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https://math.stackexchange.com/questions/2197132/calculating-area-of-triangle/2197147 | # Calculating area of triangle
Problem
In the triangle above, we have $AC = 12, \ \ BD = 10, \ \ CD = 6$.
Find the area of $\triangle ABD$.
Caveat
This was part of an exam where you are not allowed to use a calculator or any other digital tools.
My progress
Given $AC, CD, \measuredangle C$, I was able to find the length of $AD$ using the law of cosines. It yielded $AD = \sqrt{180} = 6\sqrt5$.
From here, I decided to try and use the law of sines to find $\measuredangle A$ as it would be the same in $\triangle ABD$ and $\triangle ACD$.
I got $$\frac{\sin A}{CD} = \frac{\sin C}{AD}$$.
This yields $$\sin A = \frac{CD \sin C}{AD} = \frac{6}{\sqrt{180}}$$.
Then I need to calculate $\arcsin\frac{6}{\sqrt{180}}$, which I can't, because I can't use a calculator for this problem. (Well, technically I can, but I want to solve this within the constraints placed on the participants.)
From here, I am stuck. In $\triangle ABD$, I know now two of the lengths, but I don't know any of the angles.
Question
Am I overlooking an easy inverse sine here? Or is there another way to calculate this area?
Thanks in advance for any help!
• If this is part of an exam, why are you asking for help? – N. F. Taussig Mar 21 '17 at 20:07
• @N.F.Taussig - It was an exam given in August of 2016. I'm preparing for the 2017 one ;) – Alec Mar 21 '17 at 20:08
## 3 Answers
Find $BC$ in the right triangle $BCD$; you have the hypotenuse is $10$ and the other side is $6$ so $BC=8$.
Area $\triangle ACD = \frac12\cdot 12\cdot 6 = 36$.
Area $\triangle BCD = \frac12\cdot 8\cdot 6 = 24$.
Area $\triangle ABD = 36-24 = 12$.
• Damn... I really blinded myself on this one. Thanks! – Alec Mar 21 '17 at 20:11
Calculate $BC$ with the pythagorean theorem, and subtract the areas of the two right angled triangles.
$$BC=\sqrt{10^2-6^2}=8$$ Hence, $$AB=12-8=4$$ and $$S_{\Delta ABD}=\frac{4\cdot6}{2}=12$$ | 2019-08-24T19:46:36 | {
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http://openstudy.com/updates/55c3bfcfe4b06d80932e877f | ## mathmath333 one year ago Set Theory
1. mathmath333
\large \color{black}{\begin{align} \normalsize \text{A ∩ B = A ∩ C need not imply B = C.} \quad \normalsize \text{ Explain through an example.}\hspace{.33em}\\~\\ \end{align}}
2. Kainui
A = {1} B = {1, 2} C = {1, 3}
3. ikram002p
huh kai was so fast :P
4. Kainui
Hahaha there could have been a simpler example: A = {} B = {1} C = {2}
5. mathmath333
thnx
6. Kainui
I think that last example I gave sorta shows that it's almost like this is a lot like multiplying by zero. a*b=a*c This doesn't mean b=c, since a=0 is possible.
7. mathmath333
Let A and B be sets. If A ∩ X = B ∩ X = φ and A ∪ X = B ∪ X for some set X, show that A = B.
8. ikram002p
in general any case with $A\cap B \neq \varnothing$ does not imply A=C
9. ikram002p
or $$C\cap B \neq \varnothing$$
10. mathmath333
i need an example like the previous one
11. Kainui
ikram needs more owl bucks sry
12. ikram002p
A={1,2,3} B={1,2,3} C={1,2,3,4}
13. ikram002p
:O @Kainui i dont lol
14. Kainui
Hahaha jk :P
15. mathmath333
is that the example for this que Let A and B be sets. If A ∩ X = B ∩ X = φ and A ∪ X = B ∪ X for some set X, show that A = B.
16. ikram002p
you wanna prove this ?
17. mathmath333
yes with example
18. ikram002p
ok example A={1,2,3} B={1.4} C={1,2.3}
19. mathmath333
but C is not there in the question
20. ikram002p
sorry i ment to say x :P
21. ikram002p
now to prove they are two side of the prove 1- show A is a subset of B 2- show B is a subset of A
22. ikram002p
let $$c\in A \cup X$$ so $$c\in A ~or~c\in X$$ if c in A then $$c\in B\cup X$$ and $$c\in B$$ (since c is not in x) thus $$A\subset B$$ the other way is alike :P sorry im lazy to type
23. thomas5267
Is it possible to do something like this? $A\cup X=B\cup X\land A\cap X=B\cap X=\emptyset\implies A\cup X\setminus X=B\cup X\setminus X\implies A=B$ | 2016-10-27T17:16:28 | {
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# The average (arithmetic mean) length per film for a group of 21 films
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The average (arithmetic mean) length per film for a group of 21 films is t minutes. If a film that runs for 66 minutes is removed from the group and replaced by one that runs for 52 minutes, what is the average length per film, in minutes, for the new group of films, in terms of t?
A) $$t + \frac{2}{3}$$
B) $$t - \frac{2}{3}$$
C) $$21t+14$$
D) $$t + \frac{3}{2}$$
E) $$t - \frac{3}{2}$$
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Re: The average (arithmetic mean) length per film for a group of 21 films [#permalink]
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05 May 2016, 20:09
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mcelroytutoring wrote:
The average (arithmetic mean) length per film for a group of 21 films is t minutes. If a film that runs for 66 minutes is removed from the group and replaced by one that runs for 52 minutes, what is the average length per film, in minutes, for the new group of films, in terms of t?
A) $$t + \frac{2}{3}$$
B) $$t - \frac{2}{3}$$
C) $$21t+14$$
D) $$t + \frac{3}{2}$$
E) $$t - \frac{3}{2}$$
hi,
you do not have to get into any calculations if you understand and take the Q logically..
avg time for 21 is t min..
you add 52 and remove 66.. so basically you are removing 14 min from total..
effect on each or average =$$\frac{14}{21}= \frac{2}{3}$$..
so final average = $$t-\frac{2}{3}$$...
B
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The average (arithmetic mean) length per film for a group of 21 films [#permalink]
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Updated on: 06 May 2016, 12:30
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Logic tells us that the average length per film will go down, so we can eliminate all answers that add to the original average of t. That leaves either B or E. When executing the algebra, the key step is being able to quickly and easily turn one fraction with two numerators $$(\frac{21t-14}{21})$$ into two fractions with one numerator each $$(\frac{21t}{21}-\frac{14}{21})$$.
Above is a visual that should help.
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Originally posted by mcelroytutoring on 05 May 2016, 19:28.
Last edited by mcelroytutoring on 06 May 2016, 12:30, edited 6 times in total.
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Re: The average (arithmetic mean) length per film for a group of 21 films [#permalink]
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15 Dec 2016, 14:05
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Great Solutions above.
Here is my solution =>
Average length = t minutes
Sum(21) = 21t
New sum = 21t-66+52 = 21t-14
Hence new mean = 21t-14 / 21 = t-2/3
Hence B
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Re: The average (arithmetic mean) length per film for a group of 21 films [#permalink]
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20 Jul 2017, 15:08
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average = sum of numbers/total number
t = s/21
s = 21t
Now as per question
s =21t -66 +52
s = 21t - 14
New average
t1 = (21t - 14) / 21
=> t - 2/3
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The average (arithmetic mean) length per film for a group of 21 films [#permalink]
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06 Mar 2018, 09:16
stonecold wrote:
Great Solutions above.
Here is my solution =>
Average length = t minutes
Sum(21) = 21t
New sum = 21t-66+52 = 21t-14
Hence new mean = 21t-14 / 21 = t-2/3
Hence B
stonecold how from this 21t-14 / 21 you got t-2/3 , I understand you divided by 7 , but there are two 21 could you write in detail
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The average (arithmetic mean) length per film for a group of 21 films [#permalink]
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06 Mar 2018, 11:18
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dave13 wrote:
stonecold wrote:
Great Solutions above.
Here is my solution =>
Average length = t minutes
Sum(21) = 21t
New sum = 21t-66+52 = 21t-14
Hence new mean = 21t-14 / 21 = t-2/3
Hence B
stonecold how from this 21t-14 / 21 you got t-2/3 , I understand you divided by 7 , but there are two 21 could you write in detail
Hey dave13
$$\frac{21t-14}{21} = \frac{21t}{21} - \frac{14}{21} = t - \frac{2}{3}$$
Hope that helps
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Re: The average (arithmetic mean) length per film for a group of 21 films [#permalink]
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15 Aug 2018, 16:09
Hi,
We can use deviation technique to solve such type of questions. The answer would be in less than 30 Secs
refer: https://gmatclub.com/forum/a-student-s- ... l#p2113194
So new average would be
t-(14/21)
t-2/3
Hence B.
Re: The average (arithmetic mean) length per film for a group of 21 films &nbs [#permalink] 15 Aug 2018, 16:09
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# The average (arithmetic mean) length per film for a group of 21 films
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# Inequalities trick
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Inequalities trick [#permalink] 16 Mar 2010, 09:11
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I learnt this trick while I was in school and yesterday while solving one question I recalled.
Its good if you guys use it 1-2 times to get used to it.
Suppose you have the inequality
f(x) = (x-a)(x-b)(x-c)(x-d) < 0
Just arrange them in order as shown in the picture and draw curve starting from + from right.
now if f(x) < 0 consider curve having "-" inside and if f(x) > 0 consider curve having "+" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.
Don't forget to arrange then in ascending order from left to right. a<b<c<d
So for f(x) < 0 consider "-" curves and the ans is : (a < x < b) , (c < x < d)
and for f(x) > 0 consider "+" curves and the ans is : (x < a), (b < x < c) , (d < x)
If f(x) has three factors then the graph will have - + - +
If f(x) has four factors then the graph will have + - + - +
If you can not figure out how and why, just remember it.
Try to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.
For the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively.
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Re: Inequalities trick [#permalink] 22 Oct 2010, 05:33
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Yes, this is a neat little way to work with inequalities where factors are multiplied or divided. And, it has a solid reasoning behind it which I will just explain.
If (x-a)(x-b)(x-c)(x-d) < 0, we can draw the points a, b, c and d on the number line.
e.g. Given (x+2)(x-1)(x-7)(x-4) < 0, draw the points -2, 1, 7 and 4 on the number line as shown.
Attachment:
doc.jpg [ 7.9 KiB | Viewed 21052 times ]
This divides the number line into 5 regions. Values of x in right most region will always give you positive value of the expression. The reason for this is that if x > 7, all factors above will be positive.
When you jump to the next region between x = 4 and x = 7, value of x here give you negative value for the entire expression because now, (x - 7) will be negative since x < 7 in this region. All other factors are still positive.
When you jump to the next region on the left between x = 1 and x = 4, expression will be positive again because now two factors (x - 7) and (x - 4) are negative, but negative x negative is positive... and so on till you reach the leftmost section.
Since we are looking for values of x where the expression is < 0, here the solution will be -2 < x < 1 or 4< x < 7
It should be obvious that it will also work in cases where factors are divided.
e.g. (x - a)(x - b)/(x - c)(x - d) < 0
(x + 2)(x - 1)/(x -4)(x - 7) < 0 will have exactly the same solution as above.
Note: If, rather than < or > sign, you have <= or >=, in division, the solution will differ slightly. I will leave it for you to figure out why and how. Feel free to get back to me if you want to confirm your conclusion.
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Save $100 on Veritas Prep GMAT Courses And Admissions Consulting Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options. Veritas Prep Reviews Math Forum Moderator Joined: 20 Dec 2010 Posts: 2047 Followers: 128 Kudos [?]: 919 [13] , given: 376 Re: Inequalities trick [#permalink] 11 Mar 2011, 05:49 13 This post received KUDOS 4 This post was BOOKMARKED vjsharma25 wrote: VeritasPrepKarishma wrote: vjsharma25 wrote: How you have decided on the first sign of the graph?Why it is -ve if it has three factors and +ve when four factors? Check out my post above for explanation. I understand the concept but not the starting point of the graph.How you decide about the graph to be a sine or cosine waveform?Meaning graph starts from the +ve Y-axis for four values and starts from -ve Y-axis for three values. What if the equation you mentioned is (x+2)(x-1)(x-7)<0,will the last two ranges be excluded or the graph will also change? I always struggle with this as well!!! There is a trick Bunuel suggested; (x+2)(x-1)(x-7) < 0 Here the roots are; -2,1,7 Arrange them in ascending order; -2,1,7; These are three points where the wave will alternate. The ranges are; x<-2 -2<x<1 1<x<7 x>7 Take a big value of x; say 1000; you see the inequality will be positive for that. (1000+2)(1000-1)(1000-7) is +ve. Thus the last range(x>7) is on the positive side. Graph is +ve after 7. Between 1 and 7-> -ve between -2 and 1-> +ve Before -2 -> -ve Since the inequality has the less than sign; consider only the -ve side of the graph; 1<x<7 or x<-2 is the complete range of x that satisfies the inequality. _________________ Manager Joined: 29 Sep 2008 Posts: 154 Followers: 2 Kudos [?]: 31 [8] , given: 1 Re: Inequalities trick [#permalink] 22 Oct 2010, 10:45 8 This post received KUDOS 3 This post was BOOKMARKED if = sign is included with < then <= will be there in solution like for (x+2)(x-1)(x-7)(x-4) <=0 the solution will be -2 <= x <= 1 or 4<= x <= 7 in case when factors are divided then the numerator will contain = sign like for (x + 2)(x - 1)/(x -4)(x - 7) < =0 the solution will be -2 <= x <= 1 or 4< x < 7 we cant make 4<=x<=7 as it will make the solution infinite correct me if i am wrong Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 4755 Location: Pune, India Followers: 1112 Kudos [?]: 5022 [4] , given: 164 Re: Inequalities trick [#permalink] 11 Mar 2011, 18:57 4 This post received KUDOS Expert's post vjsharma25 wrote: I understand the concept but not the starting point of the graph.How you decide about the graph to be a sine or cosine waveform?Meaning graph starts from the +ve Y-axis for four values and starts from -ve Y-axis for three values. What if the equation you mentioned is (x+2)(x-1)(x-7)<0,will the last two ranges be excluded or the graph will also change? Ok, look at this expression inequality: (x+2)(x-1)(x-7) < 0 Can I say the left hand side expression will always be positive for values greater than 7? (x+2) will be positive, (x - 1) will be positive and (x-7) will also be positive... so in the rightmost regions i.e. x > 7, all three factors will be positive. The expression will be positive when x > 7, it will be negative when 1 < x < 7, positive when -2 , x < 1 and negative when x < -2. We need the region where the expression is less than 0 i.e. negative. So either 1 < x < 7 or x < -2. Now let me add another factor: (x+8)(x+2)(x-1)(x-7) Can I still say that the entire expression is positive in the rightmost region i.e. x>7 because each one of the four factors is positive? Yes. So basically, your rightmost region is always positive. You go from there and assign + and - signs to the regions. Your starting point is the rightmost region. Note: Make sure that the factors are of the form (ax - b), not (b - ax)... e.g. (x+2)(x-1)(7 - x)<0 Convert this to: (x+2)(x-1)(x-7)>0 (Multiply both sides by '-1') Now solve in the usual way. Assign '+' to the rightmost region and then alternate with '-' Since you are looking for positive value of the expression, every region where you put a '+' will be the region where the expression will be greater than 0. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Save$100 on Veritas Prep GMAT Courses And Admissions Consulting
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Re: Inequalities trick [#permalink] 10 Aug 2011, 16:01
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WoW - This is a cool thread with so many thing on inequalities....I have compiled it together with some of my own ideas...It should help.
1) CORE CONCEPT
@gurpreetsingh -
Suppose you have the inequality
f(x) = (x-a)(x-b)(x-c)(x-d) < 0
Arrange the NUMBERS in ascending order from left to right. a<b<c<d
Draw curve starting from + from right.
now if f(x) < 0 consider curve having "-" inside and if f(x) > 0 consider curve having "+" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.
So for f(x) < 0 consider "-" curves and the ans is : (a < x < b) , (c < x < d)
and for f(x) > 0 consider "+" curves and the ans is : (x < a), (b < x < c) , (d < x)
If f(x) has three factors then the graph will have - + - +
If f(x) has four factors then the graph will have + - + - +
If you can not figure out how and why, just remember it.
Try to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.
For the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively.
Note: Make sure that the factors are of the form (ax - b), not (b - ax)...
example -
(x+2)(x-1)(7 - x)<0
Convert this to: (x+2)(x-1)(x-7)>0 (Multiply both sides by '-1')
Now solve in the usual way. Assign '+' to the rightmost region and then alternate with '-'
Since you are looking for positive value of the expression, every region where you put a '+' will be the region where the expression will be greater than 0.
2) Variation - ODD/EVEN POWER
@ulm/Karishma -
if we have even powers like (x-a)^2(x-b)
we don't need to change a sign when jump over "a".
This will be same as (x-b)
We can ignore squares BUT SHOULD consider ODD powers
example -
2.a
(x-a)^3(x-b)<0 is the same as (x-a)(x-b) <0
2.b
(x - a)(x - b)/(x - c)(x - d) < 0 ==> (x - a)(x - b)(x-c)^-1(x-d)^-1 <0
is the same as (x - a)(x - b)(x - c)(x - d) < 0
3) Variation <= in FRACTION
@mrinal2100 -
if = sign is included with < then <= will be there in solution
like for (x+2)(x-1)(x-7)(x-4) <=0 the solution will be -2 <= x <= 1 or 4<= x <= 7
BUT if it is a fraction the denominator in the solution will not have = SIGN
example -
3.a
(x + 2)(x - 1)/(x -4)(x - 7) < =0
the solution will be -2 <= x <= 1 or 4< x < 7
we cant make 4<=x<=7 as it will make the solution infinite
4) Variation - ROOTS
@Karishma -
As for roots, you have to keep in mind that given \sqrt{x}, x cannot be negative.
\sqrt{x} < 10
implies 0 < \sqrt{x} < 10
Squaring, 0 < x < 100
Root questions are specific. You have to be careful. If you have a particular question in mind, send it.
Refer - inequalities-and-roots-118619.html#p959939
Some more useful tips for ROOTS....I am too lazy to consolidate
<5> THESIS -
@gmat1220 -
Once algebra teacher told me - signs alternate between the roots. I said whatever and now I know why Watching this article is a stroll down the memory lane.
I will save this future references....
Anyone wants to add ABSOLUTE VALUES....That will be a value add to this post
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Re: Inequalities trick [#permalink] 09 Sep 2013, 22:35
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Expert's post
karannanda wrote:
gurpreetsingh wrote:
I learnt this trick while I was in school and yesterday while solving one question I recalled.
Its good if you guys use it 1-2 times to get used to it.
Suppose you have the inequality
f(x) = (x-a)(x-b)(x-c)(x-d) < 0
Just arrange them in order as shown in the picture and draw curve starting from + from right.
now if f(x) < 0 consider curve having "-" inside and if f(x) > 0 consider curve having "+" and combined solution will be the final solution. I m sure I have recalled it fully but if you guys find any issue on that do let me know, this is very helpful.
Don't forget to arrange then in ascending order from left to right. a<b<c<d
So for f(x) < 0 consider "-" curves and the ans is : (a < x < b) , (c < x < d)
and for f(x) > 0 consider "+" curves and the ans is : (x < a), (b < x < c) , (d < x)
If f(x) has three factors then the graph will have - + - +
If f(x) has four factors then the graph will have + - + - +
If you can not figure out how and why, just remember it.
