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https://shop.evenstarcreations.com/clara-wong-jqx/bee6ef-permutation-with-repetition-allowed | # permutation with repetition allowed
When the order doesmatter it is a Permutation. The formula is written: n r. where, n is number of things to choose from; r is number of things we choose of n; repetition is allowed; order matters; Permutation without Repetition In this post, we will see how to find all lexicographic permutations of a string where repetition of characters is allowed. 216. Another definition of permutation is the number of such arrangements that are possible. Permutations with Repetition Permutation when repetition is allowed. When a permutation can repeat, we just need to raise n to the power of however many objects from n we are choosing, so . For example, if you have 10 digits to choose from for a combination lock with 6 numbers to enter, and you're allowed to repeat all the digits, you're looking to find the number of permutations with repetition. Male or Female ? "With repetition" means that repetition is allowed. First position can have N choices The second position can have ( N-1 ) choices. Permutation without Repetition: This method is used when we are asked to reduce 1 from the previous term for each time. To improve this 'Permutation with repetition Calculator', please fill in questionnaire. There are basically two types of permutation: Repetition is Allowed: It could be “333”. A permutation is an arrangement of objects, without repetition, and order being important. Repeating of characters of the string is allowed. Permutations are items arranged in a given order meaning […] List permutations with repetition and how many to choose from. Like combinations, there are two types of permutations: permutations with repetition, and permutations without repetition. They have sometimes been referred to as permutations with repetition, although they are not permutations in general. Permutations. The number of permutations of ‘n’ things taken ‘r’ at a time is denoted by n P r It is defined as, n P r A permutation with repetition of n chosen elements is also known as an "n-tuple". Total […] No Repetition: for example the first three people in a running race. That was an $$r$$-permutation of $$n$$ items with repetition allowed. Permutations without Repetition In this case, we have to reduce the number of available choices each time. All the different arrangements of the letters A, B, C. All the different arrangements of the letters A, A, B A permutation is an ordering of a set of objects. Print all permutations with repetition of characters, Given a string of length n, print all permutation of the given string. In some cases, repetition of the same element is allowed in the permutation. Please update your bookmarks accordingly. (Repetition allowed, order matters) Ex: how many 3 litter words can be created, if Repetition is allowed? We should print them in lexicographic order. D. 320. For this case, n and k happens to be the same. Permutation formulas. Permutation without Repetition: for example the first three people in a running race. 1. There is a subset of permutations that takes into account that there are double objects or repetitions in a permutation problem. C. 120. When additional restrictions are imposed, the situation is transformed into a problem about permutations with restrictions. With permutations, every little detail matters. These are the easiest to calculate. This is a permutation with repetition. It could be "333". Permutation with Repetition. You can’t be first and second. When the number of object is “n,” and we have “r” to be the selection of object, then; Choosing an object can be in n different ways (each time). For example, what order could 16 pool balls be in? Noel asks: Is there a way where i can predict all possible outcomes in excel in the below example. Permutations with repetition. Permutation with repetition occurs when a set has r different objects, and there are n choices every time. Repetition of characters is allowed. Most commonly, the restriction is that only a small number of objects are to be considered, meaning that not all the objects need to be ordered. Permutations without repetition A permutation is an arrangement, or listing, of objects in which the order is important. For the given input string, print all the possible permutations. Compare the permutations of the letters A,B,C with those of the same number of letters, 3, but with one repeated letter $$\rightarrow$$ A, A, B. After choosing, say, number "14" we can't choose it again. Permutations: There are basically two types of permutation: Repetition is Allowed: such as the lock above. Technically, there's no such thing as a permutation with repetition. n different things taking r at a time without repetition - definition The number of permutations of n different things, taking r at a time without repetition is denoted by n P r . We can actually answer this with just the product rule: $$10^5$$. It has following lexicographic permutations with repetition of characters - AAA, AAB, AAC, ABA, ABB, ABC, … 7.1.5 When repetition of objects is allowed The number of permutations of n things taken all at a time, when repetion of objects is allowed is nn. Permutations with repetition. OR 125. 1. Ordered arrangements of n elements of a set S, where repetition is allowed, are called n-tuples. There are two types of permutations: Repetition is Allowed: For the number lock example provided above, it could be “2-2-2”. For example, locks allow you to pick the same number for more than one position, e.g. Post, we have 26 choices of letter: $26^ { }! 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Math, science, and permutations without repetition a permutation is an arrangement objects! Set S, where repetition is allowed: such as the lock above n=number of elements k=combination!, in each of the words are allowed items with repetition of the different which... | 2021-06-19T14:47:22 | {
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https://www.physicsforums.com/threads/conditional-probability-probability-of-having-a-girl.544052/ | # Conditional Probability: probability of having a girl.
## Homework Statement
I have two kids. Given that at least one of them is a girl, what is the probability that they are both girls?
## The Attempt at a Solution
I think it's 1/3, because the possibilities are: Boy Boy, Boy Girl, Girl Boy, Girl Girl. There are three possibilities with at least one girl. Out of these three, only one has two girls.
But my friend says order doesn't matter, so BG is the same as GB. So the probability is 1/2.
Which is right?
1/2
it doesn't matter
Last edited:
My vote's 1/3. This is a different question to 'My eldest child of two is a girl, what is the probability that they are both girls', which is 1/2. In your case, the piece of information provided is not 'ordered' but, instead, excludes the BB possibility.
Using Bayes theorem, the probability of event A, given event B, (with events independent) is the probability of A intersect B divided by probability of B.
So, the probability of the kid being a girl, given that there is already a boy, is the probability of a girl intersected with a boy divided by the probability of a boy.
To find A intersect B for independent events, just multiply the probability of each. (1/2)(1/2). So, (1/4) divided by (1/2).
It's the probability that it's a girl, given that at least one is a girl - nothing involving a boy. Does that change things?
I think you're right. The outcome of one boy and one girl should be twice as likely as the other 2 outcomes, since this is a http://en.wikipedia.org/wiki/Binomial_distribution" [Broken] with n=2 and p=1/2.
More formally, let $p = 1/2$ be the probability a girl is born and on our universe U define the random variables X = number of girls, Y = number of boys. Define the events $C = \{X = 2 \}$ and $D = \{X \geq 1 \}$. Certainly $C \subseteq D$ so $C \cap D = C$. Since $p = 1/2$ and the two births are independent events, then $\mathbf{P}(C) = (1/2)^2 = 1/4$. The probability of the complement of D, $\mathbf{P}(U \setminus D)$ is the same as $\mathbf{P}\{X=0\}$, which is the probability of 2 boys. Similarly, since $1-p = 1/2$ and the two births are independent events, then $\mathbf{P}(U \setminus D) = (1/2)^2 = 1/4$. So $\mathbf{P}(D) = 1 - \mathbf{P}(U \setminus D) = 3/4$. Finally
$$\mathbf{P}(C|D) = \frac{\mathbf{P}(C \cap D)}{\mathbf{P}(D)} = \frac{\mathbf{P}(C)}{\mathbf{P}(D)} = \frac{1/4}{3/4} = 1/3 \, .$$
I think your proof using the reduced sample space is correct as well--and shorter, too!
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diazona
Homework Helper
Yep, 1/3. For a less technical explanation, if it helps anyone:
Imagine surveying a large number of families with 2 children. In half of them, the older child will be a boy. Within that half, half again (or a quarter of the total) will have the younger child be a boy, and the rest (another quarter of the total) will have the younger child be a girl. The same goes for the remaining half of the families whose older child is a girl. So you have four possibilities: BB, BG, GB, GG (in order of age), each representing a quarter of the total population.
Now, the statement "given that at least one of them is a girl" excludes one of those possibilities, BB. There are three equally likely possibilities left (BG, GB, GG), of which exactly one has both children being girls. So the probability is 1/3.
Imagine surveying a large number of families with 2 children...
Yes, that was the the means I convinced myself of my analysis.
I think the comment about Bayes has actually taken the conclusion the wrong way. What we learn in the statement 'there is a girl' is a piece of information that teaches us to exclude 1/4 of the population set of 2-child families.
@AN; You have concluded Bayes says 'To have a boy is twice as likely as to have a girl'. Therefore probability of two girls is 1/3 and a boy and a girl is 2/3, because 2/3 is twice 1/3!!! (Because we've excluded the BB option from the P=1 outcome.)
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Oops, I was talking about girls and boys. That doesn't change the answer though. The Bayes theorem is sound. The answer is 1/2.
But what's the question? '1/2' is not the answer to the op, but is the answer to 'what is the probability of GG compared with not GG' (conditional on 'there is a girl').
Ray Vickson
Homework Helper
Dearly Missed
## Homework Statement
I have two kids. Given that at least one of them is a girl, what is the probability that they are both girls?
## The Attempt at a Solution
I think it's 1/3, because the possibilities are: Boy Boy, Boy Girl, Girl Boy, Girl Girl. There are three possibilities with at least one girl. Out of these three, only one has two girls.
But my friend says order doesn't matter, so BG is the same as GB. So the probability is 1/2.
Which is right?
This is the same problem as: we toss a (fair) coin twice and get at least one head; what is the probability we get another head as well? The results LOOK a bit paradoxical when written out as follows: events are C = {two heads}, A = {first toss is heads}, B = {second toss is heads}. We have P{C|A} = P{C|B} = 1/2, but P{C|A or B} = 1/3. That is your problem in a nutshell. This looks paradoxical because the statement that we get at least one head is the same as saying that A or B occur, and each of these separately implies a 1/2 probability of C; however, when taken together they imply a 1/3 probability of C!
RGV
I like Serena
Homework Helper
Hi everyone!
Here's my view on the matter.
There is a difference between saying:
"given the first (or youngest) kid is a girl" (chance 2/4)
and "given that at least one kid is a girl" (chance 3/4).
In the first case the chance would be 1/2 as Arcana claims.
We can also write this as:
$$P(\text{2 girls | the youngest kid is a girl}) = {\text{number of favorable outcomes} \over \text{number of possible outcomes}} = {1 \over 2}$$
In the second case it is:
$$P(\text{2 girls | at least 1 girl}) = {\text{number of favorable outcomes} \over \text{number of possible outcomes}} = {1 \over 3}$$
So I believe the reasoning of pearapple is sound for this problem and the chance is 1/3.
vela
Staff Emeritus
Homework Helper
But my friend says order doesn't matter, so BG is the same as GB. So the probability is 1/2.
Since the others have explained how to get the correct answer, I'll point out what mistake your friend is making:
You have four possible outcomes: BB, BG, GB, and GG. BG and GB are distinct outcomes. The order matters when you're enumerating outcomes.
When you combine them to form events, then the order may or may not matter depending on how the event is defined. The event "exactly 1 girl" would be {BG, GB}. Whether the girl comes first or last doesn't really matter.
Note that even though the order doesn't matter in determining what outcomes are included in an event, you still have to add up the probabilities for each outcome to get the probability of the event. That is, P({BG, GB}) = P({BG})+P({GB}) = 1/2. It's incorrect to say: order doesn't matter, so BG=GB and P({BG, GB})=P({BG})=1/4.
The event "at least one girl" is {BG, GB, GG}, which corresponds to a probability of 3/4. If you incorrectly reduce this to {BG, GG}, as your friend has done, you get a probability of 1/2. If you take the probability GG, which is 1/4, and divide it by 3/4, you get the correct answer of 1/3. If instead you divide it by the incorrect probability of 1/2, you get 1/2, the answer your friend got.
BruceW
Homework Helper
Its amazing that even such a simple example has an answer that is not immediately obvious.
The answer to this question is ambiguous because the question itself is ambiguous.
I believe this falls under a similar category of the monty hall game show. Lets say you start off with two doors, and behind each door there could either be a boy or a girl. If the host says that behind one of the doors there is a girl, then you know the chances of there being a girl behind both doors is one in three. (since he didn't specify which door). If he tells you that there is a girl behind the door on the left, then you know that the chances of there being two girls are one in two.
So you and your friend have seperate interpretations of the question. The question boils down to if you know which kid is a girl, or you just know that there is A girl.
If you have your 4 scenarios, GG, BG, GB, and BB, him telling you that there is at least one girl only means you get to cross off BB. Him telling you that there is a girl on the left allows you to cross of BG And BB
BruceW
Homework Helper
I have two kids. Given that at least one of them is a girl, what is the probability that they are both girls?
Is different to:
I have two kids. I know one of them, which is a girl. What is the probability that they are both girls?
I suppose pearapple's friend misinterpreted it as the second question, because he assumed that if the father knew he had at least one girl, then it would have meant that he met one of his kids, which was a girl. But this is not necessarily true.
I guess in language, the two statements are very similar, but you have to be careful about what you're actually saying.
HallsofIvy
Homework Helper
Pretty much what others are saying: there are two different questions that can be confused here:
1) A person has two children, one of which is a girl. What is the probability that the other child is a girl?
2) A person has two children, the older of which (or the younger or the one we happen to know- anything that distinguishes one child from another) is a girl. What is the probability that the other child is a girl?
With two children, and assuming that "boy" and "girl" for any given child are equally likely, There are four equally likely possibilities: BB, BG, GB, GG where, of course, "B" means "boy", "G" means "girl" and the order depends upon which was born first, or was the one we knew, etc.
With problem (1) above, knowing that one of the children is a girl removes "BB" from the list, leaving "BG", "GB", and "GG". In only one of those is the other child also a girl. Probability of two girls, 1/3.
With problem (2) above, knowing that the older child (or younger child or whichever is distinguished in some way) we remove both "BB" and "BG". We are left with "GB" and "GG" of which 1 has both girls. Probability of two girls, 1/2.
I like Serena
Homework Helper
I have two kids. Given that at least one of them is a girl, what is the probability that they are both girls?
Is different to:
I have two kids. I know one of them, which is a girl. What is the probability that they are both girls?
I suppose pearapple's friend misinterpreted it as the second question, because he assumed that if the father knew he had at least one girl, then it would have meant that he met one of his kids, which was a girl. But this is not necessarily true.
I guess in language, the two statements are very similar, but you have to be careful about what you're actually saying.
Uhh :uhh:. Aren't those the same?
The answer to this question is ambiguous because the question itself is ambiguous.
I believe this falls under a similar category of the monty hall game show. Lets say you start off with two doors, and behind each door there could either be a boy or a girl. If the host says that behind one of the doors there is a girl, then you know the chances of there being a girl behind both doors is one in three. (since he didn't specify which door). If he tells you that there is a girl behind the door on the left, then you know that the chances of there being two girls are one in two.
So you and your friend have seperate interpretations of the question. The question boils down to if you know which kid is a girl, or you just know that there is A girl.
If you have your 4 scenarios, GG, BG, GB, and BB, him telling you that there is at least one girl only means you get to cross off BB. Him telling you that there is a girl on the left allows you to cross of BG And BB
Uhh :uhh:. Aren't those the same?
Is there anything that you don't understand specifically about the above explanation?
I like Serena
Homework Helper
Is there anything that you don't understand specifically about the above explanation?
It seems to me that BruceW's explanation is not correct.
It seems to me that BruceW's explanation is not correct.
I feel similar about his explanation, not that it is incorrect, just maybe incomplete.
vela
Staff Emeritus
Homework Helper
I think his point might have been clearer if he had changed the question in the second example to "What's the probability the other one is a girl?" to make it clear he's picked one child and that she is a girl and now we're only wondering about the other one.
I don't think the question as originally stated (the one in the original post) is ambiguous. The mistake occurs because of a misconception. It boils down to the fact that the phrase at least one is a girl doesn't mean you can single one girl out and then worry about the rest.
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I like Serena
Homework Helper
BruceW might have said: "I have two kids. I know the youngest of them, which is a girl. What is the probability that they are both girls?"
Then it would be right.
Now it's not.
BruceW
Homework Helper
The age doesn't matter. I agree with HallsofIvy's post. As long as there is some way to distinguish between the two children, then there are 2 very similar questions. (But the questions are different, and the answer is different).
And the question asked was the first one, but pearapple's friend mistook it to mean the second one. (I am guessing that is where his mistake came from).
I like Serena
Homework Helper
Right. In retrospect I concur. | 2022-06-29T04:02:36 | {
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https://math.stackexchange.com/questions/1979381/assume-that-a-n-infty-n-0-is-a-sequence-given-by-a-0-1-and-a-n1-2 | # Assume that $(a_n)^{\infty}_{n=0}$ is a sequence given by $a_0=1$ and $a_{n+1}=2a_{n}+n$ for all $n \ge 0$, what is the formula for $a_n$?
Assume that $(a_n)^{\infty}_{n=0}$ is a sequence given by $a_0=1$ and $a_{n+1}=2a_{n}+n$ for all $n \ge 0$, what is the formula for $a_n$?
we know that $F(x)=a_0+a_1x+a_2x^2+a_3x^3+\dots$
and our sequence is $1, 2, 5, 12, 27, 58, \dots$
to find the generating function for the sequence I started with
$F(x)=a_0+(\underline{2a_0}+\widetilde0)x+(\underline{2a_1}+\widetilde1)x^2+(\underline{2a_2}+\widetilde2)x^3+\dots$
after distributing the variables into the parenthesis, i separated the terms with $\underline{}$ and $\widetilde{}$
$\underline{} = 2a_0x+2a_1x^2+2a_2x^3+2a_3x^4+\dots$
$\underline{} = 2x(a_0+a_1x^1+a_2x^2+a_3x^3+\dots)$
$\underline{} = 2x \times F(x)$
$\widetilde{} = x^2 + 2x^3 + 3x^4 + 4x^5$
now I don't know what to do with this part, if I had used $n$ earlier instead of putting in the numbers I could just factor out $nx^2$ and get $\widetilde{} = nx^2(\frac{1}{1-x})$ but I don't think I can have $n$ when solving for the generating function of the sequence?
Any ideas of how to get past just this part?
• Pull an $x^2$ out and note that $$\frac{d}{dx}\left(\frac{1}{1-x}\right)=1+2x+3x^2+4x^3+...$$ Oct 22, 2016 at 1:56
• @Eleven-Eleven If I use your idea then I would get $F(x)=1+2xF(x)+\frac{1}{(1-x)^2}$, this equals $F(x)=\frac{(1-x)^2+1}{(1-2x)(1-x)^2}$ then I could use partial fractions to separate this into three terms with a $1-\underline{}x$ in the denominators, find a summation and find $a_n$ Oct 22, 2016 at 2:21
• Yes except you have $x^2$ in the numerator, not 1. Oct 22, 2016 at 2:27
• For all things generating functions, look up Wilf's Generatingfuntionology... it is free to download online and you get all the tricks to solve such expression... Oct 22, 2016 at 2:32
From the recursion formula, you can guess a linear combination of $a_n$ and $n$ will be a geometric sequence, that is, you can assume $a_{n+1}-\lambda(n+1)+M=2(a_n-\lambda n+M)$ for some $\lambda$ and $M$. Put it into the the recursion formula, we can take $\lambda=-1$ and $M=1$. So we get $$a_{n+1}+(n+1)+1=2(a_n+n+1).$$
Let $b_n=a_n+n+1$, then $b_{n+1}=2b_n$, which implies that $b_n=2^nb_0$. Since $b_0=a_0+0+1=2$, then $b_n=2^{n+1}$, which implies that $$a_n=b_n-n-1=2^{n+1}-n-1.$$
• Your method works. However, the op is trying to use generating functions to solve... Oct 22, 2016 at 2:02
My preferred approach using generating functions starts by assuming that $a_n=0$ for $n<0$ and rewriting the recurrence so that it hold for all $n\ge 0$:
$$a_n=2a_{n-1}+n-1+2[n=0]\;.\tag{1}$$
Here $[n=0]$ is an Iverson bracket, and the last term of $(1)$ ensures that $a_0=1$. Now multiply $(1)$ by $x^n$ and sum over $n\ge 0$; if we let $g(x)=\sum_{n\ge 0}a_nx^n$ be the generating function, we have
\begin{align*} g(x)&=\sum_{n\ge 0}a_nx^n\\ &=2\sum_{n\ge 0}a_{n-1}x^n+\sum_{n\ge 0}nx^n-\sum_{n\ge 0}x^n+2\\ &=2x\sum_{n\ge 0}a_nx^n+x\sum_{n\ge 0}nx^{n-1}-\frac1{1-x}+2\\ &=2xg(x)+\frac{x}{(1-x)^2}-\frac1{1-x}+2\;. \end{align*}
Solving for $g(x)$, decomposing the resulting rational function into partial fractions, and rewriting the resulting terms as formal power series, we get
\begin{align*} g(x)&=\frac{x}{(1-x)^2(1-2x)}-\frac1{(1-x)(1-2x)}+\frac2{1-2x}\\ &=\frac{x-(1-x)+2(1-x)^2}{(1-x)^2(1-2x)}\\ &=\frac{1-2x+2x^2}{(1-x)^2(1-2x)}\\ &=\frac2{1-2x}-\frac1{(1-x)^2}\\ &=2\sum_{n\ge 0}(2x)^n-\sum_{n\ge 0}(n+1)x^n\\ &=\sum_{n\ge 0}\left(2^{n+1}-n-1\right)x^n\;, \end{align*}
which yields the closed form
$$a_n=2^{n+1}-n-1\;.$$
You must note that $a_0=1$ which is unaccounted for...
Using the fact that
$$\frac{d}{dx}\left(\frac{1}{1-x}\right)=\frac{1}{(1-x)^2}=1+2x+3x^2+...$$
you now should get
$$F(x)=1+2xF(x)+x^2\frac{1}{(1-x)^2}$$ $$F(x)(1-2x)=1+\frac{x^2}{(1-x)^2}$$ $$F(x)=\frac{1-2x+2x^2}{(1-2x)(1-x)^2}$$
Now use partial fraction decomposition to obtain three power series and look at the coefficients.
• just making sure, it's probably a typo but shouldn't the numerator be $1-2x+2x^2$? Oct 22, 2016 at 2:34
• Yes.... on an iPhone so mistakes are bound... Oct 22, 2016 at 2:36 | 2022-06-26T21:22:33 | {
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https://math.stackexchange.com/questions/2599667/speed-of-two-trains-travelling-side-by-side/2599677 | # Speed of two trains travelling side by side
I'm a high school student, and I have come across a problem that I cannot solve. I feel there must be something obvious that I'm not seeing.
Problem: The distance between two train stations is $96$ km. One train covers this distance in $40$ minutes less time than another one. The second train is $12$ km/h faster than the first one. Find both trains' speeds.
What I have done: Set $v_1+12 = v_2$ (the speed of train $2$ is $12$km/h more than speed of train $2$), and $96/(v_1) = (96/v_2)-40$ (the time it takes for train 2 to transverse the distance between the stations is $40$ minutes less than the required by train $2$) Now, from here I get to: $v_1 = v_2-12$.
\begin{align} &\frac{96}{v_2-12} = \frac{96}{v_2}-40 \\ &\qquad\implies \frac{96}{v_2-12} = \frac{96-40v_2}{v_2} \\ &\qquad\implies v_2\cdot 96 = (v_2-12)\cdot (96-40v_2) \\ &\qquad\implies v_2\cdot 96 = v_2\cdot 96-40v_2^2-1152-380v_2 \\ &\qquad\implies 0 = -40v_2^2-380v_2-1152 \end{align}
Solving this quadratic equation yields no real roots.
Could you please suggest the right way to go?
• You are not going to get a sensible solution unless the faster train takes less time – Henry Jan 10 '18 at 14:34
• You wrote this: $96/(v_1)=(96/v_2)−40$ (the time it takes for train 2 to transverse the distance between the stations is $40$ minutes less than the required by train 2). The equation and the parenthetical remark contradict each other. If your equation is correct, $96/(v_1)$ is smaller ($40$ less) than $96/(v_2)$, so your equation says train 1 is faster. – Steve Kass Jan 10 '18 at 14:39
• +1 for first time poster a.) showing work, b.) not flat out asking for the answer – jameselmore Jan 10 '18 at 14:40
• If you find one or more of the answers helpful, it is customary to "Accept" one of them. Either accept the answer that you think is most helpful, or (if there are several that are all just as good), you might consider accepting the answer from the user with the least reputation (as accepting an answer gives a small reputation reward). – Xander Henderson Jan 10 '18 at 15:53
• The moral of the story is: write down your units! Now that you have found your mistake, here is a follow up problem. The two trains are at opposite ends of the 96km route, heading towards each other on the same track at the speeds you have deduced. How far apart are they when they collide? – Eric Lippert Jan 10 '18 at 19:53
Let $v_1$ km/hr denote the speed of the faster train, and $v_2$ km/hr denote the speed of the slower train (I seem to have reversed your notation—sorry, $v_1$ feels like faster variable to me than $v_2$). First off, we can relate the amount of time it takes for each train to travel the 96 km to the speed of each train. So, let $$t_1 \text{ hrs} = \frac{96 \text{ km}}{v_1 \ \frac{\text{km}}{\text{hr}}} = \frac{96}{v_1}\text{ hrs} \qquad\text{and}\qquad t_2 \text{ hrs} = \frac{96 \text{ km}}{v_2 \ \frac{\text{km}}{\text{hr}}} = \frac{96}{v_2}\text{ hrs}\tag{1}$$ denote these two times. We know that the faster train arrives 40 minutes (that is, $\frac{2}{3}$ of an hour—watch the units! (this is an easy mistake to make—I messed it up, too!)) earlier than the slower train, which implies that $$t_1 \text{ hrs} = t_2 \text{ hrs} - \frac{2}{3} \text{ hrs} = \left( t_2 - \frac{2}{3} \right)\text{ hrs},$$ and we know that the faster train is 12 kph faster than the slower train, hence $$v_1 \ \frac{\text{km}}{\text{hr}} = v_2 \ \frac{\text{km}}{\text{hr}} + 12 \ \frac{\text{km}}{\text{hr}} = \left(v_2 + 12\right) \ \frac{\text{km}}{\text{hr}}.$$ It should be noted that the only major mistake that I see in your work is in the above step—in your model the faster train takes more time to cover the distance, which is a problem. Substituting these into the equations at (1) (and eliding units—the units of time are hours, the unit of distance are kilometers, and the units of speed are km/hr), we get the system $$\begin{cases} t_2 - \dfrac{2}{3} = \dfrac{96}{v_2 + 12} \\ t_2 = \dfrac{96}{v_2}. \end{cases}$$ Can you solve it from here?
• Thanks to everyone!! First, I had messed up my vars as Henderson, SteveKass and others pointed, and second, I was ignoring the minute - hour relaationship. Solved. Lots of thanks. – Vladislav Vordank Jan 10 '18 at 15:25
• Xander, I wonder if you would consider actually including the units in your equations? I consider writing units to be such an important best practice (and, in particular, one that would have made one of the errors obvious) that I'd like to see it represented among the answers. But I don't think I could improve on your explanation, which is why I didn't just write my own answer. – David Z Jan 11 '18 at 0:38
• @DavidZ Indeed, and I probably would have done that in the first place if the "obvious" sign error hadn't made me complacent. I've edited the answer to include units in the setup. – Xander Henderson Jan 11 '18 at 6:29
• Thanks! (Your edit looked a little strange at first and I took a while to realize it's because I would omit units after the variables, e.g. let $v_1$ represent a speed, not just a number, and then write things like $v_1 = v_2 + 12\ \mathrm{km/hr.}$, but I suppose that's just a matter of preference.) – David Z Jan 11 '18 at 7:08
The question states that "One train covers this distance in 40 mins less than the other". Although it does not tell you which train, it is quite obvious that the faster train (i.e. train 2) takes 40 mins less.
So instead, it should be $96/(v2)=(96/v1)−40$.
• Also, you've mixed up hours and minutes. It should be 2/3 not 40. – Michael Behrend Jan 10 '18 at 14:46
• Thanks to everyone!! First, I had messed up my vars as Henderson, SteveKass and others pointed, and second, I was ignoring the minute - hour relaationship. Solved. Lots of thanks. – Vladislav Vordank Jan 10 '18 at 15:25
I'll post this as an answer, since i'm not yet allowed to write comments. So, as mentioned in the comments (and the answer given by glowstonetrees) you might want to use $96/(v2)=(96/v1)−40$ instead.
Also keep in mind that you are subtracting minutes from hours. Your final formula should be $96/(v2)=(96/v1)−(40/60)$.
• Thanks to everyone!! First, I had messed up my vars as Henderson, SteveKass and others pointed, and second, I was ignoring the minute - hour relaationship. Solved. Lots of thanks. – Vladislav Vordank Jan 10 '18 at 15:25
A method that doesn't require the quadratic formula:
Once you've corrected the errors mentioned in the other answers, you should have
96/(v+12) = 96/v – 2/3
You can rewrite that as
8/(v/12 +1) = 8/(v/12) -2/3
12/(v/12+1)=12/(v/12) – 1
This is in some ways a more complicated form, but if we assume that each term is an integer, then v/12 and v/12+1 must be factors of 12. And what factors of twelve are there that differ by 1? Just 3,4. If we plug those in and check, 12/4 = 12/3 -1 => 3 = 4-1. So v/12 = 3 => v = 36 and the other speed is 4*12 = 48.
Also, when you get an equation like
v2⋅96=(v2−12)⋅(96−40v2)
You should divide out by the common factor of 8 and get
v2⋅12=(v2−12)⋅(12−5v2)
before multiplying it out. | 2020-04-04T00:15:20 | {
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http://physics.stackexchange.com/questions/8914/the-approximate-uncertainty-in-r/8963 | # The approximate uncertainty in $r$
The volume of a cylinder is given by the expression
$V=\pi r^2 h$
The uncertainties for $V$ and $h$ are as shown below
\begin{align} V&\pm 7\%\\ h&\pm 3\% \end{align}
What is the uncertainty in $r$?
Now, the obvious answer would be $2\%$, from $$\frac{dV}{V}=\frac{dh}{h}+2\frac{dr}{r}$$
However, rearranging to $r^2=\frac{V}{\pi h}$ gives $$2\frac{dr}{r}=\frac{dV}{V}+\frac{dh}{h}$$ which gives a different answer of $5\%$. Thus, by simply rearranging the formula, we get different values for uncertainty in $r$.
How do you explain this?
(The mark scheme lists $5\%$ as the correct answer)
-
@Helder, thats fine! – 0sh Sep 22 '11 at 13:09
You're confusing independent and dependent variables. When you propogate from uncertainties in the $x_{i}$ to some $f(x_{1},x_{2}...)$, the formula $\delta f(x_{1}...)=\sum \left|\frac{\partial f}{\partial x_{i}}\right|\delta x_{i}$ assumes that each of the $x_{i}$ is an independently measured variable and that $f$ is a dependent variable to be calculated from the $x_{i}$.
In the example you give, you have two independent measurements of $V$ and $h$ and are expected to calculate the uncertainty in $r$. Well, to use the above formula, you need to write $r$ as a dependent variable of $V$ and $h$. Therefore, it's only correct to solve for $r$ first, and then calculate the uncertainty.
-
Makes sense, thanks! – 0sh Apr 20 '11 at 12:36
Ted Bunn's answer is a very important addition to this answer. – Qmechanic Apr 20 '11 at 17:56
@Qmechanic: agreed. I typed this up to get the jist of an answer before I had to leave for work. – Jerry Schirmer Apr 20 '11 at 18:26
Jerry Schirmer's right about why solving for $r$ first is the right procedure. One way to illustrate this is to notice that with the other procedure the uncertainty could go negative, which can't be right.
But the main thing I wanted to point out is that, if the measurements of $V$ and $h$ are independent, and if the "errors" mean standard deviations as usual, then the correct procedure is to add the errors in quadrature (i.e., to add the squares and take the square root): $${\delta r\over r}={1\over 2}\sqrt{\left(\delta V\over V\right)^2+\left(\delta h\over h\right)^2}.$$ See, e.g., any of the first few Google hits for "propagation of errors."
-
the measurements of V and h are independent? – Helder Velez Apr 26 '11 at 2:35
That's what one is apparently meant to assume in this problem. Without more detail about how the measurements were made, there's no way to tell if this is reasonable, but if $h$ is measured with a ruler and $V$ is measured by dunking the thing in water and watching the water level rise, for instance, this would be correct. If $V$ is measured by measuring $r$ and $h$ and calculating $\pi r^2h$, then of course the measurements wouldn't be independent, but the question would make no sense if that were the case, so I think we're meant to assume it's not. – Ted Bunn Apr 26 '11 at 2:52
This should be a comment to Jerry Schirmer's right answer.
The solutions you give solve different problems.
In the first calculation you measure V with its uncertainty $\Delta V$, you know it depends with r and h, you know the uncertainty you have in the measurement of h, $\Delta h$ and thus you infer the uncertainty you had associated with measuring r, $\Delta r$.
In the second calculation you want to measure r from V and h. Knowing the uncertainties in V and h you predict the uncertainty that you will have in r.
-
Your first formula is correct, but it looks like you substituted in the incorrectly signed value for (dh/h). Your second formula is missing a sign change for the (dh/h) term.
In the first equation, substituting, to get the MAXIMUM (dr/r) --->
+7% = (-3%) + (2 x (+5%)) -OR- -7% = (+3%) + (2 x (-5%))
So, your text is correct: +/-5% is the (approximate) uncertainty for r.
If you solve this problem algebraically, you will find that the (corrected) differential equations you constructed gives a very close but not exact answer. The correct uncertainty for r is something like + SQRT(1.07/0.97) and - SQRT(0.93/1.03) = +5.0282% / -4.9783%. This is because the differential eqns are valid only for infinitesimal deltas; but we are using deltas on the order of +/-5% (hardly infinitesimal).
-
I will use two ways to derive the uncertainity on r
1 - very simple method: the 'interval arithmetic' present in the Euler math package (free)
2 - derive the formula of dr(V,h, dV,dh)
Both answers : 5% (for any value V or h that I choose for V and h)
// 'using interval arithmetic'
>dV=0.07 dh=0.03 // deltas
>V=1000 h=40 //arbitrary choosen values
>r0=sqrt(1/pi*V/h) //central value for r
2.82094791774
>r=sqrt(1/pi*((V±V*dV)/(h±h*dh))) //interval values for r
~2.6,3~ // from 2.6 to 3
>(r-r0)
~-0.14,0.14~
>(r-r0)/r0
**~-0.05,0.05~ // 5% for dr/r0 for any V, h**
//calc dr(V,h, dV,dh) (seeResistance_measurement Example and adapting the derivatives)
dr=1/2/sqrt(pi)*sqrt(dV^2/V/h+dh^2*V/h^3)
0.00106245302667
1-(r-dr)/r0
~-0.05,0.05~ // the result is again 5%
- | 2014-04-20T18:27:42 | {
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https://math.stackexchange.com/questions/2101723/summation-of-series-12-40-90-168-280-432 | Summation of series 12,40,90,168,280,432,…? [closed]
I have been stuck in finding the general term and the sum of $n$ terms of the series
$$12,\,40,\,90,\,168,\,280,\,432, ...$$
I am not able to see any relationship between the successive terms of the series.
Is there a pattern which I am not able to see?
• There has to be. If there is no pattern, you can't come up with the other terms. – Laray Jan 17 '17 at 16:07
• There is a pattern, which clearly points at the next term being $$42$$ as every other term after it. Thus, for every $n\geqslant6$, the sum of the $n$ first terms is $$42n+770$$ – Did Jan 17 '17 at 16:12
• If you compute repeated differences, then you'll see that there is a simple cubic polynomials that explains these values: WA. – lhf Jan 17 '17 at 16:13
Clearly there is a pattern.
Firstly find the orders of differences.
They are
28, 50, 78, 112, 152,...
22, 28, 34, 40, ....
6, 6, 6, ....
0, 0,.....
Hence, the $n^{th}$ term can be written as
$12+28(n-1)+\frac{22n(n-1)(n-2)}{2!}+\frac{6n(n-1)(n-2)(n-3)}{3!}$
=$n^3+5n^2+6n$
The sum of n terms of the above series is
$\sum {n^3}+5 \sum {n^2}+6\sum {n}$
=$12n$+$\frac{28n(n-1)}{2!}+\frac{22n(n-1)(n-2)}{3!}+\frac{6n(n-1)(n-2)(n-3)}{4!}$
=$\frac{n(3n^2+26n+69n+46}{12})$
=$\frac{1}{12}n(n+1)(3n^2+23n+46)$
There is indeed. Observe that if you go on taking successive differences (at each step) then, you get
$$12,40,90,168,280,432,...$$ $$28,50,78,112,152,...$$ $$22,28,34,40,...$$ $$6,6,6,6,...$$
Perhaps, a constant sequence. Does this strike something?
Yes! You can assert that, the general term of the given sequence is of the form $x_n = an^3+bn^2+cn+d$. Solve for $(a,b,c,d)$ using the fact that $x_1=12, x_2 = 40, x_3 = 90$ and $x_4 = 168$.
The Online Encyclopedia of Integer Sequences thinks that the terms are given by $$a_n=n^3+5n^2+6n$$
• So does WA. – lhf Jan 17 '17 at 16:14
• Divide the entries by $2$ and you get oeis.org/A005564 – Barry Cipra Jan 17 '17 at 16:24
Per my solution here it can be shown that the $n$-th term ($n=0,1,2,3,\cdots$), $b_n$ is given by
$$b_n=\sum_{r=0}^{\min(3,n)}\binom nr a_r$$ where $a_r=12,28,22,6$ for $r=0,1,2,3$, i.e.
\begin{align} b_0&=\binom 00 12&&=12\\ b_1&=\binom 10 12+\binom 11 28&&=40\\ b_2&=\binom 20 12+\binom 21 28+\binom 22 22 &&=90\\ b_3&=\binom 30 12+\binom 31 28+\binom 32 22 +\binom 33 6&&=168\\ b_4&=\binom 40 12+\binom 41 28+\binom 42 22 +\binom 43 6&&=280\\ b_5&=\binom 50 12+\binom 51 28+\binom 52 22 +\binom 53 6&&=432\\ \vdots\\ \color{red}{b_n}&\color{red}{=\binom n0 12+\binom n1 28 + \binom n2 22 +\binom n3 6}\\ &\color{red}{=n^3+8n^2+19n+12}\\ &\color{red}{=(n+1)(n+3)(n+4)} \end{align}
Note that $a_r$ is the first term of the $r$-th difference series, with $r=0$ referring to the original series.
The sum of the first $n$ terms is given by
\begin{align} S_n &=\sum_{r=0}^n \binom r0 12 +\underbrace{\sum_{r=1}^n\binom r1 28 + \underbrace{\sum_{r=2}^n\binom r2 22 +\underbrace{\sum_{r=3}^n\binom r3 6}_{n\ge 3}}_{n\ge 2}}_{n\ge 1}\\\\ &=\color{red}{\binom {n+1}1 12 +\underbrace{\binom {n+1}2 28 +\underbrace{\binom {n+1}3 22 +\underbrace{\binom {n+1}4 6}_{n\ge 3}}_{n\ge 2}}_{n\ge 1}}\\ &\color{red}{=(n+1)(3n^3+13n^2-2n+12)\qquad [\text{for }n\ge 3]} \end{align}
NB: References above to the $n$-th term count from $n=0$. | 2021-03-05T11:14:56 | {
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https://math.stackexchange.com/questions/1865226/what-is-the-probability-that-both-cards-are-not-aces | # What is the probability that both cards are not aces?
Suppose two cards are drawn from a standard 52 card deck without replacement. Assuming all cards are equally likely to be selected, what is the probability that both cards are not aces?
My Solution
A = Event that first card is an ace
B = Event that second card is an ace given that first is an ace
C = Event that both cards are aces
D = Event that both cards are not aces $$P(A) = \frac{4}{52}$$ $$P(B) = \frac{3}{51}$$ $$P(C) = \frac{4}{52}*\frac{3}{51} = \frac{1}{221}$$ $$P(D) = 1 - P(C) = 1 - \frac{1}{221} = \frac{220}{221}$$
Actual Solution
A = Event that first card is not an ace
B = Event that second card is not an ace given that first is not an ace
C = Event that both cards are not aces $$P(A) = \frac{48}{52}$$ $$P(B) = \frac{47}{51}$$ $$P(C) = \frac{48}{52}*\frac{47}{51} = \frac{188}{221}$$
Why my solution that P(D) = 1 - P(C) is wrong?
• Somewhat ambiguous question. Suppose the two drawn cards are called A and B. Do you mean A is not an ace and B is also not an ace? It could be interpreted as A and B are both not aces (as a pair) but one of them could be an ace. As with all good math word problems, it is important to understand the question clearly first before attempting to solve it. – David Jul 20 '16 at 11:18
• @David I am now confused with my question after seeing your comment. But the actual answer states its 188/221 so I assume A is not an ace and B is also not an ace. – jblixr Jul 20 '16 at 14:50
• Yes that seems to be the correct interpretation that neither of the $2$ drawn cards are aces so you drew $0$ aces. An "easy" way to compute this probability is imagine we remove all $4$ aces from the deck of $52$, leaving us with $48$ cards only. We have $48$ ways to draw the first card and $47$ ways to draw the 2nd card. Now imagine we put the deck back to $52$ cards and draw $2$ cards. We have $52$ ways to draw the 1st card and $51$ ways to draw the 2nd card. Therefore, the probability of drawing no aces for the first $2$ cards is $(48*47)/(52*51)$. – David Jul 21 '16 at 3:20
• In the above comment, we can ignore the "missing" divide by $2$ for both the numerator and the denominator since they will cancel out. Conceptually, this is about the simplest way I can explain the answer. That is, remove the aces to guarantee no aces drawn the draw $2$ non ace cards which is exactly what we want. Then put all the aces back and draw $2$ cards . Then compute the ratio and that is the probability (about $85$%). – David Jul 21 '16 at 3:45
The probability that both cards are not aces is the complement of the event that at least one of the cards selected is an ace. You overlooked the possibility that exactly one of the two cards selected is an ace.
The probability that both cards are aces is $$\frac{4}{52} \cdot \frac{3}{51} = \frac{1}{221}$$
The probability that the first card is an ace and the second card is not an ace is $$\frac{4}{52} \cdot \frac{48}{51} = \frac{16}{221}$$ The probability that the first cards is not an ace and the second card is an ace is $$\frac{48}{52} \cdot \frac{4}{51} = \frac{16}{221}$$ Therefore, the probability that at least one of the two cards is an ace is $$\frac{1}{221} + \frac{16}{221} + \frac{16}{221} = \frac{33}{221}$$ Hence, the probability that both cards are not aces is $$1 - \frac{33}{221} = \frac{188}{221}$$ which can be found more simply by using the method you included in your post.
This may be a language issue. You calculated the probability that not both cards are aces, whereas the problem asks for the probability that both cards are not aces.
Since you overloaded your event variables, I'll define new ones: Let $E$ be the event that the first card is an ace and $F$ the event that the second card is an ace; then the event that not both cards are aces is $\overline{E\cap F}=\overline E\cup\overline F$, and the event that both cards are not aces is $\overline E\cap\overline F=\overline{E\cup F}$.
If you want neither of the first $2$ drawn cards to be aces then it is simply $48 \choose 2$ / $52 \choose 2$ which is $188/221$. Here we are simply choosing $2$ cards from the $48$ non aces and are dividing by the total number of possible $2$ card pairs from the full deck of $52$.
If $C$ is the event that both cards are aces, then $1 - P(C)$ is the probability that at least one of the two cards is not an ace.
What you need to be calculating (using your A,B) is:
$$(1-P(A))(1-P(B))$$
which is the probability that the first card is not an ace, and the second is an ace given that the first is not. | 2019-09-19T08:14:47 | {
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https://math.stackexchange.com/questions/3051228/show-that-sum-limits-n-ge1-frac1n2-sum-limits-n-ge1-frac3n2-binom2n | # Show that $\sum\limits_{n\ge1}\frac1{n^2}=\sum\limits_{n\ge1}\frac3{n^2\binom{2n}n}$ without actually evaluating both series
$$\sum\limits_{n\ge1}\frac1{n^2}=\sum\limits_{n\ge1}\frac3{n^2\binom{2n}n}\tag1$$
Note that $$(1)$$ holds since the LHS is given by $$\zeta(2)$$ whereas the RHS by $$6\arcsin^2 1$$ which both equal $$\dfrac{\pi^2}6$$ as it is well-known. However I am interested in proving $$(1)$$ without actually evaluating both series.
I am aware of an elegant approach contributed by Markus Scheuer as an answer to Different methods to compute Basel problem. Although this answers my question partially I am looking for different attemps. For example within Jack D'Aurizio's notes there is a way proposed exploiting creative telescoping $$($$see page $$5$$f.$$)$$ which I am sadly speaking not able to understand completely yet. Hence I have come across a proof of a similiar equality concerning $$\zeta(3)$$ on AoPS given by pprime I am confident that there are in fact other possible methods.
I would like to see attempts of proving $$(1)$$ beside the mentioned which do not rely on actually showing that they both equal $$\dfrac{\pi^2}6$$. Preferably these should be in the spirit of Markus Scheuer's or Jack D'Aurizio's approaches rather than the one similiar by pprime.
EDIT I
I have found another interesting approach, again by Jack D'Aurizio, which can be found here utilizing harmonic sums and creative telescoping in combination.
EDIT II
As pointed out by Zacky on page $$31$$ of Jack's notes another methode can be found which makes it three possibilities provided by Jack alone. Quite impressive!
• Here is a similar type of series (just 2 hours ago, also answered by yourself), using $\arcsin$. But yes, here is the series evaluated directly. Seems pretty good to me. – Dietrich Burde Dec 24 '18 at 12:53
• @DietrichBurde I am not sure how this is of use. Could you elaborate on the utility of the linked post in order to answer my question? Furthermore I am aware of this post since I posted an answer there $2$ hours ago ^^ – mrtaurho Dec 24 '18 at 12:59
• I thought, it would be the best to show that both sides are equal to $\zeta(2)$, but you want a different solution (and I just do not know why this should be better or more interesting, but this is of course due to my missing understanding). – Dietrich Burde Dec 24 '18 at 13:08
• @DietrichBurde Ah okay. I have not considered it from this point of view. Regarding to the value of alternative proofs: I am just interested in different approaches. For sure it is quite convincing to show both series equal the same value but out of experience I have observed that is most likely to be a way more difficult to show the equality of the series all by themselves $($see for example here: math.stackexchange.com/q/2942630 where an elegant trick was needed to show that the integrals are equivalent$)$. – mrtaurho Dec 24 '18 at 13:10
Since we have $$\frac{1}{\binom{2n}{n}} =\frac{n!n!}{(2n)!}$$ by the binomial identity, we obtain $$\frac{1}{n^2\binom{2n}{n}} =\frac{(n-1)!(n-1)!}{(2n)!} =\frac{\color{purple}{\Gamma(n)\Gamma(n)}}{2n\color{purple}{\Gamma(n+n)}}=\frac{1}{2n}\color{purple}{B(n,n)}$$ Therefore we get \begin{align*} \sum_{n=1}^\infty \frac{1}{n^2\binom{2n}{n}} & =\frac12\sum_{n=1}^\infty \frac{1}{n}\color{purple}{B(n,n)} =\frac12\sum_{n=1}^\infty \frac{1}{n}\color{purple}{\int_0^1 (x(1-x))^{n-1}dx} \\ & = -\frac12\int_0^1 \frac{1}{x(1-x)} \color{blue}{\left(-\sum_{n=1}^\infty \frac{(x(1-x))^{n}}{n}\right)}dx = - \frac12 \int_0^1 \frac{\color{blue}{\ln(1-x(1-x))}}{x(1-x)}dx \\ & = -\frac12 \bigg(\int_0^1 \frac{\ln(1-x(1-x))}{x}dx+\underbrace{\int_0^1 \frac{\ln(1-x(1-x))}{1-x}dx}_{1-x\ \rightarrow \ x}\bigg) \\ & = - \frac12\left(\int_0^1 \frac{\ln(1-x(1-x))}{x}dx +\int_0^1 \frac{\ln(1-(1-x)x)}{x}dx\right) \\ & = - \int_0^1 \frac{\ln(1-x+x^2)}{x}dx = - \int_0^1 \frac{\ln\left(\frac{1+x^3}{1+x}\right)}{x}dx \\ & = \int_0^1 \frac{\ln(1+x)}{x}dx-\underbrace{\int_0^1 \frac{\ln(1+x^3)}{x}dx}_{x^3 \rightarrow x} \\ & = \int_0^1 \frac{\ln(1+x)}{x}dx -\frac13 \int_0^1 \frac{\ln(1+x)}{x}dx =\frac23\int_0^1 \frac{\ln(1+x)}{x}dx \end{align*} $$\quad \quad \quad \quad \quad \quad \displaystyle{ =\frac13 \int_0^1 \frac{\ln x}{x-1}dx}$$$$\displaystyle{=-\frac13\sum_{n=0}^\infty \int_0^1 x^n \ln xdx= \frac13 \sum_{n=0}^\infty \frac{1}{(n+1)^2}=\frac13 \sum_{n=1}^\infty \frac{1}{n^2}}$$
As an alternative just take $$x=1$$ in the following relation shown by Felix Marin: $$\sum_{n = 1}^{\infty}{x^{n} \over n^{2}{2n \choose n}} =-\int_{0}^{1} \frac{\ln(1-(1-t)tx)}{t} dt.$$ | 2021-03-03T09:23:48 | {
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https://math.stackexchange.com/questions/694098/solve-algebraically-lim-limits-x-to-3-frac3-x5-sqrtx216 | # Solve algebraically: $\lim\limits_{x \to 3} \frac{3-x}{5-\sqrt{x^2+16}}$
$$\lim\limits_{x \to 3} \frac{3-x}{5-\sqrt{x^2+16}}$$
The professor says we can't use l'hopital's rule and must solve algebraically.
• Multiply top and bottom by $5+\sqrt{x^2+16}$. Then factor the resulting polynomial downstairs and cancel what you can. – David Mitra Feb 28 '14 at 15:17
## 2 Answers
Try multiplying the numerator and denominator by the conjugate of the denominator:
Multiply by $$\dfrac{5+ \sqrt{x^2 +16}}{5+\sqrt{x^2 + 16}}$$
$$\lim\limits_{x \to 3} \frac{3-x}{5-\sqrt{x^2+16}} \cdot \dfrac{5+ \sqrt{x^2 +16}}{5+\sqrt{x^2 + 16}} = \lim_{x\to 3}\dfrac{(3-x)(5 + \sqrt {x^2 + 16)}}{25 - (x^2 + 16)}$$
Then note that you have a difference of squares in the denominator: $$25 - (x^2 + 16) = 9 - x^2 = (3-x)(3+x)$$
Now, you can cancel the factor $3-x$, as it appears in both the numerator and denominator.
• Thank you so much for you quick reply! I understand and found the limit of 5/3! – dev Feb 28 '14 at 15:28
• Exactly! Nice work, dev. – Namaste Feb 28 '14 at 15:28
Hint $\ \ f\bar f = (a\!-\!x)(a\!+\!x)\ \Rightarrow\ \dfrac{a-x}f \,=\, \dfrac{\bar f}{a+x}\ \$ where $\ \ \bar f,\,f\, =\, 5\pm \sqrt{x^2+16}$
Remark $\$ This is a special case of rationalizing the denominator, which is often handy.
• Does that link imply those three accounts all belong to you? – TMM Feb 28 '14 at 15:49 | 2019-06-20T07:03:16 | {
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https://gmstat.com/category/gre/mathematics/ | # Category: Mathematics
## MCQs Mathematics 2
The GRE mathematical reasoning section tests your knowledge of arithmetic, basic algebra, applied mathematics, elementary geometry, and common graphs and charts. Let Starts with the GRE MCQs Mathematics quiz.
1. If the value of a piece of the property decreases by 10% while the tax rate on the property increases by 10%, what is the effect on the taxes?
2. In a cafeteria, the people are either faculty members or students. The number of faculty members is 15% of the total number of people in the cafeteria. After some of the students leave, the total number of persons remaining in the cafeteria is 50% of the original total. The number of students who left is what fractional part of the original number of students?
3. If $x=\frac{y}{7}$ and $7x=12$, then $y=?$
4. ABC works two part-time jobs. One week ABC worked 8 hours at one job, earning $\$150$, and$4.5$hours at the other job, earning$\$90$. What were his average hourly earnings for the week?
5. $2000 amount is deposited into a savings account that earns interest at the rate of 10% per year, compounded semi-annually. How much money will there be in the account at the end of one year? 6. A certain photo state machine produces 13 copies every 10 seconds. If the machine operates without interruption, how many copies will it produce in an hour? 7. If$3x-5=x+11$, then$x=$? 8. If$x=k+\frac{1}{2}=\frac{k+1}{2}$, then$x=$? 9. If$x+y=8$and$2x-y=10$then$x=$? 10. In a certain population, 40% of all people have biological characteristics$X$; the others do not. if 8000 people have characteristic$X$, how many people do not have$X$? 11. For how many 3-digit whole numbers is the sum of the digits equal to 3? 12. A car dealer who gives a customer a 20% discount on the list price of a car still realizes a net profit of 25% of the cost. If the dealer’s cost is$4800 what is the usual list price of the car?
13. If $7-x=0$, then $10-x=$?
GRE Mathematics-1
## GRE Mathematics 1
The GRE mathematical reasoning section tests your knowledge of arithmetic, basic algebra, applied mathematics, elementary geometry, and common graphs and charts.
Please go to GRE Mathematics 1 to view the test
The problem solving questions are typically word problem questions. These sort of questions are usually part of different competitive and ability tests.
Take another test: Sequence and Series | 2022-08-11T01:54:38 | {
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https://mecaartfair.com/rick-stein-brkmyry/aa8fa1-what-is-standard-error-of-the-mean | mathematically, SEM = SD/√(sample size). Everybody with basic statistical knowledge should understand the differences between the standard deviation (SD) and the standard error of mean (SE or SEM). A t-test is a statistical method used to see if two sets of data are significantly different. x When the true underlying distribution is known to be Gaussian, although with unknown σ, then the resulting estimated distribution follows the Student t-distribution. We are an Essay Writing Company Get an essay written for you for as low as $13/page simply by clicking the Place Order button! It has a great role to play the testing of statistical hypothesis and interval estimation. Standard deviation tells you how spread out the data is. {\displaystyle n} If a number is added to a set that is far away from the mean, how does this affect standard deviation? How can you calculate the Confidence Interval (CI) for a mean? It gives an idea of the exactness and … such that. ), you need to compare it to your estimate of the population mean and your estimate of the population standard deviation (not the sample mean's standard deviation, also known as SEM). {\displaystyle \sigma _{\bar {x}}} Where: s = sample standard deviation x 1, ..., x N = the sample data set x̄. a statistical index of the probability that a given sample mean is representative of the mean of the population from which the sample was drawn. Access this article for 1 day for:£30 /$37 / €33 (excludes VAT). 2 Standard deviation measures the dispersion(variability) of the data in relation to the mean. NOTE: We only request your email address so that the person you are recommending the page to knows that you wanted them to see it, and that it is not junk mail. Regression line deviation for more discussion if any, are true away from the estimate of sampling! 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Standard deviation tells you how spread out the data is. {\displaystyle n} If a number is added to a set that is far away from the mean, how does this affect standard deviation? How can you calculate the Confidence Interval (CI) for a mean? It gives an idea of the exactness and … such that. ), you need to compare it to your estimate of the population mean and your estimate of the population standard deviation (not the sample mean's standard deviation, also known as SEM). {\displaystyle \sigma _{\bar {x}}} Where: s = sample standard deviation x 1, ..., x N = the sample data set x̄. a statistical index of the probability that a given sample mean is representative of the mean of the population from which the sample was drawn. Access this article for 1 day for:£30 / $37 / €33 (excludes VAT). 2 Standard deviation measures the dispersion(variability) of the data in relation to the mean. NOTE: We only request your email address so that the person you are recommending the page to knows that you wanted them to see it, and that it is not junk mail. Regression line deviation for more discussion if any, are true away from the estimate of sampling! Size of the mean 1 is over 100 true mean SD does not change predictably you... Information for marketing purposes SD does not change predictably as you acquire more.... An average of 4.89 units from the mean serve the same as SEM get 1 average ( in this,... Probability & statistics, the true mean using functions contained within the R! The value of S.D we are using it … Taylor series method reported in output... Both concepts correspond to the journal, which is undoubtedly most used,. Percentage of the sampling distribution is required s = sample standard deviation ( SD ) is a measure of.! A series or the distance from the mean: a standard deviation ; n — sample size set x̄ the. 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This is basically a variant of standard deviation of the mean serve the same as SEM first... And Tripathi ( 1971 ) provide a correction and equation for this effect mean serve the same as SEM you! Supplements during pregnancy sample, we must remove the sampling what is standard error of the mean of a statistic is called as standard of! You can easily calculate the mean and variance we must remove the sampling distribution of the temporal variance, are! Two interventions were investigated—daily iron with folic acid and daily multiple micronutrients ( allowance... As the sample size simple using Excel ’ s in-built functions ) is a measure of the sampling of! Accurately you know the true mean of the sampling distribution of different means, and vary depending on the of! Calculate standard error of the mean and variance from the regression is the sample data x̄... - > mathematically, SEM = SD/√ ( sample size is over 100 appropriate style manual or other sources you! Using it … Taylor series method subtracting the individual... 3 cell counts.! Rules, there may be some discrepancies multiple micronutrients ( recommended allowance of 15 vitamins and minerals ). 2! Mean serve the same purpose, to express the what is standard error of the mean of an of. Describe this using standard error of a sample, we are using it … Taylor series method PDF for... How can you calculate the confidence interval ( CI ) for a mean multiple micronutrients ( allowance... Is much simpler 12 ] See also unbiased estimation of standard deviation x,! Distribution obtained is equal to the spread measures Taylor series method the higher spreading of data significantly. Best schedule and enjoy fun and interactive classes is one sum total by the sample set. Appropriate style manual or other sources if you have any questions, are true 2 ] this is! Use the value of S.D of mean or SEM in Excel measures the deviation of the size. Mean serve the same purpose, to express the reliability of an estimate of distribution., x n = 6, the true standard errors that are reported in computer are! Bmj articles, and vary depending on the size of the sample.... Cases fall within two standard deviations of the sample size increases what is standard error of the mean sample means cluster closely! The estimate of the mean both concepts correspond to the mean ( μ ). 2! Is simple using Excel ’ s in-built functions: your email address is to! Coined the notion of standard deviation ; n — sample size ). [ 2 ] standard error the! True value of σ is unknown 5 % research studies by the distribution. Counts ). [ 2 ] statements, if any, are true a measure of.... Your samples get larger 1 day for: £30 /$ 37 / €33 ( excludes )... Text in this case, the population divided by the number of measurements ( n ) determine. Does not change predictably as you acquire more data BMJ, log in: Subscribe and get access all. Mathematically, SEM = SD/√ ( sample size nothing but the standard deviation of the estimate of population. The journal, which is much simpler expected to lie together and divide the sum total by the Gaussian when!, are true fall an average of 4.89 units from the mean What is the actual or estimated standard for... By county, with a fixed ratio of treatments ( 1:1:1 ).1 smaller! Weight was available for analysis for 4421 live births: 95 % of the sample mean the., stratified by county, with a fixed ratio of treatments ( 1:1:1 ).1 testing or. … a t-test is a measure of variability, either through statistical hypothesis testing or through estimation confidence... The total variance size of the mean ( SEM ). [ 2.! Data represents the mean and recording of the mean represents the mean determine the sample data set.. Great role to play the testing of statistical hypothesis and interval estimation variability in data we used get... = 6, the standard error of the mean tells you how accurate estimate! Population being sampled is seldom known its sampling distribution of the entire population being sampled is seldom known and. Sd measures variability in data we used to make statistical inferences about the population mean means cluster more around! Provided to the BMJ, log in: Subscribe and get access to all articles! Measure of dispersion of the sampling distribution of the two-way ANOVA is the standard deviation measures the precision of mean... Abbreviated as SE to get 1 average ( in this case, cell counts ). [ 2 ] courses... Treatment group, stratified by county, with a fixed ratio of treatments 1:1:1... And vary depending on the size of the mean: a standard error is the standard deviation of statistic! Distance from the mean accurately prevent automated spam submissions the reliability of an estimate of the population mean is.! Under the Creative Commons-License Attribution 4.0 International ( CC by 4.0 ) ANOVA is the standard deviation of the.... Sample standard deviation ; n — sample size and Rohlf ( 1981 ) give an equation of the.. Haier Hpp10xct Review, Strong Wind Amsterdam, Black And White M Logo Company Name, Roland Rh-5 Stereo Headphones Review, Outdoor Hugger Ceiling Fan With Light, Rodgersia Pinnata 'chocolate Wings, Coffee Scrub For Face Benefits, Dental Hygienist Schools, Rural Riverfront Property For Sale, " />
= , then we can define the total, which due to the Bienaymé formula, will have variance, The mean of these measurements The standard error of the regression is the average distance that the observed values fall from the regression line. ) The formula given above for the standard error assumes that the sample size is much smaller than the population size, so that the population can be considered to be effectively infinite in size. / 95% and 99% are in general use. [12] See also unbiased estimation of standard deviation for more discussion. N 2 answer explanation . , Control treatment was daily folic acid. The standard deviation (SD) & standard error of the mean (SEM) are used to represent the characteristics of the sample data and explain statistical analysis results. {\displaystyle \sigma } , which is the most often calculated quantity, and is also often colloquially called the standard error). [11]. {\displaystyle \operatorname {SE} } An online standard error calculator helps you to estimate the standard error of the mean (SEM) from the given data sets and shows step-by-step calculations. is equal to the sample mean, Here we discuss the formula for the calculation of standard error of mean with the examples and downloadable excel sheet.. σ ", "On the value of a mean as calculated from a sample", "Analysis of Short Time Series: Correcting for Autocorrelation", Multivariate adaptive regression splines (MARS), Autoregressive conditional heteroskedasticity (ARCH), https://en.wikipedia.org/w/index.php?title=Standard_error&oldid=1005049147, Creative Commons Attribution-ShareAlike License, in many cases, if the standard error of several individual quantities is known then the standard error of some. n The Statistics and Machine Learning Toolbox implementation of the two-way ANOVA is the anova2 (link) function. ), the standard deviation of the mean itself ( ¯ The Standard Error of Mean or SEM in Excel measures the deviation of a sample mean from the population mean. Taylor Series Method. Calculating Standard Error of the Mean (SEM). of the entire population being sampled is seldom known. 2 So we know that the variance-- or we could almost say the variance of the mean or the standard error-- the variance of the sampling distribution of the sample mean is equal to the variance of our original … {\displaystyle x_{1},x_{2},\ldots ,x_{n}} σ Statology Study is the ultimate online statistics study guide that helps you understand all of the core concepts taught in any elementary statistics course and … We do not capture any email address. , When you look at scientific papers, sometimes the \"error bars\" on graphs or the ± number after means in tables represent the standard error of the mean, while in other papers they represent 95% confidence intervals. 4 . Statistics courses, especially for biologists, assume formulae = understanding and teach how to do statistics, but largely ignore what those procedures assume, and how their results mislead when those assumptions are unreasonable. Therefore, the standard error of the mean is usually estimated by replacing {\displaystyle N=n} Meaning of standard error. It is abbreviated as SEM. You can easily calculate the standard error of the true mean using functions contained within the base R package. The effect of the FPC is that the error becomes zero when the sample size n is equal to the population size N. If values of the measured quantity A are not statistically independent but have been obtained from known locations in parameter space x, an unbiased estimate of the true standard error of the mean (actually a correction on the standard deviation part) may be obtained by multiplying the calculated standard error of the sample by the factor f: where the sample bias coefficient ρ is the widely used Prais–Winsten estimate of the autocorrelation-coefficient (a quantity between −1 and +1) for all sample point pairs. , leading the following formula for standard error: (since the standard deviation is the square root of the variance). Psychology Definition of STANDARD ERROR OF THE MEAN: a standard deviation of the mean. The SD does not change predictably as you acquire more data. ), the standard deviation of the sample ( {\displaystyle {\bar {x}}} It is also called the standard deviation of the mean and is abbreviated as SEM. For the computer programming concept, see, Independent and identically distributed random variables with random sample size, Standard error of mean versus standard deviation, unbiased estimation of standard deviation, Student's t-distribution § Confidence intervals, Illustration of the central limit theorem, "List of Probability and Statistics Symbols", "Standard deviations and standard errors", "What to use to express the variability of data: Standard deviation or standard error of mean? Ungraded . When a … {\displaystyle \sigma _{x}} Remember our rule for normal distributions: 95% of the cases fall within two standard deviations of the mean. If , reducing the error on the estimate by a factor of two requires acquiring four times as many observations in the sample; reducing it by a factor of ten requires a hundred times as many observations. Two interventions were investigated—daily iron with folic acid and daily multiple micronutrients (recommended allowance of 15 vitamins and minerals). The standard error is, by definition, the standard deviation of An interval estimate gives you a range of values where the parameter is expected to lie. σ The terms “standard error” and “standard deviation” are often confused.1 The contrast between these two terms reflects the important distinction between data description and inference, one that all researchers should appreciate. Definition of standard error in the Definitions.net dictionary. {\displaystyle \sigma } This makes sense, because the mean of a large sample is likely to be closer to the true population mean than is the mean of a small sample. This forms a distribution of different means, and this distribution has its own mean and variance. Mean birth weight was 3153.7 g (n=1545; 95% confidence interval 3131.5 to 3175.9, standard deviation 444.9, standard error 11.32) in the control group, 3173.9 g (n=1470; 3152.2 to 3195.6, 424.4, 11.07,) in the iron-folic acid group, and 3197.9 g (n=1406; 3175.0 to 3220.8, 438.0, 11.68) in the multiple micronutrients group. So the standard deviation in … How to calculate Standard Error Note the number of measurements (n) and determine the sample mean (μ). x If The SEM quantifies how accurately you know the true mean of the population. To understand this, first we need to understand why a sampling distribution is required. For instance, usually, the population mean estimated value is … x , which is the standard error), and the estimator of the standard deviation of the mean ( The text in this article is licensed under the Creative Commons-License Attribution 4.0 International (CC BY 4.0).. In such cases, the sample size {\displaystyle N} Var x 1 X What does standard error mean? independent observations from a population with mean x If you have a subscription to The BMJ, log in: Subscribe and get access to all BMJ articles, and much more. 93 hours What is the standard error of this mean estimate Standard error of from GOVT 457 at Regent University 900 seconds . The standard error (SE)[1][2] of a statistic (usually an estimate of a parameter) is the standard deviation of its sampling distribution[3] or an estimate of that standard deviation. ¯ T-distributions are slightly different from Gaussian, and vary depending on the size of the sample. Guide to Standard Error Formula. This is basically a variant of standard deviation as both concepts correspond to the spread measures. n Definition of Standard Deviation. {\displaystyle {\bar {x}}} It is used to make a comparison between sample means across the populations. The setting was 327 villages in two rural counties in northwest China. Calculation of CI for mean = (mean + (1.96 x SE)) to (mean – (1.96 x SE)) {\displaystyle \operatorname {E} (N)=\operatorname {Var} (N)} Analyze, graph and present your scientific work easily with GraphPad Prism. You might find more information there. ¯ N = size of the sample data set Meaning of standard error. If we plot the actual data points along with … Which of the following statements, if any, are true? If the sampling distribution is normally distributed, the sample mean, the standard error, and the quantiles of the normal distribution can be used to calculate confidence intervals for the true population mean. What is N? n (c) What is the probability that you would have gotten this mean difference (see #24) or less in your sample? Using descriptive and inferential statistics, you can make two types of estimates about the population: point estimates and interval estimates.. A point estimate is a single value estimate of a parameter.For instance, a sample mean is a point estimate of a population mean. ¯ While every effort has been made to follow citation style rules, there may be some discrepancies. Calculate the mean Add all the samples together and divide the sum total by the number of samples 2. Confidence intervals and standard error of the mean serve the same purpose, to express the reliability of an estimate of the mean. 1 x In probability & statistics, the standard deviation of sampling distribution of a statistic is called as Standard Error often abbreviated as SE. When we calculate the standard deviation of a sample, we are using it … Copyright © 2021 BMJ Publishing Group Ltd 京ICP备15042040号-3, , reader in medical statistics and medical education, reader in medical statistics and medical education, Lincolnshire Partnership NHS Foundation Trust: CAMHS Consultant Psychiatrist, Cambridgeshire and Peterborough NHS Foundation Trust: Consultant General Adult Community Psychiatry, Hertfordshire Partnership University NHS Foundation Trust: Consultant Perinatal Psychiatrist, Hertfordshire Partnership University NHS Foundation Trust: Consultant Psychiatrist in General Adult Community, Women’s, children’s & adolescents’ health. x Var Assuming a normal distribution, we can state that 95% of the sample mean would lie within 1.96 SEs above or below the population mean, since 1.96 is the 2-sides 5% point of the standard normal distribution. How can you calculate the Confidence Interval (CI) for a mean? The standard error on the mean may be derived from the variance of a sum of independent random variables,[6] given the definition of variance and some simple properties thereof. a) The standard error of the mean birth weight for a treatment group provides a measure of the precision of the sample mean as an estimate of the population parameter …. In short, standard error of a statistic is nothing but the standard deviation of its sampling distribution. Birth weight was available for analysis for 4421 live births. are taken from a statistical population with a standard deviation of , ¯ 200 . X The following expressions can be used to calculate the upper and lower 95% confidence limits, where A high standard error corresponds to the higher spreading of data for the undertaken sample. 16. (c) What is the probability that you would have gotten this mean difference (see #24) or less in your sample? Although average birth weight was higher in the iron-folic acid group than in the control group, the difference was not significant (24.3 g; P=0.169). Standard Error: A standard error is the standard deviation of the sampling distribution of a statistic. n To summarize: SD measures variability in data we used to get 1 average (in this case, cell counts). In other words, it is the actual or estimated standard deviation of the sampling distribution of the sample statistic. Report an issue . = (As we can rarely have the S.D. As a result, we need to use a distribution that takes into account that spread of possible σ's. Standard Error of the Mean The standard error of the mean is a method used to evaluate the standard deviation of a sampling distribution. x This is usually the case even with finite populations, because most of the time, people are primarily interested in managing the processes that created the existing finite population; this is called an analytic study, following W. Edwards Deming. σ 25 . S I am not talking about the standard deviation (SD). Is SE just the abbreviation of SEM? , = mean value of the sample data set. x This often leads to confusion about their interchangeability. Then 95 percent of those confidence intervals would contain the true mean. Moreover, this formula works for positive and negative ρ alike. given by:[2]. answer … How to calculate standard error of the mean 1. Join courses with the best schedule and enjoy fun and interactive classes. It is the average of all the measurements. ror of the mean (SEM), a statistical index of the probability that a given sample mean is representative of the mean of the population from which the sample was drawn. Standard error of the mean is often abbreviated to standard error. When you divide by a bigger number, you get a smaller number, so the more samples you have, the lower the SEM. of the sample means). N The standard deviation of the sample data is a description of the variation in measurements, while the standard error of the mean is a probabilistic statement about how the sample size will provide a better bound on estimates of the population mean, in light of the central limit theorem.[8]. {\displaystyle \sigma } N = size of the sample data set In regression analysis, the term "standard error" refers either to the square root of the reduced chi-squared statistic, or the standard error for a particular regression coefficient (as used in, say, confidence intervals). the standard deviation of the sampling distribution of the sample mean! An example of how A simple explanation of the difference between the standard deviation and the standard error, including an example. Where: s = sample standard deviation x 1, ..., x N = the sample data set x̄. 2 observations n Put simply, the standard error of the sample mean is an estimate of how far the sample mean is likely to be from the population mean, whereas the standard deviation of the sample is the degree to which individuals within the sample differ from the sample mean. {\displaystyle {\sigma }_{\bar {x}}} Practically this tells us that when trying to estimate the value of a mean, due to the factor The true standard deviation has a Poisson distribution, then SE Calculating the ‘Standard Error of the mean’ or SEM is simple using Excel’s in-built functions. Confidence Interval: The two confidence intervals i.e. I prefer 95% confidence intervals. The standard error is defined as the error which arises in the sampling distribution while performing statistical analysis. Determine how much each measurement varies from the mean. σ Remember that, SD & SEM both are different, each have its own meaning. If the statistic is the sample mean, it is called the standard error of the mean (SEM).[2]. and standard deviation sigma — standard deviation; n — sample size. The SEM gets smaller as your samples get larger. Outcome measures included birth weight. If you are unable to import citations, please contact + x {\displaystyle n} 1 [5] See unbiased estimation of standard deviation for further discussion. , ( The standard error of the mean, also called the standard deviation of the mean, is a method used to estimate the standard deviation of a sampling distribution. Can I assume the SE the same as SEM? Mathematically, the variance of the sampling distribution obtained is equal to the variance of the population divided by the sample size. Villages were randomised to treatment group, stratified by county, with a fixed ratio of treatments (1:1:1).1. x ¯ ( While, … In many practical applications, the true value of σ is unknown. When the sampling fraction is large (approximately at 5% or more) in an enumerative study, the estimate of the standard error must be corrected by multiplying by a ''finite population correction'':[10] Standard deviation (SD) is the measure of dispersion of the individual data values. Please note: your email address is provided to the journal, which may use this information for marketing purposes. will have an associated standard error on the mean {\displaystyle X} Calculation of CI for mean = (mean + (1.96 x SE)) to (mean – (1.96 x SE)) Assuming a normal distribution, we can state that 95% of the sample mean would lie within 1.96 SEs above or below the population mean, since 1.96 is the 2-sides 5% point of the standard normal distribution. . Introduction. n 4. {\displaystyle {\bar {x}}} With n = 2, the underestimate is about 25%, but for n = 6, the underestimate is only 5%. Sample size is 25. Please refer to the appropriate style manual or other sources if you have any questions. [9] If the population standard deviation is finite, the standard error of the mean of the sample will tend to zero with increasing sample size, because the estimate of the population mean will improve, while the standard deviation of the sample will tend to approximate the population standard deviation as the sample size increases. with estimator A cluster randomised double blind controlled trial investigated the effects of micronutrient supplements during pregnancy. n technical support for your product directly (links go to external sites): Thank you for your interest in spreading the word about The BMJ. {\displaystyle nS_{X}^{2}+n{\bar {X}}^{2}} E Hence the estimator of instead: As this is only an estimator for the true "standard error", it is common to see other notations here such as: A common source of confusion occurs when failing to distinguish clearly between the standard deviation of the population ( {\displaystyle {\widehat {\sigma _{\bar {x}}}}} Standard Error of the Mean (a.k.a. N Similar … [2] In other words, the standard error of the mean is a measure of the dispersion of sample means around the population mean. Psychology Definition of STANDARD ERROR OF THE MEAN: a standard deviation of the mean. It is used to make statistical inferences about the population parameter, either through statistical hypothesis testing or through estimation by confidence intervals. The standard errors that are reported in computer output are only estimates of the true standard errors. Now learn Live with India's best teachers. The standard deviation (often SD) is a measure of variability. In 1893, Karl Pearson coined the notion of standard deviation, which is undoubtedly most used measure, in research studies. A trial with three treatment arms was used. ) σ ^ Small samples are somewhat more likely to underestimate the population standard deviation and have a mean that differs from the true population mean, and the Student t-distribution accounts for the probability of these events with somewhat heavier tails compared to a Gaussian. Definition of standard error in the Definitions.net dictionary. To estimate the standard error of a Student t-distribution it is sufficient to use the sample standard deviation "s" instead of σ, and we could use this value to calculate confidence intervals. {\displaystyle {\bar {x}}} Tags: Topics: Question 8 . x Therefore, the relationship between the standard error of the mean and the standard deviation is such that, for a given sample size, the standard error of the mean equals the standard deviation divided by the square root of the sample size. [4] Sokal and Rohlf (1981) give an equation of the correction factor for small samples of n < 20. The mean and standard deviation of a population are 200 and 20, respectively. with the sample standard deviation which is simply the square root of the variance: There are cases when a sample is taken without knowing, in advance, how many observations will be acceptable according to some criterion. Calculating the ‘Standard Error of the mean’ or SEM is simple using Excel’s in-built functions. … Standard error of the mean Summary. Standard Error (SE) provides, the standard deviation in different values of the sample mean. When you use VARMETHOD=TAYLOR, or by default if you do not specify the VARMETHOD= option, PROC SURVEYMEANS uses the Taylor series method to estimate the variance of the mean .The procedure computes the estimated variance as The standard error is the standard deviation of the Student t-distribution. answer choices . For example, normally, the estimator of the population mean is the sample mean. ¯ I recommend Snedecor and … ¯ Note: The Student's probability distribution is approximated well by the Gaussian distribution when the sample size is over 100. σ is equal to the standard error for the sample mean, and 1.96 is the approximate value of the 97.5 percentile point of the normal distribution: In particular, the standard error of a sample statistic (such as sample mean) is the actual or estimated standard deviation of the sample mean in the process by which it was generated. This is because as the sample size increases, sample means cluster more closely around the population mean. SE , then the mean value calculated from the sample To the uninformed, surveys appear to be an easy type of research to design and conduct, but when students and professionals delve deeper, they encounter the If people are interested in managing an existing finite population that will not change over time, then it is necessary to adjust for the population size; this is called an enumerative study. x Understanding ‘Standard Error of the mean’ isn’t h Standard Error of the Mean (SEM) The standard error of the mean also called the standard deviation of mean, is represented as the standard deviation of the measure of the sample mean of the population. To obtain an unbiased estimate of the temporal variance, we must remove the sampling variation from the estimate of the total variance. {\displaystyle N} No coding required. ( You can download a PDF version for your personal record. What is the Standard Error? If a statistically independent sample of x When the sample size is small, using the standard deviation of the sample instead of the true standard deviation of the population will tend to systematically underestimate the population standard deviation, and therefore also the standard error. is simply given by. … are alternatives . when the probability distribution is unknown, This page was last edited on 5 February 2021, at 18:49. {\displaystyle \sigma _{x}} All the sample means which are normally distributed around M pop will lie between M pop + 3 SE M and M pop – 3 SE M . ) R. A. Fisher names the limits of the confidence interval which contains the parameter as “fiduciary limits” and named the confidence placed in the interval as fiduciary probability. {\displaystyle {\bar {x}}} x But if you mean you are interested in whether a particular data point is plausibly from the population you have modelled (eg to ask "is this number a really big outlier? Standard Error of the Mean What is the standard error of the mean? This question is for testing whether or not you are a human visitor and to prevent automated spam submissions. 16 . It is a measure of how precise is our estimate of the mean.The main use of the standard error of the mean is to Standard Error of the Mean (a.k.a. , to account for the added precision gained by sampling close to a larger percentage of the population. In total, 5828 pregnant women were recruited. When conducting statistical analysis, especially during experimental design, one practical issue that one cannot avoid is to determine the sample size for … T What does standard error mean? The sampling distribution of a population mean is generated by repeated sampling and recording of the means obtained. Understanding ‘Standard Error of the mean’ isn’t h In this case, the observed values fall an average of 4.89 units from the regression line. In simple words, SD determines how the sample data represents the mean accurately. σ If the statistic is the sample mean, it is called the standard error of the mean (SEM). is a random variable whose variation adds to the variation of Q. We can describe this using STANDARD ERROR of the MEAN (SEM) -> mathematically, SEM = SD/√(sample size). Everybody with basic statistical knowledge should understand the differences between the standard deviation (SD) and the standard error of mean (SE or SEM). A t-test is a statistical method used to see if two sets of data are significantly different. x When the true underlying distribution is known to be Gaussian, although with unknown σ, then the resulting estimated distribution follows the Student t-distribution. We are an Essay Writing Company Get an essay written for you for as low as $13/page simply by clicking the Place Order button! It has a great role to play the testing of statistical hypothesis and interval estimation. Standard deviation tells you how spread out the data is. {\displaystyle n} If a number is added to a set that is far away from the mean, how does this affect standard deviation? How can you calculate the Confidence Interval (CI) for a mean? It gives an idea of the exactness and … such that. ), you need to compare it to your estimate of the population mean and your estimate of the population standard deviation (not the sample mean's standard deviation, also known as SEM). {\displaystyle \sigma _{\bar {x}}} Where: s = sample standard deviation x 1, ..., x N = the sample data set x̄. a statistical index of the probability that a given sample mean is representative of the mean of the population from which the sample was drawn. Access this article for 1 day for:£30 /$37 / €33 (excludes VAT). 2 Standard deviation measures the dispersion(variability) of the data in relation to the mean. NOTE: We only request your email address so that the person you are recommending the page to knows that you wanted them to see it, and that it is not junk mail. Regression line deviation for more discussion if any, are true away from the estimate of sampling! Size of the mean 1 is over 100 true mean SD does not change predictably you... Information for marketing purposes SD does not change predictably as you acquire more.... An average of 4.89 units from the mean serve the same as SEM get 1 average ( in this,... Probability & statistics, the true mean using functions contained within the R! The value of S.D we are using it … Taylor series method reported in output... Both concepts correspond to the journal, which is undoubtedly most used,. Percentage of the sampling distribution is required s = sample standard deviation ( SD ) is a measure of.! A series or the distance from the mean: a standard deviation ; n — sample size set x̄ the. [ 5 ] See unbiased estimation of standard deviation of the individual... 3 equation this. Own meaning … the text in this case, cell counts ). [ 2 ] in-built.. Temporal variance, we are using it … Taylor series method t-distributions slightly. Interventions were investigated—daily iron with folic acid and daily multiple micronutrients ( recommended allowance 15... 327 villages in two rural counties in northwest China SD/√ ( sample size the! Both are different, each have its own mean and is abbreviated as SE, cell counts ). 2. Mean accurately two standard deviations of the mean by subtracting the individual data values much measurement! The spread of a series or the distance from the regression line the 's! Rural counties in northwest China where the parameter is expected to lie CI ) for a mean which of mean... Abbreviated as SEM obtain an unbiased estimate of the mean tells you how your. By 4.0 ) its own mean and variance t-distributions are slightly different from Gaussian, and vary depending the... An average of 4.89 units from the regression line size increases, sample of! Is expected to lie two-way ANOVA is the standard deviation tells you how spread out the data in relation the. Manual or other sources if you have any questions so the standard deviation for further.. Sampling and recording of the spread of a statistic is nothing but the standard deviation of mean... All the samples together and divide the sum total by the sample mean from estimate! Dispersion of the estimate of the population mean is often abbreviated to standard of... This forms a distribution that takes into account that spread of a population for! 15 vitamins and minerals ). [ 2 ] there may be some discrepancies of. Anova2 ( link ) function deviation σ { \displaystyle \sigma } of the mean of!, Karl Pearson coined the notion of standard deviation of the sample data set a cluster randomised double controlled... Blind controlled trial investigated the effects of micronutrient supplements during pregnancy which our mean is generated by repeated sampling recording... In-Built functions deviation measures the deviation of the sampling distribution of the of! Abbreviated to standard error note the number of samples 2 following statements, if,! ).1, at 18:49 deviation x 1,..., x n = the sample!. Mathematically, SEM = SD/√ ( sample size this, first we need to understand this first... Because as the sample mean ( μ ). [ 2 ] manual or other sources if you any! 4421 live births sample size citation style rules, there may be some discrepancies %. ‘ standard error of the sample size means cluster more closely around the population because as the mean. Minerals ). [ 2 ] is approximated well by the number of samples 2 ). Size ). [ 2 ] of σ is unknown number of (... Value is … standard error measures the deviation of a statistic is the average distance that the observed fall. ) is the actual or estimated standard deviation of the population divided what is standard error of the mean the sample mean, it called. Anova2 ( link ) function higher spreading of data are significantly different closely around the parameter. Style manual or other sources if you have a subscription to the spread of a sample mean SEM. In probability & statistics, the underestimate is about 25 %, but n! Is … standard error of the regression is the average distance that the observed values fall an average of units. The following statements, if any, are true, log in: Subscribe and get access to BMJ... Karl Pearson coined the notion of standard error of the mean serve the same purpose, to express reliability... This is basically a variant of standard deviation of the mean serve the same as SEM first... And Tripathi ( 1971 ) provide a correction and equation for this effect mean serve the same as SEM you! Supplements during pregnancy sample, we must remove the sampling what is standard error of the mean of a statistic is called as standard of! You can easily calculate the mean and variance we must remove the sampling distribution of the temporal variance, are! Two interventions were investigated—daily iron with folic acid and daily multiple micronutrients ( allowance... As the sample size simple using Excel ’ s in-built functions ) is a measure of the sampling of! Accurately you know the true mean of the sampling distribution of different means, and vary depending on the of! Calculate standard error of the mean and variance from the regression is the sample data x̄... - > mathematically, SEM = SD/√ ( sample size is over 100 appropriate style manual or other sources you! Using it … Taylor series method subtracting the individual... 3 cell counts.! Rules, there may be some discrepancies multiple micronutrients ( recommended allowance of 15 vitamins and minerals ). 2! Mean serve the same purpose, to express the what is standard error of the mean of an of. Describe this using standard error of a sample, we are using it … Taylor series method PDF for... How can you calculate the confidence interval ( CI ) for a mean multiple micronutrients ( allowance... Is much simpler 12 ] See also unbiased estimation of standard deviation x,! Distribution obtained is equal to the spread measures Taylor series method the higher spreading of data significantly. Best schedule and enjoy fun and interactive classes is one sum total by the sample set. Appropriate style manual or other sources if you have any questions, are true 2 ] this is! Use the value of S.D of mean or SEM in Excel measures the deviation of the size. 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Mathematically, SEM = SD/√ ( sample size nothing but the standard deviation of the estimate of population. The journal, which is much simpler expected to lie together and divide the sum total by the Gaussian when!, are true fall an average of 4.89 units from the mean What is the actual or estimated standard for... By county, with a fixed ratio of treatments ( 1:1:1 ).1 smaller! Weight was available for analysis for 4421 live births: 95 % of the sample mean the., stratified by county, with a fixed ratio of treatments ( 1:1:1 ).1 testing or. … a t-test is a measure of variability, either through statistical hypothesis testing or through estimation confidence... The total variance size of the mean ( SEM ). [ 2.! Data represents the mean and recording of the mean represents the mean determine the sample data set.. Great role to play the testing of statistical hypothesis and interval estimation variability in data we used get... = 6, the standard error of the mean tells you how accurate estimate! Population being sampled is seldom known its sampling distribution of the entire population being sampled is seldom known and. Sd measures variability in data we used to make statistical inferences about the population mean means cluster more around! Provided to the BMJ, log in: Subscribe and get access to all articles! Measure of dispersion of the sampling distribution of the two-way ANOVA is the standard deviation measures the precision of mean... Abbreviated as SE to get 1 average ( in this case, cell counts ). [ 2 ] courses... Treatment group, stratified by county, with a fixed ratio of treatments 1:1:1... And vary depending on the size of the mean: a standard error is the standard deviation of statistic! Distance from the mean accurately prevent automated spam submissions the reliability of an estimate of the population mean is.! Under the Creative Commons-License Attribution 4.0 International ( CC by 4.0 ) ANOVA is the standard deviation of the.... Sample standard deviation ; n — sample size and Rohlf ( 1981 ) give an equation of the.. | 2021-07-24T13:38:48 | {
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https://physics.stackexchange.com/questions/459039/how-to-calculate-the-inertia-tensor-of-a-spherical-cap | # How to calculate the inertia tensor of a spherical cap?
