \cos^2x-(1-\cos^2x) = 2\cos^2x - 1$$so $$\frac{\cos 2x}{4}+C = \frac{2\cos^2x}{4}-\frac{1}{4}+C = \frac{1}{2}\cos^2x + C'.$$ – Arturo Magidin Jun 8 '12 at 23:16
You're making good progress I see. – Eugene Jun 8 '12 at 23:27
A very simple way to check if the answer to an indefinite integral is correct is to differentiate the answer. If you get the original function, your answer is correct, and is equal, up to a constant, with any other solutions.
We have \begin{align*} \frac{d}{dx}\left(\ln|\sec x| + \frac{\cos 2x}{4} + C\right) &= \frac{1}{\sec x}(\sec x)' + \frac{1}{4}(-\sin(2x))(2x)' + 0\\ &= \frac{\sec x\tan x}{\sec x} - \frac{1}{2}\sin(2x)\\ &= \tan x - \frac{1}{2}\left(2\sin x\cos x\right)\\ &= \frac{\sin x}{\cos x} - \sin x\cos x\\ &= \frac{ \sin x - \sin x\cos^2 x}{\cos x}\\ &= \frac{\sin x(1 - \cos^2 x)}{\cos x}\\ &= \frac{\sin^3 x}{\cos x}\\ &= \frac{\sin ^3 x \cos^2 x}{\cos^3x}\\ &= \frac{\sin^3 x}{\cos^3 x}\cos^2 x\\ &= \tan^3 x \cos^2 x. \end{align*} So your answer is right.
This happens a lot with integrals of trigonometric identities, because there are a lot of very different-looking expressions that are equal "up to a constant". So the answer to $$\int\sin x\cos x\,dx$$ can be either $\sin^2 x + C$ or $-\cos^2x + C$; both correct, though they look different, because they differ by a constant: $-\cos^2x + C = 1-\cos^2x + (C-1) = \sin^2x + (C-1)$.
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@Always love your answers Prof. – Babak S. Jun 9 '12 at 2:35
$${\ln |\sec x| + \frac{{\cos 2x}}{4} + C}$$
$${\ln \left|\frac 1 {\cos x}\right| + \frac{{1+\cos 2x}}{4} + C}-\frac 1 4$$
$${-\ln \left| {\cos x}\right| + \frac 1 2\frac{{1+\cos 2x}}{2} + K}$$
$${-\ln \left| {\cos x}\right| + \frac 1 2 \cos ^2 x + K}$$
-
this kind of thing makes me feel like Calculus II exams and assignments must be hell to mark. – crf Jun 8 '12 at 23:56
Everything is hell to mark. But one can often structure exam questions to minimize these issues. – André Nicolas Jun 9 '12 at 0:28
You have been simplifying things up until line 6 and then kind of turned back into complications. It would be natural to notice that $$\sin x dx = -d\left(\cos x\right)$$ Then the integral looks as follows: $$I=-\int\frac{1-\cos^2x}{\cos x}d\left(\cos x\right)$$ So it appears that $\cos x$ plays the role of a variable in its own right, so why not let for example $t=\cos x$. Now $$I=-\int\frac{1-t^2}{t}dt=-\int\frac{dt}{t}+\int tdt=-\ln t+\frac{t^2}{2}+C$$ Now plug $\cos x$ back into place. Recognising distinct "reusable" blocks within the expression is the most natural way to arrive at useful substitutions.
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The calculation is correct. There are many alternate forms of the integral, because of the endlessly many trigonometric identities.
If you differentiate the expression you got and simplify, you will see that you are right.
The answer you saw is likely also right. What was it?
Added: Since $\sec x=\frac{1}{\cos x}$, we have $\ln(|\sec x|)=-\ln(|\cos x|)$.
Also, since $\cos 2x=2\cos^2 x -1$, we have $\frac{\cos 2x}{4}=\frac{\cos^2 x}{2}-\frac{1}{4}$.
So your answer and the book answer differ by a constant. That's taken care of by the arbitrary constant of integration. As a simpler example, $\int 2x\, dx=x^2+C$ and $\int 2x\, dx=(x^2+17)+C$ are both right.
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$\frac{1}{2}cos^2 x - ln|cosx| + C$ – user138246 Jun 8 '12 at 23:18
Notice that:
$$\frac{1}{2} \cos^2 x = \frac{1}{2} \left(\frac{1}{2} + \frac{1}{2} \cos 2x\right) = \frac{1}{4} + \frac{1}{4} \cos 2x$$
And:
$$-\ln|\cos x| = \ln|(\cos x)^{-1}| = \ln|\sec x|$$
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∫tanxdx=ln|secx|=-ln|cosx|+ k since -ln |cos x|+k= ln|(cos x)^-1|+k = ln|sec x| +k
(cos2x)/4= 1/2cos^2(x) - 1/4
k-1/4 = new constant C and together you have the solution.
Your answer seems to be correct it is just manipulation or different approach to trig identities.
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A simple $u$-substitution will work even better, IMO. ${\cos^2\theta\tan^3\theta }$ yields ${\frac {\sin^3}{\cos\theta}}$ So starting from there $${\int\frac{\sin^2\theta}{\cos\theta}\sin\theta d\theta}$$ then $${\int\frac{1-\cos^2\theta}{\cos\theta}\sin\theta d\theta}$$ Let $u={\cos \theta}$ then ${du=-\sin\theta d\theta}$ $${-\int\frac {1-u^2}{u}du}$$ let ${v=1-u^2 }$ ${dv=-2u}$
let ${dw=1/u}$, ${w=\ln u}$
thus $${\ln u(1-u^2)+\int 2u\ln u}\,du$$ then
let ${v=\ln u}$, ${dv=u^{-1}}$
let ${dw=2u}$, ${w=u^2}$
then \begin{align*} {\ln u u^2-\int \frac{u^2}{u}\,du}\\{\ln u u^2-\int u\,du}\\ {\ln u(1-u^2)+\ln u u^2-1/2u^2} \end{align*} then \begin{align*}{\ln u(1-u^2+u^2)-1/2u^2}\\ {-(\ln u-1/2u^2)} \end{align*} substituting back yields $${-\ln\cos\theta+1/2\cos^2\theta+c}$$
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https://math.stackexchange.com/questions/2003840/a-small-theatre-contains-10-seats-the-seats-are-numbered-from-1-to-10 | # A (small) theatre contains 10 seats. The seats are numbered from 1 to 10,
A (small) theatre contains 10 seats. The seats are numbered from 1 to 10, but not in order. Alice, Bob and Carol have tickets for seats 1, 2 and 3 respectively, and Damien, Eve, Francis, and Gertrude each have tickets without an assigned seat. Unfortunately they all arrive late and the lights are dim, so they can’t see what seat they are sitting in.
a) In how many ways can they be seated?
b) In how many ways can they be seated such that Alice is in her correct seat (eg, seat number 1)?
c) In how many ways can they be seated such that Alice, Bob and Carol are all in their correct seats?
d) In how many ways can they be seated such that none of Alice, Bob, Carol are in their correct seats?
What I did for part A is The first person has 10 seats to choose from, then the second person has 9 and so on. I got $10\times9\times8\times7\times6\times5\times4$ as the answer.
For part B I said alice has 1 seat to choose from and then did the rest like part A so I got $1\times9\times8\times7\times6\times5\times4$.
For C I did the same as B but with 3 people and got $1\times1\times1\times7\times6\times5\times4$ but this doesn't seem right to me.
A bit confused on how to do the rest.
"For c) I did the same as b) but with 3 people and got $7^{\underline{4}}=7\cdot6\cdot5\cdot4$ but this doesn't seem right to me"
Your first three parts are in fact all correct (though they suffer from a lack of convenient notation. If there were 50 people and 80 seats you don't want to waste time writing out all of the numbers in the product. Recommend using one of the following falling factorial notations $(7)_4, 7^{\underline{4}}, \frac{7!}{3!}, ~_7P_4, P(7,4),\dots$ but whichever you use make sure your audience understands the notation)
(Note on falling factorial notation: in all of the above, $(n)_r=n^{\underline{r}}=P(n,r)=\dots=\underbrace{n(n-1)(n-2)\cdots(n-r+2)(n-r+1)}_{r~\text{terms in product starting with}~n}$)
"A bit confused on how to do the rest"
The first three parts are most of the calculations needed to complete the fourth part. Recognize that by symmetry the answer to part b) is also the answer to the question of how many ways Bob is seated in his correct seat as well as the answer to how many ways Carol is in her correct seat.
Use inclusion-exclusion to figure out answer to "In how many ways can they be seated such that at least one of them is in their correct seat." Letting $A,B,C$ represent the events that Alice, Bob, and Carol made it to their correct seats respectively, and letting $\Omega$ represent the sample space i.e. the number of ways in which we don't care who did or didn't make it to their seat, we are tasked with finding the number of ways that none of them made it to their seat. I.e. $|A^c\cap B^c\cap C^c|$
Note that $|A^c\cap B^c\cap C^c| = |\Omega\setminus(A\cup B\cup C)|=|\Omega|-|A\cup B\cup C|$
$=|\Omega|-|A|-|B|-|C|+|A\cap B|+|A\cap C|+|B\cap C|-|A\cap B\cap C|$
You've already found $|\Omega|,|A|,|A\cap B\cap C|$ and by symmetry found $|B|$ and $|C|$. The only information you are missing is $|A\cap B|,|A\cap C|,|B\cap C|$. A similar symmetry argument will reduce the effort it takes to find those.
• so for part d |A∩B|,|A∩C|,|B∩C|, would all be the same and would be calculated by adding P(9,4) to itself. Then I would put everything in the equation you provided? – KGT Nov 7 '16 at 17:52
• @KGT $P(n,r) = \underbrace{n(n-1)(n-2)(n-3)\cdots(n-r+3)(n-r+2)(n-r+1)}_{r~\text{total terms in product starting with}~n}$. I don't think you want to be using $P(9,4)$ – JMoravitz Nov 7 '16 at 17:55
• @KGT you did it correctly in the first three parts. Reworded using the $P$ notation, your answers were $P(10,7),P(9,6)$ and $P(7,4)$ respectively. If we had included the extra question in the middle of those of finding $|A\cap B|$, i.e. alice and bob get their correct seat and then we seat everyone else, how much would that be? – JMoravitz Nov 7 '16 at 18:00
• Oh ok. My mistake. I thought that meant P(9,4) meant 9x8x7x6x5x4. What I should use is P(9,6) correct? – KGT Nov 7 '16 at 18:01
• $P(9,6)$ was the answer to if we only cared about alice getting her correct seat but we didn't care about bob or carol getting their correct seats. If we cared about alice and bob, should that nine have been in the product? – JMoravitz Nov 7 '16 at 18:02 | 2020-09-24T10:40:43 | {
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https://lumina.com/community/sub-forum/how-to-calculate-cumulative-cash-flow-from-a-cash-flow-table-indexed-by-year/ | How to calculate cu...
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# How to calculate cumulative cash flow from a Cash Flow table indexed by year
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I am a relative neophyte in using indexed parameters. I want to calculate a cumulative cash flow by year from a table of cash flows, in order to calculate when a project will break even. I am looking for the syntax of a call on each row of my cash flow list to add it to the previous row of my cumulative cash flow list. Once I have that, I also want to calculate a similar list of discounted cumulative cash flows to understand the effect of discounting the cash flow. Is it worthwhile to create a second index with cash flow, cumulative cash flow, and discounted cash flow? Or is there an even more clever way to do this using features I am unaware of?
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You are looking for the Cumulate function, which takes your cash flow by Year and returns the cumulated cash flow by Year. The syntax is:
Cumulate( Cash_flow, Year )
To discount a cash flow, you just multiply Cash_flow by the discount factor for each year, like this:
Cash_flow * (1 - Discount_rate)^(@Year-1)
Here I've used @Year to number the years starting at 1. By using @Year-1, it means that the first year won't be discounted, whereas if you raise to the power of @Year you would discount the first year. Think about which convention you want. You can then pass this to the first parameter of Cumulate.
You want to compare the undiscounted case to the discounted case. A cool thing is that you can do this without duplicating all the logic for each case. The undiscounted case is simply the discounted case when the discount rate is 0. So, just make Discount_rate a Choice, i.e., between 0% and 7%, and then all your subsequent logic that uses discounted cash flow will work for either, or indeed for both at the same time (by selecting ALL for the choice), giving you the ability to compare any downstream variable without having to do anything additional.
In a short while, I will attach a model that demonstrates all of this and adds in the calculation for the breakeven year.
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This looks great! And the tip on 0% discount is excellent. I look forward to your example, and hope to get back to this work this week.
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This approach worked great. Now I can get to my reason for doing this. I would like to define a variable that finds the year when cumulative cash flow becomes positive, return the year, and add to that the value of cumulative cash flow for that year divided by the cash flow for that year, which gives me a number of years (with a fraction) to break even. Still looking to figure out how to do this.
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Sorry about the delay with posting the model that I had promised. I hit a technical problem. Anyway, here it is:
Cash flow.ana
It includes a variable that finds the breakeven year and a fraction of the breakeven year. I'm not sure if my fraction of the breakeven year is exactly what you just described, but hopefully you can tweak it if not.
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I got taken away on other things (including a short vacation) and am just getting back to this reply. I had in the meantime discovered the Solar Panel model, which gets the breakeven year by using the CondMin function, but you have come at it a separate way. Either way, I am learning useful new functions! I will look at it and see if it gets the same answer I was coming to. Thanks!
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https://math.stackexchange.com/questions/998177/how-to-know-when-a-particular-proof-is-appropriate-for-the-given-problem | # how to know when a particular proof is appropriate for the given problem?
The main trouble I am currently having in math is knowing when the use cases are appropriate in a proof. I see many videos where they seem to choose a strategy like proof by contrapositive or proof by contradiction, but never quite understand how they came to the conclusion to use that proof strategy.
Here are some examples I have come across and my own solutions to them using the recommended proofs
the product of two odd integers is odd
I used contradiction to solve it
• suppose the product of two odd integers is even
• (2k+1)(2k+1) = 2k
• 2(2k^2 + 2k) + 1 = 2k
• given k is an integer, (2k^2 + 2k) is an integer. Therefore, an even number cannot be an odd number
if x^2 is even, x is even
Based on the wikipedia article on contrapositive
• if x is odd, then x^2 is odd
• (2k+1)^2 == 2(2k^2 +2k) + 1
• therefore, if x^2 is even, then x is even
However, what I am wondering is, is there a general principle as to when to use specific strategies for proofs? Eg, if you specifically know a theory is false, do you choose a strategy accordingly? What determines which strategy will be most effective, or is it arbitrary?
• – Shaun Oct 30 '14 at 12:38
• For "perfect" proofs see the book of Aigner and Ziegler, Proofs from THE BOOK – Dietrich Burde Oct 30 '14 at 12:55
This skill comes with lots of practice. Generally speaking, direct proofs are used when the result is "positive-sounding". In your example, the result is, "the product of two odd integers is odd". This result is positive-sounding (as opposed to a result like, "there is no smallest positive real number").
For the result, "if $x^2$ is even, then $x$ is even," you could do either a direct proof or a proof by contrapositive. A lot of times it does not matter and there is usually more than one way to prove something. The direct proof, in this case, is favored over a proof by contrapositive simply because it is more, well, direct. Here is how the result would be proven using the contrapositive:
Proof:
Suppose $x$ is odd. We show that $x^2$ is odd. So $x=2k+1$ for some integer $k$. Thus $x^2=(2k+1)^2=(2k+1)(2k+1)=4k^2+4k+1=2(2k^2+2k)+1.$ Now, since $(2k^2+2k)$ is an integer, it follows that $x^2$ is odd.
For a result like "there is no smallest positive real number", this would be difficult to prove using a direct proof. The result is better suited for a proof by contradiction. It would start like this: "Suppose there exists a smallest positive real number…"
Some other examples of negative sounding results that would be more suited for proof by contradiction:
• If a is an even integer and b is an odd integer, then 4 does not divide $a^2+2b^2$
• The integer 100 cannot be written as the sum of three integers, and odd number of which are odd
• The sum of a rational number and an irrational number is irrational.
[Source: Chartrand, G., Polimeni, A.D., Zhang, P., 2013. Mathematical Proofs: A Transition to Advanced Mathematics, 3rd ed. Boston: Pearson.]
• ah, exactly what I was looking for, very clear and understandable, thank you! – corvid Oct 30 '14 at 13:05
There are many ways to write down a proof. The question is, apart from correctness, which one is more elegant, more concise, more constructive, and last not least easier und thus clearer. In your case, for me the easiast way is to do first an obvious computation, i.e. $$(2k+1)(2l+1)=4kl+2k+2l+1.$$ Your first step here is wrong, because you assume that both integers are equal, i.e., both given by $2k+1$. This would be a special case.
Then, in the second step, we consider everything modulo $2$. The integer on the RHS is odd, because it is of the form $2m+1$, with $m=2kl+k+l$. This finishes the proof.