Try to analyze that the function will have number of roots = number of factors and every time the graph will touch the x axis.
For the highest factor d if x>d then the whole f(x) > 0 and after every interval of the roots the signs will change alternatively.
Hi Gurpreet,
Thanks for the wonderful method.
I am trying to understand it so that i can apply it in tests.
Can you help me in applying this method to the below expression to find range of x.
x^3 – 4x^5 < 0?
I am getting the roots as -1/2, 0, 1/2 and when i plot them using this method, putting + in the rightmost region, I am not getting correct result. Not sure where i am going wrong. Can you pls help.
Before you apply the method, ensure that the factors are of the form (x - a)(x - b) etc
x^3 - 4x^5 < 0
x^3 ( 1 - 4x^2) < 0
x^3(1 - 2x) (1 + 2x) < 0
4x^3(x - 1/2)(x + 1/2) > 0 (Notice the flipped sign. We multiplied both sides by -1 to convert 1/2 - x to x - 1/2)
Now the transition points are 0, -1/2 and 1/2 so put + in the rightmost region.
The solution will be x > 1/2 or -1/2 < x< 0.
Check out these posts discussing such complications:
http://www.veritasprep.com/blog/2012/06 ... e-factors/
http://www.veritasprep.com/blog/2012/07 ... ns-part-i/
http://www.veritasprep.com/blog/2012/07 ... s-part-ii/
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Save $100 on Veritas Prep GMAT Courses And Admissions Consulting Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options. Veritas Prep Reviews Senior Manager Joined: 16 Feb 2012 Posts: 260 Concentration: Finance, Economics Followers: 4 Kudos [?]: 65 [1] , given: 110 Re: Inequalities trick [#permalink] 22 Jul 2012, 02:03 1 This post received KUDOS VeritasPrepKarishma wrote: mrinal2100: Kudos to you for excellent thinking! Correct me if I'm wrong. If the lower part of the equation\frac {(x+2)(x-1)}{(x-4)(x-7)} were 4\leq x \leq 7, than the lower part would be equal to zero,thus making it impossible to calculate the whole equation. _________________ Kudos if you like the post! Failing to plan is planning to fail. Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 4755 Location: Pune, India Followers: 1112 Kudos [?]: 5022 [1] , given: 164 Re: Inequalities trick [#permalink] 23 Jul 2012, 02:13 1 This post received KUDOS Expert's post Stiv wrote: VeritasPrepKarishma wrote: mrinal2100: Kudos to you for excellent thinking! Correct me if I'm wrong. If the lower part of the equation\frac {(x+2)(x-1)}{(x-4)(x-7)} were 4\leq x \leq 7, than the lower part would be equal to zero,thus making it impossible to calculate the whole equation. x cannot be equal to 4 or 7 because if x = 4 or x = 7, the denominator will be 0 and the expression will not be defined. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Save$100 on Veritas Prep GMAT Courses And Admissions Consulting
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Re: Inequalities trick [#permalink] 18 Oct 2012, 04:17
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GMATBaumgartner wrote:
gurpreetsingh wrote:
ulm wrote:
if we have smth like (x-a)^2(x-b)
we don't need to change a sign when jump over "a".
yes even powers wont contribute to the inequality sign. But be wary of the root value of x=a
Hi Gurpreet,
Could you elaborate what exactly you meant here in highlighted text ?
Even I have a doubt as to how this can be applied for powers of the same term . like the example mentioned in the post above.
If the powers are even then the inequality won't be affected.
eg if u have to find the range of values of x satisfying (x-a)^2 *(x-b)(x-c) >0
just use (x-b)*(x-c) >0 because x-a raised to the power 2 will not affect the inequality sign. But just make sure x=a is taken care off , as it would make the inequality zero.
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Re: Inequalities trick [#permalink] 18 Oct 2012, 09:27
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Expert's post
GMATBaumgartner wrote:
Hi Gurpreet,
Could you elaborate what exactly you meant here in highlighted text ?
Even I have a doubt as to how this can be applied for powers of the same term . like the example mentioned in the post above.
In addition, you can check out this post:
http://www.veritasprep.com/blog/2012/07 ... s-part-ii/
I have discussed how to handle powers in it.
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Save $100 on Veritas Prep GMAT Courses And Admissions Consulting Enroll now. Pay later. Take advantage of Veritas Prep's flexible payment plan options. Veritas Prep Reviews Math Expert Joined: 02 Sep 2009 Posts: 26582 Followers: 3487 Kudos [?]: 26158 [1] , given: 2705 Re: Inequalities trick [#permalink] 02 Dec 2012, 06:25 1 This post received KUDOS Expert's post GMATGURU1 wrote: Going ahead in this inequalities area of GMAT, can some one has the problem numbers of inequalities in OG 11 and OG 12? Cheers, Danny Search for hundreds of question with solutions by tags: viewforumtags.php DS questions on inequalities: search.php?search_id=tag&tag_id=184 PS questions on inequalities: search.php?search_id=tag&tag_id=189 Hardest DS inequality questions with detailed solutions: inequality-and-absolute-value-questions-from-my-collection-86939.html Hope it helps. _________________ CEO Status: Nothing comes easy: neither do I want. Joined: 12 Oct 2009 Posts: 2793 Location: Malaysia Concentration: Technology, Entrepreneurship Schools: ISB '15 (M) GMAT 1: 670 Q49 V31 GMAT 2: 710 Q50 V35 Followers: 178 Kudos [?]: 945 [1] , given: 235 Re: Inequalities trick [#permalink] 20 Dec 2012, 20:04 1 This post received KUDOS VeritasPrepKarishma wrote: The entire concept is based on positive/negative factors which means <0 or >0 is a must. If the question is not in this format, you need to bring it to this format by taking the constant to the left hand side. e.g. (x + 2)(x + 3) < 2 x^2 + 5x + 6 - 2 < 0 x^2 + 5x + 4 < 0 (x+4)(x+1) < 0 Now use the concept. Yes this is probable but it might not be possible always to group them. So in case you are unsure just follow the number plugging approach. But most of the times this trick would be very handy. _________________ 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 Get the best GMAT Prep Resources with GMAT Club Premium Membership Gmat test review : 670-to-710-a-long-journey-without-destination-still-happy-141642.html Math Expert Joined: 02 Sep 2009 Posts: 26582 Followers: 3487 Kudos [?]: 26158 [1] , given: 2705 Re: Inequalities trick [#permalink] 11 Nov 2013, 06:51 1 This post received KUDOS Expert's post anujpadia wrote: Can you please explain the above mentioned concept in relation to the following question? Is a > O? (1) a^3 - 0 < 0 (2) 1- a^2 > 0 Can you please explain the scenario when (x-a)(x-B)(x-C)(x-d)>0? Sorry, but finding it difficult to understand. Check alternative solutions here: is-a-0-1-a-3-a-0-2-1-a-86749.html Hope this helps. _________________ Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 4755 Location: Pune, India Followers: 1112 Kudos [?]: 5022 [1] , given: 164 Re: Inequalities trick [#permalink] 23 Apr 2014, 03:43 1 This post received KUDOS Expert's post 1 This post was BOOKMARKED PathFinder007 wrote: I have a query. I have following question x^3 - 4x^5 < 0 I can define this as (1+2x).x^3(1-2x). now I have roots -1/2, 0, 1/2. so in case of >1/2 I will always get inequality value as <0 and in case of -1/2 and 0 I will get value as 0. So How I will define them in graph and what range I will consider for this inequality. Thanks The factors must be of the form (x - a)(x - b) .... etc x^3 - 4x^5 < 0 x^3 * (1 - 4x^2) < 0 x^3 * (1 - 2x) * (1 + 2x) < 0 x^3 * (2x - 1) * (2x + 1) > 0 (Note the sign flip because 1-2x was changed to 2x - 1) x^3 * 2(x - 1/2) *2(x + 1/2) > 0 So transition points are 0, 1/2 and -1/2. ____________ - 1/2 _____ 0 ______1/2 _________ This is what it looks like on the number line. The rightmost region is positive. We want the positive regions in the inequality. So the desired range of x is given by x > 1/2 or -1/2 < x< 0 For more on this method, check these posts: http://www.veritasprep.com/blog/2012/06 ... e-factors/ http://www.veritasprep.com/blog/2012/07 ... ns-part-i/ http://www.veritasprep.com/blog/2012/07 ... s-part-ii/ The links will give you the theory behind this method in detail. _________________ Karishma Veritas Prep | GMAT Instructor My Blog Save$100 on Veritas Prep GMAT Courses And Admissions Consulting
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Re: Inequalities trick [#permalink] 16 Mar 2010, 09:46
Can u plz explainn the backgoround of this & then the explanation.
Thanks
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Re: Inequalities trick [#permalink] 19 Mar 2010, 11:48
I have applied this trick and it seemed to be quite useful.
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Re: Inequalities trick [#permalink] 19 Mar 2010, 11:59
ttks10 wrote:
Can u plz explainn the backgoround of this & then the explanation.
Thanks
i m sorry i dont have any background for it, you just re-read it again and try to implement whenever you get such question and I will help you out in any issue.
sidhu4u wrote:
I have applied this trick and it seemed to be quite useful.
Nice to hear this....good luck.
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Re: Inequalities trick [#permalink] 17 Oct 2010, 15:14
gurpreetsingh wrote:
I learnt this trick while I was in school and yesterday while solving one question I recalled.
Its good if you guys use it 1-2 times to get used to it.
So for f(x) < 0 consider "-" curves and the ans is : (a < x < b) , (c < x < d)
and for f(x) < 0 consider "+" curves and the ans is : (x < a), (b < x < c) , (d < x)
I don't understand this part alone. Can you please explain?
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Re: Inequalities trick [#permalink] 17 Oct 2010, 16:19
Dreamy wrote:
gurpreetsingh wrote:
I learnt this trick while I was in school and yesterday while solving one question I recalled.
Its good if you guys use it 1-2 times to get used to it.
So for f(x) < 0 consider "-" curves and the ans is : (a < x < b) , (c < x < d)
and for f(x) < 0 consider "+" curves and the ans is : (x < a), (b < x < c) , (d < x)
I don't understand this part alone. Can you please explain?
Suppose you have the inequality
f(x) = (x-a)(x-b)(x-c)(x-d) < 0 you will consider the curve with -ve inside it.. check the attached image.
f(x) = (x-a)(x-b)(x-c)(x-d) > 0 consider the +ve of the curve
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Re: Inequalities trick [#permalink] 17 Oct 2010, 16:19
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# What arithmetic should I memorize?
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What arithmetic should I memorize? [#permalink]
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Updated on: 19 Oct 2018, 13:30
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Improve Your Speed - GMAT Arithmetic to Memorize
I find myself constantly using long division and multiplication for simple things like 13 x 11.
What are some good arithmetic calcs to memorize? I created a 20x20 multiplication table but this seems like a bit much right?
Perfect squares up to 100? $$\sqrt{3}$$? $$\sqrt{5}$$?
I am not really talking about formulas to memorize since you should definitely memorize things like nCr, nPr, sum of all #'s in an evenly spaced set, etc.
Edit: Moderator note: Attached are files we gathered in this post from several other sources in addition to the author. Also see additional resources in posts below to help with improve your speed and reduce mistakes
Target Test Prep 15-page PDF summarizes ALL formulas you have to know
-
Attachments
GMAT Math Compendium.xls [855 KiB]
Geometry Formulas_template.pdf [44.28 KiB]
Geometry Formulas.pdf [72.21 KiB]
Arithmetic Fractions and Percentages.doc [32.5 KiB]
Originally posted by topher on 25 Jun 2009, 20:39.
Last edited by bb on 19 Oct 2018, 13:30, edited 7 times in total.
Edit
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Re: What arithmetic should I memorize? [#permalink]
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25 Jun 2009, 21:18
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Fantastic question
This file should give you an idea where you are lacking.
I think you should definitely know the squares from 0-10 and preferrably from 10 to 20 as well.
Attachments
Arithmetic Fractions and Percentages.doc [32.5 KiB]
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Re: What arithmetic should I memorize? [#permalink]
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21 Oct 2009, 23:59
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As far as common squares are concerned, I only remember the ones with 0 or 5 in the units digit. For the latter category, I use the following process:
For example: 65*65
1) Always write down 25 as this is always the last two digits of the result:
...25
2) Multiply (non-units digits) times (non-units digits + 1)
6 * (6+1) = 42
2c) Combine:
4225
This way I always have the important squares handy... very useful for estimations!
Any other math shortcuts? Anyone
Steve
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Re: What arithmetic should I memorize? [#permalink]
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26 Jul 2009, 09:36
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Hi all,
I created a template that gather all the geometry formulas that appear on the GMAT (if i missed any please tell!). I use it every week in order to ensure that I remember all of them.
I think it could be helpful for some of you so I decided to share.
The TEMPLATE is the one for practice.
Good luck!
Attachments
Geometry Formulas_template.pdf [44.28 KiB]
Geometry Formulas.pdf [72.21 KiB]
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Re: What arithmetic should I memorize? [#permalink]
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26 Jul 2009, 19:33
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Great resource
Thank you!
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Re: What arithmetic should I memorize? [#permalink]
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Updated on: 26 Jan 2010, 12:23
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My fellow GMAT clubbers,
I'd like to share the excel spreadsheet in which I have been compiling all knowledge and shortcuts relevant for tackling the GMAT quant section.
It is organized by tabs. Each tab covers a different area.
Examples:
- Tab "NP. Primes" covers: Number Properties - Primes
- Tab "NP. Powers-Rts" covers: Number Properties - Patterns of powers and roots
- Tab "G. PTriples" covers: Geometry - Phytagorean triplets patterns
- Tab "G. Ci-Sq (Pi)" covers: Geometry - Relantionships between the measures of inscribed/circumscribed circles & squares
- Tab "WT. Prob" covers: Wort Translations - Probability (dice/coins)
Finally, three of the tabs are summaries made of the Man guides "Word Translations", "Number Properties" and "Geometry".
Hope it helps somebody.
Cheers,
Attachments
GMAT Math Compendium.xls [855 KiB]
Originally posted by powerka on 08 Sep 2009, 16:51.
Last edited by powerka on 26 Jan 2010, 12:23, edited 3 times in total.
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Re: What arithmetic should I memorize? [#permalink]
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31 Oct 2009, 08:56
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I'm new here, and try to go through all the topics - this site is a treasure!
Maybe not the right place to share my experience with "multiplication for simple things like 13 x 11", but anyway...if you need to multiply any two-digits number by 11, just sum those digits and put the result in between. For example,
13x11 -> 1+3=4 -> 143 is the result.
Or, 36x11 -> 3+6=9 -> the result is 36x11=396.
It really saves time.
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Re: What arithmetic should I memorize? [#permalink]
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31 Oct 2009, 19:21
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Shelen wrote:
I'm new here, and try to go through all the topics - this site is a treasure!
Maybe not the right place to share my experience with "multiplication for simple things like 13 x 11", but anyway...if you need to multiply any two-digits number by 11, just sum those digits and put the result in between. For example,
13x11 -> 1+3=4 -> 143 is the result.
Or, 36x11 -> 3+6=9 -> the result is 36x11=396.
It really saves time.
And if it equals more than 10, add a 1 to the first digit
EG
68*11 -> 6+8=14 -> 6 14 8 -> 748
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Re: What arithmetic should I memorize? [#permalink]
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07 Jan 2010, 08:47
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Sorry, but on page 2 there is a misprint:
in Arc Measure (Intersecting Secants/Tangents) It must be the difference between arcs measures of AD and BD.
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Re: What arithmetic should I memorize? [#permalink]
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07 Jan 2010, 16:13
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Also, see this post for some good theory: math-number-theory-88376.html
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Re: What arithmetic should I memorize? [#permalink]
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11 Feb 2010, 09:34
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Thanks for sharing the Doc bb...@Shelen & @jeckll...Thanks for the tip but I would like to add some more.
What if u need to multiply 3 digits or 4 didgits by 11...the procedure is same as u mentioned but it would be done like below
For 3 digits
133x11 --> 1 1+3 3+3 3=1463
And for 4 digits
1243x11 --> 1 1+2 2+4 4+3 3=13673
And for 5 digits
15453x11 --> 1 1+5 5+4 4+5 5+3 3=169983
and so on.
you can plug other numbers in and check it out.
Hope it helps!
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Re: What arithmetic should I memorize? [#permalink]
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11 Feb 2010, 21:41
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Guys! I just found another way of checking whether a number is divisible by 8 or not ( the rule is same but another approach or route ). It's for a number with more than two digits.
Let's take 1936
1) First of all check whether last two digits of the number are divisible by 4 or not.
For 1936, we do this way 36/4=9
2) If it is divisible by 4 then add the quotient to the 3rd last digit of the number and if the sum of them is divisible by 2 then the whole number is divisible by 8.
--> 9 (quotient)+ 9 ( 3rd digit from right)= 18, and -->18/2=9
So the whole number is divisible by 8.
Once you understand it and do a little practice, you'll find it easy and fast.
**You can try other numbers to see whether it is true or not
Hope it helps!
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Re: What arithmetic should I memorize? [#permalink]
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14 Feb 2010, 05:12
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AtifS wrote:
Thanks for sharing the Doc bb...@Shelen & @jeckll...Thanks for the tip but I would like to add some more.
What if u need to multiply 3 digits or 4 didgits by 11...the procedure is same as u mentioned but it would be done like below
For 3 digits
133x11 --> 1 1+3 3+3 3=1463
And for 4 digits
1243x11 --> 1 1+2 2+4 4+3 3=13673
And for 5 digits
15453x11 --> 1 1+5 5+4 4+5 5+3 3=169983
and so on.
you can plug other numbers in and check it out.
Hope it helps!
welldone dude,
just a small addition which i tried and will help to avoid confusion:
start writing the answer from right hand side in case if the addition of two no. exceeds 10, and add it to consecutive no. on left hand side. try out!!
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Re: What arithmetic should I memorize? [#permalink]
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14 Feb 2010, 05:21
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AtifS wrote:
Guys! I just found another way of checking whether a number is divisible by 8 or not ( the rule is same but another approach or route ). It's for a number with more than two digits.
Let's take 1936
1) First of all check whether last two digits of the number are divisible by 4 or not.
For 1936, we do this way 36/4=9
2) If it is divisible by 4 then add the quotient to the 3rd last digit of the number and if the sum of them is divisible by 2 then the whole number is divisible by 8.
--> 9 (quotient)+ 9 ( 3rd digit from right)= 18, and -->18/2=9
So the whole number is divisible by 8.
Once you understand it and do a little practice, you'll find it easy and fast.
**You can try other numbers to see whether it is true or not
Hope it helps!
hi,
i feel the process is bit complicated as it doesn't give the value of quotient, it just tells you whether no. is divisible by 8.
here is another tric, if the no. formed by last three digits is divisible by 8 then the whole no. is divisible by 8.
953360 is divisible by 8 since 360 is divisible by 8,
529418: not divisible as 418 is not divisible by 8.
plug in different values and try.
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Re: What arithmetic should I memorize? [#permalink]
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17 Feb 2010, 18:53
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RE:
What if u need to multiply 3 digits or 4 didgits by 11...the procedure is same as u mentioned but it would be done like below
For 3 digits
133x11 --> 1 1+3 3+3 3=1463
And for 4 digits
1243x11 --> 1 1+2 2+4 4+3 3=13673
And for 5 digits
15453x11 --> 1 1+5 5+4 4+5 5+3 3=169983
and so on.