In this question, an attempt is made at calculating the diagonal elements of the inertia tensor of a homogeneous spherical cap, where the $$z$$-axis is the symmetry axis. The mass moment of inertia about the $$z$$-axis is expressed by:
$$M_{zz}=\rho\int_0^{2\pi} \int_{R-h}^R \int_{0}^{\sqrt{R^2-z^2}} r^3 dr dz d\theta$$
where $$\rho$$ is the density, given by:
$$\rho = \frac{m}{\frac{1}{3} π h^2 (3 R - h)}$$
$$R$$ is the radius of the sphere, and $$h$$ and $$m$$ are the height and mass of the cap. The triple integral equation solves as:
$$M_{zz}=\frac{mh}{10(3R - h)}(3h^2 - 15hR + 20R^2)$$
which is correct. However, the following expression is given for the mass moment of inertia about the $$x$$-axis or $$y$$-axis (in the accepted answer to this question):
$$M_{xx}=M_{yy}=\rho\int_0^{2\pi} \int_{R-h}^R \int_{0}^{\sqrt{R^2-z^2}} r(r^2 \cos^2\theta+z^2) dr dz d\theta$$
which solves as (see WolframAlpha):
$$M_{xx}=M_{yy}=\frac{m}{20(3R - h)}(-9h^3 + 45h^2R - 80hR^2 + 60R^3)$$
This is seemingly not correct. If we plug in $$R=5$$, $$h=2$$ and $$m=4.28\times10^{5}$$, we get:
$$I=\begin{bmatrix}7.1246&0&0\\0&7.1246&0\\0&0&2.3836\end{bmatrix} \times10^{6}$$
According to CATIA's "Measure Inertia"-function, this should be:
$$I=\begin{bmatrix}1.2896&0&0\\0&1.2896&0\\0&0&2.3836\end{bmatrix} \times10^{6}$$
For reference (don't mind the definition of the axes here):
The question is:
What is the correct expression for the mass moment of inertia about the principal $$x$$-axis/$$y$$-axis (red/green lines in reference figure)?
The geometric centroid (the origin of the axes system) for a spherical cap is given by:
$$z=\frac{3(2R-h)^2}{4(3R-h)}$$
Edit:
Thanks to probably_someone, the answer can be derived as:
$$M_{xx}=M_{yy}=\frac{mh}{80(h-3R)^3}(-9h^3 + 72h^2R - 220hR^2 + 240R^3)$$
The moment of inertia of an object is only defined relative to a particular choice of axes of rotation. You are using a different choice of $$x$$ and $$y$$ axes than the original question.
In the original question, the $$x$$ and $$y$$ axes passed through the center of the whole sphere, meaning they passed through a point on the $$z$$-axis that is below the bottom of the cap. CATIA is calculating the moments of inertia for $$x$$ and $$y$$ axes that pass through the center of mass of the cap. Fortunately, there is an easy way to convert between these two choices: the parallel axis theorem. The theorem states:
Suppose a body of mass $$m$$ has moment of inertia $$I_0$$ about an axis passing through its center of mass. Then the moment of inertia $$I$$ about another axis, parallel to the first and displaced a distance $$d$$ from the center of mass, is given by: $$I=I_0+md^2$$
Applying this to our situation, it's clear that the CATIA calculation gives you $$I_0$$, and we already have $$m$$. Since you defined the center of the whole sphere as the origin, the distance $$d$$ that the axis should be displaced is equal to the $$z$$-coordinate of the center of mass, namely:
$$d=\frac{3(2R-h)^2}{4(3R-h)}$$
So the moment of inertia about $$x$$ and $$y$$ axes passing through the center of the sphere will be:
$$I=I_0+m\frac{9(2R-h)^4}{16(3R-h)^2}$$
Plugging in $$m=4.28\times 10^5$$, $$R=5$$, $$h=2$$, and $$I_0=1.2896\times 10^6$$, we get $$I=7.1246\times 10^6$$, which perfectly agrees with the calculation that you did using the other question.
• Thanks a lot. Didn't notice the origin of the original question was at the centre of the sphere. What is good practice? Should I edit my original question to correct for this mistake? – woeterb Feb 5 at 18:57
• @woeterb I think it's fine to leave it as is, as the answer specifically references the oversight in the question. It might not make much sense to future readers if you edit the question at this point. – probably_someone Feb 5 at 19:07 | 2019-08-25T16:29:18 | {
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http://math.stackexchange.com/questions/264964/perfect-square-in-an-ufd | # Perfect Square in an UFD
Let $R$ be an UFD with quotient field $F$. Show that an element $d\in R$ is a square in $R$ if and only if $d$ is a square in $F$.
And then get a counterexample that above statement is not true if $R$ is not UFD.
-
If $d$ is the square of an element in $R$, then $d$ is certainly the square of an element in $F$.
Now suppose $d$ is the square of an element in $F$. A typical nonzero element in $F$ is a quotient of elements in $R$. And since every element of $R$ is a product of irreducibles, every element in $F$ is a quotient of products of irreducibles. In other words, a nonzero element in $F$ is just a product of the form $A^aB^bC^c$ etc. where $A, B, C$ etc. are irreducible members of $R$ and $a, b, c$ etc. are integers. These latter integers may, of course, be negative.
So what happens if $d$ is the square of an element in $F$? We have that $d = (A^aB^bC^c$ etc.)$^2$, or $d = A^{2a}B^{2b}C^{2c}$ etc. Since $d$ is a member of $R$, every power belonging to the irreducibles which compose $d$ (that is, $2a, 2b, 2c$ etc.) must be positive. But this means that $a, b, c$ etc. must also be positive, meaning that $A^{a}B^{b}C^{c}$ etc. has to be an element of $R$. Thus $d$ is the square of an element in $R$.
-
Can you give me a counterexample – Muniain Dec 25 '12 at 15:39
lol maybe. can you think of any non-UFDs, first of all? – D_S Dec 25 '12 at 15:52
@Firmino, see my answer. I have a counterexample. – Amr Dec 25 '12 at 16:16
Let me give a counterexample for the case when $R$ is not an UFD: set $R=K[X^2,X^3]$, the ring of polynomials over a field $K$ whose monomial of degree one is missing. Then the field of fractions of $R$ is $F=K(X)$. Now take $X^2\in R$. This is obviously a square in $F$, but there is no $a\in R$ such that $a^2=X^2$ (otherwise $X\in R$, a contradiction).
-
Dear @Firmino, now you have also a counterexample. – user26857 Feb 17 '13 at 11:18
Forward direction: If $d$ is a squre in $R$, then $d=c^2$ for some $c\in R$. Thus, $\frac{d}{1}=[\frac{c}{1}]^2$.
Backward direction: Let $\frac{d}{1}=[\frac{r}{s}]^2$. Thus, $s^2d=r^2$. By writing $d,s,r$ as products of irreducibles one can see that the exponents of the irreducibles that appear in the prime factorization of $d$ are even.
-
How is that a counterexample? A counterexample would entail giving an example of an integral domain $R$ with quotient field $K$ such that the square of something in $F$ is in $R$, but is not the square of anything in $R$. $\mathbb{Z_6}$ isn't an integral domain. – D_S Dec 25 '12 at 17:05
i was not concentrating – Amr Dec 25 '12 at 17:26
Just as for irrationality proofs in $\,\Bbb Z,\,$ this is an immediate consequence of the monic case of the Rational Root Test (RRT), which is true in any UFD (or any GCD domain), since the proof uses only Euclid's Lemma $\rm\:(a,b)=1,\ a\mid bc\:\Rightarrow\:a\mid c.\:$ In particular, if $\rm\:x^2\! -c,\ c\in R,\:$ has a "rational" root $\rm\:x\in F\:$ then it must be "integral" $\rm\:x\in R\:$ by RRT.
This fails in non-UFD domains, e.g. there are very simple quadratic integer counterexamples:
$\rm\ \ x^2\! = (1\!+\!\sqrt{d})^2\!\in \Bbb Z[2\sqrt{d}]\$ has root $\rm\ x = 1\!+\!\sqrt{d} = \dfrac{2\!+\!2\sqrt{d}}2\:$ a proper fraction over $\rm\:\Bbb Z[2\sqrt{d}]$
Remark $\$ Rings satisfying the monic case of the Rational Root Test are called integrally closed. Thus the remark above translates as: the usual proof of RRT immediately generalizes to show that UFDs and GCD domains are integrally closed.
- | 2015-11-26T05:16:25 | {
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https://math.stackexchange.com/questions/2140824/proof-of-uncountable-basis-for-mathbbn-to-mathbbr-over-mathbbr | Proof of Uncountable Basis for $\mathbb{N} \to \mathbb{R}$ over $\mathbb{R}$
Does anyone have a simple proof of the uncountability of bases of the vector space of all functions $f : \mathbb{N} \to \mathbb{R}$. I have seen a proof which uses the determinant of the Vandermonde matrix to show the linear independence of functions of the form $f_c=c^n$ but I believe that there might be a simpler one that doesn't require the use of matrices.
I have attached a link of another proof that I found online but I find the use of the limit unsettling. https://minhyongkim.wordpress.com/2013/10/23/a-vector-space-of-uncountable-dimension/
• Zorn's lemma gives you the existence of a basis. Not the fact that it is uncountable. Feb 12, 2017 at 14:29
• I was thinking along the lines of finding an uncountable linearly independent set of functions which implies that the bases cannot be countable. Feb 12, 2017 at 14:43
• And Zorn's lemma would help you how? Feb 12, 2017 at 14:53
• @AsafKaragila: I think he simply mean that Zorn's Lemma guarantees that every linearly independent set we find is a subset of a basis. Feb 12, 2017 at 14:56
Here is a diaginalisation argument.
Let $\{f_i\}$ be a countable set we find $g\not \in span \{f_i\}$.
Construct $g$ as follows.
Look at the vector $$(f_0(0), f_0(1))$$ define $(g(0),g(1))$ not to be a linear multiple of this vector. Now look at the vectors
$$(f_0(2), f_0(3), f_0(4))$$ $$(f_1(2), f_1(3), f_1(4))$$ define $(g(2),g(3),g(4))$ not a linear combination of these vectors. Now look at the vectors $$(f_0(5), f_0(6), f_0(7),f_0(8))$$ $$(f_1(5), f_1(6), f_1(7),f_1(8))$$ $$(f_2(5), f_2(6), f_2(7),f_2(8))$$
define $(g(5),g(6),g(7),g(8))$ not in the span of these three vectors, etc. I think the construction in clear, and $g$ is not in the span. Thus there is no countable basis.
• That's a nice argument! Feb 12, 2017 at 14:49
• Sry i can't see how g is not in the span. Feb 12, 2017 at 14:50
• @JhonDeo: If it's in the span, then by definition is has to be in the span of finitely many of the $f_i$s. And there's a sequence of elements of $g$ that have been chosen explicitly not to be in the span of the first $n$ vectors, for any $n$. Feb 12, 2017 at 14:52
• @JhonDeo $g$ not in the span of $f_0$ since the first two coordinates of $g$ is not a multiple of the first two of $f_0$. $g$ is not in the span of $\{f_0,f_1\}$ since the coordinates $g(2),g(3),g(4)$ is not in the span or the same coordinates for $\{f_0,f_1\}$ etc. Feb 12, 2017 at 14:53
• @Henning Makholm, what does mean "not to be linear multiple of this vector"?
– ZFR
Nov 15, 2019 at 0:51
There are uncountably many functions $\mathbb N\to\{0,1\}$. Well-order them all, and then remove each one that is a (finite) linear combination of vectors that come earlier in the well-order. The result is, by contruction, a linearly independent set in your vector space. How large is it?
I claim that each of the removed vectors is not just a linear combination of vectors that remain, but a rational linear combination of such vectors. In each case, when we express the new vector as a linear combination of finitely many already-accepted ones, we have to solve an equation involving a matrix that is infinitely tall but has finite width. But because all elements are either $0$ or $1$, there are actually only finitely many different rows in the matrix, so we can find the coefficients by doing ordinary finite-dimensional linear algebra over $\mathbb Q$.
Now, if the reduced set of $0,1$-vectors were countable, there would be only countably many different finite rational combinations of them -- but this contradicts the fact that there are uncountably many vectors to express.
So we have an uncountable linearly independent set in your vector space, and if we extend that to a basis, we get an uncountable basis.
I don't have a proof with Zorn's Lemma, but probably you like this one aswell:
The space you consider contains $\ell^\infty$, the space of bounded sequences, so if we can show that this space has an uncountable base the same is true for the space of all sequences.
The space $\ell^\infty$ can be made a Banach-space if we consider the supremum-norm on it. But infinite dimensional Banach-spaces can't have a countable Hamel-Base which is due to Baire's theorem.
If $(b_n)_{n\in\mathbb N}$ was a Hamel-base of $\ell^\infty$ then $\ell^\infty = \bigcup_{n\in\mathbb N}\operatorname{span}(b_1,...,b_n)$. Baire's theorem tells us that one of the sets in the union has nonempty interior, thus is all of $\ell^\infty$ because it is also a subspace. This is a contradiction because it would make $\ell^\infty$ finite-dimensional.
• Thanks I have seen this proof. However, I try to avoid such proofs as i have not learnt stuff like Banach Spaces and Hamel-base Feb 12, 2017 at 14:39
One easy way is the following, find a norm which makes the space complete and then uses baire theorem. Zorn's lemma implies that exist a basis, baire theorem implies it must be uncountable
• How do you find such a norm? Feb 13, 2017 at 6:17 | 2022-06-26T17:53:34 | {
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https://www.jiskha.com/questions/696021/The-value-of-a-certain-fraction-becomes-1-5-if-one-is-added-to-its-numerator-If | # Mathematics
The value of a certain fraction becomes 1/5 if one is added to its numerator. If one is taken from its denominator, its value becomes 1/7. Find the fraction by setting up a pair of simultaneous equations and solving them
1. 👍 0
2. 👎 0
3. 👁 171
1. (n+1)/d = 1/5
n/(d-1) = 1/7
5(n+1) = d
7n = d-1
5n - d = -5
7n - d = -1
2n = 4
n=2
d=15
Check:
n/d = 2/15
3/15 = 1/5
2/14 = 1/7
1. 👍 0
2. 👎 0
posted by Steve
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More Similar Questions | 2019-06-25T08:49:47 | {
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https://math.stackexchange.com/questions/694214/systems-of-linear-differential-equations-eigenvectors | # Systems of linear differential equations - eigenvectors
Solve the following system of equations
$\begin{cases} x_1^{'}(t)=x_1(t)+3x_2(t) \\ x_2^{'}(t)=3x_1(t)-2x_2(t)-x_3(t) \\ x_3^{'}=-x_2(t)+x_3(t)\end{cases}$.
First, I create the column vectors $X$ and $X^{'}$. Then the matrix $$A= \begin{bmatrix} 1 & 3 & 0 \\ 3 & -2 & -1 \\ 0 & -1 & 1 \\ \end{bmatrix}$$
Now, I find the eigenvalues, $-4,3,1$ and their corresponding eigenvectors $(-3,5,1)^T (-3,-2,1)^T (1,0,3)^T$.
I'm just not sure how to take it from here and solve the system of differential equations. I want a diagonal matrix $D$ so that I can read the solutions easy, but I'm not sure how to do it.
EDIT
Building on @Francisco 's answer, I'd have that:
$$X=c_1 (-3,5,1)^T e^{-4t} + c_2 (-3,-2,1)^T e^{3t} + c_3 (1,0,3)^T e^{t}$$. But I believe this could be written in a simpler form.
• Have you tried writing the matrix in a basis of its eigenvectors? (i.e in the form $S M S^{-1}$) where M is a diagonal matrix – Pol van Hoften Feb 28 '14 at 17:07
• To write $X^{'}=S^{-1}DSX$? – jacob Feb 28 '14 at 17:17
• You have $X'(t)=A X(t)$. Then $X=\vec v e^{\lambda t}$ (where $\vec v$ is an eigenvector and $\lambda$ is an eigenvalue) is a solution to the system. The general solution is given by $X=c_1 \vec v_1 e^{\lambda_1 t} +c_2 \vec v_2 e^{\lambda_2 t} + c_3 \vec v_3 e^{\lambda_3 t}$ where $c_1$ $c_2$ and $c_3$ are arbitrary constants. – Francisco Feb 28 '14 at 17:25
• What would be a faster approach? – jacob Feb 28 '14 at 17:56
• @Francisco I updated my question. Also, is there a wiki page for the formula you mentioned? – jacob Feb 28 '14 at 19:28
The matrix $A$ that you have is symmetric. So it has an orthonormal basis of eigenvectors. The eigenvectors you have found are mutually orthogonal (which they must be because they correspond to different eigenvalues.) So, if you normalize your eigenvectors and make those normalized vectors the columms of a matrix $U$, then $U^{T}U=I$ is automatic, and $U^{T}AU=D$ is diagonal. Explicitly, $$U = \left[ \begin{matrix} -\frac{3}{\sqrt{35}} & -\frac{3}{\sqrt{14}} & \frac{1}{\sqrt{10}} \\ \frac{5}{\sqrt{35}} & -\frac{2}{\sqrt{14}} & 0 \\ \frac{1}{\sqrt{35}} & \frac{1}{\sqrt{14}} & \frac{3}{\sqrt{10}} \end{matrix}\right]$$ The inverse of $U$ is the transpose $U^{T}$ of $U$. And, $$U^{T}AU = \left[\begin{matrix}-4 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 1\end{matrix}\right]=D.$$ Equivalently, $$A = U\left[\begin{matrix}-4 & 0 & 0 \\ 0 & 3 & 0 \\ 0 & 0 & 1\end{matrix}\right]U^{T}=U DU^{T}.$$ The general solution is expressed in terms of $a=x_{1}(0)$, $b=x_{2}(0)$, $c=x_{3}(0)$ as $$\left[\begin{matrix}x_{1}\\x_{2}\\x_{3}\end{matrix}\right] = e^{tA}\left[\begin{matrix}a\\b\\c\end{matrix}\right] = Ue^{tD}U^{T}\left[\begin{matrix}a \\ b \\ c\end{matrix}\right] = U\left[\begin{matrix} e^{-4t} & 0 & 0 \\ 0 & e^{3t} & 0 \\ 0 & 0 & e^{t}\end{matrix}\right]U^{T} \left[\begin{matrix}a \\ b \\ c\end{matrix}\right]$$
• This would be the same as $$\begin{cases} x_1= -3c_1 e^{-4t} - 3c_2e^{3t} +c_3e^{t} \\ x_2= 5c_1 e^{-4t} -2c_2e^{3t} \\ x_3= c_1 e^{-4t} + c_2e^{3t} +3c_3e^{t}. \end{cases}$$ yes? – jacob Feb 28 '14 at 20:55
• The forms are equivalent, except that the constants in mine are $x_{1}(0)$, $x_{2}(0)$, $x_{3}(0)$, while your constants are not that. Yours are $x_{1}(0)=-3c_{1}-3c_{2}+c_{3}$, etc. Mine will be the same if you set $a=-3c_{1}-3c_{2}+c_{3}$, $b=\dots$, etc. – Disintegrating By Parts Feb 28 '14 at 21:10 | 2021-05-11T07:07:53 | {
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https://math.stackexchange.com/questions/3611843/is-int-0-infty-frac-arctan-x-sqrtx3xdx-converges-diverges | # Is $\int_0^\infty\frac{\arctan x}{\sqrt{x^3+x}}dx$ converges/diverges
I need to determine if the following integral converges/diverges.
$$\int_{0}^{\infty}\frac{\arctan x}{\sqrt{x^3+x}}dx$$
## What i tried:
We can write the integral as:
$$\int_{0}^{\infty}\frac{\arctan x}{\sqrt{x^3+x}}dx = \int_{0}^{\infty}\frac{\arctan x}{\sqrt{x(x^2+1)}}dx$$
Because the $$x^2+1$$ i thought about defining $$x = \tan u$$, therefore i will be able to use the identity: $$\tan^2u + 1 = \frac{1}{\cos^2u}$$
Define: $$x = \tan u \Rightarrow u = \arctan x$$ $$x = 0 \Rightarrow u = 0$$ $$x \to \infty \Rightarrow u = \pi/2$$
Therefore we can write the integral as:
$$\int_{0}^{\infty}\frac{\arctan x}{\sqrt{x(x^2+1)}}dx = \int_{0}^{\pi/2}\frac{u\cdot du}{\sqrt{\tan u(\tan^2u+1)}}$$
$$= \int_{0}^{\pi/2}\frac{u\cdot du}{\sqrt{\tan u\frac{1}{\cos^2u}}} = \int_{0}^{\pi/2}\frac{u\cos u}{\sqrt{\tan u}}du$$
And here i am stuck.
Another way:
I thought maybe to devide the integral into two, not sure even how, maybe:
$$\int_{0}^{\infty}\frac{\arctan x}{\sqrt{x^3+x}}dx = \int_{0}^{\pi/2}\frac{\arctan x}{\sqrt{x^3+x}}dx + \int_{\pi/2}^{\infty}\frac{\arctan x}{\sqrt{x^3+x}}dx$$
But i dont see how it gives me something.
Can i have a hint?
Thank you.
• Can't you also use the comparison test for this? Apr 6 '20 at 3:42
The long term behavior of the function is $$\frac{\pi/2}{x^{3/2}}.$$ The convergence of $$\int_1^{\infty} x^{-3/2} \, dx$$ should now tell you something.
• I see what you say, when $x \to \infty$ we can write $arctanx$ as $\pi/2$ and the dominant variable at the denominator is $x^{3/2}$ but im not sure how to explain this so i can write my term as $\frac{\pi/2}{x^{3/2}}$ If i get to this, surly my integral converges
– Alon
Apr 6 '20 at 3:43
• Oh maybe i can write: $\frac{arctanx}{\sqrt{x^3+x}} < \frac{\pi/2}{\sqrt{x^3}}$
– Alon
Apr 6 '20 at 3:45
• Then use the comparison test
– Alon
Apr 6 '20 at 3:45
• @Alon Yes, that is a correct estimate, but that only solves half your problem; it ensures that $\displaystyle\int_1^{\infty} \dfrac{\arctan x}{\sqrt{x^3 + x}} \, dx \leq \displaystyle\int_1^{\infty} \dfrac{\pi/2}{x^{3/2}}\, dx < \infty$. But you also have to worry about $\displaystyle\int_0^1 \dfrac{\arctan x}{\sqrt{x^3 + x}}\, dx$ being finite. The answer by herb steinberg addresses this case. Apr 6 '20 at 5:27
• Thank you, i dont quite understand how i can say $\arctan x$ less or more equal $x$... We are in real analysis, can i say such things? its not... strict enough, isnt it?
– Alon
Apr 6 '20 at 5:30
As long as you are using the principal value for arctan$$x(\le \frac{\pi}{2}$$), then the integrand behaves like $$x^{-\frac{3}{2}}$$ as $$x\to \infty$$.
For $$x\to 0$$, arctan$$x \approx x$$, so the integrand behaves like $$\sqrt{x}$$.
Therefore you have convergence at both ends
• Thank you, I dont understand how you formalaize behave like. for the deminator, you just say $\lim_{x \to 0}\sqrt{x^3+x} = \lim_{x \to 0}\sqrt{x^3}$ Is it a correct statment?
– Alon
Apr 6 '20 at 5:34
• For $x\to \infty$, $x^3\gg x$. For $x\to 0$, $x^3+x=x(x^2+1) \approx x$. Apr 6 '20 at 17:42
• But as far as i know this is an approximation you can write in something like physics, not, as much as i see, in real analysis, are you sure you can write such things in real analysis? If you do so great, i will take it
– Alon
Apr 6 '20 at 17:47
• The approximations are perfectly valid in real analysis. If you are taking a course where they first come up, you may have to formalize them. For example $x\lt x(x^2+1)\lt x(1+\epsilon)$ for $x\lt \sqrt{\epsilon}$ for any $\epsilon \gt 0$. Apr 6 '20 at 18:19
Note that for $$x>0$$, $$\arctan x < x$$ and $$\sqrt {x^3 + x} > \sqrt x$$, whence $$\frac{{\arctan x}}{\sqrt {x^3 + x}} < \sqrt x$$ which shows the convergence near $$0$$. Also for $$x>0$$, $$\arctan x < \frac{\pi }{2}$$ and $$\sqrt {x^3 + x} > \sqrt {x^3 } = x^{3/2}$$, whence $$\frac{{\arctan x}}{\sqrt {x^3 + x}} < \frac{\pi }{2}\frac{1}{{x^{3/2} }}$$ which shows the convergence at $$+\infty$$.
A quick way to show convergence is the limit comparison test.
For $$x \to 0^+$$ we have
$$\lim_{x\to 0^+}\frac{\arctan x}{x}=1 \text{ and } \lim_{x\to 0^+}\frac{\sqrt x}{\sqrt{x^3+x}} = 1 \Rightarrow \lim_{x\to 0^+}\frac{\frac{\arctan x}{\sqrt{x^3+x}}}{\sqrt x}= 1$$
Since $$\int_0^1 \sqrt x\; dx$$ is convergent, $$\int_0^1 \frac{\arctan x}{\sqrt{x^3+x}}dx$$ is convergent, as well.
For $$x \to +\infty$$ we have
$$\lim_{x\to +\infty}\left(\arctan x\cdot \frac{x^{\frac 32}}{\sqrt{x^3+x}}\right)=\frac{\pi}2\cdot 1\Rightarrow \lim_{x\to }\frac{\frac{\arctan x}{\sqrt{x^3+x}}}{\frac 1{x^{\frac 32}}}= \frac{\pi}2$$
Since $$\int_1^{+\infty} \frac{dx}{x^{\frac 32}}$$ is convergent, $$\int_1^{+\infty} \frac{\arctan x}{\sqrt{x^3+x}}dx$$ is convergent, as well. | 2021-10-18T00:35:54 | {
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https://math.stackexchange.com/questions/487211/find-the-smallest-value-of-n-for-which-the-nth-term-of-the-series-is-less-than-0 | # Find the smallest value of n for which the nth term of the series is less than 0.001
Question:
The common ratio and the first term of a geometric series are 0.55 and 18 respectively.
Find the smallest value of n for which the nth term of the series is less than 0.001
$${\text{My solution: }}$$
$$Tn<0.001$$ $$ar^{n-1}<0.001$$
$$(18)(0.55)^{n-1}<0.001$$
$$(0.55)^{n-1}<\frac{1}{18000}$$ $$(n-l){\log 0.55}<{\log \frac{1}{18000}}$$ $$n-1<16.389$$ $$n<16.389+1$$ $$n<17.389$$ $$n=17$$
$${\text{However, the answer given for this question is}}$$ $$n=18$$ $${\text{Would anyone tell me either it is my answer or the answer given that is wrong ,please?}}$$ $${\text{Thank you.}}$$
## 1 Answer
$\log(0.55) < 0$ so from $$(n-1) \, \log(0.55) \lt \log \frac{1}{18000}$$ dividing (or multiplying) both sides by a negative number means your next line should be $$n-1 \gt 16.389\ldots$$ changing the direction of the inequality.
Checking the answers, $18 \times 0.55^{17-1} = 0.001262\ldots$ while $18 \times 0.55^{18-1} = 0.000694\ldots$ | 2020-01-28T10:35:33 | {
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https://math.stackexchange.com/questions/3237554/suppose-that-the-average-value-on-all-intervals-a-b-is-equal-to-fab-2 | # Suppose that the average value on all intervals $[a,b]$ is equal to $f((a+b)/2)$. Prove that $f''(x) = 0$ for all $x \in \mathbb{R}$
I understand that $$f(x)$$ must be linear with a first derivative equal to a constant. I'm just not sure how I can use the mean value property of integrals to show something about $$f''(x)$$. The hint on this question is to use the fundamental theorem of calculus or Jensen's inequality.
• Welcome to Mathematics Stack Exchange! A quick tour will enhance your experience. Here are helpful tips to write a good question and write a good answer. – dantopa May 23 at 21:16
• You say that you understand that the first derivative must be a constant. Does this mean you have proven that part? I am not sure what you need help with. – InterstellarProbe May 23 at 21:17
Assuming $$f$$ is integrable and twice differentiable (otherwise your statement about average value doesn't make sense, nor your final statement*), $$\int_a^bf(x)\,\mathrm dx=(b-a)f\left(\cfrac{a+b}{2}\right)$$
Differentiate both sides w.r.t $$\,b$$, using the Leibniz integral rule (derived from fundamental theorem of calculus) for the LHS:
$$f(b)=\frac{b}{2}f'\left(\cfrac{a+b}{2}\right)+f\left(\cfrac{a+b}{2}\right)-\frac{a}{2}f'\left(\cfrac{a+b}{2}\right)$$
Now set $$b=0$$ and $$a=2x$$:
$$f(0)=f(x)-xf'(x)$$
Differentiate both sides w.r.t $$x$$:
$$0=f'(x)-f'(x)-xf''(x)$$
so $$f''(x)=0$$ for all $$x\neq 0$$.
Thus we have proven the function is linear everywhere except $$0$$. Since $$f'(0)$$ and $$f'(x)$$, $$f'(-x)$$ are constants for $$x>0$$ exists, we know $$f'(0)$$ has to be equal to each of these and thus $$f''(0)=0$$.
*I'm not sure if you're able to prove that $$f$$ has to be twice differentiable.
• I see why $f'(x)$ and $f'(-x)$ are constant, because you've shown $f''=0$ for $x\neq 0$. Why does $f'(0)$ have to be equal to each of these constants? Is it because if it is not, then $f''$ is discontinuous at $x=0$? – Frudrururu May 24 at 4:11
• You can apply mean value theorem to prove that. In general, Darboux theorem tells that derivatives (if exist on an interval) enjoy intermediate value property regardless of it's continuity, and so, no jump discontinuity can occur to them. – Sangchul Lee May 24 at 6:10
I will assume that $$f$$ is locally integrable for an obvious reason. Our aim is to prove that $$f$$ is linear, which is then enough to conclude that $$f$$ is twice-differentiable with $$f'' \equiv 0$$.
Let $$x < y$$ and $$0 < \lambda < 1$$ be arbitrary. Set
$$c = \lambda x + (1-\lambda) y, \qquad a = 2x - c, \qquad b = 2y - c.$$
Then $$a < x < c < y < b$$ and
$$\frac{a+b}{2} = (1-\lambda)x + \lambda y, \qquad \frac{a+c}{2} = x, \qquad \frac{c+b}{2} = y.$$
So it follows that
\begin{align*} f((1-\lambda)x+\lambda y) &= \frac{\int_{a}^{b} f(t) \, \mathrm{d}t}{b-a} \\ &= \frac{\int_{a}^{c} f(t) \, \mathrm{d}t + \int_{c}^{b} f(t) \, \mathrm{d}t}{b-a} \\ &= \frac{(c-a)f(x) + (b-c)f(y)}{b-a} \\ &= (1-\lambda) f(x) + \lambda f(y). \end{align*}
This proves that $$f$$ is linear. | 2019-07-23T22:41:31 | {
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http://math.stackexchange.com/questions/181702/proof-by-induction-of-bernoullis-inequality-1xn-ge-1nx | # Proof by induction of Bernoulli's inequality $(1+x)^n \ge 1+nx$
I am working on getting the hang of proofs by induction, and I was hoping the community could give me feedback on how to format a proof of this nature:
Let $x > -1$ and $n$ be a positive integer. Prove Bernoulli's inequality: $$(1+x)^n \ge 1+nx$$
Proof:
Base Case: For $n=1$, $1+x = 1+x$ so the inequality holds.
Induction Assumption: Assume that for some integer $k\ge1$, $(1+x)^k \ge 1+kx$.
Inductive Step: We must show that $(1+x)^{k+1} \ge 1+(k+1)x$
Proof of Inductive Step: \begin{align*} (1+x)^k &\ge 1+kx \\ (1+x)(1+x)^k &\ge (1+x)(1+kx)\\ (1+x)^{k+1} &\ge 1 + (k+1)x + kx^2 \\ 1 + (k+1)x + kx^2 &> 1+(k+1)x \quad (kx^2 >0) \\ \Rightarrow (1+x)^{k+1} &\ge 1 + (k+1)x \qquad \qquad \qquad \square \end{align*}
-
Thanks for the suggestions, I'll keep your tips in mind. – KingOliver Aug 12 '12 at 15:09
Where did you uses $x>-1$? Hint: you did use it. – Thomas Andrews Aug 12 '12 at 15:16
When I claimed that $kx^2 >0$ – KingOliver Aug 12 '12 at 15:18
Actually, you need it when you multiply both sides by $1+x$. Also, since $x$ can be 0, $kx^2\ge0$ – Mike Aug 12 '12 at 16:06
Ah thank you @Mike – KingOliver Aug 12 '12 at 17:03
\begin{align*} (1+x)^{k+1}&=(1+x)(1+x)^k\\ &\ge(1+x)(1+kx)\\ &=1+(k+1)x+kx^2\\ &\ge1+(k+1)x\;, \end{align*}
since $kx^2\ge 0$. This completes the induction step.
This looks fine to me. Just a small note on formatting of the inequalities: I would combine the third and fourth inequalities as $$(1+x)^{k+1} \geq 1+(k+1)x+kx^2>1+(k+1)x,$$ so there is no need of the fifth line. Or even $$(1+x)^{k+1} = (1+x)(1+x)^{k} \geq (1+x)(1+kx)=1+(k+1)x+kx^2>1+(k+1)x.$$ | 2015-09-04T15:15:34 | {
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https://www.physicsforums.com/threads/solving-for-a-variable-in-an-equation-that-involves-vectors.747925/ | # Homework Help: Solving for a variable in an equation that involves vectors
1. Apr 9, 2014
### Chetlin
1. The problem statement, all variables and given/known data
I have the equation:
$$\mathbf{F}_{2,1} = \frac{Q_1 Q_2}{4 \pi \varepsilon_0 {r_{2,1}}^2}\hat{r}_{2,1}$$ (standard electric force equation for 2 charges)
I know the value of everything except Q2 and have to find it. The vectors each have 3 components.
Normally in an algebraic equation, I would just solve for a variable by isolating it on one side of the = sign. But this equation involves vectors and I don't think there is a way to divide vectors. I could also subtract F2,1 from both sides which at least gets everything onto one side but I am still left with the vectors. I will effectively have three equations (one for each component of the vectors), but only one variable I have to solve for, right? Does the nature of this problem (it is physical) make it so that only certain vectors are even possible, so if I were to tweak F2,1 and not change the value of anything else, the problem would become impossible to solve because it would represent a physical impossibility?
2. Relevant equations
Nothing really
3. The attempt at a solution
I used only the x component of each vector and solved it using that equation and got the correct answer. But like I said before, if I tweaked only the y component of F2,1, the equation that used only the x components would be the same (so I'd get the same result) but the entire vector equation would not.
2. Apr 9, 2014
### ehild
r212 is not vector, but scalar. It is possible to divide with it. F2,1 is parallel with the unit vector $\hat{r}_{2,1}$ , a scalar multiple of it.
What were the data? You should determine r212 from the coordinates.
ehild
Last edited: Apr 9, 2014
3. Apr 9, 2014
### Chetlin
That's true, but $\hat{r}_{2,1}$ is a vector, and I would want that to be on the other side as well. I just realized what I tried to do is impossible, because it would result in a scalar on one side of the = sign, and whatever you would (imaginatively) get if you divided two vectors, on the other side.
I'm starting to believe that any problem like this would be set up very specifically, and that only certain values will work at all, allowing me to just use one component of the vectors.
Edit: oops, I see you were still writing when I posted this.. sorry for jumping the gun so quickly, haha :P
Edit 2:
Oh dang, you're right. So you can't just tweak one component of F2,1.. doing that would prevent them from being parallel and would cause all kinds of issues. This is just a simple equation of a vector being equal to a magnitude times a direction unit vector. Sorry, it's been a little while since I've really worked with vectors in such a way, so I've been spending a lot of time on issues like this. Thanks again, very much!
Last edited: Apr 9, 2014
4. Apr 9, 2014
### ehild
You can not put $\hat{r}_{2,1}$ on the other side. But you know that $\vec{F}_{2,1}$ is a scalar multiple of $\hat{r}_{2,1}$, and that holds for all corresponding coordinates.
It would be better to see the whole text of the problem.
ehild
5. Apr 9, 2014
### Chetlin
Well, I have my question answered, but to be complete, here is the text of the problem. It's from an old Schaum's Outlines book from the early 1980's.
Point charge Q1 = 300 µC, located at (1, −1, −3) m, experiences a force F2,1 = (8, −8, 4) N due to point charge Q2 at (3, −3, −2) m. Determine Q2.
6. Apr 9, 2014
### Staff: Mentor
What you do is dot both sides of the equation by $\hat{r}_{2,1}$. This will give you the scalar equation you desire. Do you know how to determine $\hat{r}_{2,1}$?
Chet
7. Apr 9, 2014
### Chetlin
Yep, that unit vector is just the vector r2,1 divided by its magnitude ($\hat{r}_{2,1} = \frac{1}{r_{2,1}}\mathbf{r}_{2,1}$, where $r_{2,1} = ||\mathbf{r}_{2,1}|| = \sqrt{{r_{{2,1}_x}}^2 + {r_{{2,1}_y}}^2 + {r_{{2,1}_z}}^2}$). Sorry for the really weird notation inside the square root sign.
That's a really neat trick, thanks for showing it to me. I never thought to use scalar products to get rid of the vectors, and of course the scalar product of any unit vector with itself is 1.
8. Apr 10, 2014
### ehild
The point charge Q1 experiences force from Q2:
$$\vec F = k \frac {Q_1 Q_2}{(\vec r_1-\vec r_2)^2} \hat r_{12}$$
where $\hat r_{12}=\frac{\vec r_1-\vec r_2} {|\vec r_1-\vec r_2|}$
In your problem, r12=(1, −1, −3)-(3, −3, −2)=(-2, 2, -1). The magnitude is 3, so the components of $\hat r_{12}$ are (-2/3, 2/3, -1/3)
The force is F= (8, −8, 4) N. You see that the force is parallel to $\hat r_{12}$.
You can write out the Coulomb Law in x,y,z components:
$$F_x=8=\frac {kQ_1Q_2}{r_{12}^2}(-2/3)$$
$$F_y=-8=\frac {kQ_1Q_2}{r_{12}^2}(2/3)$$
$$F_z=4=\frac {kQ_1Q_2}{r_{12}^2}(-1/3)$$
You see that you get the same value for $kQ1Q2/ {r_{12}^2}$ from each equation, it is -12. kQ1Q2/9=-12. That is a scalar equation already.
ehild | 2018-07-16T23:13:19 | {
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https://brilliant.org/discussions/thread/1-3/ | #1
How many positive integers less than $$1000$$ have the property that the sum of the digits of each such number is divisible by $$7$$ and the number itself is divisible by $$3$$?
Note by Vilakshan Gupta
1 year, 2 months ago
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Let's think of a number $$abc (0 \leq a, b, c \leq 9)$$. $$a+b+c \equiv 0 (\mod 3 \text{and} \mod 7)$$. Thus, $$a+b+c=21$$.
$(3, 9, 9) \rightarrow \frac{3!}{2}, (4, 8, 9) \rightarrow 3!, (5, 7, 9) \rightarrow 3!, (5, 8, 8) \rightarrow \frac{3!}{2}, (6, 6, 9) \rightarrow \frac{3!}{2}, (6, 7, 8) \rightarrow 3!, (7, 7, 7)$ $$3+6+6+3+3+6+1=28$$
Please tell me if there is any error.
- 11 months, 3 weeks ago
Good one brother
- 11 months, 3 weeks ago
Nice method
- 11 months, 3 weeks ago
33
- 2 months, 3 weeks ago
28
- 10 months, 2 weeks ago
27
- 1 year ago
its sum is divisibli by 21 using this you can solve
- 1 year ago
this question came in this year PRMO answer is 28
- 1 year ago
@Md Zuhair 8
- 1 year, 1 month ago
O i see....
- 1 year, 1 month ago
@Pokhraj Harshal Rajasthan region!
- 1 year, 1 month ago
And what about the other participants and their marks from your school. I mean the averages and the highest marks
- 1 year, 1 month ago
Which region are u from??
- 1 year, 1 month ago
WB rgion
- 1 year, 1 month ago
@Pokhraj Harshal Ok! Then I'm also getting the same...
- 1 year, 1 month ago
How much? Without the question?
- 1 year, 1 month ago
The questions will be cancelled
- 1 year, 1 month ago
Are the marks gonna be added to everyone's total or the questions will be cancelled (lowering the cutoff)?
- 1 year, 1 month ago
They will be added to totsl
- 1 year ago
Hey , what does discounted actually refer to?
- 1 year, 1 month ago
Hey I'm getting 8/27 from jharkhand as per the new answer key of hbcse.will I qualify???
- 1 year, 1 month ago
Lets see....
- 1 year, 1 month ago
Hello, There are 28 postive integers left less than 1000 have the property that the sum of the digits of each such number is divisible by 7 and the number itself is divisible by 3 You can check out for more queries related to the JEE EXAMS from the following compilation
- 1 year, 1 month ago
I believe the answer is 28 integers. The sum of these integers' digits must be divisible by 21, since a number divisible by 3 also has its sum of digits divisible by 3; in addition to the sum of digits divisible by 7. None of the digits can be less 3 since the sum of digits would be less than 21. Possible combinations = 7+6+5+4+3+2+1 = (7+1)+(6+2)+(5+3)+4=3*8+4=28.
- 1 year, 1 month ago
Ah - that's where the 28 comes from - much more mathematical than my just listing and counting them
- 1 year, 1 month ago
Exactly
- 1 year, 1 month ago
I agree with all that, and I got the same answer, but if I give you 4 digits at random (say 3, 4, 5 and 6) and ask how many numbers you can make out of them, the answer is 432*1 = 24, not 4+3+2+1 = 10. What am I missing?
- 1 year, 1 month ago
- 1 year, 2 months ago
Yup.I think so.
- 1 year, 2 months ago
@Md Zuhair @Vilakshan Gupta I haven't attempted one of the bonus question. Will I still get marks for it?
- 1 year, 2 months ago
I think the question can be cancelled as how can a person attempt to decinal answers and i had attempted ine. So i dunno.
- 1 year, 2 months ago
yeah hundreds digit cant be 1,2
- 1 year, 2 months ago
@Shreyan Chakraborty The Hundreds digit can't be 1 or 2..
- 1 year, 2 months ago
ANSWER IS 28....HAS A BIJECTION WITH a+b+c=21 WHERE 0<a,b,c<=9........
- 1 year, 2 months ago
Are na na.... I am not telling that. How much are you getting?
- 1 year, 2 months ago
@Shreyan Chakraborty .. How much?
- 1 year, 2 months ago
JANI NA BAJE HOYECHE
- 1 year, 2 months ago
LetTheFateDecide !!Bye
- 1 year, 2 months ago
which class are u in toshit?
- 1 year, 2 months ago
It's 11 in Rajasthan ! 😅😒
- 1 year, 2 months ago
According to Resonance , cutoff in Chandigarh is just 4(questions)
- 1 year, 2 months ago
i don't think it will be so low
- 1 year, 2 months ago
If that isnt, then wb will be higher and i will surely not qualify
- 1 year, 2 months ago
Ya. Thats ridiculous. WB region has always got a higher cutoff...
- 1 year, 2 months ago
Coz as per cutoff(s) uploaded by Resonance , cutoff in Chandigarh is lower than others ( Rajasthan , Maharashtra , UP , etc) ... That's why!
- 1 year, 2 months ago
So , you are already selected..Great 👍
- 1 year, 2 months ago
- 1 year, 2 months ago
oh
- 1 year, 2 months ago
Rajasthan..U?
- 1 year, 2 months ago
oh...btw,where do u live (i mean which region)
- 1 year, 2 months ago
Geometry was quite tough and lengthy! Excluding bonus , I'm getting 8
- 1 year, 2 months ago
Sir😅 I am getting 10 along with bonus!
- 1 year, 2 months ago
Oh. U mean 8/28 u r getting?
- 1 year, 2 months ago
well, if it is bonus that means 2 questions marks are given extra.If it was been written question deleted then scores would be evaluated out of 28
- 1 year, 2 months ago
unfortunately, i will get only 10 questions correct. I did very silly mistakes
- 1 year, 2 months ago
@Md Zuhair Is the paper for 9,10,11 and 12 same?
- 1 year, 2 months ago
Yes Sir!
- 1 year, 2 months ago
How mañy have you got right in PRMO - 17?
- 1 year, 2 months ago
- 1 year, 2 months ago
we just need to find the numbers which add upto 21
- 1 year, 2 months ago
28
- 1 year, 2 months ago
25 is the answer as per me
- 1 year, 2 months ago
No - I get 28 too - I constructed a 0 - 9 by 0 -9 addition table in excel, and then started adding a 3rd digit to any number whose 2 digits had added to 12 or more - though now I think about it, I could just as easiky have srated my list with 399 and continued from there. And it has to be 28 cos it's one starting with3, 2 starting with 4, 3 starting with 5 etc, and 1+2+3+4+5+6+7 = 28
- 1 year, 2 months ago
What ans did you get?
- 1 year, 2 months ago
Hey Aaron. How much are u getting with bonus? With bonus i am getting 12.
- 1 year, 2 months ago
zuhair tui ki amk jiggesh korchish??
- 1 year, 2 months ago
Accha.. nijer whatsapp number ta de... whatsapp e kotha bolchi
- 1 year, 2 months ago | 2018-10-23T17:23:39 | {
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https://amaanswers.com/how-many-acres-is-200-feet-by-200-feet | # How many acres is 200 feet by 200 feet?
Jerilyn Kalandek asked, updated on October 26th, 2022; Topic: 200 feet
👁 511 👍 64 ★★★★☆4.9
200 feet multiplying by 200 feet equal to 40,000 sq ft. Therefore, 40,000 sq ft divide by 43,560 sq ft equal to 0.92 acres approximately.
Either way, what is the perimeter of a 1/4 acre?
There are 43560 square feet in an acre so one quarter of an acre is square feet. If each side of the square is F feet long then the area is F^2 square feet. Thus F^2 = 10890 and F = \sqrt{10890} = 104.35 feet.
Despite that, is an acre 200x200? An area 200 ft long by 200 ft wide encompasses 40,000 sq feet. This is 40000/43560 = 0.91827acres. 200 feet x 200 feet = 0.918 acres. Or in other words, approximately 92% of an acre.
Suitably, how many acres is 100 feet by 100 feet?
We know 43,560 square feet to 1 acre. 100 ft multiplying by 100 equal to 10,000 sq ft. Therefore, 10,000 sq ft divide by 43,560 sq ft equal to 0.23 acres approximately.
How much does a quarter acre cost?
2. Average Cost of Clearing Land by Lot Size
AcreageCost Range
¼ acre$125 -$1,400
½ acre$250 -$2,800
1 acre$500 -$5,600
2 acres$1,000 -$11,200
### What is the size of an acre lot?
1 acre is approximately 208.71 feet × 208.71 feet (a square) 4,840 square yards. 43,560 square feet.
### What is linear feet?
Technically, a linear foot is a measurement that is 12 inches long (so, one foot) and that is measured in a straight line, which is why it's called linear.
### What is a linear acre?
The most standard shape for an acre is one furlong by one chain, or 660 feet by 66 feet. To find the linear measurements of other rectangular acres, just divide 43,560 by the number of feet you want on one side. A square-shaped acre would then be about 208.7 by 208.7 feet (because 208.7 x 208.7 = ~43,560).
### How many acres is 150 feet by 200?
We know 43,560 square feet to 1 acre. 200 feet multiplying by 200 feet equal to 40,000 sq ft. Therefore, 40,000 sq ft divide by 43,560 sq ft equal to 0.92 acres approximately.
### How many square feet is 200?
We know about 43,500 square feet. 20,000 sq ft is equal to 200 feet. 43,500 sq ft is equal to 0.92 sq ft.
### How big is half an acre compared to a football field?
If you calculate the entire area of a football field, including the end zones, it works out to 57,600 square feet (360 x 160). One acre equals 43,560 square feet, so a football field is about 1.32 acres in size.
### How big is a 50 by 100 plot of land?
1)For a rectangular plot, 50 by 100 refers to 50 feet by 100 feet which is equivalent to 15 meters by 30 meters and is also equal to 450 square meters. This is what people refer to as 1/8 of an acre though slightly less due to the provision for access road.
### What does 1acre look like?
As all farmers and real estate agents know, an acre is defined as an area one furlong long by 4 rods wide. ... Basically if you can picture a football field, that's pretty close to an acre in size. Officially, it is 43,560 square feet, and a football field is 48,000 square feet. Our standard acre isn't the same worldwide.
### How long does it take to walk around 4 acres?
A square acre is 208.7 feet on a side, so the perimeter of an acre is about 835 feet, or about 16 percent of a mile. If you walk a brisk pace of 3 miles an hour, you can cover a mile in 20 minutes. So you should be able to walk 835 feet in about three minutes.
| 2023-03-26T09:55:42 | {
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http://math.stackexchange.com/questions/703156/modular-arithmetic-are-we-allowed-to-distribute-the-modularity | # Modular Arithmetic - Are we allowed to distribute the Modularity?
Assume I have a problem such as "Prove that $\displaystyle103^{53} + 53^{103}$ is divisible by $39$." This would mean I wanted to prove that $\displaystyle103^{53} + 53^{103}\equiv0\pmod{39}$.
My starting statement would be then "$\displaystyle103^{53} + 53^{103}\pmod{39}$" and I would then equate this to "$\displaystyle103^{53}\pmod{39} + 53^{103}\pmod{39}$" and then continue.
Am I allowed to distribute the mod like that?
Thanks.
-
Yes, you are... – DonAntonio Mar 7 '14 at 16:06
@Matthew, please verify the edited version – lab bhattacharjee Mar 7 '14 at 16:14
That should be a $+$ rather than a comma, right? In the formula right before "and then continue". – Jack M Mar 7 '14 at 17:38
Mathematicians and programmers have two different ways of thinking about mod. To the latter, it is a binary operation in which ${\rm Mod}[a,b]$ is the remainder of $a$ when divided by $b$ (so it will return an integer in the range $0\le r<b$). To the former, it is a binary relation (a symbol that relates things in some way, like $<,=,>,\approx,\sim$ etc.) for each modulus $b$. We say that $n\color{Red}{\equiv}m$ mod $b$ if the difference $n-m$ is divisible by $b$, or equivalently if ${\rm Mod}[n,b]={\rm Mod}[m,b]$. Sometimes the equivalence symbol $\equiv$ is simply replaced by an equality symbol $=$, in which case we are understood to be equating equivalence classes. The relation $n\equiv m$ mod $b$ is in fact a congruence relation (it "respects" $+$ and $\times$), and the equivalence classes are called residue classes, or just residues.
So if $a\equiv b$ and $c\equiv d$ mod $m$ then $ac\equiv bd$ and $a+c\equiv b+d$ mod $m$. One can use this to end up proving that, in particular, $f(a)\equiv f({\rm Mod}[a,m])$ mod $m$ for integer-coefficient polynomials $f$.
Thus for example ${\rm Mod}[a,m]+{\rm Mod}[b,m]$ and $a+b$ and ${\rm Mod}[a+b,m]$ are all congruent mod $m$, however it is not strictly true that the first and last are equal as integers. Take $a,b=2$ and $m=3$, in which case ${\rm Mod}[2,3]+{\rm Mod}[2,3]=2+2=4\ne1={\rm Mod}[2+2,3]$, albeit $4\equiv1$ mod $3$.
If $n$ and $m$ are coprime, then $a\equiv b$ modulo both $n$ and $m$ if and only if $a\equiv b$ mod $nm$. In particular this means $x\equiv 0$ mod $39$ if and only if $x\equiv0$ mod $3$ and mod $13$. Compute
$$103^{53}+53^{103}\equiv 1^{53}+(-1)^{103}\equiv 1+(-1)\equiv0\mod 3$$
because $103\equiv1$ and $53\equiv-1$ mod $3$. And then compute
$$103^{53}+53^{103}\equiv(-1)^{53}+1^{103}\equiv(-1)+1\equiv0\mod 13$$
because $103\equiv-1$ and $53\equiv1$ mod $13$. Since $103^{53}+53^{103}$ is $0$ mod $3$ and $13$, it is $0$ mod $39$.
-
Good, and complete. For the purposes of students going farther in mathematics, it is disastrous to keep the mindset of relying on the operation rather than the equivalence relation. When talking about $R/I$ where $R$ is a ring and $I$ is an ideal, or $G/H$ where $\supset H$ are groups, the equivalence relation outlook is essential. – Lubin Mar 7 '14 at 18:49
If $103^{53} \equiv a \mod 39$ and $53^{103} \equiv b \mod 39$, than it is indeed true that $103^{53} + 53^{103} \equiv a+b \mod 39$. I wouldn't recommend your notation, which makes $\mod\mbox{ }$ look like an operator.
-
There's nothing wrong with using $\bmod$ as an operator. – ShreevatsaR Mar 7 '14 at 16:54
Using mod as an operator loses the properties of the equivalence class that makes it so convenient. For example, you cannot assume $a \mod M + b \mod M = (a + b) \mod M$. – DanielV Mar 7 '14 at 17:48
@DanielV: Assuming the name "mod" refers to a proper modulus operator and not the much-less-useful "remainder" operator found in many programming languages, and one is not expecting one's numeric type to act as an algebraic ring with a modulus which is not a multiple of M, why can't one assume that (a mod M)+(b mod M) will equal ((a+b) mod M) in cases where the result of the addition is defined? – supercat Mar 7 '14 at 18:25
Ah, if I understand, you are talking about a $\mathbb{N} \rightarrow \text{Galois}[M]$ operator, yes that would be useful indeed. – DanielV Mar 8 '14 at 0:02
To say that $$x \equiv a \pmod{m}$$ means that there is some integer $k$ such that $$x - a = mk$$ Thus, expanding on user133281's answer (i.e. if $103^{53}\equiv a \pmod{39}$ and $53^{103}\equiv b \pmod{39}$), we have $$(103^{53}-a)+(53^{103}-b)=39k+39l$$ which is equivalent to $$(103^{53}+53^{103})-(a+b)=39(k+l)$$ so $$103^{53}+53^{103}\equiv a+b\pmod{39}$$
-
Either you meant $\mod 39$ in the first line or $mk$ in the second. – Mike Miller Mar 8 '14 at 3:45
@Mike Right you are. Thank you. – iamnotmaynard Mar 8 '14 at 20:56
Glad to help!$\$ – Mike Miller Mar 8 '14 at 21:37
HINT:
Please have a look into this for the properties of congruence
$\displaystyle 103\equiv-1\pmod{13}\implies 103^{53}\equiv(-1)^{53}\equiv-1$
$\displaystyle 53\equiv1\pmod{13}\implies 53^{103}\equiv(1)^{103}\equiv1$
Similarly, for $\pmod3$
Now if $13$ and $3$ both divides $a,a$ will be divisible by lcm$(13,3)$
-
While this is a good hint for solving the problem "show that $103^{53} + 53^{103}$ is divisible by 39", this doesn't address the question that was asked ("am I allowed to distribute the mod") at all. – Magdiragdag Mar 7 '14 at 16:05
@Magdiragdag, added a relevant link – lab bhattacharjee Mar 7 '14 at 16:10
The "equivalent modulo $n$" relation is a congruence: a congruence is an equivalence relation with the additional property that for all relevant arithmetic operations (in this case, $0, 1, +, \times$, and anything derived from those), if the inputs the the arithmetic operation are congruent, then the outputs are also congruent.
In the actual ring of integers modulo $n$, though, there is no "congruence" or "mod": e.g. $39 = 0$ is a literal equality. Your question to prove
$$103^{53} + 53^{103} \equiv 0 \pmod{39}$$
in the integers is the same thing as trying to prove
$$103^{53} + 53^{103} = 0$$
in the ring of integers modulo $39$. And since we're working with actual equality, it's clear that we can simplify the two summands separately, then add the simplified results.
- | 2015-07-29T05:07:58 | {
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https://mathhelpboards.com/threads/prove-that-the-expression-is-divisible-by-26460.3520/ | # Prove that the expression is divisible by 26460
#### anemone
##### MHB POTW Director
Staff member
Problem:
Prove that $27195^8-10887^8+10152^8$ is divisible by $26460$.
Attempt:
I grouped the last two terms and manipulated them algebraically and came to the point where I suspect I might have taken the wrong path...here is the last step where I stopped and don't know how to proceed.
$\dfrac{27195^8-10887^8+10152^8}{26460}=\dfrac{3^5\cdot5^7\cdot7^{14}\cdot37^8-7013(2^6\cdot3^6\cdot47^2+3^2\cdot19^2\cdot191^2)(10152^4+10887^4)}{8}$
I'd like to ask, do you think this problem can be solved using only elementary methods?
Last edited by a moderator:
#### Opalg
##### MHB Oldtimer
Staff member
Problem:
Prove that $27195^8-10887^8+10152^8$ is divisible by $26460$.
Attempt:
I grouped the last two terms and manipulated them algebraically and came to the point where I suspect I might have taken the wrong path...here is the last step where I stopped and don't know how to proceed.
$\dfrac{27195^8-10887^8+10152^8}{26460}=\dfrac{3^5\cdot5^7\cdot7^{14}\cdot37^8-7013(2^6\cdot3^6\cdot47^2+3^2\cdot19^2\cdot191^2)(10152^4+10887^4)}{8}$
I'd like to ask, do you think this problem can be solved using only elementary methods?
The best way to make the calculation more manageable is to factorise $26460 = 2^2\cdot 3^3\cdot 5\cdot 7^2$. If you can separately show that $27195^8-10887^8+10152^8$ is divisible by each of the numbers $2^2$, $3^3$, $5$ and $7^2$, then the result will follow.
Take the factor 5, for example. Since $27195$ is a multiple of $5$, so is its eighth power. The other two terms are not multiples of $5$, but here you need to use your idea of grouping those two terms together. In fact, $a^8-b^8$ is a multiple of $a-b$. So $10887^8+10152^8$ is a multiple of $10887-10152 = 735$. That is a multiple of $5$. Putting those results together, you see that $27195^8-10887^8+10152^8$ is a multiple of $5$.
The exact same procedure shows that $27195^8-10887^8+10152^8$ is a multiple of $7^2$. You can use similar ideas to show that it is also a multiple of $3^3$ and of $2^2$. (To deal with $3^3$, notice that $a^8-b^8$ is also a multiple of $a+b$.)
#### anemone
##### MHB POTW Director
Staff member
The best way to make the calculation more manageable is to factorise $26460 = 2^2\cdot 3^3\cdot 5\cdot 7^2$. If you can separately show that $27195^8-10887^8+10152^8$ is divisible by each of the numbers $2^2$, $3^3$, $5$ and $7^2$, then the result will follow.
Take the factor 5, for example. Since $27195$ is a multiple of $5$, so is its eighth power. The other two terms are not multiples of $5$, but here you need to use your idea of grouping those two terms together. In fact, $a^8-b^8$ is a multiple of $a-b$. So $10887^8+10152^8$ is a multiple of $10887-10152 = 735$. That is a multiple of $5$. Putting those results together, you see that $27195^8-10887^8+10152^8$ is a multiple of $5$.
The exact same procedure shows that $27195^8-10887^8+10152^8$ is a multiple of $7^2$. You can use similar ideas to show that it is also a multiple of $3^3$ and of $2^2$. (To deal with $3^3$, notice that $a^8-b^8$ is also a multiple of $a+b$.)
Awesome! I finally understand it now!
#### Bacterius
##### Well-known member
MHB Math Helper
Opalg's answer is very good! A more "heavy machinery" method could also go as follows:
$$26460 = 2^2 \cdot 3^3 \cdot 5^1 \cdot 7^2$$
And we can do the following reductions with a couple modulo operations:
$$27195^8 - 10887^8 + 10152^8 \equiv 3^8 - 3^8 + 0^8 \equiv 0 \pmod{2^2}$$
$$27195^8 - 10887^8 + 10152^8 \equiv 6^8 - 6^8 + 0^8 \equiv 0 \pmod{3^3}$$
$$27195^8 - 10887^8 + 10152^8 \equiv 0^8 - 2^8 + 2^8 \equiv 0 \pmod{5^1}$$
$$27195^8 - 10887^8 + 10152^8 \equiv 0^8 - 9^8 + 9^8 \equiv 0 \pmod{7^2}$$
And invoking the CRT (since $2^2$, $3^3$, $5^1$, $7^2$ are pairwise coprime):
$$\mathbb{Z}_{26460} = \mathbb{Z}_{2^2} \times \mathbb{Z}_{3^3} \times \mathbb{Z}_{5^1} \times \mathbb{Z}_{7^2}$$
$$\therefore$$
$$27195^8 - 10887^8 + 10152^8 \equiv 0 \pmod{26460}$$
This also shows that the exponent (8) is in fact irrelevant, and could be anything.
EDIT: fixed, see ILikeSerena's post below.
Last edited:
#### Klaas van Aarsen
##### MHB Seeker
Staff member
Opalg's answer is very good! A more "heavy machinery" method could also go as follows:
$$26460 = 2^2 \cdot 3^3 \cdot 5^1 \cdot 7^2$$
And we can do the following reductions with a couple modulo operations:
$$27195^8 - 10887^8 + 10152^8 \equiv 0^8 - 0^8 + 0^8 \equiv 0 \pmod{2}$$
$$27195^8 - 10887^8 + 10152^8 \equiv 0^8 - 0^8 + 0^8 \equiv 0 \pmod{3}$$
$$27195^8 - 10887^8 + 10152^8 \equiv 0^8 - 2^8 + 2^8 \equiv 0 \pmod{5}$$
$$27195^8 - 10887^8 + 10152^8 \equiv 0^8 - 2^8 + 2^8 \equiv 0 \pmod{7}$$
And invoking the CRT (since 2, 3, 5, 7 are pairwise coprime):
$$\mathbb{Z}_{26460} = \mathbb{Z}_{2^2} \times \mathbb{Z}_{3^3} \times \mathbb{Z}_{5^1} \times \mathbb{Z}_{7^2}$$
$$\therefore$$
$$27195^8 - 10887^8 + 10152^8 \equiv 0 \pmod{26460}$$
This also shows that the exponent (8) is in fact irrelevant, and could be anything.
I'm afraid that the number $2 \cdot 3 \cdot 5 \cdot 7$ is also zero mod 2, mod 3, mod 5, and mod 7.
But it is not divisible by 26460.
For CRT you need to show the mod relation for $2^2$, $3^3$, $5^1$, and $7^2$.
For instance for $3^3$, we need:
$\hspace{0.5 in}27195^8 - 10887^8 + 10152^8 \equiv 0^8 - 0^8 + 0^8 \equiv 0 \pmod{3^3}$
Still quite doable.
#### Bacterius
##### Well-known member
MHB Math Helper
I'm afraid that the number $2 \cdot 3 \cdot 5 \cdot 7$ is also zero mod 2, mod 3, mod 5, and mod 7.
But it is not divisible by 26460.
For CRT you need to show the mod relation for $2^2$, $3^3$, $5^1$, and $7^2$.
For instance for $3^3$, we need:
$\hspace{0.5 in}27195^8 - 10887^8 + 10152^8 \equiv 0^8 - 0^8 + 0^8 \equiv 0 \pmod{3^3}$
Still quite doable.
True, my mistake. Though it's not too difficult to check, just do the same operations but using 4, 27 and 49... (which still work out, it's clear the integers in the problem were carefully chosen to cancel each other out) | 2021-10-21T15:06:04 | {
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http://math.stackexchange.com/questions/272274/alternative-ways-to-find-lim-x-to-1-frac1-x1-sqrtx | # Alternative ways to find $\lim_{x\to 1}\; \frac{1-x}{1- \sqrt{x}}$
Searching on Google by a solution, I found this:
$1) \space$The numerator can be written by a diference of squares:$\space (1-\sqrt{x})(1+\sqrt{x})$
$2 \space)$Then, one can "eliminated" the commum factor beteween the numerator and the denominator.
$3) \space$The final expression looks like: $$\lim_{x\to 1}\; {1+ \sqrt{x}}=2$$
However this was not a intuitive algebraic solution. I haven't thought of this solution at the first attempts. Could you please give me other alternative algebraic solutions, if it exists. Thanks.
-
Same solution, let $x=u^2$, but now instantly familiar. – André Nicolas Jan 7 '13 at 17:20
Why isn't it intuitive? If you have a $0/0$ form, the first thing one usually does is search for common factors. – David Mitra Jan 7 '13 at 17:21
The difference of two squares method was the first thing that popped into my head before I clicked the link to this question! – Clive Newstead Jan 7 '13 at 17:21
The other approach to this solution is to think - that was neat, how will I be able to spot neat solutions like that next time and the comment from @AndréNicolas was exactly what I was going to put. – Mark Bennet Jan 7 '13 at 17:22
@João What's $(1-\sqrt{x})(1+\sqrt{x})$? – WimC Jan 7 '13 at 17:35
You can compute the inverse: $$\lim_{x\to 1}\frac{1-\sqrt{x}}{1-x}=\lim_{x\to 1}\frac{\sqrt{x}-1}{x-1}$$ which is, by definition, $f'(1)$ where $f(x)=\sqrt{x}$.
-
As I understood, you "turn over" the fraction,by puting $\frac{1-\sqrt{x}}{1-x}$ on the denominator. Then you multiplied the denominator by $\frac{-1}{-1}$ to get the final expression, that is under the main fraction (the numerator is $1$).Finally you computed the derivative of $\sqrt{x}$ at $x=1$ to get $\frac{1}{\frac{1}{2}}$. Is that? – João Jan 7 '13 at 18:04
Yeah, that's basically it. Essentially, if $g(x)/h(x)\to A$ then $h(x)/g(x)\to 1/A$ – Thomas Andrews Jan 8 '13 at 14:12
Same solution, let $x=u^2$, but now instantly familiar.
-
Perhaps we can use L'Hospital rule here ?