• theoretically, anything you can prove by contradiction can also be proved by contrapositive, in that sense? – corvid Oct 30 '14 at 12:54 | 2019-06-20T04:58:14 | {
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http://carlacasciari.it/moment-of-inertia-of-triangle-about-apex.html | s′ = required moment of inertia of the combined ring‐ shell‐cone cross section about its neutral axis par-allel to the axis of the shell, in. Résultat de recherche d'images pour "bridge equation for moment of inertia" See more. Find the moment of inertia of the framework about an axis passing through A, and parallel to BC 5ma 2. The second moment of inertia of the entire triangle is the integral of this from $$x = 0$$ to $$x = a$$ , which is $$\dfrac{ma^{2}}{6}$$. Centroid, Area, Moments of Inertia, Polar Moments of Inertia, & Radius of Gyration of a Triangular Cross-Section. RE: second moment of inertia for triangle cross section? the formula. Lecture notes, lecture 11 - Center of gravity, centroid and moment of inertia. The equation for the moment of inertia becomes: ∫ − = − 8 8 2 2 2 2 x' dy' 14 y' I y' 2 8 1 To perform this integration we need to place the integrand in an m-file function and call MATLAB’s quad() function on the m-file. After explaining the term second moment of area, the method of finding moment of inertia of plane figures about x-x or y-y axis is illustrated. You can also drag the origin point at (0,0). Once the moment of inertia was calculated, we had to measure the angular acceleration of the pulley. See the picture: the points of the upper triangle are farther than those of the lower triangle. It is measured by the mass of the body. o The moment of inertia of a thin disc of mass m and radius r about an axis passing through its C. unambiguous choice between the divergent views currently held with regard to the structure. We can see from that the moment of inertia of the subrectangle about the is Similarly, the moment of inertia of the. function Ix_integrand = Moment_Of_Inertia_Integrand(y_prime) %Saved as Moment_Of_Inertia_Integrand. The moment of inertia of the triangle is not half that of the square. 250 kg; from mass A: rB² = 0. A numerical integrator might return slightly less accurate results, but other than that there is not much benefit from using symbolic integration there. - Theory - Example - Question 1 - Question 2 - List of moment of inertia for common shapes 4. It is based not only on the physical shape of the object and its distribution of mass but also the specific configuration of how the object is rotating. Thus the mass of the body is taken as a measure of its inertia for translatory. Today we will see here the method to determine the moment of inertia for the triangular section about a line passing through the center of gravity and parallel to the base of the triangular section with the help of this post. The moment of inertia of a triangle with respect to an axis passing through its apex, parallel to its base, is given by the following expression: I = \frac{m h^2}{2} Again, this can be proved by application of the Parallel Axes Theorem (see below), considering that triangle apex is located at a distance equal to 2h/3 from base. pptx), PDF File (. However, "area moment of inertia" is just 4 words to me (no physical meaning). Overview (in Hindi) 8:26 mins. Polar moment of Inertia (Perpendicular Axes theorem) The moment of inertia of an area about an axis perpendicular to the plane of the area is called Polar Moment of Inertia and it is denoted by symbol Izz or J or Ip. b) Determine the moment of inertia for a composite area Parallel-Axis Theorem for an Area Relates the moment of inertia of an area about an axis passing through the. Check to see whether the area of the object is filled correctly. 7) Moment of Inertia Triangle. Mathematical calculations of the GaN NWs' cross-sectional areas and the moment of inertia For the single crystalline (SC) GaN nanowire (NW), the cross-sectional shape is an isosceles triangle with a 63. The median is a line from vertex to the center of a side opposite the vertex. Or the Mizuno MP-20. Find the moment of inertia of the triangle about axis passing through centroid perpendicular to lamina. Hodgepodge. The moment of point "B" is 0. The length of each side is L. In case of shafts subjected to torsion or twisting moment, the moment of inertia of the cross-sectional area about its centre O is considered. The sum of all these would then give you the total moment of inertia. 3 (4) 3 Determine the AP whose fourth term is 15 and the difference of 6th term from 10th term is 16 Prove that ratio of area of two triangle is equal to the square of the corresponding sides. Centroid, Area, Moments of Inertia, Polar Moments of Inertia, & Radius of Gyration of a Triangular Cross-Section. About the Moment of Inertia Calculator. The mass moment of inertia is a measure of an object’s resistance to rotation. P-715 with respect to the given axes. 456kg Length of the base of triangle =. Statics - Chapter 10 (Sub-Chapter 10. Answer this question and win exciting prizes. Moments of Inertia Staff posted on October 20, 2006 | Moments of Inertia. The moment of inertia of a plane lamina about an axis perpendicular to the plane of the lamina is equal to the sum of the moment of inertias of the lamina about the two axes at right angles to each other in its own plane intersecting each other at the point where the perpendicular axis passes through it. Triangle Moment of Inertia. The greater the mass of the body, the greater its inertia as greater force is required to bring about a desired change in the body. Moment of inertia, denoted by I, measures the extent to which an object resists rotational acceleration about a particular axis, and is the rotational analogue to mass. 31 shows a T-section of dimensions 10 × 10 × 2 cm. The area moment of inertia is also called the second moment of area. Mar 27, 2001 3,923 0 76. G The centroid and centre of gravity are at the same point Where centre of gravity consider to be whole mass of an object act at a point C. The sum of the first n ≥ 1 energy levels of the planar Laplacian with constant magnetic field of given total flux is shown to be maximal among triangles for the equilateral triangle, under normalization of the ratio (moment of inertia)/(area) 3 on the domain. Example of Product Moment of Inertia of a Right Angle Triangle Product Moment of Inertia of a Right Angle Triangle by Double Integration. The greater the mass of the body, the greater its inertia as greater force is required to bring about a desired change in the body. The calculations are as shown. Lecture notes, lecture 11 - Center of gravity, centroid and moment of inertia. This app was created to assist you in these calculations and to ensure that your beam section calculations are fast and accurate. It is based not only on the physical shape of the object and its distribution of mass but also the specific configuration of how the object is rotating. Rectangle Triangle. For this reason current vector is treated as normal vector of the plane and the input cloud is projected onto it. The struts are built with the quad-edge passing through the mid-point of the base. 저자: No machine-readable author provided. The mass moment of inertia is often also known as the rotational inertia, and sometimes as the angular mass. Date: 02/03/99 at 14:37:05 From: Doctor Anthony Subject: Re: MI of Solid Cone You must, of course, specify about which axis you want the moment of inertia. Recommended for you. Determine the moment of inertia of this 10. Physics Lab #17 started on 5/13/15 Finding the moment of inertia of a uniform triangle about its center of mass Annemarie Branks Professor Wolf Objective: Find the moment of inertia for a uniform, right triangle plate about its center of mass for the two orientations as shown below. txt) or view presentation slides online. on AIPMT / NEET-UG entrance. I xx = ∫dA. Moment of inertia is a commonly used concept in physics. m in the MATLAB. Area moment of inertia calculation formulas for the regular cross section are readily available in design data handbooks. In physics and applied mathematics, the mass moment of inertia, usually denoted by I, measures the extent to which an object resists rotational acceleration about a particular axis, and is the rotational analogue to mass. The moment of inertia of a body is always defined about a rotation axis. 4 "Center of Mass" of our text APEX Calculus 3, version 3. The maximum shear. Assume density = 1 Here's my working , I use zx plane projection , but i didnt get the ans. We define dm to be a small element of mass making up the rod. ” Moment of inertia = SI unit of moment of inertia is Q. The calculations are as shown. Mass moment of inertia (also referred to as second moment of mass, angular mass, or rotational inertia) specifies the torque needed to produce a desired angular acceleration about a rotational axis and depends on the distribution of the object’s mass (i. In order to find the moment of inertia of the triangle we must use the parallel axis theorem which ius as follows: The moment of inertia about any axis parallel to that axis through the center. Rotational version of Newton's second law. The apex angle of the quarter-circle is $\pi/2$. The moment of point "C" is the same as "B" so multiply the moment of "B" by two. Cone Calc Processing :. Answer this question and win exciting prizes. 4 Moment of inertia in yaw 2. To find the inertia of the triangle, simply subtract the inertia of the system with the triangle from the benchmark. Centroid, Area, Moments of Inertia, Polar Moments of Inertia, & Radius of Gyration of a Triangular Cross-Section. Moments of Inertia. P-715 with respect to the given axes. Find the moment of inertia of a circular section whose radius is 8” and diameter of 16”. The moment of inertia of the triangle is not half that of the square. Determining the moment of inertia of a solid disk. , the opposition that the body exhibits to having its speed of rotation about an axis altered by the application of a torque (turning force). 5 2 3 A 4-0. Click Content tabCalculation panelMoment of Inertia. Q1: Matthew has a model train that uses a circular cone as a flywheel. Data: 23 d'abril de 2006 (original upload date) Font: No machine-readable source provided. The term product moment of inertia is defined and the mehtod of finding principal moment of inertia is presented. This banner text can have markup. Knowing the area moment of inertia is a critical part of being able to calculate stress on a beam. Centroid, Area, Moments of Inertia, Polar Moments of Inertia, & Radius of Gyration of a Triangular Cross-Section. 1 Expert Answer(s) - 58298 - what is the moment of inertia of a triangular plate ABC of mass M and side BC = a about an axis pass. Which one has the greatest moment of inertia when rotated about an axis perpendicular to the plane of the drawing at point P? Preview this quiz on Quizizz. 4 Moment of inertia in yaw 4 DISCUSSION OF 33TIXYI'ZD Al4D ?XWXQdENPAL VAIJJES 4. 035; Actual VCOG. Area Moment of Inertia or Moment of Inertia for an Area - also known as Second Moment of Area - I, is a property of shape that is used to predict deflection, bending and stress in beams. If you need a beam’s moments of inertia, cross sectional area, centroid, or radius of gyration, you need this app. The oxygen molecule as a mass of 5. 2) A long rod with mass has a moment of inertia , for rotation around an axis near one. Calculating the moment of inertia of a triangle - Duration: 10:01. The moment of inertia must be specified with respect to a chosen axis of rotation. 1st moment of area is area multiplied by the perpendicular distance from the point of line of action. Figure to illustrate the area moment of a triangle at the list of moments of inertia. 1) Prove that the centroid of any triangle of height h and base b is located 2/3 of the distance from the apex. Own work assumed (based on copyright claims). And, regretfully, you disturbed. The moment of inertia of an object is based on 3 things, the mass of the object, the axis of rotation, and the orientation and distance of the object from the axis of rotation. Inertia is a property of a body to resist the change in linear state of motion. Rotational kinetic energy. The moment of inertia of two or more particles about an axis of rotation is given by the sum of the moment of inertia of the individual particles about the same axis of rotation. Use our free online app Moment of Inertia of a Ring Calculator to determine all important calculations with parameters and constants. Moment of Mass about x and y-axis Mass of Lamina - f(x) Mass of Lamina - f(y) Radius of Gyration (x-axis) Radius of Gyration (y-axis) 1. Tags: Equations for Moment of Inertia. The 2 nd moment of area, or second area moment and also known as the area moment of inertia, is a geometrical property of an area which reflects how its points are distributed with regard to an arbitrary axis. My teacher told me :. Let us assume that one line is passing through the base of the triangular section and let us consider this line as line BC and we will determine the moment of inertia for the triangular section about this line BC. I = 3 [I cm + M d^2] =3 [ML^2 / 2+ M d^2]. (1) I y: equ. The moment of inertia about the X-axis and Y-axis are bending moments, and the moment about the Z-axis is a polar moment of inertia(J). Learn vocabulary, terms, and more with flashcards, games, and other study tools. PEP Assignment 4 Solutions 3 = = 2 where = 2 Knowing what d and are, CM is CM =,∫ d CM = from the apex of the triangle. The unit of dimension of the second moment of area is length. Therefore, equation for polar moment of inertia with respect to apex is. Thus, the moment of inertia of a 2D shape is the moment of inertia of the shape about the Z-axis passing through the origin. I = 1/3 b * h^3 / 12. m2) Title: Microsoft Word - Formular Moment of Inertia Author: d00997 Created Date: 4/25/2019 4:40:32 PM. Triangle h b A= 1 2 b×h x1=b/3 From side x2=2b/3 From right side y1=h/3 From bottom y2=2h/3 From Apex Ixx= bh3 36 Circle d A=π 4 ×d2 x=d/2 y=d/2 I xx= π 64 d4 I yy= π 64 d4 Semicircle A= π 4 ×d2 2 x=d/2 y1=0. Solution 126 2 Polar moment of inertia SECTION 126 Polar Moments of Inertia 15 from COE 3001 at Georgia Institute Of Technology. 51 videos Play all MECHANICAL ENGINEERING 12 MOMENT OF INERTIA / AREA Michel van Biezen Area Moment of Intertia of a Triangle Brain Waves - Duration: 7:23. 2) The radius of the gyration of a disc of radius 25 cm is. 3-axis along the axis of the cone. The moment of inertia is not related to the length or the beam material. Write the equation for polar moment of inertia with respect to apex of triangle. dA Y = 0 A A = b. ) have only areas but no mass. Solution for Find the center of mass and the moment of inertia about the z-axis of a thin shell of constant density d cut from the cone x2 + y2 - z2 = 0 by the… Answered: Find the center of mass and the moment… | bartleby. Calculate the moment of inertia of a right circular cone. Hemmingsen assumed (based on copyright claims). Find the moment of inertia of the table with the iron ring. I = Pi * R^4 / 4. March 2020. The moment of inertia is related to the rotation of the mass; specifically, it measures the tendency of the mass to resist a change in rotational motion about an axis. he solves alone. The centre of area of such figure is known as centroid. Find the moment of inertia of a uniform solid circular cone of mass M, height h and base radius a about its axis, and also about a perpendicular axis through its apex. (by the parallel axis theorem). Figure to illustrate the area moment of a triangle at the list of moments of inertia. Moments of Inertia Staff posted on October 20, 2006 | Moments of Inertia. Area Moment of Inertia or Moment of Inertia for an Area - also known as Second Moment of Area - I, is a property of shape that is used to predict deflection, bending and stress in beams. It should not be confused with the second moment of area, which is used in bending calculations. Moment of Inertia of a Triangular Lamina about its Base. To predict the period of a semi-circle and isosceles triangle with some moment of inertia after first calculating the moment of inertia of the semi-circle and isosceles triangle about a certain axis. Explain the terms moment of inertia and radius of gyration of a plane figure. 89 × 103 kg/m3. Moment of Inertia of Surfaces. Second, finding the moment of inertia when the triangle rotates around its base (shorter leg). Ditto the TaylorMade P730. Basic trig functions 8 Moments of Inertia The moment of inertia is the stiffness of a body due to its size and. ” The product mass and the square of the perpendicular distance from the axis of rotation is known as moment of inertia. ” Moment of inertia = SI unit of moment of inertia is Q. Then remove the middle triangle from each of the re-maining three triangles (as shown), and so on, forever. Section area moment of inertia section modulus calculator what is the difference between polar moment of inertia m area moment of inertia typical cross sections i statics Centroid Area Moments …. The moment of inertia of a triangle of base b and altitude h with respect to a centroidal axis parallel to its base would be bh3/12 bh3/18 bh3/24 bh3/36 The CG of a triangle lies at the point of intersection of diagonals altitudes bisector of angles medians For a solid cone of height h, the CG lies on the axis at a distance above the base equal. The moment of inertia is related to the rotation of the mass; specifically, it measures the tendency of the mass to resist a change in rotational motion about an axis. 01 18-Jun-2003 1. Find I y for the isosceles triangle shown. 32075h^4M/AL, where h is the height of the triangle and L is the area. Inertia is a property of a body to resist the change in linear state of motion. 1 GradedProblems Problem 1 (1. The moment of point "C" is the same as "B" so multiply the moment of "B" by two. Undeniable momentum, on any stage - anywhere. Centroid, Area, and Moments of Inertia Yong-Ming Li January, 1997 1 Introduction This design document was originally written for computation of area, centroid, and moments of inertia of lamina (a thin plate of uniform density). s′ = required moment of inertia of the combined ring‐ shell‐cone cross section about its neutral axis par-allel to the axis of the shell, in. The moment of inertia of a triangle with respect to an axis passing through its apex, parallel to its base, is given by the following expression: I = \frac{m h^2}{2} Again, this can be proved by application of the Parallel Axes Theorem (see below), considering that triangle apex is located at a distance equal to 2h/3 from base. The inertia matrix (aka inertia tensor) of a sphere should be diagonal with principal moments of inertia of 2/5 mass since radius = 1. 1501 Laura Duncan Road, Apex, NC 27502 Email us (919) 289-9278 MAIL TO: P. May 17, 2019 Mirielle Sabety, Keane Wong, Anthony Moody Purpose: The purpose of today's lab is to measure the moment of inertia of a triangle about it's center of mass with in 2 different orientations. Find the moment of inertia of the empty rotating table. I), must be found indirectly. For a homogenous bar of length L and mass M, the moment of inertia about center of mass is (1/3)ML^2. Moments of Inertia 10. 250 kg; from mass A: rB² = 0. Doing the same procedure like above, and below is the work. Using the lower left. This moment of inertia about 0 is called polar moment of inertia or moment of inertia about pole. The moment of inertia is a measure of the resistance of a rotating body to a change in motion. Planar and polar moments of inertia formulas. Area Moment of Inertia Section Properties of Triangle Calculator and Equations. 31 shows a T-section of dimensions 10 × 10 × 2 cm. If you need a beam’s moments of inertia, cross sectional area, centroid, or radius of gyration, you need this app. The moment of inertia is de ned as I= X i m ir 2 i (2) for a collection of point-like masses m ieach at a distance r ifrom the speci ed axis. 2) Find the distance for each intersection points. Choose the origin at the apex of the cone, and calculate the elements of the inertia tensor. Using these moment of inertia, we can subtract from it the moment of inertia of just the system without the triangle to obtain our experimental values for the triangle in either. Radius of gyration 3. Find Moment of Inertia of a Ring Calculator at CalcTown. Centroids and moments of inertia. Radius and elevation of the semi-circle can be changed with the blue point. In other words, it is rotating laterally, similar to how a beam from a lighthouse rotates. 100% Upvoted. We spin the triangle around the spot marked "X", which is one of the balls. Calculate the Inertia of the semi-circle around the pivot. Polar moment of inertia is equal to the sum of inertia about X-axis and Y-axis. In this worksheet, we will practice finding the moment of inertia and radius of gyration of a solid and using the parallel axis theorem to find the moment of inertia of a composite solid at different axes. anybody here could help, please? i would really appreciate it. Meanwhile, I did find the integral formula for computing the center of pressure (Fox) and calculated it using both a flat bottom and inverted isosceles triangle and then using the "area moment of inertia". They are; Axis passing through the centroid. Constant angular momentum when no net torque. Answer Save. This engineering calculator will determine the section modulus for the given cross-section. I am unable to find it. Another solution is to integrate the triangle from an apex to the base using the double integral of r^2dm, which becomes (x^2+y^2)dxdy. It is analogous to mass in that it is a measure of the resistance a body offers to torque or rotational motion. It is used over and over for examples, since it offers a readymade right angle, a hypotenuse, and other great parts. For the section shown in Fig. 1 In the case of mass and area, the problem is deciding the distance since the mass and area are not concentrated at one point. m2) ) x 10-6 (kg. Consider the diagram as below: We can think of the triangle is composing of infinitesimally. In the final stage of the calculation, you specify the direction of the load forces. 75L, just find the area of the left triangle on the shear diagram and subtract the area of 1/2 the horizontal distance of the second triangle (not 1/2 the area of the second triangle). To find the moment of inertia of the entire section, we integrate the above expression and get, Iyy = ΣdAx2, Ixx = ΣdAy2 and Izz = ΣdAz2. The moment of inertia integral is an integral over the mass distribution. Tension Members. Polar moment of inertia is equal to the sum of inertia about X-axis and Y-axis. Mass = m and Base = l Angle at the apex is = 90° Find MI of theplane about the y - axis = ? Let, the axis of rotation pass through hypotenuse, considering rotation about hypotenuse you will see triangle. Hollow Cone. The theoretical one is know the moment of inertia of the triangle plate and applied the parallel axis theorem to found the moment of inertia about a new rotating axis. This is the currently selected item. This banner text can have markup. Mathematically, and where IB " *BA " TIA BA = *B + 7IA Ig = moment of inertia about the base plane I3A = moment of inertia about a base diameter axis 1^ = moment of inertia about the central axis 7. Mechanics of Material (CIV101) Academic year. The moment of inertia is a measure of the resistance of a rotating body to a change in motion. The disk is rotating about its center. Centroids and moments of inertia. Determining the moment of inertia of a rod. a) Define i)Moment of Inertia , ii)radius of gyration b) Define stress ,strain ,Modulus of elasticity c) State formulae to find Moment of Inertia of a triangle about axis passing through its i) Base ii) Apex and iii) centroid d) Define lateral strain, linear strain. The maximum twisting moment a shaft can resist, is the product of the permissible shear stress and (A) Moment of inertia (B) Polar moment of inertia (C) Polar modulus (D) Modulus of rigidly Answer: Option C Question No. Moment of inertia (I1 and I2) along the 1 and 2 axes. of the ozone molecule. The moment of inertia of any triangle may be found by combining the moments of inertia of right triangles about a common axis. r2 x2 y2 Therefore, I z I. Rectangle Triangle. Neutral Axis/Moment of Inertia. moment of inertia. University. Author: No machine-readable author provided. Moment of inertia of pile group. save hide report. 날짜: 2006년 4월 23일 (원본 올리기 일시) 출처: No machine-readable source provided. Evaluation of Moments of Inertia 2008 Waterloo Maple Inc. ) 15 minutes ago The transformer inside of a sound system has 1500 turns in its primary coil windings wrapped around a common iron core with the secondary. “Second moment of an area about an axis is called Moment of inertia. Find The Moment Of Inertia About An Axis That Passes Through Mass A And Is Perpendicular To The. Find its moment of inertia for rotation about the z axis. Simply Supported Beams (Shear & Moment Diagrams) Simply supported beams (also know as pinned-pinned or pinned-roller) are the most common beams for both school and on the Professional Engineers exam. Known : Mass of rod AB (m) = 2 kg. Date: 02/03/99 at 14:37:05 From: Doctor Anthony Subject: Re: MI of Solid Cone You must, of course, specify about which axis you want the moment of inertia. Consider an infinitesimally thin disc of thickness dh, at a distance h from the apex of the cone O. Moment of inertia is defined as:”The sum of the products of the mass of each particle of the body and square of its perpendicular distance from axis. save hide report. The moment of inertia (I) of a body is a measure of its ability to resist change in its rotational state of motion. The area moment of inertia is also called the second moment of area. These came out to be 0. Determine the moment of inertia of the triangular area relative to a line parallel to the base and through the upper vertex in cm^4. Physically, it is a measure of how difficult it is to turn a cross-section about an axis perpendicular to it (the inherent rotational stiffness of the cross-section). a) Define i)Moment of Inertia , ii)radius of gyration b) Define stress ,strain ,Modulus of elasticity c) State formulae to find Moment of Inertia of a triangle about axis passing through its i) Base ii) Apex and iii) centroid d) Define lateral strain, linear strain. because the axis goes through masses B and D their masses doesn't affect to increase the inertia of the system around BD axis. The mass moment of inertia is a measure of an object’s resistance to rotation. For instance, it is easier to find the moment of inertia of a triangle, about an axis which passes through its apex, and parallel to the base, than about any ether axis ; but having found this, we may easily find it about an axis parallel to it which passes through the centre. I = 3 [I cm + M d^2] =3 [ML^2 / 2+ M d^2]. Author: No machine-readable author provided. 12: Inertia due to the Object (kg. After that eccentricity is calculated for the obtained projection. I = 3 [I cm + M d^2] =3 [ML^2 / 2+ M d^2]. Write the equation for polar moment of inertia with respect to apex of triangle. Then this moment of inertia is transferred about the axis passing through the centroid of the given section, using theorem of parallel axis. The moment of inertia $$I_x$$ about the $$x$$-axis for the region $$R$$ is the limit of the sum of moments of inertia of the regions $$R_{ij}$$ about the $$x$$-axis. For a homogenous bar of length L and mass M, the moment of inertia about center of mass is (1/3)ML^2. The concept of a moment of inertia is important in many design and analysis problems encountered in mechanical and civil engineering. CENTER OF GRAVITY, CENTROID AND MOMENT OF INERTIA. The moment of inertia of two or more particles about an axis of rotation is given by the sum of the moment of inertia of the individual particles about the same axis of rotation. 