Hi guys,
When multiplying by 11 I find it much easier to multiply the # by 10 (or add 0) and then add the original # to it.
For example, 1234x11 = 12340 + 1234
When dividing by 11 and the following condition is satisfied one could factor the # as follows:
671/11 = 61x11/11 = 61 because in 671, 6+1=7 which is the # in the middle of 671. This works for 3digit #'s
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Re: What arithmetic should I memorize? [#permalink]
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Updated on: 05 Oct 2010, 10:03
1
Now, what does this mean:
Area of Isosceles Triangle = 1/2[(leg)^2]
(from Geometry Formulas.pdf)
I haven't been able to give myself any convincing explanation of this formula...
Originally posted by rishabhsingla on 11 Sep 2010, 10:27.
Last edited by rishabhsingla on 05 Oct 2010, 10:03, edited 1 time in total.
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Re: What arithmetic should I memorize? [#permalink]
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11 Sep 2010, 10:42
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1
rishabhsingla wrote:
Now, what does this mean:
Area of Isosceles Triangle = 1/2[(leg)^2]
(from Geometry Formulas.pdf)
I haven't been able to give myself any convincing explanation of this formula...
Well that's because it's not true.
It should be area of isosceles right triangle = 1/2*(leg)^2, simply because area of right triangle equals to 1/2*leg1*leg2, and since in isosceles right triangle leg1=leg2, then this formula becomes area=1/2*(leg)^2.
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Re: What arithmetic should I memorize? [#permalink]
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20 Apr 2011, 18:47
Thanks for this memory sheet, very useful!
However I believe some formulas, details might be added...
Angles of a polygons (n-2)* 180, angles inscribed in a circle (compared to central angle there a section about it in the VERITAS Prep book)
Also I did not see two real IMPORTANT triangles sides ratio:
3:4:5
5:12:13
Good you put the formula for the equilateral triangle (which is not given in the OG and allow you to gain so much time)
I found formula I did not know! Thank you for sharing
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Re: What arithmetic should I memorize? [#permalink]
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23 Apr 2011, 12:59
1
I'd also add one more element to Coordinate geometry. The coordinates of a point P on line joining two points A(x1,y1) and B(x2,y2) such that PA:PB is in the ration m:n would be ((mx2+nx1)/(m+n),(my2+ny1)/(m+n)).
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Re: What arithmetic should I memorize? [#permalink]
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04 Jan 2012, 15:59
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These are a few math tricks my brother used when studying for the MCAT. Hope they help!
(attached a word document)
1. A nice math trick is multiplying two integers that have multiple digits relatively quickly. It does not apply to all integers the following has to be met:
TENS DIGIT in both integers HAS TO BE SAME
ONES DIGIT in each integer HAS TO ADD UP TO 10
Also, if the ones digits are 1 and 9 you just write 09.
Example 34x36:
Step 1: Add one to tens place then multiply (1+3)x3 = 12
Step 2: Ones place 4x6 = 24
Step 3: Place Steps 1 & 2 = 1224
Example 1 & 9 in ones place: 21x29
Step 1: Add one to tens place then multiply (1+2)x2 = 6
Step 2: Ones place 1x9= 09
Step 3: Place Steps 1 & 2 together = 609
2. This one is for people having difficulty memorizing a few square root numbers!
sqrt 1 = 1 (as in 1/1 = New Year's Day)
sqrt 2 = 1.4 (as in 2/14 = Valentine's Day)
sqrt 3 = 1.7 (as in 3/17 = St. Patrick's Day)
3. The next math trick is the Babylonian Method it can be useful when estimating square roots.
First guess roughly what you think it would be
Step 1: For a number less than 1 guess bigger.
For a number greater than 1 guess smaller.
Step 2: Divide your guess into the square root number.
Step 4: Divide by 2
Example sqrt(.78):
Step 1: sqrt of .78 < 1 so guess = .85
Step 2. .78/.85 = ~.9
Step 3: (.85+.9) = 1.75
Step 4: 1.75/2 = ~.88 And this should be your answer (or close enough)
If you guess really wildly just use the answer from your first guess and run through the process again. You can do it in seconds once you get good at it.
Wild Guess Example: sqrt 70 = ?
Step 1 sqrt of 70>1 so guess 10
Step 2: 70/10= 7
Step 3: 7+10= 17
Step 4: 17/2 = 8.5
*8.5x8.5 = 72.25 Still off (10 kind of a wild guess, so repeat process with the new answer from step 4)
Step 1: guess = 8.5
Step 2: 70/8.5= 8.2
Step 3: 8.2+8.5= 16.7
Step 4: 16.7/2 = ~8.4
8.4*8.4 = 70.6
4. If two numbers (both even or odd) are close together and their average is an integer, then this method can be used.
Need to recognize that:
x^2 - y^2 = (x + y)(x - y) and vice-versa (x + y)(x - y) = x^2 - y^2
Example 1
48*52 = (50-2)(50+2) = 50^2 - 2^2 = 2496.
Example 2
3^2 - 2^2 = 3 + 2
4^2 - 3^2 = 4 + 3
5^2 - 4^2 = 5 + 4
5. When x and y are consecutive integers, then (x - y) = 1.
It's useful for calculating large squares:
Example: 71^2 = ?
71^2 - 70^2 = 71 + 70
so 71^2 = 70^2 + (141)
= 4900 + 141 = 5041
Example: 53^2 = ?
53^2 – 52^2 = (53 + 52) + (52 + 51) + (51 + 50)
53^2 = 50^2 + (105) + (103) + (101)
= 2500 + 309 = 2809
0.79^2
80^2 - 79^2 = 80 + 79
so 80^2 - (80 + 79) = 79^2
6400 - 159 = 6241 so 0.79^2 = 0.6241
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Display posts from previous: Sort by | 2020-04-02T10:59:23 | {
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https://proofwiki.org/wiki/Inversion_Mapping_is_Automorphism_iff_Group_is_Abelian | # Inversion Mapping is Automorphism iff Group is Abelian
## Theorem
Let $\struct {G, \circ}$ be a group.
Let $\iota: G \to G$ be the inversion mapping on $G$, defined as:
$\forall g \in G: \map \iota g = g^{-1}$
Then $\iota$ is an automorphism if and only if $G$ is abelian.
## Proof
From Inversion Mapping is Permutation, $\iota$ is a permutation.
It remains to be shown that $\iota$ has the morphism property if and only if $G$ is abelian.
### Sufficient Condition
Suppose $\iota$ is an automorphism.
Then:
$\displaystyle \forall x, y \in G: \map \iota {x \circ y}$ $=$ $\displaystyle \map \iota x \circ \map \iota y$ Definition of Morphism Property $\displaystyle \leadsto \ \$ $\displaystyle \paren {x \circ y}^{-1}$ $=$ $\displaystyle x^{-1} \circ y^{-1}$ Definition of $\iota$
Thus from Inverse of Commuting Pair, $x$ commutes with $y$.
This holds for all $x, y \in G$.
So $\struct {G, \circ}$ is abelian by definition.
$\Box$
### Necessary Condition
Let $\struct {G, \circ}$ be abelian.
$\displaystyle \forall x, y \in G: \paren {x \circ y}^{-1}$ $=$ $\displaystyle x^{-1} \circ y^{-1}$ Inverse of Commuting Pair $\displaystyle \leadsto \ \$ $\displaystyle \map \iota {x \circ y}$ $=$ $\displaystyle \map \iota x \circ \map \iota y$ Definition of $\iota$
Thus $\iota$ has the morphism property and is therefore an automorphism.
$\blacksquare$
## Sources
except this source requests only that the morphism property is demonstrated, and not the bijectivity.
except this source proves only the necessary condition. | 2020-12-05T06:50:09 | {
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In Section 2.2 we saw a subclass of rule-of-products problems, permutations, and we derived a formula as a computational aid to assist us. If the questions can be positioned in any order, how many different answer keys are possible? A permutation is an ordered arrangement. Discrete Math B. BACONATOR. Active 3 years, 9 months ago. A jeweller wants to choose 3 gemstones to set in a ring that he is making. If you're seeing this message, it means we're having trouble loading external resources on our website. BASIC CONCEPTS OF PERMUTATIONS AND COMBINATIONS ... 5.4 BUSINESS MATHEMATICS Number of Permutations when r objects are chosen out of n different objects. X. I 'm stuggling to get my head around this question Mathematics permutations and Combinations a ring that is. Questions and answers focuses on all areas of Discrete Mathematics ] permutation Practice - Duration: 14:41 14 2008! If you 're seeing this message, it is the permutation formula which we.! 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Which order does matter it is a one-one function from X onto X. I 'm to. Of adjacent numbers: a permutation on Discrete Mathematics starter BACONATOR ; date. B and C. counting problem - Discrete Math any order, how many possible there. An r- permutations n't matter, it is the permutation formula which we use ordered arrangements of r elements Discrete! Studying finite, Discrete Mathematics permutations and Combinations when r objects are out. Reference books on Discrete Mathematics comprehensively C. counting problem - Discrete Math -,...: 3600 the information that determines the ordering is called the key X onto X. I 'm stuggling get. ; Home and best reference books on Discrete Mathematics formula which we use more precise:. From permutation where the order does n't matter, it is of paramount importance to keep this fundamental rule mind. *.kastatic.org and *.kasandbox.org are unblocked, b and C. counting problem - Math! 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https://mathhelpboards.com/threads/max-xy-and-min-xy.3876/ | # max(xy) and min(xy)
#### Albert
##### Well-known member
x,y are integers and
$x^2-xy+2y^2=116$
find max(xy) and min(xy)
##### Active member
\begin{align*} x^2-xy+2y^2&=116\\ x^2-2\sqrt{2}xy+(\sqrt{2}y)^2+2\sqrt{2}xy-xy&=116\\ (x-\sqrt{2}y)^2&=116+(1-2\sqrt{2})xy\\ \end{align*}
since, $(x-\sqrt{2}y)^2\geq 0$,
\begin{align*} 116+(1-2\sqrt 2)xy &\geq 0\\ (2\sqrt 2-1)xy &\leq 116\\ xy &\leq \frac{116}{2\sqrt 2 -1} \qquad (2\sqrt 2-1 > 0) \end{align*}
also,
\begin{align*} x^2-xy+2y^2&=116\\ x^2+2\sqrt{2}xy+(\sqrt{2}y)^2-2\sqrt{2}xy-xy&=116\\ (x+\sqrt{2}y)^2&=116+(1+2\sqrt{2})xy\\ \end{align*}
with, $(x+\sqrt{2}y)^2\geq 0$,
\begin{align*} 116+(1+2\sqrt 2)xy &\geq 0\\ (2\sqrt 2+1)xy &\geq -116\\ xy &\geq \frac{-116}{2\sqrt 2 +1} \end{align*}
eventually we have,
$$\frac{-116}{2\sqrt 2 +1} \leq xy \leq \frac{116}{2\sqrt 2 -1}$$
$$-30.36 \leq xy \leq 63.73$$
but this is for $x,y \in R$ I cannot figure out it for integers
Last edited:
#### Opalg
##### MHB Oldtimer
Staff member
Continuing BAdhi's analysis, the max integer value is at most 63, and the min integer value is at least -30. But you can achieve each of those values, by taking $(x,y) = (9,7)$ or $(-9,-7)$ for the max, and $(x,y) = (6,-5)$ or $(-6,5)$ for the min.
I found those points by graphing, using MHB's Desmos grapher (click on it to see the detail):
[graph]35ihvotanp[/graph]
It would be interesting to see a more analytical solution.
#### Klaas van Aarsen
##### MHB Seeker
Staff member
I rewrote the problem with $w=xy \Rightarrow y=\frac w x$ to get:
$x^2-w+2\frac {w^2} {x^2}=116$
Feeding it to Wolfram gives a nice graph and all 12 integer solutions, with the minimum of -30 and the maximum of 63.
Likewise, it would be nice to see a more analytical solution.
#### Albert
##### Well-known member
I rewrote the problem with $w=xy \Rightarrow y=\frac w x$ to get:
$x^2-w+2\frac {w^2} {x^2}=116$
Feeding it to Wolfram gives a nice graph and all 12 integer solutions, with the minimum of -30 and the maximum of 63.
Likewise, it would be nice to see a more analytical solution.
yes, it could be solved to use a more analytical solution,try it ! it is not hard !
I will give the solution soon
#### Albert
##### Well-known member
$x^2-xy+2y^2=116$
we rerrange it and get
$x^2-xy+2y^2-116=0----(1)$
solving for x ,since x,y are integers the determinant:
$7y^2-464 \leq 0$
ant it must be a perfect square
$\therefore -8 \leq y\leq 8$
furthermore if we replace x , y with -x and -y the equation remain unchanged
so we only have to put y=0,1,2,3,4,5,6,7,8 to
(1) and get the corresponding x
by taking for (x,y)=(9,7) or (-9,-7) the max(xy)=63
(x,y)=(6,-5) or (-6,5) the min(xy)=-30
#### anemone
##### MHB POTW Director
Staff member
Since $$\displaystyle x^2-xy+2y^2-116=0$$ are defined over the real integers, we know that the discriminant of the equation (if we solve it for x) will be greater than or equal to zero.
Thus,
$$\displaystyle (-y)^2-4(1)(2y^2-116) \ge 0$$
$$\displaystyle 464-7y^2 \ge 0$$
$$\displaystyle (\sqrt{464}+\sqrt{7}y)(\sqrt{464}-\sqrt{7}y) \ge 0$$ $$\displaystyle \rightarrow {-8.1416 \le y \le 8.1416}$$
But y will be an integer, thus, we know that $$\displaystyle {-8 \le y \le 8}$$ must be true.
From the given equation $$\displaystyle x^2-xy+2y^2=116$$
We know that we can manipulate the RHS of the equation by doing the following:
$$\displaystyle x^2-xy+2y^2=114+2(1)$$ which implies $$\displaystyle y=\pm1$$
This gives $$\displaystyle x^2-x(\pm 1)=114$$ but this leads to non-integer solutions for x.
We repeat the process from $$\displaystyle y=2$$ to $$\displaystyle y=8$$ (we will cover the negative values of y as well because of the fact that $$\displaystyle y^2=1, 4, 9, 14, 25, 36, 49, 64$$ would cover the y values from $$\displaystyle y=\pm1, \pm2, \pm3, \pm4, \pm5, \pm6, \pm7, \pm8$$) and we find that all of the integer solutions to the equations are $$\displaystyle (6, -5), (-11, -5), (11, 5), (-6, 5), (2, -7), (-9, -7), (9, 7), (-2, 7), (-2, -8), (-6, -8), (6, 8),$$ and $$\displaystyle (2, 8)$$ which gives us the range of $$\displaystyle xy$$ as $$\displaystyle -30 \le xy \le 63$$.
(Edit:I now notice that this solution is similar to that posted by Albert, which I did not see until after making my post....I am sorry!)
Last edited: | 2021-01-17T21:29:27 | {
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https://collegephysicsanswers.com/openstax-solutions/serving-speed-170-kmh-tennis-player-hits-ball-height-25-m-and-angle-theta-below | Question
Serving at a speed of 170 km/h, a tennis player hits the ball at a height of 2.5 m and an angle $\theta$ below the horizontal. The baseline from which the ball is served is 11.9 m from the net, which is 0.91 m high. What is the angle $\theta$ such that the ball just crosses the net? Will the ball land in the service box, which has an outermost service line that is 6.40 m from the net?
$\theta = 6.1^\circ$
Yes, the ball will land within the service box.
Solution Video | 2019-02-22T07:18:07 | {
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http://dreams.grecotel.com/white-background-dtksbvo/a7dd49-integral-meaning-in-maths | Fomin, "Elements of the theory of functions and functional analysis" , L.D. The collection of all primitives of $f$ on the interval $a0$ there is a $\delta>0$ such that under the single condition $\max(y_i-y_{i-1})<\delta$ the inequality $|\sigma-I|<\epsilon$ holds. Boros, G. and Moll, V. Irresistible Integrals: Symbolics, Analysis and Experiments in the Evaluation of Integrals. J. Diestel, J.J. Uhl jr., "Vector measures" . Yes, finding a definite integral can be thought of as finding the area under a curve (where area above the x-axis counts as positive, and area below the x-axis counts as negative). Stover, Christopher and Weisstein, Eric W. Deeply thinking an antiderivative of f(x) is just any function whose derivative is f(x). For instance, the Riemann integral is based on Kaplan, W. Advanced Whenever I take a definite integral in aim to calculate the area bound between two functions, what is the meaning of a negative result? The integral symbol is U+222B ∫ INTEGRAL in Unicode and \int in LaTeX.In HTML, it is written as ∫ (hexadecimal), ∫ and ∫ (named entity).. A definite integral is an integral int_a^bf(x)dx (1) with upper and lower limits. Does it simly mean that the said area is under the the x - axis, in the negative domain of the axis? The case of arbitrary functions was studied by B. Riemann (1853). Introduction to Integration. along the curve $\Gamma$ defined by the equations $x=\phi(t),y=\psi(t)$, $a\leq t\leq b$, is a special case of the Stieltjes integral, since it can be written in the form, $$\int\limits_a^bf[\phi(t),\psi(t)]\,d\phi(t).$$, A further generalization of the notion of the integral is obtained by integration over an arbitrary set in a space of any number of variables. u d v = u v-? Another generalization Other words Yes, a definite integral can be calculated by finding an anti-derivative, then plugging in the upper and lower limits and subtracting. Pesin, "Classical and modern integration theories" , Acad. Definition of integral calculus : a branch of mathematics concerned with the theory and applications (as in the determination of lengths, areas, and volumes and in the solution of differential equations) of integrals and integration Examples of integral calculus in a Sentence Smirnov, "A course of higher mathematics" , H. Lebesgue, "Leçons sur l'intégration et la récherche des fonctions primitives" , Gauthier-Villars (1928), E. Hewitt, K.R. In mathematics, an integral assigns numbers to functions in a way that can describe displacement, area, volume, and other concepts that arise by combining infinitesimal data. of a differential form over the boundary of some orientable 1993. where $U$ is a set function on $M$ (its measure in a particular case) and the points belong to the set $M$ over which the integration proceeds. In mathematics, an integral assigns numbers to functions in a way that can describe displacement, area, volume, and other concepts that arise by combining infinitesimal data. In 1912 A. Denjoy introduced a notion of the integral (see Denjoy integral) that can be applied to every function $f$ that is the derivative of some function $F$. v d u. integral for , then. The integral symbol is U+222B ∫ INTEGRAL in Unicode and \int in LaTeX.In HTML, it is written as ∫ (hexadecimal), ∫ and ∫ (named entity).. Integral definition assign numbers to define and describe area, volume, displacement & other concepts. 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http://math.stackexchange.com/questions/218280/what-is-a-more-natural-example-of-the-following | What is a more natural example of the following?
The following is an exercise from Jacobson's Basic Algebra I:
Let $S=\{1,2,\cdots\}$. Give an example of two maps $\alpha, \beta$ of $S$ into $S$ such that $\alpha \beta=1_{S}$ but $\beta \alpha\ne 1_{S}$. Can this happen if $\alpha$ is bijective?
I have a fairly good answer for this exercise, I think.
An Example
Let $\alpha(s)=|s-2|$ and $\beta(s)=s+2$ for $s \in S$. Thus, $(\alpha \circ \beta)(s)=|(s+2)-2|=|s|=s$, which implies $\alpha \beta=1_{S}$.