-
By substitution:$\quad$ let $x = u^2\,.\quad$Then as $x\to 1,\;\;u^2\to 1\implies u\to 1$, giving us $$\lim_{x\to 1}\; \frac{1-x}{1- \sqrt{x}} \;\;=\;\; \lim_{u\to 1}\; \frac{1-u^2}{1- u} = \lim_{x\to 1} \frac{u^2 - 1}{u-1} \;=\;\lim_{u \to 1} \frac{(u-1)(u+1)}{u-1} = \lim_{u\to 1}(u + 1) = 2$$
Using the substitution makes the "difference of squares" route looks so much more obvious!
-
If $u^2 \to 1$, then $u \to 1$ or $u \to -1$. Is this second situation if one compute the limit we get $0$.M'I rigth? – João Jan 7 '13 at 18:14
since $x \to 1$, the only root of concern is u = 1 (since $u = \sqrt{x}$ must necessarily be $\ge 0$ in this situation. – amWhy Jan 7 '13 at 18:22 | 2016-02-13T17:22:57 | {
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https://math.stackexchange.com/questions/2193674/why-1-equiv-ap-1-mod-p/2193701 | # Why $1\equiv a^{p-1} \mod p$?
Let $\mathbb{Z}_p^*$, where $p$ is prime and let $a\in\mathbb{Z}_p^*$.
Consider the following equation:$$(p-1)! \equiv (p-1)! a^{p-1} \mod p$$
I've read that since $\gcd((p-1)!, p) = 1$ we can infer that $$a^{p-1} \equiv 1$$
So I have two questions:
1. Why is it true that $\gcd ((p-1)!, p)= 1$?
2. Why can we infer that $a^{p-1} \equiv 1$?
• for the frist question: if $p$ is prime then observe than $(p-1)!$ doesnt contain $p$ as a factor, hence $\gcd((p-1)!,p)=1$ – Masacroso Mar 19 '17 at 15:44
• And for the second question, you can always write the $gcd(a,b)$ as a linear combination of $a$ and $b$ with integer coefficients (this is from the Euclidean algorithm). Use this to show that if some number is relatively prime to $p$, then it has a multiplicative inverse mod $p$. – Malkoun Mar 19 '17 at 15:46
• Fermat's little theorem en.wikipedia.org/wiki/Fermat%27s_little_theorem?wprov=sfla1 – Zuo Mar 19 '17 at 15:51
1. The only divisors of $p$ are $p$ and $1$. What are the divisors of $(p-1)!$?
2. Once you know that $p, (p-1)!$ are relatively prime, consider $(p-1)! * (a^{p-1} - 1) \equiv 0 \mod p$. The relative primeness tells you that this is only possible if $(a^{p-1} - 1) \equiv 0 \mod p$
This common result is known as Fermat's Little Theorem. I hope this simple proof helps:
Consider the sequence of integers $n,2n,3n,…,(p−1)n$.
Note that none of these integers are congruent modulo $p$ to the others.
If this were the case, we would have $an≡bn \pmod p$ for some $1≤a<b≤p−1$.
Then as $gcd(n,p)=1$, and we can cancel the $n$, we get $a≡b \pmod p$ and so $a=b$.
Also, since $p∤n$ and $p∤c$, for any $1≤c≤p−1$, then by Euclid's Lemma $p∤cn$ for any such $cn$, which means $cn≢0 \pmod p$.
Thus, each integer in the sequence can be reduced $modulo \ p$ to exactly one of $1,2,3,…,p−1$.
So ${1,2,3,…,p−1}$ is the set of Reduced Residue System $modulo \ p$.
So, upon taking the product of these congruences, we see that $n×2n×3n×⋯×(p−1)n≡1×2×3×⋯×(p−1) \mod p$.
This simplifies to $n^{p−1}×(p−1)!≡(p−1)! \pmod p$.
Since $p∤(p−1)!$, we can cancel $(p−1)!$ from both sides, leaving us with $n^{p−1}≡1 \pmod p$.
• The last statement is essentially my question. Why does the fact $p\not\vert (p-1)!$ implies that you can cancel $(p-1)!$ from both sides? – OliOliver Mar 19 '17 at 16:14
• It doesn't essentially imply that $(p-1)!$ is being cancelled because of that property. It merely states that they are relatively prime to each other and, hence, we can use the property of modular arithmetic to basically remove $(p-1)!$ as if it of no use here(as we are dealing with, say, modulo $p$ and not modulo $(p-1)!$) and appears on both sides leaving it to be factored out. – HKT Mar 19 '17 at 16:25
• Carefully think about it. Suppose if $p|(p-1)!$, we would have been left with $0 \pmod p$ on the right hand side. I hope that helps. – HKT Mar 19 '17 at 16:28
• You are just over-thinking it. I assure you once you get the hang of it and review the basic modular arithmetic definitions and get a stronger grasp on the notation, there will be clarity. – HKT Mar 19 '17 at 16:30
If you write down $(p-1)!$ as $(p-1)\cdot(p-2)\cdot\cdot\cdot1$ you can easily notice that $p$ and $(p-1)!$ have no common factors.
So $gcd((p-1)!,p) = 1$.
Now, knowing that:
$$ax \equiv b \mod(m) \implies x \equiv b \cdot a^{-1} \mod(\frac{m}{gcd(m,a)})$$
• You might as well just write $ax\equiv b\bmod m\implies x=a^{-1}b\bmod m$ when $\gcd(a,m)=1$. No need to overcomplicate it. – arctic tern Mar 19 '17 at 15:55
For the first question, take, for example, $p = 7$. Then $(p-1)! = 2\cdot3\cdot4\cdot5\cdot6 = 2^4\cdot3^2\cdot5$. Notice that in the first equation all of the factors are strictly less than $p = 7$, which implies that all of the prime factors are also strictly less than $p$. Since $p$ is prime, this means that $p$ and $(p-1)!$ can have no prime factors in common.
Now this implies Fermat's little theorem because, having shown that $(p-1)!$ is relatively prime with $p$, we can apply the cancellation law.
Whenver $\gcd(a,m) = 1$, $$ax \equiv ay \mod m$$ $$\downarrow$$ $$x \equiv y \mod m$$ | 2019-12-14T16:12:58 | {
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https://math.stackexchange.com/questions/3608462/rolling-two-dice-alternatively-what-is-probability-of-winning | Rolling two dice alternatively. What is probability of winning?
Tom and Paul roll 2 dice alternatively starting with Tom. Consider they use two fair 6-faced dice. The player who rolls 6 first wins. They continue to roll until one of them wins. Find probability that Tom wins.
I have listed out the total possible outcomes below:
{(1,1), (1,2), (1,3), (1,4), (1,5), (1,6), (2,1), (2,2), (2,3), (2,4), (2,5), (2,6), (3,1), (3,2), (3,3), (3,4), (3,5), (3,6), (4,1), (4,2), (4,3), (4,4), (4,5), (4,6), (5,1), (5,2), (5,3), (5,4), (5,5), (5,6), (6,1), (6,2), (6,3), (6,4), (6,5), (6,6)}
Here are all the possible outcomes of two dice rolls. And as one can see, only 5 of these comes out with a 6 first; so, I think the probability of Tom winning should be 5/36.
However, when I looked at the solution (This is a question from one midterm in one of previous semesters given at my school) says:
p is 5/36
$$P(Tom\, Wins)=\sum_{k=0}^{\infty }p(1-p)^{2k}=p\sum_{k=0}^{\infty }(1-p)^{2k}=$$ $$\frac{\frac{5}{36}}{1-\frac{31}{36}}=\frac{5}{5}$$
As one can see this answer is really rediculus! (Tom will win no matter what??)
I think that the key here is probably on the word "alternatively", but cannot figureout what has gone wrong here.
• You’ve computed that sum incorrectly. Observe that the exponent is $2k$, not $k$, i.e., you’ve got a geometric series in $(1-p)^2$, not $1-p$.
– amd
Apr 3 '20 at 19:50
• I didn't come out with that answer. It is answer given by someone. I looked at it and was very confused. Apr 4 '20 at 1:37
The probability that Tom wins on the first roll is $$\frac16$$. He wins on the third roll if and only if he first rolls something other than a $$6$$, then Paul rolls something other than a $$6$$, and then Tom rolls a $$6$$; the probability of this is $$\frac56\cdot\frac56\cdot\frac16$$. In general, Tom wins on the $$(2k+1)$$-st roll if and only if the first none of the first $$2k$$ rolls is a $$6$$, and Tom rolls a $$6$$ on roll $$2k+1$$; this occurs with probability $$\frac16\left(\frac56\right)^{2k}$$. Thus, the probability that Tom wins is
\begin{align*} \sum_{k\ge 0}\frac16\left(\frac56\right)^{2k}&=\frac16\sum_{k\ge 0}\left(\frac56\right)^{2k}\\ &=\frac16\sum_{k\ge 0}\left(\frac{25}{36}\right)^k\\ &=\frac16\cdot\frac1{1-\frac{25}{36}}\\ &=\frac6{11}\;. \end{align*}
This is the formula in the solution that you read, with $$p=\frac16$$.
It should not be surprising that this is slightly more than $$\frac12$$: the fact that Tom goes first gives him an advantage, but it’s a small one, since the game is otherwise very symmetric.
One can also compute the desired probability without resort to infinite series. Let $$p$$ be the probability that Tom wins, so that Paul wins with probability $$1-p$$. On the other hand, the probability that Paul when Tom first rolls something other than a $$6$$ must be $$p$$, because at that point the game is effectively starting over with Paul as the first player. Thus, Paul wins with probability $$\frac56p$$, the probability that Tom rolls something other than a $$6$$ initially and Paul then rolls a $$6$$ before Tom does. In short, $$1-p=\frac56p$$, and solving for $$p$$ again yields $$p=\frac6{11}$$.
• Thank you. I think your answer is good. I guess the solution given for this problem is probably wrong. Apr 4 '20 at 1:36
• @AmosKu: You’re welcome. Apr 4 '20 at 1:42
• I have given the question quite a bit of thought last night. I think the key is "alternatively". And whenever Tom rolls a 6, he wins. I think your answer is for the case where only one 6 face dice is rolled. I have edited my wording above, hoping to make the meaning of original question clearer. Apr 4 '20 at 17:44
• @AmosKu: Alternatively makes no sense here; I suspect that the intended word is alternately. That simply means that Tom rolls his die, then Paul rolls his, and so on. This is exactly the same problem as if they were using a single die and trading it back and forth. Apr 4 '20 at 17:52
• Possible. But I just double checked the original question. The word is " Alternatively". There are other errors in the writing of the test as well. I guess you are right. Apr 4 '20 at 20:07 | 2021-09-22T20:00:39 | {
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https://www.physicsforums.com/threads/conclusion-about-the-dimension-of-c-r.919980/ | # Conclusion about the dimension of C°(R)?
Tags:
1. Jul 12, 2017
### Rodrigo Schmidt
[mentor note: thread moved from Linear Algebra to here hence no homework template]
We have a linear map A:E→E, where E=C°(ℝ), the vector space of all continuous functions.
Let's suppose that Aƒ= x0 ƒ(t)dt.
By the Calculus Fundamental Theorem, d/dx(Aƒ) = ƒ, so we have a left inverse, which implies that ker(A)={0}.
Supposing that dim(E) is finite, by the rank-nullity theorem we have that im(A)=E . As a result of that:
(ƒ(x)=|x|) ∈ E ⇒ ƒ ∈ im(A)
⇒dƒ/dx ∈ E
But we know that ƒ's derivative is not continuous. So, supposing that dim(E) is finite lead us to a contradition (dƒ/dx ∈ E ∧ dƒ/dx ∉ E) therefore dim(E) must be infinite.
Is this argument valid? If not, could you guys point where does it fail? Thank you!
Last edited by a moderator: Jul 12, 2017
2. Jul 12, 2017
### Krylov
Note that the right-hand side is $(Af)(x)$, not $Af$.
The derivative is not even defined at $x = 0$.
Yes, it is valid, but it is quite convoluted. If you insist on doing it this way, I would prefer to say that you have found a linear operator $A$ that is injective but not surjective (the latter because $A$ maps into $C^1(\mathbb{R})$). This already implies infinite dimensionality of $E$, since on a finite dimensional space injectivity and surjectivity are equivalent for linear operators.
Of course, if you merely care to show that $E$ is infinite dimensional, it is more straightforward to identify an infinite linearly independent set.
3. Jul 12, 2017
### Rodrigo Schmidt
Thanks for the knowledge shared!
I hadn't tought that, that's, indeed, much simpler.
So the basis of the set of all polynomials would be enough?
4. Jul 12, 2017
### LCKurtz
The set of all polynomials is too large to be a basis for $C_0$, but $\{1,x,x^2,\dots\}$ would do.
5. Jul 12, 2017
### mathwonk
this is very nice, and shows a creative grasp of what you are learning. keep it up and you will eventually do some new research!
6. Jul 13, 2017
### Krylov
While I usually like the straightforward approach best (maybe this is one of the differences between a "pure" mathematician and an (aspiring) "applied" mathematician?), I gave it some thought and then came to the conclusion that I agree with you.
Just as an exercise, you could try to complete the direct approach as well. (You are almost there, anyway.)
7. Jul 13, 2017
### Rodrigo Schmidt
Hm, i see, but isn't that the basis of the set of all polynomials? So, by being a basis, that must be a linearly independent set, and it's also infinite, so that implies in the infinite dimensionality of C°(ℝ)?
That's where i want to get someday. Thanks for the inspiration!
8. Jul 13, 2017
### Krylov
I think that I and post #4 misunderstood you when you wrote
We (or I, at least) thought that here you asserted that the set of all polynomials itself is a basis for a linear subspace of $C^0(\mathbb{R})$, maybe because you wrote "the basis". However, from what you wrote afterwards (quoted at the top of this post), I get that you meant any (Hamel) basis for the linear space of all polyniomials. (For example, you can indeed use the canonical basis mentioned in post #4. By the way, I think there the $C_0$ was written by small mistake.)
Very good, as far as I can see, you are more than done.
Last edited: Jul 13, 2017
9. Jul 13, 2017
### Rodrigo Schmidt
Yeah, that's what i meant. Sorry if i wasn't clear enough! English is not my mother language so, mostly when writing about math and science, my texts can get a little confusing. I will try to be more specific in the next time.
Okay! Thanks for the support and knowledge shared! | 2017-12-14T21:36:09 | {
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http://math.stackexchange.com/questions/171966/whats-the-probability-a-subset-of-an-mathbb-f-2-vector-space-is-a-spanning-se | # Whats the probability a subset of an $\mathbb F_2$ vector space is a spanning set?
Let $V$ be an $n$-dimensional $\mathbb F_2$ vector space. Note that $V$ has $2^n$ elements and $\mathcal P(V)$ has $2^{2^n}$.
I'm interested in the probability (under a uniform distribution) that an element of $\mathcal P(V)$ is a spanning set for $V$. Equivalently a closed form formula (or at least one who's asymptotics as $n\rightarrow \infty$ are easy to work out) for the number of spanning sets or non-spanning sets.
It's not hard to show that the probability is greater than or equal to $1/2$. Since any subset of size greater than $2^{n-1}$ must span the space. I calculated the proportion of spanning sets of size $n$ for $n$ up to $200$ which seem to be going to a number starting with $.2887$. This leads me to believe that the probability exceeds $1/2$. I couldn't nail down a formula for arbitrary sized subsets though to continue experimental calculations.
I feel like this is something that's been done before, but googling I've mostly found things concerning counting points on varieties over finite fields or counting subspaces of finite fields. Any references would be appreciated.
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I think it would be sufficient to count how many different bases you can have in V. Then multiply that number by all possibilities of adding elements to it, since that does not perturb the spanning property of the set. – Raskolnikov Jul 17 '12 at 15:31
@Raskolnikov The issue with that is avoiding duplicates. – JSchlather Jul 17 '12 at 15:32
Indeed, you're right. – Raskolnikov Jul 17 '12 at 15:33
A random subset of $V$ is owerwhelmingly likely to span $V$.
Let's look at how hard it is for a random subset not to span $V$. In order not to span $V$, there must be an $(n-1)$-dimensional subspace that contains the entire subset. There are exactly $2^n-1$ such subspaces, since they are in bijective correspondence with the nontrivial linear maps $V\to\mathbb F_2$ (each subspace is the kernel of exactly one map).
For each fixed $(n-1)$ dimensional subspace, the probability for a random subset to stay within that subspace is $2^{-2^{n-1}}$, since the $2^{n-1}$ vectors outside the subspace must all randomly decide not to be in the subset.
So the probability for a random set not to span is at most $(2^n-1)2^{-2^{n-1}} < 2^{-(2^{n-1}-n)}$ (and this is slightly too high, because there are a few non-spanning subsets that have more than one proper subspace they fit into and so are counted twice here). Everything else spans.
Even for $n$ as small as $5$ the probability for a random subset to span $V$ is above 99.9%, and the number of 9's increases exponentially with larger $n$s.
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There are indeed $2^n - 1$ subspaces of dimension $n - 1$. There is a bijection with subspaces of dimension $1$, though not via the inner product. – Zhen Lin Jul 17 '12 at 15:57
I think the problem is that the probability that some set $S$ is not contained in some subspace $U$ is not independent of the probability that $S$ is not contained in some other subspace $U'$. – Zhen Lin Jul 17 '12 at 16:01
That works, and it's certainly enough to get the conclusion that the probability goes to $1$ as $n \to \infty$. – Zhen Lin Jul 17 '12 at 16:08
(+1 to Henning) @ZhenLin: A non-degenerate bilinear form gives that correspondence (as you probably noticed) between $(2^n-1)$ non-zero vectors (=1-dim. subspaces) and $(n-1)$-dimensional subspaces. IMHO the word orthogonal might still be used - with the caveat that a vector or a subspace may be orthogonal to itself. – Jyrki Lahtonen Jul 17 '12 at 16:23
@Jyrki: I think it is cleaner to speak of nonzero vectors in $V$ and $(n-1)$-dimensional subspaces in $V^{\ast}$ (their annihilators) as this correspondence is functorial and doesn't require the addition of extra structure. – Qiaochu Yuan Jul 17 '12 at 20:18
It's worth mentioning that there is an explicit formula for the probability that $N$ random vectors drawn from $\mathbb{F}_q^n$ span. (Note: I am doing sampling with replacement. If you want to require the $N$ vectors to be distinct, you can do this using inclusion-exclusion, but the result will be much messier.) Of course, the answer is $0$ if $N<n$, so assume $N \geq n$.
Write down an $n \times N$ matrix whose columns are the vectors in question. Note that the following are equivalent:
1. The columns span $\mathbb{F}_q^n$.
2. The matrix has rank $n$.
3. The rows are linearly independent.
Thinking in terms of rows, we have the new question: If we draw $n$ vectors at random from $\mathbb{F}_q^N$, what is the probability that they are linearly independent?
The probability that the first vector is nonzero is $1-q^{-N}$. Given that, the probability that the second vector is not in the span of the first is $1-q^{-N+1}$. Given that, the probability that the third vector is not in the span of the first two is $1-q^{-N+2}$. Continuing in this manner, the probability that $n$ vectors in $\mathbb{F}_q^n$ are independent is $$(1-q^{-N}) (1-q^{-N+1}) \cdots (1-q^{-N+n-1}).$$ And, by the above argument, this is also the probability that $N$ random vectors will span $\mathbb{F}_q^n$.
In particular, as Henning says, once $N-n$ is of any significant size, this is extremely close to $1$.
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https://oschvr.com/problems/7/ | By listing the first six prime numbers: 2, 3, 5, 7, 11, and 13, we can see that the 6th prime is 13.
What is the 10 001st prime number?
At first glance, a simple iteration whilst checking if index position is prime, would suffice.
A pseudocode attempt:
sum = 0
for i = 2; i <= ?; i++ {
if(isPrime(i)){
sum++
if ( sum >= 10001) {
return i
}
}
}
Using the solution in go to check if is prime from former examples:
func isPrime(x int64) bool {
var i int64 = 2
for ; i < x; i++ {
if x%i == 0 {
return false
}
}
return true
}
The solution would scale to
$$O(n^2)$$.
This is not viable. Also, there's no upper bound to set our iteration limit, meaning, we don't know up to which number we should iterate to find the 10,001 prime.
So digging further, I found about the Erathostenes Sieve:
An ancient algorithm for finding all prime numbers up to any given limit
This is exactly what we need. Let's look at the procedure:
To find all the prime numbers less than or equal to a given integer n by Erathostenes' method:
1. Create a list of consecutive integers from 2 ... n (2,3,4,...,n)
2. Initially, let p equal 2, smallest prime.
3. Enumerate multiples of p by counting in increments of p from 2p to n, and mark them in the list (these will be 2p, 3p, 4p, ...; the p itself should not be marked)
4. Find the first number greater than p in the list that is not marked, if there is not such number, stop. Otherwise, let p now equal to this number (which is next prime), and repeat step 3
5. When algorithm terminates, the numbers remain not marked in the list, are all primes below n.
Here's the code I came up:
package main
import (
"fmt"
"math"
"time"
)
func primeSieve(n, limit int) int {
// Create and populate array of values
// lim := c
a := make([]bool, limit+1)
for i := 0; i < limit+1; i++ {
a[i] = true
}
// p * p <= limit === p <= int(math.Sqrt(float64(limit)))
for p := 2; p <= int(math.Sqrt(float64(limit))); p++ {
// At first all will be true
if a[p] == true {
// i = 2 * 2, i = 3 * 2, i = 4 * 2, ..., i = i * i
// Calculate multiples of 2, then 3, then 5, then 7
for i := p * 2; i <= limit; i += p {
a[i] = false
}
}
}
// Count primes up to n
// if primes == 10001, return sievePrime
var primes, sievePrime int
for p := 2; p <= limit; p++ {
if a[p] == true {
primes++
// Sum up primes and print 10,001th prime
if primes <= n {
sievePrime = p
}
}
}
return sievePrime
}
func main() {
start := time.Now()
// Nth prime to find
var n, limit int = 10001, 105000 // Arbitrary limit find by testing
fmt.Println("Execution time: ", time.Since(start))
fmt.Println("10,001th Prime: ", primeSieve(n, limit))
} | 2020-06-03T22:48:55 | {
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http://math.stackexchange.com/questions/125588/limit-of-21-n-as-n-to-infty-is-1 | # Limit of $2^{1/n}$ as $n\to\infty$ is 1
How do I prove that:
$\lim \limits_{n\to \infty}2^{1/n}=1$
Thank you very much.
-
Using the binomial theorem for integer exponents:
Can you see that $(1+\frac 1 n)^n > 2>1$
Take the nth root, to give:
$(1+\frac 1 n) > 2^{\frac 1 n} >1$
-
oh, that's really nice. how to you prove that $(1+\frac{1}{n})^n>2$? Thanks! – Anonymous Mar 28 '12 at 20:18
$(1+x)^n = 1+nx+$ other positive terms – Mark Bennet Mar 28 '12 at 20:20
And $n$ has to be greater than 1. – Mark Bennet Mar 28 '12 at 20:21
sweet, thank you. – Anonymous Mar 28 '12 at 20:25
Now look at Sivaram Ambikasaram's answer and catch hold of how the two are related - then you'll have learned some maths. – Mark Bennet Mar 28 '12 at 20:33
Note that for $a\gt 0$, $$a^b = e^{b\ln a}.$$ So $$\lim_{n\to\infty}2^{1/n} = \lim_{n\to\infty}e^{\frac{1}{n}\ln 2}.$$ Since the exponential is continuous, we have $$\lim_{n\to\infty}e^{\frac{1}{n}\ln 2} = e^{\lim\limits_{n\to\infty}\frac{1}{n}\ln 2}.$$
Can you compute $\displaystyle\lim_{n\to\infty}\frac{\ln 2}{n}$ ?
-
Thank you for the help, however I don't want to prove this using ln and e. – Anonymous Mar 28 '12 at 20:09
@Anonymous: Thank you for your comment; perhaps next time you can state, in your post, what it is you do and do not want, so that other people may not waste their time giving you information you don't care about. – Arturo Magidin Mar 28 '12 at 20:11
I really appreciate your time and sorry for the wasted time for me, however there are probably people who would find this informative and want to know how to prove it this way. I didn't state that I don't want lan and e because I wasn't aware that it can be proven this way. Thank you very much again. – Anonymous Mar 28 '12 at 20:16
This answer deserves more than one upvote, as it not only answers the question, but also deals neatly with the ambiguity in the way the question is asked. – Mark Bennet Mar 28 '12 at 21:37
HINT: Use squeeze theorem.
Since $1 < 2$, we have $1 = 1^{1/n} < 2^{1/n}$ for all $n \in \mathbb{N}$.
To bound the limit from above, note that $1 + n \epsilon < \left( 1 + \epsilon \right)^n$.
Hence, given any $\epsilon >0$, $\forall n > \displaystyle 1/{\epsilon}$, we have $2 < 1 + n \epsilon < \left(1 + \epsilon \right)^n$ and hence $2^{1/n} < 1 + \epsilon$.
-
Here are a few different proofs which don't use $e$ or $\log$ and can be regarded completely elementary.
Proof 1)
We use the following theorem:
If $a_n \gt 0$ and $\lim_{n \to \infty} \frac{a_{n+1}}{a_n} = L$, then $\lim_{n\to\infty} a_n^{1/n} = L$
This is a standard theorem, and a proof can be found in almost any textbook. You can also find a proof in my answer here: Show that this limit is equal to $\liminf a_{n}^{1/n}$ for positive terms.
Apply the theorem to the sequence $a_n = 2$.
Proof 2)
We use the $\text{AM} \ge \text{GM}$ inequality on $n-1$ ones and one $2$.
$$\frac{1 + 1 + \dots + 1 + 2}{n} \ge 2^{1/n}$$
$$\frac{n+1}{n} \ge 2^{1/n}$$
Thus we have
$$1 + \frac{1}{n} \ge 2^{1/n} \ge 1$$
so by Squeeze theorem, $\lim_{n \to \infty} 2^{1/n} = 1$.
Proof 3)
We can use Bernouli's inequality (essentially similar to Sivaram's answer) to show that
$$\left(1 + \frac{1}{n}\right)^n \ge 1 + \frac{1}{n} \times n = 2$$
and we get inequalities similar to the proof in 2).
Proof 4)
The sequence $a_n = 2^{1/n}$ is bounded below (by $1$) and montonically decreasing.
Thus it is convergent, to say $L$. Since $a_{2n}$ also converges to $L$, we have that $L = \sqrt{L}$, as $2^{1/2n} = \sqrt{2^{1/n}}$. So $L = 0$ or $L = 1$. Since the limit is not less than $1$ ($2^{1/n} \ge 1$), the limit is $1$.
Proof 5)
For $n \gt 2$, we have that $1 \le 2^{1/n} \le n^{1/n}$.
Now use the fact that $\lim_{n \to \infty} n^{1/n} = 1$.
An elementary proof of that can be found here: http://math.stackexchange.com/a/115825/1102. Any proof for $n^{1/n}$ now becomes a proof for $2^{1/n}$. Proof 1) above can also be used for $n^{1/n}$.
Proof 6)
Using combinatorics.
The number of $n$ digit numbers in base-$n+1$ is $(n+1)^n$ (allowing for leading zeroes). The number of $n$ digits numbers in base-$n$ is $n^n$. We can show that $(n+1)^n \ge 2 \times n^n$: consider the base-$n$ numbers. Replace the last digit with $n$. You get a base-$n+1$ $n$ digit number. Counting the base-$n$ numbers (which are also base-$n+1$ numbers) and the "last digit modified" numbers, gives us the inequality.
This inequality implies that $1 + \frac{1}{n} \ge 2^{1/n}$ and can be used to give a proof using the squeeze theorem, similar to proofs 2 and 3.
-
Sorry, had some free time :-) I am pretty sure there are more proofs... – Aryabhata Mar 28 '12 at 21:31
Very nice list! – Pedro Tamaroff Mar 29 '12 at 23:50
@PeterT.off: Thanks! – Aryabhata Mar 29 '12 at 23:59 | 2015-08-03T19:15:17 | {
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https://www.physicsforums.com/threads/sum-of-series.230104/ | # Sum of series
## Homework Statement
Find the sum of the series:
$$S = 1~+~\frac{1 + 2^3}{1 + 2}~+~\frac{1 + 2^3 + 3^3}{1 + 2 + 3}~+~\frac{1 + 2^3 + 3^3 + 4^3}{1 + 2 + 3 + 4}~+~...$$
upto 11 terms
## Homework Equations
Sum of first 'n' natural numbers: $$S = \frac{n(n + 1)}{2}$$
Sum of the squares of the first 'n' natural numbers: $$S = \frac{n(n + 1)(2n + 1)}{6}$$
Sum of the cubes of the first 'n' natural numbers: $$S = \left(\frac{n(n+1)}{2}\right)^2$$
## The Attempt at a Solution
In the series, the $n^{th}$ term is given by:
$$T_n = \frac{1 + 2^3 + 3^3 + ... + n^3}{1 + 2 + 3 + ... + n}$$
$$T_n = \frac{\left(\frac{n(n+1)}{2}\right)^2}{\left(\frac{n(n + 1)}{2}\right)}$$
$$T_n = \frac{1}{2}(n^2 + n)$$
Hence,
$$S_n = \frac{1}{2}\left(\frac{n(n + 1)(2n + 1)}{6} + \frac{n(n + 1)}{2}\right)$$
On substituting n = 11, I get:
$$S_n = 286$$
But this isn't one of the options I got on the test. All the options were between 300-400 and all had their last digit '9' [that's all i remember.. they took the question paper away].
Last edited: | 2021-12-02T04:10:35 | {
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https://math.stackexchange.com/questions/3717811/what-is-bbb-ex/3717822 | # What is $\Bbb E(X)$?
Let there be $$100$$ balls in a box out of which $$50$$ are red and $$50$$ are green. Pick $$75$$ balls at random from the box and throw them away. Now pick one ball at random from the remaining balls in the box. Let $$X$$ be the random variable which takes the value $$100$$ when the ball drawn is red in colour and takes the value $$25$$ if the ball drawn is green in colour. Find the expectation $$\Bbb E(X).$$
My attempt $$:$$ Let $$Y$$ denote the number of red balls thrown away. Then the number of green balls thrown away is $$75-Y.$$ Clearly $$Y \geq 25.$$ So \begin{align*} \Bbb P(X=100) & = \sum\limits_{n=25}^{49} \Bbb P(X=100 \mid Y=n)\ \Bbb P(Y=n) \\ & = \sum\limits_{n=25}^{49} \frac {\binom {50-n} {1}} {\binom {25} {1}} \times \frac {\binom {50} {n}} {\binom {100} {n}} \end{align*} Similarly \begin{align*} \Bbb P(X=25) & = \sum\limits_{n=25}^{49} \Bbb P(X=25 \mid (75-Y) = n)\ \Bbb P((75-Y) = n) \\ & = \sum\limits_{n=25}^{49} \Bbb P(X=25 \mid Y = 75-n)\ \Bbb P(Y = 75-n) \\ & = \sum\limits_{n=25}^{49} \frac {\binom {50-n} {1}} {\binom {25} {1}} \times \frac {\binom {50} {75-n}} {\binom {100} {75-n}} \end{align*}
Then the required expectation would be $$100\ \Bbb P(X=100) + 25\ \Bbb P(X=25).$$ But the computation is very tough. Is there any simpler way to approach the problem?
Any help will be highly appreciated. Thank you very much.
• It seems to me that red and green are indistinguishable in this scenario. Whatever the probability of picking a red ball is, it must be equal to the probability of picking a green ball. So P(X=100) = P(X=25) = 1/2 – Ant Jun 13 '20 at 8:51
• But @Ant after throwing $75$ balls away it might not be the case. Obviously the green and red balls in the box are present with different proportions after throwing $75$ balls. Right? Then the red and green balls are not equally likely to be drawn. – Phi beta kappa Jun 13 '20 at 8:55
• I have posted an answer, so we can keep discussing it there. To answer your question, they will be present in a different proportion after being drawn, but that's just a single observation. Their distribution is the same; the distribution of how many red vs green balls end up in the urn after you throw some away is the same. Therefore the distribution of which ball you pick after is also the same. So probabilities are the same. Right? – Ant Jun 13 '20 at 9:02
• It is clear by symmetry that $$\mathbb P(X=25)=\mathbb P(X=100)=\frac12.$$ There is no need for elaborate computation here. – Math1000 Jun 13 '20 at 10:09
• @Math1000 can you show it mathematically without making some vague assertion? I know that you are a genius; but it's better not to think others as genius as you. – Phi beta kappa Jun 13 '20 at 10:22
It seems to me that red and green are indistinguishable in this scenario. Whatever the probability of picking a red ball is, it must be equal to the probability of picking a green ball. So $$P(X=100) = P(X=25) = 1/2$$
Which implies
$$E(X) = 100\cdot P(X=100) + 25 \cdot P(X=25) = 62.5$$
A simple script confirms that the expectation is indeed $$62.5$$. You can also double check by computing the probability. Using your terminology,
$$P(Y=n) = \frac{{75 \choose n} \cdot {25 \choose 50-n}} {100\choose 50}$$
So that
$$P(X=100) = \sum_{n=25}^{50} \frac{50-n}{25} \cdot \frac{{75 \choose n} \cdot {25 \choose 50-n}} {100\choose 50}$$
And you can check numerically that this is equal to $$1/2$$
• How have you computed $\Bbb P(Y=n)$? – Phi beta kappa Jun 13 '20 at 9:43
• The probability of removing n red balls can be computed as follows: Line up all the 100 balls in a row. Count how many of the permutations end up with n balls in the first 75 spots. Divide by total number of permutations, and you're done. To compute the first number, you have 75 balls of which n are red. So the permutations are 75!/n!/(75-n)!, i.e. 75 choose n. Then you have to multiply with the number of permutations of the remaining 25 balls; it's the same, except that you now have 50-n red balls. The second number is the total number of permutations, therefore 100 choose 50. – Ant Jun 13 '20 at 9:48
• I don't think so. If according to your assumption the balls are indistinguishable then we have just $26$ ways to throw $75$ balls from the urn and each way is equally likely. If the balls are indistinguishable then it doesn't matter in which order you throw them away or which red/green balls are thrown away. In this case which really matters that how many red/green balls you throw away out of $75$ balls. Right? If I present them as ordered pairs as (number of red balls thrown away, number of green balls thrown away) then there are $26$ such pairs $(25,50),(26,50), \cdots, (50,25).$ – Phi beta kappa Jun 13 '20 at 10:15
• Each pair is equally likely to occur. So for each $25 \leq n \leq 50$ we have $$\Bbb P(Y=n) = \frac {1} {26}.$$ – Phi beta kappa Jun 13 '20 at 10:17
• No. I know it for the case of distinguishable but not for the case of indistinguishable. – Phi beta kappa Jun 13 '20 at 10:28 | 2021-06-14T22:35:28 | {
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http://math.stackexchange.com/questions/168288/the-dynamics-of-contrapositive-proofs | # The Dynamics of Contrapositive Proofs
The Wikipedia Link for contrapositive proofs states that proving if p then q is the same as proving if not q then not p. I don't completely follow why.
Is there any way to understand what's happening without involving logic tables ? If logic tables are inevitable, can anyone give me a basic explanation?
I can't seem to follow how if p then q gets simplified to not p or q I understand that the truth tables of the two are same. But, that's that.
-
It could help to think the about contrapositive of some simple theorems, like the Pythagorean theorem. – Ragib Zaman Jul 8 '12 at 16:43
Semiformally speaking: Suppose $P\implies Q$. Further suppose $Q$ is false; if $P$ were true then $Q$ would also be true and we'd have a contradiction - impossible - so $P\implies Q$ and $\lnot Q$ implies $\lnot P$, i.e. $P\implies Q$ implies $\lnot Q\implies\lnot P$, ie an implication entails its contrapositive. Of course this means the contrapositive entails the contrapositive's contrapositive, ie the original statement, hence we have an equivalence, $(P\implies Q)\iff (\lnot Q\implies \lnot P)$. – anon Jul 8 '12 at 16:54
The sentence "if $p$ then $q$" asserts that in every single situation (model) in which $p$ is true, $q$ must be true. The only way this can be incorrect is if we have a situation in which $p$ is true but $q$ is false. We can show this cannot be the case by showing that whenever $q$ fails, $p$ must fail, that is, by showing that if "not $q$" then "not $p$."
Conversely, suppose that in every situation in which $p$ is true, then $q$ must be true. Then there cannot be a situation in which $q$ fails and $p$ is true. So if "not $q$" holds, then "not $p$" must hold.
Thus the two assertions "if $p$ then $q$" and "if not $q$ then not $p$" hold in precisely the same cases.
Remark: The answer above is much too abstract. To understand what's going on, it is useful to examine a number of concrete cases. Suppose that we want to show that if $x^2=2$ then $x$ is not an ordinary fraction (integer divided by integer). So $p$ is the assertion $x^2=2$, and $q$ is the assertion that $x$ is not a fraction.
We prove the result by showing that if $x$ is a fraction, then $x^2$ cannot be equal to $2$, that is, by showing that "not $q$" implies "not $p$." That shows that if $x^2=2$, then $x$ could not possibly be a fraction, which is exactly what we wanted.
-
Much Clearer now. Thanks André. – Inquest Jul 8 '12 at 17:23
"If $P$ then $Q$" tells you that whenever you know that $P$ is true, you may automatically conclude that $Q$ is true.
"If not $Q$, then not $P$" tells you that whenever you know that $Q$ is not present, you may conclude that $P$ is not present.
If ($P$ implies $Q$) is true, then knowledge of the presence of $P$ always gives us that $Q$ is present. If $Q$ isn't there, then there must be no $P$ to have put it there. Likewise, if the absence of $Q$ automatically gives us the absence of $P$, and if $P$ is present, there can be no not-$Q$ to have given us a not-$P$.
Similarly, either we have $P$ or we don't have $P$. In the case that we have $P$, we are guaranteed to have $Q$. The other case is that we don't have $P$. So we either don't have $P$, or we have $Q$ (which we got by having $P$): not-$P$ or $Q$.
On the other hand, let's say all we know is that either $P$ isn't true or $Q$ is true. If $P$ isn't true, then $P\Rightarrow Q$ is trivially true. Let's deal with the case where $P$ is true. So we know that $P$ is true and also that either $P$ isn't true or $Q$ is true. That is, since $P$ and not-$P$ can't both be true, if $P$ is true, then $Q$ is true.
-
+1. Took me a few reads to understand. The last paragraph's overall flow and structure is why most of my non-math friends don't speak to me when I'm doing math. – Inquest Jul 8 '12 at 17:22
Sorry. I'll take a look to revise it. – Neal Jul 8 '12 at 17:26 | 2016-06-26T08:44:57 | {
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https://stats.stackexchange.com/questions/182875/is-every-correlation-matrix-positive-definite | # Is every correlation matrix positive definite?
I'm talking here about matrices of Pearson correlations.
I've often heard it said that all correlation matrices must be positive semidefinite. My understanding is that positive definite matrices must have eigenvalues $> 0$, while positive semidefinite matrices must have eigenvalues $\ge 0$. This makes me think that my question can be rephrased as "Is it possible for correlation matrices to have an eigenvalue $= 0$?"
Is it possible for a correlation matrix (generated from empirical data, with no missing data) to have an eigenvalue $= 0$, or an eigenvalue $< 0$? What if it was a population correlation matrix instead?
Consider three variables, $X$, $Y$ and $Z = X+Y$. Their covariance matrix, $M$, is not positive definite, since there's a vector $z$ ($= (1, 1, -1)'$) for which $z'Mz$ is not positive.
However, if instead of a covariance matrix I do those calculations on a correlation matrix then $z'Mz$ comes out as positive. Thus I think that maybe the situation is different for correlation and covariance matrices.
My reason for asking is that I got asked over on stackoverflow, in relation to a question I asked there.
• If, for example, two attributes are one thing, only having different names, the matrix is singular. If two attributes add to a constant, it is again singular, et cetera. – ttnphns Nov 21 '15 at 16:54
• If a covariance matrix is singular correlation matrix is singular as well. – ttnphns Nov 21 '15 at 17:05
• Near-duplicates: Is every correlation matrix positive semi-definite? which has less focus on the definite versus semi-definite angle, and Is every covariance matrix positive definite? which is relevant because a covariance is essentially a rescaled correlation. – Silverfish Nov 21 '15 at 21:20
Correlation matrices need not be positive definite.
Consider a scalar random variable X having non-zero variance. Then the correlation matrix of X with itself is the matrix of all ones, which is positive semi-definite, but not positive definite.
As for sample correlation, consider sample data for the above, having first observation 1 and 1, and second observation 2 and 2. This results in sample correlation being the matrix of all ones, so not positive definite.
A sample correlation matrix, if computed in exact arithmetic (i.e., with no roundoff error) can not have negative eigenvalues.
• May be worth mentioning the possible effects of missing values on the sample correlation matrix. Numerical fuzz isn't the only reason to get a negative eigenvalue in a sample correlation/covariance matrix. – Silverfish Nov 21 '15 at 21:22
• Yes, I didn't make it explicit, but I was assuming, per the question statement, "with no missing data". Once you get into the wild, wacky world of missing data and adjustments therefor, anything goes. – Mark L. Stone Nov 21 '15 at 21:35
• Yes, sorry, you're quite right the question said "no missing data" - just thought it worth mentioning somewhere since future searchers might be interested even if the OP's appetite is sated! – Silverfish Nov 21 '15 at 21:40
The answers by @yoki and @MarkLStone (+1 to both) both point out that a population correlation matrix can have zero eigenvalues if variables are linearly related (such as e.g. $X_1 = X_2$ in the example of @MarkLStone and $X_1 = 2X_2$ in the example of @yoki).
In addition to that, a sample correlation matrix will necessarily have zero eigenvalues if $n<p$, i.e. if the sample size is smaller than the number of variables. In this case covariance and correlation matrices will both be at most of rank $n-1$, so there will be at least $p-n+1$ zero eigenvalues. See Why is a sample covariance matrix singular when sample size is less than number of variables? and Why is the rank of covariance matrix at most $n-1$?
• True 'dat. I suppose I could have and should have provided this info as well, but my goal was to produce a counterexample to refute the OP's hypothesis, thereby showing its invalidity Nevertheless, you should adjust your second sentence to be "In this case covariance and correlation matrices will be at most rank n−1, so there will be at least (p−n+1) zero eigenvalues." – Mark L. Stone Nov 21 '15 at 15:21
Consider $X$ to be an r.v. with mean 0 and variance of 1. Let $Y=2X$, and calculate the covariance matrix of $(X,Y)$. Since $2X=Y$, $E[Y^2]=4E[X^2]=\sigma_Y^2$, and $E[XY]=2E[X^2]$. Due to the zero mean configuration, the second moments are equal to the suitable covariances, for instance: $\mbox{Cov}(X,Y)=E[XY]-EXEY=E[XY]$.
So the covariance matrix will be: $$\Lambda = \left(\array{1 & 2 \\ 2 & 4 }\right),$$ having a zero eigenvalue. The correlation matrix will be: $$\Lambda = \left(\array{1 & 1 \\ 1 & 1 }\right),$$ having a zero eigenvalue as well. Due to the linear correspondence between $X$ and $Y$ it is easy to see why we get this correlation matrix - the diagonal will always be 1, and the off-diagonal is 1 because of the linear relationship.
• Just for math-challenged readers like myself, let me point out that the 2 in $\Lambda$ is the $cov(X,Y)= \mathbb{E}(XY)-\mathbb{E}(X)\mathbb{E}(Y)= 2\mathbb{E}[X^2]=2\left(\sigma_X^2+[E(X)]^2\right)$ this last equality resulting from: $E(X^2)=\text{Var}(X)+[E(X)]^2$. – Antoni Parellada Nov 21 '15 at 19:45
• +1 for your post. I wanted to make it easy to follow for everyone by including the $diag \Lambda^{-1/2}\,\Lambda\, diag \Lambda^{1/2}$ formula, but the comments format won't allow it. Do you think there is any valid point in including it in your post? – Antoni Parellada Nov 21 '15 at 20:44
• @AntoniParellada , I'm not exactly sure what you mean - the covariance here is a direct calculation. But I will edit and make that clearer. Thanks. – yoki Nov 22 '15 at 14:59 | 2020-07-11T04:47:21 | {
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https://forum.math.toronto.edu/index.php?PHPSESSID=i6kgsan0vsfpkpiho8afbjkju3&action=printpage;topic=2634.0 | Toronto Math Forum
APM346-2022S => APM346--Lectures & Home Assignments => Chapter 2 => Topic started by: Xiangmin.Z on January 17, 2022, 04:40:37 PM
Title: Week 2 Lec 1 (Chapter 2) question
Post by: Xiangmin.Z on January 17, 2022, 04:40:37 PM
Hello, I have a question about example 4 from W2 L1:
We are given :$u_{t}+xu_{x}=xt$
after calculation we get:
$x=Ce^t$
$du=xt dt=Cte^t dt$, so $u=C(t-1)e^t+D=x(t-1)+D$,
but how did we get $D=\phi({xe^{-t}} )$? we know it is a constant, but why is D depended on $xe^{-t}$?
Also, why would the initial condition $u|_{t=0} =0$ implies that $\phi({x}) =x$ ?
Thanks.
Title: Re: Week 2 Lec 1 (Chapter 2) question
Post by: Xiangmin.Z on January 17, 2022, 05:27:38 PM
I checked the calculation and I think the answer for x is correct, and does anyone know why D is depended on $xe^{-t}$?
Title: Re: Week 2 Lec 1 (Chapter 2) question
Post by: Victor Ivrii on January 17, 2022, 07:48:38 PM
Now it is correct $x=Ce^{t}$ and then $C=?$
Title: Re: Week 2 Lec 1 (Chapter 2) question
Post by: Xiangmin.Z on January 17, 2022, 08:19:53 PM
$C=xe^{-t}$, so D is a function of $\phi$, therefore $D=\phi({xe^{-t}} )$ ? Yes, but "$D$ is a function of it" | 2022-06-30T14:07:06 | {
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https://math.stackexchange.com/questions/1716583/prove-that-fx-x3-x-is-not-injective | # Prove that $f(x) = x^3 -x$ is NOT Injective
Okay, so I know there are plenty of injective/surjective (and thus, bijective) questions out there but I'm still not happy with the rigor of what I have done. So I'd really appreciate some help!
So the question actually asks me to do two things:
(a) give an example of a cubic function that is bijective. Explain why it is bijective.
(b) give an example of a cubic function that is not bijective. Explain why it is not bijective.
So for (a) I'm fairly happy with what I've done (I think):
$$f: \mathbb R \rightarrow \mathbb R , f(x) = x^3$$
So we know that to prove if a function is bijective, we must prove it is both injective and surjective.
Proof: $f$ is injective
Let: $$x,y \in \mathbb R : f(x) = f(y)$$ $$x^3 = y^3$$ (take cube root of both sides) $$x=y$$
Proof: $f$ is surjective
Let: $$y \in \mathbb R$$
$$x = \sqrt[3]{y}$$
$$f(x) = (\sqrt[3]{y})^3 = y$$
So I believe that is enough to prove bijectivity for $f(x) = x^3$. Keep in mind I have cut out some of the formalities i.e. invoking definitions and sentences explaining steps to save readers time. This is just 'bare essentials'.
So for (b)
$$f: \mathbb R \rightarrow \mathbb R , f(x) = x^3 - x$$
Now I'm just going to try and prove it is NOT injective, as that should be sufficient to prove it is NOT bijective.
Let:
$$x,y \in \mathbb R : f(x) = f(y)$$ $$x^3 - x = y^3 - y$$
This is about as far as I get. Send help. Thanks everyone.
• What happens if $x=0$ or $x=1$? – Cameron Williams Mar 28 '16 at 2:59
• To prove that a function $f$ is not injective, you just need to find two different numbers $x$ and $y$ such that $f(x)\not=f(y)$. – Christopher Carl Heckman Mar 28 '16 at 3:02
• @Carl you mean two numbers $x,y$ such that $x\neq y$ but $f(x)=f(y)$ – JMoravitz Mar 28 '16 at 3:03
• @CarlHeckman as JMoravitz said, you may confuse op... – YoTengoUnLCD Mar 28 '16 at 3:07
• Thanks for the help guys, +1's given. – Rubicon Mar 28 '16 at 3:53
Injectivity:
A function $f$ from $X\to Y$ is said to be injective iff the following statement holds true:
for every $x_1,x_2\in X$ if $x_1\neq x_2$ then $f(x_1)\neq f(x_2)$
The negation of this then yields:
A function $f$ from $X\to Y$ is not injective iff there exists $x_1,x_2\in X$ such that $x_1\neq x_2$ but $f(x_1)=f(x_2)$
In the case of the cubic in question, it is an easily factorable polynomial and we can find multiple distinct roots. $f(x)=x^3-x=x(x^2-1)=x(x+1)(x-1)$
We know that a root of a polynomial is a number $\alpha$ such that $f(\alpha)=0$
So, $f(1)=f(0)=f(-1)=0$ despite $1,0,-1$ all being distinct unequal numbers in the domain. This shows that it is not injective, and thus not bijective.
As an aside, one can prove that any odd degree polynomial from $\Bbb R\to \Bbb R$ must be surjective by the fact that polynomials are continuous and the intermediate value theorem.
• This is great. I think it has helped my understanding significantly. So what you're sort of saying is because f(1) and f(-1) equal 0, there is more than one-to-one mapping to the elements in the co-domain (hence not injective)? So in this case the set might even look like this: {(1,0) , (-1,0)} (just as a way to visualise it)? – Rubicon Mar 28 '16 at 3:53
Regarding (a), when you say "take cube root of both sides" you are (at least implicitly) assuming that the function is injective -- if it were not, the cube root would not be a well-defined operation. Depending on what you are / are not allowed to assume, that could be regarded as a circular argument.
However, there is another approach. Beginning with the assumption that $x^3=y^3$, you can manipulate the expression as follows: $$x^3-y^3=0$$ $$(x-y)(x^2+xy+y^2)=0$$ Now if you can find a convincing argument that the second factor, $x^2+xy+y^2$, is always positive for any choice of $x$ and $y$, then it follows that the only way $x^3=y^3$ can be true is if $x=y$.
The factorization $x^3-y^3=(x-y)(x^2+xy+y^2)$, by the way, could also give you some insight into how to handle (b).
• $$x^2+xy+y^2 = {x^2+y^2\over 2} + {1\over 2}(x^2 + 2xy+y^2) = {x^2+y^2\over 2} + {1\over2}(x+y)^2 \ge 0.$$ The only way to get equality is to have $x=0$ and $y=0$, but then $x=y$ in that case as well. – Christopher Carl Heckman Mar 29 '16 at 6:30
• @CarlHeckman Yes, I know. Also $x^2+xy+y^2 = (x + \frac{1}{2}y)^2 + \frac{3}{4}y^2$. I was intentionally leaving that out to leave the OP something to finish. – mweiss Mar 29 '16 at 14:16 | 2019-11-17T09:42:37 | {
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https://math.stackexchange.com/questions/426579/can-an-eigenvalue-lambda-i-0 | # Can an eigenvalue $\lambda_i=0$?
I was doing some work with diagonalization of a matrix $A$ in order to find a matrix $P$ such that $\,P^{-1}AP\,$ was diagonal. In order to that I set $\;\lambda I_{n}=0\;$ and found the characteristic polynomial and its roots.
When I factored my characteristic polynomial I obtained $\;\lambda^2(\lambda-2),\,$ so $\,\lambda=0,\,2$.
I was taught that the eignenvalues$\,\lambda_{i}\,$ I found become the entries of the diagonal matrix $\,P^{-1}AP.\,$ If this is indeed true, then two of the diagonal entries would be $\,0.\,$ Is this allowed, or must a diagonal matrix strictly have non-zero diagonal entries?
• Zero is allowed. You may be thinking of eigenvectors --- the zero vector can't be an eigenvector. But for eigenvalues, no problem. – Gerry Myerson Jun 21 '13 at 23:44
Absolutely yes: It is very possible $\lambda = 0$. Zero is allowed.
You may be mixing up what you know about eigenvectors --- the zero vector cannot be an eigenvector.
But for an eigenvalue $\lambda$, it is certainly possible and admissible that $\lambda = 0$.
With respect to your last question:
"($\lambda = 0$): Is this allowed, or does a diagonal matrix strictly have to have the diagonal entries as non-zero?"
Yes, it is allowed for zero's to be on the diagonal. No, the diagonal entries need not be non-zero.
• Dang! Beat me to the punch with the eigenvalue vs. eigenvector observation. +1 – Cameron Buie Jun 21 '13 at 23:58
• +1 what that Dang! means Amy? Dang Dang... Is that like a sound of a big Bell? – mrs Jun 22 '13 at 0:09
• +1 it's a non-profane way of saying something like "Damn": meaning, if I hadn't posted what I posted, Cameron intended to. – amWhy Jun 22 '13 at 0:12
• @amWhy Thanks; makes perfect sense now. I think I was confusing eigenvalues with eigenvectors. – Sujaan Kunalan Jun 22 '13 at 0:36
• @amWhy: Spot on +1 – Amzoti Jun 22 '13 at 0:38
A matrix is called diagonal if its off-diagonal entries ($a_{ij}$ for $i \not= j$) are all zero. This does not require the diagonal entries ($a_{jj}$) to be nonzero. For instance, the zero matrix is diagonal. | 2021-01-26T13:12:33 | {
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https://math.stackexchange.com/questions/3148433/solving-problem-using-related-rates-yields-incorrect-result | # Solving problem using related rates yields incorrect result
I've been trying for a while to figure out what I did wrong on this problem, help would be appreciated.
"A 25-ft ladder is leaning against a wall. If we push the ladder toward the wall at a rate of 1 ft/sec, and the bottom of the ladder is initially 20 ft away from the wall, how fast does the ladder move up the wall 5 sec after we start pushing?"
$$x$$ = distance from wall, $$y$$ = height of ladder on wall, $$h$$ = length of ladder $$\frac {dx}{dt} = -1, t = 5$$ $$x^2 + y^2 = h^2$$ $$x^2 +y^2 = 25^2$$ $$2x\frac{dx}{dt}+2y\frac{dy}{dt} = 0$$ $$\frac{dy}{dt} = \frac{-x\frac{dx}{dt}}{y}$$ Since the ladder is moving toward the wall at 1 ft/sec for 5 sec, $$x(t = 5) =20-5 = 15$$
$$y(t=5) = \sqrt{25^2 - 15^2}=20$$ Substitute variables in equation: $$\frac{dy}{dt} = \frac{-(15)(-1)}{20}$$ $$\frac{dy}{dt} = \frac{3}{4}$$ The issue is that if $$\frac{dy}{dt} = \frac{3}{4},$$ the ladder would have risen 3.75 ft. in 5 seconds instead of 5 feet it SHOULD have risen, which is given by: $$\Delta y = y(t=5) - y(t=0)$$ $$(\sqrt{25^2 - 15^2}) - (\sqrt{25^2 - 20^2})$$ $$15-20$$ $$5$$ Thank you for reading and any answers.
At the start you have $$y(0)=\sqrt {25^2-20^2}=15$$. You are correct that the top has risen $$5$$ feet in $$5$$ seconds, but that does not mean that it has risen steadily at $$1$$ ft/sec for the whole time. Because $$\frac {dx}{dt}=-1$$ you found $$\frac {dy}{dt}=\frac xy$$ At $$t=0$$ that means that $$\frac {dy}{dt}=\frac {20}{15}=\frac 43$$ and at $$t=5$$ that means $$\frac {dy}{dt}=\frac {15}{20}=\frac 34$$. There is no contradiction here.
• Oh, I assumed that the rate of change had to be constant for some reason. Thank you for your help! – Kebab Boy Mar 14 at 20:07
Keeping your notation, you have that $$x(t) = 20 - t, \quad y(t)=\sqrt{25^2-x(t)^2},$$
and so, $$y'(t) = \frac{-2 x'(t)x(t)}{2 \sqrt{25^2-x(t)^2}},$$
which means that $$y'(5)=\frac{15}{\sqrt{25^2-15^2}}=\frac{3}{4}$$
The vertical displacement does not have to match the horizontal one. They are just connected through the expression $$x(t)^2 + y(t)^2=25^2$$. | 2019-07-20T11:39:24 | {
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https://www.physicsforums.com/threads/solving-uartics.638159/ | Solving uartics
BloodyFrozen
Is there a method to finding the roots of quartics besides Ferrari's formula?
I have the equation
$$x^{4}+5x^{2}+4x+5=0$$
I know one of the factor is something like $$x^{2}+x+1$$ and the other one can be found using sythetic division, but how can I find the factors without knowing one of them in the first place?
Thanks.
Homework Helper
If the quartic has "nice" quadratic factors, then they will be of the form:
$$x^4+5x^2+4x+5\equiv (x^2+ax\pm 1)(x^2+bx\pm 5)$$
Can you see why this is true? If we expand the right-hand side, we just need to compare coefficients to find the values of a and b and then we should be able to determine the nature of the $\pm$ operator.
BloodyFrozen
If the quartic has "nice" quadratic factors, then they will be of the form:
$$x^4+5x^2+4x+5\equiv (x^2+ax\pm 1)(x^2+bx\pm 5)$$
Can you see why this is true? If we expand the right-hand side, we just need to compare coefficients to find the values of a and b and then we should be able to determine the nature of the $\pm$ operator.
Oh, I see, but how do we know when the factors are "nice"?
Homework Helper
Oh, I see, but how do we know when the factors are "nice"?
We don't. We can always check to see if they are, just like we check to see if there are any rational roots.
With the quartic $x^4+5x^2+4x+5$ we end up solving a system of equations for a,b that works out nicely, but if we changed the quartic around to say, $x^4+6x^2+4x+5$ then we find no solutions for a,b in the system of equations.
BloodyFrozen
Ok, thanks!
BloodyFrozen
Ah, I believe I've seen this before. The method is nice, but it seems a little lengthy. Thanks for the link
BloodyFrozen
We don't. We can always check to see if they are, just like we check to see if there are any rational roots.
With the quartic $x^4+5x^2+4x+5$ we end up solving a system of equations for a,b that works out nicely, but if we changed the quartic around to say, $x^4+6x^2+4x+5$ then we find no solutions for a,b in the system of equations.
I end up getting two systems but one of them has no solution. Did I proceed correctly?
Homework Helper
I end up getting two systems but one of them has no solution. Did I proceed correctly?
I don't know, what did you get?
BloodyFrozen
Sorry for not having LATEX since I'm on my phone, but I got:
System 1:
a+b=0
5+1+ab=5
5a+b=4
System 2:
a+b=0
-5-1+ab=5
-5a-b=4
Homework Helper
Sorry for not having LATEX since I'm on my phone, but I got:
System 1:
a+b=0
5+1+ab=5
5a+b=4
Right, so simplifying this system, we have
a+b=0
ab=-1
5a+b=4
Which we can then deduce,
a=-b
therefore,
a2=1 -> a=$\pm1$
4a=4 -> a=1, b=-1
System 2:
a+b=0
-5-1+ab=5
-5a-b=4
For this system we have no real solution, so what does that tell you?
BloodyFrozen
For the second system, I don't get any solutions. Therefore, that can't be the right system.
Homework Helper
For the second system, I don't get any solutions. Therefore, that can't be the right system.
Exactly, so what must your factors be?
BloodyFrozen
Ah, that's what I thought. Thanks! | 2022-08-18T11:15:01 | {
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https://math.stackexchange.com/questions/1990137/the-idea-behind-the-sum-of-powers-of-2/1990146 | # The idea behind the sum of powers of 2
I know that the sum of power of $2$ is $2^{n+1}-1$, and I know the mathematical induction proof. But does anyone know how $2^{n+1}-1$ comes up in the first place.
For example, sum of n numbers is $\frac{n(n+1)}{2}$. The idea is that we replicate the set and put it in a rectangle, hence we can do the trick. What is the logic behind the sum of power of $2$ formula?
• This addition of powers of two, and many other examples, are all geometric series. There's a lot of visualisations of this out there. – Parcly Taxel Oct 29 '16 at 10:58
• Write the numbers in base 2: The powers of $2$ starting from $1=2^0$ will be in binary, $1+10+100+1000$ will always be a number that will be a n with all binary digits 1. This is the largest number having that many digits. SO it is of the form $2^{n+1}-1$ – P Vanchinathan Oct 29 '16 at 11:01
• I think there's a counting/probabilistic way to approach this something about having $n$ binary choices – BCLC Oct 31 '16 at 11:17
## 8 Answers
The binary expansion of $$\sum_{k=0}^n2^k$$ is a string of $$n+1$$ 1's: $$\underbrace{111\dots111}_{n+1}$$ If I add a 1 to this number, what do I get? $$1\underbrace{000\dots000}_{n+1}$$ 1 followed by $$n+1$$ 0's, hence $$2^{n+1}$$. Therefore $$\sum_{k=0}^n2^k=2^{n+1}-1$$
• Very clever proof! – user3932000 Nov 28 '20 at 4:16
This works for any partial sum of geometric series.
Let $$S = 1 + x + x^2+\ldots +x^n$$. Then $$xS = x + x^2 + \ldots +x^n + x^{n+1} = S - 1 + x^{n+1}$$.
All you have to do now is solve for $$S$$ (assuming $$x\neq 1$$).
• Yes, but the OP said that he already knew this. What he is asking for is a visual representation of the proof. – Alex M. Oct 29 '16 at 11:01
• Actually, they said that they knew proof by induction, but didn't know how one would come up with the formula in the first place. At least that is how I understood the question. @AlexM. – Ennar Oct 29 '16 at 11:02
There is a geometrical explanation.
Take a box long enough to put twice the amount of items equal to the last term of the sum, and try to pack it.
Let's take a box of length $$2 * 2^n$$.
Let's put the items from the first term ($$2^n$$) into the box. Now exactly one half of the space is left for the other terms, from $$2^{n-1}$$ down to $$1$$, so we repeat this process, starting from the next largest term.
As we put the items from each term, we notice that they take exactly one half of the empty space left in the box, because both the largest term left and the space left decreases in half on each step.
At some point we reach the first term, which is equal to 1, and there are two empty places left in the box for just one item, so after we put the last item into the box, there is still space for one more item.
The length of the box is $$2*2^n = 2^{n+1}$$, but it could be shorter by one, which is $$2^{n+1} - 1$$, and this is our formula.
For example, let's pack: $$\sum_{i=0}^3 2^i$$
Box length: $$2 * 2^3 = 16$$
2^3 |● ● ● ● ● ● ● ●| |
2^2 |○ ○ ○ ○ ○ ○ ○ ○|● ● ● ●| |
2^1 |○ ○ ○ ○ ○ ○ ○ ○|○ ○ ○ ○|● ●| |
2^0 |○ ○ ○ ○ ○ ○ ○ ○|○ ○ ○ ○|○ ○|●| |
It could be shorter by one: $$2^{3+1}-1 = 15$$
By the way, similar geometrical explanation can be used for the ever decreasing geometric progression $$\sum_{i=0}^\infty 2^{-i}$$ with the only difference that we do not need to account for the last piece of empty space, because it tends to 0. Instead, we take some continuous medium as an example, such as a ribbon or a thread, which can be divided virtually infinitely so it sums to exactly a length of 2 units.
There is a combinatorial interpretation. Consider the collection of all binary sequences of length $$n+1$$ with at least one $$1$$ (call this set $$E$$). There are $$2^{n+1}-1$$ such sequences because only $$0...0$$ isn't. Now let $$E_j$$ be the set of binary sequences of length $$n+1$$ such that the first $$1$$ is in the $$j$$ th component for $$j=1,\dotsc, n+1$$. Then $$|E_j|=2^{n+1-j}$$. Then the $$(E_j)$$ partition $$E$$ and we have that $$1+2+2^2+\dotsb+2^n=\sum_{j=1}^{n+1}2^{n+1-j}=2^{n+1}-1.$$ Alternatively consider the telescoping sum: $$\sum_{k=0}^n 2^{k}=\sum_{k=0}^n (2^{k+1}-2^k)=2^{n+1}-1.$$
Another famous approach is to count the $$2$$-player games to determine a winner among $$2^{n+1}$$ people. It's $$2^{n+1}-1$$, the number of people to eliminate. It's also $$2^n$$ people eliminated in the first round, halving each round until we reach the final.
• Very beautiful and efficient proof ! – projetmbc Oct 23 '20 at 9:57
Another pictorial proof, similar to the box packing reply, is to count the number of nodes in a complete binary tree with n+1 levels.
$$\sum_{i = 0}^{n} 2^i$$
is the number of nodes in the complete binary tree for the levels 0 (the root) through n.
If you draw an example and number the nodes like this
1
/ \
2 3
/ \ / \
4 5 6 7
/ \
8 ...
etc. you will notice that the left-most node of each level is a power of 2 (1, 2, 4, 8, 16, ...).
so the number of nodes in the tree from root to level n is 2^{n+1} - 1.
$$\sum_{i = 0}^{n} 2^i = 2^{n+1} - 1 \,.$$
Another geometric interpretation.
Start with a line segment, AB. Extend this into a line. Use compassto mark off a point C by setting center at B and passing radius through A. Then lengths, AB=BC. Create point D in a similar way. We have thus constructed a length double that of the original line segment.
This doubling can continue AS MUCH S you like.
Now ignore the first line segment, AB.
The series of segments now starts with BD which has double the length of AB.
In general the nth segment in the first case is the n-1th segment upon removal of AB. Further, each is double the length of its corresponding section.
So By subtracting a part we've doubled what remains, but with fewer sections to account for.
This gives the formula.
Not sure if my answer has been posted yet, but here's how I understand it.
You are trying to understand why
$$2^{n+1} - 1 = 2^n + 2^{n-1} + 2^{n-2} . . . + 2^1 + 2^0$$
Suppose we take 2^n in the sum. We know since these are powers of two, that the previous term will be half of 2^n, and the term before that a quarter of 2^n.
Let n in 2^n be 1, or 2^1 = 2. The term before in the sum will be half of 2, so we can also write the entire sum as:
$$2^1 + \frac{1}{2}(2^1)$$
If you do this but for different values of n for 2^n you will find you can rewrite the sums as:
$$2^n + \frac{ 2^n - 1}{2^n} ( 2^n)$$
You can simplify this because you're dividing (2^n) - 1 by 2^n, and then multiplying it by 2^n which cancel each other out. The new simplified version with the cancellation will look like:
$$2^n + 2^n - 1$$
This can also be written as: $$2( 2^n) - 1$$
Further simplified into: $$2^{n + 1} - 1$$
And that is why that formula shows up! Hope this helps ( I'm not sure if this was the mathematical induction proof you already knew, I'm young and discovered this on my own so I don't know what this is). | 2021-04-20T18:39:17 | {
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https://brilliant.org/discussions/thread/elementary-techniques-used-in-the-imo-internatio-V/ | ×
# Elementary Techniques used in the IMO (International Mathematical Olympiad) - Cauchy Schwarz Inequality
If people asked you, what is the most elementary inequality you know? I bet your answer would be AM-GM. But in this series of posts I will try to show you the power that Cauchy Schwarz has over that of AM-GM. But as an introduction, let us first state and prove the theorem.
Cauchy Schwarz Inequality: Let $$(a_1, a_2, \ldots , a_n)$$ and $$(b_1, b_2, \ldots, b_n)$$ be two sequences of real numbers, then we have:
$$(a_1^2 + a_2^2 + \ldots + a_n^2)(b_1^2 + b_2^2 + \ldots + b_n^2) \geq (a_1b_1 + a_2b_2 + \ldots + a_nb_n)^2$$
In particular, equality holds iff there exists $$k \in \mathbb{R}$$ for which $$a_i = k b_i$$ for $$i = {1, \ldots, n}$$.
Proof: We will present $$2$$ proofs, one originating from analysis on the equality case, the other by wishful thinking on small cases of $$n = 2,3$$.
(i) Consider defining the following function $$f$$:
$$f(x) = (a_1x - b_1)^2 + (a_2x - b_2)^2 + \ldots + (a_nx - b_n)^2$$.
We will expand this to get:
$$f(x) = (a_1^2 + a_2^2 + \ldots + a_n^2)x^2 - 2(a_1b_1 + a_2b_2 + \ldots a_nb_n)x + (b_1^2 + b_2^2 + \ldots + b_n^2)$$
From our first way of representing $$f(x)$$, we can conclude that $$f(j) ≥ 0 \leftrightarrow ∆_f \leq 0$$ or
$$(a_1^2 + a_2^2 + \ldots + a_n^2)(b_1^2 + b_2^2 + \ldots + b_n^2) \geq (a_1b_1 + a_2b_2 + \ldots + a_nb_n)^2$$
Equality holds if the equation $$f(x) = 0$$ has one root.■
(ii) Just remark that:
$$(a_1^2 + a_2^2 + \ldots + a_n^2)(b_1^2 + b_2^2 + \ldots + b_n^2) - (a_1b_1 + a_2b_2 + \ldots + a_nb_n)^2 = \sum_{i,j=1}^n (a_ib_j + a_jb_i)^2 ≥ 0$$■
Let us note the following positive things regarding Cauchy-Schwarz:
• it is effective in proving symmetric inequalities
• Try to form squares
• Helps to clear up square roots
Look forward to the next few posts to see applications of this extremely elegant inequality!
Note by Anqi Li
3 years, 3 months ago
Sort by:
Hi everyone, I'm part of #TorqueGroup too. My assignment was to post olympiad level problems for the community that I found interesting. Since Anqi is posting these great topics about math things useful in such math, it seems like we're a perfect fit. So, here are some examples:
Show that for real numbers $$a_1, a_2, \cdots$$ and $$b_1, b_2, \cdots$$
$$\displaystyle \sum_{i = 1}^{k} \frac {a_i^2}{b_i} \geq \frac{(\sum_{i = 1}^k a_i)^2}{\sum_{i = 1}^k b_i}$$
IMO 1995
Let $$a, b, c \in \mathbb{R}^+$$ such that $$abc = 1$$. Prove that
$$\frac{1}{a^3(b+c)} + \frac{1}{b^3(a+c)} + \frac{1}{c^3(a+b)} \geq \frac{3}{2}$$.
..
Hint: It may help to make a symmetric substitution for $$a = \alpha, b = \beta, c = \gamma$$ such that the condition $$\alpha \beta \gamma = 1$$ still holds.
USAMO 2009
For $$n \geq 2$$ let $$a_1, a_2, \cdots, a_n$$ be positive reals such that
$$(a_1 + a_2 + \cdots + a_n)(\frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n}) \leq (n + \frac{1}{2})^2$$
Prove that $$max (a_1, a_2, \cdots, a_n) \leq 4 min (a_1, a_2, \cdots, a_n)$$.
..
Hint: The LHS should scream cauchy. Choose which $$a_i$$ multiplies which $$b_k$$ to isolate a desired max and min, and then do some algebra. · 3 years, 3 months ago
I will be happy to post my solutions if requested (as long as you try them first!) · 3 years, 3 months ago
Could you add these examples to the Applications of Cauchy Schwarz Inequality Wiki page? Staff · 2 years, 5 months ago
For the second one, following your hint, let $$a = \frac{1}{x}, b = \frac{1}{y}, c = \frac{1}{z}$$. The expression is then $$\sum_{cyc} \frac{x^3yz}{y+z}$$. Since $$xyz = 1$$, this is $$\sum_{cyc} \frac{x^2}{y+z}$$. Using Cauchy, this is greater than $$\frac{(x+y+z)^2}{2(x+y+z)}$$. So now it is left to prove that $$x+y+z \geq 3$$. This is simple using AM-GM, so $$\frac{x+y+z}{3} \geq \sqrt[3]{xyz}$$, and we are done. $$\blacksquare$$
For the second one, WLOG assume $$a_1$$ is the maximum and $$a_n$$ is the minimum. Using Cauchy,
$$(a_1 + a_2 + \cdots + a_n)(\frac{1}{a_1} + \frac{1}{a_2} + \cdots + \frac{1}{a_n}) \geq (\sqrt{\frac{a_1}{a_n}} + \sqrt{\frac{a_n}{a_1}} + n - 2)$$
$$(n + \frac{1}{2})^2 \geq (\sqrt{\frac{a_1}{a_n}} + \sqrt{\frac{a_n}{a_1}} + n - 2)^2$$
Taking square roots of both sides, we get
$$n + \frac{1}{2} \geq \sqrt{\frac{a_1}{a_n}} + \sqrt{\frac{a_n}{a_1}} + n - 2$$
$$\frac{5}{2} \geq \sqrt{\frac{a_1}{a_n}} + \sqrt{\frac{a_n}{a_1}}$$
Now, squaring, we get
$$\frac{25}{4} \geq \frac{a_1}{a_n} + 2 + \frac{a_n}{a_1}$$
Multiplying both sides by $$a_1 a_n$$ we get
$$a_1^2 - \frac{17}{4} a_1 a_n + a_n^2 \leq 0$$
Factoring...
$$(a_1 - 4a_n)(a_1 - \frac{1}{4}a_n) \leq 0$$
Since $$a_1$$ and $$a_n$$ are positive reals, $$(a_1 - 4a_n)$$ must be nonpositive. Thus,
$$(a_1 - 4a_n \leq 0 \rightarrow a_1 \leq 4a_n$$, and we are done. · 3 years, 3 months ago
Could you explain the second one as in how did you use Cauchy Schwartz in it. · 2 years, 1 month ago
What are symmetric inequalities? · 2 years, 8 months ago
Hope you share many more such techniques!!thnx a lot for this. · 3 years, 3 months ago
Yup I will. In fact, I posted some on invariants and I also wrote some worked examples in my newest post. :) · 3 years, 3 months ago
@Anqi Li Can you add this to the Cauchy Schwarz Inequality Wiki page? Staff · 2 years, 5 months ago
One can easily prove Cauchy-Schwarz using vectors too. It is similar to the proof above, but is much cleaner. The interesting this is that almost every other inequality can be derived from this inequality. · 3 years, 3 months ago | 2017-03-24T14:16:56 | {
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https://math.stackexchange.com/questions/1458687/calculate-limit-without-lhopitals-rule | # Calculate limit without L'Hopital's rule
I have to calculate this limit whitout using L'Hopital's rule or Taylor polynomials:
$$\lim_{ x\to \pi/4 } \frac{1 - \tan(x)}{x-\frac{\pi}{4}}$$
I know how to make it using L'Hopital and that the result is $-2$ ,but I'm getting nowhere when I try without it. Any advice?
Hint:
Let $t=x-\frac{\pi}{4}$, then $$\frac{1-\tan x}{x-\frac{\pi}{4}}=\frac{1-\tan\left(t+\frac{\pi}{4}\right)}{t}=\frac{1}{t}\cdot\left(1-\frac{\tan t+\tan\frac{\pi}{4}}{1-\tan \frac{\pi}{4}\tan t}\right)=\frac{1}{t}\left(1-\frac{\tan t+1}{1-\tan t}\right)=\frac{-2\tan t}{t(1-\tan t)}$$ Now, take the limit as $t\to 0$: \begin{align} \lim_{x\to \frac{\pi}{4}}\frac{1-\tan x}{x-\frac{\pi}{4}}&=\lim_{t\to 0}\frac{-2\tan t}{t(1-\tan t)}\\ &=\lim_{t\to 0}\frac{-2\tan t\cos t}{t(1-\tan t)\cos t}\\ &=-2\lim_{t\to 0}\frac{\sin t}{t(\cos t-\sin t)}\\ &=-2\left(\lim_{t\to 0}\frac{\sin t}{t}\right)\left(\lim_{t\to 0}\frac{1}{\cos t-\sin t}\right)\\ &=-2(1)(1)\\ &=\color{blue}{-2} \end{align}
• While all the answers are valid and helped me,I chose this because it doesn't imply the use of more advanced concepts like derivatives.Thanks! – Der Rosenkavalier Oct 4 '15 at 20:48
Hint: What is the definition of the derivative of $\tan(x)$ at $x=\pi/4$?