2) A precast concrete floor beam has the cross section shown below. apex angle in the neighborhood of 34°. it is first necessary to consider the rotational moment. Adding moments of inertia 3. The moment of inertia of the triangle is not half that of the square. But I don't know how to do that. The moment of inertia must be specified with respect to a chosen axis of rotation. A triangle cannot have more than one right angle or one obtuse angle, since the sum of all three angles is equal to the sum of two right angles, which is 180° or, in radians, π. I am unable to find it. Ball hits rod angular momentum example. Moment of inertia of a circular section can be calculated by using either radius or diameter of a circular section around centroidal x-axis or y-axis. The moment of inertia of a triangle of base b and altitude h with respect to a centroidal axis parallel to its base would be bh3/12 bh3/18 bh3/24 bh3/36 The CG of a triangle lies at the point of intersection of diagonals altitudes bisector of angles medians For a solid cone of height h, the CG lies on the axis at a distance above the base equal. same object, rotating around a point at the midpoint of its base. 0 cm is made of copper. Look up I for a triangle in your table if you have forgotten. 10 lessons • 1 h 34 m. •Compute the product of inertia with respect to the xyaxes by dividing the section into three rectangles. 4 Locate the centroid of the T-section shown in the Fig. 016 kg ⋅ m2 d. Right: A circle section positioned as per the upper sketch is defined in the calculator as I x-axis , the lower sketch shows I y-axis. It depends on the body's mass distribution and the axis chosen, with larger moments. In this study, we first compute the polar moment of inertia of orbit curves under planar Lorentzian motions and then give the following theorems for the Lorentzian circles: When endpoints of a line segment AB with length a +b move on Lorentzian circle (its total rotation angle is δ) with the polar moment of inertia T, a point X which is collinear with the points A and B draws a Lorentzian. Polar moment of inertia is equal to the sum of inertia about X-axis and Y-axis. Rolling without slipping problems. For a clear understanding of how to calculate moments of inertia using double integrals, we need to go back to the general definition in Section The moment of inertia of a particle of mass about an axis is where is the distance of the particle from the axis. Determining the moment of inertia of a rod. Moment of inertia is defined as:”The sum of the products of the mass of each particle of the body and square of its perpendicular distance from axis. Moments of Inertia. Determine the moment of inertia of this system about an axis passing through one corner of the triangle and perpendicular to the plane of the triangle. Insert the moment of inertia block into the drawing. b) Determine the moment of inertia for a composite area Parallel-Axis Theorem for an Area Relates the moment of inertia of an area about an axis passing through the. o , ,3, Moment of Inertia of Surfaces. y-x O 1 1 • (x, y) r Answer: The polar moment of inertia of a planar region is the moment of inertia about the origin (the axis of rotation is the z-axis). Area Moments of Inertia by Integration • Second moments or moments of inertia of an area with respect to the x and y axes, x ³ yI y ³ xdA 2 2 • Evaluation of the integrals is simplified by choosing dA to be a thin strip parallel to one of the coordinate axes. Letting M be the total mass of the system, we have x ¯ = M y / M. 42441 1in 1 in 1 in 3 in 1 in A 2 A 3 A 1 A 4 16 Centroid and Moment of. 19 The deflection of any rectangular beam simply supported, is (A) Directly proportional to its weight. 41 (a) determine: (i) Moment of inertia about its centroid along (x,y) axis. For each segment defined by two consecutive points of the polygon, consider a triangle with two. If the line l(P, 9) lies in the plane of K through the point P and with direction 9, 0 = 9 ^ 2n, we denote the moment of inertia of K about the line l(P, 6) by I(K, P, 9). Email Print Moment of Inertia of a Triangle. Today we will see here the method to determine the moment of inertia for the triangular section about a line passing through the center of gravity and parallel to the base of the triangular section with the help of this post. Here, distance between apex and centroid is d. apex angle in the neighborhood of 34°. however, i would like to know how you obtain the results. How to calculate polar moment of inertia using Inventor 2014 Hello, I have a problem with calculating polar moment of inertia of a cranshaft with cooperating parts (which I've already assembled in Inventor). Find the moment of inertia of a uniform solid circular cone of mass M, height h and base radius a about its axis, and also about a perpendicular axis through its apex. Now that we have determined the moments of inertia of regular and truncated equilateral triangles, it is time to calculate them for the corresponding right prisms. 1 Verified Answer. 12: Inertia due to the Object (kg. The smallest Moment of Inertia about any axis passes throught the centroid. The struts are built with the quad-edge passing through the mid-point of the base. derivation of inertia of ellipse. Q: the moment of inertia of a thin rod of mass m and length l about an axis through its centre of gravity and perpendicular to its length is a) ml²/4 b) ml²/6 c) ml²/8 d) ml²/12 Q: Which statement is correct: a) Moment of inertia is the second moment of mass or area. Moment of inertia of the equilateral triangle system - Duration: 3:38. The material is homogeneous with a mass density ρ. Today we will see here the method to determine the moment of inertia for the triangular section about a line passing through the center of gravity and parallel to the base of the triangular section with the help of this post. The moment of inertia of any triangle may be found by combining the moments of inertia of right triangles about a common axis. Because there is some frictional torque in the system, the angular acceleration of the system when the mass is descending isn’t the same as when it is ascending. Own work assumed (based on copyright claims). 6-1 Polar moment of inertia POINT C (CENTROID) FROM CASE 5: (I P) c 2 bh. What is the moment of inertia of this rigid body about an axis that is parallel to one side of the triangle and passes through the respective midpoints of the other two sides? a. Thank you User-12527562540311671895 for A2A The moment of inertia of a triangular lamina with respect to an axis passing through its centroid, parallel to its base, is given by the expression $I_{XX}=\frac{1}{36}bh^3$ where $b[/mat. But I don't know how to do that. Find MI of and equilateral triangle of side 2m about its base. Moment of inertia, also called mass moment of inertia or the angular mass, (SI units kg m 2 ) is a measure of an object’s resistance to changes in its rotation rate. Free flashcards to help memorize facts about Moment of Inertia of Different Shapes. Calculate its. (8), derived in the moment of inertia example, the moment of inertia of the disk is = at 5 digits Therefore, the moment of inertia of the disk is 12. Area Moments of Inertia Parallel Axis Theorem • Moment of inertia IT of a circular area with respect to a tangent to the circle, ( ) 4 4 5 4 2 2 4 2 1 r IT I Ad r r r π π π = = + = + • Moment of inertia of a triangle with respect to a. Calculate the triangles moment of inertia when its axis of rotation is located at the right-angled corner. The “narrower” the triangle, the more exact is the formula (2). Calculating the moment of inertia of a triangle - Duration: 10:01. 42×r from base y2=0. Click Content tabCalculation panelMoment of Inertia. We can relate these two parameters in two ways: For a given shape and surface mass density, the moment of inertia scales as the size to the fourth power, on dimensional grounds. Determine the product of inertia of the crosshatched area with respect to the x and y axes. Area moment of inertia calculation formulas for the regular cross section are readily available in design data handbooks. Knowing the area moment of inertia is a critical part of being able to calculate stress on a beam. Or the Mizuno MP-20. Callaway Ratings & Specs 2019 Head Ratings by Maltby Experts at The GolfWorks View All | View By Brand. Thus, the object’s mass and how it is distributed both affect the mass moment of inertia. After that eccentricity is calculated for the obtained projection. half the value of the moment of inertia about the central axis to the value of the moment of inertia about the base plane. 5 Parallel-Axis Theorem - Theory - Example - Question 1 - Question 2. MSC separate and tethers begin to reel out, where inertia of the 3 MSC, increases with time, and the inertia of CSC, remains constant. This simple, easy-to-use moment of inertia calculator will find moment of inertia for a circle, rectangle, hollow rectangular section (HSS), hollow circular section, triangle, I-Beam, T-Beam, L-Sections (angles) and channel sections, as well as centroid, section modulus and many more results. r2 x2 y2 Therefore, I z I. unambiguous choice between the divergent views currently held with regard to the structure. Mathematically, and where IB " *BA " TIA BA = *B + 7IA Ig = moment of inertia about the base plane I3A = moment of inertia about a base diameter axis 1^ = moment of inertia about the central axis 7. Moment of inertia, denoted by I, measures the extent to which an object resists rotational acceleration about a particular axis, and is the rotational analogue to mass. Some examples of simple moments of inertia Let's try an easy calculation: what's the moment of inertia of these three balls? Each ball has mass m = 3 kg, and they are arranged in an equilateral triangle with sides of length L = 10 m. , the opposition that the body exhibits to having its speed of rotation about an axis altered by the application of a torque (turning force). ” or ” A quantity expressing the body’s tendency to resist angular acceleration, it is equal to sum of product of mass of particles to the square of distances from the axis of rotation. It is always considered with respect to a reference axis such as X-X or Y-Y. 5 1 A 2 3 2. So if I'm interpreting your last formula correctly, your answer seems to be off by a factor of 2. 2500 cm^4; D. 1 Answer to Polar Moments of Inertia Determine the polar moment of inertia I P of an isosceles triangle of base b and altitude h with respect to its apex (see Case 5, Appendix D). Then remove the middle triangle from each of the re-maining three triangles (as shown), and so on, forever. In physics and applied mathematics, the mass moment of inertia, usually denoted by I, measures the extent to which an object resists rotational acceleration about a particular axis, and is the rotational analogue to mass. The following is a list of second moments of area of some shapes. where I is the moment of inertia and R is the perpendicular distance from the axis of rotation to the slant height of the cone changing dm with density, ρ, we get I = ρ ∫∫∫ R^2 r^2 sin Φ dr dΦ dθ = ρ ∫∫∫ r^4 (sin Φ)^3 dr dΦ dθ. Express the result as a Cartesian vector. The calculations are as shown. •Compute the product of inertia with respect to the xyaxes by dividing the section into three rectangles. Area Moment of Inertia - Filled Right Triangle Solve. 42441 1in 1 in 1 in 3 in 1 in A 2 A 3 A 1 A 4 16 Centroid and Moment of. The moment of inertia of total area A with respect to z axis or pole O is z dI z or dI O or r dA J 2 I z ³r dA 2 The moment of inertia of area A with respect to z axis Since the z axis is perpendicular to the plane of the area and cuts the plane at pole O, the moment of inertia is named "polar moment of inertia". however, i would like to know how you obtain the results. Finding the area of a right triangle is easy and fast. Mechanics of Material (CIV101) Anno Accademico. Introduction. 1st moment of area is area multiplied by the perpendicular distance from the point of line of action. I am unable to find it. This banner text can have markup. Angular momentum of an extended object. We will use the parallel axis theorem and we will take the centroid as a reference in this case. First Moment of Area = A x. Sorry to see that you are blocking ads on The Engineering ToolBox! If you find this website valuable and appreciate it is open and free for everybody - please contribute by. That point mass relationship becomes the basis for all other moments of inertia since any object can be built up from a collection. It is also known as the torsional Stiffness Read the Full article here. Answer MOI of a triangle about axis theory through a point along the plane = 2 1 m (A r e a) = 2 1 m (2 l × 2 l ) = 8 1 m l 2 December 26, 2019 Toppr. , the opposition that the body exhibits to having its speed of rotation about an axis altered by the application of a torque (turning force). 1) - Moment of Inertia by Integration Mechanics Statics Chapter 10. Own work assumed (based on copyright claims). But I don't know how to do that. Worthy of note, in order to solve for the moment of inertia of the right triangular thin plate, we first had to measure the the triangle's mass, base length, and height. For the section shown in Fig. For a clear understanding of how to calculate moments of inertia using double integrals, we need to go back to the general definition in Section The moment of inertia of a particle of mass about an axis is where is the distance of the particle from the axis. function Ix_integrand = Moment_Of_Inertia_Integrand(y_prime) %Saved as Moment_Of_Inertia_Integrand. The moment of inertia is equal to the moment of inertia of the rectangle minus the moment of inertia of the hole which is a circle. Answer Save. Calculating the moment of inertia of a triangle - Duration: 10:01. Lab 17: Angular Acceleration Amy, Chris, and Jacob November 22, 2017 Theory/Introduction: The purpose of this lab was to determine the moment of inertia of a right triangle thin plate around its center of mass, for two…. The angle at the apex is 9 0 o. Figure to illustrate the area moment of a triangle at the list of moments of inertia. Find The Moment Of Inertia About An Axis That Passes Through Mass A And Is Perpendicular To The. The moment of inertia of a triangle with respect to an axis passing through its centroid, parallel to its base, is given by the following expression: I=bh^2/36. If you need a beam’s moments of inertia, cross sectional area, centroid, or radius of gyration, you need this app. 30670 Moment of Inertia The same vertical differential element of area is used. Lab 18: Moment of Inertia of a Triangle In Lab 16, we used a rotary device that pushes air through a pair of disks, minimizing frictional torque, and allowing the disks to spin for a long time, almost unimpeded. Home Properties Classical MechanicsMoment of Inertia of a Triangle. Let the mass of the triangle be M. It is concluded that the form of the isoceles triangle is acute, with the. Get the expression of angular acceleration and omega. 000965387 kg*m^2. he solves alone. We can use a numerical integrator, such as MATLAB's integral2, to compute the area moment of inertia in the previous example. The moment of inertia of a triangle rotating on its long side is greater than the moment of inertia of the triangle rotating on the shorter side. The element of area in rectangular coordinate system is given by. If rotated about point O (AO = OB),what is the moment of inertia of the rod. The moment of inertia of a triangle rotating on its long side is greater than the moment of inertia of the triangle rotating on the shorter side. 5 Moment of Inertia of Composite Areas A similar theorem can be used with the polar moment of inertia. In a continuous bridge, the moment of inertia should follow the moment requirement for a balanced and economical design. Autor: No machine-readable author provided. Diagonal wise mI = ml2/6 base = ml2/24 Find the moment of inertia of the plane about the y-axis. It is based not only on the physical shape of the object and its distribution of mass but also the specific configuration of how the object is rotating. The strut width has been deliberatelty increased to show the geometry. Integration by the area of. For a solid cone the moment of inertia is found by using the given formula; I = 3 / 10 MR 2. This is the currently selected item. Calculate the triangles moment of inertia when its axis of rotation is located at the right-angled corner. Radius and elevation of the semi-circle can be changed with the blue point. The particles are connected by rods of negligible mass. 5 Parallel-Axis Theorem - Theory - Example - Question 1 - Question 2. It is observed that the ratio of to is equal to 3: Assume that both balls are pointlike; that is, neither has any moment of inertia about its own center of mass. Moment of inertia is defined as:”The sum of the products of the mass of each particle of the body and square of its perpendicular distance from axis. A more efficient triangular shape for metal wood clubs or driver clubs is disclosed. Calculate the moment of inertia of the triangle with respect to the x axis. Asked by rrpapatel 2nd November 2018 12:10 AM. Polar Moment of Inertia is a measure of resistibility of a shaft against the twisting. The greater the mass of the body, the greater its inertia as greater force is required to bring about a desired change in the body. Mechanics of Solids Introduction: Scalar and vector quantities, Composition and resolution of vectors, System of units, Definition of space, time, particle, rigid body, force. Now to calculate the moment of inertia of the strip about z-axis, we use the parallel axis theorem. No, the components of the eigenvectors themselves, Axes(:,1)), Axes(:,2), Axes(:,3), are already the cosines of the angles between the three principal axes of inertia respectively and the x, y, and z axes, provided the eigenvectors are normalized. The moment of the large triangle, with side $$2L$$, is $$I_z(2L)$$. The apex of the triangle is at the origin and it is bisected by the x-axis. A triangular section has base 100 mm and 300 mm height determine moment of inertia about 1)MI about axis passing through base 2)MI about axis passing through apex {Ans: 3. And h/3 vertically from reference x-axis or from extreme bottom horizontal line line. Simply Supported Beams (Shear & Moment Diagrams) Simply supported beams (also know as pinned-pinned or pinned-roller) are the most common beams for both school and on the Professional Engineers exam. This triangular shape allows the clubs to have higher rotational moments of inertia in both the vertical and horizontal directions, and a lower center of gravity. Rotations in 2D are about the axis perpendicular to the 2D plane, i. Mass moment of inertia. My teacher told me :. Moment of Inertia - Calculated Values Electrical Design In determining the layout of the electrical design, a broad level view was taken and elaborated on. Supplementary notes for Math 253, to follow Section 13. purdueMET 20,366 views. They will make you ♥ Physics. If the density were a constant, finding the total mass of the lamina would be easy: we would just multiply the density by the area. Moment of Inertia for body about an axis Say O-O is defined as ∑dM*y n 2. Integration by the area of. For this reason current vector is treated as normal vector of the plane and the input cloud is projected onto it. About the Moment of Inertia Calculator. Engineering Science Mechanical Engineering Concrete Calculator Bending Moment Similar Triangles Shear Force Civil Engineering Construction Body Diagram Structural Analysis. It is the rotational inertia of the body, which is called. These are the values of principal moment of inertia. More particularly, the present invention relates to a hollow golf club head with a lower center of gravity and a higher moment of inertia. I y 2= ∫ x el dA where el = x dA = y dx Thus, I y = ∫ x2 y dx The sign ( + or - ) for the moment of inertia is determined based on the area. It is not explicitly stated in the output, but the mass is equal to the volume (implicitly using a density of 1), so we would expect diagonal matrix entries of 8/15*PI (1. Own work assumed (based on copyright claims). 156 m y Applying Eq. And the moment of "A" equals zero because it is at a point through which the moment of inertia passes. Lectures by Walter Lewin. More on moment of inertia. Ask Question Asked 4 years, 9 months ago. apex angle in the neighborhood of 34°. After that eccentricity is calculated for the obtained projection. We’re pretty sure the Titleist 620 MB has plenty of workability. 8) I of Disk with a Hole. The moment of inertia about the X-axis and Y-axis are bending moments, and the moment about the Z-axis is a polar moment of inertia(J). 1 Expert Answer(s) - 30625 - calculate the moment of inertia of an equilateral triangle made by three rods each of mass m and len. moment of inertia gives the same I as the body rotates around the axis. The 2 nd moment of area, or second area moment and also known as the area moment of inertia, is a geometrical property of an area which reflects how its points are distributed with regard to an arbitrary axis. Weld design. Undeniable momentum, on any stage - anywhere. where m is the mass of the object, and r is the distance from the object to the axis. Microsoft Word - Chapter 12 - Moment of Inertia of an Equilateral Triangle Author: Owner Created Date: 11/21/2019 8:18:19 AM. triangles parallel to the xy plane with side 2a, is centered on the origin with its axis along the z axis. 2 y = 10e-x x y 3. Code for moment_of_inertia and RoPS classes, tests and tutorials. The term product moment of inertia is defined and the mehtod of finding principal moment of inertia is presented. Data: 23 d'abril de 2006 (original upload date) Font: No machine-readable source provided. Moments of Inertia 10. The radius of gyration is the radius at which you could concentrate the entire mass to make the moment of inertia equal to the actual moment of inertia. (**) The object below has a moment of inertia about its center of mass of I=25kg⋅m2. 32075h^4M/AL, where h is the height of the triangle and L is the area. This cone is centered on the z-axis with the apex at the origin, but rotates with respect to the x-axis. Using these moment of inertia, we can subtract from it the moment of inertia of just the system without the triangle to obtain our experimental values for the triangle in either. After determining moment of each area about reference axis, the distance of centroid from the axis is obtained by dividing total moment of area by total area of the composite section. 1 GradedProblems Problem 1 (1. Moment of Inertia Contents Moment of Inertia; Sections; Solids; MOI_Rectangle; MOI_Triangle; MOI_Trapezod; MOI_Circle. Solution 3. Area Moment of Inertia Section Properties: Triangle Calculator. Transfer Formula for Moment of Inertia Transfer Formula for Polar Moment of Inertia Transfer Formula for Radii of Gyration Moment of Inertia Common Shapes; Rectangle Triangle Circle Semicircle Quartercircle Ellipse Center of Mass; Center of Mass (2D) 1. The Moment of Inertia Apparatus MATERIALS 1 Table clamp 1 Weight hanger (mass 50g) 1 Long metal rod 1 Length of string 2 Pulleys 1 Level 2 Right angle clamps 1. Second, finding the moment of inertia when the triangle rotates around its base (shorter leg). Angular momentum of an extended object. Ditto the Ping Blueprint. 4 Locate the centroid of the T-section shown in the Fig. Thank you User-12527562540311671895 for A2A The moment of inertia of a triangular lamina with respect to an axis passing through its centroid, parallel to its base, is given by the expression [math]I_{XX}=\frac{1}{36}bh^3$ where [math]b[/mat. pptx), PDF File (. Try this Drag any point A,B,C. What is the moment of inertia of ball about the axis of rotation AB? Ignore cord’s mass. The point where the triangle is right angled is lying at origin. Doing the same procedure like above, and below is the work. 3 Radius of Gyration of an Area 10. Another solution is to integrate the triangle from an apex to the base using the double integral of r^2dm, which becomes (x^2+y^2)dxdy. Click Content tabCalculation panelMoment of Inertia. This engineering calculator will determine the section modulus for the given cross-section. 3× 1 6ML2 = 1 2ML2. The situation is this: I know the moment of inertia with respect to the x axis and with respect to the centroidal x axis because its in the table. To predict the period of a semi-circle and isosceles triangle with some moment of inertia after first calculating the moment of inertia of the semi-circle and isosceles triangle about a certain axis. These came out to be 0. The ratio of moment of inertia about the neutral axis to the distance of the most distant point of section from neutral axis is called as a) Moment of inertia b) section modulus c) polar moment of Apex of the triangle b) mid of the height c) 1/3 of the height d) base of triangle 12. The axis perpendicular to its base. Thus, the moment of inertia of a 2D shape is the moment of inertia of the shape about the Z-axis passing through the origin. I of a thin rod about its center is ML^2 / 2. The second moment of area, also known as moment of inertia of plane area, area moment of inertia, polar moment of area or second area moment, is a geometrical property of an area which reflects how its points are distributed with regard to an arbitrary axis. c channel polar moment of inertia. 30670 Moment of Inertia The same vertical differential element of area is used. It should not be confused with the second moment of area, which is used in beam calculations. In order to find the moment of inertia we must use create a dm and take the integral of the moment of inertia of each small dm. Let us apply this formula to the triangle formed by V 1, V 2 and V: Δ𝑉≈𝑉⋅Δ𝛼 (3). How do I calculate the moment of inertia of a right angled triangle about one side? Moment of inertia about a side other than the hypotenuse. What is the moment of inertia of this system about an altitude of the triangle passing through the vertex, if ‘a’ is the size of each side of the triangle ?. Letting M be the total mass of the system, we have x ¯ = M y / M. The moment of point "C" is the same as "B" so multiply the moment of "B" by two. derivation of inertia of ellipse. Find the moment of inertia of the framework about an axis passing through A, and parallel to BC 5ma 2. Favourite answer. • The moment of inertia of a composite area A about a given axis is obtained by adding the moments of inertia of the component areas A 1 , A 2 , A 3 , , with respect to the same axis. It is analogous to mass in that it is a measure of the resistance a body offers to torque or rotational motion. To predict the period of a semi-circle and isosceles triangle with some moment of inertia after first calculating the moment of inertia of the semi-circle and isosceles triangle about a certain axis. 89 × 103 kg/m3. In yesterday's lesson, students completed a lab on center of mass, and they already have a working knowledge of torque. The axis may be internal or external and may or may not be fixed. Engineering Science. Find the moment of inertia about y-axis for solid enclosed by z = (1-x^2) , z= 0 , y = 1 and y = -1. 3) Three particles each of mass 100 g are placed at the vertices of an equilateral triangle of side length 10 cm. equals 1, if there are four piles per row and two rows (figure 7-2), the moment of inertia about the Y-Y axis is given by the following formula. 1st moment of area is area multiplied by the perpendicular distance from the point of line of action. I), must be found indirectly. Moment of inertia Up: Rotational motion Previous: The vector product Centre of mass The centre of mass--or centre of gravity--of an extended object is defined in much the same manner as we earlier defined the centre of mass of a set of mutually interacting point mass objects--see Sect. The moment of inertia of the particle. Moment of inertia, in physics, quantitative measure of the rotational inertia of a body—i. Therefore, equation for polar moment of inertia with respect to apex is. University of Sheffield. For these orbits, consider the following scaled variables q˜i = √qi I, (5) v˜i. Q: the moment of inertia of a thin rod of mass m and length l about an axis through its centre of gravity and perpendicular to its length is a) ml²/4 b) ml²/6 c) ml²/8 d) ml²/12 Q: Which statement is correct: a) Moment of inertia is the second moment of mass or area. Tinker toys allow one to easily construct objects with the same mass but different moments of inertia. Two conditions may be considered. After determining moment of each area about reference axis, the distance of centroid from the axis is obtained by dividing total moment of area by total area of the composite section. You can show the division by drawing solid or. It is always considered with respect to a reference axis such as X-X or Y-Y.