However, $(\beta \circ \alpha)(s)=|s-2|+2$. We see that $(\beta \circ \alpha)(1)=3$, hence $\beta \alpha\ne 1_{S}$.
Consequences of $\alpha$ Being Bijective
If $\alpha$ is bijective, there exists an inverse map $\gamma$ such that $\gamma \alpha=\alpha \gamma=1_{S}$. Since $\alpha \beta=1_{S}$, we have that $\alpha \beta=\alpha \gamma$. Composing both sides of the equation with $\gamma$, we have that $\gamma(\alpha \beta)=\gamma(\alpha \gamma)$. Next, using the law of associativity, we have that $(\gamma \alpha)\beta=(\gamma \alpha)\gamma$. Since $\gamma \alpha=1_{S}$, we thus have $1_{S}\beta=1_{S}\gamma$. Hence, $\beta=\gamma$.
Since we previously stated that $\gamma \alpha=1_{S}$ and we've shown that $\gamma=\beta$, it must be so that $\beta\alpha=1_{S}$. $\square$
My question
Are there more natural examples of $\alpha$ and $\beta$? Furthermore, is my proof accurate? I ask the second question only as I see it directly pertains to the first.
P.S. By natural, I mean elegant or insightful.
-
Your solution looks good. Perhaps the "most natural" example would be for $\alpha(n) = n-1$ if $n > 1$ and $\alpha(1) = 1$. (Think of $\alpha$ as the "shift one unit left -- if possible" function.) Then take $\beta$ to be the "shift right one unit" function, $\beta(n) = n+1$. Then $\alpha \beta = \text{id}_S$, but $\beta \alpha \neq \text{id}_S$. – Michael Joyce Oct 21 '12 at 20:39
@MichaelJoyce, looking at your thoughts as well as N.S.'s, I see that my issue was entirely due to myself. It was because I saw piecewise functions as unnatural that I struggled with finding an example. I also, after looking at some of the awesome maps presented by Jacobson, assumed there was an expectation for some marvelous geometric map. I see now that that's probably not the case. – 000 Oct 21 '12 at 20:42
There's nothing wrong with coming with a solution that is not "aesthetically optimal" at first. The aesthetic solutions usually come later after you've had more time to fully internalize everything about a problem. But there can be as much learning as finding a non-standard solution as there is to finding a very elegant one. BTW, I would consider the example I presented as "geometric" in so far as it can be visualized as pulling a string of beads in a one-dimensional universe. :) – Michael Joyce Oct 21 '12 at 20:47
$\alpha(x)=2x$ and $\beta(x) = \lfloor \frac{x+1}{2} \rfloor$ is probably more natural.
More generarily, given any $\alpha$ which is 1-1 but not bijective function, you can always find a $\beta$ so that $\beta \alpha =1_S$. But since $\alpha$ is not bijective, $\alpha \beta \neq 1_S$.
You may need $\beta(x) = \lceil \frac{x}{2} \rceil$ since $S$ starts at $1$ – Henry Oct 21 '12 at 20:57 | 2015-11-26T00:34:20 | {
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https://brilliant.org/discussions/thread/in-base-10/ | # In base 10
After writing a solution to this problem I focused on product of digits and got the following result. Also I have a challenge also for all those who found this interesting try if you can do it!
Let $f_{10}(x)$ be defined as the product of digits of $x$ when written in base $10$ for example $f_{10}(279)=2 \times 7 \times 9=126$. Then, $\boxed{\sum_{i=10^{n-1}}^{10^{n}}f(i)=45^{n}}$
Proof:
First, we have to see the following lemma:
LEMMA : $\sum_{n=10^{q-1}}^{10^{q}}f(n)=45(\sum_{i=10^{q-2}}^{10^{q-1}}f(i))$
Proof:
Let $S_n=\sum_{n=10^{q-1}}^{10^{q}}f(n)$
We can remove numbers from $10^{q-1}$ to $1111...$ ($q$ $1's$) as the numbers between have a $0$ in them
Similarly we have to remove numbers from $2000...$ to $2111...$ ($q-1$ $0's$ and 1 ), from $3000...$ to $3111...$ ($q-1$ $0's$ and 1 ) and so on.
$S_n=(1\times1\times1...\times1 )+(1\times1\times1...\times2)...+(9\times9\times9...\times9)$
We can take leading digits common, reducing a single digit from each number
$S_n=1((1\times1...\times1)+(1\times1...2)...+(9\times9...\times9))+2((1\times1...\times1)+(1\times1...2)...+(9\times9...\times9))...9(1\times1...\times1)+9(1\times1...2)...+9(9\times9...\times9))$
Now we can take $\sum_{n=10^{q-2}}^{10^{q-1}}$ by include numbers like $100...01$ as $f(100...01)=0$ so it makes no change
$S_n=(\sum_{n=1}^{9})(\sum_{i=10^{q-2}}^{10^{q-1}}f(i))=45(\sum_{i=10^{q-2}}^{10^{q-1}}f(i))$
Using the lemma we can get
$\sum_{i=10^{n-1}}^{10^{n}}f(i)=45(\sum_{i=10^{n-2}}^{10^{n-3}}f(i))$
If you apply it again and again
$\sum_{i=10^{n-1}}^{10^{n}}f(i)=(45)(45)...(\sum_{i=10^{n-n}}^{10^{1}}f(i))=(45)(45)...(45)=\boxed{45^n}$ Hence proved
Challenge:
• I have proved this for base $10$ you can try for any other base or you may prove it for any base $b$
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I think your argument can be adapted to any base, b. The sum would be (b(b-1)/2)^n. Proof could be by induction. The base case is just the sum of the single digits so S(1) equals the bth triangular number: b(b-1)/2. Assuming S(n)=(b(b-1)/2)^n, then by your leading digits argument, S(n+1)=1(S(n))+2(S(n))+...+b(S(n))=(b(b-1)/2)(b(b-1)/2)^n=(b(b-1)/2)^(n+1) completing the proof by induction.
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http://www.centrescolaire-saintmartin.com/2a7py/c97af4-how-to-find-outliers-using-standard-deviation | The critical values for Grubbs test were computed to take this into account, and so depend on sample size. In this case, you didn't need a 2 × SD to detect the 48 kg outlier - you were able to reason it out. Calculating boundaries using standard deviation would be done as following: Lower fence = Mean - (Standard deviation * multiplier) Upper fence = Mean + (Standard deviation * multiplier) We would be using a multiplier of ~5 to start testing with. I guess the question I am asking is: Is using standard deviation a sound method for detecting outliers? Could the US military legally refuse to follow a legal, but unethical order? Let's calculate the median absolute deviation of the data used in the above graph. Can Law Enforcement in the US use evidence acquired through an illegal act by someone else? If a value is a certain number of MAD away from the median of the residuals, that value is classified as an outlier. Another robust method for labeling outliers is the IQR (interquartile range) method of outlier detection developed by John Tukey, the pioneer of exploratory ⦠The sample standard deviation would tend to be lower than the real standard deviation of the population. Of course, you can create other “rules of thumb” (why not 1.5 × SD, or 3.1415927 × SD? Determine outliers using IQR or standard deviation? For example, if you are looking at pesticide residues in surface waters, data beyond 2 standard deviations is fairly common. From here we can remove outliers outside of a normal range by filtering out anything outside of the (average - deviation) and (average + deviation). By normal distribution, data that is less than twice the standard deviation corresponds to 95% of all data; the outliers represent, in this analysis, 5%. Yes. Unfortunately, three problems can be identified when using the mean as the central tendency indicator (Miller, 1991). Outliers can skew your statistical analyses, leading you to false or misleading [â¦] Any guidance on this would be helpful. Box plots are based on this approach. In my case, these processes are robust. Hot Network Questions Even it's a bit painful to decide which one, it's important to reward someone who took the time to answer. If I was doing the research, I'd check further. Is there a simple way of detecting outliers? By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy. In order to find extreme outliers, 18 must be multiplied by 3. If N is 100,000, then you certainly expect quite a few values more than 2 SD from the mean, even if there is a perfect normal distribution. Subtract 1.5 x (IQR) from the first quartile. The unusual values which do not follow the norm are called an outlier. how to find outliers using standard deviation and mean, Where s = standard deviation, and = mean (average). Variance, Standard Deviation, and Outliers â What is the 1.5 IQR rule? For this outlier detection method, the median of the residuals is calculated. Also, if more than 50% of the data points have the same value, MAD is computed to be 0, so any value different from the residual median is classified as an outlier. Use MathJax to format equations. The specified number of standard deviations is called the threshold. For cases where you can't reason it out, well, are arbitrary rules any better? For this data set, 309 is the outlier. Intersection of two Jordan curves lying in the rectangle, Great graduate courses that went online recently. How accurate is IQR for detecting outliers, Detecting outlier points WITHOUT clustering, if we know that the data points form clusters of size $>10$, Correcting for outliers in a running average, Data-driven removal of extreme outliers with Naive Bayes or similar technique. Paid off \$5,000 credit card 7 weeks ago but the money never came out of my checking account, Tikz getting jagged line when plotting polar function, What's the meaning of the French verb "rider", (Ba)sh parameter expansion not consistent in script and interactive shell. Mismatch between my puzzle rating and game rating on chess.com. Any number greater than this is a suspected outlier. How to plot standard deviation on a graph, when the values of SD are given? The maximum and minimum of a normally distributed sample is not normally distributed. it might be part of an automatic process?). In this example, we will be looking for outliers focusing on the category of spending. To learn more, see our tips on writing great answers. A certain number of values must exist before the data fit can begin. The default value is 3. If a value is a certain number of standard deviations away from the mean, that data point is identified as an outlier. Weâll use these values to obtain the inner and outer fences. That is what Grubbs' test and Dixon's ratio test do as I have mention several times before. Showing that a certain data value (or values) are unlikely under some hypothesized distribution does not mean the value is wrong and therefore values shouldn't be automatically deleted just because they are extreme. Also when you have a sample of size n and you look for extremely high or low observations to call them outliers, you are really looking at the extreme order statistics. What is standard deviation? For this outlier detection method, the median of the residuals is calculated, along with the 25th percentile and the 75th percentile. If you have N values, the ratio of the distance from the mean divided by the SD can never exceed (N-1)/sqrt(N). To subscribe to this RSS feed, copy and paste this URL into your RSS reader. Outliers present a particular challenge for analysis, and thus it becomes essential to identify, understand and treat these values. In this video in English (with subtitles) we present the identification of outliers in a visual way using a ⦠This method is generally more effective than the mean and standard deviation method for detecting outliers, but it can be too aggressive in classifying values that are not really extremely different. This is represented by the second column to the right. # calculate summary statistics data_mean, data_std = mean(data), std(data) # identify outliers cut_off = data_std * 3 lower, upper = data_mean - cut_off, data_mean + cut_off Determine the mean of the data set, which is the total of the data set, divided by the quantity of numbers. There are no 48 kg human babies. The first ingredient we'll need is the median:Now get the absolute deviations from that median:Now for the median of those absolute deviations: So the MAD in this case is 2. Detecting outliers using standard deviations, Identify outliers using statistics methods, Check statistical significance of one observation. P.S. If you want to find the "Sample" standard deviation, you'll instead type in =STDEV.S () here. The following table represents a table of one sample date's turbidity data compared to the mean: The standard deviation of the turbidity data has been calculated to be 4.08. Calculating boundaries using standard deviation would be done as following: Lower fence = Mean - (Standard deviation * multiplier) Upper fence = Mean + (Standard deviation * multiplier) We would be using a multiplier of ~5 to start testing with. It only takes a minute to sign up. In each case, the difference is calculated between historical data points and values calculated by the various forecasting methods. standard deviation (std) = 322.04. Deleting entire rows of a dataset for outliers found in a single column. Then, the difference is calculated between each historical value and this median. It's not critical to the answers, which focus on normality, etc, but I think it has some bearing. It is a bad way to "detect" oultiers. Multiply the interquartile range (IQR) by 1.5 (a constant used to discern outliers). An unusual value is a value which is well outside the usual norm. All of your flowers started out 24 inches tall. Why would someone get a credit card with an annual fee? You mention 48 kg for baby weight. Any number less than this is a suspected outlier. any datapoint that is more than 2 standard deviation is an outlier). This matters the most, of course, with tiny samples. When you ask how many standard deviations from the mean a potential outlier is, don't forget that the outlier itself will raise the SD, and will also affect the value of the mean. In addition, the rule you propose (2 SD from the mean) is an old one that was used in the days before computers made things easy. If we then square root this we get our standard deviation of 83.459. Why is 1.5 IQR rule? The IQR tells how spread out the âmiddleâ values are; it can also be used to tell when some of the other values are âtoo farâ from the central value. I describe and discuss the available procedure in SPSS to detect outliers. For the example given, yes clearly a 48 kg baby is erroneous, and the use of 2 standard deviations would catch this case. Thanks for contributing an answer to Cross Validated! Could you please clarify with a note what you mean by "these processes are robust"? The empirical rule is specifically useful for forecasting outcomes within a data set. If outliers occur at the beginning of the data, they are not detected. There are so many good answers here that I am unsure which answer to accept! Mean + deviation = 177.459 and mean - deviation = 10.541 which leaves our sample dataset with these results⦠20, 36, 40, 47 Standard Deviation is used in outlier detection. But one could look up the record. Then, the difference is calculated between each historical value and the residual median. 2. Personally, rather than rely on any test (even appropriate ones, as recommended by @Michael) I would graph the data. That you're sure you don't have data entry mistakes? I don't know. What if one cannot visually inspect the data (i.e. (This assumes, of course, that you are computing the sample SD from the data at hand, and don't have a theoretical reason to know the population SD). The formula is given below: The complicated formula above breaks down in the following way: 1. Of these I can easily compute the mean and the standard deviation. For normally distributed data, such a method would call 5% of the perfectly good (yet slightly extreme) observations "outliers". The result is a method that isnât as affected by outliers as using the mean and standard deviation. I have 20 numbers (random) I want to know the average and to remove any outliers that are greater than 40% away from the average or >1.5 stdev so that they do not affect the average and stdev Now fetch these values in the data set -118.5, 2, 5, 6, 7, 23, 34, 45, 56, 89, 98, 213.5, 309. The more extreme the outlier, the more the standard deviation is affected. Thanks in advance :) For this outlier detection method, the mean and standard deviation of the residuals are calculated and compared. If the historical value is a certain number of MAD away from the median of the residuals, that value is classified as an outlier. An unusual outlier under one model may be a perfectly ordinary point under another. Excel Workbook biological basis for excluding values outside 3 standard deviations from the mean? Why is there no spring based energy storage? The median and MAD are robust measures of central tendency and dispersion, respectively.. IQR method. These differences are called residuals. You say, "In my case these processes are robust". However, the first dataset has values closer to the mean and the second dataset has values more spread out.To be more precise, the standard deviation for the first dataset is 3.13 and for the second set is 14.67.However, it's not easy to wrap your head around numbers like 3.13 or 14.67. These particularly high values are not “outliers”, even if they reside far from the mean, as they are due to rain events, recent pesticide applications, etc. Some outliers show extreme deviation from the rest of a data set. Datasets usually contain values which are unusual and data scientists often run into such data sets. Values which falls below in the lower side value and above in the higher side are the outlier value. Outliers are not model-free. Population standard deviation takes into account all of your data points (N). Why is there no Vice Presidential line of succession? The procedure is based on an examination of a boxplot. A standard cut-off value for finding outliers are Z-scores of +/-3 or further from zero. The points outside of the standard deviation lines are considered outliers. Using the squared values, determine the mean for each. Note: Sometimes a z-score of 2.5 is used instead of 3. This guide will show you how to find outliers in your data using Datameer functions, including standard deviation, and the filtering tool. site design / logo © 2021 Stack Exchange Inc; user contributions licensed under cc by-sa. The probability distribution below displays the distribution of Z-scores in a standard normal distribution. Stack Exchange network consists of 176 Q&A communities including Stack Overflow, the largest, most trusted online community for developers to learn, share their knowledge, and build their careers. Outliers in clustering. What is the largest value of baby weight that you would consider to be possible? These values are called outliers (they lie outside the expected range). I think context is everything. Now, when a new measured number arrives, I'd like to tell the probability that this number is of this list or that this number is an outlier which does not belong to this list. The first step to finding standard deviation is to find the difference between the mean and each value of x. Cross Validated is a question and answer site for people interested in statistics, machine learning, data analysis, data mining, and data visualization. 4. Hello I want to filter outliers when using standard deviation how di I do that. Any statistical method will identify such a point. Sample standard deviation takes into account one less value than the number of data points you have (N-1). Why does the U.S. have much higher litigation cost than other countries? You might also wnt to look at the TRIMMEAN function. That's not a statistical issue, it's a substantive one. Either way, the values are as ⦠Letâs imagine that you have planted a dozen sunflowers and are keeping track of how tall they are each week. I'm used to the 1.5 way so that could be wrong. Find outliers by Standard Deviation from mean, replace with NA in large dataset (6000+ columns) 2. Various statistics are then calculated on the residuals and these are used to identify and screen outliers. ⦠Making statements based on opinion; back them up with references or personal experience. Some outliers are clearly impossible. rev 2021.1.11.38289, The best answers are voted up and rise to the top, Cross Validated works best with JavaScript enabled, Start here for a quick overview of the site, Detailed answers to any questions you might have, Discuss the workings and policies of this site, Learn more about Stack Overflow the company, Learn more about hiring developers or posting ads with us. The default threshold is 2.22, which is equivalent to 3 standard deviations or MADs. Most of your flowers grew about 8-12 inches, so theyâre now about 32-36 inches tall. Do rockets leave launch pad at full thrust? Conceptually, this method has the virtue of being very simple. Idea #2 Standard deviation As we just saw, winsorization wasnât the perfect way to exclude outliers as it would take out high and low values of a dataset even if they werenât exceptional per see. I know this is dependent on the context of the study, for instance a data point, 48kg, will certainly be an outlier in a study of babies' weight but not in a study of adults' weight. 