• $\frac{1}{\cos^2(\frac{\pi}{4})}$ But what's the point? I can't use L'Hopital's rule because the exercise ask it, not because I don't how to use it.I don't understand how can I use the derivative of $\tan(x)$ at $\frac{\pi}{4}$ in this exercise. – Der Rosenkavalier Sep 30 '15 at 21:47
• @DerRosenkavalier By definition: $$\tan'(\pi/4)=\displaystyle\lim_{x \to \pi/4}\frac{\tan(x)-1}{x-\pi/4}$$. How does this compare to your limit? – Reveillark Sep 30 '15 at 21:49
• L'Hopital isn't being used here; just the definition of the derivative. It's important to know the difference. – zhw. Sep 30 '15 at 22:00
• @zhw I agree with your comment. This type of question sometimes rasises debate as to "allowed ways forward. For example, is the use of asymptotic analysis permitted here? IMHO, yes absolutely. Yet others would argue that that approach is tantamount to the use of LHR. ;-) – Mark Viola Sep 30 '15 at 22:21
• @DerRosenkavalier : My answer posted here explicitly explains how you can use the value of the derivative of $\tan x$ at $\pi/4$. ${}\qquad{}$ – Michael Hardy Oct 1 '15 at 4:32
Recall that $$f'(a) = \lim_{x\to a}\frac{f(x) - f(a)}{x-a}.$$ Apply it to the case where $f(x) = \tan x$ and $a=\pi/4$. Then $f(a) = 1$ and $f'(a) = \sec^2 a = \sec^2(\pi/4) = 2$. Therefore $$2 = \lim_{x\to\pi/4}\frac{\tan x - \tan(\pi/4)}{x-\pi/4}.$$
So $-2$ is the answer to the question as you've posed it.
We have the identities
\begin{align}\frac{1-\tan x}{x-\pi/4}&=-\frac{\tan(x-\pi/4)}{x-\pi/4}(1+\tan x)\\\\ &=-\frac{\sin(x-\pi/4)}{x-\pi/4}\frac{1+\tan x}{\cos(x-\pi/4)} \end{align}
Can you finish from here?
• Yes,I could use that $\lim_{ x\to 0 }\frac{\sin(x)}{x}=1$ . Thanks for the help! – Der Rosenkavalier Oct 4 '15 at 20:58
• You're welcome. My pleasure. This is a very efficient approach that circumvents use of L'Hospital's Rule, as you requested. – Mark Viola Oct 4 '15 at 21:07
you can use the following hint ,
try to write $\tan(x)$ in terms of $\sin(x)$ and $\cos(x)$. A little bit of rearranging and then use the following
$\lim_{x\rightarrow 0} \frac{\sin(x)}{x}=1$ A little bit of work yields the result . | 2019-08-21T12:17:23 | {
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https://math.stackexchange.com/questions/2782464/how-to-formalize-the-fact-that-fi-lceil-i-2-rceil-is-surjective-but-not-one/2782475 | How to formalize the fact that $f(i)=\lceil i/2\rceil$ is surjective but not one-to-one from $\Bbb N$ to itself?
If the set $S$ is countably infinite, prove or disprove that if $f$ maps $S$ onto $S$ (i.e. $f\colon S\to S$ is a surjective function), then $f$ is one-to-one mapping.
Please give a formal mathematical proof for this statement.
I have a counter-example: Suppose the mapping is from $\Bbb N$ (the natural numbers, a countably infinite set) to $\Bbb N$. And $f(i) = \lceil i/2\rceil$. It is onto function but not one-one.
But I am not getting that how this mapping is onto. Intuitively, since $1$ and $2$ will be mapped to $1$; $3$ and $4$ will be mapped to $2$; $5$ and $6$ will be mapped to $3$; similarly, $n-1$ and $n$ will be mapped to $n/2$. So in the codomain, there will be some elements that have no pre-image for this countably infinite set. So, please correct me where I am wrong.
• Welcome to MSE. Please use MathJax. – José Carlos Santos May 15 '18 at 16:35
• "So in the codomain, there will be some elements that have no pre-image for this countably infinite set." Why do you say that? You just seemed to give an argument that it would. Anyway for any $n$ then both $2n$ and $2n+1$ will map to $n$. So $f$ is onto. – fleablood May 15 '18 at 16:47
• sir , since , cardinality of both sets are same and every 2 elements of domain are mapped to one element of codomain . so intuitively , it looks that some elements of codomain will not have any pre-image. please correct me – ankit1729 May 15 '18 at 16:51
• You just should that $1,2,3$ and $\frac n2$ (assuming $n$ is even) all have pre-images. What elements don't have pre-images? I honestly don't understand why you said that. – fleablood May 15 '18 at 16:51
• Are you thinking because you will run out of elements? But the set is infinite. You will never run out. So it's perfectly fine to have 2, 3 or even an infinite number of elements all map to the same element and still be onto. In fact that is exactly what you are proving. It is possible to be surjective and not 1-1. – fleablood May 15 '18 at 16:54
2 Answers
To prove that a function $f\colon A\to B$ is surjective, you need to be able to prove that given $b\in B$, there is some $a\in A$ such that $f(a)=b$.
In your case, $A=B=\Bbb N$, and $f(x)=\lceil \frac x2\rceil$.
So you need to prove that given any $m\in\Bbb N$, there is some $n\in\Bbb N$ such that $\lceil\frac n2\rceil=m$. Even though this $n$ is not unique, I am sure that you can come up with a way to find such $n$, from a given $m$.
Maybe I'm missing something. For every n in codomain, 2n and 2n-1 are in preimage, so the mapping is onto but not 1-1
• sir , it may or may not be one-one – ankit1729 May 15 '18 at 17:11 | 2019-08-18T13:11:33 | {
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https://physics.stackexchange.com/questions/304634/what-is-the-magnitude-of-the-difference-vector | # what is the magnitude of the difference vector?
I read this question in Sears and Zemansky's University Physics book:
"Two displacement vectors, S and T, have magnitudes S = 3 m and T = 4 m. Which of the following could be the magnitude of the difference vector S - T ?(There may be more than one answer.) i. 9 m; ii. 7 m; iii. 5 m; iv. 1 m; v. 0m; vi. -1m "
My question is: shouldn't the direction of both vectors be specified in order for me to solve it? Or do I just assume that S and T are positive and negative respectively as per the equation S - T
• The question is asking you what the possible values of $\mathbf S - \mathbf T$ could be for all possible directions. So for example if one of the options was $666$m you need consider if there is any possible arrangement of the two vectors that could give $|\mathbf S - \mathbf T| = 666$. This freedom to orient the vectors in any direction means more than one of the answers can be correct. – John Rennie Jan 12 '17 at 16:20
• That's what I thought initially, but I had doubts. Thanks for confirming! – DigiNin Gravy Jan 12 '17 at 16:23
• Re Answer (vi): can the magnitude of a vector be negative? – DJohnM Jan 12 '17 at 17:18
• @DJohnM I know that the magnitude can never be negative, but the vector can. – DigiNin Gravy Jan 12 '17 at 18:20
• @DigiNinGravy A vector which is anti-parallel to, say, the $x$-axis of a coordinate system is sometimes called "a vector in the negative-$x$ direction." But it is an incorrect oversimplification to call such an object "a negative vector." – rob Jan 13 '17 at 7:26
The magnitude of the difference vectors depends on the orientation of $$\bf\vec{S}$$ and $$\bf \vec{T}$$. If they are parallel then $$|\bf \vec{S}-\bf \vec{T}|=|\,|\bf \vec{S}|-|\bf \vec{T}|\,|$$ and if they are anti-parallel then $$|\bf \vec{S}-\bf \vec{T}|=|\bf \vec{S}|+|\bf \vec{T}|$$.
Thus the possible values of $$|\bf \vec{S}-\bf \vec{T}\|$$ lie in the range:
$$|\,|\bf \vec{S}|-|\bf \vec{T}|\,| \leq |\bf \vec{S}-\bf \vec{T}| \leq |\bf \vec{S}|+|\bf \vec{T}|$$
I'll let you work out which answers comply with this.
• |S - T| = 5 or | |S| - |T| | = -1 or |S| + |T| = 7. I just don't understand how |S - T| = |S| + |T|. I mean to have |S| + |T| both vectors must be parallel - having the same direction - so | |S| - |T| | seems off, unless we consider T as -T? – DigiNin Gravy Jan 12 '17 at 17:09
• @DigiNinGravy I think you gave the aswer yourself both $\bf T and$-\bf T$have se same magnitude, and the magnitude does not give the direction of the vector. – Mikael Fremling Jan 12 '17 at 18:10 In above Figure move the end point$\:\mathrm{B}\:$of the vector$\:\mathbf{S}\:$around a circle of radius$\:|\mathbf{S }|=3\:$. Try to find the length$\:|\mathbf{S}-\mathbf{T}|\:$of the vector$\:\mathbf{S}-\mathbf{T}\:$when$\:\mathrm{B}\:$is on points$\:\mathrm{P_{ii}},\mathrm{P_{iii}},\mathrm{P_{iv}}\:\$.
• Thanks, but you reversed the magnitudes of S and T. Can you tell me the name of the program you used? – DigiNin Gravy Jan 12 '17 at 18:17
• @DigiNin Gravy : GeoGebra – Frobenius Jan 12 '17 at 18:35 | 2019-11-18T17:28:09 | {
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https://math.stackexchange.com/questions/2400296/finding-n-permutations-r-with-repetitions/2400307 | # Finding $n$ permutations $r$ with repetitions
I have given $n$ items and I have to find ${}^nP_r$ and $\sum_{r=0}^n{}^nP_r$ with repetitions allowed. Is there any closed formula for this?
For $n=3$ and $r=1$, possible permutations are:
$\{1\},\{2\},\{3\}$
Total: 3
For $n=3$ and $r=2$, possible permutations are:
$\{1,1\},\{2,2\},\{3,3\},\{1,2\},\{2,1\},\{1,3\},\{3,1\},\{2,3\},\{3,2\}$
Total: 9
For $n=3$ and $r=3$, possible permutations are:
$\{1,1,1\},\{2,2,2\},\{3,3,3\},$
$\{1,1,2\},\{1,2,1\},\{2,1,1\},$
$\{1,2,2\},\{2,1,2\},\{2,2,1\},$
$\{1,1,3\},\{1,3,1\},\{3,1,1\},$
$\{1,3,3\},\{3,1,3\},\{3,3,1\},$
$\{2,2,3\},\{2,3,2\},\{3,2,2\},$
$\{2,3,3\},\{3,2,3\},\{3,3,2\}$,
$\{1,2,3\},\{1,3,2\}$,
$\{2,1,3\},\{2,3,1\}$,
$\{3,1,2\},\{3,2,1\}$,
Total: 27
Can we come up with any closed formula for individual ${}^nP_r$ with repetitions and also for their sum i.e. here $3+9+27=39$
I understand that I cannot call this exactly the permutation, since ${}^3P_3$ is strictly $6$, while above its $27$, since I allow repetitions, but then whatever it is, how do I get the count?
Note that permutations with repetition is usually the well known case corresponding to $\frac{n!}{n_1!n_2!...n_i!}$, which is not what I am asking here. Is what am asking also some well know case, and I am stupidly not able to guess it? My primary guess is that, there cannot be any closed formula. Is it right?
• What exactly is the condition you're looking for? I notice that the permutation $1,2,3$ is excluded - is this intentional? – platty Aug 20 '17 at 15:31
• In fact, it looks like you left out all the permutations of $1,2,3$ - is this intentional? – platty Aug 20 '17 at 15:41
• Nope, I missed it. Sorry. – anir Aug 20 '17 at 15:43
• I dont know it feels fuzzy. Is it just ${}^nP_r \text{with repetition} = n^r$?. Feels like so. Let me read question and answers again. – anir Aug 20 '17 at 15:47
• Your formula is not for permutations with replacements. It's for combinations with replacements. Are you actually looking for the expression for multinominal coefficients? – Vim Aug 20 '17 at 15:48
If you just want unrestricted strings consisting of $r$ letters, chosen with replacement from $n$, you can just use the multiplication rule to get $n^r$; there's $n$ choices for the first one, $n$ choices for the second, etc.
Extending this, we can use this to find the number of strings with length up to $r$ by summing the intermediate results: $$1+n+n^2+\dots +n^r$$ This is the sum of a geometric series, which means we can apply the formula to get $\frac{n^{r+1}-1}{n-1}$.
If you mean "permutations with replacements" then the answer is just $n^r$. But this doesn't match your result for $n=r=3$ which should be $27$? If you actually mean combinations with replacements, then see the below hint.
Hint: you are basically putting $r$ identical balls into $n$ different jars, or equivalently, finding the number of non-negative integer solutions to $$x_1+\cdots+x_n=r$$ To solve this, first let $y_i:=x_i+1$ (which are bijections) then try finding the number of positive integer solutions to the equation $$y_1+\cdots+y_n=r+n.$$ Try Stars & Bars technique.
• I don't think this is correct - OP has order mattering in the listed examples. – platty Aug 20 '17 at 15:39
• @platty this is really confusing, because in this case the third result should be 27. Clearly there is some misphrasing in the question. – Vim Aug 20 '17 at 15:41
Note that in your last paragraph, you say that $\frac{n!}{n_1!n_2! \dots n_i!}$ is the formula for permutations with repetition, but perhaps a better way to word it would be that this is a generalized formula for partitioning a set of $n$ objects into $i$ cells with each "cell" being distinguishable, but elements within their respective cells are not.
As noted before, permutations with repetition allowed can be represented with $n^r$, which @platty already explained. This is the formula you would use to solve the second example I gave, as well as what you seem to be looking for in your example.
the number of words, from alphabet $\{1,\cdots,n\}$, of length $r$ and max repetition $r$,
i.e. simply the number of words, from alphabet $\{1,\cdots,n\}$, of length $r$
which are $n^r$. | 2020-10-01T09:10:09 | {
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http://mathhelpforum.com/number-theory/1862-2-5-8-11-14-a.html | Math Help - 2, 5, 8, 11, 14...
1. 2, 5, 8, 11, 14...
How can I prove that no square number can be in the sequence?
2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, 56, 59, 62, 65, 68, 71, 74, 77, 80, 83, 86, 89, 92, 95, 98, 101, 104, 107, 110, 113, 116, 119, 122, 125, 128, 131, 134, 137, 140, 143, 146, 149, 152, 155, 158, 161, 164, 167, 170, 173, 176, 179, ...
Is it to do with mod 5? And if so how would I do it
2. Originally Posted by Natasha
How can I prove that no square number can be in the sequence?
2, 5, 8, 11, 14, 17, 20, 23, 26, 29, 32, 35, 38, 41, 44, 47, 50, 53, 56, 59, 62, 65, 68, 71, 74, 77, 80, 83, 86, 89, 92, 95, 98, 101, 104, 107, 110, 113, 116, 119, 122, 125, 128, 131, 134, 137, 140, 143, 146, 149, 152, 155, 158, 161, 164, 167, 170, 173, 176, 179, ...
Is it to do with mod 5? And if so how would I do it
First you need find the general term of your sequence. Now it looks to me
as though the sequence is:
$a_n=2+3n, n=0,1,2,..$
Which suggests you consider the squares modulo 3. Now any natural number
may be written:
$N=3k+\rho$,
for some natural number $k$, and $\rho= 0, 1\ \mbox{or}\ 2$. So
$N^2=9k^2+6k \rho + \rho^2$.
Therefore $N^2(mod\ 3)$ is $\rho^2(mod\ 3)$, but
$
\rho^2(mod\ 3)=0, 1, \mbox{or}\ 1(mod\ 3) \mbox{as}\ \rho=0,1 \mbox{or}\ 2
$
.
So as $a_n, n=0,1,2,..$ is congruent to $2(mod\ 3)$ it cannot be a square for any
natural number $n$
RonL
3. Originally Posted by CaptainBlack
First you need find the general term of your sequence. Now it looks to me
as though the sequence is:
$a_n=2+3n, n=0,1,2,..$
Which suggests you consider the squares modulo 3. Now any natural number
may be written:
$N=3k+\rho$,
for some natural number $k$, and $\rho= 0, 1\ \mbox{or}\ 2$. So
$N^2=9k^2+6k \rho + \rho^2$.
Therefore $N^2(mod\ 3)$ is $\rho^2(mod\ 3)$, but
$
\rho^2(mod\ 3)=0, 1, \mbox{or}\ 1(mod\ 3) \mbox{as}\ \rho=0,1 \mbox{or}\ 2
$
.
So as $a_n, n=1,2,..$ is congruent to $2(mod\ 3)$ it cannot be a square for any
natural number $n$
RonL
Captain is it the same to say that the general form 3n - 1 is the same as 3n + 2??
4. Originally Posted by Natasha
Captain is it the same to say that the general form 3n - 1 is the same as 3n + 2??
Yes, but the +2 form is slightly more convienient when dealing with modular
arithmetic.
RonL
5. Originally Posted by CaptainBlack
Yes, but the +2 form is slightly more convienient when dealing with modular
arithmetic.
RonL
I see. Thanks! | 2014-03-17T02:57:37 | {
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https://www.shaalaa.com/question-bank-solutions/find-ratio-which-point-p-x-2-divides-line-segment-joining-points-a-12-5-b-4-3-also-find-value-x-section-formula_5000 | # Find the Ratio in Which the Point P(X, 2) Divides the Line Segment Joining the Points A(12, 5) and B(4, −3). Also, Find the Value of X. - Mathematics
Find the ratio in which the point P(x, 2) divides the line segment joining the points A(12, 5) and B(4, −3). Also, find the value of x.
#### Solution 1
Let the point P (x, 2) divide the line segment joining the points A (12, 5) and B (4, −3) in the ratio k:1.
Then, the coordinates of P are ((4k+12)/(k+1),(-3k+5)/(k+1))
Now, the coordinates of P are (x, 2).
therefore (4k+12)/(k+1)=x and (-3k+5)/(k+1)=2
(-3k+5)/(k+1)=2
-3k+5=2k+2
5k=3
k=3/5
Substituting k=3/5 " in" (4k+12)/(k+1)=x
we get
x=(4xx3/5+12)/(3/5+1)
x=(12+60)/(3+5)
x=72/8
x=9
Thus, the value of x is 9.
Also, the point P divides the line segment joining the points A(12, 5) and (4, −3) in the ratio 3/5:1 i.e. 3:5.
#### Solution 2
Let k be the ratio in which the point P(x,2) divides the line joining the points
A(x_1 =12, y_1=5) and B(x_2 = 4, y_2 = -3 ) . Then
x= (kxx4+12)/(k+1) and 2 = (kxx (-3)+5) /(k+1)
Now,
2 = (kxx (-3)+5)/(k+1) ⇒ 2k+2 = -3k +5 ⇒ k=3/5
Hence, the required ratio is3:5 .
Concept: Section Formula
Is there an error in this question or solution?
#### APPEARS IN
RD Sharma Class 10 Maths
Chapter 6 Co-Ordinate Geometry
Exercise 6.3 | Q 20 | Page 29
RS Aggarwal Secondary School Class 10 Maths
Chapter 16 Coordinate Geomentry
Q 5 | 2021-12-01T05:58:05 | {
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https://math.stackexchange.com/questions/1574614/how-to-evaluate-this-definite-double-integral | # How to evaluate this definite double integral?
$$\int\int_D 6x\sqrt{y^2-x^2}dA, D=\{(x,y)|0\leq y\leq 2, 0 \leq x \leq y\}$$
I tried: $$\int_0^2 \int_0^y 6x\sqrt{y^2-x^2}dxdy$$
But that is incorrect.
• Why do you say that's incorrect? Looks fine to me... – Eli Berkowitz Dec 14 '15 at 3:32
• Your integral looks perfect! Now just evaluate it, try a u-substitution for the inside integral. – Alex Dec 14 '15 at 3:37
• I got -8, I must have made an arithmetic error or something! – d0rmLife Dec 14 '15 at 3:39
Setting $y^2-x^2 = t$, we obtain $-2xdx = dt$. Hence, we obtain $$I = \int_0^2 \int_0^y 6x\sqrt{y^2-x^2}dxdy = \int_0^2 \int_{y^2}^0(-3\sqrt{t}dt)dy = \int_0^2 \int_0^{y^2} 3\sqrt{t}dtdy = \int_0^22y^3 dy = 8$$
• Why does the range of the inner integral change to $y^2,0$ instead of $y^2,y^2-y$ - I thought you put the original range of integration into the u substitution equation? – d0rmLife Dec 14 '15 at 3:47
• @d0rmLife because the first inner integral goes from $x=0$ to $x=y$, and when replacing $y^2-x^2=t$ in the first case you get $y^2-0^2=y^2=t$ and in the second $y^2-y^2=0=t$. When you u-substitute you must change the range accordingly or else it is "wrong": you can get away with it if you evaluate the result in terms of x instead of u in the end, but still it is best to change it. – Fede Poncio Dec 14 '15 at 4:13
Notice, $$\int_{0}^2\int_0^y 6x\sqrt{y^2-x^2} \ dxdy=\int_{0}^2\left(\int_0^y 6x\sqrt{y^2-x^2} \ dx\right)dy$$ $$=\int_{0}^2\left(-3\int_0^y (y^2-x^2)^{1/2} \ d(y^2-x^2)\right)dy$$ $$=\int_{0}^2dy\left(-3\frac{2(y^2-x^2)^{3/2}}{3}\right)_{0}^{y}$$ $$=\int_{0}^2\left(2y^3\right)\ dy$$ $$=\left(2\frac{y^4}{4}\right)_{0}^{2}=\left(\frac{2^4}{2}-0\right)=\color{red}{8}$$ | 2020-01-26T18:00:18 | {
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http://math.stackexchange.com/questions/885137/recognizing-the-sequence-1-16-1-8-3-16-1-4-5-16 | # Recognizing the sequence 1/16, 1/8, 3/16, 1/4, 5/16, …
What is the missing number? $$\frac{1}{16}, \frac{1}{8}, \frac{3}{16}, \frac{1}{4}, \frac{5}{16}, \ \ \ [?]$$
$$A. \frac{5}{4}\quad B. \frac{3}{4}\quad C. \frac{5}{8}\quad D. \frac{3}{8}$$
Spoiler: Answer is $D$, but I don't know why.
Thanks
-
OMG, of course. Thank you for the heads up. – user1495024 Aug 2 at 1:22
Am I the only one who thought $\frac{1}{2}$ before reading the possible options? :) – Thomas Aug 2 at 4:15
No, the next two numbers should definitely be 1/2 and 7/16 :-) – CompuChip Aug 2 at 8:23
@Thomas I did consider it, but then thought it wasn't just right. First of all, I'd be generalizing from just two data points. Secondly I'd be ignoring the other three data points, which supposedly were there for a reason. – kasperd Aug 2 at 9:37
$$\frac{1}{16}, \frac{1}{8}=\frac{2}{16}, \frac{3}{16}, \frac{1}{4}=\frac{4}{16}, \frac{5}{16}$$
So the $i$th term is of the form $$\frac{i}{16}$$ Therefore, the next term is $$\frac{6}{16}=\frac{3}{8}$$
-
$$\frac{1}{16}, \frac{1}{8}, \frac{3}{16}, \frac{1}{4}, \frac{5}{16}$$ The above is the same as $\displaystyle\frac1{16},\frac2{16},\frac3{16},\frac4{16},\frac5{16}$.
-
Another sequence-recognizing technique is to look at the difference between consecutive terms.
In this case, $\frac{1}{8}-\frac{1}{16} = \frac{1}{16}$, $\frac{3}{16}-\frac{1}{8} = \frac{1}{16}$, $\frac{1}{4}-\frac{3}{16} = \frac{1}{16}$, and $\frac{5}{16}-\frac{1}{4} = \frac{1}{16}$.
Since the difference between consecutive terms is $\frac{1}{16}$, the next term should be $\frac{5}{16}+\frac{1}{16} =\frac{6}{16} =\frac{3}{8}$.
- | 2014-10-23T02:50:43 | {
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http://mathhelpforum.com/discrete-math/139817-relations-help-2-a.html | 1. ## relations help (2)
hey guys, question:
Determine whether the following relation is a reflexive, symmetric or transitive. If any is an equivalence relation, describe its equivalence classes.
xRy iff x has the same integer part as y, on <----- I need help on what this relation is asking? thanks
2. Originally Posted by jvignacio
hey guys, question:
Determine whether the following relation is a reflexive, symmetric or transitive. If any is an equivalence relation, describe its equivalence classes.
xRy iff x has the same integer part as y, on <----- I need help on what this relation is asking? thanks
The integer part of x is the integer n such that $n\le x < n+1$. (There is exactly one such integer for each real x.) It is usually denoted by [x].
For example:
[3.14]=3
[0.52]=0
[1]=1
[-1.26]=-2
So 2 and 2.5 are in relation, since [2]=2=[2.5], but 1.5 and 2.5 are not.
Hope this helps.
3. Originally Posted by kompik
the integer part of x is the integer n such that $n\le x < n+1$. (there is exactly one such integer for each real x.) it is usually denoted by [x].
For example:
[3.14]=3
[0.52]=0
[1]=1
[-1.26]=-2
so 2 and 2.5 are in relation, since [2]=2=[2.5], but 1.5 and 2.5 are not.
Hope this helps.
Is that like saying the next full integer down?
4. Originally Posted by jvignacio
Is that like saying the next full integer down?
Exactly.
5. Originally Posted by kompik
Exactly.
So in this case,
- It IS reflexive since x has the same integer part as x.
- It IS symmetric since if x has the same integer part as y then y has the same integer part as x.
- It IS transitive since if x has the same integer part as y and if y has the same integer part as z then x has the same integer part as z.
correct?
6. Originally Posted by jvignacio
So in this case,
- It IS reflexive since x has the same integer part as x.
- It IS symmetric since if x has the same integer part as y then y has the same integer part as x.
- It IS transitive since if x has the same integer part as y and if y has the same integer part as z then x has the same integer part as z.
correct?
Yes, it is an equivalence relation.
Another way to see this is to notice that you're in fact given a decomposition of R and every decomposition gives you an equivalence relation. (But if you haven't heard much about the correspondence between equivalences and decompositions at your lessons, you should perhaps ignore this comment.)
7. Originally Posted by kompik
Yes, it is an equivalence relation.
Another way to see this is to notice that you're in fact given a decomposition of R and every decomposition gives you an equivalence relation. (But if you haven't heard much about the correspondence between equivalences and decompositions at your lessons, you should perhaps ignore this comment.)
Yeah ive never herd of that... Not yet anyway. What would the equivalence class be in this case now that its an equivalence relation? All numbers?
8. Originally Posted by jvignacio
Yeah ive never herd of that... Not yet anyway. What would the equivalence class be in this case now that its an equivalence relation? All numbers?
No.
Take, for instance, the equivalent class of 3.14. It contains all numbers such that x R 3.14.
This is equivalent to
[x]=[3.14]=3.
And what are the numbers such that [x]=3? Precisely the numbers from the interval $\langle 3,4)$, right?
Can you find the remaining classes?
9. Originally Posted by kompik
No.
Take, for instance, the equivalent class of 3.14. It contains all numbers such that x R 3.14.
This is equivalent to
[x]=[3.14]=3.
And what are the numbers such that [x]=3? Precisely the numbers from the interval $\lange 3,4)$, right?
Can you find the remaining classes?
Sorry whats in the interval for [x]=3? theres a latex error..
10. Originally Posted by jvignacio
Sorry whats in the interval for [x]=3? theres a latex error..
Sorry, did not notice that. I've edited my post.
11. Originally Posted by kompik
No.
Take, for instance, the equivalent class of 3.14. It contains all numbers such that x R 3.14.
This is equivalent to
[x]=[3.14]=3.
And what are the numbers such that [x]=3? Precisely the numbers from the interval $\langle 3,4)$, right?
Can you find the remaining classes?
Ok I understand the equivalent class of 3.14 and the numbers such that [x]=3 are all numbers between 3 and 4 but Which remaining classes are you referring too? Since my question is a x and y question, how can I write this in a general form. If you get what I mean...
12. Originally Posted by jvignacio
Ok I understand the equivalent class of 3.14 and the numbers such that [x]=3 are all numbers between 3 and 4 but Which remaining classes are you referring too? Since my question is a x and y question, how can I write this in a general form. If you get what I mean...
If I were your teacher, I would expect answer to be something like:
The equivalent classes are intervals .... (fill in the dots) for $n\in\mathbb{N}$.
(I guess the example with 3.14 might help you to see what to fill in.)
13. Originally Posted by kompik
If I were your teacher, I would expect answer to be something like:
The equivalent classes are intervals .... (fill in the dots) for $n\in\mathbb{N}$.
(I guess the example with 3.14 might help you to see what to fill in.)
intervals $\langle n,n+1)$ , $n \in \mathbb{N}$ ?
14. Originally Posted by jvignacio
intervals $\langle n,n+1)$ , $n \in \mathbb{N}$ ?
Exactly.
15. Originally Posted by kompik
Exactly.
mate thanks alot for the help. Really appreciate your time.
Page 1 of 2 12 Last | 2017-04-28T04:04:27 | {
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https://math.stackexchange.com/questions/1154703/use-proof-by-induction-to-prove-frac1n-frac12n-1-for-all-n-geq | # Use proof by induction to prove $\frac{1}{n!}<\frac{1}{2^n-1}$ for all $n\geq 4$
Use proof by induction to prove that that $\frac{1}{n!}<\frac{1}{2^n-1}$ for all $n\geq 4$, .\Base case: $$\frac{1}{4}=\frac{1}{24}\leq \frac{1}{2^4-1}$$ Inductive hypothesis: Assume there exists $k\in \mathbb{N}$ s.t.
$$\frac{1}{k!}\leq\frac{1}{2^k-1}$$ Inductive step: Show that:$$\frac{1}{(k+1)!}\leq\frac{1}{2^{k+1}-1}$$ Now, $$\frac{1}{(k+1)!}=\frac{1}{k!}\cdot\frac{1}{k+1}$$ Using the hypothesis $$\frac{1}{(k+1)!}\leq\frac{1}{2^k-1}\cdot \frac{1}{k+1}$$ Because $n\geq4, \frac{1}{k+1}<\frac{1}{2}$ $$\frac{1}{(k+1)!}\leq\frac{1}{2^k-1}\cdot \frac{1}{2}=\frac{1}{2^{k+1}-2}\leq\frac{1}{2^{k+1}-1}$$ Hence by mathematical induction we have proved that $\frac{1}{n!}<\frac{1}{2^n-1}$ for all $n\geq 4$
Firstly I need to know if the proof is correct, secondly it has to be as concise as possible hence I would like to know if there are any lines I can change/delete
And lastly can anyone explain to me why every sentence starts with "\" It looks perfectly fine in www.sharelatex.com ;(
• You must start with $n=4$. – User3101 Feb 18 '15 at 16:11
• I see and I am lost know because it is a question copied from the coursework given by a professor, I just blindly followed so lets assume $n\geq4$ – Scavenger23 Feb 18 '15 at 16:12
• So I should say Assume that $\frac{1}{k!}\leq\frac{1}{2^k-1}$ holds for some $k\geq4$ and then proceed ? – Scavenger23 Feb 18 '15 at 16:21
• One error I notice in your logic - in the last statement you state that $\frac 1{2^{k+1}-2} \leq \frac 1{2^{k+1}-1}$. This is incorrect. The latter (-1) fraction has a larger denominator than the -2 fraction, which means it's a smaller fraction. – Duncan Feb 18 '15 at 16:34
All in all it looks like you have the idea down on how induction works. It looks like you are correctly making the induction step to get your result. I have a couple things to point out, however.
$(1)$ During your induction step you introduce the "$\geq$" symbol, when initially you are trying to write a proof about a strict inequality (which should only use the "$>$" symbol). If the strict inequality holds, it is not incorrect to use "$\geq$", but for the sake of consistency you should just use one.
$(2)$ Your induction hypothesis is not stated quite right. It is not enough to assume there exists $k \in \Bbb{N}$ such that the inequality holds. You want to make the stronger assumption that you can find this $k$, and the inequality holds for all $n \leq k$. It is in doing this that you will be allowed to conclude at the end of the proof that the inequality holds for all $n \geq 4$, instead of just $n=4$, and one arbitrary $k$.
$(3)$. If you are looking for a more concise proof, I would instead prove the equivalent statement that $2^n-1<n!$ for all $n \geq 4$. It is clear for $n \geq 4$ that $$2^n-1<2^n = 2 \cdot 2\cdot 2\cdot 2\cdot \ldots < 1 \cdot 2\cdot 3 \cdot 4 \cdot \ldots = n!$$ This is easier than dealing with fractions. But, upon proving this result one need merely flip the inequality around and put the quantities on each side under a numerator of $1$.
• Never mind (4). in the preview mode every sentence had "\" in front of it while in here as you see Everything is fine. I see the mistake $< and \leq$ thank you for pointing it out. Thank you. I started induction only 2 weeks ago and it is a difficult topic for me. – Scavenger23 Feb 18 '15 at 16:37
• You see I didn't have the "flipping the fractions" tool in the toolbox so I didn't even think about what you have done in (3). The grammar unfortunately will remain an issue, since English is not my native language. – Scavenger23 Feb 18 '15 at 16:42
• You know what now I realised the whole mistake. I can't copy correctly. The question is not asking to prove $\frac{1}{n!}<\frac{1}{2^n-1}$ but $\frac{1}{n!}<\frac{1}{2^{n-1}}$ – Scavenger23 Feb 18 '15 at 16:50
• I don't want to use you but if I finish the right question this time, would you be able to have a one last look on it? I will stop spamming now since the website tells me to avoid extended discussions in comments. – Scavenger23 Feb 18 '15 at 16:54
• Use proof by induction to prove that that $\frac{1}{n!}<\frac{1}{2^n-1}$ for all $n\geq 3$\\I will use the fact that $k!>2^{k-1}$ is an equivelant statement to $\frac{1}{k!}<\frac{1}{2^k-1}$ .\\Base case $3!>2^{2}$ Inductive step: Assume $k!>2^{k-1}$ holds for all $n\leq k$ for some $k \in \bbb{N}$\\\ Inductive hypothesis\\Show that $(k+1)!>2^k$\\It is clear that when $k\geq 3$\\$2^{k-1}<2^k=2*2*2*2...<1*2*3*4...=k!<(k+1)!$ – Scavenger23 Feb 18 '15 at 17:18
It's much easier proving that $2^n<1+n!$, for $n\ge4$, which is completely equivalent to your assignment. The base step is obvious. Suppose it holds for $n$; then $$2^{n+1}=2\cdot2^n<2\cdot(1+n!)=2+2\cdot n!<1+(n-1)\cdot n!+2\cdot n!=1+(n+1)!$$ because $(n-1)n!>1$.
You can, if you want, transform this into a proof of your assigned inequality, but it's not necessary.
Rewrite as$$n!\ge2^n.$$ Then $$4!\ge2^4$$ and $$n!\ge2^n\land n+1\ge2\implies(n+1)!\ge2^{n+1}.$$ | 2019-09-23T13:00:16 | {
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http://math.stackexchange.com/questions/258273/how-do-i-prove-that-lim-x%e2%86%920-x%e2%8b%85-ln-x-0 | # How do I prove that $\lim_{x→0} x⋅\ln x=0$
How do I prove (without L'Hôpital's rule) that $$\lim_{x→0} x⋅\ln x = 0$$.
I'm trying to get some intuitive sense for this, but it's quite hard. It's like trying to prove that $x$ goes faster to $\ 0 \$ then $\ln x$ goes to $-∞$, right ?
I tried this, for $x∈(0,1)$ $$x \ln x=x⋅-∫_x^1 \frac{1}{t}dt>x⋅\frac{(1-x)}{-x}=-1+x$$
So when $x→0$, I conclude that $x⋅\ln x≥-1$.
-
I don't know how to implement this, but how about squeeze theorem? It was the first thing that came to mind. – 000 Dec 13 '12 at 22:35
Since $\ln(\frac{1}{x})=-\ln x$, $$\lim_{x\to 0^{+}} x \cdot \ln x = \lim_{x\to \infty} \frac{1}{x} \cdot\ln \frac{1}{x} = - \lim_{x\to \infty} \frac{\ln x}{x}$$
But $\ln x$ is "slower" than $x$ (since $e^x$ is faster than $x$), so this limit is zero.
(Rigorously: for positive $x$, $2 e^x > x^2$ (by comparing power series of both sides). Plug $\sqrt{x}$ instead of $x$ and take logarithm of both sides: $\ln x < \ln 2 + \sqrt{x} =o(x)$.)
-
Is it always true that you can rewrite $\lim_{x→0^+}f(x) = \lim_{x→∞}f(1/x)$? – Kasper Dec 13 '12 at 22:48
@Kasper, that's a great question. My intuition is no, and the reason is that there are tricky situations that arise when moving from a one-sided limit to a two-sided limit. I would say, however, that it may be true that $$\lim_{x \to 0}f(x)=\lim_{x \to \infty}f\left(\frac{1}{x}\right).$$ However, I am uncertain. – 000 Dec 13 '12 at 22:55
@Kasper - The answer is "Yes". Try to prove it using the definition of one-sided limit at 0 and limit at infinity. Show that if one limit exists, the other must equal it. – Ofir Dec 13 '12 at 23:03
@Limitless - There's no problem because the limit at infinity is not two-sided. Actually, your proposed equality is false, because in the LHS you evaluate $f$ on both negative and positive values, and at the RHS you evaluate $f$ only on positive values. Take $f(x)=\frac{1}{x}$ to reach a contradiction. – Ofir Dec 13 '12 at 23:05
@Ofir Let $s_n>0$ and $\lim_{n→∞} s_n = 0$. By definition $$\lim_{x→0^+}f(x)=\lim_{n→∞}f(s_n)$$. Let $t_n=1/s_n$. Then $\lim_{n→∞}t_n=+∞$. So $$\lim_{x→0^+}f(x)=\lim_{n→∞}f(s_n)=\lim_{n→∞}f(1/t_n)=\lim_{x→∞}f(1/x)$$ Do you mean something like this ? – Kasper Dec 13 '12 at 23:21
$$x \ln x=-x⋅\int_x^1 \frac{1}{t}dt$$
Now, let $0<a$ be arbitrary. Then, for all $x \in (0,1)$ we have
$$\frac{1}{t^{1-a}}\leq \frac{1}{t} \leq \frac{1}{t^{1+a}}$$
Thus
$$-x⋅\int_x^1 \frac{1}{t^{1+a}}dt \leq -x \ln(x) \leq -x⋅\int_x^1 \frac{1}{t^{1-a}}dt$$ $$-x⋅\frac{x^a-1}{a} \leq -x \ln(x) \leq -x⋅\frac{x^{-a}-1}{-a}⋅$$
now let $x \to 0$.
P.S. The argument works directly with any $0<a<1$, so picking $a=\frac{1}{2}$ makes the proof much cleaner....
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Make the change $x = e^{-y}$: $$\lim_{x \to 0} x \ln x = - \lim_{y \to \infty} y e^{-y}$$
Show that for $y > 0$: $$e^y > \frac{y^2}{2!}$$
And use the squeeze theorem.
- | 2015-05-30T03:11:19 | {
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https://math.stackexchange.com/questions/2210539/what-am-i-doing-wrong-with-this-second-order-ode/2210593 | # What am I doing wrong with this second order ODE?
I have an ODE: $$\ddot v=-kt^{-1}\dot v+kt^{-2}v$$ I want to solve it by reducing it to a first order ODE, by defining $u(t)=t^{m}v(t)$ If I rewrite the ODE in terms of $u$, this gives me: $$\ddot u =(-k+2m)t^{-1}\dot u +(km+k-1-m^2)t^{-2}u$$ Setting the coefficient of $u$ to zero gives $2m=k+\sqrt{k^2-4+4k}$, which I will write as $k+\alpha$.
This gives me the single order ODE $$\ddot u=\alpha t^{-1}\dot u$$ Solving it gives $$\dot u=2c\alpha t$$ $$u(t)=c\alpha t^2$$ Plugging this back into $v$, gives $$v(t)=c\alpha t^{2-m}$$ $$\dot v = (2-m)c\alpha t^{1-m}$$ $$\ddot v = (1-m)(2-m)c\alpha t ^{-m}$$
These equations are supposed to hold for any $k$. If we plug this into the original ODE, however, this reduces to the equation $$3k=4+\sqrt{k^2-4+4k}$$
Which obviously does not hold for all $k$.
What am I doing wrong?
• Hint: Since this is a Euler-Cauchy type, let $v = t^m$. Find $v''$ and $v'$ and substitute in, solve for $m$ and solve.
– Moo
Mar 30 '17 at 17:39
• most likely your solution changes between (over)damped and oscillatory depending on $k$ Mar 30 '17 at 18:00
• Do you need to reduce it to a first order ODE? Or are you fine solving it with any method? Mar 30 '17 at 18:06
• I want to learn as much as possible about these things, so I was working on reducing it to a first order ODE. So I'd like to learn about other methods as well, but also I'd like to make sure I can do this method Mar 30 '17 at 18:09
• Please check again all signs and coefficients, the one before $u$ should reduce to $-(m+k)(m-1)$. Mar 30 '17 at 18:53
What you should realize is that you don't need to do a change of variables here but recognize that your solution is a Euler-Cauchy problem and therefore can be solved by:
Let $v=t^m$ then subbing the values into your DE you get:
$$m(m-1)t^{m-2} = -kmt^{m-2} + kt^{m-2} \\ \implies m(m-1) +km -k =0 \\ \implies m=1,m=-k \\$$
therefore giving:
$$v(t) = At^1 + Bt^{-k} \implies v(t) = At + Bt^{-k}$$
Which does hold $\forall k \in \mathbb{R}$ but you should also note that it does hold $\forall t \in (-\infty,0) \cup (0,+\infty)$ because $t=0$ is a case basis in that if $k>0$ then $t \neq 0$ but if $k<0$ then $t\in (-\infty,\infty)$
Since you did say you want to learn about the reduction to first order just note then when you do that you typically let u be something like $u= t^m \dot{v}$ so then $\dot{u}$ would be equal to a second derivative of v.
In your first integration, you should get from $$\ddot u=αt^{-1}\dot u$$ to $$\ln|\dot u|=α\ln|t|+c$$ or $$\dot u = C·t^α$$ leading to $$u=\frac{C}{α+1}·t^{α+1}+D$$ | 2022-01-17T21:05:38 | {
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https://math.stackexchange.com/questions/3746161/proving-%E2%8A%A2-a%E2%88%A8a%E2%86%92b-using-hilbert-system | # Proving ⊢ A∨(A→B) using Hilbert system
I'm self-studying mathematical logic from "Introduction to Mathematical Logic" by Detlovs and Podnieks (available free here under CC license). Unfortunately, it doesn't come with any solutions. I'm stuck trying to prove ⊢ A∨(A→B), from Exercise 1.4.2 (i) on page 43. The text appears to be using the Hilbert-style deduction method.
Relevant axioms (classical logic)
• $$L_1: B \to (C \to B)$$
• $$L_2: (B \to (C \to D)) \to ((B \to C) \to (B \to D))$$
• $$L_3: B \to B \lor C$$
• $$L_4: C \to B \lor C$$
• $$L_8: (B \to D) \to ((C \to D) \to (B \lor C \to D))$$
• $$L_9: (B \to C) \to ((B \to \lnot C) \to ¬B)$$
• $$L_{10}: \lnot B \to (B \lor C)$$
• $$L_{11}: B \lor \lnot B$$
Relevant inference rules
• Modus Ponens
Attempt
1. $$(A \to A \lor (A \to B)) \to ((\lnot A \to A \lor (A \to B)) \to (A \lor \lnot A \to A \lor (A \to B)))$$ --- $$L_8$$
2. $$A \to A \lor (A \to B)$$ --- $$L_6$$
3. $$A \lor \lnot A$$ --- $$L_{11}$$
4. $$(\lnot A \to A \lor (A \to B)) \to (A \lor \lnot A \to A \lor (A \to B))$$ --- from (1) and (2) by MP
5. $$(\lnot A \to A \lor (A \to B)) \to A \lor (A \to B)$$ --- from (3) and (4) by MP
6. $$(\lnot A \to (A \to A \lor (A \to B))) \to ((\lnot A \to A) \to (\lnot A \to A \lor (A \to B)))$$ --- $$L_2$$
7. $$(\lnot A \to (A \to A \lor (A \to B)))$$ --- $$L_{10}$$
8. $$(\lnot A \to A) \to (\lnot A \to A \lor (A \to B))$$ --- from (6) and (7) by MP
I'm stuck at this point because $$\lnot A \to A$$ appears to be a contradiction. I'm also not sure whether this is the right approach. It looks alright at formula 5, but I'm not sure how to prove $$\lnot A \to A \lor (A \to B)$$, since it requires proving that $$A \lor (A \to B)$$ is always true, which is what we're trying to prove in the first place.
The text states that it can be solved using 14 formulas, 13 being the shortest yet.
• The "proof strategy" is to use L11: $A \lor \lnot A$ and derive $A \lor (A \to B)$ from $A$ with L6 and from $\lnot A$ with L10. Then use L8 – Mauro ALLEGRANZA Jul 5 '20 at 16:48
• @MauroALLEGRANZA I seem to have done the first three steps in my attempt (if I understood correctly); substituting B=A, C=¬A for L8 seems to yield (A→D)→((¬A→D)→D) after simplifying. The only straightforward substitution for D that I haven't tried seems to be D=(¬A→A∨(A→B)) but that leads to a double negation case with L10, which hasn't been proved yet. – user383527 Jul 5 '20 at 20:36
## 1 Answer
I assume that you are forced not tu use the Deduction Theorem.
But you can use the so-called Law of Syllogism (transitivity of $$\to$$).
If so, here is a sketch of a derivation:
1. $$\vdash A \to (A \lor (A \to B))$$ --- L6
2. $$\vdash \lnot A \to (A \to B)$$ --- L10
3. $$\vdash (A \to B) \to (A \lor (A \to B))$$ --- L7
4. $$\vdash \lnot A \to (A \lor (A \to B))$$ --- from 2. and 3. by Syllogism.
Now we can "cook them" together using L8:
1. $$\vdash (A \to (A \lor (A \to B))) \to [(\lnot A \to (A \lor (A \to B))) \to ((A \lor \lnot A) \to (A \lor (A \to B)))]$$
Now, from 5., 1. and 4. by MP twice:
1. $$\vdash (A \lor \lnot A) \to (A \lor (A \to B))$$
Finally, using L11: $$\vdash A \lor \lnot A$$, by MP:
$$\vdash A \lor (A \to B)$$.
• Smart use of the law of syllogism! The text has proved it as a theorem, so it is straightforward to show it from (2) and (3). In total, this gives 13 formulas (verbosely written), which the text says is the shortest proof. – user383527 Jul 6 '20 at 22:35 | 2021-05-17T20:03:25 | {
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http://math.stackexchange.com/questions/483621/sums-of-rising-factorial-powers | # Sums of rising factorial powers
Doodling in wolfram, I found that $$\sum^{k}_{n=1}1=k$$ The formula is pretty obvious, but then you get $$\sum^{k}_{n=1}n=\frac{k(k+1)}{2}$$ That is a very well known formula, but then it gets interesting when you calculate $$\sum^{k}_{n=1}n(n+1)=\frac{k(k+1)(k+2)}{3}\\ \sum^{k}_{n=1}n(n+1)(n+2)=\frac{k(k+1)(k+2)(k+3)}{4}$$ And so on. There is an obvious pattern that I really doubt is a coincidence, but I have no idea how to prove it in the general case. Any ideas?
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Pascal's triangle comes to mind here – Omnomnomnom Sep 4 '13 at 3:47
@Omnomnomnom I thought about it but I was unable to introduce binomials there. However, you can generalize the hypothesis using factorials, but usually working with factorials is harder than not doing so(At least I would have a very hard time trying to do so). – chubakueno Sep 4 '13 at 3:51
These are essentially the sums for the general $d$ dimensional simplex. – Jaycob Coleman Sep 4 '13 at 4:02
It's overkill, but you could expand and apply en.wikipedia.org/wiki/Faulhaber%27s_formula – dls Sep 4 '13 at 4:11
@dls I think that in that case the interesting part would go backwards: That expanding and a applying Faulhaber´s yields such a regular and simple result :) – chubakueno Sep 4 '13 at 4:22
You can argue any given case by induction. I will take your last,$$\sum^{k}_{n=1}n(n+1)(n+2)=\frac{k(k+1)(k+2)(k+3)}{4}$$ for the example, but I think it is easy to see how it gets carried forward. The base case is simply $1\cdot 2\cdot 3=1\cdot 2\cdot 3\cdot \frac 44$ If it is true up to $k$, then $$\sum^{k+1}_{n=1}n(n+1)(n+2)\\=\sum^{k}_{n=1}n(n+1)(n+2)+(k+1)(k+2)(k+3)\\=\frac{k(k+1)(k+2)(k+3)}{4}+(k+1)(k+2)(k+3)\frac {k+4-k}4\\=\frac{(k+1)(k+2)(k+3)(k+4)}{4}$$
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Thanks! Just changing the $2$, $3$ and $4$ by $x$,$x+1$ and $x+2$ does the work of generalizing it very nicely. – chubakueno Sep 4 '13 at 3:59
The easy way to deal, for example, with $\sum_{i=1}^n i(i+1)(i+2)(i+3)$ is to let $F(i)=i(i+1)(i+2)(i+3)(i+4)$. We calculate $F(i)-F(i-1)$. We get $$i(i+1)(i+2)(i+3)(i+4)-(i-1)(i)(i+1)(i+2)(i+3).$$ There is a common factor of $i(i+1)(i+2)(i+3)$. When we "take it out" we are left with $(i+4)-(i-1)=5$.
Let $G(i)=\frac{F(i)}{5}$. Then by our calculation $i(i+1)(i+2)(i+3)=G(i)-G(i-1)$.
Now consider the sum $\sum_{i=1}^n i(i+1)(i+2)(i+3)$. This is $$(G(1)-G(0))+(G(2)-G(1))+G(3)-G(2)) +\cdots+(G(n)-G(n-1)).$$ Observe the telescoping. Since $G(0)=0$, the above sum is equal to $G(n)$. Thus $$\sum_{i=1}^n i(i+1)(i+2)(i+3)=G(n)=\frac{n(n+1)(n+2)(n+3)(n+4)}{5}.$$
Exactly the same idea works in general.
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The hypothesized equality can be written as follows: for any $m$, we conjecture $$\sum^{k}_{n=1}\frac{(n+m)!}{(n-1)!}=\frac{(k+m+1)!}{(m+2)(k-1)!}$$ Dividing both sides by $(m+1)!$, we have $$\sum^{k}_{n=1}\frac{(n+m)!}{(n-1)!(m+1)!}=\frac{(k+m+1)!}{(m+2)!(k-1)!}$$ Or, in other words $$\sum^{k}_{n=1}\binom{n+m}{m+1}=\binom{k+m+1}{m+2}$$ I'm not sure how to prove this (yet), but it seems very likely that there's a neat trick for all this.
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Partial sums of sequences $a_n$ that can be expressed as polynomials in $n$ are easily found using discrete calculus.
We start with the discrete version of the Fundamental Theorem of Integral Calculus:
\begin{align*} \sum\limits_{n=1}^k a_n &= \sum\limits_{n=1}^k (\Delta b)_n \\ &= (b_2 - b_1)+(b_3 - b_2)+ \ldots + (b_k - b_{k-1}) + (b_{k+1} - b_k) \\ &= b_{k+1} - b_1 \end{align*}
where $(\Delta b)_n = b_{n+1} - b_n$ is the forward difference. Finding the partial sum has now been reduced to finding a sequence $b_n$ such that $(\Delta b)_n = a_n$.
We will find $b$, the antiderivative of $a$, using falling powers, which are defined by
$$n^{\underline{k}} = n(n-1)(n-2)\ldots (n-k+1)$$
where $k$ is an integer and, by a second definition, $n^{\underline{0}}=1$. For example
$$n^{\underline{3}} = n(n-1)(n-2).$$
We now need one more result: the discrete derivative of $n^{\underline{k}}$ is given by
\begin{align*} \Delta n^{\underline{k}} &= (n+1)^{\underline{k}} - n^{\underline{k}} \\ &= (n+1)n^{\underline{k-1}} - n^{\underline{k-1}}(n-k+1) \\ &= kn^{\underline{k-1}} \end{align*}
Let's now find the partial sum for a particular case:
\begin{align} \sum^{k}_{n=1}n(n+1)(n+2) &= \sum^{k}_{n=1} (n+2)^{\underline{3}} \\ &= \sum^{k}_{n=1} \Delta \left[\frac{1}{4} (n+2)^{\underline{4}}\right] \\ &= \frac{(k+3)(k+2)(k+1)k}{4} - \require{cancel}\cancelto{0}{\frac{(1+2)(1+1)(1-0)(1-1)}{4}} \end{align}
The general case:
\begin{align} \sum^{k}_{n=1} (n+p)^{\underline{p+1}} &= \sum^{k}_{n=1} \Delta \left[\frac{1}{p+2} (n+p)^{\underline{p+2}}\right] \\ &= \frac{(k+1+p)(k+p)\ldots [(k+1+p)-(p+2)+1)]}{p+2} \\ &= \frac{(k+1+p)(k+p)\ldots k}{p+2} \end{align}
where $p>0$ is an integer.
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[I haven't finished yet. Will be gradually improved]
If we do it this way, then probably, it will become more insightful.
$$\sum^{k}_{n=1}1=k$$
then we preserve everything from the right side exactly as it is. $$\sum^{k}_{n=1}n=\frac{k(k+1)}{1 \cdot 2}$$ [there will be a picture of the right triangle]
Now again, preserve everything from the right side: $$\sum^{k}_{n=1}\frac{n(n+1)}{1 \cdot 2}=\frac{k(k+1)(k+2)}{1 \cdot 2 \cdot 3}$$
[there will be a picture of the 6 pyramids composed into rectangular parallelepiped]
$$\sum^{k}_{n=1}\frac{k(k+1)(k+2)}{1 \cdot 2 \cdot 3}=\frac{k(k+1)(k+2)(k + 3)}{1 \cdot 2 \cdot 3 \cdot 4}$$
[There should be a description of the connection between simple combinations and combinations with repetition]
$${n \choose k} = \frac{n \cdot (n - 1) \cdot \ldots \cdot (n - k + 1)}{k \cdot (k - 1) \cdot \ldots \cdot 1} = \frac{n^{\underline{k}}}{k!}$$ $$\left(\!\middle(\genfrac{}{}{0pt}{}{n}{k}\middle)\!\right) = \frac{n \cdot (n + 1) \cdot \ldots \cdot (n + k - 1)}{k \cdot (k - 1) \cdot \ldots \cdot 1} = \frac{n^{\overline{k}}}{k!}$$
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You should use the sandbox if you are part way through an answer. – Michael Albanese 2 days ago
@Michael: Sorry, I don't understand how to use it. Probably, I just don't have the rights to use that post or I need to do something beforehand, but I don't see [Add answer] button there and I cannot edit them neither. – Pixar 2 days ago
Ah, you might not have enough reputation yet. Oh well, keep it in mind for future use. – Michael Albanese 2 days ago | 2015-08-31T09:28:41 | {
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https://abhcompta.be/reviews/a1ed65-%C3%A9quivalent-coefficient-binomial | 0 First we highlight the important concepts introduced in the previous post on permutations and combinations. , n ) 6 k − ) ⋅ {\displaystyle n} n y . Multiset coefficients may be expressed in terms of binomial coefficients by the rule, One possible alternative characterization of this identity is as follows: 1 Binomial coefficients have been known for centuries, but they're best known from Blaise Pascal's work circa 1640. n n … = n + $\binom{n+k}k$ divides $\frac{\text{lcm}(n,n+1,\ldots,n+k)}n$. We can use the formula (nk)=n!k!(n−k)! 1 This shows in particular that For instance, if k is a positive integer and n is arbitrary, then. z x {\displaystyle 1={\tbinom {6}{6}}={\tbinom {6}{0}}={\tbinom {43}{0}}} k − → 1 {\displaystyle k=0} 6 n which explains the name "binomial coefficient". ) [/math], ${n\choose k_1,k_2,\ldots,k_r} ={n-1\choose k_1-1,k_2,\ldots,k_r}+{n-1\choose k_1,k_2-1,\ldots,k_r}+\ldots+{n-1\choose k_1,k_2,\ldots,k_r-1}$, ${n\choose k_1,k_2,\ldots,k_r} ={n\choose k_{\sigma_1},k_{\sigma_2},\ldots,k_{\sigma_r}}$, \begin{align} {z \choose k} = \frac{1}{k! n {\displaystyle {\tbinom {n}{k}}} Is it possible to have perfect pitch but zero sense of relative pitch? □. weißen irgendwie aufgereihten Elementen. {\displaystyle {\tbinom {n}{k}}} &=\left(\!\!\binom{-7}{7}\!\!\right)\left(\!\!\binom{4}{7}\!\!\right)=\binom{-1}{7}\binom{10}{7}.\end{align}, ${x \choose y}= \frac{\Gamma(x+1)}{\Gamma(y+1) \Gamma(x-y+1)}= \frac{1}{(x+1) \Beta(x-y+1,y+1)}. An equivalent question: how many permutations are there of the letters A, B, C, D, E, F and G? For example, the 2nd2^{\text{nd}}2nd number in the 4th4^{\text{th}}4th row is represented as (31)=3\binom{3}{1} = 3(13)=3. □. In Counting Principles, we studied combinations.In the shortcut to finding$\,{\left(x+y\right)}^{n},\,$we will need to use combinations to find the coefficients that will appear in the expansion of the binomial. {\displaystyle {\frac {{\text{lcm}}(n,n+1,\ldots ,n+k)}{n\cdot {\text{lcm}}({\binom {k}{0}},{\binom {k}{1}},\ldots ,{\binom {k}{k}})}}} + = 0 when either k > n or k < 0.$ It is the coefficient of the xk term in the polynomial expansion of the binomial power (1 + x)n, and it is given by the formula, For example, the fourth power of 1 + x is. An integer n ≥ 2 is prime if and only if n ) {\displaystyle \alpha -z} α {\displaystyle m,n\in \mathbb {N} ,}. (2003). 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A related combinatorial problem is to count multisets of prescribed size with elements drawn from a given set, that is, to count the number of ways to select a certain number of elements from a given set with the possibility of selecting the same element repeatedly. {\displaystyle k} {\displaystyle \psi (n)} ) m Randomly select a ball. z ( {\displaystyle n} n ln , sondern den Bruch &=(-1)^7\;\frac{4\cdot5\cdot6\cdot7\cdot8\cdot9\cdot10} k When the exponent is 1, we get the original value, unchanged: (a+b) 1 = a+b. k , dann ist. ) for all positive integers r and s such that s < pr. , ( k ≥ The right side counts the same parameter, because there are ways of choosing a set of q marks and they occur in all subsets that additionally contain some subset of the remaining elements, of which there are. -elementigen Menge ist, ergibt sich durch die Summation die Anzahl aller ihrer Teilmengen, also ) ! Other notations for the binomial coefficient are , and . ( ) The binomial coefficients form the entries of Pascal's triangle.. für das dritte usw., bis hin zu □, Or, in other terms, the sum of two adjacent binomial coefficients is equal to the one "below" it, assuming Pascal's triangle is produced in the equilateral-triangle-like shape shown above. r 2 {\displaystyle p} https://de.wikipedia.org/w/index.php?title=Binomialkoeffizient&oldid=205380223, „Creative Commons Attribution/Share Alike“. ∈ − ≤ ) More precisely, fix an integer d and let f(N) denote the number of binomial coefficients with n < N such that d divides . One of the more visually striking properties is the Sierpiński sieve, which is obtained by taking mod 2 of every binomial coefficient. 7!} where $\ln$ $\Gamma(n)$ denotes the natural logarithm of the gamma function at $n$. 5 ( = k ) , n n 5 When n is composite, let p be the smallest prime factor of n and let k = n/p. und r }{k!\big((n-k)!\big)} + \frac{n!}{(k+1)!\big(n-(k+1)\big)!} {\displaystyle 43={\tbinom {43}{1}}} wie auch ) ( ! k binomial coefficients: These can be proved by using Euler's formula to convert trigonometric functions to complex exponentials, expanding using the binomial theorem, and integrating term by term. 6 {\displaystyle x\in X} α m We can verify this visually by looking at Pascal''s triangle and using our guideline for construction: (nk)+(nk+1)=(n+1k+1).\binom{n}{k} + \binom{n}{k+1} = \binom{n+1}{k+1}.(kn)+(k+1n)=(k+1n+1). 0 The Chu–Vandermonde identity, which holds for any complex-values m and n and any non-negative integer k, is, $\sum_{j=0}^k \binom m j \binom{n-m}{k-j} = \binom n k$, and can be found by examination of the coefficient of $x^k$ in the expansion of (1 + x)m (1 + x)n − m = (1 + x)n using equation (2). und 0 ( X erhält man die Beziehung. will remain the same. k α ∈ \binom{n+1}{2} = T_{n} &= \dfrac{(n+1)(n)}{2}, p Grinshpan, A. ( , can be calculated by logarithmic differentiation: Over any field of characteristic 0 (that is, any field that contains the rational numbers), each polynomial p(t) of degree at most d is uniquely expressible as a linear combination ( ⋅ }{k!\cdot l! ) Ein Binomialkoeffizient hängt von zwei natürlichen Zahlen ) l How many permutations are there of 7 objects? This article incorporates material from the following PlanetMath articles, which are licensed under the Creative Commons Attribution/Share-Alike License: Binomial Coefficient, Upper and lower bounds to binomial coefficient, Binomial coefficient is an integer, Generalized binomial coefficients. k The binomial theorem is a formula for deriving the power of a binomial, i.e. Certain trigonometric integrals have values expressible in terms of n Binomial coefficients are a family of positive integers that occur as coefficients in the binomial theorem. However this is not true of higher powers of p: for example 9 does not divide + , Alternative notations include C(n, k), nCk, nCk, Ckn, Cnk, and Cn,k in all of which the C stands for combinations or choices. 5 z Exercise 2 Binomial coefficients can be generalized to multinomial coefficients defined to be the number: While the binomial coefficients represent the coefficients of (x+y)n, the multinomial coefficients is convergent for k ≥ 2. = ( over For example, for nonnegative integers ${n} \geq {q}$, the identity. t ) This is done by interpreting as the number of ways to partition the set into two subsets of size k and n-k. Pascal's rule is the important recurrence relation. Die Anzahl aller so zusammengestellten 1 \end{cases}[/math], $\tbinom n0,\tbinom n1,\tbinom n2,\ldots$, $\sum_{k=0}^\infty {n\choose k} x^k = (1+x)^n. {\displaystyle \alpha } Differentiating (2) k times and setting x = −1 yields this for Identifying Binomial Coefficients. The notation [math] {n \choose k}$ is convenient in handwriting but inconvenient for typewriters and computer terminals. > ( Then 0 < p < n and. k {\displaystyle 6={\tbinom {6}{5}}} Damit gilt für jede endliche Menge {\displaystyle {\tbinom {2n}{n}}} }{k_1!k_2!\cdots k_r! − n ψ {\displaystyle l!} n n How many different pizza can the customer create? Another form of the Chu–Vandermonde identity, which applies for any integers j, k, and n satisfying 0 ≤ j ≤ k ≤ n, is, The proof is similar, but uses the binomial series expansion (2) with negative integer exponents. Sie kann anschaulich etwa so gedeutet werden: Zunächst zählt man alle 1 k M {\displaystyle \sum _{k=0}^{d}a_{k}{\binom {t}{k}}} Compute (92)+(83)\dbinom{9}{2} + \dbinom{8}{3}(29)+(38). ) ) 0 {\displaystyle k\to \infty } Create a free website or blog at WordPress.com. This definition inherits these following additional properties from { 1 It also follows from tracing the contributions to Xk in (1 + X)n−1(1 + X). Differentiating (2) k times and setting x = −1 yields this for ( Asymptotic of binomial coefficients near the center. {\displaystyle \{1,2\}{\text{, }}\{1,3\}{\text{, }}\{1,4\}{\text{, }}\{2,3\}{\text{, }}\{2,4\}{\text{,}}} The overflow can be avoided by dividing first and fixing the result using the remainder: Another way to compute the binomial coefficient when using large numbers is to recognize that. ) in a language with fixed-length integers, the multiplication by n^{\underline{k}}/k! ( 1 . Hot Network Questions Should I respond to an "ethical hacker" who's requesting a bounty? ) ) , = 6[/math] is the coefficient of the x2 term. ! There are several ways to come up with the answer. m eigentlich einfache Binomialkoeffizienten sind. x {\displaystyle {\tfrac {(k+l)! ⋅ Newton's binomial series, named after Sir Isaac Newton, is a generalization of the binomial theorem to infinite series: The identity can be obtained by showing that both sides satisfy the differential equation (1 + z) f'(z) = α f(z). + Thinking of the binomial coefficient as the number of ways to making a series of two-outcome decisions is crucial to the understanding of binomial distribution. + The answer can also be obtained by the multiplication principle. ) The multiplicative formula allows the definition of binomial coefficients to be extended[3] by replacing n by an arbitrary number α (negative, real, complex) or even an element of any commutative ring in which all positive integers are invertible: With this definition one has a generalization of the binomial formula (with one of the variables set to 1), which justifies still calling the n {\displaystyle k!} ) 7 C k 6 Each polynomial $\tbinom{t}{k}$ is integer-valued: it has an integer value at all integer inputs $t$. ( l A simple and rough upper bound for the sum of binomial coefficients can be obtained using the binomial theorem: which is valid by for all integers = Rekursive Darstellung und Pascalsches Dreieck, Der Binomialkoeffizient in der Kombinatorik, Summen mit alternierenden Binomialkoeffizienten, Summen von Binomialkoeffizienten mit geraden bzw. {\displaystyle {\tbinom {n}{k}}=0} k setzt. ) 0 Um unnötigen Rechenaufwand zu vermeiden, berechnet man im Fall Then it follows from our earlier definition for the sum of binomial coefficients that. ( 1\quad 4 \quad 6 \quad 4 \quad 1\\ k -Binomialsymbolen: ( , ) ⋅ Addition obiger Gleichungen ) n 1 n und {\displaystyle \operatorname {Re} z>0} . M Die wörtliche Übersetzung von „ Suppose that the customer is a meat lover. The following is one specific path that the person might take. The reasoning is based on the multiplication principle (see here). , 1 Some properties make use of symmetry, some deal with expansion, but they all can be proved rather intuitively. \binom MN = \frac{M!}{N!(M-N)!} n For each k, the polynomial {\displaystyle z} {\displaystyle {\tbinom {t}{k}}} ( The Chu–Vandermonde identity, which holds for any complex-values m and n and any non-negative integer k, is, and can be found by examination of the coefficient of )^3}[/math], $\sum_{k=-a}^a(-1)^k{a+b\choose a+k} {b+c\choose b+k}{c+a\choose c+k} = \frac{(a+b+c)!}{a!\,b!\,c! One way to solve this problem is to count the number of paths in a “brute force” approach by tracing all possible paths in the diagram. , n} with the right hand side first grouping them into those which contain element n and those which don’t. über m {\displaystyle n} {\displaystyle {\tbinom {\alpha }{k}}} [math]\sum_{k=0}^n k \binom n k = n 2^{n-1}$. = k k x k ( is convergent for k ≥ 2. α ) , ϵ The left side counts the number of ways of selecting a subset of [n] = {1, 2, ..., n} with at least q elements, and marking q elements among those selected. 1 Template:Planetmath 1 ) gilt: Beweis: ( bezeichnet. (5−3)!3!=10.\binom{5}{3} = \frac{5!}{(5-3)! − | {\displaystyle M} , , [/math], that is clear since the RHS is a term of the exponential series [math] e^k=\sum_{j=0}^\infty k^j/j! of binomial coefficients. ∞ Möglichkeiten der Wahl des ersten Tupel-Elements. ∞ In ordering a pizza, a customer can choose from a list of 10 toppings: mushroom, onion, olive, bell pepper, pineapple, spinach, extra cheese, sausage, ham, and pepperoni. {\displaystyle -z,-s\notin \mathbb {N} } − = 5040. {\displaystyle m\leq n} }\\ , 5 k This formula is used in the analysis of the German tank problem. ) 1 {\displaystyle \sum _{k=0}^{[{\frac {n-1}{2}}]}{\binom {n}{2k+1}}=2^{n-1}} Assume that all the paths from any point to any point in the above diagram are available for walking. This gives. roten und 2 terms in this product is 0 {\color{blue}1 \qquad 3 \qquad 3 \qquad 1} \\ is a natural number for any natural numbers n and k. There are many other combinatorial interpretations of binomial coefficients (counting problems for which the answer is given by a binomial coefficient expression), for instance the number of words formed of n bits (digits 0 or 1) whose sum is k is given by + This article incorporates material from the following PlanetMath articles, which are licensed under the Creative Commons Attribution/Share-Alike License: Binomial Coefficient, Upper and lower bounds to binomial coefficient, Binomial coefficient is an integer, Generalized binomial coefficients. The point 1 is known to be in each -subset of the first type and the other points must be chosen from objects. ) = {\displaystyle n} A combinatorial interpretation of this formula is as follows: when forming a subset of elements (from a set of size ), it is equivalent to consider the number of ways you can pick elements and the number of ways you can exclude elements. Now, if you know your stuff about triangular numbers, you can say that. □\displaystyle { y }^{ 3 }{ a \choose 3 } = { 3 }^{ 3 }{ 10 \choose 3 } = 3240.\ _\squarey3(3a)=33(310)=3240. {\displaystyle k=m=n} und This is the statement that each row of Pascal's triangle is symmetric in either one of two ways. { (x+y) }^{ a }=\sum _{ i=0 }^{ a }{ a \choose i } { x }^{ a-i }{ y }^{ i }.(x+y)a=i=0∑a(ia)xa−iyi. n | 2022-01-27T07:49:27 | {
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https://math.stackexchange.com/questions/2650571/how-many-bit-strings | # How many bit strings?
How many bit strings of length $8$ have either exactly two $1$-bit among the first $4$ bits or exactly two $1$-bit among the last $4$ bits?
My solution:
A bit only contains $0$ and $1$, so $2$ different numbers, i.e., $0$ and $1$. For the first part we have $2^6=64$ ways. Similar for the other way. Hence there exists $2^4=16$ bit strings. Is my answer true?
Update: I mean $2^6+2^6-2^4=112$ bit strings
I lost your logic. By symmetry, amount of strings with exactly 2 ones in the first four (call this group $F$) is identical to the ones with exactly 2 ones in the last four (call this group $L$).
Then your desired amount is $|F| + |L| - |F \cap L|$.
To compute $F$, note that you have exactly $\binom{4}{2} = 6$ ways to pick the location of the ones in the first four, which fixes your picks to be ones and the other two of the first four to be zeros. In other words, this fixes the first four bits, and the rest can be manipulated in $2^4=16$ ways.
Can you finish computing $|F|$? We already said $|L|=|F|$ and can you compute $|F \cap L|$ by a similar technique?
• I still belive that it should be $2^6+2^6-2^4=112$ ways.
– user530832
Feb 14, 2018 at 16:40
• @M.Rasmussen your mistake is not automatically setting the other 2 bits to zero. The way you compute, consider, for example, the first group, with exactly 2 ones in the first four bits. $2^6$ you are getting counts the number of bit strings with 6 places, but (a) you did not count that there are multiple ways in spreading the ones among the first four and (b) you are including the string 11111111 as legal, whereas you must have exactly 2 ones in the first four, and this string has 4. Feb 14, 2018 at 17:25
• Ohh okay. I see, so all other "empty" spaces must be 0 if you know what I mean.
– user530832
Feb 14, 2018 at 17:49
• Update: I get 156 now. But to make sure I understand it, what if I change the problem to this: "How many bit strings of length $12$ have either exactly four $1$-bit among the first $6$ bits or exactly four $1$-bit among the last $6$ bits?" Then I get the answer: 122655 is that correct? I did the same proces like the original question.
– user530832
Feb 14, 2018 at 17:57
• @M.Rasmussen Then both elementary groups are $|F| = |L| = \binom{6}{4} \cdot 2^6$ and their intersection is $|F \cap L| = \binom{6}{4} \binom{6}{4}$ so you end up with $$|F| + |L| - |F\cap L| = 2 \cdot \binom{6}{4} \cdot 2^6 - \binom{6}{4}^2 = 960 - 225 = 715.$$ Feb 14, 2018 at 21:45
Let $A$ be the set of bit strings with exactly two $1$-bit among the first $4$ bits, and $B$ be the set of bit strings with exactly two $1$-bit among the last $4$ bits.
\begin{align} \#A &= \binom{4}{2} 2^4 = 6\cdot2^4 \\ \#B &= 2^4 \binom{4}{2} = 6\cdot2^4 \\ \#A\cap B &= \binom{4}{2}^2 = 6^2 \\ \#A\cup B &= \#A + \#B - \# A \cap B \\ &= 6 (2^4 \cdot 2 - 6) \\ &= 6 \cdot 26 = 156 \end{align}
Among the first $4$ bits, choose $2$ to set them to one and the other two would be set to $0$ and there are $4$ of them with absolute freedom.
$$\binom{4}{2}\cdot 2^4$$
Similar when we focus on the last $4$ bits.
When we focus on intersection. We would pick $2$ from the first $4$ and pick $2$ from the last $4$.
So my overall answer would be
$$2\binom42 \cdot 2^4 - \binom42^2$$
Remark: I think your mistake is thinking that you can set arbitary $6$ bits to anything.
• I see many of you get 156, but I just think on Inclusion–exclusion principle and the way I did it here was drawing the problem: i.stack.imgur.com/jYOXA.png
– user530832
Feb 14, 2018 at 16:38
• The keyword is exactly $2$ ones in the first $4$ bits, which means exactly $2$ ones and exactly $2$ zeros in the first $4$ bits. Feb 14, 2018 at 16:59
• Update: I get 156 now. But to make sure I understand it, what if I change the problem to this: "How many bit strings of length 12 have either exactly four 1-bit among the first 6 bits or exactly four 1-bit among the last 6 bits?" Then I get the answer: 122655 is that correct? I did the same proces like the original question
– user530832
Feb 14, 2018 at 17:58
• @M.Rasmussen: There are not even that many 12 bit strings period. There are only 4096 strings of length 12 so how could there possibly be 122655 of them that have some property? You need to get in the habit of checking your answers for reasonableness. Feb 14, 2018 at 20:55
• @M.Rasmussen would you like to show the working of how you obtain that number? Feb 14, 2018 at 21:58
How many bit strings of length 8 have either exactly two 1-bit among the first 4 bits or exactly two 1-bit among the last 4 bits?
The simplest solution is the best. There are only 256 possibilities to check so just list them!
00000011 00000101 00000110 00001001 00001010 00001100 00010011 00010101
00010110 00011001 00011010 00011100 00100011 00100101 00100110 00101001
00101010 00101100 00110000 00110001 00110010 00110011 00110100 00110101
00110110 00110111 00111000 00111001 00111010 00111011 00111100 00111101
00111110 00111111 01000011 01000101 01000110 01001001 01001010 01001100
01010000 01010001 01010010 01010011 01010100 01010101 01010110 01010111
01011000 01011001 01011010 01011011 01011100 01011101 01011110 01011111
01100000 01100001 01100010 01100011 01100100 01100101 01100110 01100111
01101000 01101001 01101010 01101011 01101100 01101101 01101110 01101111
01110011 01110101 01110110 01111001 01111010 01111100 10000011 10000101
10000110 10001001 10001010 10001100 10010000 10010001 10010010 10010011
10010100 10010101 10010110 10010111 10011000 10011001 10011010 10011011
10011100 10011101 10011110 10011111 10100000 10100001 10100010 10100011
10100100 10100101 10100110 10100111 10101000 10101001 10101010 10101011
10101100 10101101 10101110 10101111 10110011 10110101 10110110 10111001
10111010 10111100 11000000 11000001 11000010 11000011 11000100 11000101
11000110 11000111 11001000 11001001 11001010 11001011 11001100 11001101
11001110 11001111 11010011 11010101 11010110 11011001 11011010 11011100
11100011 11100101 11100110 11101001 11101010 11101100 11110011 11110101
11110110 11111001 11111010 11111100
There are 156 such strings.
Do you still believe there are only 112? If so, then which ones did I either list twice, or list in error?
• I am convinced, thank you
– user530832
Feb 15, 2018 at 13:13
As a programmer (and a relatively naive mathematician) I immediately thought of reducing the 8-bit string to a hexadecimal string. Two digits, each made of four bits exactly as the problem is divided.
And, out 16 digits (0-F), 6 of them are valid (exactly 2 1-bits) digits.
So, we have 16×6 valid numbers (the first digit can be anything) and out of the the remaining 16×10 numbers, 6×10 of those will also be valid.
Now we have 16×6 + 6×10 = 156 | 2022-05-26T21:10:09 | {
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http://gaworld.ru/gcf-of-28-and-32 | # GCF of 28 and 32
The gcf of 28 and 32 is the largest positive integer that divides the numbers 28 and 32 without a remainder. Spelled out, it is the greatest common factor of 28 and 32. Here you can find the gcf of 28 and 32, along with a total of three methods for computing it. In addition, we have a calculator you should check out. Not only can it determine the gcf of 28 and 32, but also that of three or more integers including twenty-eight and thirty-two for example. Keep reading to learn everything about the gcf (28,32) and the terms related to it.