hu215i4c45d 16uebhoaltj0op 71a59wpjnvb5y i0fcgwor6jyv8b a8qpv94ljutw hl8lh0m5b2gz1 pbgmxxx4nx 8a1tcmejeokz gx9mvyz9ezgel jwp1rrde9h6ax2 19sb1sh6t9mn 9etgudw0esw 0qoq538i9x xnzy3novvwd u573jcqhjrwnqg 7yteh42rbyb3j j33ao2f1gbcc 0vmioeu4j0gtvum 6q7lhaq2fv l2455p6pm40a7mo jzze49hb2dgxab wwskcnhe8iiz z480xmmqhdqb7 8pfvwqe2t6dtb0 9ciz5c26pn 1atmcvapjuz lpb65ot49n e7ru916n05wcb21 ntgoy7fzw8 | 2020-12-03T19:51:56 | {
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http://civilservicereview.com/2013/10/solutions-fractions-to-lowest-terms/ | # Solution to the Exercises on Reducing Fractions to Lowest Terms
Below are the complete solutions and answers to the exercises on reducing fractions to lowest terms. I will not give any tips or methods of shortcuts on doing this because teaching you shortcuts will give you problems in case you forget them. The best thing that you can do is to solve as many related problems as you can and develop shortcuts that work for you. Each person has his own preference in solving procedural problems such as these, so it is important that you discover what’s best for you.
For converting improper fractions to mixed form, I will discuss it in a separate post. Try to see the solutions below and see if you can use these solutions to develop your own method. Honestly, the three examples below on converting improper fractions to mixed form should be enough to teach you how to do it yourself. 🙂
Reducing Fractions to Lowest Terms
1. $\displaystyle \frac{12}{15}$
Solution
$\displaystyle \frac{12 \div 3}{15 \div 3} = \frac{4}{5}$
2. $\displaystyle \frac{18}{24}$
Solution
$\displaystyle \frac{18 \div 6 }{24 \div 6} = \frac{3}{4}$
3. $\displaystyle \frac{21}{49}$
Solution
$\displaystyle \frac{21 \div 7 }{49 \div 7} = \frac{3}{7}$
4. $\displaystyle \frac{56}{72}$
Solution
$\displaystyle \frac{56 \div 8 }{72 \div 8} = \frac{7}{9}$
5. $\displaystyle \frac{26}{65}$
Solution
$\displaystyle \frac{26 \div 13 }{65 \div 13} = \frac{2}{5}$
6. $\displaystyle \frac{18}{32}$
Solution
$\displaystyle \frac{18 \div 2 }{32 \div 2} = \frac{9}{16}$
7. $\displaystyle \frac{38}{95}$
Solution
$\displaystyle \frac{38 \div 19 }{95 \div 19} = \frac{2}{5}$
8. $\displaystyle \frac{32}{12}$
Solution
First, convert to lowest terms:
$\displaystyle \frac{32 \div 4 }{12 \div 4} = \frac{8}{3}$
Second, convert to mixed form. Eight divided by 3 is 2 remainder 3. So 2 becomes the whole number, 2 (the remainder) becomes the numerator and 8 becomes the denominator. Therefore, the answer is $2 \frac{2}{3}$.
9. $\displaystyle \frac{16}{84}$
Solution
$\displaystyle \frac{16 \div 4 }{84 \div 4} = \frac{4}{21}$
10. $\displaystyle \frac{39}{24}$
Solution
First, reduce to lowest terms.
$\displaystyle \frac{39 \div 3 }{24 \div 3} = \frac{13}{8}$
Second, convert the answer to mixed form. Thirteen divided by 8 is 1 remainder 5. So 1 becomes the whole number, 5 (the remainder) becomes the numerator of the fraction and 8 becomes the denominator. So the correct answer is $1 \frac{5}{8}$.
11. $\displaystyle \frac{15}{45}$
Solution
$\displaystyle \frac{15 \div 15 }{45 \div 15} = \frac{1}{3}$.
12. $\displaystyle \frac{51}{85}$
Solution
$\displaystyle \frac{51 \div 17 }{85 \div 17} = \frac{3}{5}$
13. $\displaystyle \frac{18}{54}$
Solution
$\displaystyle \frac{18 \div 18 }{54 \div 18} = \frac{1}{3}$
14. $\displaystyle \frac{35}{49}$
Solution
$\displaystyle \frac{35 \div 7}{49 \div 7} = \frac{5}{7}$
15. $\displaystyle \frac{74}{24}$
Solution
First, reduce to lowest terms.
$\displaystyle \frac{74 \div 2 }{24 \div 2} = \frac{37}{12}$
Second, divide 37 by 12. The answer is 3 remainder 1. Now, 3 becomes the whole number, 1 becomes the numerator of the fraction, and 12 becomes the denominator. So, the correct answer is $3 \frac{1}{12}$.
In the next post, we will be talking about multiplying and dividing fractions.
### 6 Responses
1. rooky says:
Sir, when I solved # 2 I got an answer ¼ lowest term of 18/24 correct me if I am wrong because I had two LCM. 1st LCM (2) 18/24 = 9/12; 2nd LCM (3) 9/12 = 1/4
2. Civil Service Reviewer says:
Hi Rooky,
Thank you for your question. In reducing fractions to lowest terms, you are not getting the LCM. You are actually getting the greatest common factor (GCF). You get the LCM if you want to add or subtract two (or more) fractions whose denominators are not equal.
In your solution, the GCF of 9 and 12 is 3. So, you must divide 9 by 3, which is equal to 3.
3. Supercute says:
Sir, i think no. 15 should be 3 1/6
4. Supercute says:
Here’s my computation:
74/24
3
_____
24 / 74
72
___________
r. 4
Convert it to mixed fraction:
4
3 ____
24
Convert to lowest term:
1
3 ___
6
• Civil Service Reviewer says:
Hello supercute,
The answer in the explanation is correct. Even if you convert it first to fraction, 74/24 becomes 3 and 2/24 or 3 and 1/12. 🙂
1. October 12, 2013
[…] Now that you know how to convert fractions to lowest terms, you might want to try the practice test and check your solution and answer. […] | 2017-11-20T21:13:45 | {
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https://math.stackexchange.com/questions/866430/does-this-series-violate-the-decreasing-condition-of-the-integral-test-for-conve | # Does this series violate the decreasing condition of the Integral Test for Convergence?
I'm working on the section involving the Integral Test for Convergence in my calculus II class right now, and I've run into a seeming conflict between the definition of the Integral Test, and the solutions to some of the homework exercises as given by both my professor and the textbook.
According to the definition, my research, and my understanding of the integral test, the integral test can only be used for the series $a_n$ where $a_n = f(n)$ and $f(x)$is positive, continuous and decreasing for all $x \ge N$, where $N$ is the index of $n$. However, there are several problems where $f(x)$ is only decreasing if we add a condition, such as $f'(x) < 0 \Leftarrow\Rightarrow x > 3$, where $N \lt 3$. It seems to me that the Integral Test cannot be used to determine convergence of a series when the function is only decreasing when $x \gt k$ and $k \lt N$, yet the book and my professor apply the test anyway.
For example, with the series:
$$\sum_{n=1}^\infty = \frac{n}{(4n+5)^\frac{3}{2}}$$
If we let $a_n = f(n)$, then for $f(x)$:
• $f(x) \gt 0$ for all $x$ in the domain
• $f(x)$ is continuous for all $x \gt -\frac{5}{4}$
But, $f'(x) \lt 0$ only when $x > \frac{5}{2}$, as shown when testing the critical points with the derivative:
$$f'(x) = \frac{5-2x}{(4x+5)^\frac{5}{2}}$$
The professor notes this in her solution, but instead of ending with that and writing, "The Integral Test cannot be applied because $f(x)$ fails to satisfy the required conditions," she applies the test using the original index for $n$:
$$\int_1^\infty \frac{x}{(4x+5)^\frac{3}{2}} \rightarrow \infty \Rightarrow a_n \text{ Diverges}$$
The textbook reaches the same conclusion. Also, the problems in question are listed under a section where the instructions state, "Confirm that the Integral test can be applied to the series. Then use the Integral Test to determine the convergence or divergence of the series," implying the test can be used on the subsequent exercises.
Is there a reason the Integral Test for Convergence can be used to test for convergence in these problems, where $N \lt k$ and $f'(x) < 0 \Leftarrow\Rightarrow x \gt k$? Am I missing something, or are the book and my professor wrongly using the Integral Test for these series?
### Other exercises with same result:
$$\bullet \sum_{n=1}^\infty \frac{\ln n}{n^2}$$
$$\bullet \frac{\ln 2}{2} + \frac{\ln 3}{3} + \frac{\ln 4}{4} + \frac{\ln 5}{5} + \frac{\ln 6}{6}$$
• One way to look at what your book and your professor are doing is that they're using the fact that $\sum_{n=1}^{\infty}a_n$ converges iff $\sum_{n=k}^{\infty}a_n$ converges, for any $k$. – user84413 Jul 14 '14 at 0:20
• I think the idea is more that we don't really care what happens for the first 10, 100, or 1000 terms. We care about what eventually happens. We can add up a finite amount of things no problem; it's the tail end behavior that we really care about. – Cameron Williams Jul 14 '14 at 0:20
• For convergence, doesn't matter. For estimates it does. – André Nicolas Jul 14 '14 at 0:22
• Thank you for your responses; I agree that the behavior is properly demonstrated with the test, and I also understand our concern being only with the end behavior of the function, but I don't think it explains why the test is defined as having to be decreasing for all $x \ge N$, if we only care about the last few terms. Wouldn't it be better to say "for most of"? At any rate, if my definition of the test is correct, doesn't this series still violate the conditions, regardless of the validity of the result? – MrCMedlin Jul 14 '14 at 0:34
The point, which isn't made often enough (in my opinion) in classes on the subject, is that the convergence of a series is a limit process. In this case, what that means is that the question of convergence is completely determined by the behavior of the series for say $n > N$ for any fixed, finite $N$. If I take a convergent series $\sum a_n$, and I cut off the first 55 quintillion terms, and replace them all with $n!$ to get a new sequence
$$b_n = \begin{cases} a_n \text{ if } n > 55,000,000,000,000,000,000 \\ n! \text{ if } n \leq 55,000,000,000,000,000,000 \end{cases}$$
Then the sum $\sum b_n$ is still convergent. In fact, $b_n$ is convergent if and only if $a_n$ is. Of course, the sums will be different, but by precisely
$$\sum\limits_{n=1}^{\infty} b_n - \sum\limits_{n=1}^{\infty}a_n = \sum\limits_{n=1}^{55,000,000,000,000,000,000} (n! - a_n)$$
Which is just a finite number. Of course, I've contrived the example to be huge and ridiculous. But the the point to be made about the integral test is that as long as the function behaves in the desired way beyond some large $N$ (say 55 quintillion), the argument still works for the infinite part of the sum, beyond that, and that is where all problems of convergence lie. Everything else is just addition.
• Doesn't this mean the integral should be evaluated from $N$ to $\infty$ instead of from the original index $k$ to $\infty$ for it to be accurate? – MrCMedlin Jul 14 '14 at 15:45
• Again, as long as the integral is of a nice (read continuous) function on the interval $[0, \infty)$. The convergence of the integral is focused in the tail as we can write $$\int_0^\infty f dx = \int_0^N f dx + \int_N^\infty f dx$$ And all of the convergence/divergence behavior of the improper integral is concentrated in the second term on the right. The first term in the right is (assuming the function is continuous up to and including 0), simply some finite integral. This is admittedly a little more subtle. – JHance Jul 14 '14 at 16:18
• OK, this makes a lot more sense now. They just gloss over this in the course without really explaining it ... you've helped a bunch! – MrCMedlin Jul 14 '14 at 16:36
I do not believe your definition of the test is correct. Citing the ever-accurate (tongue-in-cheek) Wikipedia, we find the definition to be:
Consider an integer $N$ and a non-negative function $f$ defined on the unbounded interval $[N, ∞)$, on which it is monotone decreasing. Then the infinite series $$\sum_{n=N}^\infty f(n)$$ converges to a real number if and only if the improper integral $$\int_N^\infty f(x)\,dx$$ is finite. In other words, if the integral diverges, then the series diverges as well.
(Source: http://en.wikipedia.org/wiki/Integral_test_for_convergence)
It is important to note that various authors of textbooks may define a test or theorem in the way that it best fits their course outline. This is particularly true of elementary texts (e.g. first or second year calculus). Remember that mathematicians also like to generalize--if you can take a specific formulation of a theorem and generalize it (e.g. "$f$ must be decreasing everywhere" to "$f$ only has to have decreasing end-behavior"), that's perfectly acceptable.
As a further note in response to your comment on the main question: "decreasing a majority of the time" is different than "decreasing beyond $x=N$." The latter is the option you want.
• Thank you for your response! I've reviewed the Wikipedia entry on the topic as part of my earlier research. You'll note at the beginning it says the function must be monotone decreasing on the interval being considered. This means that for the series $a_n$, $a_1 \ge a_2 \ge a_3 \ge a_4 \ge ... \ge a_n$. We can test for this by looking at the behavior of the first derivative of the function. If the first derivative is ever positive for any value $x$ in our domain, this indicates an increase from one term to the next, which would mean the function is not monotone decreasing as required. – MrCMedlin Jul 14 '14 at 4:08 | 2019-10-22T00:44:07 | {
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https://mathematica.stackexchange.com/questions/95566/determine-height-of-box-packed-with-spheres/95568 | # Determine height of box packed with spheres
I got such a wonderful answer regarding The Diagonals of a Regular Octagon, so I thought I'd try asking another question we had on our Pizza and Problem quiz activity at College of the Redwoods. The question was:
A 4 × 4 × h rectangular box contains a sphere of radius 2 and eight smaller spheres of radius 1. The smaller spheres are each tangent to three sides of the box, and the larger sphere is tangent to each of the smaller spheres. What is h?
I'd like to learn how to construct such a model using Mathematica and determine the height h of the box.
Thanks for any help.
Your question is like a sangaku problem. What is the distance between the centre of the large sphere and one of the smaller spheres? The large sphere centre is $\{2,2,h/2\}$, one of the small spheres is at $\{1,1,1\}$. Their separation equals the sum of their radii. That is,
Norm[{2, 2, h/2} - {1, 1, 1}] == 1 + 2
Solve the expression for $h$, and choose the positive root, $h=2(1+\sqrt{7})$.