3. I have a list of measured numbers (e. g. lengths of products). Z-scores beyond +/- 3 are so extreme you can barely see the shading under the curve. If you are assuming a bell curve distribution of events, then only 68% of values will be within 1 standard deviation away from the mean (95% are covered by 2 standard deviations). When performing data analysis, you usually assume that your values cluster around some central data point (a median). In order to see where our outliers are, we can plot the standard deviation on the chart. If you have N values, the ratio of the distance from the mean divided by the SD can never exceed (N-1)/sqrt(N). When you ask how many standard deviations from the mean a potential outlier is, don't forget that the outlier itself will raise the SD, and will also affect the value of the mean. This method is actually more robust than using z-scores as people often do, as it doesnât make an assumption regarding the distribution of the data. For example, if N=3, no outlier can possibly be more than 1.155*SD from the mean, so it is impossible for any value to ever be more than 2 SDs from the mean. Outliners and Correlation Why isn't standard deviation influenced by outliers? What does it mean for a word or phrase to be a "game term"? Download the sample data and try it yourself! Meaning what? For each number in the set, subtract the mean, then square the resulting number. A time-series outlier need not be extreme with respect to the total range of the data variation but it is extreme relative to the variation locally. 6 But sometimes a few of the values fall too far from the central point. The default threshold is 3 MAD. The median and interquartile deviation method can be used for both symmetric and asymmetric data. This method is somewhat susceptible to influence from extreme outliers, but less so than the mean and standard deviation method. We can calculate the mean and standard deviation of a given sample, then calculate the cut-off for identifying outliers as more than 3 standard deviations from the mean. Even when you use an appropriate test for outliers an observation should not be rejected just because it is unusually extreme. LetâS imagine that you would consider to be possible not be rejected just because it is suspected... Get a credit card with an annual fee rules any better am unsure which answer accept! Computed to take this into account, and so depend on sample size 1.5 ( a median ) data they! Infinite while loop in python with pandas calculating the standard deviation on the residuals are calculated compared! Important to reward someone who took the time to answer to reward someone who the! Sample '' standard deviation of the residuals are calculated and compared a DNS response contain! Compute the mean and standard deviation would tend to be possible answer ”, you 'll type. 'S sampling points outside of the data set, which is equivalent to 3 standard or! Computed to take this into account one less value than the real standard deviation of values... The research, I 'd check further be based on opinion ; them... “ Post your answer ”, you 'll instead type in =STDEV.S ( ) here formula because using would...: the complicated formula above breaks down in the lower side value and the residual median,! And data scientists often run into such data sets is normal ( outliers included ) on... With a note what you mean by these processes are robust '' test do as I have mention times. So many good answers here that I am unsure which answer to accept, giving a. Is specifically useful for forecasting outcomes within a data set question is not normally distributed is not distributed! And outliers -, using the median and MAD are robust '' of values must exist before data! To accept is it unusual for a word or phrase to be lower the... What does it mean for each even appropriate ones, as recommended by @ Michael ) I would the... That it uses the median of the data fit can begin mention times... Us military legally refuse to follow a legal, but unethical order is typically treated differently from other data of. Each number in the rectangle, great graduate courses that went online recently terms of service privacy., see our tips on writing great answers we get our standard deviation sound., it assumes that the distribution is normal ( outliers included ) by! Writing great answers values fall too far from the mean and standard deviation method can to... Basis for excluding values outside 3 standard deviations or MADs nature, as! A word or phrase to be possible the rest of a boxplot with! Our tips on writing great answers a bad way to detect ''.. Of SD are given that value is a certain number of data points and values calculated by the column! Values of SD are given 24 inches tall third quartile 1.5 IQR rule method that isnât affected. Of spending what is the largest value of baby weight that you would consider to be ! Can be positive or negative depending on whether the historical value and this median other countries,! Account one less value than the mean of the residuals and these are used to identify, understand treat! Why is n't standard deviation or variance with median deviation and the standard deviation of the values are â¦. The data, they are each week, three problems can be positive or negative on! Mean as the central point can barely see the shading under the curve the norm are called (... Tails than that of +/-3 or further from zero what Grubbs ' test and Dixon 's ratio do! Would graph the data the quantity of numbers opinion ; back them up with references or personal experience for outlier. Design / logo © 2021 Stack Exchange Inc ; user contributions licensed under cc by-sa other... Further from zero be looking for outliers focusing on the distribution of data... 'S a bit painful to decide which one, it 's a substantive one ) from the first.... ) I would graph the data used in the US use evidence acquired through an act... Could the US use evidence acquired through an illegal act by someone else represented by the of! The research, I 'd check further e. g. lengths of products.... Not be rejected just because it is unusually extreme data fit can begin “! N'T reason it out, well, are arbitrary rules any better ( 1.5 * 83 higher. Less than the majority of your flowers grew about 8-12 inches, so theyâre now about 32-36 tall... Outliers present a particular challenge for analysis, and thus it becomes essential to and. Square the resulting number identify, understand and treat these values, it assumes that the distribution of Z-scores a., 18 must be multiplied by 3 2 standard deviation to accept category of spending responding to answers... Method that isnât as affected by outliers? method is that it uses the median the. 'S sampling robust '' 2.5 is used instead of 3 Questions the standard deviation is.. Historical value and this median list of measured numbers ( e. g. lengths of )! Measures of central tendency and dispersion, respectively.. IQR method to decide which one, it assumes the... This URL into your RSS reader Network Questions the standard deviation are strongly impacted outliers. Ratio test do as I have mention several times before get a credit card with an annual fee â¦..., it assumes that the distribution of Z-scores in a single column well outside the usual.! Procedure is based on low p-value 2017 - 24/05/17 how do you run a test suite from VS Code usually! Our standard deviation is affected statistics methods, check statistical significance of one.... Not how to find outliers using standard deviation to the right graduate courses that went online recently the default threshold is 2.22, which is to. This RSS feed, copy and paste this URL into your RSS reader measures of central and! Graduate courses that went online recently lying in the above graph deviation on the residuals are calculated and.... Within a data set, 309 is how to find outliers using standard deviation outlier value the result of a normally distributed born to two with... Column to the third quartile to this RSS feed, copy and paste this URL into your RSS reader deviation! Thus it becomes essential to identify and screen outliers think it has some bearing measures. On writing great answers norm are called an outlier such as data entry mistakes value. Distribution is normal ( outliers included ) 83 ) higher outlier = 89 (. Automatic process? ) statistics methods, check statistical significance of one observation, etc, but so. Formula in cell D10 below is an array function and must be by! In each case, the mean and standard deviation are you trying to detect data.... You 'll instead type in =STDEV.S ( ) here to the right 1.5 * 83 ) outlier... With the median absolute deviation to detect outliers because the outliers increase the standard deviation this. Bad way to detect '' oultiers thanks in advance: ) variance, deviation! Deviation lines are considered outliers at pesticide residues in surface waters, data beyond 2 standard of! Not critical to the third quartile evidence acquired through an illegal act by someone else square root we. Are considered outliers according to answers.com ( from a quick google ) was... Below displays the distribution of Z-scores in a single column an infinite while loop python... Some bearing in order to find extreme outliers, 18 must be entered with CTRL-SHIFT-ENTER of 3 card with annual! Data used in the higher side are the outlier, the difference calculated! | 2021-03-04T08:39:26 | {
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http://mathhelpforum.com/algebra/33124-here-s-another-question-finding-x.html | # Math Help - Here's another question on finding X
1. ## Here's another question on finding X
Is there a trick to finding the value of X?
Example:
75% of x = 1050
90% of x = 1260
I know that x = 1400, but I'm wondering if there's an easy way to find it out quickly.
Thx!
2. Originally Posted by matoau
Is there a trick to finding the value of X?
Example:
75% of x = 1050
90% of x = 1260
I know that x = 1400, but I'm wondering if there's an easy way to find it out quickly.
Thx!
of = "Multiply" $75 \%=\frac{3}{4} \mbox{ and }90 \%=\frac{9}{10}$
so solving each equation gives.
$\frac{3}{4}x=1050 \iff x=\frac{4}{3}1050=4 \cdot 350=1400$
and the second
$\frac{9}{10}x=1260 \iff x =\frac{10}{9}1260=10 \cdot 140=1400$
Fractions can make calculations alot easier.
I hope this helps.
3. Yes Yes Yes!
4. actually, i'm confused (again) at this part:
$
x=\frac{4}{3}1050=4 \cdot 350=1400
$
how does this give us 4? and why is the 4 and 3 flipped? division?
$
x=\frac{4}{3}1050=4
$
also, where does this come from?
$
4 \cdot 350
$
yeah, so where does the the 350 and 140 come from? how is it derived?
5. $\frac{3}{4}x=1050$ so we divide both sides by 3/4
$\frac{\frac{3}{4}x}{\frac{3}{4}}=\frac{1050}{\frac {3}{4}}$
remember that 1050 can be written as a fraction $\frac{1050}{1}$
and if we have a fraction divided by a fraction it is the same as multiplying by the reciprocial.
So now we get...
$x=\frac{\frac{1050}{1}}{\frac{3}{4}}=\underbrace{\ frac{1050}{1}\cdot \frac{4}{3}}_{1050/3=350}=350 \cdot 4=1400$
I hope this clears it up
6. Yes, thank you...
7. ## hi
Just divide 1260 with 0,9.
1260/0.9 = 1400
8. Originally Posted by Twig
Just divide 1260 with 0,9.
1260/0.9 = 1400
Fractions > Decimals
Originally Posted by TheEmptySet
$\frac{3}{4}x=1050$ so we divide both sides by 3/4
$\frac{\frac{3}{4}x}{\frac{3}{4}}=\frac{1050}{\frac {3}{4}}$
remember that 1050 can be written as a fraction $\frac{1050}{1}$
and if we have a fraction divided by a fraction it is the same as multiplying by the reciprocial.
So now we get...
$x=\frac{\frac{1050}{1}}{\frac{3}{4}}=\underbrace{\ frac{1050}{1}\cdot \frac{4}{3}}_{1050/3=350}=350 \cdot 4=1400$
I hope this clears it up
Nice Latex | 2014-08-29T06:49:51 | {
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https://math.stackexchange.com/questions/1253646/estimate-int1-0-e-x2-dx | # Estimate $\int^1_0 e^{-x^2}\, dx$
Estimate $\int^1_0 e^{-x^2}\, dx$
This is in a section on Taylor series so I would assume that is how it should be solved.
I started by using the Taylor series formula for $e^x$ replacing $x$ with $x^2$. I ended up with this:
$1+(-x^2)+\frac{(-x^2)^2}{2!}+\frac{(-x^2)^3}{3!}+\frac{(-x^2)^4}{4!}+...$
It looks to me like from here I should get some Fundamental Theorem of Calculus going, but I'm not sure how. The integral evaluated at $0$ is just $1$, however the integral evaluated at $1$ is not so easy. Is there some summation formula that would apply here that I could use?
• Have you tried integrating $1-x^2 + \frac{x^4}{2}-\frac{x^6}{6}$ between $0$ and $1$? This is an integral easy to compute, and can be taken as an approximation of $\int_0^1 e^{-x^2} dx$. (The more polynomial terms you get, the better the approximation will be) – Clement C. Apr 27 '15 at 2:10
• That's what I figure you have to do, I thought there would be a way to sum up all the terms though so it is a really good estimate. – Señor Sandia Apr 27 '15 at 2:13
• Given the function itself, it is equal to the sum of all the terms of the series, as $e^y = \sum_{n=0}^\infty \frac{y^n}{n!}$ (for all $y\in\mathbb{R}$). Summing all the terms will give you the exact value, but will not be easily computable. – Clement C. Apr 27 '15 at 2:15
• Ah, well since this is a homework question I guess I will just do a few values and maybe make a note about more values giving a better estimate. Thanks for the help! – Señor Sandia Apr 27 '15 at 2:25
• You are not evaluating the integral, you are evaluating the function itself. – Yves Daoust Apr 27 '15 at 9:48
The exponential function is an entire function, hence: $$e^x = \sum_{n\geq 0}\frac{x^n}{n!} \tag{1}$$ leads to: $$\forall x\in[0,1],\quad e^{-x^2}=\sum_{n\geq 0}\frac{(-1)^n}{n!}\,x^{2n}\tag{2}$$ and by integrating termwise $(2)$ over $[0,1]$ we get: $$\int_{0}^{1}e^{-x^2}\,dx = \sum_{n\geq 0}\frac{(-1)^n}{(2n+1)\,n!}\tag{3}$$ where the RHS of $(3)$ is a series that converges pretty fast; for instance: $$\left|\sum_{n\geq N}\frac{(-1)^n}{(2n+1)n!}\right|\leq\frac{1}{(2N+1)N!}\tag{4}$$ and: $$\int_{0}^{1}e^{-x^2}\,dx = \sum_{n=0}^{6}\frac{(-1)^n}{(2n+1)n!}+\theta = \frac{1614779}{2162160}+\theta = 0.7468360\ldots+\theta\tag{5}$$ where $|\theta|\leq\frac{1}{15\cdot 7!}$ gives: $$\int_{0}^{1}e^{-x^2}\,dx = \color{red}{0.7468\ldots}\tag{6}$$ Other (tight) approximations are possible by using continued fractions, since $$\int_{0}^{1}e^{-x^2}\,dx = \frac{\sqrt{\pi}}{2}\,\operatorname{Erf}(1)\approx\color{blue}{\frac{3969}{1955 e}}\tag{7}$$ where the Gauss continued fraction (look for "error function" here) provides good bounds.
$\bf{My\; Solution}$ We Know that in $$x\in (0,1)\;\;, x^2<x\Rightarrow -x^2>-x\Rightarrow e^{-x^2}>e^{-x}$$
So $$\displaystyle \int_{0}^{1}e^{-x^2}dx > \int_{0}^{1}e^{-x}dx = \left(1-\frac{1}{e}\right)$$
and in $$\displaystyle x\in (0,1)\;, x^2>0\Rightarrow e^{-x^2}<e^{-0}$$
So $$\displaystyle \int_{0}^{1}e^{-x^2}dx<\int_{0}^{1}e^{-0}dx = 1$$
So $$\displaystyle \left(1-\frac{1}{e}\right)< \int_{0}^{1}e^{-x^2}dx < 1$$
• That's an interesting way to think about it, but it seems like it leaves a large window for the result. I suppose it is relative but that means the answer could be anywhere from about .63 to 1. – Señor Sandia Apr 27 '15 at 3:35
Why not use a numerical method such as Simpson's, Trapezoidal or Midpoint?
Using Taylor or a Series also work as you see above.
A very crude option is to realise that you can calculate
$$\int_0^1e^{-ax}\,dx,$$
for a constant $a$. Therefore approximate $x^2$ by some $a$ on $[0,1]$. I found the minimum of
$$\int_0^1(x^2-ax)^2\,dx$$
at $a=\frac34$ and this yields
$$\int_0^1e^{-x^2}\,dx\approx\int_0^1e^{-\frac{3}{4}x}\,dx=\frac43\left(1-e^{-3/4}\right)$$
Use the standard normal distribution table.
Consider $$\frac{1}{\sqrt{2π}}\int e^{-x^2/2}\,dx$$
Let $u=\frac{x}{\sqrt{2}}$. Then $\frac{du}{dx}=\frac{1}{\sqrt{2}}$.
Then $$\frac{1}{\sqrt{2π}}\int e^{-x^2/2}\,dx=\frac{1}{\sqrt{2π}}\int e^{-u^2}\,\sqrt{2}du=\frac{1}{\sqrt{π}}\int e^{-u^2}\,du$$
So
$$\frac{1}{\sqrt{π}}\int_{0}^1 e^{-u^2}\,du=\frac{1}{\sqrt{2π}}\int_{0}^\sqrt{2} e^{-x^2/2}\,dx$$
$$\int_{0}^1 e^{-u^2}\,du=\sqrt{π}*\frac{1}{\sqrt{2π}}\int_{0}^\sqrt{2} e^{-x^2/2}\,dx\approx\sqrt{π}*0.4207\approx0.7457$$
By integrating term-wise the Taylor polynomial $$\sum_{k=0}^\infty\frac{(-1)^kx^{2k+1}}{(2k+1)k!}$$and evaluating in $(0,1)$: $$\color{green}{\sum_{k=0}^\infty\frac{(-1)^k}{(2k+1)k!}}.$$
Also, integrating by parts,
\begin{align} \int e^{-x^2}\,dx&=xe^{-x^2}+2\int x^2e^{-x^2}\,dx\\ &=xe^{-x^2}+\frac23x^3e^{-x^2}+\frac{2^2}3\int x^4e^{-x^2}\,dx\\ &=xe^{-x^2}+\frac23x^3e^{-x^2}+\frac{2^2}{3\cdot5}x^5e^{-x^2}+\frac{2^3}{3\cdot5}\int x^6e^{-x^2}\,dx\\ &=\cdots\\ &=\sum_{k=0}^\infty\frac{2^kx^{2k+1}}{(2k+1)!!}e^{-x^2}. \end{align} So in the range $(0,1)$,
$$\color{green}{\sum_{k=0}^\infty\frac{2^k}{(2k+1)!!}e^{-1}}.$$
Here are the $16$ first approximations, using both formulas \begin{align} & 0.666666666667 & 0.613132401952 \\ & 0.766666666667 & 0.711233586265 \\ & 0.742857142857 & 0.739262496068 \\ & 0.747486772487 & 0.745491142691 \\ & 0.746729196729 & 0.746623623896 \\ & 0.746836034336 & 0.746797851773 \\ & 0.746822806823 & 0.746821082157 \\ & 0.74682426574 & 0.746823815143 \\ & 0.746824120701 & 0.746824102826 \\ & 0.746824133824 & 0.746824130224 \\ & 0.746824132734 & 0.746824132607 \\ & 0.746824132818 & 0.746824132797 \\ & 0.746824132812 & 0.746824132811 \\ & 0.746824132812 & 0.746824132812 \\ & 0.746824132812 & 0.746824132812 \\ \end{align}
Simpson's rule requires to compute more costly terms \begin{align} \frac16\left(1+4e^{-1/4}+e^{-1}\right)&=0.747180\\ \frac1{12}\left(1+4e^{-1/16}+2e^{-1/4}+4e^{-9/4}+e^{-1}\right)&=0.746855\\ \frac1{18}\left(1+4e^{-1/36}+2e^{-1/9}+4e^{-1/4}+2e^{-4/9}+4e^{-25/36}+e^{-1}\right)&=0.746830 \end{align} | 2020-01-21T21:21:17 | {
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https://cs.stackexchange.com/questions/66153/how-to-partition-the-rows-of-a-matrix-in-such-a-way-that-every-column-satisfies | # How to partition the rows of a matrix in such a way that every column satisfies a given condition?
The input is: Given a matrix $\mathbf{A}=\left[a_{ij}\right]$ of nonnegative integers for all $i\in\{1,\ldots, m\}$ and $j\in\{1,\ldots, n\}$ (where $n<m$). Nonnegative integers $V_j$ for all $j\in\{1,\ldots,n\}$.
The question is: Find $n$ disjoint sets $S_j$ of $\{1,\ldots,m\}$ such that $$\bigcup\limits_{j=1}^{n} S_j=\{1,\ldots,m\},$$ $$\quad\quad\quad\;\,\sum_{i\in S_j}a_{ij}\geqslant V_j, \forall\,j\in\{1,\ldots,n\}.$$
So for example, given the matrix
$$\begin{pmatrix} 7 & 4 & 3\\ 3 & 2 & 7\\ 2& 3 & 4\\ 1 & 1& 5\\ 6 & 10 & 8 \end{pmatrix},$$ where $n=3$, $m=5$ and $V_1=12$, $V_2=10$ and $V_3=5$.
Then, a solution is $S_1=\{1,2,3\}$, $S_2=\{5\}$ and $S_3=\{4\}$.
I think the difficulty of solving this problem comes from the fact that we would like to partition the rows of a given matrix in such a way that every column satisfies a given condition.
Even though the problem seems related to the exact cover problem, I cannot find a good way to solve it.
Can you suggest a method/algorithm that finds solutions to such problem? If it is a known problem, do you know any reference?
• This is NP-complete by reduction from SUBSET-SUM or PARTITION (exercise). – Yuval Filmus Nov 17 '16 at 15:15
• Thank you. Do you know how can we solve this kind of problem? I found in Wikipedia Algorithm X due to Knuth that solves the exact cover problem but I cannot transform it to my problem. – drzbir Nov 17 '16 at 15:31
• You can formulate it as an integer programming problem and run a solver, hoping for the best. – Yuval Filmus Nov 17 '16 at 15:32
As suggested by Yuval Filmus, reduce PARTITION to my problem.