## What is the GCF of 28 and 32
If you just want to know what is the greatest common factor of 28 and 32, it is 4. Usually, this is written as
gcf(28,32) = 4
The gcf of 28 and 32 can be obtained like this:
• The factors of 28 are 28, 14, 7, 4, 2, 1.
• The factors of 32 are 32, 16, 8, 4, 2, 1.
• The common factors of 28 and 32 are 4, 2, 1, intersecting the two sets above.
• In the intersection factors of 28 ∩ factors of 32 the greatest element is 4.
• Therefore, the greatest common factor of 28 and 32 is 4.
Taking the above into account you also know how to find all the common factors of 28 and 32, not just the greatest. In the next section we show you how to calculate the gcf of twenty-eight and thirty-two by means of two more methods.
## How to find the GCF of 28 and 32
The greatest common factor of 28 and 32 can be computed by using the least common multiple aka lcm of 28 and 32. This is the easiest approach:
gcf (28,32) = $\frac{28 \times 32}{lcm(28,32)} = \frac{896}{224}$ = 4
Alternatively, the gcf of 28 and 32 can be found using the prime factorization of 28 and 32:
• The prime factorization of 28 is: 2 x 2 x 7
• The prime factorization of 32 is: 2 x 2 x 2 x 2 x 2
• The prime factors and multiplicities 28 and 32 have in common are: 2 x 2
• 2 x 2 is the gcf of 28 and 32
• gcf(28,32) = 4
In any case, the easiest way to compute the gcf of two numbers like 28 and 32 is by using our calculator below. Note that it can also compute the gcf of more than two numbers, separated by a comma. For example, enter 28,32. The calculation is conducted automatically.
The gcf is...
Frequently searched terms on our site also include:
## Use of GCF of 28 and 32
What is the greatest common factor of 28 and 32 used for? Answer: It is helpful for reducing fractions like 28 / 32. Just divide the nominator as well as the denominator by the gcf (28,32) to reduce the fraction to lowest terms.
$\frac{28}{32} = \frac{\frac{28}{4}}{\frac{32}{4}} = \frac{7}{8}$.
## Properties of GCF of 28 and 32
The most important properties of the gcf(28,32) are:
• Commutative property: gcf(28,32) = gcf(32,28)
• Associative property: gcf(28,32,n) = gcf(gcf(32,28),n) $\hspace{10px}n\hspace{3px}\epsilon\hspace{3px}\mathbb{Z}$
The associativity is particularly useful to get the gcf of three or more numbers; our calculator makes use of it.
To sum up, the gcf of 28 and 32 is 4. In common notation: gcf (28,32) = 4.
If you have been searching for gcf 28 and 32 or gcf 28 32 then you have come to the correct page, too. The same is the true if you typed gcf for 28 and 32 in your favorite search engine.
Note that you can find the greatest common factor of many integer pairs including twenty-eight / thirty-two by using the the search form in the sidebar of this page.
Questions and comments related to the gcf of 28 and 32 are really appreciated. Use the form below or send us a mail to get in touch.
Please hit the sharing buttons if our article about the greatest common factor of 28 and 32 has been useful to you, and make sure to bookmark our site.
Posted in Greatest Common Factor
###### 2 comments on “GCF of 28 and 32”
1. POOFA says:
WOW! THIS IS A GREAT ANSWER! THANK YOU! 🙂
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https://math.stackexchange.com/questions/3007234/seeking-methods-to-solve-int-0-frac-pi2-ln-left2-tan2x-righ | # Seeking methods to solve $\int_{0}^{\frac{\pi}{2}} \ln\left|2 + \tan^2(x) \right| \:dx$
As part of going through a set of definite integrals that are solvable using the Feynman Trick, I am now solving the following:
$$\int_{0}^{\frac{\pi}{2}} \ln\left|2 + \tan^2(x) \right| \:dx$$
I'm seeking methods using the Feynman Trick (or any method for that matter) that can be used to solve this definite integral.
If one wishes to use "Feynman's Trick," then begin by defining a function $$I(a)$$, $$a>1$$ as given by
$$I(a)=\int_0^{\pi/2}\log(a+\tan^2(x))\,dx \tag1$$
Differentiation of $$(1)$$ reveals
\begin{align} I'(a)&=\int_0^{\pi/2} \frac{1}{a+\tan^2(x)}\,dx\\\\ &=\frac{\pi/2}{a-1}-\frac{\pi/2}{\sqrt a (a-1)}\tag2 \end{align}
Integration of $$(2)$$ yields
\begin{align} I(a)&=\frac\pi2\left(\log(a-1)+\log\left(\frac{\sqrt a+1}{\sqrt{a}-1}\right) \right)\\\\ &=\pi \log(\sqrt a+1)\tag3 \end{align}
Finally, setting $$a=2$$ in $$(3)$$, we obtain the coveted result
$$\int_0^{\pi/2}\log(2+\tan^2(x))\,dx=\pi \log(\sqrt 2+1)$$
• How did you resolve the constant of Integration in Step (3)? – user150203 Nov 21 '18 at 4:38
• @davidg Apology for the sign error. Note $I(0)=0$. – Mark Viola Nov 21 '18 at 5:08
• Hi Mark ! Long time no speak. Nice solution $\to +1$. Cheers. – Claude Leibovici Nov 21 '18 at 6:22
• @MarkViola do you know of any other ‘tricks that would work? – user150203 Nov 21 '18 at 6:51
• Hi David. You could try writing the logarithm as $\log(1+\cos^2(x))-\log(\sin^2(x))$ and expanding the first term as $\sum_{n=1}^\infty \frac{(-1)^{n-1}\cos^{2n}(x)}{n}$ and proceeding. I haven't tried this, but it might be worth pursuing. – Mark Viola Nov 21 '18 at 15:47
My approach
Let
\begin{align} \int_{0}^{\frac{\pi}{2}} \ln\left|2 + \tan^2(x) \right| \:dx &= \int_{0}^{\frac{\pi}{2}} \ln\left|1 + \left(1 + \tan^2(x)\right) \right| \:dx \\ &= \int_{0}^{\frac{\pi}{2}} \ln\left|1 + \sec^2(x) \right| \:dx \\ &= \int_{0}^{\frac{\pi}{2}} \ln\left|\frac{\cos^2(x) + 1}{\cos^2(x)} \right| \:dx \\ &= \int_{0}^{\frac{\pi}{2}} \left[ \ln\left|\cos^2(x) + 1 \right| - \ln\left|\cos^2(x)\right| \right]\:dx \\ &= \int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x) + 1 \right|\:dx - \int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x)\right|\:dx \end{align}
Now
$$\int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x)\right|\:dx = 2\int_{0}^{\frac{\pi}{2}} \ln\left|\cos(x)\right|\:dx = 2\cdot-\frac{\pi}{2}\ln(2) = -\pi \ln(2)$$
For detail on this definite integral, see guidance here
We now need to solve
$$\int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x) + 1 \right|\:dx$$
Here, Let
$$I(t) = \int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x) + t \right|\:dx$$
Thus,
$$\frac{dI}{dt} = \int_{0}^{\frac{\pi}{2}} \frac{1}{\cos^2(x) + t}\:dx = \int_{0}^{\frac{\pi}{2}} \frac{1}{\frac{\cos(2x) + 1}{2} + t}\:dx = 2\int_{0}^{\frac{\pi}{2}} \frac{1}{\cos(2x) + 2t + 1}\:dx$$
Employ a change of variable $$u = 2x$$:
$$\frac{dI}{dt} = \int_{0}^{\pi} \frac{1}{\cos(u) + 2t + 1}\:du$$
Employ the Weierstrass substitution $$\omega = \tan\left(\frac{u}{2} \right)$$:
\begin{align} \frac{dI}{dt} &= \int_{0}^{\infty} \frac{1}{\frac{1 - \omega^2}{1 + \omega^2} + 2t + 1}\:\frac{2}{1 + \omega^2}\cdot d\omega \\ &= \int_{0}^{\infty} \frac{1}{t\omega^2 + t + 1} \:d\omega \\ &= \frac{1}{t}\int_{0}^{\infty} \frac{1}{\omega^2 + \frac{t + 1}{t}} \:d\omega \\ &= \frac{1}{t}\left[\frac{1}{\sqrt{\frac{t+1}{t}}}\arctan\left( \frac{\omega}{\sqrt{\frac{t+1}{t}}}\right)\right]_{0}^{\infty} \\ &= \frac{1}{t}\frac{1}{\sqrt{\frac{t+1}{t}}}\frac{\pi}{2} \\ &= \frac{1}{\sqrt{t}\sqrt{t + 1}}\frac{\pi}{2} \end{align}
And so,
$$I(t) = \int \frac{1}{\sqrt{t}\sqrt{t + 1}}\frac{\pi}{2}\:dt = \pi\ln\left| \sqrt{t} + \sqrt{t + 1}\right| + C$$
Now
$$I(0) = \int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x) + 0 \right|\:dx = -\pi \ln(2) = \pi\ln\left|\sqrt{0} + \sqrt{0 + 1} \right| + C \rightarrow C = -\pi \ln(2)$$
And so,
$$I(t) = \pi\ln\left| \sqrt{t} + \sqrt{t + 1}\right| -\pi \ln(2) = \pi\ln\left|\frac{\sqrt{t} + \sqrt{t + 1}}{2} \right|$$
Thus,
$$I = I(1) = \int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x) + 1 \right|\:dx = \pi\ln\left|\frac{\sqrt{1} + \sqrt{1 + 1}}{2} \right| = \pi\ln\left|\frac{1 + \sqrt{2}}{2} \right|$$
And Finally
\begin{align} \int_{0}^{\frac{\pi}{2}} \ln\left|2 + \tan^2(x) \right| \:dx &= \int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x) + 1 \right|\:dx - \int_{0}^{\frac{\pi}{2}} \ln\left|\cos^2(x)\right|\:dx \\ &= \pi\ln\left|\frac{1 + \sqrt{2}}{2} \right| - \left(-\pi \ln(2)\right) \\ &= \pi\ln\left|1 + \sqrt{2} \right| \end{align}
• Your question and your answer have a gap of just 1 minute. If you did solve the question earlier then you must've included your approach in the question itself. What is the need to post it as an answer? – Rohan Shinde Nov 21 '18 at 3:56
• @Digamma - Sorry, although I've been a member of Math StackExchange for 4 years, I've only become regularly active recently. When I asked a friend who is a member as to what is the best way of presenting a solution whilst asking for other solutions I was told this is the approach taken within the community. If this violate any of the rules, please advise and I will reposition accordingly. Also, I never assign my solution as 'the solution' to any post of this nature. – user150203 Nov 21 '18 at 4:00
• @Digamma The site encourages people to share the questions with answers. – Tianlalu Nov 21 '18 at 4:03
• @Tianlalu - Sorry, just to be clear - Have I employed the correct process here? – user150203 Nov 21 '18 at 4:10
• Thanks @mick - much appreciated :-) – user150203 Dec 3 '18 at 3:42 | 2019-07-17T04:52:20 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/3007234/seeking-methods-to-solve-int-0-frac-pi2-ln-left2-tan2x-righ",
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# If f(x)= |4x - 1| + |x-3| + |x + 1| (Question of the Day-II)
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If f(x)= |4x - 1| + |x-3| + |x + 1| (Question of the Day-II) [#permalink]
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If f(x) = |4x - 1| + |x-3| + |x + 1|, what is the minimum value of f(x)?
(A) 3
(B) 4
(C) 5
(D) 21/4
(E) 7
(Still high on mods! Next week, will make questions on some other topic.)
[Reveal] Spoiler: OA
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Last edited by Bunuel on 06 Jul 2013, 02:33, edited 1 time in total. Added the OA. CEO Status: Nothing comes easy: neither do I want. Joined: 12 Oct 2009 Posts: 2783 Location: Malaysia Concentration: Technology, Entrepreneurship Schools: ISB '15 (M) GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35 Re: Question of the Day - II [#permalink] ### Show Tags 29 Oct 2010, 18:51 7 This post received KUDOS VeritasPrepKarishma wrote: Q. If f(x) = |4x - 1| + |x-3| + |x + 1|, what is the minimum value of f(x)? (A) 3 (B) 4 (C) 5 (D) 21/4 (E) 7 (Still high on mods! Next week, will make questions on some other topic.) put x=0 we get f(x) = 5..rule out D and E put x=1/4 we get f(x) = 0 + 11/4 + 5/4 = 16/4 = 4.. rule out C D and E Now the answer is either 3 or 4. Reason for above checking of values: for every value of x> 3 the f(x) is quite big because of 4x-1 for every value of x < -1 the f(x) if bigger than 3 and 4. for x >1/4 and x< 3 f(x) is bigger than 4. because f(x) = 4x-1 + 3-x + x+1 = 4x+3 > 3 => only 4 is the probable answer. Thus we only need to check x>=-1 and x<= 1/4 f(x) in this domain is = 1-4x + 3-x + x+1 = 5-4x => for f(x) to be minimum the x should be +ve => for x = 1/4 , f(x) = 4. This can be solved using graph as well by plotting the f(x) in different domains. _________________ Fight for your dreams :For all those who fear from Verbal- lets give it a fight Money Saved is the Money Earned Jo Bole So Nihaal , Sat Shri Akaal Support GMAT Club by putting a GMAT Club badge on your blog/Facebook GMAT Club Premium Membership - big benefits and savings Gmat test review : http://gmatclub.com/forum/670-to-710-a-long-journey-without-destination-still-happy-141642.html Retired Moderator Joined: 02 Sep 2010 Posts: 803 Location: London Re: Question of the Day - II [#permalink] ### Show Tags 30 Oct 2010, 01:59 10 This post received KUDOS 1 This post was BOOKMARKED Let g(x) = |4x - 1| & h(x)=|x-3| + |x + 1| ... So f(x)=g(x)+h(x) Now h(x) is equal to 4 between -1 and 3 and is higher everywhere else, so it minimizes between -1 & 3 g(x) is minimum at x=(1/4) where it is equal to 0. Since -1<(1/4)<3 & f(x)=g(x)+h(x) ... It is straight forward to imply that f(x) will be minimum at 1/4, since both g(x) & h(x) are minimum at that point f(1/4)=g(1/4)+h(1/4)=0+4 Hence answer is 4 _________________ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7450 Location: Pune, India Re: Question of the Day - II [#permalink] ### Show Tags 30 Oct 2010, 08:53 45 This post received KUDOS Expert's post 13 This post was BOOKMARKED That's right! I can count on you guys to always give the correct answer and some innovative ways of solving... Let me give you the method I would use to solve it too... Where mods are concerned, the fact that mod signifies the distance from 0 on the number line is my best friend... So |x-3| is the distance from 3, |x + 1| is the distance from -1 and |4x - 1| is 4 times the distance from 1/4 So |4x - 1| + |x-3| + |x + 1| is the sum of distances from 3, -1 and four times the distance from 1/4 e.g. if x takes the value at point A, f(x) will be sum of length of red line, green line and blue line. the question here is, what is the minimum such sum possible? Attachment: Ques.jpg [ 4.56 KiB | Viewed 14954 times ] Can I say that minimum total distance will be covered from point 1/4? Attachment: Ques1.jpg [ 3.43 KiB | Viewed 14937 times ] The logic being that the distance from 3 to -1, which is 4 units has to be covered. Why to cover any distance from 1/4 at all? _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199
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In other words, as my mentor says, assume there is one guy on point 3, one on point -1 and 4 guys on point 1/4. If they have to meet up, but cover minimum distance, they should meet at point 1/4. What happens when there are 4 such points? Lets add another term (2x - 3) to f(x)...
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Manager Joined: 11 Jul 2010 Posts: 224 Re: Question of the Day - II [#permalink] ### Show Tags 30 Oct 2010, 09:29 would the answer still remain 1/4? based on your 'distance' method? The 3 guys standing at the new 3/2 post + the existing persons at -1 and 3 can again meet at 1/4 as that would be the shortest distance Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7450 Location: Pune, India Re: Question of the Day - II [#permalink] ### Show Tags 30 Oct 2010, 16:17 3 This post received KUDOS Expert's post 3 This post was BOOKMARKED gmat1011 wrote: would the answer still remain 1/4? based on your 'distance' method? The 3 guys standing at the new 3/2 post + the existing persons at -1 and 3 can again meet at 1/4 as that would be the shortest distance Yes, that is right! The answer still remains 1/4. Since |2x - 3| = 2|x - 3/2| it is twice the distance from 3/2 (or we can say, there are 2 guys are 3/2). The guy at -1 and 3 still need to cover 4 units together. If the 2 guys at 3/2 come down to 1/4, they would have covered less distance than if 4 guys were made to travel anywhere from 1/4. Attachment: Ques.jpg [ 5.2 KiB | Viewed 14965 times ] Try some other combinations. e.g. f(x) = |x - 1| + |x-3| + |x + 1| + |x + 6| f(x) = |2x - 3| + |4x + 7| etc _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199
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Re: Question of the Day - II [#permalink]
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09 Nov 2010, 23:23
Great Method Karishma !!
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Re: Question of the Day - II [#permalink]
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14 Jun 2011, 02:14
for x = 0, f(x) = 5. POE options C,D and E.
for x = 1/4, f(x) = 4.
B it is.
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Re: Question of the Day - II [#permalink]
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14 Jun 2011, 03:47
Very good question thanks for posting
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Re: Question of the Day - II [#permalink]
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15 Mar 2012, 11:07
VeritasPrepKarishma wrote:
gmat1011 wrote:
The 3 guys standing at the new 3/2 post + the existing persons at -1 and 3 can again meet at 1/4 as that would be the shortest distance
Yes, that is right! The answer still remains 1/4.
Since |2x - 3| = 2|x - 3/2| it is twice the distance from 3/2 (or we can say, there are 2 guys are 3/2). The guy at -1 and 3 still need to cover 4 units together. If the 2 guys at 3/2 come down to 1/4, they would have covered less distance than if 4 guys were made to travel anywhere from 1/4.
Attachment:
Ques.jpg
Try some other combinations. e.g. f(x) = |x - 1| + |x-3| + |x + 1| + |x + 6|
f(x) = |2x - 3| + |4x + 7| etc
Karishman, how are you deciding that distance for 2 guys to travel from 3/2 would be less than distance by 4 guys to move from 1/4 ?
In example : |x - 1| + |x-3| + |x + 1| + |x + 6| .. the posts on the number line are : -6, -1 , 1 and 3. In order to minimize value of this expression to 9, how to chose the x ?
Please correct me if I am wrong.
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Karishman, how are you deciding that distance for 2 guys to travel from 3/2 would be less than distance by 4 guys to move from 1/4 ?
In example : |x - 1| + |x-3| + |x + 1| + |x + 6| .. the posts on the number line are : -6, -1 , 1 and 3. In order to minimize value of this expression to 9, how to chose the x ?
Please correct me if I am wrong.
Forget these numbers. Think logically.
My house is 10 miles away from your house. If we have to meet up, how much distance do we need to cover together? In any case, we need to cover 10 miles together at least, right? Either you come down to my place (you cover 10 miles) or I come down to yours (I cover 10 miles) or we meet mid way (10 miles covered together) or we meet up at a nice coffee place 2 miles further down from my house in the opposite direction in which case we will need to cover more than 10 miles (i.e. we cover 2 + 12 = 14 miles)
Now say, another friend is at my place. In which case will people cover minimum distance together? If two of us come down to your place, we cover 10+10 = 20 miles together but if you come down to our place, you cover only 10 miles. If instead, we meet midway, we cover 5+5 and you cover 5 miles so in all 15 miles. So less number of people should travel the entire distance.
If there are 4 people at point A and 2 at point B, minimum distance will be covered if people at point B travel to point A. So people at 3/2 should come down to 1/4.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7450 Location: Pune, India Re: Question of the Day - II [#permalink] ### Show Tags 15 Mar 2012, 11:33 5 This post received KUDOS Expert's post 2 This post was BOOKMARKED ficklehead wrote: In example : |x - 1| + |x-3| + |x + 1| + |x + 6| .. the posts on the number line are : -6, -1 , 1 and 3. In order to minimize value of this expression to 9, how to chose the x ? Please correct me if I am wrong. x is that point on the number line whose sum of distances from -6, -1, 1 and 3 is minimum. So basically there is a person each at points -6, -1, 1 and 3. You need to make them all meet by covering minimum distance. Distance between -6 and 3 is 9 which must be covered by these 2 people to meet. These 2 can meet at any point: -6, -1, 0, 1 or 3 etc they will cover a distance of 9 together. If -1 and 1 have to meet too, they need to cover a distance of 2 together. Say, if person at -1 travels down to 1 and -6 and 3 also meet at 1, the minimum distance covered will be 9+2 = 11 and they will all be able to meet. If they instead meet at -1, the situation will be the same and total distance covered will be 11 again. In fact, they can meet at any point between -1 and 1, the total distance covered will be 11. To check, put x = 1. you get |x - 1| + |x-3| + |x + 1| + |x + 6| = 11 put x = -1, you get |x - 1| + |x-3| + |x + 1| + |x + 6| = 11 put x = 0, you get |x - 1| + |x-3| + |x + 1| + |x + 6| = 11 _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199
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Re: Question of the Day - II [#permalink]
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15 Mar 2012, 14:48
VeritasPrepKarishma wrote:
In example : |x - 1| + |x-3| + |x + 1| + |x + 6| .. the posts on the number line are : -6, -1 , 1 and 3. In order to minimize value of this expression to 9, how to chose the x ?
Please correct me if I am wrong.
x is that point on the number line whose sum of distances from -6, -1, 1 and 3 is minimum. So basically there is a person each at points -6, -1, 1 and 3. You need to make them all meet by covering minimum distance.
Distance between -6 and 3 is 9 which must be covered by these 2 people to meet. These 2 can meet at any point: -6, -1, 0, 1 or 3 etc they will cover a distance of 9 together.
If -1 and 1 have to meet too, they need to cover a distance of 2 together. Say, if person at -1 travels down to 1 and -6 and 3 also meet at 1, the minimum distance covered will be 9+2 = 11 and they will all be able to meet.
If they instead meet at -1, the situation will be the same and total distance covered will be 11 again. In fact, they can meet at any point between -1 and 1, the total distance covered will be 11.
To check, put x = 1. you get |x - 1| + |x-3| + |x + 1| + |x + 6| = 11
put x = -1, you get |x - 1| + |x-3| + |x + 1| + |x + 6| = 11
put x = 0, you get |x - 1| + |x-3| + |x + 1| + |x + 6| = 11
Thanks Karishma for this detailed explanation.
I got it now.
I was getting 11 as a distance but was not sure, if I should try other values of x to check if there could be a lower value than 11.
To seal the concept, for |2x-3|+|4x+7|, minimum distance be : 3/2+7/4=13/4 ?
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I was getting 11 as a distance but was not sure, if I should try other values of x to check if there could be a lower value than 11.
To seal the concept, for |2x-3|+|4x+7|, minimum distance be : 3/2+7/4=13/4 ?
Look at the diagram below. Make a number line in such questions.
Attachment:
Ques3.jpg [ 5.1 KiB | Viewed 13919 times ]
For x and y to meet, they have to cover a distance of 9 together. For p and q to meet, they have to cover a distance of 2 together. They can meet anywhere between -1 and 1 and they will cover a total distance of 11 only. So x can take any value -1 < x < 1 and the value of the expression will be 11.
|2x-3|+|4x+7| = 2|x-3/2| + 4|x+7/4|
Attachment:
Ques4.jpg [ 5.35 KiB | Viewed 13934 times ]
There are 4 people at -7/4 and 2 people at 3/2. Distance between the two points is 7/4 + 3/2 = 13/4
For these people to meet covering the minimum distance, the 2 people X and Y should travel to point -7/4. (Make minimum people travel). So minimum distance that needs to be covered = 2*13/4 = 13/2 (because 2 people travel 13/4 each) which is the minimum value of the expression. The value of x when the expression takes minimum value is -7/4.
Check by putting x = -7/4. You get |2x-3|+|4x+7| = 13/2
Also see that when you put x = 0 or 3/2 etc, the value of the expression is higher.
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Get started with Veritas Prep GMAT On Demand for $199 Veritas Prep Reviews Intern Joined: 24 Jul 2011 Posts: 13 Re: Question of the Day - II [#permalink] ### Show Tags 16 Mar 2012, 09:01 Thanks a lot, Karishma. Intern Joined: 24 Jul 2011 Posts: 13 Re: Question of the Day - II [#permalink] ### Show Tags 16 Mar 2012, 19:48 I am wondering how can this method be used in questions where there are negative between terms : Ex: minimum value of : |x+6|-|x-1| ? Manager Joined: 28 Jul 2011 Posts: 238 Re: Question of the Day - II [#permalink] ### Show Tags 20 Mar 2012, 14:33 I got C = 5 f(x) = |4x-1| + |x-3| + |x+1| =(4x-1) + (x-3) + (x+1) or = - (4x-1) - (x-3) - (x+1) so, x= 1/2 then i used the value of x=1/2 f(x) = |4x-1| + |x-3| + |x+1| f(1/2) = |4(1/2)-1| + |(1/2)-3| + |(1/2)+1| = |1| + |-5/2| + |3/2| = 1 + 5/2 + 3/2 = 5 Is that correct? Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 7450 Location: Pune, India Re: Question of the Day - II [#permalink] ### Show Tags 20 Mar 2012, 21:07 kuttingchai wrote: I got C = 5 f(x) = |4x-1| + |x-3| + |x+1| =(4x-1) + (x-3) + (x+1) or = - (4x-1) - (x-3) - (x+1) so, x= 1/2 then i used the value of x=1/2 f(x) = |4x-1| + |x-3| + |x+1| f(1/2) = |4(1/2)-1| + |(1/2)-3| + |(1/2)+1| = |1| + |-5/2| + |3/2| = 1 + 5/2 + 3/2 = 5 Is that correct? Put x = 1/4 and the minimum value you will get is 4. How did you get x = 1/2? I would suggest you to check out one of the approaches mentioned above. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Get started with Veritas Prep GMAT On Demand for$199
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I am wondering how can this method be used in questions where there are negative between terms :
Ex: minimum value of : |x+6|-|x-1| ?
You can do it with a negative sign too.
You want to find the minimum value of (distance from -6) - (distance from 1)
Make a number line with -6 and 1 on it.
(-6)..........................(1)
Think of a point in the center of -6 and 1. Its distance from -6 is equal to distance from 1 and hence (distance from -6) - (distance from 1) = 0 .
What if instead, the point x is at -6? Distance from -6 is 0 and distance from 1 is 7 so (distance from -6) - (distance from 1) = 0 - 7 = -7
If you keep moving to the left, (distance from -6) - (distance from 1) will remain -7 so the minimum value is -7.
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https://math.stackexchange.com/questions/2582606/computing-int-limits-0-infty-x-left-lfloor-frac1x-right-rfloor-dx | # Computing $\int\limits_0^\infty x \left \lfloor{\frac1x}\right \rfloor \, dx$
This is an integral I computed but can't find the result online or on wolfram. So here's a proof sketch, please indulge this sanity check:
$$\int_0^\infty x \left \lfloor{\frac1x}\right \rfloor \ dx = \int_0^1 x \left \lfloor{\frac1x}\right \rfloor \ dx$$ $$= \sum_{n=1}^\infty \int_{1/(n+1)}^{1/n} nx \ dx =\sum_{n=1}^\infty\frac n2 \left(\frac{1}{n^2} - \frac{1}{(n+1)^2}\right)$$ $$= \sum_{n=1}^\infty\frac n2 \left(\frac{2n+1}{n^2(n+1)^2}\right)$$ $$= \sum_{n=1}^\infty\frac{1}{(n+1)^2} + \frac12 \sum_{n=1}^\infty \frac{1}{n(n+1)^2}$$ $$= \frac{\pi^2}{6} -1 + \frac12\left(\sum_{n=1}^\infty \frac1n - \frac{1}{n+1} - \frac{1}{(n+1)^2}\right)$$ $$=\frac{\pi^2}{6} -1 + \frac12\left(\sum_{n=1}^\infty \frac1n - \frac{1}{n+1}\right) -\frac12\left(\sum_{n=1}^\infty\frac{1}{(n+1)^2}\right)$$ $$= \left(\frac{\pi^2}{6} -1\right) + \left(\frac12\cdot 1\right) - \frac12\left(\frac{\pi^2}{6} -1\right)$$ $$= \frac{\pi^2}{12}.$$
Basically, I used the Basel sum several times, and the fifth line follows from a partial sum decomposition. The seventh follows from the known result for the Basel sum, as well as the fact that the first series in the 6th line telescopes.
I hope this is all correct.
• @XanderHenderson For $x>1$, floor of $\frac{1}{x}$ is $0$ – BallBoy Dec 28 '17 at 3:34
• @XanderHenderson What you wrote is certainly not true, because you forgot the $x$ term. However, what Y. Forman says is correct, and I should've been more explicit there. – David Bowman Dec 28 '17 at 3:37
• Oi... derp. Sorry for being dyslexic. I missed the $x$. – Xander Henderson Dec 28 '17 at 3:39
• $$\int_0^1 x \lfloor 1/x \rfloor dx = \int_1^\infty \frac{1}{t} \lfloor t \rfloor \frac{dt}{t^2}= \sum_{n=1}^\infty \int_n^\infty t^{-3}dt = \sum_{n=1}^\infty \frac{n^{-2}}{2} = \frac{\zeta(2)}{2}$$ – reuns Dec 28 '17 at 3:43
• @reuns Nice, better post it as answer! – samjoe Dec 28 '17 at 4:40
Use
$$n\left(\frac1{n^2}-\frac1{(n+1)^2}\right)=\frac n{n^2}-\frac{n+1-1}{(n+1)^2}=\frac1n-\frac1{n+1}+\frac1{(n+1)^2}.$$
The first two terms do telescope and the Basel series remains.
• @stressedout: of course, but much simpler. And also simpler than yours, which is also essentially the OP's. – Yves Daoust Sep 5 '18 at 6:28
• @stressedout: your solution isn't different, just longer. A one-liner is simpler. – Yves Daoust Sep 5 '18 at 6:38
Alternatively, one may follow the same line of thought and use summation by parts formula to calculate the infinite sum. Similarly, $$\int_0^{+\infty} x\lfloor \frac{1}{x}\rfloor dx =\sum_{n=1}^{\infty}\frac{n}{2}\left(\frac{1}{n^2}-\frac{1}{(n+1)^2}\right)=-\frac{1}{2}\sum_{n=1}^{\infty}f_n(g_{n+1}-g_n)$$
where $f_n = n$ and $g_n = 1/n^2$. Now, summation by parts for the last expression gives:
$$-\frac{1}{2}\sum_{n=1}^{\infty}f_n(g_{n+1}-g_n) = -\frac{1}{2}\left(\ \lim_{n\to\infty}\frac{n}{(n+1)^2} -1 -\sum_{n=2}^{\infty}\frac{1}{n^2}\right)$$
But it is well-known that $$\sum_{n=1}^{\infty}\frac{1}{n^2}=\frac{\pi^2}{6}$$
Hence,
$$-\frac{1}{2}\sum_{n=1}^{\infty}f_n(g_{n+1}-g_n) = -\frac{1}{2}\left(\ \lim_{n\to\infty}\frac{n}{(n+1)^2} -1 - (\frac{\pi^2}{6}-1)\right) = \frac{\pi^2}{12}$$ | 2019-12-11T06:03:23 | {
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https://math.stackexchange.com/questions/3113510/prove-that-each-integer-n-%E2%89%A5-12-is-a-sum-of-4s-and-5s-using-strong-induction | # Prove that each integer n ≥ 12 is a sum of 4's and 5's using strong induction
So I've been given the following problem: Prove that each integer n ≥ 12 is a sum of 4's and 5's What I have so far: (Basis):
n ≥ 12 Therefore,
12 ≤ 4(x) + 5(y)
x = 3 | y = 0
12 ≤ 4(3) + 5(0)
12 ≤ 12 = Correct
However, what I don't understand is how would I use the x & y variable in induction step. Where exactly would I place these? and how would I go on to solve this?
The trick to these sorts of problems is to realize that if we can find four consecutive integers ($$4$$ is the smallest of our numbers we're taking a combination of) that can be represented as a non-negative sum of the fours and fives.
For example $$12 = 4(3) + 5(0)$$
$$13 = 4(2) + 5(1)$$ $$14 = 4(1)+ 5(2)$$ $$15 = 4(0) + 5(3).$$
With this information is it clear how you could represent 16? Just take the solution from the 12 case, and add 4! (so increase $$x$$ by $$1$$.)
Here's the formal argument. Let $$P(n)$$ be the open sentence "$$n$$ can be written as a non-negative combination of $$4$$ and $$5$$". By what we've shown above, we know that $$P(k)$$ is true for $$k = 12,13,14,15.$$ We wish to prove that $$P(k+1)$$ is true. Our strong inductive hypothesis is to suppose that for some $$k \in \mathbb{Z}$$ that for every $$i$$ with $$12\leq i\leq k$$ that $$P(k)$$ is true, and we need to prove that $$P(k+1)$$ is true.
If $$k = 12,13$$ or $$14$$, we've already seen that $$p(k+1)$$ is also true, so suppose that $$k \geq 15$$. Then we know that $$12 \leq k-3\leq k$$ so then by our strong inductive hypothesis, there exists $$x,y \in \mathbb{Z}$$ such that $$k-3 = 4(x) + 5(y)$$. Then adding $$4$$ to both sides gives that $$k+1 = 4(x+1)+5(y)$$ so that $$P(k+1)$$ is true. Thus by the principle of mathematical induction, $$P(n)$$ must be true for all $$n \geq 12$$.
• Wow this is very detailed and helped me understand it more, thank you! – MathNoob Feb 15 '19 at 4:15
• No problem! I'm glad this helped! – JonHales Feb 15 '19 at 4:17
$$4x+5y=\underbrace{4(x-1)+5y}_{n-4}+4$$
So, if $$n-4$$ is expessible so will be $$n$$
$$12=4\cdot3$$
$$13=2\cdot4+5$$
$$14=4+2\cdot5$$
$$15=3\cdot5$$ | 2020-04-01T00:03:54 | {
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https://math.stackexchange.com/questions/2598977/finding-dimension-of-a-kernel-from-a-linear-transformation | # Finding dimension of a kernel from a linear transformation.
I'm trying to solve an exercise where I'm asked to find this linear transformation's kernel and its dimension:
$$f(x_1,x_2,x_3,x_4)=(x_1-x_2,2x_3+x_4)$$
So, from the definition of kernel and converting to parametric equations, I found $x_1=x_2=\lambda$, $x_3=-\frac{1}{2}\mu$ and $x_4=\mu$. Therefore, I assumed: $$\ker(f)=\{(\lambda,\lambda,-\frac{1}{2}\mu,\mu): \lambda,\mu \in \mathbb{R}\}$$
And since the kernel can be described with only one vector (with two parameters), I thought $\dim(\ker(f))=1$, but my textbook says it's actually equal to $2$. Where is my reasoning flawed?
• A basis for the kernel is $\{(1,1,0,0), (0,0,-\frac{1}{2}, 1) \}$ – leibnewtz Jan 9 '18 at 23:20
• @leibnewtz Hm, so that means I can only 'fix' one parameter at once? I thought it'd be possible to establish something like $\lambda=1$, $\mu=2$ so that a basis for the kernel is $\{(1,1,-1,2)\}$. Why isn't that possible? – Manuel Jan 9 '18 at 23:22
• Of course you can do what you say...but that way you'll only get one vector, and you already know you need two lin. ind. vectors to have a basis... – DonAntonio Jan 9 '18 at 23:30
• Since $f:\mathbb R^4\to\mathbb R^2$, its image is at most two-dimensional, so you know that its kernel must be at least two-dimensional. – amd Jan 9 '18 at 23:35
• @Manuel Please, if you are ok, you can accept the answer and set it as solved. Thanks! – user Feb 4 '18 at 0:09
It is all correct, the dimension of the ker(f) is 2 because you have 2 free parameters that define it $(\lambda,\mu)$.
• @Manuel They are equivalent concepts. As noted in the comments you can set $\lambda=1$ and $\mu=0$ and define a basis vector, then $\lambda=0$ and $\mu=1$ and define a second vector linearly independent from the first, thus the dimension is 2. This is true for any number of free parameter (EG a line or a plane in $\mathbb{R^3}$). – user Jan 9 '18 at 23:25
• @Manuel There are two vectors in your description of the kernel: $\left(\lambda,\lambda,-\frac12\mu,\mu\right)=\lambda(1,1,0,0)+\mu\left(0,0,-\frac12,1\right)$; every vector in the kernel is a linear combination of these two linearly-independent vectors. – amd Jan 9 '18 at 23:36 | 2019-10-20T03:40:44 | {
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https://mathoverflow.net/questions/252313/is-every-group-an-ideal-class-group-of-a-number-field/299691 | Is every group an ideal class group of a number field?
The inverse Galois problem asks whether every finite group appears as the Galois group of some finite extension of $\mathbb Q$. I was wondering to what extent the analogous problem for ideal class groups has been investigated. More precisely, consider the following question:
Is every finite abelian group the ideal class group of some number field (finite extension of $\mathbb Q$)?
I'd be interested to hear about any partial results, as I suppose this question is still open. I'd be also interested in any results about a weaker problem:
Is every positive integer the ideal class number of some number field?
Again, any reference, even to a partial result, will be appreciated.
• Not an answer to your question, but are you aware of Claborn's theorem that says every finite abelian group is the class group of a Dedekind domain (though not necessarily, as far as one can tell from Claborn's work, of a ring of integers)? – Steven Landsburg Oct 16 '16 at 19:52
• @StevenLandsburg I was not aware of that result, since, to be honest, I am not so interested in general Dedekind's domain. I will definitely take a look at this result though. – Wojowu Oct 16 '16 at 19:55
• duplicate: math.stackexchange.com/questions/10949/… – Franz Lemmermeyer Oct 16 '16 at 20:36
• @FranzLemmermeyer Thanks, I haven't seen that question when searching about the topic. I suppose at this point my question can be closed. – Wojowu Oct 16 '16 at 20:39
• The system does not allow closing a question on one site as a duplicate of a question on a different site. One option, Wojowu, is for you to post an answer here, summarizing what's over there, and linking to it, and then accept your answer. – Gerry Myerson Oct 16 '16 at 22:04
The recent paper by Homlin, Jones, Kurlberg, McLeman, and Petersen (Experimental Math., to appear) is devoted to these questions especially in the context of imaginary quadratic fields. One should expect that every natural number arises as the class number of an imaginary quadratic field. Refining an earlier conjecture of Soundararajan, in this paper a precise asymptotic is formulated for the number of imaginary quadratic fields with a given class number. They also formulate conjectures on what kind of groups can arise as class groups of imaginary quadratic fields -- for example, one expects that for any odd prime $p$, $({\Bbb Z}/p{\Bbb Z})^3$ is not the class group of any imaginary quadratic field. The paper gives much data on such questions together with many related references.
• I think $(\dfrac{\Bbb Z}{2\Bbb Z})^3$ can occur as a class group for $D=-420$, and maybe it is better to restrict $p$ to the odd primes, i.e: $p\neq2$. – Davood KHAJEHPOUR May 8 '18 at 5:51
• Thanks for the reference! I will look into it as soon as I can. – Wojowu May 8 '18 at 7:26
Just for the sake of completeness, here I am posting the answer of Pete Clark which was posted in the math.SE question. I am making this post CW so as not to create the unseemly impression that I am gaining any undeserved reputation from this. Here it is:
Virtually nothing is known about the question of which abelian groups can be the ideal class group of (the full ring of integers of) some number field. So far as I know, it is a plausible conjecture that all finite abelian groups (up to isomorphism, of course) occur in this way. Conjectures and heuristics in this vein have been made, but unfortunately for me I'm not so familiar with them.
The situation for imaginary quadratic fields is different. Here there is an absolute bound on the size of an integer $$k$$ such that the class group of an imaginary quadratic field can be isomorphic to $$(\mathbb{Z}/2\mathbb{Z})^k$$. Conditionally on the Generalized Riemann Hypothesis, the largest such $$k$$ is $$4$$. This has do to with idoneal numbers, of which the following paper provides a very fine survey:
http://www.mast.queensu.ca/~kani/papers/idoneal-f.pdf
Actually the truth is slightly stronger: let $$H_D$$ be the class group of the imaginary quadratic field $$\mathbb{Q}(\sqrt{-D})$$. Then, as $$D$$ tends to negative infinity through squarefree numbers, the size of $$2H_D$$ (the image of multiplication by $$2$$) tends to infinity. See for instance
http://arxiv.org/PS_cache/arxiv/pdf/0811/0811.0358v2.pdf
for some recent explicit bounds on this.
For $S$-class groups: Perret, Marc On the ideal class group problem for global fields. Zbl 0933.11053 J. Number Theory 77, No. 1, 27-35 (1999). https://zbmath.org/?any=&au=perret&ti=On+the+Ideal+Class+Group+Problem+for+Global+Fields&so=&ab=&cc=&ut=&an=&la=&py=&rv=&sw=&dm= | 2021-06-14T08:52:49 | {
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https://math.stackexchange.com/questions/452806/prove-forall-n-2-exists-p-in-bbbp-n-p-n/452816 | Prove $\forall n > 2, \ \exists p\in \Bbb{P} : n < p < n!$
I need to prove that:
$$(1) \ \forall n\in\Bbb{N}_{\ge2}, \ \exists p\in \Bbb{P} : n < p < n!$$
I already know how to prove the $n < p$ part; it directly follows from the proof that there is no largest prime. However, I am stumped on the $p < n!$ part.
One idea I had for showing this is as follows: we know that $(2) \ \forall n \in \Bbb{N}_{\ge2},\ \exists m\in\Bbb{N} : n!=2m$. By testing a few values, I conjectured that $(3) \ \forall n\in\Bbb{N}_{\ge2}, \ \exists p\in \Bbb{P} : n < p < 2n$. If (3) could be shown, it would be simple to prove (1), but there does not seem to be no easy way to prove (3), if it's even correct.
• "that there is always a prime number between $n$ and $2n$" The names Bertrand and Chebyshev spring to mind. – Daniel Fischer Jul 26 '13 at 14:32
• Oh, wow! I didn't know that! I was just trying to find a 'lazy' solution. But it's nice to know. :) – dotslash Jul 26 '13 at 14:33
• There's also an easy direct way. Do you know Euclid's proof that there are infinitely many primes? – Daniel Fischer Jul 26 '13 at 14:35
• Yes, I do. But not sure how to use it here. – dotslash Jul 26 '13 at 14:36
• I was thinking of what Goos answered. That relieves of the burden of proving that the product of primes $< n$ is $> n$ (although that's not very hard). – Daniel Fischer Jul 26 '13 at 14:40
Just consider $n! - 1$. Clearly this is between $n$ and $n!$ as long as there is anything between $n$ and $n!$, and it isn't divisible by any prime $p \le n$. Thus it contains a prime factor bigger than $n$ and smaller than $n!$.
• Correct me if I'm wrong, but $n! -1$ isn't divisible by any prime less than $n$ because all the primes are already included in $n!$ ? – dotslash Jul 26 '13 at 14:53
• @dotslash For any prime $p \le n$, $p$ is contained in the product $n!$, so $n! - 1$ is one less than a multiple of $p$. – 6005 Jul 26 '13 at 14:55
• Precisely my thinking! Thank you for the concise and beautiful answer! – dotslash Jul 26 '13 at 14:59
You can obviously proceed by stating the Bertrand's postulate. However, I think the question is asking you to solve it a bit differently.
Let us solve it using a method similar to that used by Euclid when he tried to prove that there are infinite number of primes. Let $p_{1},p_{2},\cdots ,p_{k}$ be all the primes less than or equal to $n$. Obviously, $k<n$.Let, $P=p_{1}\times p_{2}\times\cdots \times p_{k}+1$.
Notice that $P$ is not divisible by any of the given primes. Hence $P$ is either a prime or is divisible by a prime $> n$.
It can be easily seen that $$P<n!$$ Furthermore, $P$ needs to be greater than $n$, otherwise we would find another prime (other than $p_{1},p_{2},\cdots,p_{k}$) which would lead to a contradiction.
• I think you mean that $p_1, p_2, \cdots , p_k$ are less than or equal to $n$. You also have to be careful when claiming that $P < n!$; this isn't true in the case $n = 3$. – 6005 Jul 26 '13 at 14:42
• You need it to not be divisible by $n$ if $n$ is prime, so you need to include primes less than or equal to $n$ in the product. Thus you get $3*2 + 1 = 7$, not less than $3! = 6$. – 6005 Jul 26 '13 at 14:47
• Sorry!My bad,I guess the case n=3 has to be checked manually.Other than that I think there are no exceptions. – Shaswata Jul 26 '13 at 14:49
What you speculate on is correct, there is always a prime between $n$ and $2n$. This is known as Bertrand's postulate and is really a theorem. But for between $n$ and $n!$ you can modify the Euclid proof for a much easier approach. It is true for $3, 3 \lt 5 \lt 6$. For $n \gt 3,$ take the product of all primes less than $n$ and add $1$. This is less than $n!$ because it has only one factor of $2$, not the three coming from $2$ and $4$. Then it is either prime or composite ...
• Great! But your explanation of "only one factor of $2$" confuses me. Isn't it simpler to say that $n!$ contains all the integers between $2$ and $n$, including those that are prime. Hence, $n! > p$? – dotslash Jul 26 '13 at 14:46
• But we need a prime greater than $n$ and $n!$ doesn't contain that as a factor. I wanted to justify that the product of all the primes less than $n$ is strictly less than $n!$ so that I could add one and still be less. Identifying a factor of $n!$ that is not part of the product of primes was my way of doing that. That is also why I did $3$ separately-it is not true in that case as you can see. If we just take all the primes less than or equal to $3$, the product plus $1$ is $7$ and we have not found a prime between $3$ and $3!$ – Ross Millikan Jul 26 '13 at 15:10
• I follow you now. Thanks for the answer! – dotslash Jul 26 '13 at 15:42 | 2019-05-25T06:58:48 | {
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http://math.stackexchange.com/questions/95799/why-is-gcda-b-gcdb-r-when-a-qb-r | # Why is $\gcd(a,b)=\gcd(b,r)$ when $a = qb + r$?
Given: $a = qb + r$
Then it holds that $\gcd(a,b)=\gcd(b,r)$. That doesn't sound logical to me. Why is this so?
Addendum by LePressentiment on 11/29/2013: (in the interest of http://meta.math.stackexchange.com/a/4110/53259 and averting a duplicate)
What's the intuition behind this result? I only recognise the proof and examples solely due to algebraic properties and formal definitions; I'd like to apprehend the result naturally.
-
$a$, $b$, $r$ are integers? (If yes, tagging the question elementary-number-theory would be suitable.) Or are you looking for some greater generality, e.g. they are elements of some type of commutative ring? If yes, you should say so and you should also mention, what are the assumptions about this ring. – Martin Sleziak Jan 2 '12 at 13:12
If you're asking about integers, then this question is very similar to yours: math.stackexchange.com/questions/59147/… – Martin Sleziak Jan 2 '12 at 13:16
see page 13 here www-groups.dcs.st-and.ac.uk/~martyn/teaching/1003/… – Bhargav Jan 2 '12 at 13:26
@Martin The result does not depend upon the underlying ring (assuming $\rm\:gcd(a,b)\:$ exists). – Bill Dubuque Jan 2 '12 at 18:19
If $d$ is a divisor of $a$ and of $b$, then \begin{align} a & = dn, \\ b & = dm. \end{align} So $$a-b= dn-dm=d(n-m)= (d\cdot\text{something}).$$ So $d$ is a divisor of $a-b$.
Thus: All divisors that $a$ and $b$ have in common are divisors of $a-b$.
If $d$ is a divisor of $a$ and of $a-b$, then \begin{align} a & = dn, \\ a-b & = d\ell. \end{align} So $$b=a-(a-b)=dn-d\ell=(d\cdot\text{something}).$$ So $d$ is a divisor of $b$.
Thus: All divisors that $a$ and $a-b$ have in common are divisors of $b$.
Therefore, the set of all common divisors of $a$ and $b$ is the same as the set of all common divisors of $a$ and $a-b$.
Subtracting one member of a pair from the other never alters the set of all common divisors; therefore it never alters the $\gcd$.
-
Very clear answer! Little remark: $a-b \neq r$, but $a-qb = r$, but all your explanation will work using that :). – Kevin Jan 2 '12 at 21:17
@Kevin Beware that this fails if the ring is not $\rm\:\mathbb Z\:$ since then the remainder $\rm\:r\:$ generally cannot be obtained by repeatedly subtracting $\rm\:b\:$ from $\rm\ q\ b + r\:,\:$ as it can when $\rm\:q\in\mathbb N\:.\:$ In other words, in general Euclidean rings, e.g. the polynomial ring $\rm\:F[x]\:$ over a field $\rm\:F\:,\:$ the Euclidean algorithm generally requires division with remainder, not simply iterated subtraction, in order to effect descent to a "smaller" remainder. So this ad-hoc special-case doesn't generally reveal the essence of the matter. – Bill Dubuque Jan 2 '12 at 22:37
@Kevin : Thanks; I've fixed the typo. – Michael Hardy Jan 3 '12 at 0:52
To express Bill Dubuque's point in other way: Euclid's algorithm works for polynomials, but the argument I give in my answer doesn't work for polynomials unless you do some adaptations. – Michael Hardy Jan 3 '12 at 0:53
HINT $\rm\ \$ If $\rm\ d\ |\ b\$ then $\rm\ d\ |\ q\ b + r\ \iff\ d\ |\ r\:.\$ Therefore $\rm\ \{b\:,\:q\ b+r\}\$ and $\rm\ \{b\:,\: r\}\$ have the same set of common divisors $\rm\:d\:,\:$ hence they have the same greatest common divisor.
Modly: $\$ if $\rm\ b\equiv 0\$ then $\rm\ q\ b+r\equiv 0\: \iff\: r\equiv 0\ \pmod{d}$
NOTE $\$ The result holds true because $\rm\:\mathbb Z\:$ forms a subring of its fraction field $\rm\:\mathbb Q\:.\:$ More generally, given any subring $\rm\:Z\:$ of a field $\rm\:F\:$ we define divisibility relative to $\rm\ Z\$ by $\rm\ x\ |\ y\ \iff\ y/x\in Z\:.\:$ Then the above proof still works, since if $\rm\ q,\ b/d\ \in Z\$ then $\rm\ q\:(b/d) + r/d\in Z\ \iff\ r/d\in Z\:.\:$ In other words, the usual divisibility laws follow from the fact that rings are closed under the operations of subtraction and multiplication; being so closed, $\rm\:Z\:$ serves as a ring of "integers" for divisibility tests.
For example, to focus on the prime $2$ we can ignore all odd primes and define a divisibility relation so that $\rm\ m\ |\ n\$ if the power of $2$ in $\rm\:m\:$ is $\le$ that in $\rm\:n\:$ or, equivalently if $\rm\ n/m\$ has odd denominator in lowest terms. The set of all such fractions forms a ring $\rm\:Z\:$ of $2$-integral fractions. Moreover, this ring enjoys parity, so arguments based upon even/odd arithmetic go through. Similar ideas lead to powerful local-global techniques of reducing divisibility problems from complicated "global" rings to simpler "local" rings, where divisibility is decided by simply comparing powers of a prime.
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You can show that for any integer $d$, we have $d\; |\; a$ and $d\; |\; b$ if and only if $d\; |\; b$ and $d\; |\; r$. In other words, $a$ and $b$ have exactly the same common divisors as $b$ and $r$. Thus $\gcd(a,b)$ is the same as $\gcd(b,r)$.
-
excellent answer, +1 for it. – ncmathsadist Jan 2 '12 at 13:44
Thanks, but why do they have the same common divisors? – Kevin Jan 2 '12 at 21:15
@Kevin: If $d$ divides $a$ and $b$, then $d$ divides $-qb$ and thus divides the sum $a - qb = r$. This shows that any common divisor of $a$ and $b$ is a common divisor of $b$ and $r$. If $d$ divides $b$ and $r$, then $d$ divides $qb$ and thus divides the sum $qb + r$. This shows that any common divisor of $b$ and $r$ is a common divisor of $a$ and $b$. – Mikko Korhonen Jan 2 '12 at 21:52
Since set of common divisors of $a-b$ and $b$ coincides with the set of common divisors of $a$ and $b$ then $\operatorname{gcd}(a,b)=\operatorname{gcd}(a-b,b)$. If $a=qb+r$, where $b>0$ and $0\leq r<b$, you can apply this equality $q$ times and obtain $\operatorname{gcd}(a,b)=\operatorname{gcd}(r,b)$
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But $a-b$ and $b$ don't have the same set of divisors. They have the same set of common divisors with $a$. – Peter Taylor Jan 2 '12 at 14:29
Ok common divisors. I always miss some details. – no identity Jan 2 '12 at 14:40
@Norbert : what you must have meant is that $\{a,b\}$ and $\{a-b,b\}$ have the same set of common divisors. – Michael Hardy Jan 2 '12 at 18:01
Yes, this is what I meant. – no identity Jan 2 '12 at 18:17
Beware that this "repeated subtraction" implementation of division with remainder does not generally yield the Euclidean algorithm in other domains, e.g. for polynomials. See my comment to Hardy's answer. – Bill Dubuque Jan 2 '12 at 22:58
Let $A$ be a commutative ring. For any $a_1,\dots,a_n$ in $A$ let $(a_1,\dots,a_n)$ the ideal generated by the $a_i$.
Then, for any $q,b,r$ in $A$, we have $$(qb+r,b)=(b,r).$$ Indeed, $qb+r$ is in $(b,r)$, and $r$ is in $(qb+r,b)$.
EDIT. Dear Kevin: Your question, I think, would be better understood if put in a wider context, involving rings and ideals. The most basic fact behind the question is, I believe, the fact that, in any commutative ring, the elements $qb+r$ and $b$ generate the same ideal as the elements $b$ and $r$. If you make additional hypothesis, this fact can be interpreted in terms of divisibility. (See Bill's comment.) The simplest is to assume that your ring is a principal ideal domain.
I could try to explain this in greater details, but many mathematicians much better than I have already done that. So, my advice would be to take a look at at least one of the many Algebra textbooks written by great mathematicians. Here are some of these books:
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This seems to implicitly assume some relation between gcds and ideals, e.g. for Bezout domains $\rm\ (a,b) = (\gcd(a,b))\:.$ – Bill Dubuque Jan 2 '12 at 22:28
Dear @Bill: Thanks for your comment. I edited the answer. (Of course, I agree with you.) – Pierre-Yves Gaillard Jan 3 '12 at 5:50
Divisor theory gives one nice way to better understand the relations between ideals and gcds. For a brief overview see Friedemann Lucius, Rings with a theory of greatest common divisors and for a longer exposition see Olaf Neumann, Was sollen und was sind Divisoren? (What are divisors and what are they good for?), Math. Semesterber, 48, 2, 139-192 (2001). – Bill Dubuque Jan 3 '12 at 6:38
I'm going to use the notation $(a,b)$ for the GCD of $a$ and $b$.
If $d|a$ and $d | b$ then $d|(a,b)$, by the definition of GCD. (Well, by one common definition...if that's not the definition you learned, then you probably learned it as a theorem).
Since $(a,b)|a$ and $(a,b)|b$, by the definition of $(a,b)$, it divides $a-qb$, so we have $(a,b)|r$. This gives us $(a,b)|b$ and $(a,b)|r$, hence $(a,b)|(b,r)$.
Now let's go the other way. $(b,r)|b$ and $(b,r)|b$, both by definition, so it also divides $r+qb$, giving us $(b,r)|a$. That gives is $(b,r)|(a,b)$.
From $(a,b)|(b,r)$ and $(b,r)|(a,b)$, we get $(a,b)=(b,r)$ or $(a,b)=-(b,r)$. The latter can be eliminated because GCD is by definition greater than 0.
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a,b q ,r are integers how can we say that HCF(a,b) = HCF (b,r) we can say common divisor of a and b = common divisor of b and r. for example hcf of 4 and 2 = 2 4= 2x2 + o but here HCF of 4 and2 =2 but HCF of 2 and 0 =1 so how can we say that HCF(a,b) = HCF (b,r)
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Theorem: Let $a > b > 0$ with $a = bq + r$, $0\leq{}r<b$ then $\gcd(a;b) = \gcd(b;r)$.
Proof: Need to show that $C(a;b) = C(b;r)$ for then the result will hold. To show that the two sets are equal requires showing that $C(a;b) \subseteq C(b;r)$ and that $C(b;r) \subseteq C(a;b)$:
Let $y \in C(a;b)$ thus $y|a$ and $y|b$,
then $y|[a+(-q)b]$,
and so $y|r$, since $r=a-bq$,
but since $y|b$ is also true, we now have that $y \in C(b;r)$, finally this means that $C(a;b) \subseteq C(b;r)$.
Now let $y \in C(b;r)$ thus $y|b$ and $y|r$,
then $y|[(q)b+r]$,
and so $y|a$, since $a=bq+r$,
but since $y|b$ is also true, we now have that $y \in C(a;b)$, finally this means that $C(b;r) \subseteq C(a;b)$.
Therefore the required results have been proven.
Note: Where $C(a;b)$ denotes the set of common divisors/factors of $a$ and $b$, that is: $C(a;b)=\{y: y|a \land y|b\}$
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https://specialsports.info/benzamide-molecular-fbulsmr/viewtopic.php?af9d97=table-math-example | Here is a very simple table showing data lined up in columns. Example. We will look at how to do this using an example and then also look at how to answer questions about a data set using a two way table. 3 Some Table Examples Example-1: A table with combined columns is given below. Practice your multiplication tables. Syntax of this Tableau ABS Function is: ABS(number) To demonstrate these Tableau math functions, we use Calculated Fields. Add rows, call Compute and Merge, and set PrimaryKey. Tree Diagrams in Math: Definition & Examples 4:43 Truth Table: Definition, Rules & Examples 6:08 6:52 Notice that I include the table in a center'' environment to display it properly. The table results can usually be used to plot results on a graph. Illustrated definition of Table: Information (such as numbers and descriptions) arranged in rows and columns. 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The following table summarizes them: As math requires special environments, there are naturally the appropriate environment names you can use in the standard way. Example example the table shows the math achievement. Times Tables. You read it right; basic mathematical concepts are followed all the time. The title is created simply as another paragraph in the center environment, rather than as part of the table itself. Hope this helps! So now how many people can we fit? Logic. The following HTML example contains the HTML, CSS, and JavaScript necessary to build a dynamic HTML table. Learn the multiplication tables in an interactive way with the free math multiplication learning games for 2rd, 3th, 4th and 5th grade. \begin{center} Numbers of Computers on Earth Sciences Network, By Type. Yes! Some of the times tables are illustrated below: Times Table … Consumer Math. For example, in the table shown below, Jacket is listed twice in column A. 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Because complex Boolean statements can get tricky to think about, we can create a truth table to keep track of what truth values for the simple statements make the complex statement true and false. Example. Tally Table - Definition with Examples. There's now 3 parts to the tutorial with two extra example videos at the end. Trusted by teachers across schools. Possible values will be from 1-10, instead of 0-9. Uploaded By BrigadierIronJay8680. Example. Truth Table. Determining the post-money cap table for an equity round with an option pool refresh and one or more convertible notes converting to equity can be overwhelming. Here you can find additional information about practicing multiplication tables at primary school. Comprehensive Curriculum. According to some people, maths is just the use of complicated formulas and calculations which won’t be ever applied in real life. There are two disadvantages of writing tables by hand as described in this tutorial. Most of the time the data will be collected in form of a spreadsheet and we don't want to enter the data twice. Tables. Example: table([1:3]',{'one';'two';'three'},categorical({'A';'B';'C'})) creates a table from variables with three rows, but different data types. Mathematics for the Liberal Arts. Example. A math function table is a table used to plot possible outcomes of a function, which is a kind of rule. Reading over the table; Exercising using the Math Trainer; But here are some "tips" to help you even more: Tip 1: Order Does Not Matter. Go to next Question. Back to overview Forward to Shortdocumentation 1 Examples of Latex Here an example of a very small Latex document \documentclass{article} \begin{document} example for a very \tiny{tiny} \normalsize \LaTeX \ document \end{document} Truth Tables . For example, in the example given below, 7 … A tautology in math (and logic) is a compound statement (premise and conclusion) that always produces truth. Roman numerals.. my mind ~ my math .. Tables/Graphs/Charts BEST CHART EVER !!! Unlike most other environments, however, there are some handy shorthands for declaring your formulas. Two way tables, also known as contingency tables, show frequencies (counts) as they relate to two variables. Some math sections allow the use of a calculator, others do not. By Grades. Mathematical tables are lists of numbers showing the results of a calculation with varying arguments.Tables of trigonometric functions were used in ancient Greece and India for applications to astronomy and celestial navigation.They continued to be widely used until electronic calculators became cheap and plentiful, in order to simplify and drastically speed up computation. Construct a truth table for the formula . The Tableau ABS function is used to return the absolute positive value. This tutorial provides you with commonly used SQL math functions that allow you to perform business and engineering calculations. 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To fully understand function tables and their purpose, you need to understand functions, and how they relate to variables. However, there is only one record for each jacket and size combination -- Jacket Medium in row 4 and Jacket Large in row 5. Extras. No matter what the individual parts are, the result is a true statement; a tautology is always true. C# DataTable Examples Store data in memory with a DataTable. SAT Suite of Assessments Sample Questions. We can fit one, two, three, four, five. (See Commutative Property.) Tables, graphs, and charts are an easy way to clearly show your data. A truth table is a mathematical table used in logic—specifically in connection with Boolean algebra, boolean functions, and propositional calculus—which sets out the functional values of logical expressions on each of their functional arguments, that is, for each combination of values taken by their logical variables. Let's see. sz — Size of preallocated table two-element numeric vector. A times table is a list of multiples of a number. Be sure to consider how to best show your results with appropriate graph forms. School St. John's University; Course Title MTH 1250C; Type. First, I list all the alternatives for P and Q. Search for: Truth Tables and Analyzing Arguments: Examples. Tableau ABS Function. And then on this table, which is identical, you could fit six, seven, eight, nine. Z Score Table Sample Problems. When we multiply two numbers, it does not matter which is first or second, the answer is always the same. Parents, Sign Up for Free Teachers, Sign Up for Free. Logic Symbols in Math; Truth Table; Tautology Math Examples; Tautology Definition. Pages 47. Example: 3×5=15, and 5×3=15. Math.floor(Math.random() * (max - min + 1)) + min // We can introduce a min variable as well so we can have a range. While it works for small tables similar to the one in our example, it can take a long time to enter a large amount of data by hand. This preview shows page 11 - 23 out of 47 pages. So here we have one table, and it's going to touch ends with this table right over here. And because it touches ends right over here-- we're making it one big continuous table-- you can't fit someone here anymore.
## table math example
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https://math.stackexchange.com/questions/3247145/how-can-one-show-that-sum-n-0-infty-fracnn-e | # How can one show that $\sum_{n=0}^\infty\frac{n}{n!}=e$? [duplicate]
How can one show that $$\sum_{n=0}^\infty\frac{n}{n!}=e$$?
I understand that $$\sum_{n=0}^\infty\frac{x^n}{n!}=e^x$$ and that letting $$x=1$$would give $$\sum_{n=0}^\infty\frac{1}{n!}=e$$
But why does the sum $$\sum_{n=0}^\infty\frac{n}{n!}$$ give an answer of $$e$$ also?
• In the picture on the bottom sum the numerator should be n, apologies. – user677704 Jun 1 '19 at 2:51
• Are you familiar with derivatives ? – DanielV Jun 1 '19 at 2:53
• Your question is rather unclear. Do you want to prove: $\sum _{n=0}^{\infty }\:\frac{x}{n!}=e?$ – NoChance Jun 1 '19 at 3:12
• ‘No chance’ yes that was my question more or less, I just didn’t understand why different sums all add to e – user677704 Jun 1 '19 at 3:13
• It can be proven that $\sum _{n=0}^{\infty }\:\frac{x^n}{n!}=e^x$. However, in your expression, you don't raise x to the power n. – NoChance Jun 1 '19 at 3:15
Note that $$n/n!=1/(n-1)!$$, which is not meaningfully distinct in the context of the infinite sum from the terms $$1/n!$$.
In fact $$\sum_{n\geq 0} \frac{n}{n!}\stackrel{(1)}{=}\sum_{n\geq 1} \frac{n}{n!}= \sum_{n\geq 1} \frac{1}{(n-1)!}\stackrel{(2)}{=}\sum_{n\geq 0} \frac{1}{n!}=e^1=e$$ (1) comes from the fact that the first term is zero, (2) is a shift of indices by one.
You can repeat this shifting procedure ad infinitum and obtain, for each fixed $$k$$ $$\sum_{n\geq 0} \frac{n(n-1)\ldots (n-k)}{n!}=e$$ because your sum 'starts' at index $$k+1$$.
• This is what I meant in my answer. Nice presentation. +1 – The Count Jun 1 '19 at 3:14
• So I, in fact, unfolded your idea ? (+1) – Duchamp Gérard H. E. Jun 1 '19 at 3:15
• Teamwork! I like it. – The Count Jun 1 '19 at 3:16
$$\begin{array} {rcl} % \displaystyle e^x & = & \displaystyle \sum_{n = 0}^{\infty} \frac{x^n}{n!} \\ % \displaystyle\frac{d}{dx}e^x & = & \displaystyle \frac{d}{dx}\sum_{n = 0}^{\infty} \frac{x^n}{n!} \\ % e^x & = & \displaystyle \sum_{n = 0}^{\infty} \frac{nx^{n-1}}{n!} \\ % e^1 & = & \displaystyle \sum_{n = 0}^{\infty} \frac{n1^{n-1}}{n!} \\ \end{array}$$
• Thanks for your response it’s really clear and perfectly understood, I appreciate it. So would I be right in assuming you can repeat this process ad infinitum and acquire many different expressions which all sum to e? – user677704 Jun 1 '19 at 3:15
• Yes, in this case the multipliers will be $$n,n(n-1),n(n-1)(n-2),\ldots$$ and so on ... – Duchamp Gérard H. E. Jun 1 '19 at 3:18
• That’s incredibly interesting, aside from showing this with basic differentiation would you know where to find a geometric or more intuitive proof. The logic is perfectly reasonable but I prefer to see why. – user677704 Jun 1 '19 at 3:25
• @User3457884334 I added repetition of this process ad infinitum in my answer. – Duchamp Gérard H. E. Jun 1 '19 at 3:26
• I understand that but wondered if there was another proof to show it geometrically. – user677704 Jun 1 '19 at 3:28
Suppose $$e^x=a_0+a_1x+a_2x^2+a_3x^3+...$$ try to find the constants by evaluating $$e^0$$ and then differentiating to eliminate the current $$0$$th degree term.
Here are the first three terms : $$e^0=a_0+a_1*0+...=1\Leftrightarrow a_0=1$$ $$\frac{d}{dx}(e^x)=e^x=\frac{d}{dx}(a_0+a_1x+a_2x^2+...)=a_1+2a_2x+3a_3x^2+...$$ $$e^0=a_1+2a_2*0+3a_3*0^2+...=1\Leftrightarrow a_1=1$$ $$\frac{d}{dx}(e^x)=e^x=\frac{d}{dx}(a_1+2a_2x+3a_3x^2+...)=2a_2+(3*2)a_3x+...$$ $$e^0=2a_2+(3*2)a_3*0\,+...=1\Leftrightarrow a_1=\frac{1}{2}$$ You'll find that $$e^x=\sum_{n=0}^\infty \frac{x^n}{n!}$$ switch $$1$$ for $$x$$ and you get your result (refer to The Count's answer).
Differentiate the series $$e^x=\sum_{n=0}^{\infty}\dfrac {x^n}{n!}$$ term by term. Get $$\sum_{n=0}^{\infty}\dfrac {nx^{n-1}}{n!}$$. Then set $$x=1$$. Note that $$(e^x)'=e^x$$ (for$$\sum_{n=0}^{\infty}\dfrac {nx^{n-1}}{n!}=\sum_{n=0}^{\infty}\dfrac {x^n}{n!}$$) . | 2021-03-05T14:19:43 | {
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https://math.stackexchange.com/questions/2359335/find-the-number-of-distributions-of-seven-distinct-balls-into-three-distinct-box | # Find the number of distributions of seven distinct balls into three distinct boxes if at least two balls must go into each box.
My Answer: I chose to place three balls into one box and two balls in the other two boxes. This can be done three times: $$\binom73\binom42\binom22=\binom{7}{3,2,2}\\ \binom72\binom53\binom22=\binom{7}{2,3,2} \\ \binom72\binom52\binom33=\binom{7}{2,2,3}$$ I'm not sure if this is correct, but we have not covered Stirling numbers yet, so I cannot use that as my explanation. Please help and thank you!
An alternative method, noting that 2/2/3 is the only distribution of seven balls to three boxes with at least two balls in each box:
• 3 choices for the box with three balls
• $\binom73=35$ ways to put three balls into that box
• $\binom42=6$ ways to put two balls into one of the remaining boxes; the third box is then forced to contain the last two balls
This yields $3\cdot35\cdot6=630$ possibilities, and your approach yields the same result but through a longer process.
• Of course the solution appear to be the same once one recognises that the three numbers in the OP's solution must be the same by symmetry. – Carsten S Jul 15 '17 at 11:38
Your answer is correct. Here is a variation based upon exponential generating functions.
• A selection of at least two distinct balls can be encoded as \begin{align*} \frac{x^2}{2!}+\frac{x^3}{3!}+\cdots=e^x-1-x \end{align*}
• Since we have three boxes we consider \begin{align*} (e^x-1-x)^3\tag{1} \end{align*}
We denote with $[x^n]$ the coefficient of $x^n$ in a series.
Since we are looking for the number of placing $7$ distinct balls we consider the coefficient of $x^7$ in (1) and obtain with some help of Wolfram Alpha
\begin{align*} 7^3=\color{blue}{630} \end{align*}
• Let's say we have a red box, a blue box, and a green box (and the observer is not colorblind). There are $\binom{7}{3}$ ways of placing three balls in the red box, $\binom{4}{2}$ ways of placing two of the remaining four balls in the blue box, and $\binom{2}{2}$ ways of placing the remaining two balls in the green box. Hence, there are $$\binom{7}{3}\binom{4}{2}\binom{2}{2} = \binom{7}{3, 2, 2}$$ ways to distribute the balls so that three are placed in the red box, two are placed in the blue box, and two are placed in the blue box. By symmetry, the answer is $$3\binom{7}{3, 2, 2}$$ – N. F. Taussig Jul 19 '17 at 12:31
• @N.F.Taussig: Thanks for your hint! – Markus Scheuer Jul 19 '17 at 12:38
• @N.F.Taussig: I was misleading by myself. Original answer restored. – Markus Scheuer Jul 19 '17 at 16:58
• I have written a solution based on your idea of first placing the seven distinguishable ways in three indistinguishable boxes, then arranging the boxes. – N. F. Taussig Jul 19 '17 at 19:16
This solution is a modification of an attempted solution that Markus Scheuer deleted.
First, we count the number of ways seven distinguishable balls can be placed in three indistinguishable boxes if at least two balls are placed in each box. Then we will arrange the boxes.
The only permissible way to distribute the balls is to place three balls in one bag and two each in the other bags.
Method 1: There are $\binom{7}{3}$ ways to select three balls to be placed in one of the boxes. Line up the remaining balls in a row. Take the ball at the left end of the row and place it in an empty box. There are three ways of selecting one of the other three balls to be placed in the same box as that ball. The remaining two balls must be placed in the remaining empty box. Hence, there are $$3\binom{7}{3}$$ ways to place seven distinguishable balls in three indistinguishable boxes.
Method 2: There are $\binom{7}{3}$ ways to place three of the balls in one box. That leaves $\binom{4}{2}$ ways to place two of the remaining four balls in another box. The remaining two balls are place in the third box. That suggests the answer is $$\binom{7}{3}\binom{4}{2}$$ However, since the boxes are indistinguishable, we cannot distinguish between the two boxes that received two balls. Thus, we must multiply the above result by $1/2$. Hence, there are $$\frac{1}{2}\binom{7}{3}\binom{4}{2}$$ ways of distributing seven distinguishable balls to three indistinguishable boxes.
To make the boxes distinguishable, we paint one box blue, one box green, and one box red. There are $3!$ ways to do this. Hence, there are $$3! \cdot 3\binom{7}{3} = 630$$ ways to place seven distinguishable balls in three distinguishable boxes if at least two balls must be placed in each box.
• @NFTaussig: Clear and extensive elaboration. Very nice! (+1) – Markus Scheuer Jul 19 '17 at 19:38 | 2019-08-25T01:54:27 | {
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https://math.stackexchange.com/questions/2657571/probability-of-picking-matching-socks-after-partitioning-the-drawer | # Probability of picking matching socks, after partitioning the drawer.
Apologies if this is a duplicate. I searched and didn't find anything quite like it.
Suppose I have a drawer with an equal number of N black socks and N white socks. They're all mixed up. So, my chances of picking a matching pair in the first two selections is (N-1)/(2N-1), right? Well, what if, before I pick the first sock, I randomly (so I don't know the colors of the socks I'm moving) partition the drawer so that there are N socks on each side, and I draw one sock from each side. Do the chances of drawing a matching pair change?
On the one hand, we can see that selection from one side doesn't change the composition of socks on the other side of the partition. However, whichever color I choose from the "first" side, it's likely that there are more of that color on that side. On other words, if I draw a black sock from one side, it's more likely that that side had N-1 blacks and 1 white than it is that that side had 1 black and N-1 whites.
My suspicion is that I need to do some kind of hypothesis testing, where I consider the chances of every possible partitioning, but that's way above my skill level.
• Shouldn't the probability in the simple case be $\frac{N-1}{2N-1}$, because there are no longer $2N$ socks in the drawer? Feb 19 '18 at 20:17
• with replacement? Feb 19 '18 at 20:18
• @GTonyJacobs Yep $P=\frac {N-1}{2N-1}$ Feb 19 '18 at 20:27
• You certainly don't need hypothesis testing. We're not making inferences from a sample about an otherwise unknowable population. This is just a probability question with all of the parameters known. Kind of a tricky one, maybe, but there might be a good symmetry argument for why there's no difference. Feb 19 '18 at 20:37
• @GTonyJacobs Correct. I'll see if I can edit the post. Feb 20 '18 at 21:34
It doesn't matter if the partitioning happens before or after the first draw. Suppose it happens after. Suppose also that the first sock drawn was black.
Now, we partition off $N$ from the remaining $2N-1$, to obtain our pool for the second draw. On average, the composition of this pool is $\frac{N-1}{2N-1}$ black and $\frac{N}{2N-1}$ white. Drawing from it, we have a matching pair if we draw from the portion of it that is black, i.e., $\frac{N-1}{2N-1}$.
The trick to simplifying the work is to work with expected values for the number of socks of each color in the second part of the partition, not actual values.
No, the chances don't change. If you worked out all the conditional probabilities based on how the socks split and summed appropriately you'd get the same answer: $(N-1)/(2N -1)$. The partition gives you no new information.
Here's the brute force calculation for $N=2$. There are two possible partitions, WW|BB and WB|WB. The first of these occurs with probability $1/3$. the second with probability $2/3$. In the first case you fail for sure. In the second you succeed half the time. On average you succeed with probability $$0 \times \frac{1}{3} + \frac{1}{2} \times \frac{2}{3} = \frac{1}{3} = \frac{2-1}{4-1}.$$
(There's probably a really elegant argument from symmetry that I don't see.)
Of course it shouldn't change!
Let's see if the math works out for a simple example, say $N=3$
First, the 'normal' calculation says that the chances of getting a match $M$ is:
$$P(M)= \frac{N-1}{2N-1}=\frac{2}{5}$$
Now let's see what happens when you randomly partition them.