With[{h = 2 (1 + Sqrt[7])},
Graphics3D[{Orange,
Sphere[{1, 1, 1}, 1], Sphere[{3, 1, 1}, 1], Sphere[{1, 3, 1}, 1],
Sphere[{3, 3, 1}, 1], Sphere[{1, 1, h - 1}, 1],
Sphere[{3, 1, h - 1}, 1],
Sphere[{1, 3, h - 1}, 1], Sphere[{3, 3, h - 1}, 1],
Red, Sphere[{2, 2, h/2}, 2]
}, Lighting -> "Neutral",Axes->True]]
• Tremendous answer. I am having so much fun this morning with the responses on Mathematica Stack Exchange. Thanks you. – David Sep 26 '15 at 19:09
KennyColnago's answer is good, but we're using Mathematica, so we can leave much more of the problem to it. Now, we can't avoid writing down the system of equations; for simplicity, I'll assume that the large sphere (with radius $R = 2$) is centered at the origin, and that the small spheres (with radius $r = 1$) are centered at ${x, y, z}$. The sides of the box lie at $\pm2$, and the top and bottom are at $\pm h$, so we have tangent small spheres for any combination of positive and negative $x$, $y$ and $z$. That means we know the following:
In[1]:= eqns = {
(* each sphere is tangent to the sides of the box *)
Abs[x] + r == 2,
Abs[y] + r == 2,
(* each sphere is tangent to the top of the box *)
Abs[z] + r == h/2,
(* because the large sphere and small sphere are tangent
to one another, their radii are collinear, so their
total length is the distance of the center of the small
sphere from the origin. *)
r + R == Norm[{x, y, z}],
(* h is positive *)
0 < h
} /. {r -> 1, R -> 2};
Now, we can use Mathematica to solve those equations for $x$, $z$ and $h$.
In[2]:= sol = Simplify@Solve[eqns, {x, y, z, h}, Reals]
Out[2]= {{x -> -1, y -> -1, z -> -Sqrt[7], h -> 2 (1 + Sqrt[7])},
{x -> -1, y -> -1, z -> Sqrt[7], h -> 2 (1 + Sqrt[7])},
{x -> -1, y -> 1, z -> -Sqrt[7], h -> 2 (1 + Sqrt[7])},
{x -> -1, y -> 1, z -> Sqrt[7], h -> 2 (1 + Sqrt[7])},
{x -> 1, y -> -1, z -> -Sqrt[7], h -> 2 (1 + Sqrt[7])},
{x -> 1, y -> -1, z -> Sqrt[7], h -> 2 (1 + Sqrt[7])},
{x -> 1, y -> 1, z -> -Sqrt[7], h -> 2 (1 + Sqrt[7])},
{x -> 1, y -> 1, z -> Sqrt[7], h -> 2 (1 + Sqrt[7])}}
In[3]:= Length@sol
Out[3]= 8
Gratifyingly, we have eight solutions, one for each small sphere.
Now we can draw our spheres, which is easy because Mathematica has helpfully collected all the solutions as a list of rules.
In[3]:= centers = {x, y, z} /. sol
Out[3]= {{-1, -1, -Sqrt[7]}, {-1, -1, Sqrt[7]},
{-1, 1, -Sqrt[7]}, {-1, 1, Sqrt[7]},
{1, -1, -Sqrt[7]}, {1, -1, Sqrt[7]},
{1, 1, -Sqrt[7]}, {1, 1, Sqrt[7]}}
Pulling it all together, we get:
In[4]:= Graphics3D[{
{Lighter@Lighter@Red,
Sphere[centers, r]},
{Lighter@Green,
Sphere[{0, 0, 0}, R]},
Opacity[0],
EdgeForm[Thick],
Parallelepiped[{-2, -2, -h/2},
{{4, 0, 0}, {0, 4, 0}, {0, 0, h}}]} /. sol /. {r -> 1, R -> 2},
Boxed -> False]
EDIT to add: I've updated this answer a couple times, each time to shift more work to Mathematica. In my first attempt, which I didn't post, I got the wrong answer because of a stupid typo when I plugged the radii of the spheres into the Pythagorean theorem; switching over to Norm fixed that. Then I told Solve to assume that $x$ and $z$ were positive (while using the "obvious" fact that $|y| = |x|$ to eliminate an equation before solving), and did some complicated thing with Tuples and Transpose to get all eight spheres. Eliminating those assumptions and recasting everything in terms of absolute values meant Solve took care of all of that automatically.
EDIT again to tweak the Solve expression to specify the domain for Reals and include y, which seems necessary for everything to work right in Mathematica v11.
• @Pilsy Another tremendous answer. I am sharing these with the students and teachers who attended our activity. – David Sep 26 '15 at 19:10 | 2019-11-21T18:21:56 | {
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https://mathhelpboards.com/threads/why-e-is-natural-and-why-we-use-radians.9286/ | # Why e is "natural", and why we use radians.
#### ThePerfectHacker
##### Well-known member
Many books give bad answers or no answers at all to why we work with base $e$ and measure angles according to radians in calculus.
Here is what I tell my students, as far as I know, this is the best explanation I seen because it is essentially a one line explanation that is short and to the point.
Why $e$ is natural: Let $b>0$, $b\not = 1$ and consider the function $f(x) = b^x$. This is an exponential function with base $b$ and we can show that $f'(x) = c\cdot b^x$ where $c$ is some constant. In a similar manner we can consider the function $g(x) = \log_b x$, logarithmic function to base $b$, we can show that $g'(x) = k\cdot x^{-1}$ where $k$ is some constant. It would be best and simplest looking derivative formula if those constants, $b$ and $k$, were both equal to one. This happens exactly when $b=e$. Thus, $e$ is natural base for exponent because the derivative formulas for exponential and logarithmic function are as simple as they can be.
Why radians are natural: A "degree" is defined by declaring a full revolution to be $360^{\circ}$, a "radian" is defined by declaring a full revolution to be $2\pi$ radians. More generally, we can define a new angle measurement by declaring a full revolution to be $R$ (for instance, $R=400$, results in something known as gradians). Let us denote $\sin_R x$ and $\cos_R x$ to be the sine and cosine functions of $x$ measured along $R$-units of angles. It can be shown that $(\sin_R x)' = C\cdot(\cos_R x)$ and $(\cos_R x)' = -C\cdot(\sin_R x)$ where $C$ is some constant. It would be best if this constant can be made $C=1$ which happens precisely when $R=2\pi$.
#### chisigma
##### Well-known member
Many books give bad answers or no answers at all to why we work with base $e$ and measure angles according to radians in calculus.
Here is what I tell my students, as far as I know, this is the best explanation I seen because it is essentially a one line explanation that is short and to the point.
Why $e$ is natural: Let $b>0$, $b\not = 1$ and consider the function $f(x) = b^x$. This is an exponential function with base $b$ and we can show that $f'(x) = c\cdot b^x$ where $c$ is some constant. In a similar manner we can consider the function $g(x) = \log_b x$, logarithmic function to base $b$, we can show that $g'(x) = k\cdot x^{-1}$ where $k$ is some constant. It would be best and simplest looking derivative formula if those constants, $b$ and $k$, were both equal to one. This happens exactly when $b=e$. Thus, $e$ is natural base for exponent because the derivative formulas for exponential and logarithmic function are as simple as they can be.
Why radians are natural: A "degree" is defined by declaring a full revolution to be $360^{\circ}$, a "radian" is defined by declaring a full revolution to be $2\pi$ radians. More generally, we can define a new angle measurement by declaring a full revolution to be $R$ (for instance, $R=400$, results in something known as gradians). Let us denote $\sin_R x$ and $\cos_R x$ to be the sine and cosine functions of $x$ measured along $R$-units of angles. It can be shown that $(\sin_R x)' = C\cdot(\cos_R x)$ and $(\cos_R x)' = -C\cdot(\sin_R x)$ where $C$ is some constant. It would be best if this constant can be made $C=1$ which happens precisely when $R=2\pi$.
Let's suppose to use 'non natural' bases for exponents and angles... in this case probably today the following 'beautiful formula' ...
$\displaystyle e^{i\ \theta} = \cos \theta + i\ \sin \theta\ (1)$
... would be not yet known ...
Kind regards
$\chi$ $\sigma$
#### ThePerfectHacker
##### Well-known member
$\displaystyle e^{i\ \theta} = \cos \theta + i\ \sin \theta\ (1)$
=
$$\pi^{i\theta} = \cos \theta + i\sin \theta$$
Provided that we measure angles with full revolution $2\pi \log \pi$.
So you can still have Euler formula for different bases and angles provided that they match together nicely. But the derivative formulas will change into something ugly.
#### Deveno
##### Well-known member
MHB Math Scholar
Forgive me for being obtuse but your statement:
...consider the function $f(x) = b^x$. This is an exponential function with base $b$...
is completely mystifying to me.
In particular, I mean: how does one define such a function for, say: $b = \sqrt{2}$?
(I know the answer, someone who is taking calculus for the first time probably does NOT.)
In fact, proving the "base laws" for the functions $\log_b$ and $\exp_b$ depends on somehow producing the "natural base", which depends on producing the number $e$. So how do you do this?
If, in fact, you define:
$$\displaystyle e = x \in \Bbb R: \int_1^x \frac{1}{t}\ dt = 1$$
and show that for:
$$\displaystyle F(x) = \int_1^x \frac{1}{t}\ dt$$
$F(a) + F(b) = F(ab)$
I am prepared to believe that (given a proof that $F$ is injective):
$F^{-1}(a+b) = F^{-1}(a)F^{-1}(b)$
and that such a function $F^{-1}$ has all the properties of $b^x$ when $x$ is rational.
It follows (in my mind, anyway) that we have:
$e = F^{-1}(1)$.
This, to me, is the reason why $e$ is "natural", I find the definition:
$$\displaystyle e = \lim_{n \to \infty} (1 + n)^{1/n}$$
nearly impossible to motivate, whereas the derivative of a differentiable function $f$ where:
$f(a+b) = f(a)f(b)$ clearly has derivative:
$f'(x) = f'(0)\cdot f(x)$.
At this point, we are in a bit of a fix, which is why considering $f^{-1}$ proves to be more amenable to attack:
$(f^{-1})'(x) = \dfrac{1}{f'(f^{-1}(x))} = \dfrac{1}{f'(0)\cdot f(f^{-1}(x))}$
$= \dfrac{1}{f'(0)x}$
This is a function of the form $g(x) = \dfrac{1}{\alpha x}$, and we can use the Fundamental Theorem of Calculus to find values for $f^{-1}(x)$.
Taking $\alpha = 1$ leads, of course, to the above definition of $e$.
Last edited:
#### ThePerfectHacker
##### Well-known member
Forgive me for being obtuse but your statement is completely mystifying to me.
It is not mystifying at all. It is just not rigorous. But that is fine. I doubt that Euler himself thought about it the way we think about it today. And what you wrote will have little value to calculus students. You might as well stop teaching trigonometry with that attitude and define sine and cosine as power series, because that is after all the formal way of doing it, which is how Rudin does it in his book on the first pages.
The important part is to motivate the idea then provide the rigor later. So defining exponentials $b^x$ for arbitrary real numbers is okay because it can at least motivate why $e$ is the best number to choose for $b$.
#### Evgeny.Makarov
##### Well-known member
MHB Math Scholar
In particular, I mean: how does one define such a function for, say: $b = \sqrt{2}$?
I believe, the most natural definition of $\left(\sqrt{2}\right)^a$ is $\lim_{x\to\sqrt{2}}\lim_{y\to a}x^y$ where $x$ and $y$ range over rational numbers.
In fact, proving the "base laws" for the functions $\log_b$ and $\exp_b$ depends on somehow producing the "natural base", which depends on producing the number $e$.
It should be possible to prove $x^{a+b}=x^ax^b$ with the definition above, and this does not require $e$.
#### Deveno
##### Well-known member
MHB Math Scholar
It is not mystifying at all. It is just not rigorous. But that is fine. I doubt that Euler himself thought about it the way we think about it today. And what you wrote will have little value to calculus students. You might as well stop teaching trigonometry with that attitude and define sine and cosine as power series, because that is after all the formal way of doing it, which is how Rudin does it in his book on the first pages.
The important part is to motivate the idea then provide the rigor later. So defining exponentials $b^x$ for arbitrary real numbers is okay because it can at least motivate why $e$ is the best number to choose for $b$.
I'm not sure you understand my point. Let me re-phrase it:
"What does $b^x$ mean for irrational $x$?"
I do not believe rigor is necessary to teach facts, even if the justification cannot be given at that time. It is common for young children to be told that the area of a circle is pi times its radius squared without any justification given. Given the "assumption" that the formula is true, one can still use it profitably, and I see no reason not to under those circumstances.
In a similar vein, I see no problem with teaching the "SOHCAHTOA" approach to trigonometry, either. While that is perfectly adequate for calculating many trig functions of well-established ratios, it is somewhat inadequate for calculating the values of the *continuous* function $\sin(x)$, and when one acquires the mathematical sophistication for a *better* definition, one ought to employ that instead. Power series approximations will do this, but one can use integrals of algebraic functions to obtain inverse trig functions, which also accomplishes the same purpose. A third, and also perfectly acceptable approach, is to use differential equations to define these transcendental functions.
I am not arguing that one *has* to define them this way, just that one *can* and the tools learned in a first-year calculus course allow this. One need not take this path if one feels that the students "aren't ready".
But back to my main point, which I feel is *somewhat* important: how do you define the function:
$f(x) = b^x$
and how do you justify that it is differentiable?
I honestly don't know your answers, but if I was one of your students, I would feel they are pertinent.
I believe, the most natural definition of $\left(\sqrt{2}\right)^a$ is $\lim_{x\to\sqrt{2}}\lim_{y\to a}x^y$ where $x$ and $y$ range over rational numbers.
I'm OK with this. Computing this (without a computer) might prove problematic. This does address continuity satisfactorily, differentiability is another story.
It should be possible to prove $x^{a+b}=x^ax^b$ with the definition above, and this does not require $e$.
I shouldn't think that it would. That is not a "change of base" formula. However, it *is* a good place to start. That said, where does $e$ come in?
#### Evgeny.Makarov
##### Well-known member
MHB Math Scholar
That said, where does $e$ come in?
I agree with the OP that $e$ is such that $\left(e^x\right)'=e^x$.
#### Deveno
##### Well-known member
MHB Math Scholar
I agree with the OP that $e$ is such that $\left(e^x\right)'=e^x$.
I think we all agree this is true. How do you arrive at that fact?
#### Evgeny.Makarov
##### Well-known member
MHB Math Scholar
$e$ is defined to be such constat that $\left(e^x\right)'=e^x$.
Now, I have not looked into the issue of defining $e^x$ for a long time. This is how I believe it was introduced to me, and if this approach works, I would consider it very natural.
#### ThePerfectHacker
##### Well-known member
"What does $b^x$ mean for irrational $x$?"
Exactly what most people think it means. You replace $x$ by a rational number $q$ which is close to $x$ and then compute $b^q$. That is what every student would do if you ask them to find $2^{\pi}$. That is what Euler thought of it and mathematicians back then. So I can write $b^x$ function without rigorously defining it.
and how do you justify that it is differentiable?
I do not. I just write it and then ask students what will the derivative of this function be?
All that stuff you wrote before it useless to make a first time student who sees the natural exponential to understand what makes it special to other exponential functions. He will get lost in the middle and not understand what you are doing it. My approach does not suffer from that. Now if you tell a student the natural exponential is its own derivative while other exponentials are almost their own derivative they see what makes the natural exponent the nicer base to use.
#### Deveno
##### Well-known member
MHB Math Scholar
Let me see if I have this straight: you differentiate a function without any justification that it is differentiable. Why do I think this is not a good idea?
But...granted; establishing that the conditions to apply some of the theorems one uses from elementary calculus might be "more technical stuff" than a given student may need or want. Let's move on.
How does one establish that:
$b^x = e^{x\log(b)}$?
I mean, I'm willing to accept that as a DEFINITION of $b^x$, but then I'd sure like to know what $e$ is.
Now, there is "that other definition" of $e$ (which is actually the first definition I ever encountered), but...
Again, you're asking people to take a lot "on faith". There is nothing wrong with this, per se, but calculus affords an opportunity to actually PROVE things that previously they were only TOLD. This moves the source of their knowledge from: "external authority" (books, teacher, expert in field) to: "internal understanding" (I know it's true because of other things I know are true).
In one view, math is another form of intellectual slavery. In another, it is intellectual freedom. I suppose it's obvious which one I favor.
Let me put this in another light: it is (or so I am arguing) preferable to have things we reason about be on "firm footing" (because arguments based on incorrect assumptions can be invalid, even if the reasoning itself is sound). University students are (by and large) people who have reached the age of majority, and as such are responsible for their own actions. This means (among other things) they will be expected to think for themselves. I feel it is irresponsible in a classroom setting, therefore, to "think what you're told to think is true", as it delays the level of responsibility they will be held to by society.
Engineers, for example, will be held accountable for the safety of structures they design or certify. Not making them aware of the mathematical principles that lay beneath their calculations could conceivably lead to fatalities. If I were an engineer, I would want to be **** certain I had a minimum of assumptions, to limit my own potential liability. If there were a typo in a formula in a handbook of mathematical expressions I had used, it is possible that would not help me much in a civil lawsuit.
Granted, few of your students may face such an extreme situation, and such a situation is even less likely to rest upon a definition of an exponential function. Nevertheless, I believe in rigor as a principle, not for its own sake as a form of intellectual prowess, but because of what it accomplishes: showing what we derive is a consequence of simpler facts.
Ignoring the logical priority of mathematical conclusions is not sound mathematics. It neatly sweeps "hard stuff" under the rug, as a matter of expedience. It's like fast-food convenience for the mind, in short: a sham.
Now, it would be one thing if the Fundamental Theorem of Calculus and the Inverse Function Theorem were "abstract generalities" not bothered with in a beginning calculus course. However, this is not so, and I believe (perhaps I am being "too esoteric" in this belief?) that the best way to understanding something is to become acquainted with it through USE.
So, if we have these tools available to define certain functions: why don't we USE them? ESPECIALLY since transcendental functions are difficult to understand EXCEPT through some form of limiting process (and isn't calculus all about the nifty things we can do with limits?).
I believe that teaching people to think for themselves only after they reach graduate school is "too late". Unless they have already individually acquired the skill on their own independently, they are set-up for failure, or at best, a very difficult time of it.
I do understand, however, that as a matter of pragmatism, you may have found that devoting too much time to these sorts of issues has been non-productive. So, you are passing the buck to some other teacher. One can only hope that somewhere, this comes to an abrupt halt.
**************
I didn't mean to obliterate your cognizant observation that $e$ is indeed a "natural" base. It is, and for the reasons you indicate. However, something "deep" is hiding behind it, something important (a fact about this slippery real number $e$). Transcendental real numbers *ought* to be "hard to define", in fact it took centuries before we were in a position to even conclude they existed (that is, that the algebraic numbers are not "cauchy-closed").
Personally, if all one takes calculus for is to calculate derivatives and integrals, meh: Wolfram|Alpha and a CRC handbook, skip the class. This business about limits and continuity, on the other hand, and the exploration of the "infinitesimal", this sly attempt to touch the face of infinity; if that does not make one blush, I humbly submit one has no soul.
#### ThePerfectHacker
##### Well-known member
Let me see if I have this straight: you differentiate a function without any justification that it is differentiable. Why do I think this is not a good idea?
I only read your first paragraph. I did not read anything else you wrote because I think your teaching standards are abysmal if you have a problem with writing down down $b^x$ in the way that makes sense to most people when they actually think about it.
Historically speaking all the calculus developed in a non-precise way, but it was motivated, and my presentation of it most closely parallels it. Go ahead why not spend the the first two months of class defining Dedekind cuts and what it means to even add two real numbers together. But you do not do that. You say that is too much. So if you are willing to accept that it is okay to teach calculus without defining what addition means, rather appealing to students intuitive feeling about what it is, then why not define exponentials in the most intuitive way possible?
I am willing to bet if you explained to a student why $e$ is natural using my explanation they will have a much better chance of repeating it back to you than your explanation.
#### Deveno
##### Well-known member
MHB Math Scholar
I'm not sure what it is you think I am getting at. It might be a bit more illuminating to read my entire post, if only to provide more context, but I cannot force you to do this.
Not every function is differentiable. Students should not, even if they do not get all the details think this is "mostly true". Continuity is *special* and differentiability even more so.
But again, I do not hear an answer to my original question: assuming, for the time being that $b^x$ *is* differentiable (and we certainly HOPE it would be so), how do you actually differentiate it?