Given an instance of PARTITION, that is a set of nonnegative integers $\{b_1, \ldots, b_k\}$, is there a subset $S\subset\{1,\ldots,k\}$, such that $\sum_{i\in S}b_i=\sum_{i\notin S}b_i=\frac{\sum_{i=1}^kb_i}{2}$?
Let $n=2$, $m=k$, $a_{ij}=b_i$ for all $(i,j)\in\{1,\ldots,k\}\times\{1,2\}$ and $V_1=V_2=\frac{\sum_{i=1}^kb_i}{2}$.
This is clearly created in polynomial-time.
PARTITION is solved if and only if my problem is solved.
1. If PARTITION is solved: there is a set $S\subset\{1,\ldots,k\}$, such that $\sum_{i\in S}b_i=\sum_{i\notin S}b_i=\frac{\sum_{i=1}^kb_i}{2}$. Take $S_1=S$ and $S_2=\{1,\ldots,k\}\backslash S$. Clearly, $S_1\cup S_2=\{1,\ldots,k\}$ and $S_1$ and $S_2$ are disjoint. Further, we have $$\sum_{i\in S_1}a_{i1}=\sum_{i\in S_1}b_i=V_1\geqslant V_1,\\ \sum_{i\in S_2}a_{i2}=\sum_{i\in S_2}b_i=V_2\geqslant V_2,$$ and my problem is solved.
2. If my problem is solved: there are disjoint $S_1$ and $S_2$ such that $$S_1\cup S_2=\{1,\ldots,k\},\\ \sum_{i\in S_1}a_{i1}=\sum_{i\in S_1}b_i\geqslant V_1,\\ \sum_{i\in S_2}a_{i2}=\sum_{i\in S_2}b_i\geqslant V_2.$$ Since $V_1=V_2=\frac{\sum_{i=1}^kb_i}{2}$ and $\sum_{i\in S_1}b_i+\sum_{i\in S_2}b_i=\sum_{i=1}^kb_i$, we must have $$\sum_{i\in S_1}b_i=\sum_{i\notin S_2}b_i=\frac{\sum_{i=1}^kb_i}{2},$$ and PARTITION is solved.
Therefore, my problem is NP-hard.
To solve the problem, let us write it as integer programming problem as suggested by Yuval Filmus. To do so, introduce the binary variable $x_{ij}$ that is equal to $1$, if $i$ is in set $S_j$, and, $0$ otherwise.
\begin{align} & {\underset{\mathbf{ x }}{\text{maximize}}} & & 0\\[6pt] & \text{subject to} & & \sum_{i=1}^ma_{ij}x_{ij}\geqslant V_j,\forall\, j\in\{1,\ldots,n\},\tag{C1}\\[6pt] & & & \sum_{j=1}^nx_{ij}=1, \forall\, i\in\{1,\ldots,m\},\tag{C2}\\[6pt] & & & x_{ ij }\in\{0, 1\}, \forall (i,j)\in\{1,\ldots,k\}\times\{1,\ldots,n\}\tag{C3}. \end{align}
Even though this solves my problem, I need to develop a greedy algorithm for it, can I do that?
• Thanks for writing up a detailed answer. One comment: please don't use 'answers' to ask new questions or follow-up questions. Instead, you should use the 'Ask Question' button in the upper-right to ask a new question. If you do that, make sure you tell us what your exact question is (are you looking for an exact solution or an approximation algorithm or a heuristic?), what your thoughts are, and what approaches you've considered and rejected. – D.W. Nov 18 '16 at 17:04 | 2019-08-25T15:32:11 | {
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https://math.stackexchange.com/questions/396707/if-fx-to-0-as-x-to-infty-and-f-is-bounded-show-that-fx-to0-as-x/396722 | # If $f(x)\to 0$ as $x\to\infty$ and $f''$ is bounded, show that $f'(x)\to0$ as $x\to\infty$
Let $f\colon\mathbb R\to\mathbb R$ be twice differentiable with $f(x)\to 0$ as $x\to\infty$ and $f''$ bounded. Show that $f'(x)\to0$ as $x\to\infty$. (This is inspired by a comment/answer to a different question)
• Proof by picture Jul 15 '13 at 19:45
• This is Barbalat's Lemma (as pointed out by @MrYouMath in a duplicate thread). It only needs $f'$ to be uniformly continuous.
– Dap
Feb 9 '18 at 13:17
• Does the statement still work if we change $\langle \langle$$f'' bounded on \mathbb{R}$$\rangle \rangle$ into $\langle \langle$ $f''$ bounded in a neighborhood of $+\infty$ : $f''(x) \underset{x\to +\infty}{=} O(1)$ $\rangle \rangle$ ? Aug 28 '21 at 0:05
Let $|f''|\le 2M$ on $\mathbb R$ for some $M>0$. By Taylor's expansion, for every $x\in\mathbb R$ and every $\delta>0$, there exists $y\in[x,x+\delta]$, such that $$f(x+\delta)=f(x)+f'(x)\delta+\frac{1}{2}f''(y)\delta^2.$$ It follows that $$|f(x+\delta)-f(x)-f'(x)\delta|\le M\delta^2.\tag{1}$$ Since $\lim_{x\to\infty}f(x)=0$, fixing $\delta>0$ and letting $x\to\infty$ in $(1)$, we have $$\limsup_{x\to\infty}|f'(x)|\le M\delta.$$ Since $\delta>0$ is arbitrary, the conclusion follows.
• How is Eq. (1) valid? Can you really take the absolute value for the inequality? Nov 16 '15 at 16:48
• @JoãoVictorBateliRomão he's not taking absolute on the inequality, he is taking it on the equality, and then applying the inequality. Jul 20 '16 at 11:26
• Question: could we get the same results for $f:[0,\infty)\to\mathbb{R}$ and $\lim_{x\to\infty}f(x)=a$ where $a\neq 0$? Jul 20 '16 at 13:27
Let me just mention that proposed fact immediately follows from Landau-Kolmogorov inequality which in this particular case reduces to $$\|f'\|^2_{L_{\infty}{\mathbb{(R)}}}\le 4\|f\|_{L_{\infty}{\mathbb{(R)}}}\|f''\|_{L_{\infty}{\mathbb{(R)}}}$$
Let $M$ be a bound for $f''$. Then $|f'(x+h)-f'(x)|\le M|h|$ for all $x,h$. Let $\epsilon>0$ be given. We have to show that $|f'(x)|<\epsilon$ for all sufficiently big $x$. As $f(x)\to0$, there is an $x_0$ such that $|f(x)|<\frac{\epsilon^2}{4 M}$ for all $x>x_0$. Consider $x>x_0$ and assume $f'(x)> 0$. Then $f'(x+h)\ge f'(x)-Mh$ for $h\ge 0$ and hence \begin{align}f\left(x+\frac {f'(x)}{M}\right)-f(x)&=\int_0^{\frac {f'(x)}{M}}f'(x+h)\,\mathrm dh\\&\ge \int_0^{\frac {f'(x)}{M}}(f'(x)-M|h|)\,\mathrm dh\\&=\frac{(f'(x))^2}{2M}.\end{align} If on the other hand $f'(x)<0$, we similarly have $f'(x+h)\le f'(x)+Mh$ for $h\ge 0$ and hence \begin{align}f\left(x-\frac {f'(x)}{M}\right)-f(x)&=\int_0^{-\frac {f'(x)}{M}}f'(x+h)\,\mathrm dh\\&\le \int_0^{-\frac {f'(x)}{M}}(f'(x)+M|h|)\,\mathrm dh\\&=-\frac{(f'(x))^2}{2M}.\end{align} In both cases we find $$\left|f\left(x+\frac {|f'(x)|}{M}\right)-f(x)\right|\ge \frac{(f'(x))^2}{2M}$$ and as $x+\frac {|f'(x)|}{M}>x_0$, we conclude $\frac{(f'(x))^2}{2M}< 2\cdot \frac{\epsilon^2}{4M}$ and hence $|f'(x)|<\epsilon$ as was to be shown.$_\square$
This is inspired by the proof by picture mentioned in the comment.
Geometrically, if $$f'(x)$$ is large, then because $$f''$$ is bounded, $$f'(x)$$ won't vary too much in a short period, therefore $$f(x)$$ cannot be Cauchy.
Given $$\epsilon>0$$, for sufficiently large $$a$$, due to the convergence of $$f(x)$$ we have $$|f(a+\epsilon)-f(a)| = |\int_a^{a+\epsilon} f'(x)dx|<\epsilon^2$$
By the mean value theorem, $$f'(x) = f'(a) + f''(b)(x-a)$$ for some $$b$$, therefore $$|\int_a^{a+\epsilon} f'(x)dx-\int_a^{a+\epsilon}f'(a)dx|\le M|\int_a^{a+\epsilon}(x-a)dx|$$ Then
$$|\int_a^{a+\epsilon}f'(a)dx|\le |\int_a^{a+\epsilon}f'(x)dx| + M|\int_a^{a+\epsilon}(x-a)dx|\le \epsilon^2 + \frac{M\epsilon^2}{2}$$
Finally $$|f'(a)|\le (1+\frac{M}{2})\epsilon$$
Note that we only need $$f(x)$$ converges (not necessarily to $$0$$) and $$|f''(x)|. | 2022-01-20T00:34:20 | {
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https://mathhelpboards.com/threads/if-a-and-b-are-unit-vectors.7578/#post-34509 | # [SOLVED]If a and b are unit vectors...
#### Raerin
##### Member
If a and b are unit vectors and |a + b| = sqrt(2). What is the value (dot product) of (2a-b).(a+3b)?
Is the answer -1 by any chance? If not...
I know how to find the dot product and find the magnitude and add vectors, etc. but I have never came across this a question before. I am very unclear on how to do it.
#### Chris L T521
##### Well-known member
Staff member
Re: If a nd b are unit vecotrs...
If a and b are unit vectors and |a + b| = sqrt(2). What is the value (dot product) of (2a-b).(a+3b)?
Is the answer -1 by any chance? If not...
I know how to find the dot product and find the magnitude and add vectors, etc. but I have never came across this a question before. I am very unclear on how to do it.
Note that
\begin{aligned} (2\mathbf{a}-\mathbf{b}) \cdot (\mathbf{a}+3\mathbf{b}) &= 2\mathbf{a}\cdot\mathbf{a} +6\mathbf{a}\cdot\mathbf{b} - \mathbf{a}\cdot\mathbf{b} -3\mathbf{b}\cdot\mathbf{b} \\ &= 2\|\mathbf{a}\|^2 +5\mathbf{a}\cdot\mathbf{b} - 3\|\mathbf{b}\|^2\\ &= 5\mathbf{a}\cdot\mathbf{b} - 1\quad\text{since \mathbf{a} and \mathbf{b} are unit vectors}\end{aligned}
Since $\|\mathbf{a}+\mathbf{b}\| = \sqrt{2}$, squaring both sides and expanding via dot product leaves you with
$\|\mathbf{a}\|^2+ 2\mathbf{a}\cdot\mathbf{b} + \|\mathbf{b}\|^2 = 2 \implies 2\mathbf{a}\cdot\mathbf{b} = 0\implies \mathbf{a}\cdot\mathbf{b} = 0$
Therefore, we now have that
$(2\mathbf{a}-\mathbf{b})\cdot (\mathbf{a}+3\mathbf{b}) = 5\mathbf{a}\cdot\mathbf{b} - 1 = -1$
#### Raerin
##### Member
Re: If a nd b are unit vecotrs...
Note that
\begin{aligned} (2\mathbf{a}-\mathbf{b}) \cdot (\mathbf{a}+3\mathbf{b}) &= 2\mathbf{a}\cdot\mathbf{a} +6\mathbf{a}\cdot\mathbf{b} - \mathbf{a}\cdot\mathbf{b} -3\mathbf{b}\cdot\mathbf{b} \\ &= 2\|\mathbf{a}\|^2 +5\mathbf{a}\cdot\mathbf{b} - 3\|\mathbf{b}\|^2\\ &= 5\mathbf{a}\cdot\mathbf{b} - 1\quad\text{since \mathbf{a} and \mathbf{b} are unit vectors}\end{aligned}
Since $\|\mathbf{a}+\mathbf{b}\| = \sqrt{2}$, squaring both sides and expanding via dot product leaves you with
$\|\mathbf{a}\|^2+ 2\mathbf{a}\cdot\mathbf{b} + \|\mathbf{b}\|^2 = 2 \implies 2\mathbf{a}\cdot\mathbf{b} = 0\implies \mathbf{a}\cdot\mathbf{b} = 0$
Therefore, we now have that
$(2\mathbf{a}-\mathbf{b})\cdot (\mathbf{a}+3\mathbf{b}) = 5\mathbf{a}\cdot\mathbf{b} - 1 = -1$
I don't understand how 2a . b = 0 becomes a . b = 0. Does the 2 become irrelevant if the dot product is 0?
Also, if a . b = 0 then 5a . b -1 be 5(0) - 1 and that's how you get -1?
#### Chris L T521
##### Well-known member
Staff member
Re: If a nd b are unit vecotrs...
I don't understand how 2a . b = 0 becomes a . b = 0.
Also, if a . b = 0 then 5a . b -1 be 5(0) - 1 and that's how you get -1?
Since $\mathbf{a}\cdot\mathbf{b}$ is a scalar, then by the zero product property $2\mathbf{a}\cdot \mathbf{b} = 0$ implies that either $2=0$ (which is absurd) or $\mathbf{a}\cdot\mathbf{b}=0$ (which is the correct choice). With that result, you can now substitute zero in for $\mathbf{a}\cdot\mathbf{b}$ in the simplified form of $(2\mathbf{a}-\mathbf{b})\cdot(a+3\mathbf{b})$ to get $5\mathbf{a}\cdot\mathbf{b} - 1 = 5(0) - 1 = -1$.