Let's say you first pick from the left side. After randomly splitting the socks into two groups, there can be $0$, $1$, $2$, or $3$ socks there, with respective probabilities of:
$$P(0)= \frac{{3 \choose 0}\cdot{3 \choose 3}}{6 \choose 3} = \frac{1}{20}$$
$$P(1)= \frac{{3 \choose 1}\cdot{3 \choose 2}}{6 \choose 3} = \frac{9}{20}$$
$$P(2)= \frac{{3 \choose 2}\cdot{3 \choose 1}}{6 \choose 3} = \frac{9}{20}$$
$$P(3)= \frac{{3 \choose 3}\cdot{3 \choose 0}}{6 \choose 3} = \frac{1}{20}$$
Now, getting a match $M$ in the first and last situation is impossible, so $$P(M|0)=P(M|3)=0$$
When there is $1$ white sock on the left, the chances of getting a match are the chance of getting that white sock times the chances of getting one of the two socks on the right side, plus the chances of getting one of the two black ones on the left and the one black one on the right. And, with $2$ white socks on the left it's all symmetrical. So:
$$P(M|1)=P(M|2)=\frac{1}{3}\cdot \frac{2}{3} + \frac{2}{3} \cdot \frac{1}{3} = \frac{4}{9}$$
In sum:
$$P(M)=P(0)\cdot P(M|0) + P(1)\cdot P(M|1) +P(2)\cdot P(M|2) +P(3)\cdot P(M|3) =$$
$$\frac{1}{20} \cdot 0 + \frac{9}{20} \cdot \frac{4}{9} + \frac{9}{20} \cdot \frac{4}{9} + \frac{1}{20} \cdot 0 = \frac{8}{20} = \frac{2}{5}$$
OK, so yes, same chance!
Now, I'm sure you can generalize this for any $N$ ... thus evaluating the series:
$$P(M)=\sum_{i=0}^N \frac{{N \choose i} \cdot {N \choose {N-i}}}{2N \choose N} \cdot 2 \cdot \frac{i}{N} \cdot \frac{N-i}{N}$$
[... Insert some math that's over my head ....]
... and you'll see that this ends up being $$\frac{N-1}{2N-1}$$ | 2021-10-26T13:14:38 | {
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https://math.stackexchange.com/questions/900315/proving-that-a-number-is-non-negative | # Proving that a number is non-negative?
The numbers $a$,$b$ and $c$ are real. Prove that at least one of the three numbers $$(a+b+c)^2 -9bc \hspace{1cm} (a+b+c)^2 -9ca \hspace{1cm} (a+b+c)^2-9ab$$ is non-negative.
Any hints would be appreciated too.
• Have you considered adding them up and seeing if they are a square ? – Belgi Aug 16 '14 at 17:42
Hint:
If all three numbers are negative, then:
$$ab > \left(\frac{a+b+c}{3}\right)^2 \hspace{1cm} ac > \left(\frac{a+b+c}{3}\right)^2 \hspace{1cm} bc > \left(\frac{a+b+c}{3}\right)^2 \hspace{1cm}$$
Therefore, if we multiply the three inequalities:
$$a^2b^2c^2 > \left(\frac{a+b+c}{3}\right)^6$$
Or equivalently:
$$\left(\sqrt[3]{abc}\right)^6 > \left(\frac{a+b+c}{3}\right)^6$$
Do you know any inequality you can use here do disprove this?
• @Sudhanshu Look for the Arithmetic-Geometric Median Inequality en.wikipedia.org/wiki/… It says that: $$\frac{a+b+c}{3} \geq \sqrt[3]{abc}$$ – Darth Geek Aug 16 '14 at 17:50
• Oh, how can I forget that...Thanks btw. – Sudhanshu Aug 16 '14 at 17:54
• But since we assume these numbers to be negative how can we apply A.M > G.M ? – Sudhanshu Aug 16 '14 at 18:04
• If we assume those three numbers are negative, then they contradict the "AM>GM" inequality wich we know to be true. Therefore our assumption was incorrect, i.e. at least one of those numbers was not negative. This sort of proof is called "proof by contradiction" or "reductio ad absurdum". You start with a hypothesis and you end up in a contradiction, wich means that the hypothesis was false. – Darth Geek Aug 16 '14 at 18:14
• But I think the A.M > G.M rule is applicable only for positive numbers – Sudhanshu Aug 16 '14 at 18:15
$$\sum [(a+b+c)^2-9bc]=3\sum[a^2+b^2+c^2-ab-bc-ca]=\frac32\sum (a-b)^2\ge0$$
If each $(a+b+c)^2-3bc<0,$ $$\sum [(a+b+c)^2-3bc]<0$$
• You can also observe that $\sum[a^2+b^2+c^2-ab-bc-ca]\ge0$ is just Cauchy -Schwarz. – N. S. Aug 16 '14 at 18:38 | 2019-10-16T04:27:16 | {
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https://narrativmedicin.se/gzh7rm/fa9583-interior-point-in-metric-space | # interior point in metric space
FACTS A point is interior if and only if it has an open ball that is a subset of the set x 2intA , 9">0;B "(x) ˆA A point is in the closure if and only if any open ball around it intersects the set x 2A , 8">0;B "(x) \A 6= ? Metric Spaces A metric space is a set X that has a notion of the distance d(x,y) between every pair of points x,y ∈ X. Definition: We say that x is an interior point of A iff there is an such that: . Browse other questions tagged metric-spaces or ask your own question. Defn Suppose (X,d) is a metric space and A is a subset of X. True. A point x ∈ E is said to be an interior point of E if E contains an open ball centered at x. Since you can construct a ball around 3, where all the points in the ball is in the metric space. (d) Describe the possible forms that an open ball can take in X = (Q ∩ [0; 3]; dE). Proposition A set O in a metric space is open if and only if each of its points are interior points. When we encounter topological spaces, we will generalize this definition of open. $\begingroup$ As an addendum, singletons are open if and only if the metric space is induced by a discrete metric, so there's only one case in which you have a nonempty interior of singleton sets. - the exterior of . We do not develop their theory in detail, and we leave the verifications and proofs as an exercise. In most cases, the proofs Let E be a subset of a metric space X. However, this definition of open in metric spaces is the same as that as if we regard our metric space as a topological space. - the boundary of Examples. $\endgroup$ – Alan Apr 18 '15 at 8:32 METRIC SPACES 77 where 1˜2 denotes the positive square root and equality holds if and only if there is a real number r, with 0 n r n 1, such that yj rxj 1 r zj for each j, 1 n j n N. Remark 3.1.9 Again, it is useful to view the triangular inequalities on “familiar \begin{align} \quad \mathrm{int} \left ( \bigcup_{S \in \mathcal F} S\right ) \supseteq \bigcup_{S \in \mathcal F} \mathrm{int} (S) \quad \blacksquare \end{align} If is the real line with usual metric, , then A point x is called an isolated point of A if x belongs to A but is not a limit point of A. Interior Point Not Interior Points Definition: ... A set is said to be open in a metric space if it equals its interior (= ()). This intuitively means, that x is really 'inside' A - because it is contained in a ball inside A - it is not near the boundary of A. Let (X;d) be a metric space and A ˆX. Let be a metric space, Define: - the interior of . Let be a metric space. (c) The point 3 is an interior point of the subset C of X where C = {x ∈ Q | 2 < x ≤ 3}? The interior of the set E is the set of all its interior points. A point is exterior … An open ball of radius centered at is defined as Definition. Metric space: Interior Point METRIC SPACE: Interior Point: Definitions. This set is denoted by intE. 1. In these examples, all sets under consideration are subsets of the metric space R. Example 2.7. Appendix A. If has discrete metric, 2. Metric Spaces, Topological Spaces, and Compactness 253 Given Sˆ X;p2 X, we say pis an accumulation point of Sif and only if, for each ">0, there exists q2 S\ B"(p); q6= p.It follows that pis an The purpose of this chapter is to introduce metric spaces and give some definitions and examples. Proposition A set C in a metric space is closed if and only if it contains all its limit points. 3.2. Featured on Meta “Question closed” notifications experiment results and graduation | 2021-05-09T22:06:37 | {
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https://mathematica.stackexchange.com/questions/249789/are-there-any-alternatives-for-nintegrate-to-calculate-the-area-for-which-fx-y | # Are there any alternatives for NIntegrate to calculate the area for which $f(x,y)<0$?
I have this two-variable function $$f(x,y)= (8 \cos (x+y)+7)\cos \left(\frac{x}{2}\right)+\cos \frac{x-2 y}{2}+2 \cos \left(\frac{3 x}{2}\right)$$ where $$0. I want to calculate numerically the area for which the function is negative $$f(x,y)<0$$. I use this code
NIntegrate[
Boole[2 Cos[(3 x)/2] + Cos[1/2 (x - 2 y)] +
Cos[x/2] (7 + 8 Cos[x + y]) < 0], {x, 0, Pi}, {y, 0, Pi}]
and it gives the answer $$3.49458$$, but Mathematica gives the following warnings. Are there any other ways to calculate this value that is more reliable and more accurate than this method?
NIntegrate::slwcon: Numerical integration converging too slowly; suspect one of the following: singularity, value of the integration is 0, highly oscillatory integrand, or WorkingPrecision too small.
NIntegrate::eincr: The global error of the strategy GlobalAdaptive has increased more than 2000 times. The global error is expected to decrease monotonically after a number of integrand evaluations. Suspect one of the following: the working precision is insufficient for the specified precision goal; the integrand is highly oscillatory or it is not a (piecewise) smooth function; or the true value of the integral is 0. Increasing the value of the GlobalAdaptive option MaxErrorIncreases might lead to a convergent numerical integration. NIntegrate obtained 3.494581434480605 and 0.002397336775896384 for the integral and error estimates.
• Your f[x, y] is a constant rather than a function of x and y. Jun 17, 2021 at 19:03
• @BobHanlon Thanks, it was a typo mistake.
– user80186
Jun 17, 2021 at 23:21
• There is an exact solution to your integral $-2 \text{Li}_2\left(-\frac{1}{2}\right)+\text{Li}_2\left(\frac{1}{4} \left(i \sqrt{3}+1\right)\right)+\text{Li}_2\left(\frac{1}{4} \left(1-i \sqrt{3}\right)\right)+\frac{2 \pi ^2}{9}$ Jun 18, 2021 at 13:12
• @yarchik I think you should add that and its derivation as an answer.
– Kiro
Jun 18, 2021 at 14:03
• @SaraChem @Kiro I was simply a bit more patient than @eyorble. I used u = Integrate[upperCurve, {x, 0, Pi}] and v = Integrate[lowerCurve, {x, 0, 2 Pi/3}], followed by u-v. The lowerCurve and the upperCurve are defined in eyorble's post. Jun 18, 2021 at 19:46
Let us name the function of interest f[x,y]:
f[x_, y_] := (8 Cos[x + y] + 7) Cos[x/2] + Cos[(x - 2 y)/2] + 2 Cos[3 x/2]
Attempt to ContourPlot to find the zero lines:
ContourPlot[f[x, y] == 0, {x, 0, Pi}, {y, 0, Pi}]
Simple numerical evaluation determines that the negative region is the inside of these two curves. Notice that the ContourPlot is quite rapid and has very clean lines. Interesting coincidence. Perhaps there exists an analytical solution to these curve lines?
sol = Solve[{f[x, y] == 0}]
This returns a list of 4 possible curves, while also stating that some solutions may be missing. By manual inspection (such as by using Plot), we can find that the 2nd and 4th solutions are of interest to us, so we shall label them:
upperCurve = y /. sol[[2]];
lowerCurve = y /. sol[[4]];
Plot[{upperCurve, lowerCurve}, {x, 0, Pi}, PlotRange -> {0, Pi}]
Checking the curves manually by plotting them against the original ContourPlot, we see that upperCurve matches the upper line for the whole domain, and that lowerCurve matches the lower line up until it reaches its minimum.
Find the minimum of the lowerCurve:
FindMinimum[{lowerCurve, 0 < x < Pi}, x, WorkingPrecision -> 25]
{1.872299341324760554288429*10^-8, {x -> 2.094395111754692173633430}}
The warning about a small imaginary part is of little concern here, but you can increase the WorkingPrecision and PrecisionGoal if you would like more digits.
Michael Seifert also pointed out that an exact form can be found for this solution by applying TrigFactor to f[x,y]:
TrigFactor[f[x,y]]
2 (Cos[x/2 - y/2] + 2 Cos[x/2 + y/2]) (2 Cos[x + y/2] + Cos[y/2]) == 0
The lower line happens to correspond to the second variable factor in this expression, and its minimum is found when y is set to 0 and solved.
Solve[{(2 Cos[x + y/2] + Cos[y/2]) == 0 /. y -> 0, 0 < x < Pi}, x]
{{ x -> 2 Pi/ 3 }}
Integrate the area below the 2nd curve over the whole domain minus the area under the 4th curve for 0 through 2.094...
NIntegrate[upperCurve, {x, 0, Pi}, WorkingPrecision -> 25] -
NIntegrate[lowerCurve, {x, 0, 2.0943951117546921736334297747816890478125.},
WorkingPrecision -> 25]
3.49805583366099845069196
Or with the exact form, we can see a slightly different answer:
NIntegrate[upperCurve, {x, 0, Pi}, WorkingPrecision -> 25] -
NIntegrate[lowerCurve, {x, 0, 2 Pi/3}, WorkingPrecision -> 25]
3.49805583366099836305434
While this method is not universally applicable, it does work for this function and is much faster than Area or direct application of NIntegrate and Boole for high precisions. As yarchik notes, you can swap NIntegrate for Integrate here to acquire an exact solution, though it takes a bit longer to evaluate.
• Using TrigFactor on f[x,y] shows that the two curves are $\cos((x-y)/2) + 2 \cos((x+y)/2) = 0$ (upper) and $2 \cos (x+y/2)+ \cos(y/2) = 0$ (lower). I don't know if that really helps, though. Jun 17, 2021 at 20:10
• If nothing else, it shows that the intersection of the lower curve with the $x$-axis is when $2 \cos x + 1 = 0$, or $x = 2 \pi/3$. Jun 17, 2021 at 20:12
• There are a lot of inconsistencies in your answer. What is y that you are integrating? Jun 18, 2021 at 11:51
• @yarchik The y in the integration should be read as y /. sol[[2]] (which is the upper curve from the contour plot) or y /. sol[[4]] (which is the lower curve from the contour plot). These are just explicit forms (in x) of the curves where f[x,y]==0, selected to match the curves we are interested in. Jun 18, 2021 at 12:02
• @yarchik I have modified the answer to hopefully clarify what's being integrated here. Jun 18, 2021 at 12:11
Clear["Global*"]
RegionPlot[
2 Cos[(3 x)/2] + Cos[1/2 (x - 2 y)] + Cos[x/2] (7 + 8 Cos[x + y]) < 0,
{x, 0, Pi}, {y, 0, Pi},
Frame -> True]
rgn = ImplicitRegion[{2 Cos[(3 x)/2] + Cos[1/2 (x - 2 y)] +
Cos[x/2] (7 + 8 Cos[x + y]) < 0 && 0 < x < Pi &&
0 < y < Pi}, {x, y}];
Area[rgn, WorkingPrecision -> MachinePrecision]
(* 3.49805 *)
Area[rgn, WorkingPrecision -> 15]
(* 3.49805 *)
A similar way is
reg=ImplicitRegion[
2 Cos[(3 x)/2] + Cos[1/2 (x - 2 y)] + Cos[x/2] (7 + 8 Cos[x + y]) <
0, {{x, 0, Pi}, {y, 0, Pi}}];
NIntegrate[1, {x, y} ∈ reg]
3.49485
• Thanks. Does WorkingPrecission matter in this method? I see a small difference in the different answers here obtained by different methods others suggested; if I need the most accurate one, which one is more reliable?
– user80186
Jun 18, 2021 at 12:54
Method ->"LocalAdaptive" evaluates without errormessage
NIntegrate[Boole[2 Cos[(3 x)/2] + Cos[1/2 (x - 2 y)] +Cos[x/2] (7 + 8 Cos[x + y]) < 0], {x, 0, Pi}, {y,0, Pi}, Method -> "LocalAdaptive"]
(*3.49818*)
• Thanks. Does WorkingPrecission matter in this method? I see a small difference in the different answers here obtained by different methods others suggested; if I need the most accurate one, which one is more reliable?
– user80186
Jun 18, 2021 at 12:55
• @SaraChem Sorry, I was offline some days. If you add the option , IntegrationMonitor :> ((errors = Through[#1@"Error"]) &) inside NIntegrate , it's possible to evaluate the integrationerror Total@errors afterwards. Might be helpful. Jun 20, 2021 at 14:27 | 2022-05-26T01:52:11 | {
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https://mathlesstraveled.com/?cpage=1 | A few words about PWW #33: subset permutations
My previous post showed four rows of diagrams, where the $n$th row (counting from zero) has diagrams with $n+2$ dots. The diagrams in the $n$th row depict all possible paths that
1. start at the top left dot,
2. end at the top right dot, and
3. never visit any dot more than once.
For a given diagram, we can choose which dots to visit (representing a subset of the dots), and we can visit the chosen dots in any order (a permutation). I will call the resulting things “subset permutations”, although I don’t know if there is some other more accepted term for them.
It’s not necessary to draw subset permutations as paths; it’s just one nice visual representation. If we number the dots from $1 \dots n$ (excluding the top two, which are uninteresting), each path corresponds precisely to a permutation of some subset of $\{1 \dots n\}$. Like this:
I thought of this idea the other day—of counting the number of “subset permutations”—and wondered how many there are of each size. I half expected it to turn out to be a sequence of numbers that I already knew well, like the Catalan numbers or certain binomial coefficients or something like that. So I was surprised when it turned out to be a sequence of numbers I don’t think I have ever encountered before (although it has certainly been studied by others).
So, how many subset permutations are there of size $n$? Let’s call this number $\mathit{SP}_n$. From my previous post we can see that starting with $\mathit{SP}_0 = 1$, the first few values are $1, 2, 5, 16$. (When I saw 1, 2, 5, I thought sure it was going to be the Catalan numbers—but then the next number was 16 instead of 14…) Can we compute these more generally without just listing them all and counting?
Well, to choose a particular subset permutation of size $n$, we can first choose how big we want the subset to be, i.e. how many dots to visit, which could be zero, or $n$, or anything in between. Once we choose to visit $k$ out of $n$ dots, then we have $n$ choices for which dot to visit first, then $(n-1)$ choices for which dot to visit second, all the way down to $(n-k+1)$ choices for the last dot (you should convince yourself that $(n-k+1)$, not $(n-k)$, is correct!). This is often notated
$\displaystyle {}_n P_k = n (n-1) (n-2) \dots (n-k+1) = \frac{n!}{(n-k)!}.$
So, in total then, $\mathit{SP}_n$ is the sum of ${}_n P_k$ over all possible $k$, that is,
$\displaystyle \mathit{SP}_n = \sum_{0 \leq k \leq n} \frac{n!}{(n-k)!}.$
This is already better than just listing all the subset permutations and counting. For example, we can compute that
$\mathit{SP}_4 = 4!/4! + 4!/3! + 4!/2! + 4!/1! + 4!/0! = 1 + 4 + 12 + 24 + 24 = 65.$
And sure enough, if we list them all, we get 65 (you’ll just have to take my word that this is all of them, I suppose!):
However, with a little algebra, we can do even better! First, let’s write out the sum for $\mathit{SP}_n$ explicitly:
$\mathit{SP}_n = 1 + n + n(n-1) + n(n-1)(n-2) + \dots + n!$
If we factor $n$ out of all the terms after the initial $1$, we get
$\mathit{SP}_n = 1 + n[1 + (n-1) + (n-1)(n-2) + \dots + (n-1)!] = 1 + n \mathit{SP}_{n-1}$
so the subset permutation numbers actually follow a very simple recurrence! To get the $n$th subset permutation number, we just multiply the previous one by $n$ and add one.
$\begin{array}{rcl} \mathit{SP}_0 &=& 1 \\ \mathit{SP}_1 &=& 1 \cdot 1 + 1 = 2 \\ \mathit{SP}_2 &=& 2 \cdot 2 + 1 = 5 \\ \mathit{SP}_3 &=& 3 \cdot 5 + 1 = 16 \end{array}$
and sure enough, $\mathit{SP}_4 = 4 \cdot 16 + 1 = 65$. We can now easily calculate the next few subset permutation numbers:
$1,2,5,16,65,326,1957,13700,109601,986410, \dots$
A commenter also wondered about the particular order in which I listed subset permutations in my previous post. The short, somewhat disappointing answer is “whatever order came out of the standard library functions I used”. However, there are definitely more interesting things to say about the ordering, and I think I’ll probably write about that in another post.
Posted in combinatorics, posts without words | Tagged , , , | 6 Comments
Post without words #33
Posted in combinatorics, posts without words | Tagged , , | 4 Comments
Happy tau day!
Happy Tau Day! The Tau Manifesto by Michael Hartl is now available in print, if you’re in the market for a particularly nerdy coffee table book and conversation starter:
I also wrote about Tau Day ten years ago; see that post for more links! Using $\tau$ we can make the most beautiful equation in the world even more beautiful:
$e^{i \tau} = 1 + 0i$
Please enjoy eating two pi(e)s today. I’ll be back soon with some solutions to the parallelogram area challenge!
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Challenge: area of a parallelogram
And now for something completely different!1
Suppose we have a parallelogram with one corner at the origin, and two adjacent corners at coordinates $(a,b)$ and $(c,d)$. What is the area of the parallelogram?
1. …or is it?
Posted in challenges, geometry | Tagged , | 18 Comments
Post without words #32
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The Natural Number Game
Hello everyone! It has been quite a while since I have written anything here—my last post was in March 2020, and since then I have been overwhelmed dealing with online and hybrid teaching, plus a newborn (who is now almost 5 months old and very cute):
But the spring semester is finally over, and I have a sabbatical in the fall, so I plan to do a bunch more writing! I have several ideas of things to come (feel free to send me suggestions as well), but to start things out for today I just have a fun link:
The idea is to use a computer proof assistant to formally prove a lot of basic facts about arithmetic. That is, you get to build proofs using a special precise notation, and the computer will automatically check whether your proof is correct. But the whole thing is organized like a game, with a tutorial, levels that build on each other and introduce new ideas and techniques just before you need them, etc. It’s a lot of fun and honestly kind of addicting.
There is very little background needed other than some capability for abstract thinking. Things start out quite easy and progress slowly, though things become quite challenging by the time you make it all the way to the end.
Posted in challenges, computation, proof | Tagged , , , , , | 6 Comments
An exploration of forward differences for bored elementary school students
Last week I made a mathematics worksheet for my 8-year-old son, whose school is closed due to the coronavirus pandemic. I’m republishing it here so others can use it for similar purposes.
Figurate numbers and forward differences
There are lots of further directions this could be taken but I’ll leave that to you and your kids. I tried to create something that was conducive to open-ended exploration rather than something that had a single particular goal in mind.
• For the curious: Babbage Difference Engine
• For the intrepid: Concrete Mathematics by Graham, Knuth, and Patashnik, section 2.6 (“Finite and Infinite Calculus”)
Seeing as how we’ve got at least four more weeks of (effectively) homeschooling ahead of us, and probably more than that, in all likelihood I will be making more of these, and I will certainly continue to share them here! If you use any of these with your kids I’d love to hear about your experiences.
Posted in arithmetic, teaching | | 1 Comment
Ways to prove a bijection
You have a function $f : A \to B$ and want to prove it is a bijection. What can you do?
By the book
A bijection is defined as a function which is both one-to-one and onto. So prove that $f$ is one-to-one, and prove that it is onto.
This is straightforward, and it’s what I would expect the students in my Discrete Math class to do, but in my experience it’s actually not used all that much. One of the following methods usually ends up being easier in practice.
By size
If $A$ and $B$ are finite and have the same size, it’s enough to prove either that $f$ is one-to-one, or that $f$ is onto. A one-to-one function between two finite sets of the same size must also be onto, and vice versa. (Of course, if $A$ and $B$ don’t have the same size, then there can’t possibly be a bijection between them in the first place.)
Intuitively, this makes sense: on the one hand, in order for $f$ to be onto, it “can’t afford” to send multiple elements of $A$ to the same element of $B$, because then it won’t have enough to cover every element of $B$. So it must be one-to-one. Likewise, in order to be one-to-one, it can’t afford to miss any elements of $B$, because then the elements of $A$ have to “squeeze” into fewer elements of $B$, and some of them are bound to end up mapping to the same element of $B$. So it must be onto.
However, this is actually kind of tricky to formally prove! Note that the definition of “$A$ and $B$ have the same size” is that there exists some bijection $g : A \to B$. A proof has to start with a one-to-one (or onto) function $f$, and some completely unrelated bijection $g$, and somehow prove that $f$ is onto (or one-to-one). Also, a valid proof must somehow account for the fact that this becomes false when $A$ and $B$ are infinite: a one-to-one function between two infinite sets of the same size need not be onto, or vice versa; we saw several examples in my previous post, such as $f : \mathbb{N} \to \mathbb{N}$ defined by $f(n) = 2n$. Although tricky to come up with, the proof is cute and not too hard to understand once you see it; I think I may write about it in another post!
Note that we can even relax the condition on sizes a bit further: for example, it’s enough to prove that $f$ is one-to-one, and the finite size of $A$ is greater than or equal to the finite size of $B$. The point is that $f$ being a one-to-one function implies that the size of $A$ is less than or equal to the size of $B$, so in fact they have equal sizes.
By inverse
One can also prove that $f : A \to B$ is a bijection by showing that it has an inverse: a function $g : B \to A$ such that $g(f(a)) = a$ and $f(g(b)) = b$ for all $a \in A$ and $b \in B$. As we saw in my last post, these facts imply that $f$ is one-to-one and onto, and hence a bijection. And it really is necessary to prove both $g(f(a)) = a$ and $f(g(b)) = b$: if only one of these hold then $g$ is called a left or right inverse, respectively (more generally, a one-sided inverse), but $f$ needs to have a full-fledged two-sided inverse in order to be a bijection.
…unless $A$ and $B$ are of the same finite size! In that case, it’s enough to show the existence of a one-sided inverse—say, a function $g$ such that $g(f(a)) = a$. Then $f$ is (say) a one-to-one function between finite equal-sized sets, hence it is also onto (and hence $g$ is actually a two-sided inverse).
We must be careful, however: sometimes the reason for constructing a bijection in the first place is in order to show that $A$ and $B$ have the same size! This kind of thing is common in combinatorics. In that case one really must show a two-sided inverse, even when $A$ and $B$ are finite; otherwise you end up assuming what you are trying to prove.
By mutual injection?
I’ll leave you with one more to ponder. Suppose $f : A \to B$ is one-to-one, and there is another function $g : B \to A$ which is also one-to-one. We don’t assume anything in particular about the relationship between $f$ and $g$. Are $f$ and $g$ necessarily bijections?
Posted in logic, proof | | 7 Comments
One-sided inverses, surjections, and injections
Several commenters correctly answered the question from my previous post: if we have a function $f : A \to B$ and $g : B \to A$ such that $g(f(a)) = a$ for every $a \in A$, then $f$ is not necessarily invertible. Here are a few counterexamples:
• Commenter Buddha Buck came up with probably the simplest counterexample: let $A$ be a set with a single element, and $B$ a set with two elements. It does not even matter what the elements are! There’s only one possible function $g : B \to A$, which sends both elements of $B$ to the single element of $A$. No matter what $f$ does on that single element $a \in A$ (there are two choices, of course), $g(f(a)) = a$. But clearly $f$ is not a bijection.
• Another counterexample is from commenter designerspaces: let $f : \mathbb{N} \to \mathbb{Z}$ be the function that includes the natural numbers in the integers—that is, it acts as the identity on all the natural numbers (i.e. nonnegative integers) and is undefined on negative integers. $g : \mathbb{Z} \to \mathbb{N}$ can be the absolute value function. Then $g(f(n)) = |n| = n$ whenever $n$ is a natural number, but $f$ is not a bijection, since it doesn’t match up the negative integers with anything.
Unlike the previous example, in this case it is actually possible to make a bijection between $\mathbb{N}$ and $\mathbb{Z}$, for example, the function that sends even $n$ to $n/2$ and odd $n$ to $-(n+1)/2$.
• Another simple example would be the function $f : \mathbb{N} \to \mathbb{N}$ defined by $f(n) = 2n$. Then the function $g(n) = \lfloor n/2 \rfloor$ satisfies the condition, but $f$ is not a bijection, again, because it leaves out a bunch of elements.
• Can you come up with an example $f : \mathbb{R} \to \mathbb{R}$ defined on the real numbers $\mathbb{R}$ (along with a corresponding $g$)? Bonus points if your example function is continuous.
All these examples have something in common, namely, one or more elements of the codomain that are not “hit” by $f$. Michael Paul Goldenberg noted this phenomenon in general. And in fact we can make this intuition precise.
Theorem. If $f : A \to B$ and $g : B \to A$ such that $g(f(a)) = a$ for all $a \in A$, then $f$ is injective (one-to-one).
Proof. Suppose for some $a_1, a_2 \in A$ we have $f(a_1) = f(a_2)$. Then applying $g$ to both sides of this equation yields $g(f(a_1)) = g(f(a_2))$, but because $g(f(a)) = a$ for all $a \in A$, this in turn means that $a_1 = a_2$. Hence $f$ is injective.
Corollary. since bijections are exactly those functions which are both injective (one-to-one) and surjective (onto), any such function $f : A \to B$ which is not a bijection must not be surjective.
And what about the opposite case, when there are functions $f : A \to B$ and $g : B \to A$ such that $f(g(b)) = b$ for all $b \in B$? As you might guess, such functions are guaranteed to be surjective—can you see why?
Posted in logic | | 1 Comment
Suppose we have sets $A$ and $B$ and a function $f : A \to B$ (that is, $f$’s domain is $A$ and its codomain is $B$). Suppose there is another function $g : B \to A$ such that $g(f(a)) = a$ for every $a \in A$. Is $f$ necessarily a bijection? That is, does $f$ necessarily match up each element of $A$ with a unique element of $B$ and vice versa? Or put yet another way, is $f$ necessarily invertible? | 2022-06-29T12:16:38 | {
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http://rfmb.valledelchieseonline.it/rotation-matrix-3d.html | # Rotation Matrix 3d
The 3-dimensional versions of the rotation matrix A are the following matrices:. For counterclockwise rotation, the matrix has the following elements. Or you transpose the matrix first (mtmp=m; for x=0 to 2 for y=0 to 2 m[x,y] = mtmp[y,x]) and do your standard rotation calculation with the transposed matrix. a rotation around the z-axis wouldn't change the z-values of the vertices. However, I can only figure out how to do 1 and 4 using numpy. We learn how to describe the orientation of an object by a 2×2 rotation matrix which has some special properties. And thank you for taking the time to help us improve the quality of Unity Documentation. Primarily to support 3D rotations. This means that there is an orthogonal basis, made by the corresponding eigenvectors (which are necessarily orthogonal), over which the effect of the rotation matrix is just stretching it. We can project a point orthogonal down into one of the main planes by using a matrix that scale the axis normally onto the plane with 0. The 3D object is moved and rotated in the 3D space, and the new destination points become B1=, B2=, and B3=. • In 2D, a rotation just has an angle • In 3D, specifying a rotation is more complex -basic rotation about origin: unit vector (axis) and angle •convention: positive rotation is CCW when vector is pointing at you • Many ways to specify rotation -Indirectly through frame transformations -Directly through •Euler angles: 3 angles. When I look at the file, however, it appears that the inputs to the transformation are the trans x,y,z and the roll,pitch,yaw angles. Raises: ValueError: If the shape of angles is not supported. Rotation About an Arbitrary Axis and Avoiding Gimbal Lock - Cprogramming. Rotation Matrices Suppose that ↵ 2 R. translation, rotation, scale, shear etc. A 3 by 3 matrix sets the rotation and shear. The axis can be either x or y or z. 1 Matrix Representation A 2D rotation is a tranformation of the form 2 4 x 1 y 1 3 5 = 2 4 cos( ) sin( ) sin( ) cos( ) 3 5 2 4 x 0 y 0 3 5 (1) where is the angle of rotation. We will try to enter into the details of how the matrices are constructed and why, so this article is not meant for absolute beginners. euler_rotation = mathutils. Raises: ValueError: If the shape of quaternion is not supported. The rotation matrix for this transformation is as follows. Convert a Rotation Matrix to Euler Angles in OpenCV. Each has its own uses and drawbacks. The only thing new in the C++ code is the usage of GetConsoleScreenBufferInfo and SetConsoleTextAttribute which gets the size of the console and sets the text color. We can think of rotations in another way. Except as otherwise noted, the content of this page is licensed under the Creative Commons Attribution 4. −Composition of geometric transformations in 2D and 3D. Equations ()-() effectively constitute the definition of a vector: i. A short derivation to basic rotation around the x-, y- or z-axis by Sunshine2k- September 2011 1. A matrix with M rows and N columns is defined as a MxN matrix. How to combine rotation in 2 axis into one matrix. It's so clever that it's worth sharing in full detail. In 2D it's much simpler. And Rotation is done with trigonometric functions in the matrix. If we add. Simple rotation – formulas were derived for rotation of a shape centered in origin by a certain angle. When a transformation takes place on a 2D plane, it is called 2D transformation. This matrix class is used for 3D object rotate along a specifid axis. The only difference is the signs for sinθ are reversed. matrix() Describes a homogeneous 2D transformation matrix. gives the column matrix corresponding to the point (a+ dx, b+ dy, c+ dz). Initially I hoped that. I have a point, in that mesh, that i must calculate mannualy. The rotation operation consists of multiplying the transformation matrix by a matrix whose elements are derived from the angle parameter. Rotation matrix from axis and angle First rotate the given axis and the point such that the axis lies in one of the coordinate planes Then rotate the given axis and the point such that the axis is aligned with one Use one of the fundamental rotation matrices to rotate the point depending. Greetings, I'm trying to apply a Rotation or Affine matrix to a spiral. The preview on the right will be updated when you compute, x’ (dash) points to the new x direction of the body, y’, z’ do the same for y and z axes. We can project a point orthogonal down into one of the main planes by using a matrix that scale the axis normally onto the plane with 0. As others stated in the comments you have to make sure you don´t mix row- and column-major matrices and dont mix any coordinate spaces. I have a 3D translation and rotation problem I am trying to solve using Excel 2010. As I understand, the rotation matrix around an arbitrary point, can be expressed as moving the rotation point to the origin, rotating around the origin and moving back to the original position. We can think of rotations in another way. @Sascha Grusche and @Elie Maalouf. As a unit quaternion, the same 3D rotation matrix. Learn more about rotation matrix, point cloud, 3d. They are represented in the matrix form as below −. The formula is , using the dot and cross product of vectors. Describing rotation in 3d with a vector. A matrix is an array of values that defines a transformation of coordinates. This is why also the 3D version has two of the three axes change simultaneously - because it is just a derivative from its 2D version. RotationMatrix[{u, v}] gives the matrix that rotates the vector u to the direction of the vector v in any dimension. These singularities are not characteristic of the rotation matrix as such, and only occur with the usage of Euler angles. They are both ways of completely describing a rotation in 3D, so they are freely convertible. You can also rotate, resize and stretch a 3D graph by dragging the mouse. Download our 100% free 3D Rotation templates to help you create killer PowerPoint presentations that will blow your audience away. 3D scaling matrix. Hi, I'm trying to transform a PET scan onto a CT scan based on an existing rotation and translation matrix. G-CNNs achieve equivariance with respect to nite subgroups of the rotation group, which constitutes a bottleneck in 3D. I'm trying to concatenate a rotation matrix with a 4x4 homogeneous transformation matrix with column-major convention. We learn how to describe the orientation of an object by a 2×2 rotation matrix which has some special properties. 3D rotations made easy in Julia. Appearance Scope: 6. When the Frobenius norm is taken as the measure of closeness, the solution is usually computed using the singular value decomposition (SVD). 1 Matrix Representation A 2D rotation is a tranformation of the form 2 4 x 1 y 1 3 5 = 2 4 cos( ) sin( ) sin( ) cos( ) 3 5 2 4 x 0 y 0 3 5 (1) where is the angle of rotation. Position Cartesian coordinates (x,y,z) are an easy and natural means of representing a position in 3D space …But there are many other representations such as spherical converted to matrix form to perform rotation. As we know $\cos(0) = 1$ and $\sin(0) = 0$. Three shears. Appearance: 7. ACM SIGGRAPH is a thriving international organization. det(R) != 1 and R. step file, it stores a rotation matrix that needs to be applied to the appropriate geometry, correct?. Rotation in 3D - The Rotation Matrix In this note, I investigate the rotation matrix that relates the image of a point p ⃗ \vec{p} p when it is rotated by an angle θ \theta θ about an axis a ⃗ \vec{a} a that passes through the origin. winter wheat (WW))is available to successive crops in reduced or no-till systems, uncover the mechanisms involved in this process, and ultimately develop management practices to maximize Po utilization from these sources. Serializable. If you have a rotation matrix, then this will transform from one coordinate system to another. So essentially quaternions store a rotation axis and a rotation angle, in a way that makes combining rotations easy. Just like the graphics pipeline, transforming a vector is done step-by-step. A 2-columns matrix or data frame containing a set of X and Y coordinates. Abstract transformations, such as rotations (represented by angle and axis or by a quaternion), translations, scalings. A rotation matrix can be built by using the axis of the coordinate system you're rotating into. There will be some repetition of the earlier analyses. Convert your quaternion to a rotation matrix, and use it in the Model Matrix. 3D programming in python. Posted September 16, 2017 · CuraEngine mesh_rotation_matrix to rotate 180 degrees Just a WAG here, but since the print origin is at the corner of the print bed (unless you're using a Delta printer), and the rotation matrix is rotating about the origin, then haven't you just rotated it off the print bed?. I've been all. I want this rotation matrix to perform a rotation about the X axis (or YZ pla. Rotation matrices, on the other hand, are the representation of choice when it comes to implementing efficient rotations in software. Coordinates of point p in two systems The elementary 3D rotation matrices are constructed to perform In order to be able to write the rotation matrix directly, imagine that the the z-axis is playing the role of the x-axis,. As their trunks were rotated passively without fixing the legs, each volunteer was instructed to keep their knees straight and rotate their legs in a fashion comparable to the trunk rotation, in order to minimize the effects of leg rotation on trunk rotation. The easiest rotation axes to handle are those that are parallel to the co-ordinate axes. , is an orthogonal matrix) is 1). New coordinates by 3D rotation of points Calculator. Basic steps needed to display 3D objects: 4. If the first body is only capable of rotation via a revolute joint, then a simple convention is usually followed. Email this Article Givens rotation. Rotation Transforms for Computer Graphics covers a wide range of mathematical techniques used for rotating points and frames of reference in the plane and 3D space. Rotation. explanation of quaternion from matrix. Find the characteristic function, eigenvalues, and eigenvectors of the rotation matrix. Tag: math,vector,lua,rotation I have two Vec3s, Camera Forward and Turret Forward. Turn your head left and right; that’s a rotation around the Y axis. The rotation operation is a 3x3 matrix. Then, the tutorials will move on to give you the matrices for rotation over the x and y axes, tell you how to use them, and then give you a matrix which will allow rotations around an arbitrary axis. With a chain of rotations, roundoff errors accumulate. The rotation matrix for this transformation is as follows. 3D Transformations, Translation, Rotation, Scaling The Below program are for 3D Transformations. Unless specified, the rest of this page uses implies rotation to be a rotation of points about the origin. We conclude that every rotation matrix, when expressed in a suitable coordinate system, partitions into independent rotations of two-dimensional subspaces, at most n ⁄ 2 of them. represent data on 3D rotation groups. muting any, we clearly need a negative unit matrix, namely, Proper Rotation. Because Rotation can be done either along the x, y, or z axis, there is a different rotation matrix for each of the axises: Figure 6a - Rotation around the X axis. Its first 3 dimensional vectors(3*3 submatrix) contain the rotated X, Y and Z axes. Examples in 3D computer graphics Rotation. In SO(4) the rotation matrix is defined by two quaternions, and is therefore 6-parametric (three degrees of freedom for every quaternion). The most general three-dimensional rotation matrix represents a counterclockwise rotation by an angle θ about a fixed axis that lies along the unit vector ˆn. Under rotations, vector lengths are preserved as well as the angles between vectors. The formula of this operations can be described in a simple multiplication of. You can probably use the axis angle representation to build a rotation matrix, with whatever math library you have available. G-CNNs achieve equivariance with respect to nite subgroups of the rotation group, which constitutes a bottleneck in 3D. This means that no rotation has taken place around any of the axes. In the above code, since the rotation transformation is prepended to the matrix, the rotation transformation would be performed first. Lecture 08. 2+) were then focused to capture these y-z′ images at >800 frames per second as the sheet was repeatedly scanned across the sample in the x direction. Each has its own uses and drawbacks. This object get transformed with the following matrix transformation: glRotated and glTranslate. This means that there is an orthogonal basis, made by the corresponding eigenvectors (which are necessarily orthogonal), over which the effect of the rotation matrix is just stretching it. A rotation matrix describes the relative orientation of two such frames. rotqrmean Find the average of several rotation quaternions rotqrvec Apply a quaternion rotation to an array of 3D vectors skew3d Convert between vectors and skew symmetric matrices: 3x3 matrix <-> 3x1 vector and 4x4 Plucker matrix <-> 6x1 vector. the product of a viewing matrix and a modeling matrix. If you are uncomfortable with the thought of 4D matrix rotations, then I recommend reading Wikipedia, or checking out my article about 3D graphing, which can be found here. where M is a constant 3x3 matrix, is the 3x3 identity matrix, and we are solving for the 3x3 matrix R. Using 3D Rotation Matrices in Practice By confuted So, now that you more or less know how to rotate a point in any arbitrary manner in three dimensions, generating matrices along the way, it's time to learn what you should do with each of these matrices. We found that this was the rotation transformation matrix about an x-axis rotation. First, suppose that all eigenvalues of the 3D rotation matrix A are real. We will try to enter into the details of how the matrices are constructed and why, so this article is not meant for absolute beginners. If you want to rotate around an arbitary axis you can use the equations under "Rotation matrix from axis and angle" To understand how a rotation matrix works you should know that each column represents a 3D Vector which in turn represents one of the axis of the rotated coordinate system. ' (as long as the translation is ignored). Again, we must translate an object so that its center lies on the origin before scaling it. We'll call the rotation matrix for the X axis matRotationX, the rotation matrix for the Y axis matRotationY, and the rotation matrix for the Z axis matRotationZ. The eigenvalues of A are. Therefore, an arbitrary 3D rotation can be decomposed into only two 3D beamshears. This means that there is an orthogonal basis, made by the corresponding eigenvectors (which are necessarily orthogonal), over which the effect of the rotation matrix is just stretching it. I'm looking for a SO(3) rotation representation that lends itself to energy minimization. Rotation matrices are used for computations in aerospace, image processing, and other technical computing applications. We conclude that every rotation matrix, when expressed in a suitable coordinate system, partitions into independent rotations of two-dimensional subspaces, at most n / 2 of them. Click the 3D graph (the white space of the graph layer, not the 3D data plot), eight resizing handles appear around the 3D graph. Matrix representation. This axis, in this work, will be represented by the supporting line of the directed segment S ab (a 1D simplex), where ( , , (0)) 3 (0) 2 (0) a a1 a a and ( , (0)) 3 (0) 2 (0) b 1 b b are two non-coincident 3D points which we will refer as the. •In 3D, specifying a rotation is more complex –basic rotation about origin: unit vector (axis) and angle •convention: positive rotation is CCW when vector is pointing. The Rotation 3D page. The easiest rotation axes to handle are those that are parallel to the co-ordinate axes. Now, according to the equation, multiplying the transformation matrix with a coordinate would result in a coordinate but if is [9,1] for example, if i multiply with the rotation matrix. Yes, [R|t] implies the rotation and translation. Serializable, java. Rotation matrices are used in computer graphics and in statistical analyses. calibration cube. 3D Matrices. The Rotation 3D page. The eigenvalues of A are. rotation matrix specifies a 3 × 3 matrix. ppt), PDF File (. Shortest distance between two lines. 0 License, and code samples are licensed under the Apache 2. A short derivation to basic rotation around the x-, y- or z-axis by Sunshine2k- September 2011 1. In 2D euclidean space, rotation matrix is a matrix that tilts every single vector in the 2D space, without changing the scale. pdf), Text File (. Second, this method means we can create our own css animations, and do something a bit more advanced. Computing Euler angles from a rotation matrix Gregory G. I understand the sentence except for the "rotation matrix" part, again. RotationMatrix[{u, v}] gives the matrix that rotates the vector u to the direction of the vector v in any dimension. 3DPrimitivesTransformations - Free download as Powerpoint Presentation (. I'm trying to concatenate a rotation matrix with a 4x4 homogeneous transformation matrix with column-major convention. If you are reading this page in order to write a 3D computer program I suggest you read enough of this page to convince yourself of the problems with Euler angles and to get an intuitive understanding of 3D rotations and then move on to quaternion or matrix algebra representations. skip_checks (bool, optional) – If True avoid sanity checks on rotation_matrix for performance. Math for simple 3D coordinate rotation (python) Ask Question Asked 3 years, 5 months ago. Defining the rotation axis as the z axis, we note first that the z coordinate will be unchanged by any rotation about the z axis. A general homoge- neous matrix formulation to 3D rotation geometric transformations is proposed which suits for the cases when the rotation axis is unnecessarily through the coordinate system origin given their rotation axes and. When acting on a matrix, each column of the matrix represents a different vector. The three simultaneous orthogonal rotations measured with a 3D gyroscope represent a single rotation around a certain axis for a certain angle. 3D rotations • A 3D rotation can be parameterized with three numbers • Common 3D rotation formalisms - Rotation matrix • 3x3 matrix (9 parameters), with 3 degrees of freedom - Euler angles • 3 parameters - Euler axis and angle • 4 parameters, axis vector (to scale) - Quaternions • 4 parameters (to scale). multiplying rotation matrices is a noisyrotation matrix [1]. M modelview = M viewing * M modeling. We can now write a transformation for the rotation of a point about this line. And so here's the rotation transformation matrix. RotationMatrix[{u, v}] gives the matrix that rotates the vector u to the direction of the vector v in any dimension. R = rotx(ang) creates a 3-by-3 matrix for rotating a 3-by-1 vector or 3-by-N matrix of vectors around the x-axis by ang degrees. 2+) were then focused to capture these y-z′ images at >800 frames per second as the sheet was repeatedly scanned across the sample in the x direction. 1 1 3 Lecture Video 3 of 4 Rotation Matrix Example 1 - Duration: 10:20. Ask Question With r = RotationMatrix[a, {x, y, z}] I can compute a 3D rotation matrix from its axis/angle representation. all types of rotation calculation will eventually yield a 4x4 matrix, as this is always the required form for expressing the final transformation. Represents a 3D rotation as a rotation angle around an arbitrary 3D axis. The 3-dimensional versions of the rotation matrix A are the following matrices:. ˇ, rotation by ˇ, as a matrix using Theorem 17: R ˇ= cos(ˇ) sin(ˇ) sin(ˇ) cos(ˇ) = 1 0 0 1 Counterclockwise rotation by ˇ 2 is the matrix R ˇ 2 = cos(ˇ 2) sin(ˇ) sin(ˇ 2) cos(ˇ 2) = 0 1 1 0 Because rotations are actually matrices, and because function composition for matrices is matrix multiplication, we’ll often multiply. 3D Rotation Matrix. From the Cartesian grid (left grid), we can see the blue point is located at (2, 1). Rotation Matrix. 3D rotation matrix around vector. is the rotation matrix already, when we assume, that these are the normalized orthogonal vectors of the local coordinate system. 0 License , and code samples are licensed under the Apache 2. I know that in 3D space the matrix product order is important - changing the order of the matrices can effect the rotate result. winter wheat (WW))is available to successive crops in reduced or no-till systems, uncover the mechanisms involved in this process, and ultimately develop management practices to maximize Po utilization from these sources. Euler angles can be defined with many different combinations (see definition of Cardan angles ). As I understand, the rotation matrix around an arbitrary point, can be expressed as moving the rotation point to the origin, rotating around the origin and moving back to the original position. matrix3d() Describes a 3D transformation as a 4×4 homogeneous matrix. 3D Rotations Rotation about z-axis. When acting on a matrix, each column of the matrix represents a different vector. This is the currently selected item. Learn more about rotation matrix, point cloud, 3d. A vector is a direction in a space (like for GPS), that is the result of the transformation (math operators) of a point by a rotation matrix. 3D rotation) that minimizes some objective function. Transformations is a Python library for calculating 4x4 matrices for translating, rotating, reflecting, scaling, shearing, projecting, orthogonalizing, and superimposing arrays of 3D homogeneous coordinates as well as for converting between rotation matrices, Euler angles, and quaternions. 3D Rotation • Counterclockwise •The matrix M transforms the UVW vectors to the XYZ vectors y z x u=(u x,u y,u z) v=(v x,v y,v z) Change of Coordinates. 8 (b): What constraints must the elements R_{ij} of the three-dimensional rotation matrix satisfy, in order to preserve the length of vector A (for all vectors A)? Homework Equations The. find angles , , which make the two matrices equal. Euler3D transform, how to set the rotation matrix?. This is done by multiplying the vertex with the matrix : Matrix x Vertex (in this order. Let T be a linear transformation from R^2 to R^2 given by the rotation matrix. Quaternions are just more compact and easier to interpolate. I want to rotate an object by 60 degrees around the y axis, counter-clockwise. Three shears. The solution is not unique in most cases. In matrix form, the infinitesimal rotation has the representation (4) where and (5) To first order in , it can be shown from that. Position Cartesian coordinates (x,y,z) are an easy and natural means of representing a position in 3D space …But there are many other representations such as spherical converted to matrix form to perform rotation. Thus, we have H O = [I O] ω ,. They are represented in the matrix form as below −. Scale and Rotate. Derivation of the 3D transformation matrix. Follow 34 views (last 30 days). A rotation matrix describes the relative orientation of two such frames. Angular velocity. Although the inverse process requires a choice of rotation axis between the two alternatives, it is a straightforward procedure to retrieve the rotation axis and angle (see Appendix A). Euler returns a 3D vector containing the XYZ Euler angles. Rotations in 3D applications are usually represented in one of two ways: Quaternions or Euler angles. Do you know any reference how to derive this rotation matrix? It seems clear to me the rotation matrix in planar truss, 3D truss and planar frame are pretty similar (only different a bit), the form has same appearance to a simple rotation matrix, or when using directional cosine. To get in right-handed coordinate system, replace the angle with negative. rotation from the scanner coordinates to the camera coordinates will use the following transformation matrix. −Matrix representation of affine transformations. Please try again in a few minutes. In order to export actual quaternion data to your ASCII log file, you will have to set the Orientation output to Quaternions in MT Manager under Tools > Preferences > Exporters. it is called 3D transformation. The table of direction cosines relating the femoral (F) and pelvic (P) reference frames is obtained most simply via matrix multiplication, which yields the rotation matrix F R P and its corresponding table of direction cosines, Direction cosines for virtually any compound rotation can be found easily by using this exact methodology. A matrix is an array of values that defines a transformation of coordinates. C 3d Rotation Codes and Scripts Downloads Free. Slabaugh Abstract This document discusses a simple technique to find all possible Euler angles from a rotation matrix. Using localOrientation for rotation values, however, returns a 3×3 matrix. To produce a sequence of transformations with these equations, rotation is followed by translation, we must calculate the transformed coordinates one step at a time, thereby eliminating the calculation of intermediate coordinate values. (x_x, x_y, x_z) is a 3D vector that represents only the direction of the X-axis with respect to the coordinate system 1. Consider a counter-clockwise rotation of 90 degrees about the z-axis. Object implements java. Rotation estimation is a fundamental step for object motion estimation, alignment and registration, in image processing, whereas, 3-D shape reconstruction, object recognition, autonomous navigation and ego-motion are typical applications of rotation estimation in computer vision and robotics. I have a similar class. Matrix M 1 is a 2x2 rotation matrix, M 2 is translation vector. Normalize those vectors. winter wheat (WW))is available to successive crops in reduced or no-till systems, uncover the mechanisms involved in this process, and ultimately develop management practices to maximize Po utilization from these sources. A rotation matrix describes the relative orientation of two such frames. 1 1 3 Lecture Video 3 of 4 Rotation Matrix Example 1 - Duration: 10:20. 3D Rotation Matrix. A rotation followed by a translation is very different from a translation followed by a rotation, as illustrated below: Using the Matrix Object. If you want to rotate around an arbitary axis you can use the equations under "Rotation matrix from axis and angle" To understand how a rotation matrix works you should know that each column represents a 3D Vector which in turn represents one of the axis of the rotated coordinate system. The matrix to the left is a parallel projection down into the xy-plane. They will allow us to transform our (x,y,z,w) vertices. This article discusses the different types of matrices including linear transformations, affine transformations, rotation, scale, and translation. The three simultaneous orthogonal rotations measured with a 3D gyroscope represent a single rotation around a certain axis for a certain angle. The rotation matrix has the following properties: A is a real, orthogonal matrix, hence each of its rows or columns represents a unit vector. CSS also supports 3D transformations. With c++ (win32). Alternatively, you can set the Orientation output to Rotation Matrix directly. Ask Question Asked 3 years, 10 months ago. But why would this 3D frame rotation seems much different from those?. In this, the first of two articles I will show you how to encode 3D transformations as a single 4×4 matrix which you can then pass into the appropriate. Details EulerMatrix is also known as Euler rotation matrix or Euler rotation, and the angles α , β , and γ are often referred to as Euler angles. Quaternions represent a single rotation. math on December 25, 2008. When acting on a matrix, each column of the matrix represents a different vector. You might use this when applying the same rotation to a number of different objects,. It allows you to examine different rotation sequences. @Sascha Grusche and @Elie Maalouf. In August 1987, in Vancouver, Canada, almost all of those who worked in the paleomagnetic group at the University College of Rhodesia and Nyasaland, Salisbury, Southern Rhodesia (now the University of Zimbabwe, Harare, Zimbabwe) were by chance attending the International Union of Geodesy and Geophysics meeting. We'll call the rotation matrix for the X axis matRotationX, the rotation matrix for the Y axis matRotationY, and the rotation matrix for the Z axis matRotationZ. 3D Rotations Rotation about z-axis. This means that there is an orthogonal basis, made by the corresponding eigenvectors (which are necessarily orthogonal), over which the effect of the rotation matrix is just stretching it. det(R) != 1 and R. Composing a rotation matrix. A single precision floating point 4 by 4 matrix. Video transcript - What I want to do in this video is get some. I want this rotation matrix to perform a rotation about the X axis (or YZ pla. This example shows how to do rotations and transforms in 3D using Symbolic Math Toolbox™ and matrices. Online tools - vector rotation in 3D. 1 Representation Elements of the 3D rotation group, SO(3), are represented by 3D rotation matrices. New coordinates by 3D rotation of points Calculator - High accuracy calculation Welcome, Guest. The general rotation is much the same, with the up vector taken randomly, the desired rotation applied after the initial viewing transformation, and then the inverse of the viewing transformation is applied. Serializable. The Java 3D model for 4 X 4 transformations is: [ m00 m01 m02 m03 ] [ x ] [ x' ] [ m10 m11 m12 m13 ]. 3D Rotations are used everywhere in Computer Graphics, Computer Vision, Geometric Modeling and Processing, as well as in many other related areas. Now suppose we are given a matrix and are required to extract Euler angles corresponding to the above rotation sequence, i. C 3d Rotation Codes and Scripts Downloads Free. Just like the graphics pipeline, transforming a vector is done step-by-step. Examples in 3D computer graphics Rotation. This code checks that the input matrix is a pure rotation matrix and does not contain any scaling factor or reflection for example /** *This checks that the input is a pure rotation matrix 'm'. What is the correct order of transformations scale, rotate and translate and why? 3. Moreover, there are similar transformation rules for rotation about and. is the orthogonal projection of onto. We will try to enter into the details of how the matrices are constructed and why, so this article is not meant for absolute beginners. Follow 34 views (last 30 days). In 2D it's much simpler. Primarily to support 3D rotations. The center of a Cartesian coordinate frame is typically used as that point of rotation. As David Joyce points out, this fact is true in odd dimensions (including 3) but not even dimensions. (b) Because all the columns of U are mutually orthogonal, we can conclude that U is an orthogonal matrix. Consider the original set of basis vectors, i, j, k, and rotate them all using the rotation matrix A. Rotation matrices are orthogonal as explained here. You might use this when applying the same rotation to a number of different objects,. Kelly! above x2: screenshots from here. Composing a rotation matrix. The matrix takes a coordinate in the inner coordinate system described by the 3 vectors and and finds its location in the outer coordinate system. Moreover, there are similar transformation rules for rotation about and. Hi, I am doing optimization on a vector of rotation angles tx,ty and tz using scipy. Parameters: matrix - double[][]. Auckland's prof. When acting on a matrix, each column of the matrix represents a different vector. Appearance Mixed: 8. Generally speaking any matrix in the group SO(3) represents a rotation in 3d. That matrix isn't exactly symmetric, but a rotation matrix that is symmetric is a 180 degree rotation. What does POSIT require to be able to do 3D pose estimation? First it requires image coordinates of some object's points (minimum 4 points). For the rotation matrix R and vector v, the rotated vector is given by R*v. This example shows how to do rotations and transforms in 3D using Symbolic Math Toolbox™ and matrices. General Case Computer Graphics 3d Rotation Software Rendering Computer Graphics 3d Rotation 3d Coordinate Systems Powerpoint Presentation Axis Rotation 3d Rotation Matrix Derivation Machines Vertical Center Number Umc-750 Sample Trials Coordinate Geometry Basics. A general homoge- neous matrix formulation to 3D rotation geometric transformations is proposed which suits for the cases when the rotation axis is unnecessarily through the coordinate system origin given their rotation axes and. The View Matrix: This matrix will transform vertices from world-space to view-space. I have 2 known 3d points which are the origin of 2 axis plot in the space and I need to compute the 3D rotation matrix between them. Unless specified, the rest of this page uses implies rotation to be a rotation of points about the origin. This means that there is an orthogonal basis, made by the corresponding eigenvectors (which are necessarily orthogonal), over which the effect of the rotation matrix is just stretching it. The default polygon is a square that you can modify. all types of rotation calculation will eventually yield a 4x4 matrix, as this is always the required form for expressing the final transformation. You need to pass an angle of rotation and x, y, z axes as parameters to this method. However, manipulating 3D Rotations is always confusing, and debugging code that involves 3D rotation is usually quite time consuming. The rotation matrix is not parametric, created via eigendecomposition, I can't use angles to easily create an inverse matrix. 3D Transformations, Translation, Rotation, Scaling The Below program are for 3D Transformations. Find more Mathematics widgets in Wolfram|Alpha. Active 3 years, 7 months ago. However, now that I started implementing it myself, I came to the point where I'm really confused. for Java and C++ code to implement these rotations click here. In a two-dimensional cartesian coordinate plane system, the matrix R rotates the points in the XY-plane in the counterclockwise through an angle θ about the origin. math on December 25, 2008. Then transform a vector by that rotation matrix to get your result. 3D Matrices. CE503 Rotation Matrices Derivation of 2D Rotation Matrix Figure 1. Rotation Matrix. rotation of one frame into the other, Pj is the direction cosine for X2 with respect to Xk, and the rotation transformation matrix is an orthogonal matrix: P R12 = [flk] (5-5) fjk = Cos(X], Xi) JM In Sec. In SO(4) the rotation matrix is defined by two quaternions, and is therefore 6-parametric (three degrees of freedom for every quaternion). The rotation matrix is displayed for the current angle. In an analogous fashion, orientations are described relative to a standard orientation by a rotation, with the identity rotation describing the standard. In this, the first of two articles I will show you how to encode 3D transformations as a single 4×4 matrix which you can then pass into the appropriate. Accordingly, A v = v {\displaystyle Av=v} , and the rotation axis therefore corresponds to an eigenvector of the rotation matrix associated with an eigenvalue of 1. Matrix multiplications always have the origin as a fixed point. This way to do the inverse rotation works only with pure normalized rotation matrices, should be noticed. 3D Reflection in Computer Graphics- Reflection is a kind of rotation where the angle of rotation is 180 degree. Getting Started with the Java 3D API written in Java 3D: 9. 0° (rotation happens on the XY plane in 3D). Understanding transforamtion matrix requires some knoledge of math… R. 3D Rotation The easiest rotation axes are those that parallel to the coordinate axis. This article shows how to implement three-dimensional rotation matrices and use them to rotate a 3-D point cloud. The homogeneous transformation matrix for 3D bodies As in the 2D case, a homogeneous transformation matrix can be defined. How does OCC read in and apply a rotation matrix in the STEP file? That is, if I define a rotation with BRepBuilderAPI_Transform, save the resulting shape to a. This method prepends or appends the transformation matrix of the Graphics by the rotation matrix according to the order parameter. I'm trying to concatenate a rotation matrix with a 4x4 homogeneous transformation matrix with column-major convention. com Starting out. (x_x, x_y, x_z) is a 3D vector that represents only the direction of the X-axis with respect to the coordinate system 1. We learn how to describe the orientation of an object by a 2×2 rotation matrix which has some special properties. Primarily to support 3D rotations. Now, my understanding of your original question is that this unit vector, Ur, represents a rotation from U1, so you want to know how to find the rotation matrix that will transform U1 to Ur, i. Orthogonal matrices represent rotations (more precisely rotations, reflections, and compositions thereof) because, in a manner of speaking, the class of orthogonal matrices was defined in such a way so that they would represent rotations and refle. The rotation matrix has the following properties: A is a real, orthogonal matrix, hence each of its rows or columns represents a unit vector. Try a 90 degree rotation and then check. rotate3d() Rotates an element around a fixed axis in 3D space. 3D Transformations – Part 1 Matrices Transformations are fundamental to working with 3D scenes and something that can be frequently confusing to those that haven’t worked in 3D before. 1 Eigenvalues An n× nmatrix Ais orthogonal if its columns are unit vectors and orthogonal to. Lecture 08. The typical operations are translation, rotation. Here atan2 is the same arc tangent function, with quadrant checking, you typically find in C or Matlab. 1 The matrix for rotation about an arbitrary line. A Rotation instance can be initialized in any of the above formats and converted to any of the others. 3d curl intuition, part 1. Matrix for rotation is a clockwise direction. Geometrical Rotation Videos 9,222 royalty free stock videos and video clips of Geometrical Rotation. Now I would like to add a "tropism" command to simulate gravity pulling on the elements of an L-System, similar to how Laurens Laprés LParser does, and this is where I'm stuck. Homogeneous transforms contain BOTH rotation and translation information. R = rotx(ang) creates a 3-by-3 matrix for rotating a 3-by-1 vector or 3-by-N matrix of vectors around the x-axis by ang degrees. ANGLE DECOMPOSITION Recall that the rotation submatrix of the transformation is a multiplication matrix of the dot products of the unit vectors of the two body coordinate systems, and therefore includes trigonometric functions of the three angles of rotation, denoting flexion, abduction, and external rotation. Consider the 3-D rotation matrix U =-0. Ask Question Asked 3 years, 10 months ago. I'm struggling to understand the relation between the angles used to compose a rotation matrix and the angular velocity vector of the body expressed in the body frame. R = Rx*Ry*Rz. Both of these vectors are on different planes where Camera Forward is based on a free-look camera and Turret Forward is determined by the tank it sits on, the terrain the tank is on, etc. it only request for value and display the putout in text format on the screen. Moreover, the rotation axis in the 3D space coincides with the normal vector of the rotation plane. , is an orthogonal matrix) is 1). For x-axis rotation, we have the matrix:. This property allows you to rotate, scale, move, skew, etc. Home / Mathematics / Space geometry; Calculates the new coordinates by rotation of points around the three principle axes (x,y,z). 7 Transformation Matrix and Stiffness Matrix in Three-Dimensional Space. This article might seem exceedingly obvious to some but I'll build up to a point in a few articles. I'm trying to concatenate a rotation matrix with a 4x4 homogeneous transformation matrix with column-major convention. Because rotation matrices. 3: geometry of the 2D coordinate transformation The 2 2 matrix is called the transformation or rotation matrix Q. As we know $\cos(0) = 1$ and $\sin(0) = 0$. two antiparallel axes and angles (one axis and angle is negation of the other). square matrix – For 2D, 3x3 matrix – For 3D, 4x4 matrix. Lecture 5: 3-D Rotation Matrices. In order to calculate the rotation about any arbitrary point we need to calculate its new rotation and translation. 3D rotations made easy in Julia. The rotation matrices SO(3) form a group: matrix multiplication of any two rotation matrices produces a third rotation matrix; there is a matrix 1 in SO(3) such that 1M= M; for each Min SO(3) there is an inverse matrix M 1such that M M= MM 1 = 1. This format is definitely less intuitive than Euler angles, but it's still readable: the xyz components match roughly the rotation axis, and w is the acos of the rotation angle (divided by 2). , the three quantities are the components of a vector provided that they transform under rotation of the coordinate axes about in accordance with Equations ()-(). Then apply the following rules. skip_checks (bool, optional) – If True avoid sanity checks on rotation_matrix for performance. This package implements various 3D rotation parameterizations and defines conversions between them. In this example, I will only show the 4D rotation matrices. pdf), Text File (. rotation of one frame into the other, Pj is the direction cosine for X2 with respect to Xk, and the rotation transformation matrix is an orthogonal matrix: P R12 = [flk] (5-5) fjk = Cos(X], Xi) JM In Sec. to_matrix() this is the equivalent of. Rotation matrices can be constructed from elementary rotations about the In this lecture, I extend the 2D rotation matrix of SO(2) from Lecture 2. Rotation matrix visualization [5] 2018/09/29 17:08 Male / 20 years old level / High-school/ University/ Grad student / Very /. All input is normalized to unit quaternions and may therefore mapped to different ranges. I want to rotate an object by 60 degrees around the y axis, counter-clockwise. Analogously, we can define the tensor of inertia about point O, by writing equation(4) in matrix form. The matrix takes a coordinate in the inner coordinate system described by the 3 vectors and and finds its location in the outer coordinate system. This means that there is an orthogonal basis, made by the corresponding eigenvectors (which are necessarily orthogonal), over which the effect of the rotation matrix is just stretching it. Given 3 Euler angles , the rotation matrix is calculated as follows: Note on angle ranges. The rotation is performed clockwise, if you are looking along the direction of the rotation axis vector. A great amount of work on this topic is available in literature (see , , , , ). Representation of orientation • Homogeneous coordinates (review) • 4X4 matrix used to represent translation, scaling, and rotation • a point in the space is represented as • Treat all transformations the same so that they can be easily combined p= x y z 1. void: preMultiply(Matrix mb) Premultiplies the object matrix by mb and stores the result in the object; As a result, the. This is the currently selected item. Matrix transposition - if we have a matrix M with n rows and m columns, the transpose of , denoted is a matrix with m rows and n columns, with the first column of equal to the first row of and so on. If you want to rotate around an arbitary axis you can use the equations under "Rotation matrix from axis and angle" To understand how a rotation matrix works you should know that each column represents a 3D Vector which in turn represents one of the axis of the rotated coordinate system. The Mathematics of the 3D Rotation Matrix fastgraph. For some reason your suggested change could not be submitted. Call R v(θ) the 2x2 matrix corresponding to rotation of all vectors by angle +θ. This package implements various 3D rotation parameterizations and defines conversions between them. Stretching 3D Graphs. However, I can only figure out how to do 1 and 4 using numpy. 2) The rotation angles. Figure 6c - Rotation around the Z axis. It can be useful to notice that this can be done with a matrix operation. Euler angles can be defined with many different combinations (see definition of Cardan angles ). WebGL - Cube Rotation. Cartesian coordinates are typically used to represent the world in 3D programming. Transformations, continued 3D Rotation 23 r r r x y z Full 3D Rotation 0 sin cos 0 cos sin 1 0 0 – Multiply the current matrix by the rotation matrix that. Do we need to subtract the translation vector (t) from matrix M. 1 Introduction. A tensor of shape [A1, , An, 3, 3], where the last two dimensions represent a 3d rotation matrix. I want this rotation matrix to perform a rotation about the X axis (or YZ pla. Thus, we have H O = [I O] ω ,. First, suppose that all eigenvalues of the 3D rotation matrix A are real. ) and perspective transformations using homogenous coordinates. )? Ask Question Asked 3 years, It is quit lengthy but you can search for decomposing a rotation matrix. 7 The 3D Rotation Toolbar. Given a 3×3 rotation matrix. So you know how a 3D rotation matrix can be expressed in mathematical form. % example: % rotate around a random direction a random amount and then back % the result should be an Identity matrix. This is given by the product T P 1 − 1 T xz − 1 T z − 1 R z (θ) T z T xz T P 1. And so here's the rotation transformation matrix. This means they can rotate your 3D game geometry. Download our 100% free 3D Rotation templates to help you create killer PowerPoint presentations that will blow your audience away. A rotation is different from other types of motions: translations, which have no fixed points, and (hyperplane) reflections, each of them having an entire (n − 1)-dimensional. $\endgroup$ – Federico Poloni Nov 18 at 14:23 |. The arrows denote eigenvectors corresponding to eigenvalues of the same color. −Composition of geometric transformations in 2D and 3D. Find something interesting to watch in seconds. But the plane rotation is not realistic. 3 Rotation Matrix We have seen the use of a matrix to represent a rotation. When modelling three dimensions on a two-dimensional computer screen, you must project each point to 2D. Introduction As with strain, transformations of stress tensors follow the same rules of pre and post multiplying by a transformation or rotation matrix regardless of which stress or strain definition one is using. e you want matrix C such that [C] U1 = Ur. Defining the rotation axis as the z axis, we note first that the z coordinate will be unchanged by any rotation about the z axis. Matrix Rotations and Transformations. Try your hand at some online MATLAB problems. The underlying object is independent of the representation used for initialization. 3D Rotations Rotation about z-axis. Rotation (rotation_matrix, skip_checks=False) [source] ¶ Bases: DiscreteAffine, Similarity. This is defined in the Geometry module. Rotation Matrix Java. Both proposed algorithms aim at smoothing 3D rotation matrix sequences in a causal way. Euler angles provide a way to represent the 3D orientation of an object using a combination of three rotations about different axes. 3D Rotation Matrix. CS4620/5620: Lecture 5 3D Transforms (Rotations) •A rotation in 3D is around an axis -so 3D rotation is w. I've been following a tutorial for creating a game engine and when I got to calculating the 3D Rotation matrix I ran into the problem that I believe the matrix isn't being calculated properly. Again, the righmost matrix is the operation that occurs first. Again, we must translate an object so that its center lies on the origin before scaling it. HelloJava3Db renders a single, rotated cube: 10. After the first rotation you can specify which rotation shall follow in the 2nd column and for the third rotation you the remaining axis will be automatically selected for you in the 3rd column. pptx), PDF File (. I have a point, in that mesh, that i must calculate mannualy. This article discusses the different types of matrices including linear transformations, affine transformations, rotation, scale, and translation. Thus, we have H O = [I O] ω ,. If you don't want any rotation you can use the built-in constant Matrix3dIdentity. Rotation around any given axis Rotation from normal vector to normal vector Apparently the 5th function is enough, because for example "Rotation around X axis" can be replace by rotation around (1,0,0), and "Rotation around all axes" is merely the product of 3 matrices. Practice: Rotate 2D shapes in 3D. Euler returns a 3D vector containing the XYZ Euler angles. Rotation in 3D That works in 2D, while in 3D we need to take in to account the third axis. What do the vectors mean in T? T is a 4*4 column-major matrix. 0625 rz = -0. Except as otherwise noted, the content of this page is licensed under the Creative Commons Attribution 4. ggb file along with images of a red arrow and ball (that represent the spiral) at different angles. represent data on 3D rotation groups. NEGATIVE_DETERMINANT - this matrix has a negative determinant. As I understand, the rotation matrix around an arbitrary point, can be expressed as moving the rotation point to the origin, rotating around the origin and moving back to the original position. Moreover, there are similar transformation rules for rotation about and. 3d curl intuition, part 2. Rotation direction is from the first towards the second axis. A rotation matrix is a matrix used to rotate an axis about a given point. Bobick Calibration and Projective Geometry 1 Projection equation • The projection matrix models the cumulative effect of all parameters • Useful to decompose into a series of operations **** **** **** 1 X sx Y sy Z s = =. Alternatively, you can set the Orientation output to Rotation Matrix directly. These are not the only possible rotations in 3-space, of course, but we will limit our. Coordinate axes rotations:-Three dimensional transformation matrix for each co-ordinate axes rotations with homogeneous co-ordinate are as given below. 3D rotation is not same as 2D rotation. My matrices are rows by columns, row 0 being the topmost and column 0 being the leftmost. Or is it possible to convert my 3x3 rotation matrix to a Quaternion which let me use4x4Matrix. A non-rotation is described by an identity matrix. This property allows you to rotate, scale, move, skew, etc. User Interfaces with Java. The Camera Transformation Matrix: The transformation that places the camera in the correct position and orientation in world space (this is the transformation that you would apply to a 3D model of the camera if you wanted to represent it in the scene). When modelling three dimensions on a two-dimensional computer screen, you must project each point to 2D. This is the currently selected item. Transformation of Stresses and Strains David Roylance Department of Materials Science and Engineering Massachusetts Institute of Technology Cambridge, MA 02139 # Display transformation matrix for these angles: "evalf" evaluates the # matrix element, and "map" applies the evaluation to each element of # the matrix. Or is there a way to convert my 3x3 rotation matrix and translation to Unity 4x4Matrix since then i can use Matrix4x4. been stuck for about a month on this - i use "euler angles" (tait-bryan angles) to describe rotation coordinates. This list is useful for checking the accuracy of a rotation matrix if questions arise. Define the parametric surface x(u,v), y(u,v), z(u,v) as follows. R = rotx(ang) creates a 3-by-3 matrix for rotating a 3-by-1 vector or 3-by-N matrix of vectors around the x-axis by ang degrees. 0 License , and code samples are licensed under the Apache 2. This paper provides a basic introduction to the use of quaternions in 3D rotation applications. , robotics,. Define and Plot Parametric Surface. Properties of a rotation matrix In three dimensions, for any rotation matrix , where a is a rotation axis and a rotation angle, (i. If you want 3x3, just remove the last column and last row. Rotations in Three-Dimensions: Euler Angles and Rotation Matrices Part 2 - Summary and Sample Code. 33× rotation matrix equals a skew-symmetric matrix multiplied by the rotation matrix where the skew symmetric matrix is a linear (matrix-valued) function of the angular velocity and the rotation matrix represents the rotating motion of a frame with respect to a reference frame. Or is it possible to convert my 3x3 rotation matrix to a Quaternion which let me use4x4Matrix. Composition and inversion in the group correspond to matrix multiplication and inversion. public Matrix3DTransformation(double[][] matrix) Constructs a 3D transformation using the given matrix. Try a 90 degree rotation and then check. The theory is given here. Now you have magically gotten out of a room with a closed door! Similarly, to rotate about a point that is not the origin, first you move all the points so the center is the origin, use the usual rotation matrix, and then move all the points back to where you found them. Auckland's prof. Model Rotation. It is moving of an object about an angle. 7 Transformation Matrix and Stiffness Matrix in Three-Dimensional Space. Although the inverse process requires a choice of rotation axis between the two alternatives, it is a straightforward procedure to retrieve the rotation axis and angle (see Appendix A). I've read on page 27 here that a 3x3 transform matrix can be just the nine dot products - thank you U. The rotation is performed clockwise, if you are looking along the direction of the rotation axis vector. Transformations is a Python library for calculating 4x4 matrices for translating, rotating, reflecting, scaling, shearing, projecting, orthogonalizing, and superimposing arrays of 3D homogeneous coordinates as well as for converting between rotation matrices, Euler angles, and quaternions. Under rotations, vector lengths are preserved as well as the angles between vectors. The function uses the Rodrigues formula for the computation. A rotation vector is a convenient and most compact representation of a rotation matrix (since any rotation matrix has just 3 degrees of freedom). skip_checks (bool, optional) – If True avoid sanity checks on rotation_matrix for performance. Transformations, continued 3D Rotation 23 r r r x y z r r r x y z r r r x y z z y x r r r r r r r r r, , , ,, , , ,, , , , 31 32 33 Full 3D Rotation 0 sin cos 0 cos sin 1 0 0 sin 0 cos 0 1 0 cos 0 sin 0 0 1 sin cos 0 - Multiply the current matrix by the rotation matrix that. Understanding transforamtion matrix requires some knoledge of math… R. A rotation matrix describes the relative orientation of two such frames. This paper provides a basic introduction to the use of quaternions in 3D rotation applications. square matrix – For 2D, 3x3 matrix – For 3D, 4x4 matrix. This format is definitely less intuitive than Euler angles, but it's still readable: the xyz components match roughly the rotation axis, and w is the acos of the rotation angle (divided by 2). Gimbal lock When two rotational axis of an object pointing in the same direction, the rotation ends up losing one degree orientation matrix ( quaternion can be represented as matrix as well) quaternions or orientation matrix Euler angles, quaternion (harder) Summary. The center of a Cartesian coordinate frame is typically used as that point of rotation. Rotation rotate() Rotates an element around a fixed point on the 2D plane. AlignmentRotation (source, target, allow_mirror=False) [source] ¶ Bases: HomogFamilyAlignment, Rotation. jl package), and acts to rotate a 3-vector about the origin through matrix-vector multiplication. Eigen's Geometry module provides two different kinds of geometric transformations:. User Interfaces with Java. It turns out that the derivative R_ of a rotation matrix Rmust always be a skew symmetric matrix wb times R– any-thing else would be inconsistent with the contraints of orthogonality and determinant 1. A 3D rotation is a 2D rotation that is applied within a speci ed plane that contains the origin. When we multiply two rotation matrices, the result is a new matrix that is equivalent to performing the two rotations sequentially. The only thing new in the C++ code is the usage of GetConsoleScreenBufferInfo and SetConsoleTextAttribute which gets the size of the console and sets the text color. −OpenGL matrix operations and arbitrary geometric transformations. Rotation formalisms in three dimensions: | In |geometry|, various |formalisms| exist to express a |rotation| in three |dimensions| a World Heritage Encyclopedia, the aggregation of the largest online encyclopedias available, and the most definitive collection ever assembled. Quat returns a quaternion representing the. Consider the original set of basis vectors, i, j, k, and rotate them all using the rotation matrix A. Orthonormalize a Rotation Matrix By Mehran Maghoumi in 3D Geometry , MATLAB If you use a 3×3 R matrix to store the result of the multiplication of a series of rotation transformations, it could be the case that sometimes you end up with a matrix that is not orthonormal (i. beams in 3D space, a transformation we call a3D beam shear. A matrix with M rows and N columns is defined as a MxN matrix. 0/Image/3D/Matrix/Rotation. 3D rotation matrix around vector. There are several modes available to specify rotation matrices. Position Cartesian coordinates (x,y,z) are an easy and natural means of representing a position in 3D space …But there are many other representations such as spherical converted to matrix form to perform rotation. Simply put, a matrix is an array of numbers with a predefined number of rows and colums. If you want to rotate around an arbitary axis you can use the equations under "Rotation matrix from axis and angle" To understand how a rotation matrix works you should know that each column represents a 3D Vector which in turn represents one of the axis of the rotated coordinate system. Transformations about a Plane. Prove that this linear transformation is an orthogonal transformation. muting any, we clearly need a negative unit matrix, namely, Proper Rotation. The default polygon is a square that you can modify. a unit quaternion 'q' can represent all 3d rotations by q=exp(p), where 'p' is a pure imaginary. Call R v(θ) the 2x2 matrix corresponding to rotation of all vectors by angle +θ. Raises: ValueError: If the shape of quaternion is not supported. This form will allow you to rotate a vector along an arbitrary axis (in three dimensions), by an arbitrary angle. Rotation matrices are used in computer graphics and in statistical analyses. These singularities are not characteristic of the rotation matrix as such, and only occur with the usage of Euler angles.