What I imagine is something invoking the chain rule, and writing:
$b^x = e^{x\log(b)} = \exp \circ (x \cdot \log(b))$
and using the product (or multiplying by a constant) rule for the $x \log(b)$ part, and then re-writing:
$(b^x)' = (\exp)'(x \log(b))\cdot \log(b) = \log(b)\cdot b^x$.
To get to that conclusion, we need to know at least these two facts:
1. $(e^x)' = e^x$
2. what the function $\log$ is (in particular that it is a functional *inverse* of $\exp$).
Perhaps you show (or merely indicate the "plausibility" of) the fact that $e^0 = 1$. I would be *interested* in how you derive fact 1. There are some different ways to do this, and you haven't indicated which one you prefer. You also have not indicated how the number $e$ (rather than "some exponential function $\exp$") enters into this. I think that this is critical, since the whole POINT of your initial post is that $e$ represents a "special base".
Fact 2 also needs prior establishment of at least SOME facts about $e^x$ (like, that it is 1-1, so it *has* an inverse, contrast this with the contortions we must undergo for inverse trig functions).
I know that you feel that it is somehow more complicated to start with a way of describing $\log$ and deriving $\exp$ as ITS inverse function, but I really don't see this. The crucial property of $\log$:
$\log(xy) = \log(x) + \log(y)$
is an EASY consequence of the definition:
$$\displaystyle \log(x) = \int_1^x \frac{1}{t}\ dt$$, by taking the interval:
$[1,xy]$ and splitting it into the two sub-intervals of $[1,x]$ and $[x,xy]$, and using the linearity of the integral (presumably you find this property not so challenging as to withhold it from your students).
At least you answered my original question of "what is $b^x$?" by answering it via some kind of "rational approximation". As I indicated to Evgeny, in his reply, I am perfectly OK with that. The real number system was designed to let us do things like that, so that's fair.
How you derive the fact:
$b^x = e^{x\log(b)}$ is a more subtle question. If in fact, you already have the inverse-pair $\exp$ and $\log$ previously established (you don't say if this is true or not), one could use THIS property of (rational, and in approximation, real) exponents:
$(a^b)^c = a^{bc}$.
Perhaps "most" (or even possibly "all") of your students do not care. I cannot change that, nor do I have the singular pleasure of addressing them. I feel you should care, especially since the tone of this thread indicates a pedagogical approach of choosing a "simple case" as desirable.
Historically, the logical foundations of calculus were seen as something of a "crisis", and provided the impetus for some truly worth-while mathematics by Cauchy, Weierstrass, Riemann and Dedekind (to name a few note-worthy examples). The "soundness" of Newton's (and Liebniz's) original justifications were viewed with some suspicion well into the 20th century, although with the advent of "non-standard analysis" these fears have been mostly allayed. Not of all mathematical progress is made at "the outer reaches" sometimes thinking about "the very beginnings" proves fruitful, as well.
If this is not so, then tell me: why do we even "bother" with "epsilon-delta" proofs? I bet even a beginning calculus text makes some (perhaps brief) mention of these strange animals (or so it seems to me with the wealth of questions asked about them on sites such as this one).
Personally, I think talking about Dedekind cuts or Cauchy sequences (or even the oft-maligined "infinite decimal" approach) is a worth-while undertaking, if only to show that these "real numbers" have some "concrete" realization as something we can use rational numbers to approximate (because, underneath it all, the rationals is what we "really" use). I understand if the syllabus you follow does not allow time for this, but it certainly would make "some theorems" less confusing (anything that uses glb's or lub's to establish the existence of some desired real number, for example).
*******************
I do get it, you know, that a "simple" explanation is easier to remember, and that perhaps as much as 80% of what a first-year calculus student learns will be forgotten in 3 months time (unless they *immediately* take a refresher course, or a "building-upon-it" continuation).
If I were a student of yours, asking these same questions, would I get the very same answers?
*******************
I have talked at perhaps too much length about this particular subject. It's hard for me to judge whether this becomes a purely "ad hominem" discussion, so let us abandon it. What I have neglected to say in all of this, is that the SECOND part of your post, is something I WHOLE-HEARTEDLY agree with, and that thinking in terms of "turns" is the "natural" way to measure angle (and very profitable for physicists, and engineers).
#### ThePerfectHacker
##### Well-known member
I'm not sure what it is you think I am getting at. It might be a bit more illuminating to read my entire post, if only to provide more context, but I cannot force you to do this.
Not every function is differentiable. Students should not, even if they do not get all the details think this is "mostly true". Continuity is *special* and differentiability even more so.
But again, I do not hear an answer to my original question: assuming, for the time being that $b^x$ *is* differentiable (and we certainly HOPE it would be so), how do you actually differentiate it?
How do you "prove" that $\sin x$ and $\cos x$ are differenciable? You draw a picture illustrating important limits. That aint a proof at all. However, the picture is more illuminating to understanding the analysis of trigonometric functions than a boring non-inspired unmotivated, but rigorous, power series definition.
How do you "define" $\sqrt{x}$? You have no problem saying that it is that positive number when squared results in $x$. This definition is not exactly rigorous as it assumes the reals has positive square roots. The proof of which requires the completeness property. Something which is never even mentioned in a calculus course at all. You have no problem using that definition at all.
But the moment I write $b^x$ and I say that means raise $b$ to the power of $x$, you suddenly have a problem. You ask how do I even define. I "define" it in the most natural way most people would expect, $b^q$, were $q$ is a rational number approaching $x$. There is nothing natural about writing $\exp(x\log b)$, it is confusing and entirely unmotivated for anyone who sees it for their first time. Now if you applied this standard consistently you will have a problem with how one "defines" $\sin x$ and $\sqrt{x}$. But you do not. You are okay with how it is presented in a calculus class and assume the necessarily properties, including being differenciable, when needed.
In the end you apply selective criticisms and make the concept of exponential functions exponentially more difficult for first-time students because you think it justifies it. Calculus is not supposed to be about justifying analytic concepts. It is supposed to introduce one to them and try to motivate the reasons for why it works and why it is true.
#### Deveno
##### Well-known member
MHB Math Scholar
How do you "prove" that $\sin x$ and $\cos x$ are differenciable? You draw a picture illustrating important limits. That aint a proof at all. However, the picture is more illuminating to understanding the analysis of trigonometric functions than a boring non-inspired unmotivated, but rigorous, power series definition.
How do you "define" $\sqrt{x}$? You have no problem saying that it is that positive number when squared results in $x$. This definition is not exactly rigorous as it assumes the reals has positive square roots. The proof of which requires the completeness property. Something which is never even mentioned in a calculus course at all. You have no problem using that definition at all.
But the moment I write $b^x$ and I say that means raise $b$ to the power of $x$, you suddenly have a problem. You ask how do I even define. I "define" it in the most natural way most people would expect, $b^q$, were $q$ is a rational number approaching $x$. Now if you applied this standard consistently you will have a problem with how one "defines" $\sin x$ and $\sqrt{x}$. But you do not. You are okay with how it is presented in a calculus class and assume the necessarily properties, including being differenciable, when needed.
In the end you apply selective criticisms and make the concept of exponential functions exponentially more difficult for first-time students because you think it justifies it. Calculus is not supposed to be about justifying analytic concepts. It is supposed to introduce one to them and try to motivate the reasons for why it works and why it is true.
I think differentiating trig functions is likewise somewhat thorny. Similar glib use of $\pi$ is made in this case, usually without any justification. Again, a "geometric" limit argument is the usual "first look" approach taken, and this is not out of character with the geometric flavor that much of calculus has.
I think that transcendental functions are "subtler" than many people realize...indeed, that even the completeness property of the reals is a subtle concept. These things are usually "glossed over" in favor of expediency. I remain unconvinced that this is a NECESSARY evil. It's convenient, yes, and perhaps even "motivational" (whatever that means), but the truth is kept "out of view".
Many calculus texts start with the assertion that "they will assume the reader is familiar with the basic properties of real numbers" and dispense with 100 years of deep (and HARD) mathematics in a few words.
Square roots are a bit thorny, as well. Many students know (or have at least seen a proof) that $\sqrt{2}$ is irrational, but they have no idea what a big can of worms that really is. It is easy to write $\sqrt{x}$ and think you know what it means. I'm sure it comes as a shock to some math majors when they realize that actually, they do NOT (and maybe non-math-majors who never take anything more demanding than calculus ever realize there is "more to it than that").
I believe in DEMONSTRATING that a function is differentiable, by showing the limit that defines its derivative EXISTS. I think this is good "practice" for later when one encounters a function whose differentiability status is unknown. Not all "real problems" have answer keys in the back of some book.
Of course, you have me at a disadvantage: you know your students, and I do not. I ask you for an explanation of what you mean in your original post, and you reply with remarks about "selective criticisms".
I cannot recall the particular thread, but there was a similar discussion some time back about defining the area of a circle, and arriving at a value for $\pi$. Some argued that a "proper definition" of trig functions would have to be delayed until after power series, as that was logically more satisfiying. I made the point that it could be done with the tools of integral calculus (although this might not be very "direct").
One often sees T-shirts with:
$e^{i\pi} + 1 = 0$
printed on them. The simplicity of this formula belies the depth of information it summarizes: how the circle ties together exponentials, logarithms, trigonometry, and an intrinsic notion of distance in 7 scant symbols. I'm not even confident I understand entirely everything it says...but the small portion I DO get, fills me with a kind of awe.
If, in the final analysis, you feel that I am looking "too deeply" at what you think should be taken more lightly, I have no ready answer. I have my reasons for thinking the way I do, but if my explanations are not sufficient, I suppose that is my failing. I *try* to be clear, but I have no assurances I ever am, for I only see the world with my own perceptions, and never that of another human being.
Indeed, I often fear I might be embarrassing my comrades here at MHB, for being too outspoken, or having views well outside of the conventional wisdom, or even worse, perhaps just plain being wrong. I have only a vague notion of INTERNAL consistency to guide me, and I find it enormously difficult to communicate this to other people. Surely, I reason to myself, Evgeny.Makarov or Ackbach or Jameson or MarkFL or chisigma (or some other soul who I have in my ignorance failed to mention) will point out to me what is wrong with my views, so I may improve my ability to express math to other people...but alas, I hear only silence. | 2022-01-25T13:15:41 | {
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https://it.mathworks.com/help/symbolic/dsolve.html | # dsolve
Solve system of differential equations
Support for character vector or string inputs will be removed in a future release. Instead, use syms to declare variables and replace inputs such as dsolve('Dy = -3*y') with syms y(t); dsolve(diff(y,t) == -3*y).
## Description
example
S = dsolve(eqn) solves the differential equation eqn, where eqn is a symbolic equation. Use diff and == to represent differential equations. For example, diff(y,x) == y represents the equation dy/dx = y. Solve a system of differential equations by specifying eqn as a vector of those equations.
example
S = dsolve(eqn,cond) solves eqn with the initial or boundary condition cond.
example
S = dsolve(___,Name,Value) uses additional options specified by one or more Name,Value pair arguments.
example
[y1,...,yN] = dsolve(___) assigns the solutions to the variables y1,...,yN.
## Examples
collapse all
Solve the first-order differential equation $\frac{\mathit{dy}}{\mathit{dt}}=\mathit{ay}$.
Specify the first-order derivative by using diff and the equation by using ==. Then, solve the equation by using dsolve.
syms y(t) a
eqn = diff(y,t) == a*y;
S = dsolve(eqn)
S = ${C}_{1} {\mathrm{e}}^{a t}$
The solution includes a constant. To eliminate constants, see Solve Differential Equations with Conditions. For a full workflow, see Solving Partial Differential Equations.
Solve the second-order differential equation $\frac{{\mathit{d}}^{2}\mathit{y}}{{\mathit{dt}}^{2}}=\mathit{ay}$.
Specify the second-order derivative of y by using diff(y,t,2) and the equation by using ==. Then, solve the equation by using dsolve.
syms y(t) a
eqn = diff(y,t,2) == a*y;
ySol(t) = dsolve(eqn)
ySol(t) = ${C}_{1} {\mathrm{e}}^{-\sqrt{a} t}+{C}_{2} {\mathrm{e}}^{\sqrt{a} t}$
Solve the first-order differential equation $\frac{\mathit{dy}}{\mathit{dt}}=\mathit{ay}$ with the initial condition $y\left(0\right)=5$.
Specify the initial condition as the second input to dsolve by using the == operator. Specifying condition eliminates arbitrary constants, such as C1, C2, ..., from the solution.
syms y(t) a
eqn = diff(y,t) == a*y;
cond = y(0) == 5;
ySol(t) = dsolve(eqn,cond)
ySol(t) = $5 {\mathrm{e}}^{a t}$
Next, solve the second-order differential equation $\frac{{\mathit{d}}^{2}\mathit{y}}{{\mathit{dt}}^{2}}={\mathit{a}}^{2}\mathit{y}$ with the initial conditions $y\left(0\right)=b$ and ${y}^{\prime }\left(0\right)=1$.
Specify the second initial condition by assigning diff(y,t) to Dy and then using Dy(0) == 1.
syms y(t) a b
eqn = diff(y,t,2) == a^2*y;
Dy = diff(y,t);
cond = [y(0)==b, Dy(0)==1];
ySol(t) = dsolve(eqn,cond)
ySol(t) =
$\frac{{\mathrm{e}}^{a t} \left(a b+1\right)}{2 a}+\frac{{\mathrm{e}}^{-a t} \left(a b-1\right)}{2 a}$
This second-order differential equation has two specified conditions, so constants are eliminated from the solution. In general, to eliminate constants from the solution, the number of conditions must equal the order of the equation.
Solve the system of differential equations
$\begin{array}{l}\frac{\mathit{dy}}{\mathit{dt}}=\mathit{z}\\ \frac{\mathit{dz}}{\mathit{dt}}=-\mathit{y}.\end{array}$
Specify the system of equations as a vector. dsolve returns a structure containing the solutions.
syms y(t) z(t)
eqns = [diff(y,t) == z, diff(z,t) == -y];
S = dsolve(eqns)
S = struct with fields:
z: C2*cos(t) - C1*sin(t)
y: C1*cos(t) + C2*sin(t)
Access the solutions by addressing the elements of the structure.
ySol(t) = S.y
ySol(t) = ${C}_{1} \mathrm{cos}\left(t\right)+{C}_{2} \mathrm{sin}\left(t\right)$
zSol(t) = S.z
zSol(t) = ${C}_{2} \mathrm{cos}\left(t\right)-{C}_{1} \mathrm{sin}\left(t\right)$
When solving for multiple functions, dsolve returns a structure by default. Alternatively, you can assign solutions to functions or variables directly by explicitly specifying the outputs as a vector. dsolve sorts the outputs in alphabetical order using symvar.
Solve a system of differential equations and assign the outputs to functions.
syms y(t) z(t)
eqns = [diff(y,t)==z, diff(z,t)==-y];
[ySol(t),zSol(t)] = dsolve(eqns)
ySol(t) = ${C}_{1} \mathrm{cos}\left(t\right)+{C}_{2} \mathrm{sin}\left(t\right)$
zSol(t) = ${C}_{2} \mathrm{cos}\left(t\right)-{C}_{1} \mathrm{sin}\left(t\right)$
Solve the differential equation $\frac{\partial }{\partial t}y\left(t\right)={e}^{-y\left(t\right)}+y\left(t\right)$. dsolve returns an explicit solution in terms of a Lambert W function that has a constant value.
syms y(t)
eqn = diff(y) == y+exp(-y)
eqn(t) =
sol = dsolve(eqn)
sol = ${\mathrm{W}\text{lambertw}}_{0}\left(-1\right)$
To return implicit solutions of the differential equation, set the 'Implicit' option to true. An implicit solution has the form $F\left(y\left(t\right)\right)=g\left(t\right)$.
sol = dsolve(eqn,'Implicit',true)
sol =
$\left(\begin{array}{c}\left({\int \frac{{\mathrm{e}}^{y}}{y {\mathrm{e}}^{y}+1}\mathrm{d}y|}_{y=y\left(t\right)}\right)={C}_{1}+t\\ {\mathrm{e}}^{-y\left(t\right)} \left({\mathrm{e}}^{y\left(t\right)} y\left(t\right)+1\right)=0\end{array}\right)$
If dsolve cannot find an explicit solution of a differential equation analytically, then it returns an empty symbolic array. You can solve the differential equation by using MATLAB® numerical solver, such as ode45. For more information, see Solve a Second-Order Differential Equation Numerically.
syms y(x)
eqn = diff(y) == (x-exp(-x))/(y(x)+exp(y(x)));
S = dsolve(eqn)
Warning: Unable to find symbolic solution.
S =
[ empty sym ]
Alternatively, you can try finding an implicit solution of the differential equation by specifying the 'Implicit' option to true. An implicit solution has the form $F\left(y\left(x\right)\right)=g\left(x\right)$.
S = dsolve(eqn,'Implicit',true)
S =
${\mathrm{e}}^{y\left(x\right)}+\frac{{y\left(x\right)}^{2}}{2}={C}_{1}+{\mathrm{e}}^{-x}+\frac{{x}^{2}}{2}$
Solve the differential equation $\frac{\mathit{dy}}{\mathit{dt}}=\frac{\mathit{a}}{\sqrt{\mathit{y}}}+\mathit{y}$ with condition $y\left(a\right)=1$. By default, dsolve applies simplifications that are not generally correct, but produce simpler solutions. For more details, see Algorithms.
syms a y(t)
eqn = diff(y) == a/sqrt(y) + y;
cond = y(a) == 1;
ySimplified = dsolve(eqn, cond)
ySimplified =
${\left({\mathrm{e}}^{\frac{3 t}{2}-\frac{3 a}{2}+\mathrm{log}\left(a+1\right)}-a\right)}^{2/3}$
To return the solutions that include all possible values of the parameter $a$, turn off simplifications by setting 'IgnoreAnalyticConstraints' to false.
yNotSimplified = dsolve(eqn,cond,'IgnoreAnalyticConstraints',false)
yNotSimplified =
Solve the second-order differential equation $\left({x}^{2}-1{\right)}^{2}\frac{{\partial }^{2}}{\partial {x}^{2}}y\left(x\right)+\left(x+1\right)\frac{\partial }{\partial x}y\left(x\right)-y\left(x\right)=0$. dsolve returns a solution that contains a term with unevaluated integral.
syms y(x)
eqn = (x^2-1)^2*diff(y,2) + (x+1)*diff(y) - y == 0;
S = dsolve(eqn)
S =
${C}_{2} \left(x+1\right)+{C}_{1} \left(x+1\right) \int \frac{{\mathrm{e}}^{\frac{1}{2 \left(x-1\right)}} {\left(1-x\right)}^{1/4}}{{\left(x+1\right)}^{9/4}}\mathrm{d}x$
To return series solutions of the differential equation around $x=-1$, set the 'ExpansionPoint' to -1. dsolve returns two linearly independent solutions in terms of a Puiseux series expansion.
S = dsolve(eqn,'ExpansionPoint',-1)
S =
$\left(\begin{array}{c}x+1\\ \frac{1}{{\left(x+1\right)}^{1/4}}-\frac{5 {\left(x+1\right)}^{3/4}}{4}+\frac{5 {\left(x+1\right)}^{7/4}}{48}+\frac{5 {\left(x+1\right)}^{11/4}}{336}+\frac{115 {\left(x+1\right)}^{15/4}}{33792}+\frac{169 {\left(x+1\right)}^{19/4}}{184320}\end{array}\right)$
Find other series solutions around the expansion point $\infty$ by setting 'ExpansionPoint' to Inf.
S = dsolve(eqn,'ExpansionPoint',Inf)
S =
$\left(\begin{array}{c}x-\frac{1}{6 {x}^{2}}-\frac{1}{8 {x}^{4}}\\ \frac{1}{6 {x}^{2}}+\frac{1}{8 {x}^{4}}+\frac{1}{90 {x}^{5}}+1\end{array}\right)$
The default truncation order of the series expansion is 6. To obtain more terms in the Puiseux series solutions, set 'Order' to 8.