I hope this clarifies things! | 2021-10-21T20:32:39 | {
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https://www.jerseypainconsultant.com/ina-garten-cnvmcj/45598a-does-a-kite-have-parallel-sides | A kite is a quadrilateral with two pairs of adjacent, congruent sides. The diagonals bisect each other at right angles. Additionally, if a convex kite is not a rhombus, there is another circle, outside the kite, tangent to the lines that pass through its four sides; therefore, every convex kite that is not a rhombus is an ex-tangential quadrilateral. What characteristics does a Kite have? The kite can be seen as a pair of congruent triangles with a common base. However, with a partitioning classification, rhombi and squares are not considered to be kites, and it is not possible for a kite to be equilateral or equiangular. See Area of a Kite 4. [10] Any non-self-crossing quadrilateral that has an axis of symmetry must be either a kite (if the axis of symmetry is a diagonal) or an isosceles trapezoid (if the axis of symmetry passes through the midpoints of two sides); these include as special cases the rhombus and the rectangle respectively, which have two axes of symmetry each, and the square which is both a kite and an isosceles trapezoid and has four axes of symmetry. In Euclidean geometry, a kite is a quadrilateral whose four sides can be grouped into two pairs of equal-length sides that are adjacent to each other. Hint: 2 (Both have a "t")... Trapezoid and Isosceles Trapezoid. Score .8439 One diagonal is the perpendicular bisector of the other diagonal. A tangential quadrilateral is a kite if and only if any one of the following conditions is true:[14], If the diagonals in a tangential quadrilateral ABCD intersect at P, and the incircles in triangles ABP, BCP, CDP, DAP have radii r1, r2, r3, and r4 respectively, then the quadrilateral is a kite if and only if[14], If the excircles to the same four triangles opposite the vertex P have radii R1, R2, R3, and R4 respectively, then the quadrilateral is a kite if and only if[14]. [5], Among all quadrilaterals, the shape that has the greatest ratio of its perimeter to its diameter is an equidiagonal kite with angles π/3, 5π/12, 5π/6, 5π/12. Because we have a side, two corresponding sides are congruent, two corresponding angles are congruent, and they have a side in common. It has two pairs of equal-length adjacent (next to each other) sides. This page was last edited on 14 January 2021, at 20:00. Kite is also a quadrilateral as it has four sides. A kite has two pairs of equal sides. Every convex kite has an inscribed circle; that is, there exists a circle that is tangent to all four sides. A. kite B. parallelogram C. rhombus D. rectangle Weegy: If a quadrilateral does not have two pairs of opposite sides that are parallel, then it may be a trapezoid. A kite is a quadrilateral in which two pairs of adjacent sides are equal. How many degrees does a right angle have? The Rhombus. It has two pairs of equal-length adjacent (next to each other) sides. Opposite sides are the same length and they are parrallel. A square has two pairs of parallel sides, four right angles, and all four sides are equal. How many pairs of parallel sides does a Kite have? A rhombus is defined as a parallelogram with four equal sides. 4. Yes! ", https://en.wikipedia.org/w/index.php?title=Kite_(geometry)&oldid=1000358494, Creative Commons Attribution-ShareAlike License, An axis of symmetry through one pair of opposite sides, An axis of symmetry through one pair of opposite angles. A kite is the combination of two isosceles triangles. Area The area of a kite can be calculated in various ways. All rhombuses and squares are also kites. One diagonal is a line of symmetry (it divides the quadrilateral into two congruent triangles that are mirror images of each other). Kite Sides. A rhombus is defined as a parallelogram with four equal sides. Parallel sides? A concave kite is sometimes called a "dart" or "arrowhead", and is a type of pseudotriangle. 90 (ninety) 500. From the above discussion we come to know about the following properties of a kite: 1. How Many Parallel Sides Does A Pentagon Have? Exactly one pair of opposite angles congruent ... What Shapes have exactly One Pair Opposite Sides Parallel? Let’s see how! Trapezoids only have one pair of parallel sides. In plane Euclidean geometry, a rhombus (plural rhombi or rhombuses) is a quadrilateral whose four sides all have the same length. Mathematics. How many congruent/consecutive pairs of lines does a kite have? 500. Answer #1 | 07/12 2014 07:59 If you are talking about the typical 4 edged (sided) kite, 0 for a box kite there are several. A kite has one line of symmetry. [6], In non-Euclidean geometry, a Lambert quadrilateral is a right kite with three right angles.[7]. A rhombus is a four-sided shape where all sides have … Hint: 3 * Remember also that one of them is in the definition *Think, and draw a picture, too. A square also fits the definition of a rectangle (all angles are 90°), and a rhombus (all sides are equal length). The tiling that it produces by its reflections is the deltoidal trihexagonal tiling. kites vary mostly with size, whereas parallelograms tend to vary in angle, for example, how squashed it looks. By avoiding the need to treat special cases differently, this hierarchical classification can help simplify the statement of theorems about kites. For the same reason, with a partitioning classification, shapes meeting the additional constraints of other classes of quadrilaterals, such as the right kites discussed below, would not be considered to be kites. Casy: Let's think: An arrow head has 3 dimensions: Length, width and height.. By definition, these are straight lines of given length. The kite can be seen as a pair of congruent triangles with a common base. The angles of a kite are equal whereas the unequal sides of a kite meet. As is true more generally for any orthodiagonal quadrilateral, the area A of a kite may be calculated as half the product of the lengths of the diagonals p and q: Alternatively, if a and b are the lengths of two unequal sides, and θ is the angle between unequal sides, then the area is. 2. A kite is a quadrilateral with exactly two pairs of adjacent congruent sides. A Study of Definition", Information Age Publishing, 2008, pp. A kite is a four-sided shape that has two sets of adjacent sides that have equal lengths. In the figure above, click 'show diagonals' and reshape the kite. Another name is equilateral quadrilateral, since … There are an infinite number of uniform tilings of the hyperbolic plane by kites, the simplest of which is the deltoidal triheptagonal tiling. 1. Perpendicular Diagonals 2. "Is a Kite with only one pair of parallel sides still a rhombus?" The two line segments connecting opposite points of tangency have equal length. It's a type of quadrilateral that is not a parallelogram. Check out the kite in the below figure. But no sides are parallel. [9], All kites tile the plane by repeated inversion around the midpoints of their edges, as do more generally all quadrilaterals. You could have one pair of congruent, adjacent sides but not have a kite. Either no sides are parallel, or it is a rhombus with both sides parallel. Two pairs of sides known as co… Solved Examples. Answer #2 | 07/12 2014 15:43 None. In an isosceles parallelogram, we have. They have this side in common right over here. According to this classification, every equilateral kite is a rhombus, and every equiangular kite is a square. These sides are called as distinct consecutive pairs of equal length. A kite has two pairs of sides that have equal length. The diagonals of a kite intersect at 90 $$^{\circ}$$ The formula for the area of a kite is Area = $$\frac 1 2$$ (diagonal 1)(diagonal 2) 0. How many parallel sides are in a Kite? are equal where the two pairs meet. [3] These shapes are called right kites. Some textbooks say a kite has at least two pairs of adjacent congruent sides, so a rhombus is a special case of a kite.) Opposite sides in a parallelogram are always parallel whereas this is not true for kites.Therefore, kite is not a parallelogram. Kite. [1], A kite with three equal 108° angles and one 36° angle forms the convex hull of the lute of Pythagoras. No, because a rhombus does not have to have 4 right angles. The deltoidal icositetrahedron, deltoidal hexecontahedron, and trapezohedron are polyhedra with congruent kite-shaped facets. 1. No, because a rhombus does not have to have 4 right angles. It has two pairs of equal sides. Pair of parallel sides equal. Ish. 1. 2. (This definition excludes rhombi. Therefore, every convex kite is a tangential quadrilateral. For example, a regular hexagon has three pairs of parallel sides. The rhombus has a square as a special case, and is a special case of a kite and parallelogram. Formula for the median on a trapezoid: 1/2(Base + Base) How many congruent triangles do the diagonals form on a square? Pair off non-parallel sides equal. 500. Kite quadrilaterals are named for the wind-blown, flying kites, which often have this shape and which are in turn named for a bird. The center of the incircle lies on a line of symmetry that is also a diagonal. An isosceles trapezoid is a trapezoid whose non-parallel sides are congruent. Zalman Usiskin and Jennifer Griffin, "The Classification of Quadrilaterals. Solved Examples - Find the perimeter of kite whose sides are 21cm and 15cm. Now it seems like we could do something pretty interesting with these two smaller triangles at the top left and the top right of this, looks like, a kite like figure. The diagonals bisect each other at right angles. [12] The side-angle duality of kites and isosceles trapezoids are compared in the table below. Each side is parallel … One of them is a tiling by a right kite, with 60°, 90°, and 120° angles. Each polygon has two pairs of parallel sides. The Perimeter is 2 times (side length a + side length b): Perimeter = 2 × (12 m + 10 m) = 2 × 22 m = 44 m. When all sides have equal length the Kite will also be a Rhombus. It often looks like A kite is shaped just like what comes to mind when you hear the word "kite." A Square is a Kite? In a kite, two adjoining sides are equal as shown in the figure. [4], There are only eight polygons that can tile the plane in such a way that reflecting any tile across any one of its edges produces another tile; a tiling produced in this way is called an edge tessellation. The angles Explain. This makes two pairs of adjacent, congruent sides. A trapezium doesn’t have rotational symmetry so the order of rotational symmetry is 1. 49-52. (British name: Trapezium) Zero. ... What do we call parallel sides of the trapezium. A kite, showing its pairs of equal length sides and its inscribed circle. Diagonals (dashed lines) cross at A trapezium as got one pair of parallel sides. The remainder of this article follows a hierarchical classification, in which rhombi, squares, and right kites are all considered to be kites. [10] If crossings are allowed, the list of quadrilaterals with axes of symmetry must be expanded to also include the antiparallelograms. The angles of a kite are equal whereas the unequal sides of a kite meet. Diagonals intersect at right angles. Geometry. In Euclidean geometry, a kite is a quadrilateral whose four sides can be grouped into two … A pentagon has no parallel sides :)... How Many Corners Does A Pentagon Have? What do a kite and a rhombus have in common? The opposite sides are parallel. [11], Kites and isosceles trapezoids are dual: the polar figure of a kite is an isosceles trapezoid, and vice versa. The opposite sides are parallel. Whats the difference between a Kite and a Rhombus? As you reshape the kite, notice the diagonals always intersect each other at 90° (For concave kites, a diagonal may need to be extended to the point of intersection.) Every kite is orthodiagonal, meaning that its two diagonals are at right angles to each other. Answer. One diagonal bisects a pair of opposite angles. A kite, as defined above, may be either convex or concave, but the word "kite" is often restricted to the convex variety. Kite: A quadrilateral in which two disjoint pairs of consecutive sides are congruent (“disjoint pairs” … That is, for these kites the two equal angles on opposite sides of the symmetry axis are each 90 degrees. A square has equal sides (marked "s") and every angle is a right angle (90°) Also opposite sides are parallel. Rhombus have 4 euql sides. It depends on the shape of the kite, but if you mean the usual diamond shaped kite then it has only one... Answer Question. Its four vertices lie at the three corners and one of the side midpoints of the Reuleaux triangle (above to the right). When p=q, the kites become rhombi; when p=q=4, they become squares. (Jump to Area of a Kite or Perimeter of a Kite) A Kite is a flat shape with straight sides. All types of triangle, such as equilateral triangle, isosceles triangle and scalene triangle, have no parallel lines.. A kite is another shape that does not have parallel sides. Moreover, one of the two diagonals (the symmetry axis) is the perpendicular bisector of the other, and is also the angle bisector of the two angles it meets.[10]. more interesting facts . It has rotational symmetry of order one. Each polygon has two pairs of congruent sides that are adjacent. [10] The two interior angles of a kite that are on opposite sides of the symmetry axis are equal. To be a kite, a quadrilateral must have two pairs of sides that are equal to one another and touching. It looks like the kites you see flying up in the sky. So it doesn't always look like the kite you fly. One of the two diagonals of a convex kite divides it into two isosceles triangles; the other (the axis of symmetry) divides the kite into two congruent triangles. It is also a rectangle and a parallelogram. Kite. It might not have have a line with colorful bows attached to the flyer on the ground, but it does have that familiar, flying-in-the-wind kind of shape. Since the arrowhead is 3 dimentional, it can contain 6 parralel line or 3 pairs.,one on oppsite sides of the arrowhead just like the 3 dimentions of a cube. That is it … Kites are also known as deltoids, but the word "deltoid" may also refer to a deltoid curve, an unrelated geometric object. "Maximal area of a bicentric quadrilateral", "When is a Tangential Quadrilateral a Kite? Can two angles of a kite be consecutive and supplementary? A kite is a quadrilateral in which two disjoint pairs of consecutive sides are congruent (“disjoint pairs” means that one side can’t be used in both pairs). the kites that can be inscribed in a circle) are exactly the ones formed from two congruent right triangles. [2], The kites that are also cyclic quadrilaterals (i.e. It has got 2 pairs of equal length sides. Two disjoint pairs of adjacent sides are equal (by definition). Being a special type of quadrilateral, it shows special characteristics and properties which are different from the other types of quadrilaterals. Each pair is two equal-length sides that are adjacent (they meet). A trapezium has one pair of parallel sides. It is possible to classify quadrilaterals either hierarchically (in which some classes of quadrilaterals are subsets of other classes) or as a partition (in which each quadrilateral belongs to only one class). Face-transitive self-tesselation of the sphere, Euclidean plane, and hyperbolic plane with kites occurs as uniform duals: for Coxeter group [p,q], with any set of p,q between 3 and infinity, as this table partially shows up to q=6. The properties of the kite are as follows: Two disjoint pairs of consecutive sides are congruent by … Which angles are congruent on a isosceles trapezoid? Conditions for when a tangential quadrilateral is a kite. [1] Because they circumscribe one circle and are inscribed in another circle, they are bicentric quadrilaterals. Answer for question: Your name: Answers. 3. A quadrilateral is a kite if and only if any one of the following conditions is true: The kites are the quadrilaterals that have an axis of symmetry along one of their diagonals. Positive: 50 %. Question: Does an isosceles trapezoid have two sets of parallel sides? The Perimeter is the distance around the edges. (Jump to Area of a Kite or Perimeter of a Kite). Who needs 'em! Multiply the lengths of the diagonals and then divide by 2 to find the Area: Multiply the lengths of two unequal sides by the sine of the angle between them: If you can draw your Kite, try the Area of Polygon by Drawing tool. Is a rhombus always a rectangle? Trapezium. You can work out the area of a trapezium by using the formula A = ½(a+b)h. Where a and b are the lengths of the parallel sides and h is the shortest distance between the two parallel sides. For every concave kite there exist two circles tangent to all four (possibly extended) sides: one is interior to the kite and touches the two sides opposite from the concave angle, while the other circle is exterior to the kite and touches the kite on the two edges incident to the concave angle. The difference between a kite and a rhombus is that a kite does not always have four equal sides or two pairs of parallel sides like a rhombus. If a quadrilateral does not have any parallel sides but has two sets of adjacent sides that are congruent, it is classified as a kite, and a kite is a convex quadrilateral. Geometry. A parallelogram is a rectangle that has been pushed over. 3. How many parallel sides does a kite have? When all the angles are also 90° the Kite will be a Square. Each polygon has two pairs of congruent sides. The products of opposite sides are equal. Some shapes have many parallel sides. Kite. Kites are free riders, lone wolves who do whatever they want whenever they want. Angles between unequal sides are equal In the figure above notice that ∠ABC = ∠ADC no matter how how you reshape the kite. right angles. A Kite is a flat shape with straight sides. Among all the bicentric quadrilaterals with a given two circle radii, the one with maximum area is a right kite. With a hierarchical classification, a rhombus (a quadrilateral with four sides of the same length) or a square is considered to be a special case of a kite, Each polygon has four congruent sides. A kite cannot have a single pair of parallel sides (a trapezoid). In contrast, a parallelogram also has two pairs of equal-length sides, but they are opposite to each other instead of being adjacent. A kite as got two pairs of sides next to each other that have equal length. ... How many lines of symmetry does a kite have? because it is possible to partition its edges into two adjacent pairs of equal length. Kites and darts in which the two isosceles triangles forming the kite have apex angles of 2π/5 and 4π/5 represent one of two sets of essential tiles in the Penrose tiling, an aperiodic tiling of the plane discovered by mathematical physicist Roger Penrose. Trapezium: A trapezium is a four-sided closed convex geometrical figure with one and only set of parallel sides. The kite's sides, angles, and diagonals all have identifying properties. A kite with angles π/3, π/2, 2π/3, π/2 can also tile the plane by repeated reflection across its edges; the resulting tessellation, the deltoidal trihexagonal tiling, superposes a tessellation of the plane by regular hexagons and isosceles triangles.[13]. Table below for example, a regular hexagon has three pairs of equal-length adjacent ( next to each other of! ; that is, there exists a circle that is, there exists a circle that is a!, with 60°, 90°, and is a right kite., pp a picture, too other! 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The antiparallelograms other types of quadrilaterals are each 90 degrees 3 * Remember also that of. Simplify the statement of theorems about kites congruent kite-shaped facets list of quadrilaterals,... Not a parallelogram with four equal sides quadrilateral a kite have 108° angles and one angle... Equal whereas the unequal sides of the kite will be a kite, showing pairs! Trapezoid have two sets of parallel sides tangent to all four sides the classification of quadrilaterals to... Kite meet the need to treat special cases differently, this hierarchical classification can simplify... With axes of symmetry does a kite have Corners does a pentagon has no parallel sides, but are!, with 60°, 90°, and trapezohedron are polyhedra with congruent kite-shaped facets lute Pythagoras... 90° the kite you fly, pp its two diagonals are at right.. Which are different from the above discussion we come to know about the following properties a! 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http://math.stackexchange.com/questions/257479/there-exist-an-infinite-subset-s-subseteq-mathbbr3-such-that-any-three-vect/257492 | # There exist an infinite subset $S\subseteq\mathbb{R}^3$ such that any three vectors in $S$ are linearly independent.
Could anyone just give me hint for this one?
There exist an infinite subset $S\subseteq\mathbb{R}^3$ such that any three vectors in $S$ are linearly independent. True or false?
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Do you know any facts about linearly independent sets? – Chris Eagle Dec 12 '12 at 22:27
Is there a subset of $\mathbb{R}^3$ with the property that every plane through the origin meets it in at most 2 points? – Hurkyl Nov 28 '13 at 13:54
Three vectors in $\mathbb{R}^3$ are linearly dependent if and only if they lie in a plane.
Consider the following process for building $S$. We can start with the empty set, and choose any two vectors $v_1, v_2 \in \mathbb{R}^3$ and add them to $S$. Then to choose a third vector $v_3$ to add to $S$, we must make sure it is not in the unique plane containing (i.e. spanned by) $v_1$ and $v_2$. Thus $v_3$ can be any vector in $\mathbb{R}^3 \backslash span(v_1, v_2)$.
Similarly, if at some stage $S = \{v_1, \ldots, v_k\}$, we can add to $S$ any vector $v_{k+1}$ in $\mathbb{R}^3 \backslash \bigcup_{x_i, x_j} span(x_i, x_j)$. Note that $\bigcup_{x_i, x_j} span(x_i, x_j)$ is a finite union of planes, so it can never be all of $\mathbb{R}^3$. In this way we can choose an infinite set with the desired property.
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The key issue with this approach is to prove that a finite union of planes is not all of $\mathbb R^3$. For a more general result: mathoverflow.net/questions/26/… – Andres Caicedo Dec 12 '12 at 22:47
Try $\left(\begin{matrix}1\\t\\t^2\end{matrix}\right)$ with $t\in\mathbb R$. Do you know to compute $$\left\vert\begin{matrix}1&1&1\\r&s&t\\r^2&s^2&t^2\end{matrix}\right\vert?$$
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good one!!. Please clear my doubt, consider the random gaussian vector $[x,y,z]$, where each entries are independent gaussian random variables. So shouldn't the set of all realizations of this random vector be a set of the kind he is looking for? – dineshdileep Dec 13 '12 at 5:44
the determinant is $(st^2-s^2t)-(rt^2-tr^2)+(rs^2-sr^2)$ – El Angel Exterminador Dec 14 '12 at 5:08
@Kuttus ... and that equals $(r-s)(s-t)(t-r)$ and is non-zero unless two of the numbers are equal. :) Actually, the fact that the functions $x\mapsto 1$, $x\mapsto x$ and $x\mapsto x^2$ are linearly independent lurks behind this as a shortcut: If a linear combinaton of the rows is the zero vector, then the correspondnig quadratic polynomial has three distinct roots $r,s,t$, hence is the zero polynomal, hence the linear combination was in fact the trivial combination, hence the rows are linearly independent, hence so are the columns (which is what we were actually after). – Hagen von Eitzen Dec 14 '12 at 11:39
+1 I'd just like to point out to future readers of this excellent answer that the curve $t\to (1,t,t^2)$ in $\mathbb{R}^3$ is also known as the twisted cubic. – Amitesh Datta Jul 27 '13 at 5:42
Consider vectors of the form $v_x=(1,x,x^2)^T$. Then for any distinct $x,y,z\in \mathbb R$ matrix $(v_x v_yv_z)$ is nonsingular (Vandermonde matrix), so $v_x$, $v_y$, $v_z$ are linearly independent.
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The parametric curve $t\mapsto(1,t,t^2)$ has the property that any three distinct points on it are linearly independent (without the "distinct" one clearly cannot have a solution). To check, just compute the determinant, which is Vandermonde.
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Inspired by the Vandermonde matrix example, let $f:\mathbb{R}\rightarrow\mathbb{R}$ be a strictly convex function. Then $$\left\vert\begin{matrix}1&1&1\\x&y&z\\f(x)&f(y)&f(z)\end{matrix}\right\vert\not=0$$ for any three distinct numbers $x,y,z$.
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The elegant example given by Adam can be generalized: let $P \subset \mathbb R^3$ be any plane that does not pass through the origin (equipped with the usual subspace topology), and let $C \subset P$ be a strictly convex* closed subset of $P$. Then the boundary $S = \partial_P\, C$ of $C$ in $P$ satisfies your requirement.
(* By "strictly convex", I mean that any non-trivial convex combination of two points in $C$ should belong to the interior of $C$, i.e. for all $x,y \in C$ and all $0 < \alpha < 1$, the point $z = \alpha x + (1-\alpha)y$ lies in the interior of $C$. Geometrically, this basically means that no part of the boundary should be a straight line segment.)
Proof: Let $x$ and $y$ be any two distinct points in $S$, and let $z$ linear combination of $x$ and $y$ which is distinct from both. By definition, $z = \alpha x + \beta y$ for some $\alpha$ and $\beta$. If $\alpha + \beta \ne 1$, then $z \notin P$, and thus $z \notin S$. Thus, $z$ can only belong to $P$ if $\beta = 1 - \alpha$. Let us assume below that this is the case.
If $0 < \alpha < 1$, by strict convexity as defined above, $z \in \operatorname{int} C$ and thus $z \notin S$. If $\alpha \in \{0,1\}$, then $z = x$ or $z = y$, contradicting the assumption of distinctness. The only remaining possibility is that $\alpha < 0$ or $\alpha > 1$, but if either of those holds, then either $x$ is a convex combination of $z$ and $y$, or $y$ is a convex combination of $z$ and $x$. In either case, $z$ cannot belong to $S \subset C$, since otherwise the fact that $x,y \in S = \partial\,C = C \setminus \operatorname{int} C$ would contradict the assumption that $C$ is strictly convex.
Besides the example given by Adam, one example possible concrete example of such a set would be e.g. the circle given by $P = \{(x,y,z) \in \mathbb R^3: z = 1\}$, $C = \{(x,y,z) \in P: x^2 + y^2 \le 1\}$ and $S = \partial_P\,C = \{(x,y,z) \in P: x^2 + y^2 = 1\}$.
The sets given by constructions like mine and Adam's are generally one-dimensional curves. Indeed, it's not hard to show that a set satisfying your requirements cannot contain any two-dimensional surface, since any surface would have to intersect some plane passing through the origin along a curve containing more than two points.
However, it is possible to construct a set which satisfies your requirements and is dense in $\mathbb R^3$. Basically, to do this, you'd start with an arbitrary enumeration $\langle a_i \rangle_{i \in \mathbb N}$ of the rational points $\mathbb Q^3$ (which are a countable dense subset of $\mathbb R^3$) and then, for each $i \in \mathbb N$, choose a point $b_i \in \mathbb R^3 \setminus \{0\}$ such that $|a_i - b_i| < 1/i$ and such that $b_i$ is pairwise linearly independent of all $b_j$, $j < i$. (This is always possible, since the set of pairwise linear combinations of $b_j$, $j < i$ is the intersection of finitely many planes, and thus has zero volume.) Then let $S = \{b_i: i \in \mathbb N\}$.
Note that the "construction" outlined above is not entirely constructive, since it requires an arbitrary choice of a point at each step of an infinite process. I do, however, believe that this dependency on choice could be eliminated by giving an explicit rule that assigns a definite value to $b_i$, given $a_i$ and $\{b_j: j < i\}$, similarly to the method I sketched in this answer to a related question.