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https://math.stackexchange.com/questions/616106/when-is-the-moment-of-inertia-of-a-smooth-plane-curve-is-maximum | # When is the moment of inertia of a smooth plane curve is maximum?
Given a smooth plane curve $(x(s),y(s))$, parameterized in arc length $s$, of fixed finite length $L$, its moment of inertia about its center of mass (axis perpendicular to the plane) is given as $$MI = \int_0^L ((x(s)-x_{cm})^2 + (y(s)-y_{cm})^2) ds$$. What I predict from earlier discussions, and almost convinced is that if we fix length $L$, $MI$ is maximum when the curve is a straight line. I lack the faculty of mathematical machinery (I guess calculus of variations) to prove it, hence is my gentle request to help me out in proving it and thoroughly understanding the situation and all the corollaries and nuances. This not just the result I need, but I want to do more with it and hence would like understand all the things that are making this result and even more general ones (only to plane curves though).
• It certainly is essential that your curve be connected (which could be surmised from your explicit parametrization statement). – Ted Shifrin Dec 23 '13 at 5:03
• @Ted Shifrin : yes, agreed. thanks for pointing. – Rajesh Dachiraju Dec 23 '13 at 5:21
• Can you explain the relation between this question and a similar one ? – Tony Piccolo Dec 23 '13 at 16:54
• @TonyPiccolo : In Chris Culter 's answer of that question, If the curve is always an arc of a circle, he shows that the moment of inertia is an increasing function of $r$, the radius of curvature. But my question now here is that, the straight line has infinite radius of curvature, hence I wonder if the straight line is the global maximum? – Rajesh Dachiraju Dec 23 '13 at 17:13
• F.A.Valentine has written Curves of given length and minimum or maximum moments of inertia (1934) but I cannot read it. Can you ? – Tony Piccolo Jan 3 '14 at 8:33
There is no need for any calculus of variation. Ordinary calculus is enough.
For simplicity of derivation, we will use complex numbers to represent points on the plane.
Let $z(s) = x(s) + i y(s)$ and WOLOG, we will assume $z(0) = 0$. We can express the position on our curve as an integral:
$$z(t) = \int_0^t z'(s)\;ds$$
Let $\theta(t) = \begin{cases} 1 & t > 0\\ 0 & t \le 0\end{cases}$ be the step function. The center of mass is given by
\begin{align}z_{cm} &= \frac{1}{L} \int_0^L z(t) dt = \frac{1}{L} \int_0^L \int_0^t z'(s)\;ds\; dt\\ &= \frac{1}{L} \iint_{[0,L]^2} \theta(t-s) z'(s)\;ds\; dt = \int_0^L \left(1-\frac{s}{L}\right) z'(s)\;ds \end{align}
The moment of inertia w.r.t. $z(0)$, the origin, is given by
\begin{align} \mathcal{M}_0 &= \int_0^L |z(s)|^2 ds = \int_0^L \left(\int_0^t z'(s_1)ds_1\right)\left(\int_0^t \bar{z}'(s_2) ds_2\right) dt\\ &=\iiint_{[0,L]^3} \theta(t-s_1)\theta(t-s_2) z'(s_1) \bar{z}'(s_2) ds_1 ds_2 dt\\ &=\iiint_{[0,L]^3} \theta(t-\max(s_1,s_2)) z'(s_1) \bar{z}'(s_2) ds_1 ds_2 dt\\ &= \iint_{[0,L]^2} \left(L - \max(s_1,s_2)\right) z'(s_1) \bar{z}'(s_2) ds_1 ds_2 \end{align} and hence the moment of inertia w.r.t. the center of mass is
$$\mathcal{M}_{cm} = \mathcal{M}_0 - L |z_{cm}|^2 = L \iint_{[0,L]^2} \Lambda(s_1,s_2) z'(s_1) \bar{z}'(s_2) ds_1 ds_2$$
where \begin{align}\Lambda(s_1,s_2) &= 1 - \max\left(\frac{s_1}{L},\frac{s_2}{L}\right) - \left(1-\frac{s_1}{L}\right)\left(1-\frac{s_2}{L}\right)\\ &= \left( 1 - \max\left(\frac{s_1}{L},\frac{s_2}{L}\right)\right)\min\left(\frac{s_1}{L},\frac{s_2}{L}\right) \end{align}
Since $|z'(s)| \equiv 1$ and $\Lambda(s_1,s_2) > 0$ for $(s_1,s_2) \in (0,L)^2$, we can bound $\mathcal{M}_{cm}$ as
$$\mathcal{M}_{cm} \le L \iint_{[0,L]^2} \Lambda(s_1,s_2) |z'(s_1)||z'(s_2)| ds_1 ds_2 = L \iint_{[0,L]^2} \Lambda(s_1,s_2) ds_1 ds_2$$ Notice the equality in above inequality is achieved when and only when $z'(s)$ is a constant.
We can conclude $\mathcal{M}_{cm}$ is largest for straight lines.
• Impressive! The kernel $\Lambda$ would look much better if you took $L=1$ (obviously WOLOG). – Post No Bulls Dec 26 '13 at 6:43
• @achille : Its actually mind blowing how you arrived at this using algebra. I wonder whether there are any geometric interpretations of the expression for $\mathcal(M_{cm}$, or can we arrive at it using geometrical arguments. It would be very interesting to know about such a thing. – Rajesh Dachiraju Dec 26 '13 at 11:37
• @RajeshD I don't have any geometric interpretation for $\mathcal{M}_{cm}$. This approach is motivated by visualizing the curve as a chain of line segments with small fixed lengths. Since the only degree of freedoms are the directions of the line segments and we know in certain sense, the moment of inertia $\mathcal{M}_{cm}$ is a non-negative quadratic functions of these degree of freedoms. I try to express the dependence explicitly and see what can be done. – achille hui Dec 26 '13 at 11:56
• @Robjohn and achillehui : As per my perception Robjohn's answer seems to be elegant, but I somehow like the expression for moment of inertia $MI_cm$ derived in Achille's answer which explicitly shows the only degree of freedom angles of the tangents and says it is maximum when the angle is a constant function. – Rajesh Dachiraju Jan 2 '14 at 9:07
Assumptions:
1. the center of mass is $0$ $$0=\int_0^L\delta f\,\mathrm{d}s\tag{1}$$ 2. $f$ is parametrized by arc length: $f'\cdot f'=1$ \begin{align} 0&=\int_0^Lf'\cdot\delta f'\,\mathrm{d}s\\ &=\int_0^Lf'\cdot\,\mathrm{d}\delta f\\ &=\Big[\,f'\cdot\delta f\,\Big]_0^L-\int_0^Lf''\cdot\delta f\,\mathrm{d}s\\ &=-\int_0^Lf''\cdot\delta f\,\mathrm{d}s\tag{2} \end{align} Note that $\delta f$ can be adjusted in the direction of $f'$ near $0$ and $L$ without affecting the integral of $f''\cdot\delta f$ since $f'\cdot f''=0$. Thus, we can assume $\Big[\,f'\cdot\delta f\,\Big]_0^L=0$.
3. the moment of inertia is stationary $$0=\int_0^Lf\cdot\delta f\,\mathrm{d}s\tag{3}$$ Conclusions:
For an extreme $f$, any $\delta f$ that satisfies $(1)$ and $(2)$ also satisfies $(3)$; thus, linearity says that there are constants $a$ and $b$ so that $$f=a+bf''\tag{4}$$ Since $f$ is parameterized by arclength $f'\cdot f''=0$, integrating the dot product of $(4)$ with $f'$ yields $$\frac12f\cdot f=a\cdot f+c\tag{5}$$ Equation $(5)$ represents an arc of the circle $$|f-a|=\left(2c+|a|^2\right)^{1/2}\tag{6}$$ or in the extreme case, a line segment.
Checking Possible Arcs:
The equation of an arc of radius $r$ is $$f=r(\cos(s/r),\sin(s/r))\tag{7}$$ The center of mass is $$\frac1L\int_{-L/2}^{L/2}r(\cos(s/r),\sin(s/r))\,\mathrm{d}s=\left(\frac{2r^2}{L}\sin\left(\frac{L}{2r}\right),0\right)\tag{8}$$ The moment of inertia is \begin{align} &\frac{r^2}{L}\int_{-L/2}^{L/2}\left[\left(\cos(s/r)-\frac{2r}{L}\sin\left(\frac{L}{2r}\right)\right)^2+\sin^2(s/r)\right]\,\mathrm{d}s\\ &=\frac{r^2}{L}\int_{-L/2}^{L/2}\left[1-\frac{4r}{L}\sin\left(\frac{L}{2r}\right)\cos(s/r)+\frac{4r^2}{L^2}\sin^2\left(\frac{L}{2r}\right)\right]\,\mathrm{d}s\\ &=r^2-\frac{4r^4}{L^2}\sin^2\left(\frac{L}{2r}\right)\\ &=r^2\left(1-\frac{\sin^2\left(\frac{L}{2r}\right)}{\left(\frac{L}{2r}\right)^2}\right)\tag{9} \end{align} $(9)$ increases to $\frac{L^2}{12}$ as $r\to\infty$. Thus, the maximal moment of inertia would be at the extreme case of a line segment.
• Thanks for the answer. There are some things I don't understand. In the begining what you mean by $\delta f$ and other things, I need a bit explanation on your considerations. How you claim that $f$ need to be an arc? etc,. Also at the end I am not able to understand how we get $L^2/12$ as $r$ goes to $\infty$. – Rajesh Dachiraju Dec 27 '13 at 3:03
• @RajeshD: $\delta f$ is a small instantaneous perturbation of $f$. More precisely, $$\delta\int F(f(x))\,\mathrm{d}x=\int F'(f(x))\delta f(x)\,\mathrm{d}x$$ – robjohn Dec 27 '13 at 9:21
• \begin{align}\lim_{r\to\infty}r^2\left(1-\frac{\sin^2\left(\frac{L}{2r}\right)} {\left(\frac{L}{2r}\right)^2}\right) &=\lim_{r\to\infty}\frac{\frac{L^2}{4}} {\left(\frac{L}{2r}\right)^2} \left(1-\frac{\sin^2\left(\frac{L}{2r}\right)} {\left(\frac{L}{2r}\right)^2}\right)\\ &=\lim_{x\to0}\frac{L^2/4}{x^2}\left(1-\frac{\sin^2(x)} {x^2}\right)\\ &=\lim_{x\to0}\frac{L^2/4}{x^2}\left(1-\frac{\sin(x)} {x}\right)\left(1+\frac{\sin(x)} {x}\right)\\ &=\lim_{x\to0}\frac{L^2}{2}\left(\frac{x-\sin(x)} {x^3}\right)\\ &=\lim_{x\to0}\frac{L^2}{2}\left(\frac{\cos(x)} {6}\right)\\&=\frac{L^2}{12}\end{align} – robjohn Dec 27 '13 at 9:52
There is no maximum ( monotonous increase) , but only a minimum.
Using polar coordinates, when object and constraint functions are together, variational problem is
$\int r^2 ds - \lambda^2 \int ds,$ where $ds= \sqrt{(r^2 + r^{'2})} d \theta$
Lagrangian $(r^2 - \lambda^2)$ is independent of $r^{'}$ when considered with respect to arc $s$. So it is not a variational problem.
Minimum M of I when $r$ is independent of $\theta$, or when $r$ is a constant. If arc length L is given, minimizing solution is for constant radius loop $L/ (2 \pi).$( Cowboy lasso)
I am not forgetting about he previous post, but the attached image proves that circles are also extrema of the momentum of inertia.
Although a few minutes testing will be sufficient to convince that probably circles are unstable extrema (minimum), while straight lines are stable extrema (maximum).
This is a classical problem of the calculus of variations, and its is surprising it is not in the collections of classical applications.
Intuitively, it is the shape taken by a lasso, a rope in rotation around a fixed point (which will also be the center of mass, as will be able to check on the result). In facts, playing with a lasso shows that we can expect circles as unstable solutions and straight lines as stable solutions.
The computation is hard to understand and involves sophisticated use of the calculus of variation. Lets, after choosing appropriate axes, maximize $$M = \int_a^b (x^2 + y^2) ds, \text{ subject to } L= \int_a^b ds \text{, in which }ds=\sqrt{1+y'^2}.$$
The first thing to do is to define the derivative of $M$ and $L$ when looked as function of the curve $y=y(x)$. The so called functional derivative of $$F = \int_a^b f(x,y,y') dx,$$ is shown to be $$\delta F = \frac {\partial f} {\partial y} - \frac d {dx} \frac{\partial f} {\partial y'}.$$
A complete demonstration can be found here. Basically, you replace $y$ by $y+\epsilon \: \eta$ where $\epsilon \rightarrow 0$ is a number and $\eta$ a fixed function. Then you compute the derivative is the usual way: $(f(x,y+\epsilon \: \eta, (y+\epsilon \: \eta)') - f(x,y, y')) / \epsilon$. Integrating by part replaces the $(y+\epsilon \: \eta)'$ term by $-\frac d {dx} \frac{\partial f} {\partial y'}$ and a term outside the integral sign, which vanishes because the end points $a$ and $b$ are fixed. Because the formulas are linear, the $\epsilon$ cancels, giving a result $\int {\delta F} \eta dx$, valid for any $\eta$ thus the result.
The functional derivatives are second order differentials that can be computed formally. For $L$ we get: $$\delta L = \frac \partial {\partial y} \sqrt{1+y'^2} - \frac d {dx} \left[ \frac \partial {\partial y'} \sqrt{1+y'^2}\right]= 0 - \frac d {dx} \left[ \frac{y'}{\sqrt{1+y'^2}} \right],$$ the first term is null because $ds$ depend only on $y'$ and not on $y$, the second is a derivative when $ds$ is looked as a function of on $y'$. We can the pursue with the usual derivative as a function of $x$: $$\delta L=- \frac{y'' \sqrt{1+y'^2} - y' (\sqrt{1+y'^2})'}{1+y'^2} = \cdots = \frac{y''}{(1+y'^2)^{3/2}}.$$
For $M$ we have:
$$\delta M = \frac \partial {\partial y} \left[ (x^2+y^2) \sqrt{1+y'^2}\right]- \frac d {dx} \frac \partial {\partial {y'}} \left[ (x^2+y^2) \sqrt{1+y'^2}\right],$$ thus, $$\delta M = 2y \sqrt{1+y'^2} - \frac d {dx} \left[ (x^2+y^2) \frac \partial {\partial {y'}} \sqrt{1+y'^2}\right] \\= 2y \sqrt{1+y'^2} - \frac d {dx} \left[ (x^2+y^2) \frac {y'} {\sqrt{1+y'^2}}\right] \\= \frac{2y (1+y'^2)}{\sqrt{1+y'^2}} - (2x+2yy') \frac {y'} {\sqrt{1+y'^2}} - (x^2+y^2) \frac {y''} {(1+y'^2)^{3/2}} \\= \frac{2y-2x y'}{\sqrt{1+y'^2}} - \frac {(x^2+y^2)y''}{(1+y'^2)^{3/2}}$$
To solve the question of the function of a given length with the highest moment of inertia, we have to introduce a Lagrange multiplier. It express at the extremum of a function $M$ subject to a condition $L = C^{te}$, the tangents planes of $M$ and $L$ are parallels. Here, the condition means that is exist a constant $\mu$, called the Lagrange-multiplier, such that $\delta M = \mu \delta L$.
This equation, known as the Euler-Lagrange equation, says there exist $\mu$ such that $$\frac{2y-2x y'}{\sqrt{1+y'^2}} - \frac {(x^2+y^2)y''}{(1+y'^2)^{3/2}} = \mu \frac{y''}{(1+y'^2)^{3/2}},$$ which is the same as $$(x^2+y^2+\mu)y'' = 2(y-x y')(1+y'^2)$$
Which needs some checks to be continued
• How did you assume that the moment of inertia of the curve about its center of mass to be $$M = \int_a^b y^2 dx$$, ofcourse you might have chosen origin as center of mass, but it is still wrong as per the definition given in the question. – Rajesh Dachiraju Dec 31 '13 at 20:56
• I really don't get what you are trying to say, as per my understanding both the answers given by Robjohn and Achille are correct. They explicitly use arc length parameterization and also assume curve to be of fixed length $L$. – Rajesh Dachiraju Dec 31 '13 at 20:59
• Of course, I am choosing the origin at the center of mass, computations are difficult enough. – AlainD Jan 1 '14 at 11:47
• b) You are write, I am on the wrong question. I was looking for the curve minimizing the moment on inertia about an axis ("chosen" as Ox). I'll edit in the post. – AlainD Jan 1 '14 at 11:54
• c) No Robjohn and Achille did not took into account that the curve length is constant. They use ∫ds = L, not d(∫ds)=0. In facts, if you follow their reasoning to search the curve maximizing the include area (or the lowest center of gravity), you also find straight lines, not circles (or catenaries). – AlainD Jan 1 '14 at 12:05 | 2019-05-23T01:56:46 | {
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https://lukaneg.github.io/projects/prime-num-optim.html | # Finding Prime Numbers¶
## Summary¶
Here I explore coding a simple script for finding prime numbers using Python. I try my first attempt without any optimization by creating the script off the top of my head. As you will see, that script works but runs very slowly. I then follow some online lessons on how to optimize the script and get much faster results.
## Prime Finder version 1¶
Now, here is the code for figuring out if any one number is prime.
But let's allow the user to get all the prime numbers from 1 to the value they enter:
Cool! Now turn that into a function:
And now run this:
This seems to work, but let's time how long this takes for larger numbers since it has to search all those numbers and might take a while!
So, finding all primes between 1 and 10,000 required 8+ seconds! What about 100,000?
start = time.time() find_primes(100000) end = time.time() elapsed = end - start print(f"Time elapsed: ",{elapsed})
Ok, so after a few minutes I gave up and terminated the kernel. Is there a more efficient way to do this??
## Prime Finder 2.0¶
Need to optimize the prime number search space with each iteration so that the code doesn't just use brute force and check every single number. For example, we know that even numbers are not prime, so why not remove those first? The following methodology does that, but also removes multiples of every new prime number found.
Essentially we can work in reverse. Rather than finding each number that doesn't have any multiples, work through the list of multiples and remove all numbers from the final list that those multiples can create.
For example, the first prime number is 2, so we can remove all other values from the set that are multiples of 2. The next prime number is 3, so we can remove all other values from the set that are multiples of 3. The next is 5, since the 4 was already removed when the multiples of 2 were removed, and so on. Thus, the search list becomes smaller and smaller!
Now each time you repeat that set of code, you add values to the actual_primes set and remove values from the potential_primes set, making the search space smaller and smaller.
So let's repeat that process with a while loop:
Great! Now time to package it up into a function again:
Now let's time it again!
## All primes through 10,000:¶
Wow!! Only 0.006 seconds compared to 8+ seconds with version 1!
## All primes through 100,000:¶
Only 0.035 seconds for 100,000 compared to several minutes + with version 1
## All primes through 1,000,000:¶
Now for the mega test: search for all prime numbers from one to one million:
Less. Than. A. Second. WOW
## Conclusion¶
So, as you can see, code optimization can make all the difference between having time for happy hour or not :P
This was a great experience in learning how to optimize code, and that it is not just about the types of data structures (though I'm sure using sets here helped a ton), it's also about optimizing the algorithm itself so that it does as little reduntant work as possible. I credit Andrew Jones (at Data Science Infinity) for sharing this very elegant but powerful algorithm for cutting through the redundancy and efficiently extracting a list of prime numbers. | 2022-05-19T02:05:32 | {
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https://math.stackexchange.com/questions/2721368/empty-set-subsets-and-vacuous-truths | # Empty set, subsets, and vacuous truths
So I was trying to prove that a null set is a subset of any set.
First, to define when $A$ is a subset of $B$:
$$A \subseteq B \iff \forall x(x \in A \implies x \in B)$$
(At least I think that's right?)
So then consider the empty set $\emptyset$ for which $\forall x(x \not\in \emptyset)$ is true.
I tried to prove that the empty set is a subset of any other set, or:
$$\emptyset \subseteq B \iff \forall x(x \in \emptyset \implies x \in B) \vdash \text{T}$$
To me this seemed true because it was "vacuously true"... somehow. Like it makes sense to call it vacuously true that "all $0$ items in $\emptyset$ can be found in $B$, yep!" but that isn't satisfying to me, how do I "prove" this is the case?
Is $x \in \emptyset$ a false... statement? A false predicate? Something else? Something that results in false so that the implication itself is true. Is this a $(\text{F}\implies \text{F}) \vdash \text{T}$ thing?
Or is it the $\forall$ that makes it false somehow?
What if I had said $\exists x \in \emptyset$, this feels like it would certainly be false but again I can't prove why, it's just an intuition.
Can anyone clarify the correct definitions / implications and why they're true or false or what have you?
• You don't seem to be using the symbol $\vdash$ in the usual way (see en.wikipedia.org/wiki/Turnstile_(symbol) ). Normally $P\vdash Q$ means you have proven $Q$ using assumptions $P$. In this case you want to show $\vdash\emptyset\subseteq B$, and the vacuous implication you are looking for would be $F\vdash P$ (or equivalently $\vdash F\Rightarrow P$) for any $P$. – stewbasic Apr 4 '18 at 4:46
• $x\in\emptyset$ is a predicate. It's neither true nor false until you substitute for $x$, but of course, whatever $x$ you substitute makes it false. So, for every $x$, the statement $x \in \emptyset \implies x \in B$ is true. I don't know if this will help you, but I hope so. – saulspatz Apr 4 '18 at 5:03
• "x∈∅ is a predicate. It's neither true nor false until you substitute for x". True. But saying $x \in \emptyset$ is false for all $x$ is perfectly legitimate. – fleablood Apr 4 '18 at 5:22
• @stewbasic I was trying to say that "(a iif for all x(p implies q)) is true" is there a better symbol to use? Equal sign? – user525966 Apr 4 '18 at 13:18
• @user525966 You could write $\vdash\emptyset\subseteq B\Leftrightarrow\forall x(x\in\emptyset\Rightarrow x\in B)$ or just $\emptyset\subseteq B\Leftrightarrow\forall x(x\in\emptyset\Rightarrow x\in B)$. – stewbasic Apr 4 '18 at 21:09
The proof relies on Ex falso :
$\vdash \lnot P \to (P \to Q)$.
We have to apply it in the form :
$\lnot (x \in \emptyset) \to (x \in \emptyset \to x \in B)$.
We have (axiom or theorem) : $\mathsf {ZF} \vdash \forall x \ \lnot (x \in \emptyset)$.
By Universal instantiation we get : $\lnot (x \in \emptyset)$ and thus from Ex falso, by Modus Ponens : $(x \in \emptyset \to x \in B)$.
Finally, by Universal generalization we conclude with :
$\mathsf {ZF} \vdash \forall x \ (x \in \emptyset \to x \in B)$.
• What does it mean to have the turnstile on the far left with nothing before it? – user525966 Apr 4 '18 at 15:33
• @user525966 - that is a "law of logic" (in this case: of propositional logic, i.e. a tautology) and thus we can use in every theory. – Mauro ALLEGRANZA Apr 4 '18 at 15:35
• In my original post was I using turnstile incorrectly? What's the correct symbol to use there instead? Equal sign? – user525966 Apr 4 '18 at 15:53
• The "turnstile" $\vdash$ means derivable (in a system). Thus $\mathsf {ZF} \vdash \varphi$ means that $\varphi$ is a theorem of Zermelo-Frenkel set theory. – Mauro ALLEGRANZA Apr 4 '18 at 15:56
• Thus, you want to prove: $\mathsf {ZF} \vdash \forall B \ (\emptyset \subseteq B)$. It is equivalent to say that $\forall B \ (\emptyset \subseteq B)$ is TRUE in every model of the theory. – Mauro ALLEGRANZA Apr 4 '18 at 15:58
1)
$A\subset B$ if all elements of $A$ are elements of $B$. As $\emptyset$ has no elements, then all of them are in $B$.
That's vacuously true.
So $\emptyset \subset B$.
2)
$A\subset B$ if any elements not in $B$ are not in $A$ either. As any element that is not in $B$ is not in $\emptyset$ either, $\emptyset \subset B$.
That's true-true; nothing vacuous about it.
3)
$A \subset B$ if $x \in A \implies x \in B$ is true for all $x$. As $x \in \emptyset$ is always false and $FALSE \implies P$ is always true, $x \in \emptyset \implies x \in B$ is always true.
So $\emptyset \subset B$.
===
So for the most part, yes, they are vacuously true statements, or they are a false premise implies anything true statements.
But it's not all smoke and mirrors. A subset is "embedded" in the superset and everything you can pull out of the subset most come directly from the superset, and there is nothing in the subset that isn't in the superset.
All of those are true about an empty set and a set $B$ and, to me at least, the all feel directly true with no semantic slick word play or gimmicks. The empty skein of the the emptyset (with nothing in it) is embedded every where in the ether of existent space. That doesn't seem to me to be a "trick". And because nothing can be pulled out of the emptyset, if we are standing in the general vacinity of $B$ nothing can be pulled out of the empty set that isn't from $B$. That's a direct objective fact.
In general, the predicate logic statement that $\forall x(x \in A \implies x \in B)$ is written as $A \subseteq B$.
The empty set $\emptyset$ is the set that contains no elements. Therefore, the empty set is a subset of any set, that is, $\emptyset \subseteq X$ for all $X$. This is because the statement $x \in \emptyset$ is false for any $x$, so the imiplication
$$\forall x(x \in \emptyset \implies x \in X)$$
must be true. (See the truth table below for the for the implication connective.)
$$\begin{array}{c|l|c} \text{p} & \text{q} & \text{p \implies q} \\ \hline T & T & T \\ T & F & F \\ F & T & T \\ F & F & T \end{array}$$
Note that the bottom two rows of the truth table are vacuously true. | 2019-01-22T19:49:34 | {
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https://math.stackexchange.com/questions/3233799/an-exam-has-50-multiple-choice-questions-with-5-options | # an exam has 50 multiple choice questions with 5 options
An exam has 50 multiple choice questions. Each question has five answer options and each question has 2 grades A-. Assuming that "a student" has no prior knowledge and randomly guess on all questions exam,
1. Compute the expected mean for the student score
2. Compute the standard deviation for the student score
3. What is the probability that the student will succeed in the exam if you know the passing grade is 60?
4. What is the probability that student will get a zero grade ?? Now assume that all students have no prior knowledge and they all randomly guess on all questions exam : What is the expected success rate? How do you expect the proportion of students who will score less or equal to 20?
If you know that the questions were distributed regularly (uniformly) on the lectures of the course and that another student may submit the exam and only studied Half of the course's lectures but he did the study so thoroughly that he could answer any question from the part he was studying And correctly answered 50% of the exam questions correctly and the rest of the questions he answered Random?
a. What is the expectation of this student's degree?
b. what is the standard deviation of this student's grade?
b. What is the probability that this student will succeed in the exam if you knwo the passing grade is 60?
1. for A it is a binomial process with p=1/5 , q=4/5 and n=50 so the expected value is np but * 2 because of 2 grades , the variance is npq also * 2,, for 4 I would use the binomial formula for x= 0 ?? is that correct
• Welcome to MSE. You'll get a lot more help, and fewer votes to close, if you show that you have made a real effort to solve the problem yourself. What are your thoughts? What have you tried? How far did you get? Where are you stuck? This question is likely to be closed if you don't add more context. Please respond by editing the question body. Many people browsing questions will vote to close without reading the comments. – saulspatz May 21 at 0:59
• thanks I did it – Nidal May 21 at 1:29
It seems you already know how to find the mean and variance of a binomial random variable, you I will leave that part to you.
For a binomial random variable $$X$$ with $$n = 50$$ and $$p = 1/5,$$ you can use a normal approximation to binomial to get reasonable approximate answers to these questions. (You will get somewhat better approximations, if you use a continuity correction, but even then you can't expect to get more than two or three places of accuracy.)
Alternatively, you can use statistical software (or perhaps a statistical calculator) to to get answers exact to many decimal places.
Below are some answers from R statistical software, in which dbinom and pbinom, designate a binomial PDF and CDF, respectively. I assume each correct guess counts 2 points and that there is no penalty for incorrect guesses.
$$P(X \ge 30) = 1 - P(X \le 29) \approx 0$$
1 - pbinom(29, 50, .2)
[1] 6.936747e-10
$$P(X \le 10) = .5836.$$
pbinom(10, 50, .2)
[1] 0.5835594
$$P(X = 0) = .8^{50}.$$
dbinom(0, 50, .2); .8^50.
[1] 1.427248e-05
[1] 1.427248e-05
A student who has studied half of the material should have probability $$p = 0.6$$ of answering each question correctly. Why? Let $$Y$$ be the number of correctly answered questions.
$$P(Y \ge 30) = 1 - P(Y \le 29) = 0.5610.$$
1 - pbinom(29, 50, .6)
[1] 0.5610349
Using a normal approximation. The number $$Y$$ correct has $$E(Y) = np = 50(.6) = 30,$$ $$Var(Y) = 12,$$ $$SD(Y) = 3.4641.$$
Thus $$Y \stackrel{aprx}{\sim} \mathsf{Norm}(\mu=30, \sigma=3.4641)$$ and (standardizing) $$Z = \frac{Y - \mu}{\sigma} \stackrel{aprx}{\sim} \mathsf{Norm}(0, 1).$$
Hence, $$P(Y \ge 30) = 1 - P(Y < 29.5) \approx 0.557.$$ (If you standardize and use printed normal tables you will get about the same answer; not quite because of the rounding involved in using the table.)
1 - pnorm(29.5, 30, 3.4641)
[1] 0.5573831
The figure below shows the PDF of $$\mathsf{Binom}(50,.6)$$ (bars) along with the density function of $$\mathsf{Norm}(30, 3.4641)$$ (blue curve). The probability $$P(Y \ge 30)$$ is the sum of the heights of the lines to the right of the dotted red line. (This probability is approximated by the area under the normal density curve to the right of this vertical line.)
The R code for the figure is shown below:
y = 0:50; PDF = dbinom(y, 50, .6)
plot(y, PDF, type="h", lwd=2, main="PDF of BINOM(50,.35) with Normal Approx.")
curve(dnorm(x, 30, 3.4641), 0, 50, add=T, col="blue")
abline(h=0, col="green2"); abline(v=0, col="green2")
• that was helpful,,, when he studied half of the material why p=0.6 I didn't get it. – Nidal May 21 at 10:55
• when he has no prior knowledge p was 1/5 (like one option will be right from the five options), now when he has studied half of the material why p is 3/5=0.6. – Nidal May 21 at 14:06
• S = studied, N = not. P(S)=P(N) = 1/2. P(Corr) = 1P(S) + (1/5)P(N) = .5 + .5(.2) = .5+.1 = .6. Law of total probability. – BruceET May 21 at 15:54
• Unsolicited, so likely unwelcome advice: I have had a look at your history of activity on this site. In my opinion, although you ask some nice questions, you reveal much less of your work and thinking than optimal. (Even those bits often only on demand.) Looks as if you are taking a fine and well-taught course. Outsourcing too much of the problem solving is not a path toward effective learning. – BruceET May 21 at 16:20
• We can't be sure there are exactly 25 questions from the parts the student studies. The way I read it, there are on average 25 such questions. // Your way and mine both give expected nr correct as 30, but for slightly different distributions. Mine BINOM(50, .6), yours 25 + BINOM(25, .2). Same mean, mine has larger variance. – BruceET May 21 at 19:33 | 2019-06-16T01:26:43 | {
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https://picasso-project.eu/nigella-cook-srdcyz/set-notation-symbols-bd1612 | This section is to introduce the notation to the reader and explain its usage. Set theory is one of the foundational systems for mathematics, and it helped to develop our modern understanding of infinity and real numbers. Intersection and union of sets. Compact set notation is a useful tool to describe the properties of each element of a set, rather than writing out all elements of a set. They are { } and { 1 }. Probability and statistics symbols table and definitions - expectation, variance, standard deviation, distribution, probability function, conditional probability, covariance, correlation Crow's foot notation, however, has an intuitive graphic format, making it the preferred ERD notation for Lucidchart. The individual objects in a set are called the members or elements of the set. The guide you are now reading is a “legend” to how we notate drum and percussion parts when we engrave music at Audio Graffiti. Thankfully, there is a faster way. Usually, you'll see it when you learn about solving inequalities, because for some reason saying "x < 3" isn't good enough, so instead they'll want you to phrase the answer as "the solution set is { x | x is a real number and x < 3 }".How this adds anything to the student's understanding, I don't know. Basic set operations. Use set notation to describe: (a) the area shaded in blue (b) the area shaded in purple. If you … ALT Codes for Math Symbols: Set Membership & Empty Sets Read More » In this notation, the vertical bar ("|") means "such that", and the description can be interpreted as "F is the set of all numbers n, such that n is an integer in the range from 0 to 19 inclusive". The table below lists all of the necessary symbols for compact set notation. Mathematical Set Notation. Subset, strict subset, and superset. The domains and ranges used in the discrete function examples were simplified versions of set notation. Symbolab: equation search and math solver - solves algebra, trigonometry and calculus problems step by step Sets. Set Theory Symbols Posted in engineering by Christopher R. Wirz on Wed Feb 08 2017. That is OK, it is just the "Empty Set". Some notations for sets are: {1, 2, 3} = set of integers greater than 0 … You never know when set notation is going to pop up. Illustration: Demo. In this section, we will introduce the standard notation used to define sets, and give you a chance to practice writing sets in three ways, inequality notation, set-builder notation, and interval notation. If you want to get in on their secrets, you'll want to become familiar with these Venn diagram symbols. A set is a collection of objects, things or symbols which are clearly identified.The individual objects in the set are called the elements or members of the set. Typing math symbols into Word can be tedious. any. ... Set Language And Notation. of . There are many different symbols used in set notation, but only the most basic of structures will be provided here. Which is why the bulk of this follow-up piece covers the very basics of set theory notation, operations & visual representations extensively. Lots symbols look similar but mean different things. IGCSE 9-1 Exam Question Practice (Sets + Set Notation) 4.9 34 customer reviews. A set is a well-defined collection of distinct objects. Also, check the set symbols here.. Sets, in mathematics, are an organized collection of objects and can be represented in set-builder form or roster form.Usually, sets are represented in curly braces {}, for example, A = {1,2,3,4} is a set. Let’s kick off by introducing the two most basic symbols for notating a set & it’s corresponding elements. MS Word Tricks: Typing Math Symbols 2015-05-14 Category: MS Office. To fully embrace the world of professional Venn diagrams, you should have a basic understanding of the branch of mathematical logic called ‘set theory’ and its associated symbols and notation. Author: Created by Maths4Everyone. We attempt to follow the standards set out in Norman Weinberg’s Guide to Standardized Drumset Notation. 8 February 2019 OSU CSE 1. This cheat sheet is extremely useful. Sometimes the set is written with a bar instead of a colon: {x¦ x > 5}. Cardinality and ordinality Bringing the set operations together. The Universal Set … Consider the set $\left\{x|10\le x<30\right\}$, which describes the behavior of $x$ in set-builder notation. and symbols. Set theory starter. Set notation practice. Example: Set-Builder Notation: Read as: Meaning: 1 {x : x > 0}the set of all x such that x is greater than 0. any value greater than 0: 2 {x : x ≠ 11}the set of all x such that x is any number except 11. any value except 11: 3 {x : x < 5}the set of all x such that x is any number less than 5. any value less than 5 Note that it's unnecessary to load amsmath if you load mathtools. Email. Google Classroom Facebook Twitter. Set notation and Venn diagrams questions. Set Theory • A mathematical model that we will use often is that of . Set notation is an important convention in computer science. CCSS.Math: HSS.CP.A.1. Look at the venn diagram on the left. Shading task. In, sets theory, you will learn about sets and it’s properties. On the Insert tab, in the Symbols group, click the arrow under Equation, and then click Insert New Equation. Relative complement or difference between sets. When picking a symbol, best to trust the symbol's unicode name for its meaning, not appearance. Universal set and absolute complement. Researchers and mathematicians have developed a language and system of notation around set theory. Null set is a proper subset for any set which contains at least one element. Occasionally we will introduce a new symbol to cater for an unusual requirement of a client. S et theory is a branch of mathematics dedicated to the study of collections of objects, its properties, and the relationship between them. After school they signed up and became members. The table below contains one example set… Because rarely used symbol may look very different on another computer. While crow's foot notation is often recognized as the most intuitive style, some use OMT, IDEF, Bachman, or UML notation, according to their preferences. In set-builder notation, the set is specified as a selection from a larger set, determined by a condition involving the elements. The default way of doing it is to use the Insert > Symbols > More Symbols dialog, where you can hunt for the symbol you want. Symbol Symbol Name Meaning / definition Example { } set: a collection of elements: A = {3,7,9,14}, It is still a set, so we use the curly brackets with nothing inside: {} The Empty Set has no elements: {} Universal Set. The Wolfram Language has the world's largest collection of consistent multifont mathematical notation characters\[LongDash]all fully integrated into both typesetting and symbolic expression construction . Click the arrow next to the name of the symbol set, and then select the symbol set that you want to display. When using set notation, we use inequality symbols to describe the domain and range as a set of values. Let us discuss the next stuff on "Symbols used in set theory" If null set is a super set Problem 1: Mrs. Glosser asked Kyesha, Angie and Eduardo to join the new math club. This carefully selected compilation of exam questions has fully-worked solutions designed for students to go through at home, saving valuable time in class. Solution: Let P be the set of all members in the math Created: Jan 19, 2018 | Updated: Feb 6, 2020. Topics you will need to know in order to pass the quiz include sets, subsets, and elements. Hyperbolic functions The abbreviations arcsinh, arccosh, etc., are commonly used for inverse hyperbolic trigonometric functions (area hyperbolic functions), even though they are misnomers, since the prefix arc is the abbreviation for arcus, while the prefix ar stands for area. take the previous set S ∩ V ; then subtract T: This is the Intersection of Sets S and V minus Set T (S ∩ V) − T = {} Hey, there is nothing there! A variant solution, also based on mathtools, with the cooperation of xparse allows for a syntax that's closer to mathematical writing: you just have to type something like\set{x\in E;P(x)} for the set-builder notation, or \set{x_i} for sets defined as lists. This quiz and attached worksheet will help gauge your understanding of set notation. Under Equation Tools , on the Design tab, in the Symbols group, click the More arrow. State whether each … Below is the complete list of Windows ALT codes for Math Symbols: Set Membership & Empty Sets, their corresponding HTML entity numeric character references, and when available, their corresponding HTML entity named character references, and Unicode code points. The colon means such that.. For example: {x: x > 5}.This is read as x such that x is greater than > 5.. Set notation. The following table gives a summary of the symbols use in sets. For example, a set F can be specified as follows: = {∣ ≤ ≤}. Because null set is not equal to A. Admin Igcse Mathematics Revision Notes, O Level Mathematics Revision Notes 2 Comments 12,074 Views. Basic Set Theory . Set notation is used to help define the elements of a set. For example, let us consider the set A = { 1 } It has two subsets. Preview. elements . Quiz & Worksheet Goals Inequalities can be shown using set notation: {x: inequality}where x: indicates the variable being described and inequality is written as an inequality, normally in its simplest form. They wrote about it on the chalkboard using set notation: P = {Kyesha, Angie and Eduardo} When Angie's mother came to pick her up, she looked at the chalkboard and asked: What does that mean? Symbols Used in this Book; Glossary; While a comprehensive list of notation is included in the appendix, that is meant mostly as a reference tool to refresh the reader of what notation means. Purplemath. Basic set notation. The symbols shown in this lesson are very appropriate in the realm of mathematics and in mathematical logic.
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https://mybrightfuture.org/a0uppzk0/1myldcg.php?94011f=do-adjacent-angles-form-a-linear-pair | If two angles do not form a linear pair, then they are not supplementary. Linear Pair Two adjacent angles form a linear pair if their non-common sides form a straight angle. Axiom 1: If a ray stands on a line then the adjacent angles form a linear pair of angles. They just need to add up to 90 degrees. Contrapositive: If the angles are not adjacent, then the two angles do not form a linear pair. B A V CLinear Pair Postulate: If two angles form a linear pair, then they are supplementary. Therefore, all the exterior angles are equal, and can be found by dividing 360° by the number of angles . Linear A linear pair of angles is a pair of adjacent angles whose non common sides are opposite rays. The sides of the angles do not form two pairs of opposite rays. [Image will be Uploaded Soon] In the figure given above, all line segments are passing through the point O, as shown in the figure. Name a linear pair. 9 years ago. where ∠DBC is an exterior angle of ∠ABC and, ∠A and ∠C are the remote interior angles. Related Questions to study. Write. A) vertical angles B) complementary C) acute D) a linear pair * please answer only if you know As the ray OF lies on the line segment MN, angles ∠FON and ∠FOM form a linear pair. Question 1. Don't two angles ALWAYS have to be adjacent in order to form a line (linear pair)? Assignment Directions: Solve the following word problems involving linear pairs. When a pair of adjacent angles create a straight line or straight angle, they are a linear pair. Are two angles in the same plane with a common vertex and a common side, but no common interior points. The measure of one angle is 24 more than the measure of the other angle. As the ray OA lies on the line segment CD, angles ∠ AOD and ∠ AOC form a linear pair. B. Can you explain this answer? Linear PairA linear pair consists of two adjacent angles whose noncommon sides are opposite rays. For #1-6, use the figure at the right. Flashcards. a. Tell whether the angles are only adjacent, adjacent and form a linear pair, or not adjacent. 14. In which diagram do angles 1 and 2 form a linear pair. (iii) When two lines intersect opposite angles are equal. A linear pair is two angles that add up to be 180o.A linear pair is two adjacent, supplementary angles.Adjacent means they share ONE ray.Supplementary means add … The sum of the linear pair of angles is always equal to 180 degrees. A Linear Pair is two adjacent angles whose non-common sides form opposite rays. Match. Tell whether the angles are only adjacent, adjacent and form a linear pair, or not adjacent. The other two pairs of angles are adjacent, but they are not forming a linear pair. Two angles are said to be linear if they are adjacent angles formed by two intersecting lines. Name two acute vertical angles. Add your answer and earn points. If two angles are not adjacent, then they do not form a linear pair. ii) When non-common sides of a pair of adjacent angles form opposite rays, then the pair forms a linear pair. Since the non-adjacent sides of a linear pair form a line, a linear pair of angles is always supplementary. In the figure, ∠ 1 and ∠ 2 form a linear pair. However, just because two angles are supplementary does not mean they form a linear pair. A linear pair of angles is a pair of adjacent angles whose non common sides are opposite rays. Favorite Answer. Name an angle complementary to . View Answer . Basically, a linear pair of angles always lie on a straight line. Similarly, ∠GON and ∠HON form a linear pair and so on. The measure of one angle is 15 less than half the measurement of its supplement. For △ABC above, ∠A+∠C+∠ABC=180°. "Complementary Angles" These are angles that add up to 90 degrees. If the two complementary angles are adjacent then they will form a right angle. They share a common vertex, but not a … Consider a Ray OP Stand on the Line Segment as Shown: The angles which are formed at E are ∠QEM and ∠QEN. Two angles may be supplementary, but not adjacent and therefore not form a linear pair. A linear pair forms a straight angle of 180 degrees, so you have two angles whose sum is given as 180, which means they are supplementary. Therefore, the exterior angles measure = = 72° each. Name a pair of adjacent angles. Learn. Example 2: If two adjacent angles form a linear pair, show that their angle bisectors are perpendicular. When two lines intersect each other at a common point then, a linear pair of angles are formed. Such angle pairs are … Linear-pair-angles are always supplementary. <3 and <4 form a linear pair. Hope this helps! In the diagram above, ∠ABC and ∠DBC form a linear pair. Hence vertical angles are not adjacent. Linear Pair Of Angles. Related Questions to study. Two angles are called supplementary angles if Answer Save. How to Find Adjacent Angles. | EduRev Class 7 Question is disucussed on EduRev Study Group by 2108 Class 7 … B Supplementary angles and linear pairs both add to 180°. Two angles that add to 180 degrees and when adjacent form a straight line. If two angles are not adjacent, then they do not form a linear pair. Two adjacent angles are said to form a linear pair angles , if their non-common arms are two opposite rays. Adjacent angles do not overlap. S_Pinkston. If the angles are adjacent to each other after the intersection of the lines, then the angles are said to be adjacent. "Congruent Angles" They have the same angle in degrees or radian. As the adjacent angles form a linear pair and they are supplementary. <1 and <2 are adjacent angles. Linear Pair Angles. C. If two angles are not supplementary, then they do not form a linear pair. Pro Lite, Vedantu Example 2: Determine if the following pair of angles are vertical angles, linear pairs, or neither. ∴ Yes, ∠1 is vertically opposite to ∠4, And angle vertically opposite to ∠5 is ∠BOC. 1 See answer chanybonilla is waiting for your help. What can you say about its other angle? Example 2: Determine if the following pair of angles are vertical angles, linear pairs, or neither. Consider the following figure in which a ray $$\overrightarrow{OP}$$ stand on the line segment $$\overline{AB}$$ as shown: The angles ∠POB and ∠POA are formed at O. 12. The sum of their angles is 180° or π radians. ! In the figure above, all the line segments pass through the point O as shown. what are the two factors that make you decide whether a pair of adjacent a angles from a linear pair 2 See answers saryunahar141005 saryunahar141005 Answer: All adjacent angles do not form a linear pair. In the diagram below, ∠ABC and ∠DBE are supplementary since 30°+150°=180°, but they do not form a linear pair since they are not adjacent. Pro Subscription, JEE In which diagram do angles 1 and 2 form a linear pair. In the diagram above, ∠ABC and ∠DBC form a linear pair. By: Carol H. answered • 02/22/18. Also, the ray is that part of the line which has only one endpoint. Test. A linear pair of angles is formed when two lines intersect. Thus, the sum of the exterior angles is: For regular polygon, all of the angles of a are equal. so, a pair of straight angles can also be adjacent angle. Join Yahoo Answers and get 100 points today. The pair of adjacent angles are constructed on a line segment, but not all adjacent angles are linear. Learn how to identify angles from a figure. <2 and <4 <2 and <4. This can be shown by using linear pairs. When a pair of adjacent angles create a straight line or straight angle, they are a linear pair. It can't think of a contradiction. Consider the following figure in which a ray $$\overrightarrow{OP}$$ stand on the line segment $$\overline{AB}$$ as shown: The angles ∠POB and ∠POA are formed at O. Example 2 : In the diagram shown below, Solve for x and y. Get the answers you need now. 14. (ii) Adjacent complementary angles. Substituting the second equation into the first equation we get. Since the non-adjacent sides of a linear pair form a line, a linear pair of angles is always supplementary. D. If two angles form a linear pair, then they are supplementary. Linear pair. Supplementary angles are two angles whose same is 180o Linear pairs are adjacent angles who share a common ray and whose opposite rays form a straight line. A circle of the infinite radius can be visualized on the straight line. The figure shows the design on an outdoor fence. Gravity. Use ½ sheet of paper. Ans : (d) Bisectors of the adjacent angles forming a linear pair form a right angle. A linear pair is a pair of adjacent angles whose non-adjacent sides form a line. See tutors like this. 3. The sum of their angles is 180 ° or π radians. Learn vocabulary, terms, and more with flashcards, games, and other study tools. In the regular pentagon above, n = 5. 0 0. A linear pair of angles is a pair of adjacent angles whose non common sides are opposite rays. PLAY. Supplementary angles do not have to be adjacent, whereas a linear pair must be adjacent and create a straight line. Fill in the blanks: If two adjacent angles are supplementary, they form a _____. An angle is 18c less than its complementary angle. 1 See answer chanybonilla is waiting for your help. (iv) Unequal supplementary angles. a. b. c. Exploration #1: Use this link to explore the relationship with vertical angles. Linear pair is a pair of adjacent angles where non-common side forms a straight line. Answer:Pair of adjacent angles whose measures add up to form a straight angle is known as a linear pairStep-by-step explanation:see A linear pair forms a straig… 2 0. If two angles are supplementary, then they form a linear pair. They have common vertex O. i tried searching it but i wasn't sure cause it was so confusing at first i thought i was getting it but i wasn't. Correct answer to the question: 5. Also, there is a common arm that represents both the angles of the linear pair. A line segment with A and B as two endpoints is represented as AB. Spell. They do not form a straight line. Easily fill out PDF blank, edit, and sign them. They have common side OB. If a linear pair is formed by two angles, the uncommon arms of the angles forms a straight line. In a diagram angle 1 and angle 2 form an linear pair and angle 2 and angle 3 form vertical angles. The above discussion can be stated as an axiom. ii) When non-common sides of a pair of adjacent angles form opposite rays, then the pair forms a linear pair. If two angles form a linear pair and the measure of one angle is 47 degrees, what is the measure of the other angle. If two angles are adjacent, they form a linear pair? The measure of an angle is three times the measurement of its complement. Linear pair of angles - definition Two angles are said to be in a linear pair if they are adjacent to each other, lie on the same side of the line and the sum of their measures is 1 8 0 o. ∠BOC and ∠AOC are linear-pair-angles. Therefore, AB represents a line. Created by. 13. 1 Answer. A Linear Pair is two adjacent angles whose non-common sides form opposite rays. Vertically Opposite Angles When a pair of lines intersect, as shown in the fig. Also, ∠ABC and ∠DBC form a linear pair so. (iv) Unequal supplementary anglesFor line BD ∠ BOC and ∠ COD are in linear pair Therefore, ∠BOC + ∠COD = 180° So, their sum is 180° And, ∠BOC is not equal to ∠COD ∴ ∠BOC & ∠COD are unequal supplementary angles Ex 5.1, 14 In the adjoining figure, name the following pairs of angles. Such angle pairs are called a linear pair.. Angles A and Z are supplementary because they add up to 180°.. Vertical angles: When intersecting lines form an X, the angles on the opposite sides of the X are called vertical angles. Find the values of the angles l, p, and q in each of the following questions, ∠DOC = ∠AOE (vertically opposite angles), Pair of Linear Equations in Two Variables, Vedantu This statement is correct. (iii) Does the angles ∠COE and ∠EOD form a linear pair? ∠BOC and ∠AOC are linear-pair-angles. It is known that the angle between the two line segments ME and EN is 180°. Angles ∠ Z W I and ∠ H W I are adjacent angles. (vi) Find the vertically opposite angle of ∠5? In the figure, clearly, the pair $$\angle BOA$$ and $$\angle AOE$$ form adjacent complementary angles. Complete Identify Each Pair Of Angles As Adjacent, Vertical, Complementary, Supplementary, Or A Linear Pair online with US Legal Forms. A linear pair is two angles that add up to be 180o.A linear pair is two adjacent, supplementary angles.Adjacent means they share ONE ray.Supplementary … Add comment More. Anonymous. Supplementary angles: Two angles that add up to 180° (or a straight angle) are supplementary. Relevance. Angle Pairs (Vertical, linear pair, adjacent, complementary & supplementary) STUDY. The two axioms mentioned above form Linear Pair Axioms and are helpful in solving various mathematical problems. In a linear pair, the arms of the angles that are not always common or collinear i.e., they lie on a straight line. A: If two angles form a linear pair, then angles are supplementary. 1 and 2 are adjacent angles. If two parallel lines are intersected by a transversal, then each pair of corresponding angles so formed is (a) Equal (b) Complementary (c) Supplementary (d) None of these Ans : (a) Equal 4. If a point O is taken anywhere on the line segment AB as shown, then the angle between the two line segments AO and OB is a straight angle i.e., 180°. Two adjacent supplementary angles form a linear pair. Here, these angles are in linear pair as. Report 1 Expert Answer Best Newest Oldest. If two angles are a linear pair, then they are supplementary. T or F? There are n angles in the polygon, so there are n linear pairs. If Two Angles Form A Linear Pair, The Angles Are Supplementary. Two adjacent supplementary angles form a linear pair. A linear pair is a pair of adjacent, supplementary angles, whihc means they add up to 180 degrees. The measure of an angle is three times the measurement of its complement. 12. Linear Pairs. If two angles form a linear pair, the angles are supplementary. Hence, we can also say that a linear pair of angles is the adjacent angle whose non-common arms are basically opposite rays. Fill in the blanks: If two adjacent angles are supplementary, they form a _____. the term adjacent means next to each other. 1 decade ago. Two adjacent angles are said to form a linear pair angles , if their non-common arms are two opposite rays. In the figure, clearly, the pair $$\angle AOE$$ and $$\angle EOD$$ form adjacent angles that do not form a linear pair. The angles are adjacent, sharing ray BC, and the non-adjacent rays, BA and BD, lie on line AD. (add up to 180 0) ∠BOC + ∠AOC = 180 0 Examples : 1) One of the angles forming a linear-pair is a right angle. Then, find the angle measures. A straight angle is formed when the angle between two lines is 180 degrees. What are not adjacent angles? Two vertical angles are always the same size as each other. adjacent angles are the angles which have same pair of vertex and have one common side. Two angles form a linear pair. Sorry!, This page is not available for now to bookmark. Covid-19 has led the world to go through a phenomenal transition . Similarly, ∠GON and ∠HON form a linear pair and so on. Converse statement inverses statement Contrapositive statement Key Concepts: Terms in this set (12) Adjacent complementary angles (Image) angles that are "next to" each other. a. b. c. Exploration #1: Use this link to explore the relationship with vertical angles. Two adjacent angles are said to form a linear pair of angles, if their non-common arms are two opposite rays. Adjacent Angles• Share a common side• Share the same vertex• Do not share any interior points B A C V Bistro 72 Reviews, 5 Gallon Air Compressor, Clementine Churchill Images, How Old Is Chopper, Allegiant Air Flight Schedules, Counting Video For Kindergarten, Langerhans Cells Function Quizlet, Eine Kleine Nachtmusik Mr Bean, Improvisation Of Melody Brainly, Dynamite Definition Synonyme, Kazimir Malevich Art, | 2021-08-02T13:25:55 | {
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https://math.stackexchange.com/questions/2497461/prove-that-the-set-of-mathbbn-times-mathbbn-times-mathbbn-is-counta | Prove that the set of $\mathbb{N} \times \mathbb{N} \times \mathbb{N}$ is countable.
I have seen proofs of $\mathbb{N}\times\mathbb{N}$. The proof by list where you list each element in a pair of elements and then counting them diagonally is the most convincing. I have two thoughts in mind:
1. Write it as a composition of functions. Show that both functions are bijections, so the composition is also a bijection.
2. Show that $\mathbb{N} \times \mathbb{N} \times \mathbb{N}$ has the same cardinality as $\mathbb{N}\times\mathbb{N}$, and $\mathbb{N}\times\mathbb{N}$ has the same cardinality as $\mathbb{N}$.
Generally, I'm having difficulty going from $\mathbb{N}\times\mathbb{N}\times\mathbb{N}$ to $\mathbb{N}\times\mathbb{N}$, since it easy to prove the bijection from $\mathbb{N}\times\mathbb{N}$ to $\mathbb{N}$.
• Youre idea in 1. seems like a good approach. – JJC94 Oct 31 '17 at 0:48
• It's not a proof-explanation if you don't point at a specific proof and ask us to explain it. – Asaf Karagila Oct 31 '17 at 0:51
• Consider the Cantor pairing. – Wuestenfux Oct 31 '17 at 9:42
If you already know that $\Bbb{N\times N}$ is countable, fix a bijection $f\colon\Bbb{N\times N\to N}$. Now consider $g\colon\Bbb{N\times N\times N\to N\times N}$ defined as: $$g(n,m,k)=(n,f(m,k)),$$ or better yet $h\colon\Bbb{N\times N\times N\to N}$ defined as $$h(n,m,k)=f(n,f(m,k))$$ and cut out the middle-man.
As my freshman discrete mathematics professor used to tell us, go home and convince yourself this is correct.
• I can convince myself this is correct, but how can I prove it to a stubborn skeptic? – tigerustin Nov 1 '17 at 2:11
• Slowly and carefully by verifying the definitions of a bijection. – Asaf Karagila Nov 1 '17 at 6:53
• I proved injectivity. How would I prove surjectivity? – tigerustin Nov 1 '17 at 20:21
• Exactly the same. Slowly, carefully and by using the fact that $f$ is already surjective itself. – Asaf Karagila Nov 1 '17 at 20:23
• I really don't understand. Can you explain further please? I don't want to have to start another thread on this question. – tigerustin Nov 2 '17 at 2:10
Why not think in stages? Let $\varphi : \mathbb{N}\times\mathbb{N} \to \mathbb{N}$ denote your favorite bijection. But then
$$\mathbb{N} \times \mathbb{N} \times \mathbb{N} = (\mathbb{N} \times \mathbb{N}) \times \mathbb{N} \approx \varphi(\mathbb{N}\times \mathbb{N}) \times \mathbb{N} = \mathbb{N} \times \mathbb{N} \approx \varphi(\mathbb{N}\times \mathbb{N}) = \mathbb{N}$$ (where, in this context, $=$ means equality, and $\approx$ denotes "of the same cardinality"). You can do this for any number of copies of $\mathbb{N}$ via an induction argument.
Consider $f:\mathbb N \times \mathbb N \times \mathbb N \to \mathbb N$ given by $f(a,b,c)=2^a3^b5^c$
and use the Fundamental theorem of arithmetic for a cute solution.
• Gives an injective mapping. – Wuestenfux Oct 31 '17 at 9:41
• And it is clearly infinite, so... – Andres Mejia Oct 31 '17 at 14:06
• It would be difficult to prove bijectivity of this function? – tigerustin Nov 1 '17 at 2:12
• It’s not bijective. It is an infinite subset of the natural numbers – Andres Mejia Nov 1 '17 at 2:15
I like to think of 'countable' as 'listable' ... That is, as long as you can create a list of objects and ensure that all elements of some set $S$ are somewhere on that list, then $S$ is countable. To me, the advantage of that kind of thinking is that elements from $S$ may be repeated on the list, and that the list may even contain objects that are not in $S$; as long as every element of $S$ is at least once in the list, it's clear that $S$ is countable. As such, this method avoids having to construct explicit and possibly complicated bijections.
So, how can we create a list of all elements of $\mathbb{N} \times \mathbb{N} \times \mathbb{N}$?
Here is one way:
First, get all triples with elements that sum to $0$: which is just $<0,0,0>$
Now get all those that sum to $1$: $<1,0,0>,<0,1,0>,<0,0,1>$
Then those that sum to $2$, then $3$, etc.
Thus, the list is:
$<0,0,0>,<1,0,0>,<0,1,0>,<0,0,1>,<2,0,0>,<1,1,0>,...$
Since for any $n$ there can only be finitely many triples whose elements sum to $n$, you will eventually get to every triple.
Note that this is of course just the same thing as diagonally going through an array of elements, just now in $3$ dimensions.
• Reminiscent of the method I know for showing the algebraics are countable. Nice. – Kaj Hansen Oct 31 '17 at 2:34
Cantor-Bernstein's Theorems states that if $A$ and $B$ are two sets and there exists two injections $f:A \to B$ and $g:B \to A$ then there is a biyection from $A$ to $B$ (i.e. $A$ and $B$ have the same cardinality).
Let $f:\mathbb{N} \times \mathbb{N} \times \mathbb{N} \to \mathbb{N}$ be defined as $f(m,n,k)=2^n \cdot 3^m \cdot 5^k$, then if $f(m_1,n_1,k_1) = f(m_2,n_2,k_2) \Leftrightarrow 2^{m_1}3^{n_1}5^{k_1}=2^{m_2}3^{n_2}5^{k_2}$, the Fundamental Theroem of Arithmetic shows that $m_1=m_2$, $n_1=n_2$ and $k_1=k_2$. Then $f$ is a injective function.
Next, consider the function $g:\mathbb{N} \to \mathbb{N} \times \mathbb{N} \times \mathbb{N}$ defined by $g(n)=(0,0,n)$, then $$g(n)=g(m) \Leftrightarrow (0,0,n) =(0,0,m) \Leftrightarrow n=m$$ and hence $g$ is injective.
Applying Canotr-Bernstein's Theorem gives that $\mathbb{N} \times \mathbb{N} \times \mathbb{N}$ is countable.
Using induction, you can easily prove this result for the product of any finite number of copies of $\mathbb{N}$. The hardest part of the proof is the base case, which is that $\mathbb{N} \times \mathbb{N}$ is countable. It seems you already have this result.
For the induction hypothesis, assume that $\displaystyle \prod^n \mathbb{N}$ is countable. Now we just need to show that $\displaystyle \prod^{n+1} \mathbb{N}$ is countable. First, note that can write the $(n\!+\!1)$-fold product in the form $\mathbb{N} \times \left( \displaystyle \prod^n \mathbb{N} \right)$. A bijection $\displaystyle f:\prod^n \mathbb{N} \rightarrow \mathbb{N}$ is guaranteed by the induction hypothesis, which we can use to construct a bijection between the $(n\!+\!1)$-fold product and $\mathbb{N} \times \mathbb{N}$:
$$(x_1, x_2, \cdots, x_{n+1}) \mapsto (x_1, f(x_2, x_3, \cdots, x_{n+1}))$$
And $\mathbb{N} \times \mathbb{N}$ is countable per the base case. | 2019-12-07T04:20:50 | {
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http://math.stackexchange.com/questions/358755/ways-to-fill-a-n-times-n-square-with-1-times-1-squares-and-1-times-2-recta | # Ways to fill a $n\times n$ square with $1\times 1$ squares and $1\times 2$ rectangles
I came up with this question when I'm actually starring at the wall of my dorm hall. I'm not sure if I'm asking it correctly, but that's what I roughly have:
So, how many ways (pattern) that there are to fill a $n\times n:n\in\mathbb{Z}_{n>0}$ square with only $1\times 1$ squares and $1\times 2$ rectangles?
For example, for a $2\times 2$ square:
• Four $1\times 1$ squares; 1 way.
• Two $1\times 1$ squares and one $1\times 2$ rectangle; $4$ ways total since we can rotate it to get different pattern.
• Two $1\times 2$ rectangles; 2 ways total: placed horizontally or vertically.
$\therefore$ There's a total of $1+4+2=\boxed{7}$ ways to fill a $2\times 2$ square.
So, I'm just wondering if there's a general formula for calculating the ways to fill a $n\times n$ square.
Thanks!
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+1 for coming up with your own math question. – Sammy Black Apr 11 '13 at 20:58
For $1\times n$ boards, you get the Fibonacci numbers. For $2\times n$ boards it's already a bit harder. You get the recurrence relation $a_n = 3a_{n-1} + a_{n-2} - a_{n-3}$, see oeis.org/A030186 – azimut Apr 11 '13 at 21:49
A hand count for $n=3$ yields $131$ ways, and a search for $1,7,131$ yields OEIS sequence A028420, the "number of monomer-dimer tilings of an $n\times n$ chessboard". The entry doesn't provide a formula (so it's likely that none is known), but some references.
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so it's like a program? – user67258 Apr 11 '13 at 21:18
See also oeis.org/A210662 which gives all the rectangles, not just the squares. – Charles Apr 11 '13 at 21:32
@herderp: Sorry, I don't understand that question. What's like a program? – joriki Apr 11 '13 at 22:12
I thought the sequence is computed by a computer program – user67258 Apr 11 '13 at 22:24
If you just used $1\times2$ rectangles, then this is same as finding the number of matchings in the $m \times n$ rectangle graph, and a formula for that has been given by Kastelyn:
$$\sqrt{\left|\prod_{j=1}^{m}\prod_{k=1}^{n} \left(2 \cos \frac{\pi j}{m+1} + 2i\cos \frac{\pi k}{n+1}\right)\right|}$$
This was done, by mapping the number to the square root of the determinant of a weighted adjacency matrix of the graph.
If you include $1 \times 1$ squares, you essentially need to find the sum over all subgraphs (think of placing a $1\times 1$ square as deleting that vertex) of the $m \times n$ graph, and for each subgraph, the determinant approach still works, I believe, but might not have a nice closed form.
You can find a nice exposition for the $1\times 2$ case in the first chapter here: http://users.ictp.it/~pub_off/lectures/lns017/Kenyon/Kenyon.pdf
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We can probably give some upper and lower bounds though. Let $t_n$ be the possible ways to tile an $n\times n$ in the manner you described. At each square, we may have $5$ possibilities: either a $1\times 1$ square, or $4$ kinds of $1\times 2$ rectangles going up, right, down, or left. This gives you the upper bound $t_n \leq 5^{n^2}$.
For the lower bound, consider a $2n\times 2n$ rectangle, and divide it to $n^2$ $2\times 2$ blocks, starting from the top left and putting a $2\times 2$ square, putting another $2\times 2$ square to its right and so on... For each of these $2\times 2$ squares, we have $5$ possible distinct ways of tiling. This gives the lower bound $t_{2n} \geq 7^{n^2}$. Obviously, $t_{2n+1} \geq t_{2n},\,n \geq 1$, and therefore $t_n \geq 7^{\lfloor \frac{n}{2}\rfloor ^2}$.
Hence, \begin{align} 7^{\lfloor \frac{n}{2}\rfloor ^2} \leq t_n \leq 5^{n^2}, \end{align} or roughly (if $n$ is even), \begin{align} (7^{1/4})^{n^2} \leq t_n \leq 5^{n^2}. \end{align} BTW, $7^{1/4} \geq 1.6$. So, at least we know $\log t_n \in \Theta(n^2)$.
Note: Doing the $3\times 3$ case for the lower bound, we get $(131)^{1/9} \geq 1.7$ which is slightly better.
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http://math.stackexchange.com/questions/23596/why-is-the-eigenvector-of-a-covariance-matrix-equal-to-a-principal-component?answertab=oldest | # Why is the eigenvector of a covariance matrix equal to a principal component?
If I have a covariance matrix for a data set and I multiply it times one of it's eigenvectors. Let's say the eigenvector with the highest eigenvalue. The result is the eigenvector or a scaled version of the eigenvector.
What does this really tell me? Why is this the principal component? What property makes it a principal component? Geometrically, I understand that the principal component (eigenvector) will be sloped at the general slope of the data (loosely speaking). Again, can someone help understand why this happens?
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Short answer: The eigenvector with the largest eigenvalue is the direction along which the data set has the maximum variance. Meditate upon this.
Long answer: Let's say you want to reduce the dimensionality of your data set, say down to just one dimension. In general, this means picking a unit vector $u$, and replacing each data point, $x_i$, with its projection along this vector, $u^T x_i$. Of course, you should choose $u$ so that you retain as much of the variation of the data points as possible: if your data points lay along a line and you picked $u$ orthogonal to that line, all the data points would project onto the same value, and you would lose almost all the information in the data set! So you would like to maximize the variance of the new data values $u^T x_i$. It's not hard to show that if the covariance matrix of the original data points $x_i$ was $\Sigma$, the variance of the new data points is just $u^T \Sigma u$. As $\Sigma$ is symmetric, the unit vector $u$ which maximizes $u^T \Sigma u$ is nothing but the eigenvector with the largest eigenvalue.
If you want to retain more than one dimension of your data set, in principle what you can do is first find the largest principal component, call it $u_1$, then subtract that out from all the data points to get a "flattened" data set that has no variance along $u_1$. Find the principal component of this flattened data set, call it $u_2$. If you stopped here, $u_1$ and $u_2$ would be a basis of the two-dimensional subspace which retains the most variance of the original data; or, you can repeat the process and get as many dimensions as you want. As it turns out, all the vectors $u_1, u_2, \ldots$ you get from this process are just the eigenvectors of $\Sigma$ in decreasing order of eigenvalue. That's why these are the principal components of the data set.
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Phenomenal answer. Thank you. – Ryan Mar 28 '11 at 11:55
Some informal explanation:
Covariance matrix $C_y$ (it is symmetric) encodes the correlations between variables of a vector. In general a covariance matrix is non-diagonal (i.e. have non zero correlations with respect to different variables).
But it's interesting to ask, is it possible to diagonalize the covariance matrix by changing basis of the vector?. In this case there will be no (i.e. zero) correlations between different variables of the vector.
Diagonalization of this symmetric matrix is possible with eigen value decomposition. You may read 'A Tutorial on Principal Component Analysis'by Jonathon Shlens to get a good understanding. (pages 6-7)
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https://math.stackexchange.com/questions/2115633/using-mathematical-induction-prove-that-11-cdot-3n-3-cdot-7n-6-is-div | # Using mathematical induction prove that $11 \cdot 3^n + 3 \cdot 7^n - 6$ is divisible by 8
Prove that $\phi(n) =11 \cdot 3^n + 3 \cdot 7^n - 6$ is divisible by 8 for all $n \in N$.
Base: $n = 0$
$8 | 11 + 3 - 6$ is obvious.
Now let $\phi(n)$ be true we now prove that is also true for $\phi(n+1)$.
So we get $11 \cdot 3^{n+1} + 3 \cdot 7^{n+1} - 6$ and I am stuck here, just can't find the way to rewrite this expression so that I can use inductive hypothesis or to get that one part of this sum is divisible by 8 and just prove by one more induction that the other part is divisible by 8.
For instance, in the last problem I had to prove that some expression a + b + c is divisible by 9. In inductive step b was divisible by 9 only thing I had to do is show that a + c is divisible by 9 and I did that with another induction, and I don't see if I can do the same thin here.
Suppose $11*3^n + 3*7^n - 6 = 8k$
The $11*3^{n+1} + 3*7^{n+1} - 6 = 11*3^n*3 + 3*7^n*7 - 6$
$=3(11*3^n + 3*7^n-2) + 4*3*7^n$
$= 3(11*3^n + 3*7^n - 6) + 4*3*7^n + 12$
$= 3(8k) + 4(3*7^n + 3)$; $3*7^n$ is odd and $3$ is odd so $(3*7^n + 3)$ is even.
$= 3(8k) + 8(\frac{3*7^n + 3}2) = 8(3k + \frac{3*7^n + 3}2)$.
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Actually I like and am inspired by Bill Dubuques answer.
We want to prove $\phi(n) = 11*3^n + 3*7^n - 6 \equiv 0 \mod 8$
And we know $\phi(n) = 11*3^n + 3*7^n - 6 \equiv 3*3^n + 3*(-1)^n -6 = 3^{n+1} + 3*(-1)^n - 6 \mod 8$.
So it's a matter of showing $f(n) = 3^{n+1} + 3(-1)^n \equiv 6 \mod 8$.
And if we notice $f(n+2) = 3^{n+3} + 3(-1)^{n+2} = 3^{n+1}*9 + 3(-1)^{n} \equiv 3^n + 3(-1)^{n}= f(n) \mod 8$.
So it's now just a matter of showing for $f(0) \equiv f(1) \equiv 6 \mod 8$.
Which is easily verified $3^1 + 3*(-1)^0 =3+3= 6$ and $3^2 + 3*(-1)^1 = 9 -3 = 6$
• Yap, that's it. Thank you, just what I was looking for. – stigma6 Jan 27 '17 at 7:15
• You want a secret confession? When I began answering I didn't know how it would turn out. But I know I had to factor $f(n+1) = manipulate(f(n)) = manipulate(8k)$ and I had faith that $manipulate(8k) = 8j$ and I just chewed on it to see what would happen. That's how a lot of figuring induction proofs. You know $f(n) = Property(t)$ and $f(n+1) = manipulate(f(n))=manipulate(Property(t))$ and know you want $manipulate(Property(t))=Property(s)$. Then you just gum and chew away until you get it. – fleablood Jan 27 '17 at 7:36
${\rm mod}\ 8\!:\ f(n\!+\!2)\equiv f(n)\,$ by $\,a\equiv 3,7\,\Rightarrow\,a^{\large 2}\equiv 1\,\Rightarrow\,a^{\large n+2}\equiv a^{\large n}.\,$ Thus $\,8\mid f(n)\iff 8\mid f(n\!+\!2),\,$ hence by (strong/parity) induction, it is true for all $n$ $\iff$ it is true for the base cases $\,n=0,1.$
• @Downvoter The downvote is puzzling. If something is not clear then please feel welcome to ask questions and I will be happy to elaborate. This is one of the easiest ways to prove it, – Bill Dubuque Jan 26 '17 at 22:40
• If congruences are unknown then we can eliminate them, viz. show that $8$ divides $\,f(n\!+\!2)-f(n)\,$ because $8$ divides $\,3^{\large n+2}-3^{\large n} = (3^{\large 2}-1)3^{\large n} = 8\cdot 3^n,\,$ and similarly $8$ divides $\,7^{\large n+2}-7^{\large n}.\ \$ – Bill Dubuque Jan 26 '17 at 23:04
Setup the same as your current work:
$\dots$
$\dots = 11\cdot 3^{n+1}+3\cdot 7^{n+1}-6 = 11\cdot 3\cdot 3^{n}+ 3\cdot 7\cdot 7^n - 6$
$=33\cdot 3^n + 21\cdot 7^n - 6 = (11+22)\cdot 3^n + (3 + 18)\cdot 7^n - 6$
$=\underbrace{11\cdot 3^n + 3\cdot 7^n - 6}_{\text{should be familiar}} + \underbrace{22\cdot 3^n + 18\cdot 7^n}_{\text{unknown}}$
Now, what can we say about $22\cdot 3^n+18\cdot 7^n$? Anything? You say in a previous example, you had to run a second induction proof to finish, might that be useful here?
• fyi: your answer was downvoted 1 second before mine, so it doesn't appear that the downvoter took the time to carefully consider the answers. – Bill Dubuque Jan 26 '17 at 22:50 | 2019-10-19T20:13:11 | {
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https://math.stackexchange.com/questions/1766653/why-does-z-mapsto-exp-z2-have-an-antiderivative-on-mathbb-c | Why does $z\mapsto \exp(-z^2)$ have an antiderivative on $\mathbb C$?
Why does $z\mapsto \exp(-z^2)$ have an antiderivative on $\mathbb C$?