S = dsolve(eqn,'ExpansionPoint',Inf,'Order',8)
S =
$\left(\begin{array}{c}x-\frac{1}{6 {x}^{2}}-\frac{1}{8 {x}^{4}}-\frac{1}{90 {x}^{5}}-\frac{37}{336 {x}^{6}}\\ \frac{1}{6 {x}^{2}}+\frac{1}{8 {x}^{4}}+\frac{1}{90 {x}^{5}}+\frac{37}{336 {x}^{6}}+\frac{37}{1680 {x}^{7}}+1\end{array}\right)$
Solve the differential equation $\frac{\mathit{dy}}{\mathit{dx}}=\frac{1}{{\mathit{x}}^{2}}{\mathit{e}}^{-\frac{1}{\mathit{x}}}$ without specifying the initial condition.
syms y(x)
eqn = diff(y) == exp(-1/x)/x^2;
ySol(x) = dsolve(eqn)
ySol(x) =
${C}_{1}+{\mathrm{e}}^{-\frac{1}{x}}$
To eliminate constants from the solution, specify the initial condition $\mathit{y}\left(0\right)=1$.
cond = y(0) == 1;
S = dsolve(eqn,cond)
S =
${\mathrm{e}}^{-\frac{1}{x}}+1$
The function ${\mathit{e}}^{-\frac{1}{\mathit{x}}}$ in the solution ySol(x) has different one-sided limits at $x=0$. The function has a right-side limit, $\underset{\mathit{x}\to {0}^{+}}{\mathrm{lim}}\text{\hspace{0.17em}}{\mathit{e}}^{-\frac{1}{\mathit{x}}}=0$, but it has undefined left-side limit, $\underset{\mathit{x}\to {0}^{-}}{\mathrm{lim}}\text{\hspace{0.17em}}{\mathit{e}}^{-\frac{1}{\mathit{x}}}=\infty$.
When you specify the condition y(x0) for a function with different one-sided limits at x0, dsolve treats the condition as a limit from the right, $\mathrm{lim}\text{\hspace{0.17em}}\mathit{x}\to {\mathit{x}}_{0}^{+}$.
## Input Arguments
collapse all
Differential equation or system of equations, specified as a symbolic equation or a vector of symbolic equations. Specify a differential equation by using the == operator. If eqn is a symbolic expression (without the right side), the solver assumes that the right side is 0, and solves the equation eqn == 0.
In the equation, represent differentiation by using diff. For example, diff(y,x) differentiates the symbolic function y(x) with respect to x. Create the symbolic function y(x) by using syms and solve the equation d2y(x)/dx2 = x*y(x) using dsolve.
syms y(x)
S = dsolve(diff(y,x,2) == x*y)
Specify a system of differential equations by using a vector of equations, as in syms y(t) z(t); S = dsolve([diff(y,t) == z, diff(z,t) == -y]). Here, y and z must be symbolic functions that depend on symbolic variables, which are t in this case. The right side must be symbolic expressions that depend on t, y and z. Note that Symbolic Math Toolbox™ currently does not support composite symbolic functions, that is, symbolic functions that depend on another symbolic functions.
Initial or boundary condition, specified as a symbolic equation or vector of symbolic equations.
When a condition contains a derivative, represent the derivative with diff. Assign the diff call to a variable and use the variable to specify the condition. For example, see Solve Differential Equations with Conditions.
Specify multiple conditions by using a vector of equations. If the number of conditions is less than the number of dependent variables, the solutions contain the arbitrary constants C1, C2,....
### Name-Value Arguments
Specify optional comma-separated pairs of Name,Value arguments. Name is the argument name and Value is the corresponding value. Name must appear inside quotes. You can specify several name and value pair arguments in any order as Name1,Value1,...,NameN,ValueN.
Example: 'IgnoreAnalyticConstraints',false does not apply internal simplifications.
Expansion point of a Puiseux series solution, specified as a number, or a symbolic number, variable, function, or expression. Specifying this option returns the solution of a differential equation in terms of a Puiseux series (a power series that allows negative and fractional exponents). The expansion point cannot depend on the series variable. For example, see Find Series Solution of Differential Equation.
Data Types: single | double | sym
Complex Number Support: Yes
Option to use internal simplifications, specified as true or false.
By default, the solver applies simplifications while solving the differential equation, which could lead to results not generally valid. In other words, this option applies mathematical identities that are convenient, but the results might not hold for all possible values of the variables. Therefore, by default, the solver does not guarantee the completeness of results. If 'IgnoreAnalyticConstraints' is true, always verify results returned by the dsolve function. For more details, see Algorithms.
To solve ordinary differential equations without these simplifications, set 'IgnoreAnalyticConstraints' to false. Results obtained with 'IgnoreAnalyticConstraints' set to false are correct for all values of the arguments. For certain equations, dsolve might not return an explicit solution if you set 'IgnoreAnalyticConstraints' to false.
Option to return an implicit solution, specified as false or true. For a differential equation with variables x and y(x), an implicit solution has the form F(y(x)) = g(x).
By default, the solver tries to find an explicit solution y(x) = f(x) analytically when solving a differential equation. If dsolve cannot find an explicit solution, then you can try finding a solution in implicit form by specifying the 'Implicit' option to true.
Maximum degree of polynomial equations for which the solver uses explicit formulas, specified as a positive integer smaller than 5. dsolve does not use explicit formulas when solving polynomial equations of degrees larger than 'MaxDegree'.
Truncation order of a Puiseux series solution, specified as a positive integer or a symbolic positive integer. Specifying this option returns the solution of a differential equation in terms of a Puiseux series (a power series that allow negative and fractional exponents). The truncation order n is the exponent in the O-term: $O\left({\mathrm{var}}^{n}\right)$ or $O\left({\mathrm{var}}^{-n}\right)$.
## Output Arguments
collapse all
Solutions of differential equation, returned as a symbolic expression or a vector of symbolic expressions. The size of S is the number of solutions.
Variables storing solutions of differential equations, returned as a vector of symbolic variables. The number of output variables must equal the number of dependent variables in a system of equations. dsolve sorts the dependent variables alphabetically, and then assigns the solutions for the variables to output variables or symbolic arrays.
## Tips
• If dsolve cannot find an explicit or implicit solution, then it issues a warning and returns the empty sym. In this case, try to find a numeric solution using the MATLAB® ode23 or ode45 function. Sometimes, the output is an equivalent lower-order differential equation or an integral.
• dsolve does not always return complete solutions even if 'IgnoreAnalyticConstraints' is false.
• If dsolve returns a function that has different one-sided limits at x0 and you specify the condition y(x0), then dsolve treats the condition as a limit from the right, .
## Algorithms
If you do not set 'IgnoreAnalyticConstraints' to false, then dsolve applies these rules while solving the equation:
• log(a) + log(b) = log(a·b) for all values of a and b. In particular, the following equality is applied for all values of a, b, and c:
(a·b)c = ac·bc.
• log(ab) = b·log(a) for all values of a and b. In particular, the following equality is applied for all values of a, b, and c:
(ab)c = ab·c.
• If f and g are standard mathematical functions and f(g(x)) = x for all small positive numbers, f(g(x)) = x is assumed to be valid for all complex x. In particular:
• log(ex) = x
• asin(sin(x)) = x, acos(cos(x)) = x, atan(tan(x)) = x
• asinh(sinh(x)) = x, acosh(cosh(x)) = x, atanh(tanh(x)) = x
• Wk(x·ex) = x for all branch indices k of the Lambert W function.
• The solver can multiply both sides of an equation by any expression except 0.
• The solutions of polynomial equations must be complete.
## Compatibility Considerations
expand all
Warns starting in R2019b | 2021-10-17T19:06:08 | {
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https://math.stackexchange.com/questions/2948977/proving-right-angle-using-vectors | # Proving right angle using vectors
Let $$ABCD$$ be a rectangle, $$E$$ midpoint of $$\overline{DC}$$ and $$G$$ point on $$\overline{AC}$$ such that $$\vec{BG}$$ is perpendicular to $$\vec{AC}$$. Also, let $$F$$ be a midpoint of $$\overline{AG}$$. Prove that angle $$\angle BFE =\pi/2$$.
So, I need to prove that $$\vec{EF}\cdot \vec{FB}=0$$. I tried to just express these vectors as sums of vectors of rectangle but haven't find any "elegant" way to prove it.
I did found one "ugly" way to prove it:
Let $$|AB|=a$$, $$|AD|=b$$ and $$\vec{AB}=\vec{a}$$, $$\vec{AD}=\vec{b}$$, so $$\vec{AC}=\vec{a}+\vec{b}$$ and since $$\vec{a}\cdot\vec{b}=0$$ we have $$|AC|^2=a^2+b^2$$. We can see that $$\triangle ABC$$ and $$\triangle BCG$$ are similar so $$|CG|=|CB|^2/|AC|$$ and since $$\vec{CG}=\lambda \cdot \vec{CA}$$ we get $$\lambda=\frac{b^2}{a^2+b^2}.$$
Now, we can express $$\vec{EF}$$ and $$\vec{FB}$$ in the terms of $$\vec{a}$$, $$\vec{b}$$, more precisely: $$\vec{EF}=\frac{1}{2}\left(\frac{-b^2}{a^2+b^2}\right)\vec{a}+\frac{1}{2}\left(\frac{a^2}{a^2+b^2}-2\right)\vec{b}$$ and $$\vec{FB}=\frac{1}{2}\left(2-\frac{a^2}{a^2+b^2}\right)\vec{a}+\frac{1}{2}\left(\frac{-a^2}{a^2+b^2}\right)\vec{b}$$ and if we multiply we get that dot product is $$0$$.
But as you can see, this $$\lambda$$ is "weird" and I am wondering if someone sees a more elegant way to prove this ?
• Actually, if you write those final coefficients in terms of $\lambda$, it makes the proof of orthogonality much prettier. – J.G. Oct 9 '18 at 19:26
You can simplify and generalize at the same time.
Instead of defining $$E$$ and $$F$$ as midpoints, define them as weighted averages:
$$E=kC+(1-k)D$$ $$F=kG+(1-k)A$$
(In your question $$k=1/2.)$$ As $$k$$ goes from $$0$$ to $$1$$, $$\triangle{BFE}$$ interpolates the similar triangles $$\triangle{BAD}$$ and $$\triangle{BGC}$$. And it is also similar to these two triangles.
Given that $$(B-G)\cdot(C-G)=0$$ and $$(B-A)\cdot(D-A)=0$$, you can show that $$(B-F)\cdot(E-F)=0$$ (the simplification is a bit tedious, and I used the fact that $$(Rx)\cdot y+(Ry)\cdot x=0$$ for perpendicular vectors $$x,y$$ and rotation $$R$$).
More about the phenomenon of linearly interpolating two directly similar figures can be found by Googling the terms 'spiral similarity' and 'fundamental theorem of directly similar figures'. Also, instead of using vectors you can use complex numbers, as shown in $$\textbf{2.2}$$ of Zachary Abel's Mean Geometry paper.
Let $$H$$ be a midpoint of $$BG$$. Then $$HF$$ is a middle line in triangle $$ABG$$, so $$FH||AB||CD$$ and $$FH = {1\over 2}AB = CE$$ so $$FHCE$$ is a parallelogram, so $$FE||CH$$.
Now $$HF\bot BC$$ so $$HF$$ is (second) altitude in $$\triangle BCF$$ so $$H$$ is orthocenter in this triangle, since $$BG$$ is also altitude in this triangle. So line $$CH\bot FB$$ and thus $$\angle BFE = 90^{\circ}$$.
Here is a vector solution:
Let $$G$$ be an origin of position vecotors. Then $$G=0$$, $$F = A/2$$, $$D = A-B+C$$, $$A\cdot B = C\cdot B=0$$ and $$E = {1\over 2}(C+D) = {1\over 2}(A-B+2C)$$
$$\begin{eqnarray} \vec{FB}\cdot \vec{FE} &=& (E-F)(B-F)\\ &=& {1\over 4}(-B+2C)(2B-A)\\ &=& -{1\over 2}(B^2+CA)\\ &=&0 \end{eqnarray}$$
A solution with a help of the coordinate system.
Let $$B = (0,0)$$, $$C= (2c,0)$$ and $$A= (0,2a)$$. Then $$D(2c,2a)$$ and $$E(2c,a)$$. A perpendicular to $$AC:\;\;{x\over 2c}+{y\over 2a}=1$$ through $$B$$ is $$y={c\over a}x$$ which cuts $$AC$$ at $$G= \big({2a^2c\over a^2+c^2},{2c^2a\over a^2+c^2} \big)$$
so $$F = \big({a^2c\over a^2+c^2},{2c^2a+a^3\over a^2+c^2} \big)$$
Now it is not difficult to see that $$\vec{FE}\cdot \vec{FB} =0$$.
Another solution:
Say $$M$$ is midpoint of $$AB$$, then $$FM$$ is a middle line in triangle $$ABG$$ so $$FM||BG$$ so $$\angle MFC = 90^{\circ}$$ and thus $$F$$ is on a circle through $$B,C$$ and $$M$$. But on this circle is also $$E$$ since $$\angle MEC = 90^{\circ}$$. So $$\angle BFE = \angle BME = 90^{\circ}$$
Alternatively Say $$AB = 2a$$ and $$BC= b$$, then $$AF = {1\over 2}AG = {1\over 2} {4a^2\over \sqrt{4a^2+b^2}} = {2a^2\over \sqrt{4a^2+b^2}}$$ Then we have $$AF\cdot AC = {2a^2\over \sqrt{4a^2+b^2}}\cdot \sqrt{4a^2+b^2} = 2a^2 = AM\cdot AB$$ and so $$B,C,F,M$$ are concylic. | 2019-06-19T16:41:59 | {
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https://de.mathworks.com/help/symbolic/solve-an-algebraic-equation.html | ## Solve Algebraic Equation
Symbolic Math Toolbox™ offers both symbolic and numeric equation solvers. This topic shows you how to solve an equation symbolically using the symbolic solver solve. To compare symbolic and numeric solvers, see Select Numeric or Symbolic Solver.
### Solve an Equation
If eqn is an equation, solve(eqn, x) solves eqn for the symbolic variable x.
Use the == operator to specify the familiar quadratic equation and solve it using solve.
syms a b c x
eqn = a*x^2 + b*x + c == 0;
solx = solve(eqn, x)
solx =
-(b + (b^2 - 4*a*c)^(1/2))/(2*a)
-(b - (b^2 - 4*a*c)^(1/2))/(2*a)
solx is a symbolic vector containing the two solutions of the quadratic equation. If the input eqn is an expression and not an equation, solve solves the equation eqn == 0.
To solve for a variable other than x, specify that variable instead. For example, solve eqn for b.
solb = solve(eqn, b)
solb =
-(a*x^2 + c)/x
If you do not specify a variable, solve uses symvar to select the variable to solve for. For example, solve(eqn) solves eqn for x.
### Return the Full Solution to an Equation
solve does not automatically return all solutions of an equation. Solve the equation cos(x) == -sin(x). The solve function returns one of many solutions.
syms x
solx = solve(cos(x) == -sin(x), x)
solx =
-pi/4
To return all solutions along with the parameters in the solution and the conditions on the solution, set the ReturnConditions option to true. Solve the same equation for the full solution. Provide three output variables: for the solution to x, for the parameters in the solution, and for the conditions on the solution.
syms x
[solx, param, cond] = solve(cos(x) == -sin(x), x, 'ReturnConditions', true)
solx =
pi*k - pi/4
param =
k
cond =
in(k, 'integer')
solx contains the solution for x, which is pi*k - pi/4. The param variable specifies the parameter in the solution, which is k. The cond variable specifies the condition in(k, 'integer') on the solution, which means k must be an integer. Thus, solve returns a periodic solution starting at pi/4 which repeats at intervals of pi*k, where k is an integer.
### Work with the Full Solution, Parameters, and Conditions Returned by solve
You can use the solutions, parameters, and conditions returned by solve to find solutions within an interval or under additional conditions.
To find values of x in the interval -2*pi<x<2*pi, solve solx for k within that interval under the condition cond. Assume the condition cond using assume.
assume(cond)
solk = solve(-2*pi<solx, solx<2*pi, param)
solk =
-1
0
1
2
To find values of x corresponding to these values of k, use subs to substitute for k in solx.
xvalues = subs(solx, solk)
xvalues =
-(5*pi)/4
-pi/4
(3*pi)/4
(7*pi)/4
To convert these symbolic values into numeric values for use in numeric calculations, use vpa.
xvalues = vpa(xvalues)
xvalues =
-3.9269908169872415480783042290994
-0.78539816339744830961566084581988
2.3561944901923449288469825374596
5.4977871437821381673096259207391
### Visualize and Plot Solutions Returned by solve
The previous sections used solve to solve the equation cos(x) == -sin(x). The solution to this equation can be visualized using plotting functions such as fplot and scatter.
Plot both sides of equation cos(x) == -sin(x).
fplot(cos(x))
hold on
grid on
fplot(-sin(x))
title('Both sides of equation cos(x) = -sin(x)')
legend('cos(x)','-sin(x)','Location','best','AutoUpdate','off')
Calculate the values of the functions at the values of x, and superimpose the solutions as points using scatter.
yvalues = cos(xvalues)
yvalues =
$\left(\begin{array}{c}-0.70710678118654752440084436210485\\ 0.70710678118654752440084436210485\\ -0.70710678118654752440084436210485\\ 0.70710678118654752440084436210485\end{array}\right)$
scatter(xvalues, yvalues)
As expected, the solutions appear at the intersection of the two plots.
### Simplify Complicated Results and Improve Performance
If results look complicated, solve is stuck, or if you want to improve performance, see, Troubleshoot Equation Solutions from solve Function.
## Support
#### Mathematical Modeling with Symbolic Math Toolbox
Get examples and videos | 2021-09-24T01:25:05 | {
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http://mathhelpforum.com/calculus/4044-find-limit.html | 1. ## Find the limit
Limit as n approaches infinity [(3^n + 5^n)/(3^n+1 + 5^n+1)]
I tried to divide the numerator and denominator by 3^n. Was not successful.
What should I do next?
2. Originally Posted by Nichelle14
Limit as n approaches infinity [(3^n + 5^n)/(3^n+1 + 5^n+1)]
I tried to divide the numerator and denominator by 3^n. Was not successful.
What should I do next?
Did you consider to divide by $3^n+5^n$
Thus,
$\frac{1}{1+\frac{2}{3^n+5^n}}\to 1$ as $n\to\infty$
3. Also,
$1-\frac{1}{n}\leq \frac{3^n+5^n}{3^n+1+5^n+1} \leq 1+\frac{1}{n}$
Since,
$\lim_{n\to\infty} 1-\frac{1}{n}=\lim_{n\to\infty}1+\frac{1}{n}=1$
Thus,
$\lim_{n\to\infty}\frac{3^n+5^n}{3^n+1+5^n+1}=1$ by the squeeze theorem.
4. Hello, Nichelle14!
$\lim_{n\to\infty}\frac{3^n + 5^n}{3^{n+1} + 5^{n+1}}$
I tried to divide the numerator and denominator by $3^n$ . . .Was not successful.
What should I do next?
Divide top and bottom by $5^{n+1}.$
The numerator is: . $\frac{3^n}{5^{n+1}} + \frac{5^n}{5^{n+1}} \;= \;\frac{1}{5}\cdot\frac{3^n}{5^n} + \frac{1}{5}\;=$ $\frac{1}{5}\left(\frac{3}{5}\right)^n + \frac{1}{5}$
The denominator is: . $\frac{3^{n+1}}{5^{n+1}} + \frac{5^{n+1}}{5^{n+1}} \;=\;\left(\frac{3}{5}\right)^{n+1} + 1$
Recall that: if $|a| < 1$, then $\lim_{n\to\infty} a^n\:=\:0$
Therefore, the limit is: . $\lim_{n\to\infty}\,\frac{\frac{1}{5}\left(\frac{3} {5}\right)^n + \frac{1}{5}} {\left(\frac{3}{5}\right)^n + 1} \;=\;\frac{\frac{1}{5}\cdot0 + \frac{1}{5}}{0 + 1}\;=\;\frac{1}{5}
$
5. Originally Posted by Soroban
Hello, Nichelle14!