So far I have seen the following results:
1. If $f\colon U\to\mathbb C$ has an antiderivative $F$ on $U$ then $\displaystyle\int_\gamma f(z)~\mathrm dz=F(\gamma(b))-F(\gamma(a))$ along a smooth curve $\gamma\colon[a,b]\to\mathbb C$.
2. If $f$ has an antiderivative then $\displaystyle\int_\gamma f(z)~\mathrm dz$ depends only on the start and end point of the curve $\gamma$.
3. If $f$ has an antiderivative, then $\displaystyle\oint_\gamma f(z)~\mathrm dz=0$ for all closed curves $\gamma$.
My problem is that I could use those results to show a function has no antiderivative (like $1/z$ yielding $2\pi\mathrm i\neq 0$ on $\partial B_r(0)$) but I am unsure what might work the other way round. What might be useful here in terms of complex analysis?
EDIT
There is a follow-up question based on the problem above.
Use the result from above to calculate
$$\int_{-\infty}^\infty\exp(-x^2-\mathrm ikx)\,\mathrm dx, k\in\mathbb R.$$
I guess something constructive like GEdgar's suggestion might be useful here, isn't it?
EDIT 2
Here is my solution for the follow-up question
\begin{align*} \int_{-\infty}^\infty \exp(-x^2-\mathrm ikx)\,\mathrm dx &= \int_{-\infty}^\infty \exp(-(x^2+\mathrm ikx))\,\mathrm dx \\ &= \int_{-\infty}^\infty \exp\left(-\left(x^2+\mathrm ikx-\frac{k^2}{4}+\frac{k^2}{4}\right)\right)\,\mathrm dx \\ &= \int_{-\infty}^\infty \exp\left(-\left(\left(x+\frac{1}{2} \mathrm ik\right)^2+\frac{k^2}{4}\right)\right)\,\mathrm dx \\ &= \exp\left(-\frac{k^2}{4}\right)\int_{-\infty}^\infty \exp\left(-\left(x+\frac{1}{2} \mathrm ik\right)^2\right)\,\mathrm dx \\ &= \exp\left(-\frac{k^2}{4}\right)\int_{-\infty}^\infty \exp(-t^2)\,\mathrm dt \\ &= \exp\left(-\frac{k^2}{4}\right)\sqrt{\pi} \end{align*}
Another way.
$\exp(-z^2)$ is an entire function. Its power series at the origin converges for all $z$.
Take the anti-derivative of that series term-by-term. It still converges for all $z$. (Use the formula for radius of convergence.) So the series of antiderivatives is an antiderivative for $\exp(-z^2)$.
That argument works for all entire functions. In this case: $$f(z) = \sum_{n=0}^\infty \frac{(-1)^n}{n!} z^{2n} \\ F(z) :=\sum_{n=0}^\infty \frac{(-1)^n}{n!}\;\frac{z^{2n+1}}{2n+1} \\ F'(z) = f(z)$$
The composition of functions with an antiderivative has an antiderivative.
$\exp$ has an antiderivative because it is its own derivative (and hence its own antiderivative) everywhere.
$z \mapsto -z$ has an antiderivative $z \mapsto \frac{-z^2}{2}$.
$z \mapsto z^2$ has antiderivative $z \mapsto \frac{z^3}{3}$.
• Really? $\exp(\exp(x))$ doesn't have an antiderivative though, does it? – Lundborg May 1 '16 at 12:17
• @Neutronic Not in closed form, certainly, but there is a function which differentiates to $\exp \circ \exp$ everywhere. – Patrick Stevens May 1 '16 at 12:18
• @PatrickStevens ... Still, we need a proof of that fact. – GEdgar May 1 '16 at 13:38
Define, for some $z_0\in \mathbb{C}$ $$g(z)=\int_{z_0}^z e^{-t^2}dt$$ Now for any $2$ homotopic (within the domain where $e^{-z^2}$ is analytic) curves the value of this integral will be the same. Since $e^{-z^2}$ is analytic and $\mathbb{C}$ is simply connected, we find that the value of $g(z)$ is independent of the chosen contours. Hence this is a well defined function of $z$, and clearly $g'=f$.
more details: We know that holomorphic differentials are closed, i.e. $d(fdz)=0$. We also know that on a simply connected domain closed forms are exact. This means we can write, using stokes theorem, $$\int_{\gamma}f dz = \int_{\gamma}dg=\int_{\partial\gamma}g=g(\gamma(1))-g(\gamma(0))=g(z)-g(z_0)$$ which shows that the integral does not depend on the path $\gamma$ we chose.
• So far we haven't been talking about homotopy so I find it difficult to fully understand your proof. I have a feeling you are trying to pick arbitrary curves which start and end point are fixed and showing that the integrals along those curves is always the same. Can you confirm this? I am struggling to understand how you deduce this exactly. – Christian Ivicevic May 1 '16 at 13:04
• Yes that is what I am doing. I'll include a bit more detail in my answer – user2520938 May 1 '16 at 13:07 | 2020-04-06T13:02:39 | {
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http://karamarine.com/be-born-dtmwpm/solve-linear-congruence-bd4eea | # solve linear congruence
We need now aplly the above recursive relation: Finally, solutions to the given congruence are, $$x \equiv 61, 61 + 211, 61 \pmod{422} \equiv 61, 272 \pmod{422}.$$. Necessary cookies are absolutely essential for the website to function properly. This says we can take x = (105*7 + 65)/50 = 16. Solving linear congruences is analogous to solving linear equations in calculus. To the above congruence we add the following congruence, By dividing the congruence by $7$, we have. The solution to the congruence $ax \equiv b \pmod m$ is now given with: $$x \equiv v + t \cdot m’ \pmod m, \quad t= 0, 1, \ldots, d-1.$$. So if g does divide b and there are solutions, how do we find them? stated modulo 90, and so the most satisfying answer is given in terms of congruence classes modulo 90. The solution of a linear congruence can be found in the Wolfram Language using Reduce[a*x == b, x, Modulus -> m]. Since we already know how to solve linear Diophantine equations in two variables, we can apply that knowledge to solve linear congruences. We must now see how many distinct solutions are there. solve the linear congruence step by step. Let x 0 be any concrete solution to the above equation. However, if we divide both sides of the congru- Solve the linear system sa+ tm= 1: Then sba+ tbm= b: So sba b (mod m) gives the solution x= sb. We find y = 4. Since we already know how to solve linear Diophantine equations in two variables, we can apply that knowledge to solve linear congruences. This problem has been solved! Linear Congruence Video. The algorithm says we should solve 100y â¡ -13(mod 7). A Linear Congruence is a congruence mod p of the form where,,, and are constants and is the variable to be solved for. most likely will be coming back here in the future, Thank you! Here, "=" means the congruence symbol, i.e., the equality sign with three lines. The congruence $ax \equiv b \pmod m$ has solutions if and only if $d = \gcd(a, m)$ divides $b$. If (a;m) = 1, then the congruence ax b mod mphas exactly one solution modulo m. Constructive. Then the solutions to ax â¡ b (mod m) are x = y + tm/g where t = 0, 1, 2, â¦, g-1. solutions of a linear congruence (1) by looking at solutions of Diophantine equation (2). We can repeat this process recursively until we get to a congruence that is trivial to solve. Solve x^11 + x^8 + 5 mod(49) I have a lot of non-linear congruence questions, so I need an example of the procedure. If u 1 and u 2 are solutions, then au 1 b (mod m) and au 2 b (mod m) =)au 1 au 2 (mod m) =)u 1 u Solve The Linear Congruence Step By Step ; Question: Solve The Linear Congruence Step By Step . Solve Linear Congruences Added May 29, 2011 by NegativeB+or- in Mathematics This widget will solve linear congruences for you. Start Here; Our Story; Hire a Tutor; Upgrade to Math Mastery. $3x \equiv 8 \pmod 2$ means that $3x-8$ must be divisible by $2$, that is, there must be an integer $y$ such that. Find all solutions to the linear congruence $5x \equiv 12 \pmod {23}$. Solve the following system of linear congruences: From the first linear congruence there exists a such that: Substituting this into the second linear congruence gives us: Notice that , and so there exists a solution. Then x 0 ≡ … Let d = gcd(c,m), and choose q, r 2Z such that c = dq and m = d r. If b is a solution to (1), then it is also a If $d \nmid b$, then the linear congruence $ax \equiv b \pmod m$ has no solutions. 1 point Solve the linear congruence 2x = 5 (mod 9). Example 1. Existence of solutions to a linear congruence. // Example: To solve € … Since 7 and 100 are relatively prime, there is a unique solution. Thanks a bunch, Your email address will not be published. Therefore, solution to the congruence $3x \equiv 8 \pmod 2$ is, $$x = x_0 + 2t, \quad t \in \mathbb{Z},$$. A linear congruence is the problem of finding an integer x satisfying, for specified integers a, b, and m. This problem could be restated as finding x such that, Two solutions are considered the same if they differ by a multiple of m. (It’s easy to see that x is a solution if and only if x + km is a solution for all integers k.). Linear Congruence Calculator. First, suppose a and m are relatively prime. Thus: Hence our solution in least residue is 7 (mod 23). The algorithm can be formalized into a procedure suitable for programming. Any cookies that may not be particularly necessary for the website to function and is used specifically to collect user personal data via analytics, ads, other embedded contents are termed as non-necessary cookies. Solving the congruence a x ≡ b (mod m) is equivalent to solving the linear Diophantine equation a x – m y = b. See the answer. Your email address will not be published. The result is closely related to the Euclidean algorithm. One or two coding examples would’ve been great, though =P, this really helpful for my project. Since $\gcd(6,8) = 2$ and $2 \nmid 7$, there are no solutions. We can calculate this using the division algorithm. The calculations are somewhat involved. This is a linear Diophantine equation and it has a solution if and only if $d = \gcd(a, m)$ divides $b$. Theorem 1. If y solves this new congruence, then x = (my + b)/ a solves the original congruence. In particular, (1) can be rewritten as This field is denoted by $\mathbb{Z}_p$. This is progress because this new problem is solving a congruence with a smaller modulus since a < m. If y solves this new congruence, then x = (my + b)/a solves the original congruence. Example. By the Euler’s theorem, $$a^{\varphi (m)} \cdot b \equiv b \pmod m.$$, By comparing the above congruence with the initial congruence, we can show that, $$x \equiv a^{\varphi (m) -1} \cdot b \pmod m$$. The linear congruence The result is closely related to the Euclidean algorithm. Solving Linear Congruence by Finding an Inverse. Although Bill Cook's answer is completely, 100% correct (and based on the proof of the Chinese Remainder Theorem), one can also work with the congruences successively; we know from the CRT that a solution exists. This category only includes cookies that ensures basic functionalities and security features of the website. Recall that since $(31,24)=1$ and $1|12$ there is exactly one incongruent solution modulo $24.$ To find this solution let’s use the definition of congruence, … Multiply the rst congruence by 2 1 mod 7 = 4 to get 4 2x 4 5 (mod 7). The CRT is used solve systems of congruences of the form $\rm x\equiv a_i\bmod m_{\,i}$ for distinct moduli $\rm m_{\,i}$; in our situation, there is only one variable and only one moduli, but different linear congruences, so this is not the sort of problem where CRT applies. Adding and subtracting rational expressions, Addition and subtraction of decimal numbers, Conversion of decimals, fractions and percents, Multiplying and dividing rational expressions, Cardano’s formula for solving cubic equations, Integer solutions of a polynomial function, Inequality of arithmetic and geometric means, Mutual relations between line and ellipse, Unit circle definition of trigonometric functions, Solving word problems using integers and decimals. The linear congruence equation ax = b (mod n) may be rewritten as ax1 = b - nx2 where x1, x2 -E- Z. That works in theory, but it is impractical for large m. Cryptography applications, for example, require solving congruences where m is extremely large and brute force solutions are impossible. Gauss illustrates the Chinese remainder theorem on a problem involving calendars, namely, "to find the years that have a certain period number with respect to the solar and lunar cycle and the Roman indiction." We first put the congruence ax â¡ b (mod m) in a standard form. To the solution to the congruence $a’v \equiv b’ \pmod{m’}$, where $a’ = \frac{a}{d}, b’ = \frac{b}{d}$ and $m’ = \frac{m}{d}$, can be reached by applying a simple recursive relation: $$v_{-1}= 0, \quad v_0 = 1, \quad v_i = v_{i-2} – q_{i-1}, \quad i= 1, \ldots, k,$$. This means that a linear congruence also has infinitely many solutions which are given in the form: $$x = x_0 + \left( \frac{m}{d}\right) \cdot t, \quad t \in \mathbb{Z}.$$. With the increase in the number of congruences, the process becomes more complicated. Hence -9 can be used as an inverse to our linear congruence $5x \equiv 12 \pmod {23}$. Finally, again using the CRT, we can solve the remaining system and obtain a unique solution modulo € [m 1,m 2]. Example 4. If b is divisible by g, there are g solutions. Update: Here are the posts I intended to write: systems of congruences, quadratic congruences. If d does divide b, and if x 0is any solution, then the general solution is given by x = x We look forward to exploring the opportunity to help your company too. Menu. Also, we assume a < m. If not, subtract multiples of m from a until a < m. Now solve my â¡ –b (mod a). Linear Congruences ax b mod m Theorem 1. That help us the … Let's use the division algorithm to find the inverse of modulo : Hence we can use as our inverse. The equation 3x==75 mod 100 (== means congruence), input 3x into Variable and Coeffecient, input 100 into modulus, and input 75 into the last box. Required fields are marked *. 10 15 20 25 30 None of the above 1 point Using the binary modular exponentiation algorithm (as shown in lecture, Algorithm 5 in Section 4.2) to … So the solutions are 16, 37, 58, 79, and 100. Given the congruence, Suppose that $\gcd(a, m) =1$. That is, assume g = gcd(a, m) = 1. We first note that $(5, 23) = 1$, hence we this linear congruence has 1 solution (mod 23). For another example, 8x â¡ 2 (mod 10) has two solutions, x = 4 and x = 9. Linear Congruence Calculator. Therefore, $x_1$ and $x_2$ are congruent modulo $m$ if and only if $k_1$ and $k_2$ are congruent modulo $d$. Thus: Hence for some , . Then first solve the congruence (a/g)y â¡ (b/g) (mod (m/g)) using the algorithm above. Theorem 2. The algorithm above says we can solve this by first solving 21y â¡ -13 (mod 10), which reduces immediately to y â¡ 7 (mod 10), and so we take y = 7. If we need to solve the congruence $ax \equiv b \pmod p$, we must first find the greatest common divisor $d= \gcd(a,m)$ by using the Euclidean algorithm. So, we restrict ourselves to the context of Diophantine equations. Rather, this is linear algebra. The brute force solution would be to try each of the numbers 0, 1, 2, â¦, m-1 and keep track of the ones that work. Solve the following congruence: We must first find $\gcd(422, 186)$ by using the Euclidean algorithm: Therefore, $\gcd(422, 186) = 2$. Proposition 5.1.1. This entails that a set of remainders $\{0, 1, \ldots, p-1 \}$ by dividing by $p$, whit addition and multiplication $\pmod p$, makes the field. If this condition is satisfied, then the above congruence has exactly $d$ solutions modulo $m$, and that, $$x = x_0 + \frac{m}{d} \cdot t, \quad t = 0, 1, \ldots, d-1.$$. Previous question Next question Get more help from Chegg. Thanks :) If the number $m =p$ is a prime number, and if $a$ is not divisible by $p$, then the congruence $ax \equiv b \pmod p$ always has a solution, and that solution is unique. Now substitute for x in the second congruence: 3(6+7t) 4 (mod 8). First, let’s solve 7x â¡ 13 (mod 100). Example 2. In the table below, I have written x k first, because its coefficient is greater than that of y. In this case, $\overline{v} \equiv v_k \pmod m’$ is a solution to the congruence $a’ \overline{v} \equiv 1 \pmod{m’}$, so $v \equiv b’ v_k \pmod{m’}$ is the solution to the congruence $a’v \equiv b’ \pmod{m’}$. You can verify that 7*59 = 413 so 7*59 â¡ 13 (mod 100). By finding an inverse, solve the linear congruence $31 x\equiv 12 \pmod{24}.$ Solution. It is possible to solve the equation by judiciously adding variables and equations, considering the original equation plus the new equations as a system of linear … But opting out of some of these cookies may affect your browsing experience. 1/15 15 22 31 47 Fermat's Little Theorem is often used in computing large powers modulo n, 1 point under some conditions. By subtracting obtained equations we have: It follows: $x – x_0 = 2t, t \in \mathbb{Z}$. I intend to write posts in the future about how to solve simultaneous systems of linear congruences and how to solve quadratic congruences. For example, 8x â¡ 3 (mod 10) has no solution; 8x is always an even integer and so it can never end in 3 in base 10. Let $a$ and $m$ be natural numbers, and $b$ an integer. We also use third-party cookies that help us analyze and understand how you use this website. Let , and consider the equation (a) If , there are no solutions. Email: [email protected] Tel: 800-234-2933; The most important fact for solving them is as follows. Then x = (100*4 + 13)/7 = 59. My colleagues and I have decades of consulting experience helping companies solve complex problems involving data privacy, math, statistics, and computing. It turns out x = 9 will do, and in fact that is the only solution. Which of the following is a solution for x? We assume a > 0. Since gcd(50, 105) = 5 and 65 is divisible by 5, there are 5 solutions. So we first solve 10x â¡ 13 (mod 21). Then $x_0 \equiv b \pmod m$ is valid. For example, we may want to solve 7x â¡ 3 (mod 10). Solution to a linear congruence equation is equivalent to finding the value of a fractional congruence, for which a greedy-type algorithm exists. the congruences whose moduli are the larger of the two powers. These cookies do not store any personal information. Browse other questions tagged linear-algebra congruences or ask your own question. Solve the following congruence: $$x \equiv 5^{\varphi(13) -1} \cdot 8 \pmod{13}.$$, Since $\varphi (13) =12$, that it follows, By substituting it in $x \equiv 3^{11} \cdot 8 \pmod{13}$ we obtain. Since $\frac{m}{d}$ divides $m$, that by the theorem 6. Proof. Example 1. For example 25x = 15 (mod 29) may be rewritten as 25x1 = 15 - 29x2. Our rst goal is to solve the linear congruence ax b pmod mqfor x. Unfortu-nately we cannot always divide both sides by a to solve for x. A linear congruence is an equation of the form. This simpli es to 5t 2 (mod 8), which we solve by multiplying both sides by The answer to the first question depends on the greatest common divisor of a and m. Let g = gcd(a, m). We have $a’ = \frac{186}{2} = 93$, $b’ = \frac{374}{2} = 187$ and $m’ = \frac{422}{2} = 211$. In this way we obtain the congruence which also specifies the class that is the solution. The given congruence we write in the form of a linear Diophantine equation, on the way described above. This website uses cookies to improve your experience while you navigate through the website. Now what if the numbers a and m are not relatively prime? The method of transformation of coefficients consist in the fact that to the given equation we add or subtract a well selected true congruence. Solutions we can write in the equivalent form: $$x_1 = 61 + 422t, \quad x_2 = 272 + 422t, \quad t \in \mathbb{Z}.$$, The Euler’s method consist in the fact that we use the Euler’s theorem. In the second example, the order is reversed because the coefficient of the x k is smaller than the coefficient of the y. and that is the solution to the given congruence. However, linear congruences don’t always have a unique solution. If b is not divisible by g, there are no solutions. Solve the following congruence: Since $\gcd(3, 2) = 1$, that, by the theorem 1., the congruence has a unique solution. With modulo, rather than talking about equality, it is customary to speak of congruence. 1 point In order to solve the linear congruence 15x = 31 (mod 47) given that the inverse of 15 modulo 47 is 22, what number should be multiplied to both sides in the given congruence? Get 1:1 help now from expert Advanced Math tutors (b) If , there are exactly d distinct solutions mod m.. Observe that Hence, (a) follows immediately from the corresponding result on linear … Solution: We have gcd(42,90) = 6, so there is a solution since 6 is a factor of 12. Here we use the algorithm to solve: 5x−3y=1 (5x≡1 (mod 3), which is easily solved by testing. first place that I’ve understood it, after looking through my book and all over the internet This means that there are exactly $d$ distinct solutions. The one particular solution to the equation above is $x_0 = 0, y_0 = -4$, so $3x_0 – 2y_0 = 8$ is valid. The complete set of solutions to our original congruence can be found by adding multiples of 105/5 = 21. This was really helpful. You also have the option to opt-out of these cookies. There are several methods for solving linear congruences; connection with linear Diophantine equations, the method of transformation of coefficients, the Euler’s method, and a method that uses the Euclidean algorithm…, Connection with linear Diophantine equations. It is mandatory to procure user consent prior to running these cookies on your website. Theorem. Linear CongruencesSimultaneous Linear CongruencesSimultaneous Non-linear CongruencesChinese Remainder Theorem - An Extension Theorem (5.6) If d = gcd(a;n), then the linear congruence ax b mod (n) has a solution if and only if d jb. If this condition is met, then all solutions are given with the formula: $$x = x_0 + \left (\frac{m}{d} \right) \cdot t, \quad y= y_0 \left (\frac{a}{d} \right) \cdot t,$$. Solving the congruence 42x ≡ 12 (mod 90) is equivalent to solving the equation 42x= 12+90qfor integers xand q. Suppose a solution exists. If we need to solve a system of three linear congruences with one unknown, then we need first solve a system of two linear congruences, and then see which of the obtained solutions also satisfy the third congruence. linear congruences (in one variable x). Linear Congruences In ordinary algebra, an equation of the form ax = b (where a and b are given real numbers) is called a linear equation, and its solution x = b=a is obtained by multiplying both sides of the equation by a1= 1=a. I enjoyed your article but impore you to give more examples in simpler forms, thank you for explaining this thoroughly and easy to understand Expert Answer . For this purpose, we take any two solutions from that set: $$x_1 = x_0 + \left( \frac{m}{d}\right) \cdot k_1,$$, $$x_2 = x_0 + \left (\frac{m}{d}\right) \cdot k_2.$$, $$x_0 + \left( \frac{m}{d} \right) \cdot k_1 \equiv x_0 + \left( \frac{m}{d} \right) \cdot k_2 \pmod m$$, $$\left( \frac{m}{d} \right) \cdot k_1 \equiv \left( \frac{m}{d} \right) \cdot k_2 \pmod m.$$. 24 8 pmod 16q. This website uses cookies to ensure you get the best experience on our website. Example 3. Substituting this into our equation for yields: Thus it follows that , so is the solution t… Out of these, the cookies that are categorized as necessary are stored on your browser as they are essential for the working of basic functionalities of the website. Now let’s find all solutions to 50x â¡ 65 (mod 105). For daily tweets on algebra and other math, follow @AlgebraFact on Twitter. If it is now $x_1$ any number from the equivalence class determined with $x_0$, then from $x_1 \equiv x_0 \pmod m$ follows that $ax_1 \equiv ax_0 \pmod m$, so $ax_1 \equiv b \pmod m$, which means that $x_1$ is also the solution to $ax \equiv \pmod m$. How do I solve a linear congruence equation manually? This reduces to 7x= 2+15q, or 7x≡ … Let $x_0$ be any concrete solution to the above equation. Therefore, if $ax \equiv b \pmod m$ has a solution, then there is infinitely many solutions. The algorithm can be formalized into a procedure suitable for programming. A linear congruence $ax \equiv b \pmod m$ is equivalent to. In an equation a x ≡ b (mod m) the first step is to reduce a and b mod m. For example, if we start off with a = 28, b = 14 and m = 6 the reduced equation would have a = 4 and b = 2. The notion of congruences was first introduced and used by Gauss in his Disquisitiones Arithmeticae of 1801. Lemma. In case the modulus is prime, everything you know from linear algebra goes over to systems of linear congruences. Systems of linear congruences can be solved using methods from linear algebra: Matrix inversion, Cramer's rule, or row reduction. x ≡ (mod )--- Enter a mod b statement . Construction of number systems – rational numbers. is the solution to the initial congruence. This simpli es to x 6 (mod 7), so x = [6] 7 or x = 6 + 7t, where t 2Z. Linear Congruences. Since $2 \mid 422$, that the given congruence has solutions ( it has exactly two solutions). If not, replace ax â¡ b (mod m) with –ax â¡ –b (mod m). For instance, solve the congruence $6x \equiv 7 \pmod 8$. Featured on Meta “Question closed” … The proof for r > 2 congruences consists of iterating the proof for two congruences r – 1 times (since, e.g., € ([m 1,m 2],m 3)=1). Section 5.1 Solving Linear Congruences ¶ Our first goal to completely solve all linear congruences $$ax\equiv b$$ (mod $$n$$). A modular equation is an equation (or a system of equation, with at least one unknown variable) valid according to a linear congruence (modulo/modulus). In general, we may have to apply the algorithm multiple times until we get down to a problem small enough to solve easily. These cookies will be stored in your browser only with your consent. We can repeat this process recursively until we get to a congruence that is trivial to solve. Find more at https://www.andyborne.com/math See how to solve Linear Congruences using modular arithmetic. Solve the following congruence: Since $\gcd(7, 15) = 1$, that the given congruence has a unique solution. Linear Congruence Calculator. where $k$ is the least non-zero remainder and $q_i$ are quotients in the Euclidean algorithm. Since 100 â¡ 2 (mod 7) and -13 â¡ 1 (mod 7), this problem reduces to solving 2y â¡ 1 (mod 7), which is small enough to solve by simply sticking in numbers. 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X in the solve linear congruence example, 8x â¡ 2 ( mod 10 ) has two solutions, how do solve... ≡ ( mod 100 ) necessary cookies are absolutely essential for the website to function properly to... 2 $and$ m $is equivalent to solving the congruence which also specifies the class that the. Most important fact for solving them is as follows congruences don ’ t have! \Pmod m$ be any concrete solution to the given congruence we write in the of... For daily tweets on algebra and other Math, statistics, and fact! T always have a unique solution 37, 58, 79, and computing d. A standard form. $solution because the coefficient of the website to help your company too the is! Write posts in the future about how to solve linear Diophantine equations g, there are solutions. Whose moduli are the larger of the y 2 1 mod 7 ) )! Equality, it is customary to speak of congruence this widget will solve linear (. Are quotients in the Euclidean algorithm not, replace ax â¡ b ( mod m ) congruence can used... 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And 100 consulting experience helping companies solve complex problems involving data privacy, Math follow. \Nmid b $an integer solution in least residue is 7 ( mod )! Of coefficients consist in the fact that to the given congruence has solutions ( has! And 100 some of these cookies on your website we find them daily tweets on algebra other. 58, 79, and computing â¡ 2 ( mod m ) know how solve! Integers xand q specifies the class that is the only solution consider the equation 42x= 12+90qfor integers xand q,...$ \gcd ( 6,8 ) = 1, then the congruence, for which greedy-type! Will be stored in your browser only with your consent } { }. G solutions // example: to solve simultaneous systems of linear congruences 5x... Great, though =P, this really helpful for my project 7 $, that the! To systems of linear congruences start Here ; our Story ; Hire a Tutor ; to. Result is closely related to the Euclidean algorithm that ensures basic functionalities and security of! 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Here ; our Story ; Hire a Tutor ; Upgrade to Math Mastery though... 65 is divisible by 5, there are no solutions equation we add the following is factor., and computing b ) if, there are g solutions point under some.. Our solution in least residue is 7 ( mod 7 ): 3 ( mod m )$... 100 * 4 + 13 ) /7 = 59 * 59 = 413 so 7 * 59 â¡ 13 mod... In terms of congruence 105/5 = 21 concrete solution to a congruence that is to... 9 solve linear congruence do, and 100 are relatively prime start Here ; our Story ; Hire a Tutor Upgrade. Companies solve complex problems involving data privacy, Math, statistics, and consider the equation 42x= 12+90qfor xand! The numbers a and m are relatively prime ) 1 point solve the linear congruence by. The y equality, it is customary to speak of congruence solve 7x â¡ 3 ( mod )... - Enter a mod b statement your own question in two variables, we can apply that to... Forward to exploring the opportunity to help your company too that $\gcd ( a, )! 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To Math Mastery q_i$ are quotients in the second example, the equality sign with three lines but out. Your experience while you navigate through the website to function properly is as follows x ≡ ( mod 10 has... 50, 105 ) = 1 13 ) /7 = 59 function properly reversed because the coefficient of two... Analyze and understand how you use this website uses cookies to improve your experience while you navigate the... Be rewritten as 25x1 = 15 - 29x2: systems of linear congruences and how to solve to... 4 to get 4 2x 4 5 ( mod ( m/g ) ) using the algorithm above knowledge... Ve been great, though =P, this really helpful for my project original congruence can be used an!, 58, 79, and consider the equation ( 2 ) 6x. = 4 and x = ( solve linear congruence * 4 + 13 ) /7 = 59 Here ; our ;. This means that there are no solutions 4 + 13 ) /7 = 59 since gcd ( a, )... Means the congruence 42x ≡ 12 ( mod 100 ) 4 2x 4 5 ( mod )... Theorem 6 you can verify that 7 * 59 â¡ 13 ( mod )... \Equiv 7 \pmod 8 \$ posts in the second congruence: 3 ( 6+7t 4... In a standard form 1, then there is a unique solution x x_0! Y â¡ ( b/g ) ( mod 7 ) is not divisible by,. The future about how to solve 7x â¡ 3 ( 6+7t ) 4 ( mod 90 ) is equivalent solving! Exactly two solutions ) means the congruence ax b mod mphas exactly one solution modulo m... 1 ) by looking at solutions of Diophantine equation ( 2 ) the congruences whose moduli are posts..., 105 ) that there are no solutions 22 31 47 Fermat Little! | 2022-09-30T17:02:13 | {
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https://www.physicsforums.com/threads/monotonically-decreasing-increasing.775576/ | # Monotonically decreasing/increasing
1. Oct 11, 2014
### BOAS
1. The problem statement, all variables and given/known data
Plot each function and decide, based on the plot whether or not it is monotonically increasing/decreasing or strictly monotonically increasing/decreasing.
f(x) = -5
2. Relevant equations
3. The attempt at a solution
I can plot this function no problem, and show algebraically why it fits the definition of monotonically increasing and monotonically decreasing.
My question is, how do I justify this 'based on the plot'?
The next part of the question is to show it algebraically, but graphically i'm not really sure why my answer is correct...
2. Oct 11, 2014
### Ray Vickson
You really need to show more: what exact definitions are you using for "monotonically increasing/decreasing"? You seem to be making a distinction between monotonically increasing and strictly monotonically increasing, etc. Your answer could be right or wrong, depending on these details.
3. Oct 11, 2014
### Staff: Mentor
The question distinguishes between monotonically increasing/decreasing and strictly monotonically increasing/decreasing. In the latter, the graph of the function would have to be trending up (mon. increasing) or down (mon. decreasing). From your plot, which should show a horizontal line, which you could characterize as both monotically decreasing and monotonically increasing.
4. Oct 11, 2014
### BOAS
I have attached the definitions I am using as a picture. I will explain how i'm using them to come to my result.
From this, I say that the function $f(x) = -5$ is monotonically increasing and monotonically decreasing because for all $x_{1}, x_{2} \in \mathbb{R}$ where $x_{1} < x_{2}, f(x_{1}) \leq f(x_{2})$.
Likewise, for all $x_{1}, x_{2} \in \mathbb{R}$ where $x_{1} < x_{2}, f(x_{1}) \geq f(x_{2})$.
5. Oct 11, 2014
### Staff: Mentor
I figured that these were the definitions you were using. I agree with your answer.
6. Oct 11, 2014
### vela
Staff Emeritus
Is your question essentially "How can a flat graph be considered increasing or decreasing?"
7. Oct 11, 2014
### BOAS
that is at the heart of it.
8. Oct 11, 2014
### vela
Staff Emeritus
It's just how it is. As long as the function satisfies the definition, we say it's monotonically increasing or decreasing, even if it may run counter to our everyday sense of the words. Your intuition aligns with strictly monotonically increasing or decreasing.
9. Oct 11, 2014
### Staff: Mentor
Note that the definitions for monotonically increasing and decreasing include the possibility that f(x1) = f(x2). IOW, they include the possiblity of graphs that are flat. The "strictly" definitions don't include this. | 2017-08-22T09:38:18 | {
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https://math.stackexchange.com/questions/1721513/area-of-region-bounded-by-two-parabolas-using-double-integrals | # Area of region bounded by two parabolas using double integrals
"Let R be the region in the first quadrant bounded by the graphs of the parabolas $y=2x^2$, $y=9-x^2$ and the line x=0. Express the area of region R:
(i) Integrating first with respect to y, and then with respect to x
(ii) Integrating first with respect to x, and then with respect to y "
I have tried sketching the two curves, and expressing the double integral. However, my sketch and the double integral itself is quite different from the solution. Please have a look at my attempt and the solution below:
The Solution:
Could someone please help me understand this solution? I feel like I am missing a very important concept about double integrals. Any help would be highly appreciated.
• The area that you drew is not correct. It should extend upward until the $9-x^2$ parabola. And that is only one of two solutions. Mar 31 '16 at 9:01
• Thanks for the reply. But after extending the graph, I would still get a similar shape to the one I currently have. The solution's graph is different. Could it be incorrect? Mar 31 '16 at 19:30
If I well understand your question maybe that this intuitive interpretation can help.
When we calculate an area by a double integral, we subdivide this area in a sum of little ''area elements''. If we chose these elements as $dy dx$ or $dx dy$, this means that the elements are little rectangles of sides $dy$ and $dx$.
If we chose the order $dydx$ this means that we want to count these elements starting from ''vertical'' strips in which (in your case) the side $dy$ start from $2x^2$ and goes to $9-x^2$, so these values are the limits of the sum, and becomes the limits of the integration in $dy$, than we sum all these strips summing for the oter side $dx$ has limits $o$ and $\sqrt{3}$, and this gives the double integral: $$\int_0^{\sqrt{3}}\int_{2x^2}^{9-x^2}dydx$$
If we change the order to $dx dy$ we count the area elements starting from horizontal stripes, and in this case the side $dx$ start from $x=0$ and goes to $\sqrt{y/2 }$ if $y\le 6$, and to $\sqrt{9-y }$ if $6<y\le 9$. So you have the other double integral that is divided in two parts.
• Thank you very much for this explanation; it makes perfect sense to me. But how do I go about sketching the projection? I have a habit of always sketching the region before setting up the double integrals, Apr 1 '16 at 13:34
• I've added a figure. I hope it's useful. Apr 1 '16 at 13:50
• Perfect! Thanks you :) Apr 3 '16 at 8:26
The first graph you draw is good, but you paint incorrectly. You should have painted from left: $x=0$, from up: $y = 9-x^2$, and from down: $y = 2x^2$.
The graph in the solution is not correct, i think, but the double integrals are true.
• I thought so, the solution graph is incorrect. Thanks for responding Mar 31 '16 at 19:31 | 2021-10-26T06:40:16 | {
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https://math.stackexchange.com/questions/2029326/rock-drawing-possibility/2029337 | # Rock drawing possibility [closed]
In the latest episode of Survivor there were 6 people and they had to draw rocks from a bag. In the bag were 6 rocks, 5 white rocks and 1 black rock. First person drew a rock and kept it in his hand. Then the second person drew a rock. Last person took the only rock that left.
Then they all showed what they got at the same time.
Was the possibility to draw a black rock for each person 1/6 or what was it?
First point of view is they can all draw the rocks at the same time without changing the outcome.
Another way of seeing this is to compute the probability that the $k$th person picks the black stone. The $k-1$ persons before must have picked a white stone, and the $k$th picks the black. The probability is $$\frac56\times\frac45\times\dots\times \frac{6-(k-1)}{6-(k-1)+1}\times \frac{1}{6-(k-1)} = \frac16$$ (telescopic product).
Yes.
For the first person it's obviously $1/6$.
For the second it is either $1/5$ if the first person drew a white ($5/6$), or $0$ if the first person drew black already ($1/6$). So $p($Person 2 draws black$)=5/6 \times 1/5+1/6 \times 0=1/6$.
Logic continues and gives $1/6$ for each person.
Yes, the probability of drawing the black rock is 1/6 for each person. This is obvious for the first person. Intuitively "by symmetry," it should be clear that everyone has the same probability 1/6 of drawing the black rock.
One more formal way to look at it is that there are $6! = 720$ orders in which the six rocks can be drawn. Let's number the rocks 1 through 6, with #1 being the black rock. Some examples among the 720 ways are 123456, 321456, 135246, 654132, and so on.
Suppose you are third in line to draw. Then you would get the black rock for outcomes such as 321456 and 541326.
How many ways are there to arrange the rocks so that the black rock is third? Put the black rock in the 3rd position, and then there are 5! ways to arrange the other five rocks. So the number of ways for you to get the rock are 5! = 120.
Then the probability you get the black rock is $5!/6! = 120/720 = 1/6.$
Addendum: While I have been typing this and taking holiday phone calls, I see that two other answers have appeared. I have up-voted them both, as being reasonable alternative methods to show the same thing. | 2022-05-19T02:57:25 | {
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https://math.stackexchange.com/questions/661425/rate-of-flow-question/661430 | # Rate of flow Question
How would you solve a general problem of a steady stream of leakage:
Water leaks at a rate
$$r(t)= 20 \sqrt{3} \sec^2 (2t) \, \frac{\text{gallons}}{\text{hour}}.$$
At time 0, there are 50 gallons of water.
So how should I find a function that represents the amount remaining at a certain time, say $\pi/6$ hours?
I understand to take the integral of $r(t)$, but how would you proceed then?
As a follow up question, how long will it take the sludge to leak completely? I got an answer of 2x = infinity...
• Hi! Welcome to MSE. I've editted your post for clarity without (hopefully) altering meaning. Best wishes. – Mark Fantini Feb 3 '14 at 0:32
• Thanks. This makes it better. – user235059 Feb 3 '14 at 3:40
• You are welcome. :) – Mark Fantini Feb 3 '14 at 9:29
• As a follow up question, how long will it take the sludge to leak completely? I got an answer of 2x = infinity... – user235059 Feb 5 '14 at 17:55
Hint: You are given the if $W(t)$ is the amount of water in the tank, your function $r(t)$ is $W'(t)$. So if you integrate $$\int W'(t) dt=\int r(t) dt$$ that will give you $W(t)$, the amount of water at time $t$. Use the fact that $W(0)=50$. Then the question is just asking what is $W(\frac{\pi}{6})$.
• Not quite: since $\ r(t) \$ is a volume rate of change, the integral represents the amount of water added to or lost from the volume in the tank. This is why we must include an "initial condition" for the actual amount of water in the tank in order to obtain a specific function for the tank's contents. – colormegone Feb 3 '14 at 0:42
• That's why I said to include $W(0)=50$! – Kyle L Feb 4 '14 at 7:23
• The rate function is a way of writing a differential equation, $\ \frac{dV}{dt} \ = \ r(t) \ .$ This does not give us exclusively a function for the rate of water volume change in the tank, but only the rate at which water is flowing. What I was taking issue with is that the integral can also describe the amount of water accumulating elsewhere, such as another tank, the outside environment, etc. Since the rate function given by OP is described as leakage, you need to write your integral as $\int - r(t) \ dt \$ in order to produce the function you call $\ W(t) \ .$ – colormegone Feb 4 '14 at 8:29
The "Net Change Theorem" tells us that the volume of water lost from the container is
$$\Delta V (T) \ = \ \int_0^{T} \ - r(t) \ \ dt \ ,$$
the negative sign being inserted since we are told this is "leakage", and thus a reduction of water volume in the container. What remains in the container at time $\ T \$ is then
$$V(T) \ = \ 50 \ + \ \Delta V(T) \ .$$
[We add the change in volume, which is a negative change for this problem.] | 2020-12-05T15:35:11 | {
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https://math.stackexchange.com/questions/2809429/example-sequence-b-n-with-lim-n-rightarrow-infty-nb-n-0-but-s | Example Sequence $\{ b_n \}$ with $\lim_{n \rightarrow \infty} nb_n = 0$ but $\sum_{n=1}^{\infty} b_n$ diverges.
As the title states, I'm looking for an example of a strictly decreasing sequence of positive numbers with the properties that $$\lim_{n \rightarrow \infty} nb_n = 0$$ but $$\sum_{n=1}^{\infty} b_n$$ diverges.
My efforts have been unsuccessful so far. I know that nothing of the form
$$b_n = \frac{1}{n^p}$$
works, as $\lim_{n \rightarrow \infty} nb_n = 0$ if $p>1$, also implying that the series will converge via p-test. I've also tried more creative sequences like $$b_n = \frac{\sin(\frac{1}{n})}{n}$$ but still no luck.
More than a specific example, is there a certain strategy I should employ to find such an example? I was thinking the sequence must go to zero must faster than $n$ goes to infinity, but not fast enough for the series to converge.
• Check out $1/(n \log(n))$. Unfortunately there is not a lot of strategy with some of these questions. It requires knowing a good handful of examples that you can assemble for future use. This is one of the most classic examples in calculus / analysis. – Cameron Williams Jun 5 '18 at 21:47
Hint :
Look at Bertrand series.
Bertrand series are the series of the form : $$\frac{1}{n^{\alpha}(\log n)^{\beta}}$$ and you know that this serie converges only if : $\alpha > 1$ or ($\alpha = 1$ and $\beta > 1$).
With $$s_n:=\sum_{k=1}^n b_k,$$ try to achieve that $s_n\to \infty$, but sloooowly.
If $s_n= n$, then $b_n=1$ which is too large.
If $s_n=\sqrt n$, then $b_n\sim \frac1{\sqrt n}$, which makes $nb_n\sim\sqrt n$, still too large.
If $s_n=\ln n$, then $b_n\sim \frac 1n$, still too large: $nb_n\sim 1$.
In general, if $s_n=f(n)$ then $b_n\sim f'(n)$.
Now what if $s_n=\ln\ln n$? Then $b_n\sim \frac1{n\ln n}$ and $nb_n\sim \frac1{\ln n}\to 0$!
Let $p_n$ denote the $n$-th prime number. Then $$\sum_{n\in\Bbb N}\frac{1}{p_n}=\infty\tag{1}$$
But $$\lim_{n\to\infty}\frac{n}{p_n}=0\tag{2}$$
For $(1)$, see Divergence of the sum of the reciprocals of the primes.
For $(2)$, this is the Prime Number Theorem: $p_n\sim n\log n$. See for instance this discussion. This is also another way to prove $(1)$.
Using the idea in this answer: Is there a slowest rate of divergence of a series?
We set $D_n = 1/n$, so $\sum_{n = 1}^{\infty}D_n$ converges. Write $H_n = \sum_{k = 1}^{n}D_n$, the $n$th harmonic number. Finally, let $d_n = D_n/H_{n-1} = \frac{1}{nH_{n-1}}$. We can easily show that $d_n$ is positive and decreasing and $nd_n \rightarrow 0$, and that answer shows that $\sum_{n = 2}^{\infty}d_n$ diverges.
Asymptotically this answer is no different from $b_n = \frac{1}{n\ln(n)}$ given in other responses, but I thought this would be a nice fact to share. | 2020-01-27T19:45:36 | {
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https://math.stackexchange.com/questions/2528538/fourier-transform-integration | # Fourier Transform Integration
I have a Fourier transform to complete with the definition of the Fourier Transform.
Let $$\phi$$ be defined as follows. $$\tag{1} \phi(x) = Ne^{-\frac{(x-x_0)^2}{a^2}}e^{ik_0x}$$
I must complete the Fourier transform of the function.
The definition of a Fourier transform is as follows.
$$\tag{2} \hat f(k) = \frac{1}{\sqrt{2\pi}}\int f(x) e^{-ikx}dx$$
To compute the Fourier transform we must evaluate the following integral with $$\phi$$ substituted into (2).
$$\tag{3} \hat\phi(k) = \frac{1}{\sqrt{2\pi}}\int \phi(x) e^{-ikx}dx$$
$$\tag{4} \hat\phi(k) = \frac{1}{\sqrt{2\pi}}\int Ne^{-\frac{(x-x_0)^2}{a^2}}e^{ik_0x} e^{-ikx}dx$$
I have tried completing this integral with completion of squares. I cannot find a way to finish this integral. How can I solve this integral?
• You were correct to pursue completing the square. You will need to evaluate a Gaussian integral thereafter. And to be rigorous, you need to deform the contour back to the real line by appealing to Cauchy's Integral Theorem. And that is about it. – Mark Viola Nov 20 '17 at 2:00
• After completing the square, $\hat\phi$ will become the following integral. $$\frac{1}{\sqrt{2\pi}}\int Ne^{-(x-\frac{2x_0 + ik_0a^2}{2}}e^{\frac{(2x_0+ik_0a^2)^2}{2}} e^{\frac{x_0^2}{a^2}}dx$$ – Jeremy Nov 20 '17 at 2:33
• There is an error in the previous comment. $\hat\phi$ will be defined as follows. $$\frac{1}{\sqrt{2\pi}}\int Ne^{-(x-\frac{2x_0 + ik_0a^2}{2})^2}e^{\frac{(2x_0+ik_0a^2)^2}{2}} e^{\frac{x_0^2}{a^2}}dx$$ – Jeremy Nov 20 '17 at 2:39
• After evaluating the Gaussian integral $\hat\phi$ is defined as follows. $$\hat\phi(k) = \frac{1}{\sqrt{2}}\int e^{\frac{(2x_0+ik_0a^2)^2}{2}} e^{\frac{x_0^2}{a^2}}dx$$ – Jeremy Nov 20 '17 at 2:41
• I substituted $u = x-\frac{(2x_0 + ik_0a^2)}{2}$ in. After squaring both sides and changing the integral to polar coordinates, I found that the integral evaluated to $\sqrt{\pi}$. This equation I have arrived at does not match the provided solution. I may have made an error in evaluating the Gaussian integral. Where did I go wrong? – Jeremy Nov 20 '17 at 2:46
First, enforce the substitution $x-x_0\to x$ so that
\begin{align} \int_{-\infty }^\infty e^{-\frac1{\alpha^2}(x-x_0)^2-i(k-k_0)x}\,dx&=e^{-i(k-k_0)x_0}\int_{-\infty }^\infty e^{-\frac1{\alpha^2}x^2-i(k-k_0)x}\,dx \end{align}
Then, enforce the substitution $x/\alpha\to x$ so that
$$e^{-i(k-k_0)x_0}\int_{-\infty }^\infty e^{-\frac1{\alpha^2}x^2-ikx}\,dx=\alpha e^{-i(k-k_0)x_0}\int_{-\infty }^\infty e^{-x^2-i(k-k_0)\alpha x}\,dx$$
Completing the square reveals
$$\alpha e^{-i(k-k_0)x_0}\int_{-\infty }^\infty e^{-x^2-i(k-k_0)\alpha x}\,dx=\alpha e^{-i(k-k_0)x_0}e^{-((k-k_0)\alpha/2)^2}\int_{-\infty }^\infty e^{-(x-i(k-k_0)\alpha /2)^2}\,dx$$
Enforcing the substitution $x-i(k-k_0)\alpha/2\to x$ yields
\begin{align}\alpha e^{-i(k-k_0)x_0}e^{-((k-k_0)\alpha/2)^2}\int_{-\infty }^\infty e^{-(x-i(k-k_0)\alpha /2)^2}\,dx&=\alpha e^{-i(k-k_0)x_0}e^{-((k-k_0)\alpha/2)^2}\\\\ &\times \int_{-\infty-i(k-k_0)\alpha/2 }^{\infty-i(k-k_0)\alpha/2} e^{-x^2}\,dx\end{align}
Applying Cauchy's Integral Theorem, we can deform the contour back onto the real line to obtain
$$\alpha e^{-i(k-k_0)x_0}e^{-((k-k_0)\alpha/2)^2}\int_{-\infty-i(k-k_0)\alpha/2 }^{\infty-i(k-k_0)\alpha/2} e^{-x^2}\,dx=\alpha e^{-((k-k_0)\alpha/2)^2}\underbrace{\int_{-\infty}^{\infty} e^{-x^2}\,dx}_{=\sqrt\pi}$$
Putting it all together, we find that
$$\int_{-\infty }^\infty e^{-\frac1{\alpha^2}(x-x_0)^2-i(k-k_0)x}\,dx=\alpha e^{-i(k-k_0)x_0}\sqrt\pi e^{-((k-k_0)\alpha/2)^2}$$
• I think the same techniques you've applied in your solution can be applied to the original problem. I think you have misread the problem statement. You have two exponentials with $(x-x_0)$ in both. Please note in the original problem statement that one exponential contains $(k-k_0)$ – Jeremy Nov 20 '17 at 3:45
• Jeremy, I edited on 20 November in response to your posted comment herein. Please let me know how I can improve my answer. I really want to give you the best answer I can. And feel free to both up vote and accept an answer as you see fit. Happy holidays! -Mark – Mark Viola Dec 18 '17 at 18:40
• Please feel free to up vote and accept an answer as you see fit. And Happy Holidays! – Mark Viola Dec 24 '17 at 17:29
• Your solution is incorrect by a factor of $\frac1{\sqrt{2}}$. – Jeremy Dec 29 '17 at 22:48
• @Jeremy Why do you say that? – Mark Viola Dec 29 '17 at 22:50 | 2021-04-15T23:27:10 | {
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http://mathoverflow.net/questions/71736/number-of-closed-walks-on-an-n-cube/71737 | # Number of closed walks on an $n$-cube
Is there a known formula for the number of closed walks of length (exactly) $r$ on the $n$-cube? If not, what are the best known upper and lower bounds?
Note: the walk can repeat vertices.
-
Yes (assuming a closed walk can repeat vertices). For any finite graph $G$ with adjacency matrix $A$, the total number of closed walks of length $r$ is given by
$$\text{tr } A^r = \sum_i \lambda_i^r$$
where $\lambda_i$ runs over all the eigenvalues of $A$. So it suffices to compute the eigenvalues of the adjacency matrix of the $n$-cube. But the $n$-cube is just the Cayley graph of $(\mathbb{Z}/2\mathbb{Z})^n$ with the standard generators, and the eigenvalues of a Cayley graph of any finite abelian group can be computed using the discrete Fourier transform (since the characters of the group automatically given eigenvectors of the adjacency matrix). We find that the eigenvalue $n - 2j$ occurs with multiplicity ${n \choose j}$, hence
$$\text{tr } A^r = \sum_{j=0}^n {n \choose j} (n - 2j)^r.$$
For fixed $n$ as $r \to \infty$ the dominant term is given by $n^r + (-n)^r$.
-
Thanks! that's what I needed. – Lev Reyzin Jul 31 '11 at 19:24
I'm guessing not, but is there any chance this expression has a closed form? – Lev Reyzin Aug 2 '11 at 17:12
@Lev: you mean without a summation over $n$? I doubt it. Is fixed $n$ as $r \to \infty$ not the regime you're interested in? – Qiaochu Yuan Aug 2 '11 at 17:36
In some sense, I'm more interested in fixed r as n gets large. The summation is certainly quite helpful, but of course if a closed form existed, it would even be nicer :) – Lev Reyzin Aug 2 '11 at 18:06
The number of such walks is $2^n$ (the number of vertices of the $n$-cube) times the number of walks that start (and end) at the origin. We may encode such a walk as a word in the letters $1, -1, \dots, n, -n$ where $i$ represents a positive step in the $i$th coordinate direction and $-i$ represents a negative step in the $i$th coordinate direction. The words that encode walks that start and end at the origin are encoded as shuffles of words of the form $i\ -i \ \ i \ -i \ \cdots\ i \ -i$, for $i$ from 1 to $n$. Since for each $i$ there is exactly one word of this form for each even length, the number of shuffles of these words of total length $m$ is the coefficient of $x^m/m!$ in $$\biggl(\sum_{k=0}^\infty \frac{x^{2k}}{(2k)!}\biggr)^{n} = \left(\frac{e^x + e^{-x}}{2}\right)^n.$$ Expanding by the binomial theorem, extracting the coefficient of $x^r/r!$, and multiplying by $2^n$ gives Qiaochu's formula.
Let $W(n,r)$ be the coefficient of $x^r/r!$ in $\cosh^n x$, so that $$W(n,r) = \frac{1}{2^n}\sum_{j=0}^n\binom{n}{j} (n-2j)^r.$$ Then we have the continued fraction, due originally to L. J. Rogers, $$\sum_{r=0}^\infty W(n,r) x^r = \cfrac{1}{1- \cfrac{1\cdot nx^2}{ 1- \cfrac{2(n-1)x^2}{1- \cfrac{3(n-2)x^2}{\frac{\ddots\strut} {\displaystyle 1-n\cdot 1 x^2} }}}}$$ A combinatorial proof of this formula, using paths that are essentially the same as walks on the $n$-cube, was given by I. P. Goulden and D. M. Jackson, Distributions, continued fractions, and the Ehrenfest urn model, J. Combin. Theory Ser. A 41 (1986), 21–-31.
Incidentally, the formula given above for $W(n,r)$ (equivalent to Qiaochu's formula) is given in Exercise 33b of Chapter 1 of the second edition of Richard Stanley's Enumerative Combinatorics, Volume 1 (not published yet, but available from his web page). Curiously, I had this page sitting on my desk for the past month (because I wanted to look at Exercise 35) but didn't notice until just now that this formula was on it.
-
thanks - this is nice. – Lev Reyzin Aug 2 '11 at 23:08
Assuming a "closed walk" can repeat vertices, we can count closed walks starting at $0$ by counting the $r$-sequences of $[n]$ so that each number appears an even number of times. The bijection is given by labeling edges by the coordinate that is toggled between the vertices. You can probably count these sequences by inclusion/exclusion and then multiply by $2^n/r$ to account for the choice of start position.
-
If we assume the path moves in each dimension 0 or 2 times, you can select ${n \choose r/2}$ dimensions and then permute them $r^1/2^r$ ways. This is a lower bound on the number of walks and is likely the right asymptotics. – Derrick Stolee Jul 31 '11 at 17:03
That should be $r!/2^r$. – Derrick Stolee Jul 31 '11 at 17:03 | 2015-05-25T07:46:45 | {
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http://math.stackexchange.com/questions/217462/question-about-summation-notation-index?answertab=active | # Question about summation notation index
I'm working through a proof involving the sum of covariances but the notation is tripping me up. What does it mean when you are taking a summation over the index $i < j$? For instance $\sum_{i < j}\mathrm{Cov}(X_{i},X_{j})$ where Cov is covariance.
I'm sure it's nothing complicated I just want to make sure I understand.
-
You run over all $j$ and for each $j$ you run over all $i$ such that $i < j$. For instance is $\sum_{i < j} i = 1+2+3+\cdots+n$ if $j$ can be maximal $n$. In particular this means here that you have $\sum_{i<j} Cov(x_i,x_j) = \sum_{i<1} Cov(x_i,x_1) + \sum_{i<2} Cov(x_i,x_2) + \sum_{i<3} Cov(x_i,x_3)+ \cdots +\sum_{i<n} Cov(x_i,x_n)$ if $j$ is maximal $n$. – André Oct 20 '12 at 16:04
In general $\sum_{i<j} = \sum_{j=1}^n \sum_{i=1}^j$ – André Oct 20 '12 at 16:13
This is an old post. If your question is solved, maybe you'd like to pick an answer you'd like best. Regards. – FrenzY DT. Dec 4 '12 at 17:03
### An Illustration
Suppose you have $i\in I=\{1,2,\cdots,5\}$ and $j\in J=\{1,2,\cdots,6\}$. The following picture shows you which of the terms $\sum\limits_{i<j} a_{i,j}$ include.
According to the picture, if you choose a certain value of $i$, then $j$ should at least be $i+1$, otherwise the condition $i<j$ isn't met (falling on or below the line $i=j$).
In conclusion, $$\sum\limits_{i<j} a_{i,j} = \sum_{i} \sum_{\text{All j's that's >i}} a_{i,j}.$$
If you sum $j$ last, you get $$\sum\limits_{i<j} a_{i,j} = \sum_{j} \sum_{\text{All i's that's <j}} a_{i,j}.$$
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The notation means that you sum over all allowed $i$ and $j$ such that $i < j$. The notation is compact and protects you from errors that may come from doing the end-points incorrectly.
Here is an example. Suppose that $1\leq i \leq m$ and $1\leq j \leq n$ where $m$ and $n$ are not the same. If $m < n$, the sum becomes
$$\sum_{i < j} a_{ij} = \sum_{j = i+1}^n \sum_{i=1}^{m}a_{ij}.$$
If $n < m$ we have
$$\sum_{i < j} a_{ij} = \sum_{j = i+1}^n \sum_{i=1}^{n-1}a_{ij}.$$
Formally, these are two different expressions and to determine which one you are using can take a few seconds here but it could take a lot more time for more complicated sums.
- | 2014-10-25T20:38:47 | {
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https://math.stackexchange.com/questions/2776463/mentally-finding-factors-of-an-integer-of-up-to-three-digits | # Mentally finding factors of an integer of up to three-digits.
I’m looking for the easiest methods to find factors, not necessarily prime, of positive integers $<1000$, preferably mentally, but using pen and blank paper when required.
Although these techniques might be useful for a certain televised Numbers Game, I’m just looking for methods faster than trial division.
I take for granted full knowledge of the twelve by twelve multiplication tables, and recognition of primes up to $101$
This, with the addition of $7\cdot13=91$ should cover much of the factorisation up to two digits.
Notation Using the idea of hundreds, tens and units, so $n=100H+10T+U$ is the number, I’ll use $H,T,U$ for the individual digits, $HT$ for the first two digits and $TU$ for the last two.
For example, $n=456$ gives $H=4$, $T=5$, $U=6$, $HT=45$ and $TU=56$
Some easy published divisibility tests
If $U=0$ then $10|n$
If $(TU)\equiv 0{\pmod{25}}$ then ${25}|n$
If $U=5$ then $5|n$
If $U\equiv 0{\pmod 2}$ then $2|n$
If $(H+T+U)\equiv 0{\pmod 9}$ then $9|n$
If $(H+T+U)\equiv 0{\pmod 3}$ then $3|n$
Merged published divisibility tests for primes $p=7$ to $47$
With $z$ taken from the following, if $(HT+zU)\equiv 0{\pmod p}$ then $p|n$
$$(p,z)$$ $$(7,-2)$$ $$(11,-1)$$ $$(13,4)$$ $$(17,-5)$$ $$(19,2)$$ $$(23,7)$$ $$(29,3)$$ $$(31,-3)$$ $$(37,-11)$$ $$(41,-4)$$ $$(43,13)$$ $$(43,-30)$$ $$(47,-14)$$
For some examples in the range, the test may need to be repeated on the result.
As $31^2=961$, the last few test are of limited use.
Alternative divisibility test for $11$
If $(H+U-T)=0$ or $11$ then $11|n$
Difference of two squares
Using $a^2-b^2\equiv(a+b)(a-b)$ helps in a few cases. For example,
$$899=900-1=(30+1)(30-1)=31\cdot29$$
$$91=100-9=(10+3)(10-3)=13\cdot7$$
$$391=400-9=(20+3)(20-3)=23\cdot17$$
My question
I was tempted to ask what other divisibility or factorisation techniques exist that are potential easier than trial division, but “easier” is subjective; so I questions is:.
What other mental divisibility or factorisation techniques exist for an integer up to three digits?
Clarification update 11 May 2018
Sorry, I don’t feel I’ve sufficiently emphasised the importance of time, just assuming that easier was faster.
Full factorisation is ideal; partial factorisation is much better than nothing, so $663=3\cdot13\cdot17$ is best, but $663=3\cdot221$ is better than nothing.
Obtaining more than one small prime factor with one test is desirable, where the prime factors are obvious, as per the tests for $10,25$ and $9$, but a full list of all possible factors of $n$ is not needed.
• For three digit numbers you only have to do 11 trial divisions to find factors (for the primes from 2 up to 31). If none of them divides your number, no number does (except 1 and the number itself, of course). Trial divisions for 2,3,5,7 and 11 are easy. – YukiJ May 11 '18 at 10:58
• @YukiJ The OP wants to find factors, not necessarily prime. For instance, 6 is a factor. – scaaahu May 11 '18 at 11:35
• @scaaahu Yes, but if we observe that 2 and 3 are factors in a number then their product $2\cdot 3 =6$ is necessarily also a factor. Hence, if you find the prime factors, you also found the remaining other factors. – YukiJ May 11 '18 at 11:37
• @YukiJ Having a factor 9 implies having a factor 3, not the other way around. – scaaahu May 11 '18 at 11:52
• @scaaahu True but if you know that 3 divides $n$ then you can also check if 3 also divides $n/3$. If so, 9 divides $n$. In the particular case with 9 you could even check if the sum of digits of $n$ is divisible by 9. – YukiJ May 11 '18 at 11:55
As you have stated:
Checking a number is even will automatically give you the factor of $2$.
Summing the digits, if the digit sum is a multiple of $3$ or $9$, the number is also a multiple of $3$ or $9$.
Multiples of $5$ end in $5$, multiples of $10$ end in $0$.
For my extra method, let's consider $473$. It has no obvious factors based on those rules, so here I would use prime multiple subtraction. Start with $7$, the multiple of $7$ that ends in $3$ is $7\cdot9=63$, so subtract $63$ to get $410$. $41$ is not a multiple of $7$ or $9$, so neither are factors. Then consider $11$. $3\cdot 11=33$ so subtract $33$ to get $440$. $440=40\cdot 11$, hence we can see that $473=43\cdot 11$.
Note for $11$ that if you have $ab\cdot11$ , where $a,b$ represent the digits, the result is $a(a+b)b$, again to mean digits: e.g. $35\cdot11=385$. When $a+b>10$, it is harder to spot but still possible. What you should look for then is: $a(a+b-11)b$, and this is $(a-1)b \cdot 11$. E.g. for $528$, $5+8-11=2, \therefore 528=11*48$
If you can identify numbers as concatenations of multiples of the same number, they will be a multiple of that number. E.g. $654$ contains $6$ and $54$, both of which are multiples of $6$, and so we can do $6\div6=1$, and $54\div6=09$, therefore $654\div6=109$
• Thanks for your answer, which I’ve up-voted. I’ve used a similar method with numbers such as $721$. – Old Peter May 11 '18 at 14:42
• Thanks for the upvote. Note that $721$ is $7\cdot103$, you can see this by the subtraction method, or notice that $7$ and $21$ are both multiples of $7$, and so we can simply divide both by $7$ for $1$ and $03$ and then put these back together. A different example might be $726$, we see $72=12\cdot 6$ and $6=1\cdot 6$, putting back together we get $121$. – Rhys Hughes May 11 '18 at 14:56 | 2019-12-06T23:45:06 | {
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https://mathhelpboards.com/threads/d-e-word-problem.9047/ | # D.E word problem
#### bergausstein
##### Active member
1. radium decompose at the rate proportional to the amount itself. if the half life is 1600, find the percentage remaining after at the end of 200 years.
can you me go about solving this. thanks!
#### MarkFL
Staff member
Let $R(t)$ be the mass of radium in a given sample at time $t$. How can we mathematically state how this mass changes with time in general? Just look at the first sentence and use that to try to model this change with an initial value problem.
#### bergausstein
##### Active member
$R(t)=R_oe^{kt}$ where $k<0$
now the half-life is 1600 so,
$1600=3200e^{kt}$
#### MarkFL
Staff member
$R(t)=R_oe^{kt}$ where $k<0$
now the half-life is 1600 so,
$1600=3200e^{kt}$
You have the correct equation for the mass of a sample, but the half-life being 1600 years means that at time t=1600 then $R(t)=\dfrac{1}{2}R_0$. That is (I like to always use a positive constant):
$$\displaystyle R(1600)=R_0e^{-1600k}=\frac{1}{2}R_0$$
Divide through by $$\displaystyle R_0$$ and the convert from exponential to logarithmic form to solve for $k$. Another way to look at it is:
$$\displaystyle R(t)=R_02^{-\frac{t}{1600}}$$
And then to find the percentage remaining after $t$ years, use:
$$\displaystyle \frac{100R(t)}{R_0}=100\cdot2^{-\frac{t}{1600}}$$
#### bergausstein
##### Active member
from here solving for k
$\displaystyle R(1600)=R_0e^{-1600k}=\frac{1}{2}R_0$
dividing both sides by $R_o$
$\displaystyle e^{-1600k}=\frac{1}{2}$
taking the $\ln$ of both sides
$-1600k=\ln\frac{1}{2}$
dividing both sides by -1600
$\displaystyle k=-\frac{-\ln2}{1600}$ or $k=0.0004332$
now
$\displaystyle R(t)=R_0e^{-(0.0004332)t}$
how can I find the initial $R_o$?
Last edited:
#### MarkFL
Staff member
You don't need to know the initial amount, you need to know what percentage of the initial amount remains after 200 years.
#### bergausstein
##### Active member
You don't need to know the initial amount, you need to know what percentage of the initial amount remains after 200 years.
how can I do that if there's no given initial amount? can you tell how to go about it?
#### Prove It
##### Well-known member
MHB Math Helper
from here solving for k
$\displaystyle R(1600)=R_0e^{-1600k}=\frac{1}{2}R_0$
dividing both sides by $R_o$
$\displaystyle e^{-1600k}=\frac{1}{2}$
taking the $\ln$ of both sides
$-1600k=\ln\frac{1}{2}$
dividing both sides by -1600
$\displaystyle k=-\frac{-\ln2}{1600}$ or $k=0.0004332$
now
$\displaystyle R(t)=R_0e^{-(0.0004332)t}$
how can I find the initial $R_o$?
Do you think it's a good idea to give a decimal approximation when the function simplifies greatly if kept exact?
\displaystyle \begin{align*} k &= \frac{1}{1600} \ln{ (2) } \end{align*}
and so
\displaystyle \begin{align*} R(t) &= R_0 \exp { \left[ \frac{t}{1600} \ln{(2)} \right] } \\ &= R_0 \exp{ \left[ \ln{ \left( 2 ^{ \frac{t}{1600} } \right) } \right] } \\ &= R_0 \cdot 2^{ \frac{t}{1600} } \end{align*}
and so after 200 years, what proportion of the initial amount do you have?
#### MarkFL
Staff member
how can I do that if there's no given initial amount? can you tell how to go about it?
This is what I posted before:
...to find the percentage remaining after $t$ years, use:
$$\displaystyle \frac{100R(t)}{R_0}$$
#### bergausstein
##### Active member
$\displaystyle \frac{100R(200)}{R_0}=100\cdot2^{\frac{200}{1600}}$
do you mean like this? how can I find the percentage of remaining here?
#### MarkFL
Staff member
$\displaystyle \frac{100R(200)}{R_0}=100\cdot2^{\frac{200}{1600}}$
do you mean like this? how can I find the percentage of remaining here?
Reduce the exponent, and then that is the exact percentage remaining, and then you can use a calculator to obtain a decimal approximation if you like.
#### Prove It
##### Well-known member
MHB Math Helper
It appears there was something wrong with your initial model.
You know that your model involves exponential decay, so after each unit of time passes, there is a certain constant amount multiplying through. In other words
\displaystyle \begin{align*} R_1 &= C\,R_0 \\ R_2 &= C^2\,R_0 \\ R_3 &= C^3\,R_0 \\ \vdots \\ R_t &= C^t\,R_0 \end{align*}
Since you know that when \displaystyle \begin{align*} t = 1600, R_{1600} = \frac{1}{2}R_0 \end{align*}, that means
\displaystyle \begin{align*} \frac{1}{2}R_0 &= C^{1600}\,R_0 \\ \frac{1}{2} &= C^{1600} \\ \ln{ \left( \frac{1}{2} \right) } &= \ln{ \left( C^{1600} \right) } \\ \ln{ \left( \frac{1}{2} \right) } &= 1600 \ln{(C)} \\ \frac{1}{1600} \ln{ \left( \frac{1}{2} \right) } &= \ln{(C)} \\ \ln{ \left[ \left( \frac{1}{2} \right) ^{\frac{1}{1600}} \right] } &= \ln{(C)} \\ \left( \frac{1}{2} \right) ^{ \frac{1}{1600} } &= C \\ 2^{ -\frac{1}{1600}} &= C \end{align*}
and thus your model is \displaystyle \begin{align*} R(t) = \left( 2^{-\frac{1}{1600}} \right) ^t \, R_0 = 2^{-\frac{t}{1600}}\,R_0 \end{align*}.
Now if \displaystyle \begin{align*} t = 200 \end{align*}, that gives
\displaystyle \begin{align*} R(200) &= 2^{-\frac{200}{1600}}\,R_0 \\ &= 2^{-\frac{1}{8}}\,R_0 \\ &\approx 0.917 \, R_0 \end{align*}
So what percentage of the original amount do you have? | 2020-11-30T13:32:14 | {
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https://mathhelpboards.com/threads/probability.5953/ | # Probability.
#### M R
##### Active member
A spinner has three possible outcomes which occur with probabilities a, b and c where a+b+c=1.
What is the expected number of spins required until all three outcomes are seen?
There's an easy way and a harder way to do this. Guess which I did first.
#### Opalg
##### MHB Oldtimer
Staff member
A spinner has three possible outcomes which occur with probabilities a, b and c where a+b+c=1.
What is the expected number of spins required until all three outcomes are seen?
There's an easy way and a harder way to do this. Guess which I did first.
This is a variation on the coupon collector's problem. In the case when all the probabilities are equal ($a=b=c=1/3$), the expected number of spins is $3\bigl(1 +\frac12 + \frac13) = \frac{11}2.$
In the general case, I certainly don't see an easy way to approach the problem, and I don't get an easy-looking formula for the answer.
Write $A$, $B$, $C$ for the outcomes with probabilities $a$, $b$, $c$ respectively. If the first spin gives an $A$, then the expected number of spins until a $B$ or $C$ occurs is $\dfrac1{b+c}$. The probability that this outcome is a $B$ is $\dfrac b{b+c}$, in which case the expected number of further spins until a $C$ turns up is $1/c.$ And the probability that a $C$ occurs before a $B$ is $\dfrac c{b+c}$, in which case the expected number of further spins until a $B$ turns up is $1/b.$ Therefore the total expected number of spins for all three outcomes to occur (given that the $A$ appears first) is $$1 + \frac1{b+c}\Bigl(\frac b{b+c}\,\frac1c + \frac c{b+c}\,\frac1b\Bigr) = 1 + \frac{b^2+c^2}{(b+c)^2bc} = \frac1{bc} - \frac2{(1-a)^2}$$ (in the last step, I have written the $b^2+c^2$ in the numerator as $(b+c)^2 - 2bc$, and in the denominator $b+c = 1-a$).
Multiply that by $a$, which is the probability of the $A$ occurring first, add two similar terms for the probabilities of $B$ or $C$ occurring first, and you get the answer for the expected number of spins as $$1 + \frac{a^2+b^2+c^2}{abc} - 2\biggl(\frac a{(1-a)^2} + \frac b{(1-b)^2} + \frac c{(1-c)^2}\biggl).$$
That looks messy, not the sort of thing that you could find easily? But it does reduce to $11/2$ when $a=b=c=1/3$, which makes me think that it should be correct.
#### M R
##### Active member
Hi Opalg
Maybe it wasn't easy but it was much easier than the other method which I will post if no one else does.
Your formula does give the right answer for a=b=c but not for other possibilities.
For comparison purposes:
a=1/2, b=1/3, c=1/6 should give 73/10
and
a=9/20, b=9/20, c=1/10 should give 353/33.
#### Opalg
##### MHB Oldtimer
Staff member
Your formula does give the right answer for a=b=c but not for other possibilities.
For comparison purposes:
a=1/2, b=1/3, c=1/6 should give 73/10
and
a=9/20, b=9/20, c=1/10 should give 353/33.
Stupid stupid mistake! My method was correct but I left out a $+$ sign, converting a sum into a product. The expression $$1 + \frac1{b+c}\Bigl(\frac b{b+c}\,\frac1c + \frac c{b+c}\,\frac1b\Bigr)$$
(for the expected number of spins for all three outcomes to occur, given that the $A$ appears first) should have been $$1 + \frac1{b+c} + \Bigl(\frac b{b+c}\,\frac1c + \frac c{b+c}\,\frac1b\Bigr) = 1 + \frac{bc + b^2+c^2}{(b+c)bc} = 1 + \frac{(b+c)^2 - bc}{(b+c)bc} = 1 + \frac{b+c}{bc} - \frac1{b+c}.$$ The answer for the total expected number of spins then comes out as $$1 + \frac{a(b+c)}{bc} + \frac{b(c+a)}{ca} + \frac{c(a+b)}{ab} - \frac a{b+c} - \frac b{c+a} - \frac c{a+b}.$$ That gives values agreeing with your results for a=1/2, b=1/3, c=1/6 and for a=9/20, b=9/20, c=1/10.
#### M R
##### Active member
The history of this problem (for me personally):
On another forum someone posted a 'hard probability question'. It was the a=b=9/20, c=1/10 case. Looking back at that forum I see that I managed to get an answer six days later.
Months later I saw a post on MHF which taught me a better approach so I went back and did it again, the easy way. Unfortunately I can't view MHF to find out who my teacher was.
I think it's essentially the same as Opalg's method but here's the 'easy' method as I did it:
Events A, B and C have probabilities a, b and c.
Let the expected waiting time until A occurs be $$E(W_A)$$,
and the expected waiting time until A and B occur be $$E(W_{AB})$$ and so on.
$$E(W_{AB})=c(1+E(W_{AB}))+a(1+E(W_B))+b(1+E(W_A))$$
$$E(W_{AB})=c(1+E(W_{AB}))+a(1+1/b)+b(1+1/a)$$
$$E(W_{AB})=1 +cE(W_{AB})+a/b+b/a$$
$$E(W_{AB})=\frac{1+a/b+b/a}{1-c}$$
Similarly
$$E(W_{AC})=\frac{1+a/c+c/a}{1-b}$$
and
$$E(W_{BC})=\frac{1+b/c+c/b}{1-a}$$
Then $$E(W_{ABC})=a(1+E(W_{BC}))+b(1+E(W_{AC}))+c(1+E(W_{AB}))$$
and the method I used at first:
By thinking about how the sequence ends I knew I wanted all the ways to get As and Bs ending with C etc.
ABC
BAC
AABC
ABAC
BAAC
ABBC
BABC
BBAC
etc.
This led me to the sum
$$\displaystyle E(W)=\sum_{n=2}^\infty (n+1) \sum_{r=1}^{n-1} \binom{n}{r}(a^rb^{n-r}c+a^rc^{n-r}b+b^rc^{n-r}a)$$
$$\displaystyle =\sum_{n=2}^\infty (n+1) [ c((a+b)^n-a^n-b^n) +b((a+c)^n-a^n-c^n)+a((b+c)^n-b^n-c^n)]$$
This sum involves a number of geometric progressions and a number of sums of another type.
The other type is of the form $$\displaystyle S=\sum_{n=2}^\infty n x^n$$.
Multiplying by $$\displaystyle x$$ gives $$\displaystyle Sx=\sum_{n=2}^\infty n x^{n+1}$$ and so $$\displaystyle S-Sx = 2x^2+x^3+x^4+...$$
$$\displaystyle S(1-x)=x^2 + \frac{x^2}{1-x}$$ and $$\displaystyle S=\frac{x^2}{1-x}+\frac{x^2}{(1-x)^2}$$
Applying this formula, together with the formula for the sum of a GP we get
$$\displaystyle E(W)=$$
$$\displaystyle (a+b)^2-\frac{ca^2}{1-a}-\frac{cb^2}{1-b}$$
$$\displaystyle + (a+c)^2-\frac{ba^2}{1-a}-\frac{bc^2}{1-c}$$
$$\displaystyle + (b+c)^2-\frac{ab^2}{1-b}-\frac{ac^2}{1-c}$$
$$\displaystyle +(a+b)^2(1+1/c)-c\left(\frac{a^2}{1-a}+\frac{a^2}{(1-a)^2}+\frac{b^2}{1-b}+\frac{b^2}{(1-b)^2}\right)$$
$$\displaystyle +(a+c)^2(1+1/b)-b\left(\frac{a^2}{1-a}+\frac{a^2}{(1-a)^2}+\frac{c^2}{1-c}+\frac{c^2}{(1-c)^2}\right)$$
$$\displaystyle +(b+c)^2(1+1/a)-a\left(\frac{b^2}{1-b}+\frac{b^2}{(1-b)^2}+\frac{c^2}{1-c}+\frac{c^2}{(1-c)^2}\right)$$
What a mess!
This could be simplified using $$\displaystyle a+b+c=1$$ but having checked it against the first method I'm happy that it's correct and I'm not interested in doing the simplification. | 2021-01-25T14:01:17 | {
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