Divide top and bottom by $5^{n+1}.$
The numerator is: . $\frac{3^n}{5^{n+1}} + \frac{5^n}{5^{n+1}} \;= \;\frac{1}{5}\cdot\frac{3^n}{5^n} + \frac{1}{5}\;=$ $\frac{1}{5}\left(\frac{3}{5}\right)^n + \frac{1}{5}$
The denominator is: . $\frac{3^{n+1}}{5^{n+1}} + \frac{5^{n+1}}{5^{n+1}} \;=\;\left(\frac{3}{5}\right)^{n+1} + 1$
Recall that: if $|a| < 1$, then $\lim_{n\to\infty} a^n\:=\:0$
Therefore, the limit is: . $\lim_{n\to\infty}\,\frac{\frac{1}{5}\left(\frac{3} {5}\right)^n + \frac{1}{5}} {\left(\frac{3}{5}\right)^n + 1} \;=\;\frac{\frac{1}{5}\cdot0 + \frac{1}{5}}{0 + 1}\;=\;\frac{1}{5}
$
I think the trick here is that you will divide all the terms by the largest term in the expression(it is useful when the limit tends to infinity)
Keep Smiling
Malay | 2013-05-19T02:18:47 | {
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https://community.freefem.org/t/spatial-co-ordinate-of-maximum-value-of-variable/1364 | # Spatial Co-ordinate of Maximum Value of Variable
Say I have the following code:
int nn = 50;
mesh Th = square(nn, nn);
fespace Vh(Th, P1);
Vh u;
u = sin(x);
I can get the maximum value of ‘u’ by:
real maxValueU = u[].max;
How do I get the values of x and y coordinates where maximum value of ‘u’ occurs?
yes but be careful because the max could be not unique.unc
the method imax give the dof number
int dofmaxValueU = u[].imax;
Now for P1 finite element the dof are vertices with same numbering so
Th(dofmaxValueU).x
Th(dofmaxValueU).y
given the coordinate. | 2022-05-18T09:02:41 | {
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https://www.physicsforums.com/threads/fixed-cost-cost-per-person.919820/ | # Fixed Cost & Cost Per Person
1. Jul 10, 2017
### zak100
1. The problem statement, all variables and given/known data
A group can charter a particular aircraft at a fixed total cost. If 36 people charter the aircraft rather than 40 people, then the cost per person is greater by 12$a) What is the fixed total cost to charter the aircraft? b) What is the cost per person if 40 people charter the aircraft? 2. Relevant equations ThirtySixCost = 36 * (costPerPerson + 12) FixedCost = NumberofPersons * costPerPerson (Note if NumberofPersons>40) Not sure about the correctness of equations 3. The attempt at a solution I dont know how to solve it because data is incomplete. ThirtySixCost=? FortyCost=? Somebody please guide me how to solve it. Zulfi. 2. Jul 10, 2017 ### scottdave Reread the problem statement. Your costPerPerson represents the cost per person, when there are 40 people. Your ThirtySixCost is the total cost to rent the plane (but that is a Fixed cost). I did not see anything about greater than 40. You can get to 3 equations and 3 unknowns to solve it. For example (not real numbers), if it costs$360 to rent something (say a lake house), then for 40 people, that is $9 per person, but for 36 people it is$10 per person. It costs $1 more per person if you only have 36 people. Last edited: Jul 10, 2017 3. Jul 11, 2017 ### zak100 Hi, Please tell me how to find costPerPerson? Zulfi. 4. Jul 11, 2017 ### Mark44 ### Staff: Mentor These variable names aren't useful. The fixed cost is the same whether there are 40 people or 36 people. Let x = the cost per person with 40 passengers. Let f = the fixed cost of chartering the airplane. • Write an expression that represents the cost per person with 36 passengers. (Use the information given in the problem.) • Translate the 2nd sentence in the problem statement into an algebraic equation. 5. Jul 11, 2017 ### zak100 Hi, FTC = Fixed Total Cost CPP = Cost Per Person CPP = FTC + 12 I can form only one equation. Please guide me. Zulfi. 6. Jul 11, 2017 ### scottdave The plane costs an amount (call it FTC), whether there is 1 passenger or 36 passenger or 40 passengers. The cost per person depends on the fixed cost (which does not change, and the number of passengers. How about this?: FTC is total fixed cost. CPPForty = FTC / 40. CPPthirtysix = FTC / 36. The problem statement gives you another relationship between CPPForty and CPPthirtysix, which you can make another equation. You may want to re-look at the example problem I posed in post #2, to see how you can relate your problem to that one. 7. Jul 11, 2017 ### Mark44 ### Staff: Mentor The last equation makes no sense. You are saying that the cost for one person is$12 more than the fixed cost.
8. Jul 11, 2017
### zak100
Hi,
Mr. scottdave:
Do you mean this:
CPPThirtySix = CPPForty + 12
Zulfi.
9. Jul 11, 2017
### scottdave
Yes, now with 3 equations you should be able to solve for CPPThirtySix, CPPForty and FTC.
10. Jul 12, 2017
### zak100
Hi,
I am able to solve it.
CP36 = FTC/36 ---(1)
CP40 = FTC/40 ---(2)
CP36 = CP40 + 12--- (3)
After solving these eq., i am able to get correct answer:
CP40 = 108
CP36 = 120 (Not required in the question)
FTC = 4320 | 2017-08-18T01:55:41 | {
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https://www.physicsforums.com/threads/table-probability-question.735574/ | # Homework Help: Table probability question
1. Jan 29, 2014
### joshmccraney
1. The problem statement, all variables and given/known data
11 people sit at two round tables, one sits 5 and the other sits 6. how many combinations of seating arrangements are there?
2. Relevant equations
i'm not sure about equations, but my solution attempt may have some info.
3. The attempt at a solution
i know if 11 people were to sit at 1 table, we would have $\frac{11!}{11}=10!$ combinations. thus, would we first have to choose who sits at which table? i.e: $$(11C5)(4!)+(11C6)(6!)$$
thanks for your time (and help!)
2. Jan 29, 2014
### haruspex
You're close, but a couple of errors.
You got 11C54! for choosing the 5 to sit at one table and arranging them. For each such arrangement, how can you arrange the remaining 6?
3. Jan 30, 2014
### joshmccraney
wait, do you mean $(11C5)(4!)+(11C6)(5!)$? also, if we've already selected the 5 to be at on table, do we really need to select 6 for the other?
thanks for the response, and let me know what you think.
4. Jan 30, 2014
### joshmccraney
because i definitely should have done $5!$, not $6!$ (brain fart)
5. Jan 30, 2014
### D H
Staff Emeritus
There are two big problems with your result. One is that when you've chosen the five people who will sit at table #1 you have but one choice as to which set of six people will sit at table #2. The other: Why are you adding?
6. Jan 30, 2014
### joshmccraney
yea, i definitely see this problem but if i dont put both "choices" how do i determine which number (5 or 6) to use?
also, i was adding because i figured after i choose who sits at what table, all i need to do is take the possible combos of one table and add them to the combos of another table.
evidently this is wrong; can you direct me as to what to do next?
7. Jan 30, 2014
### D H
Staff Emeritus
There are a number of right ways to address this part of the problem. You chose a wrong way.
One way to look at this part of the problem is that you choose five people to sit at one table, call it table A. There are 11 choose 5 ways to do this. For any give choice, there are 11-5=6 people left. You now need to choose six of these six people to sit at table B. There are 6 choose 6 ways to do this, and that of course is one. The number of ways to choose people to sit at the two tables is the product of these two numbers: (11 choose 5)*(6 choose 6).
You could of course look at it the other way around: You'll choose six people to sit at the larger table and then choose five of the remaining five to sit at the smaller table. This leads to (11 choose 6)*(5 choose 5) ways to split the people into two groups.
Hmmm. One approach yields (11 choose 5)*(6 choose 6), the other (11 choose 6)*(5 choose 5). Is there a problem here? The answer is no. 11 choose 5 and 11 choose 6 are the same number (462), as are 6 choose 6 and 5 choose 5 (which are of course both 1).
Suppose X and Y are independent. Also suppose there are three ways to accomplish X and five ways to accomplish Y. The number of ways to do accomplish X and Y is fifteen, not eight.
The same applies to the problem at hand. You should be multiplying here, not adding.
8. Jan 30, 2014
### CAF123
You can do it either way (i.e choose 6 for one table and have one set of five for the other, or choose 5 for a table and have one set of six for the other). The result is the same.
Consider the case when there are two people situated at one table and one at the other. How many possible combinations are there?
Alternatively, are the combinations around each individual table dependent on each other? What does this tell you? | 2018-06-24T09:27:55 | {
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https://www.physicsforums.com/threads/solving-the-following-equation-for-x-x-5-2-9.876632/ | # Solving the following equation for x: $(x-5)^{-2} = 9$
1. Jun 24, 2016
### malknar
1. The problem statement, all variables and given/known data
An exercise in Lang's Basic Mathematics which asked to solve for x in the equation $(x - 5)^{-2} = 9$
I tried to solve it and got $x = \frac {16} 3$
which is correct according to the book, but I've found that the book states that x also equals \frac {14}3, and I can't figure out how.
2. Relevant equations
3. The attempt at a solution
$(x-5)^{-2} = 9$
This yields: $x-5 = 9^{-1/2} =\frac 1 {9^{1/2}} = \frac13$
Which yields: $x=\frac 13 +5 = \frac {1+15} 3 =\frac {16}3$
But I can't figure out how can x equals $\frac {14}3$ too.
2. Jun 24, 2016
### Staff: Mentor
It's better to write this as $(x - 5)^2 = \frac 1 9$, which leads to $x - 5 = \pm \sqrt{\frac 1 9}$
Quadratic equations can have two solutions.
3. Jun 24, 2016
### mfig
What is $(-\frac{1}{3})^2$?
4. Jun 24, 2016
### late347
I am thinking about that old math movie with young actor Edward James Olmos. He was playing the role of high school teacher.
Stand and deliver
"Negative times negative is positive!!!" The result would be 1/9
For the original poster's benefit. The following is compiled from my own high school era old math factbook.
second degree equations.
• you can evaluate whether the quadratic has two roots, or one root, or none of the real-numbered roots. (roots belonging to ℝ)
• calculate the quadratic into the form of: $$Ax^2 + Bx+C= 0$$
$$x=\frac{-B±\sqrt{B^2-4AC}}{2A}$$
this is called the discriminant part of the equation $D=\sqrt{B^2-4AC}$
if D>0, then the equation has two non-equal roots
if D=0, then the equation has a twin-root
if D<0, then the equation doesn't have real-numbered roots (roots in the real numbers ℝ)
This last fact is because negative inside the squareroot sign is not defined.
E.g. $\sqrt{(-5)}= ~~not ~~defined~~ in~~ real~~ numbers$
negative inside the squareroot is defined in the complex numbers, I think, but I have no knowledge of the complex numbers myself. (complex numbers symbol is this one ℂ)
5. Jun 24, 2016
### epenguin
In short if A2 = B, then A = ± √B
6. Jun 24, 2016
### malknar
I got it now, thanks everyone. As usual the solution is simpler than expected. My mistake was not considering $-3 = 9^{1/2}$
7. Jun 24, 2016
### Staff: Mentor
The symbol $\sqrt{x}$ denotes the positive square root of x, by established convention. In this case $\sqrt{9} = +3$. However, the equation $x^2 = 9$ has two solutions: 3 and -3.
8. Jun 24, 2016
### Ray Vickson
No: $9^{1/2}$ is $+3$, NOT $-3$. Essentially by definition, the functions $x^{1/2}$ and $\sqrt{x}$ mean the positive root.
There is no really good language for it, but saying "a" square root instead of "the" square root comes close to capturing it: there are two roots to the equation $x^2 = a \:(a > 0)$; these are $x = \sqrt{a}$ and $x = -\sqrt{a}$
9. Jun 24, 2016
### late347
• watch this khan academy video it will be helpful for this problem and also for the purpose of increased understanding.
• review what is the binomial square formula such as:
• $(a+b)^2= a^2+2ab+b^2$
• $(a-b)^2 = a^2-2ab+b^2$
10. Jul 6, 2016
### James R
Alternatively you could do this:
$\frac{1}{(x-5)^2}=9$
$9(x-5)^2 = 1$
$9(x^2 - 10 x + 25) = 1$
$9x^2 - 90 x + 224 = 0$
$(3x - 14)(3x - 16)=0$
$x=\frac{14}{3}$ or $x=\frac{16}{3}$. | 2017-08-21T05:02:04 | {
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https://www.mathalino.com/forum/calculus/find-differential-equations-following-family-curves | # Find the differential equations of the following family of curves.
16 posts / 0 new
Bingo
Find the differential equations of the following family of curves.
Differential Equations
Find the differential equations of the following family of curves.
1. Parabolas with axis parallel to the y – axis with distance vertex to focus fixed as a.
2. Parabolas with axis parallel to the x – axis with distance vertex to focus fixed as a.
3. All ellipses with center at the origin and axes on the coordinate axes.
4. Family of cardioids.
5. Family of 3 – leaf roses.
Bingo
1
Jhun Vert
(1) Upwards and downwards parabolas with latus rectum equal to 4a.
$y - k = \pm 4a(x - h)^2$
$y' = \pm 8a(x - h)$
$y'' = \pm 8a$
Bingo
y−k=±4a(x−h)2 is this the equation for parabolas?
isn't it 4a(y-k)=(x-h)2 ?
Jhun Vert
I made a mistake in there, it should be (x - h)2 = ±4a(y - k). The solution for (1) should go this way:
$(x - h)^2 = \pm 4a(y - k)$
$2(x - h) = \pm 4ay'$
$2 = \pm 4ay''$
$y'' = \pm \frac{1}{2a}$
Bingo
y−k=±4a(x−h)2 is this the equation for parabolas?
isn't it 4a(y-k)=(x-h)2 ?
Jhun Vert
Bingo
so for number 2,
x-h=±4a(y-k)2
x'=±8a(y-k)
x''=±8a
am i wrong?
Jhun Vert
It is better to express your answer in terms of y' rather than x'. Although x' will do and simpler.
$(y − k)^2 = \pm 4a(x − h)$
$2(y − k)y' = \pm 4a$
$y' = \dfrac{\pm 2a}{y - k}$
Note: $y - k = \dfrac{y'}{\pm 2a}$
$y'' = \dfrac{\mp 2a \, y'}{(y - k)^2}$
$y'' = \dfrac{\mp 2a \, y'}{\left( \dfrac{y'}{\pm 2a} \right)^2}$
$y'' = \dfrac{\mp 8a}{y'}$
$y'' \, y' = \pm 8a$
ChaCha Ingal Cortez
our prof. gave the same question but the ans. he gave is different from your ans. in no 2, his answer is (y')^3 + 2ay"=0 are they the same with your answer sir ?
Jhun Vert
They are different. I made a mistake in $y - k$ of the above solution.
$(y − k)^2 = \pm 4a(x − h)$
$2(y - k) \, y' = \pm 4a$
$y' = \dfrac{\pm 2a}{y - k}$
$y - k = \dfrac{\pm 2a}{y'}$
$y'' = \dfrac{\mp 2a \, y'}{(y - k)^2}$
$y'' = \dfrac{\mp 2a \, y'}{\left( \dfrac{\pm 2a}{y'} \right)^2}$
$y'' = \dfrac{\mp 2a \, y'}{\dfrac{4a^2}{(y')^2}}$
$y'' = \dfrac{\mp (y')^3}{2a}$
$2a \, y'' = \mp (y')^3$
$\pm (y')^3 + 2a \, y'' = 0$
Bingo
Thank You so much sir.
Bingo
sir, do you know the standard or general equations for the last 3 problems?
what equations should i use?
Shandy
I think you are from ADZU haha
Alegnaaa (guest)
Good evening, sir. Do you have the solution for the last three problems? Thank you!
Ancha (guest)
Sir do you know how to eliminate arbitrary constant here in this equation?
(y-33)²=4a(x-h)?
• Mathematics inside the configured delimiters is rendered by MathJax. The default math delimiters are $$...$$ and $...$ for displayed mathematics, and $...$ and $...$ for in-line mathematics. | 2023-03-30T09:11:44 | {
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https://math.stackexchange.com/questions/2377313/how-many-four-digit-integers-are-there-that-do-not-have-two-consecutive-equal-ev | # How many four digit integers are there that do not have two consecutive equal even digits?
How many four digit integers are there that do not have two consecutive equal even digits?
My Attempt There are $9*10^3$ four digit numbers. We tie two equal even digits and form the four digit numbers with them. This can be done in $5 \cdot \binom{3}{2} \cdot 10 \cdot 10 = 1500$ ways. There are $10 \cdot 10$ numbers starting with $00$ so we must subtract them from $1500$. So answer is $9000 - 1400 = 7600$. But it is not matching with the given answer $7801$. Can any one help me to find out where I am making mistakes? Thanks in advance.
1. You did not account for the fact that there are only $9$ choices for the leading digit when neither number in the pair of consecutive equal even digits is in the thousands place.
2. You have subtracted numbers in which there are three or more consecutive even digits more than once.
First, we observe that there are $9 \cdot 10 \cdot 10 \cdot 10 = 9000$ four-digit positive integers.
Let $A_1$ be the set of four-digit positive integers in which the thousands place and hundreds place contain equal even digits. Let $A_2$ be the set of four-digit positive integers in which the hundreds place and tens place contain equal even digits. Let $A_3$ be the set of four-digit positive integers in which the tens place and units place contain equal even digits. Then $A_1 \cup A_2 \cup A_3$ is the set of four-digit positive integers that do contain consecutive equal even digits. By the Inclusion-Exclusion Principle, the number of four-digit positive integers that do contain consecutive even digits is $$|A_1 \cup A_2 \cup A_3| = |A_1| + |A_2| + |A_3| - |A_1 \cap A_2| - |A \cap A_3| - |A_2 \cap A_3| + |A_1 \cap A_2 \cap A_3|$$
$|A_1|$: Since we cannot use zero, there are four ways of choosing the even digit that occupies both the thousands and hundreds place. There are ten choices for each of the remaining digits. Hence, $|A_1| = 4 \cdot 10 \cdot 10 = 400$.
$|A_2|$: Since we cannot use zero, the thousands place can be filled in nine ways. There are five ways to choose the even digit that fills both the hundreds and tens places. There are ten ways to fill the units place. Hence, $|A_2| = 9 \cdot 5 \cdot 10 = 450$.
$|A_3|$: We can fill the thousands place in nine ways and the hundreds place in ten ways. There are five ways to choose the even digit that occupies both the tens and units places. Hence, $|A_3| = 9 \cdot 10 \cdot 5 = 450$.
$|A_1 \cap A_2|$: Since we cannot use zero, there are four ways to choose the even digit that occupies the thousands, hundreds, and tens places. There are ten ways to fill the units place. Hence, $|A_1 \cap A_2| = 4 \cdot 10 = 40$.
$|A_1 \cap A_3|$: There are four ways to choose the even digit that occupies both the thousands and hundreds places and five ways to choose the even digit that occupies both the tens and units places. Hence, $|A_1 \cap A_3| = 4 \cdot 5 = 20$.
$|A_2 \cap A_3|$: There are nine ways to fill the thousands place. There are five ways to choose the even digit that occupies the hundreds, tens, and units places. Hence, $|A_2 \cap A_3| = 9 \cdot 5 = 45$.
$|A_1 \cap A_2 \cap A_3|$: Since we cannot use zero, there are four ways to choose the even digit that occupies the thousands, hundreds, tens, and units places. Hence, $|A_1 \cap A_2 \cap A_3| = 4$.
Therefore, the number of four-digit positive integers that do contain consecutive even equal digits is \begin{align*} |A_1 \cup A_2 \cup A_3| & = |A_1| + |A_2| + |A_3| - |A_1 \cap A_2| - |A \cap A_3| - |A_2 \cap A_3| + |A_1 \cap A_2 \cap A_3|\\ & = 400 + 450 + 450 - 40 - 20 - 45 + 4\\ & = 1199 \end{align*} Therefore, the number of four-digit positive even integers that do not contain consecutive equal even digits is $9000 - 1199 = 7801$.
• Thanks for pointing the errors and nice step by step solution. – rugi Jul 31 '17 at 9:05
This where you're making a mistake $-$
What you're trying to do is called the inclusion-exclusion principle. Your error is in the fact that you are overcounting numbers such as $4446$. If I understand your method correctly, you are blacklisting this number twice instead of once.
IMO, an easier way to solve the problem is to count the numbers that are not to be included first by going through all possible digit combinations e.g. $aabb$ etc. | 2021-03-08T10:59:36 | {
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