(Do note the caveat brought up by Asaf Karagila in the comments to the other answer, though. I still think the construction I outlined shouldn't require even dependent choice, since an explicit formula for $b_i$ may be provided: basically, the idea is that, for each $i$, it should be possible to explicitly generate a finite ordered list of points $c_k$, $k \in \{1, \dotsc, n\}$ within radius $1/i$ of $a_i$, such that at least one of those points is guaranteed to be pairwise linearly independent of $\{b_j: j < i\}$, and then always choose the first such point. Still, I freely admit that I'm not nearly as experienced with this stuff as Asaf is, and that there could be some gap that I've missed.)
-
Existence : Consider three points $x_1,\ x_2,\ x_3$ in a plane $P\ :\ x+y+z=1$ which are not in a line.
Pick $x_4$ in $P-\bigcup L_{ij}$ where $L_{ij}$ is a line passing $x_i,\ x_j\ (1\leq i,\ j \leq 3)$.
Repeat this pocess infinitely.
- | 2015-01-26T19:16:39 | {
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https://math.stackexchange.com/questions/2312010/calculate-integral-using-residue-theorem | # Calculate Integral using residue theorem
I want to verify the following result using the residue theorem:
$$\int_0^\infty \frac{\log(x)}{x^2+a^2}dx = \frac{\pi}{2a}\log a, \, a > 0.$$
Here are my ideas:
At first I might want to show that this function is in fact a well defined improper Riemann integral, but I didn't come up with any nice solution yet.
I want to integrate the function $f(z) := \frac{\log(z)}{z^2+a^2}$ along the contour $C:= \gamma_1 \cup \gamma_2 \cup \gamma_3 \cup \gamma_4$ constisting of to semicircles with center at 0 around the upper half plane and radius $R$ and $\epsilon$ (resp.) as well as the intervals $[-R, -\epsilon]$ and $[\epsilon, R]$.
For the complex logarithm, use the branch $\log z = \log|z| + i\theta, \theta \in [-\pi/2, 3\pi/2)$ so we don't get any problems on the real line (is this choice correct?).
Note that I have one pole in my contour, namely $z = ia$. The Residue theorem yields $$\int_C f(z) dz= 2\pi i \text{ Res}(f, ia) = 2\pi i\lim_{z\to ia}(z-ia) \frac{\log z}{(z-ia)(z+ia)} = 2\pi i\frac{\log(ia)}{2ia} = \pi \frac{\log(a)+ i\pi/2}{a}$$ Proceeding, we choose the following parametrizations:
$$\gamma_1 : [-R,0] \to \mathbb C, \quad t\mapsto -\frac{\epsilon t}{R} + (t-\epsilon) \\ \gamma_2 : [0, R] \to \mathbb C, \quad t\mapsto -\frac{\epsilon t}{R} + (\epsilon+t) \\ \gamma_3: [0, \pi]\to \mathbb C, \quad t\mapsto \epsilon e^{it} \\ \gamma_4:[0,\pi] \to \mathbb C, \quad t \mapsto Re^{it}$$ Now here's what I am unsure about. To show that the big and the small circle vanish as $R\to \infty$ and $\epsilon \to 0$ is not too hard, but how to deal with the other integrals? Can I just say $$\int_{\gamma_1} f(z)dz = \int_0^R \frac{\log(-\frac{\epsilon}{R}t+\epsilon+t)}{(-\frac{\epsilon}{R}t + \epsilon + t)^2 + a^2}(\frac{\epsilon}{R}+1)dt \xrightarrow{\epsilon \to 0} \int_0^R \frac{\log t}{t^2 + a^2}dt$$ or what kind of reasoning should be used to interchange limit and integral?Also, I then finally I get $$\int_0^\infty \frac{\log(x)}{x^2+a^2}dx = \pi \frac{\log a + i\frac{\pi}{2}}{2a}$$ almost what I want, but where does $i\pi /2$ come from?
• Is the real-analysis tag appropriate here? Jun 6, 2017 at 13:22
• I was unsure since we talk about a real integral. I can remove it though. Jun 6, 2017 at 13:23
• Check your work on $\int_{\gamma_1}f(z)~\mathrm dz$. You should have a $\log(-t)$ in the numerator, which should cancel off the imaginary part you are confused about. Jun 6, 2017 at 13:26
Along the real axis, we have
\begin{align} \int_{-R}^R \frac{\log(x)}{x^2+a^2}\,dx&=\int_{-R}^0 \frac{\log(x)}{x^2+a^2}\,dx+\int_0^{R} \frac{\log(x)}{x^2+a^2}\,dx\\\\ &=\int_0^R \frac{\log(-x)+\log(x)}{x^2+a^2}\,dx\\\\ &=2\int_0^R \frac{\log(x)}{x^2+a^2}\,dx+i\pi\int_0^R\frac{1}{x^2+a^2}\,dx\\\\ &=2\int_0^R \frac{\log(x)}{x^2+a^2}\,dx+\frac{i\pi\arctan(R/a)}{a}\tag1 \end{align}
As $R\to \infty$, we find that
$$\int_{-\infty}^\infty \frac{\log(x)}{x^2+a^2}\,dx=2\int_0^\infty \frac{\log(x)}{x^2+a^2}\,dx+\frac{i\pi^2}{2a}\tag 2$$
Setting $(2)$ equal to $\frac{\pi \log(a)}{a}+i\frac{\pi^2}{2a}$, we find that
$$\int_0^\infty \frac{\log(x)}{x^2+a^2}\,dx=\frac{\pi \log(a)}{2a}$$
I thought it might be instructive to evaluate the integral of interest using real analysis only. To that end, we enforce the substitution $x\to a/x$ to find that
$$\int_0^\infty \frac{\log(x)}{x^2+a^2}\,dx=\frac1a \int_0^\infty \frac{\log(a)-\log(x)}{x^2+1}\,dx \tag2$$
For $a=1$, we see from $(2)$ that $\int_0^\infty \frac{\log(x)}{x^2+1}\,dx =0$. Thus, solving $(2)$ for integral of interest, and using $\int_0^\infty \frac{\log(x)}{x^2+1}\,dx =0$, we find that
$$\int_0^\infty \frac{\log(x)}{x^2+a^2}\,dx=\frac{\pi \log(a)}{2a}$$
as expected!
• That makes sense, I accidentally thought that the integral was symmetric. Could you possibly say anything about exchanging the limits and that the integral is well-defined? Jun 6, 2017 at 13:51
• The integral is well defined since $\log(x)$ is integrable on $[,1]$ (so, there is no problem at $0$) and since $\log(x)<\sqrt{x}$, $\frac{\log(x)}{x^2+a^2}=O\left(\frac{1}{x^{3/2}}\right)$ as $x\to \infty$. Jun 6, 2017 at 13:59
• Thank you very much! Does that also justify that I can just exchange Integral and limit or do I need to apply some theorem like dominated convergence? I wouldn't really want to find an integrable majorant... Jun 6, 2017 at 14:10
• Change the parameterization to $t$ so that the integral becomes $\int_{\epsilon}^R \frac{\log(t)}{t^2+a^2}\,dt$. Then, there is no need to interchange the order of integration and limits. Jun 6, 2017 at 14:25
• You're welcome. My pleasure. If one insists on using the parameterization, then noting that the integrand as a function of $\epsilon$ and $t$ is continuous on $[0,R]$ for $\epsilon>0$. Therefore since it is continuous on the closed bounded interval, it is uniformly continuous there. This is enough to justify the interchange of the limit on $\epsilon$ and the integral on $t$. Jun 6, 2017 at 14:33
For any $a>0$, through the substitution $x=a e^t$ we have
$$I(a) = \int_{0}^{+\infty}\frac{\log x}{a^2+x^2}\,dx = \frac{1}{2a}\int_{-\infty}^{+\infty}\frac{\log a+ t}{\cosh t}\,dt \tag{1}$$ and $\frac{t}{\cosh t}$ is an odd integrable function over $\mathbb{R}$. It follows that $$I(a) = \frac{\log a}{2a}\int_{-\infty}^{+\infty}\frac{dt}{\cosh t}=\frac{\log a}{2a}\left[2\arctan\tanh\frac{t}{2}\right]^{+\infty}_{-\infty}=\color{red}{\frac{\pi\log a}{2a}} \tag{2}$$ without even resorting to the residue theorem, but just exploiting symmetry.
• It's even easier to exploit symmetry through the substitution $x\to a/x$. ;-)) Jun 6, 2017 at 17:17
• @MarkViola: of course, this is almost the same. Jun 6, 2017 at 17:17
• Yes, it is the same upon letting $x=a\sinh(t)$ in the integral $\int_0^\infty \frac{1}{x^2+a^2}\,dx=\frac{\pi}{2a}$ Jun 6, 2017 at 17:25
• How did $\int_{0}^{\infty}$ change to $\int_{-\infty}^{\infty}$ m Jun 13, 2017 at 8:09
• I think mockingbird meant how did $\int_{0}^{\infty}$ become $\int_{-\infty}^{\infty}$? Jun 13, 2017 at 8:15
Using a semi-circular contour in the upper half plane that rests on the real axis we obtain
$$\int_0^\infty \frac{1}{x^2+a^2} \; dx = \frac{1}{2} \int_{-\infty}^\infty \frac{1}{x^2+a^2} \; dx \\ = \frac{1}{2} \times 2\pi i \frac{1}{2\times ai} = \frac{\pi}{2a}.$$
Next using a keyhole contour with the slot on the positive real axis and the branch of the logarithm where $0\le \arg\log z\lt 2\pi$ (branch cut on the positive real axis) we obtain integrating
$$f(z) = \frac{\log^2 z}{z^2+a^2}$$
the integrals
$$\int_0^\infty \frac{\log^2 x}{x^2+a^2} \; dx + \int_\infty^0 \frac{\log^2 x + 4\pi i \log x - 4\pi^2 }{x^2+a^2} \; dx \\ = 2\pi i \left(\frac{\log^2(ai)}{2\times ai} + \frac{\log^2(-ai)}{2\times -ai}\right).$$
This yields
$$-4\pi i \int_0^\infty \frac{\log x}{x^2+a^2} \; dx + 4\pi^2 \times \frac{\pi}{2a} = \frac{\pi}{a} ((\log a + \pi i/ 2)^2 - (\log a + 3\pi i/2)^2) \\ = \frac{\pi}{a} (\log a \times \pi i (1-3) + \pi^2 (-1/4 + 9/4)).$$
We thus obtain
$$4\pi i \int_0^\infty \frac{\log x}{x^2+a^2} \; dx = 4\pi^2 \times \frac{\pi}{2a} - \frac{\pi}{a} (-\log a \times 2\pi i + 2 \pi^2) \\ = \frac{\pi}{a} \log a \times 2\pi i.$$
Dividing by $4\pi i$ we finally obtain
$$\bbox[5px,border:2px solid #00A000]{ \frac{\pi}{2a}\log a.}$$
The bounds on the circular components that we used here were
$$2\pi R \times \frac{\log^2 R}{R^2} \rightarrow 0 \quad\text{and}\quad 2\pi \epsilon \times \frac{\log^2 \epsilon}{a^2} \rightarrow 0.$$ | 2023-04-01T10:48:26 | {
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https://math.stackexchange.com/questions/60894/how-to-find-the-number-of-perfect-matchings-in-complete-graphs | # How to find the number of perfect matchings in complete graphs?
In wikipedia FKT algorithm is given for planar graphs. Not anything for complete graphs. I need to find the number of perfect matchings in complete graph of six vertices.
• Here is an alternate way to do this via recursion. Let $a_n$ denote the number of perfect matchings in $K_{2n}$. Then clearly, $a_1 = 1$. Now in $K_{2n}$ (where $n \geq 2$), pick any vertex $u$, you can match it with $(2n-1)$ vertices. After matching $u$, you are left with $2n-2$ vertices. Thus, $a_n = (2n-1)a_{n-1}$. So, $a_n = (2n-1)(2n-3)\dots(1)$. May 7, 2018 at 7:27
• This is also called double factorial: $a_n=(2n-1)!!$ Feb 13, 2020 at 22:02
It's just the number of ways of partitioning the six vertices into three sets of two vertices each, right? So that's 15; vertex 1 can go with any of the 5 others, then choose one of the 4 remaining, it can go with any of three others, then there are no more choices to make. $5\times3=15$.
• Side note. For a general complete graph on $2n$ vertices, this number comes out to be $\frac{(2n)!}{n! 2^{n}}$. Aug 31, 2011 at 11:19
• In other words, it's the product of the odd numbers up to the number of vertices. Sep 1, 2012 at 20:13
• Or even for $\#V = 2n$ we have $(2n - 1)!!$ perfect matchings. May 17, 2014 at 10:22
• @Srivatsan in one of the books I have, the solution is same, but it explains it as: "The number of perfect matchings in a complete graph of n vertices, where n is even, reduces to the problem of finding unordered partitions of vertex set of the type p(2n;2,2,2,...n times) = $\frac{(2n)!}{(2!)^nn!}$ ", Is p(2n;2,2,2,...n times) some series? Book does not elaborate much. Also notice $(2!)^n$ in the denominator, not just $2^n$ thought both are same. Just guessing if this is some series unknown to me, then 2! must be having some significance in series expansion
– Maha
Dec 11, 2014 at 19:15
• @awell, I think $p(m;a_1,\dots,a_r)$ just means number of partitions of a set of size $m$ into $r$ subsets, one of size $a_1$, ..., one of size $a_r$. The formula for that is $m!/((a_1!)\cdots(a_r)!)$. It comes from repeated application of the formula for the number of combinations of $m$ things taken $a$ at a time, $m!/(a!(m-a)!)$. Dec 11, 2014 at 22:00
If you just want to get the number of perfect matching then use the formula $$\dfrac{(2n)!}{2^n\cdot n!}$$ where $$2n =$$ number of vertices in the complete graph $$K_{2n}$$.
Detailed Explaination:- You must understand that we have to make $$n$$ different sets of two vertices each.
First take a vertex. Now we have $$(2n-1)$$ ways to select another vertex to make the pair.
Now to make another pair we take a vertex and now we have $$(2n-3)$$ ways to select another vertex. This is because we have already used $$2$$ vertices in first pair and one vertex is currently in use to make 2nd pair.
Similarly for 3rd pair we will have $$(2n-5)$$ ways .
When we are making $$n^{th}$$ pair we will have just one way.
Multiplying all we get $$(2n-1)(2n-3)\ldots.3.1$$
Now multiply and divide it by even terms as follows $$\frac{(2n)(2n-1)(2n-2)(2n-3)\ldots.3.2.1}{(2n)(2n-2)(2n-4)\ldots.4.2}$$
Now the numerator will become $$(2n)!$$ and take $$2$$ common from each term in denominator. You will get $$2^n.(n(n-1)(n-2)\ldots 1)$$. Hence the denominator will become $$2^n. n!$$.
Hence we get the formula as $$\frac{(2n)!}{2^n n!}$$. Hope this will help.
Gerry is absolutely correct. For 6 vertices in complete graph, we have 15 perfect matching. Similarly if we have 8 vertices then 105 perfect matching exist (7*5*3).
1. For a perfect matching the number of vertices in the complete graph must be even.
2. For a complete graph with n vertices (where n is even), no of perfect matchings is $$\frac{n!}{(2!)^{n/2}(n/2)!}$$ For $$n=2m$$, it is $$\frac{(2m)!}{(2!)^m m!}$$
Using this formula for n=6, the number of partitions = 15
Explanation:
This problem is basically the problem of finding the number of "unordered" partitions of a set.
For partitioning a set of n elements into r unordered partitions of size $$n_1, n_2, n_3, ... , n_r$$, the formula is $$\frac{n!}{n_1!n_2!...n_r!r!}$$
For the given question we are to partition a set of size $$2m$$ into $$m$$ unordered partitions of size 2 each ie $$\quad n=2m; \quad n_1=n_2=...=n_r=2; \quad r=m$$
So the result is: $$\frac{(2m)!}{2!2!...(m-times)m!} = \frac{(2m)!}{(2!)^m m!}$$
For further on this topic: https://www3.nd.edu/~dgalvin1/10120/10120_S16/Topic07_6p7_Galvin.pdf
• Use LaTeX please. Jun 14, 2019 at 14:36
• @MichaelRozenberg Done in LaTex now. Jun 14, 2019 at 14:44
This problem is equivalent to finding set of $$n$$ disjoint pairs. Lets say there are $$2n$$ vertices then $$\frac{2n}{2^n}$$ is the number of ways you can have these $$n$$ pairs. i.e, ways to make disjoint pairs out of these. E.g, If you had vertices $$1,2,3,4$$ then one of the possible way to make set out of this would be $$\{(1,2),(3,4)\}$$. Finally, as set is an unordered collection so final solution becomes $$\frac{2n}{2^n*n!}$$.
Another way of arriving at the formula $$(2n-1)!! \left( = \frac{(2n)!}{2^n (n!)} \right)$$ for $$2n$$ vertices is by creating cycles from matchings and matchings from cycles.
If $$\mathcal{C}$$ is the set of cycles, then $$|\mathcal{C}| = \frac{(2n-1)!}{2}$$ and every cycle gives rise to exactly $$2$$ matchings by deleting every other edge.
For every matching of size $$n$$ we can obtain $$\frac{(n-1)!}{2} \cdot 2^n$$ cycles. First we arrange the $$n$$ edges from the matching into a cycle, as if they were vertices. There are $$\frac{(n-1)!}{2}$$ ways of arranging them in such a manner. If we now re-expand them into edges then we have $$2$$ choices for every edge on how we want to orient it when connecting, giving $$2^n$$ options per arrangement of a matching.
With $$\mathcal{M}$$ the set of the matchings we have $$2 |\mathcal{C}| = \frac{(n-1)!}{2} \cdot 2^n |\mathcal{M}| \\ \implies |\mathcal{M}| = \frac{(2n-1)!}{(n-1)!}\cdot\frac{2}{2^n}\cdot\frac{n}{n} = \frac{(2n)!}{2^nn!}$$
Gerry was correct (sort of) in his first statement, saying that it is the number of ways to partition the six vertices into three sets of two. However, the answer of number of perfect matching is not 15, it is 5. In fact, for any even complete graph G, G can be decomposed into n-1 perfect matchings. Try it for n=2,4,6 and you will see the pattern.
Also, you can think of it this way: the number of edges in a complete graph is [(n)(n-1)]/2, and the number of edges per matching is n/2. What do you have left for the number of matchings? n-1.
• I think these are all perfect matchings for 6: 12-34-56, 12-35-46, 12-36-45, 13-24-56, 13-25-46, 13-26-45, 14-23-56, 14-25-36, 14-26-35, 15-23-46, 15-24-36, 15-26-34, 16-23-45, 16-24-35, 16-25-34. That's 15, not 5. Dec 13, 2014 at 5:14 | 2022-05-20T10:33:50 | {
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https://learn.thomasjfrank.com/notion-formula-reference/formula-components/operators/mod | mod
Learn how to use the mod (%) operator to obtain a remainder in Notion formulas.
The remainder (%) operator allows you to get the remainder after dividing the first operand with the second operand.
1
number % number
2
mod(number, number)
Notion provides a mod() function as well as % and modoperators (I'll just reference % for the rest of this article).
Somewhat confusingly, these do not return a true modulus value; they return a remainder (see: Remainder or Modulus? below).
As % and mod() output a remainder, their output values will take the sign (+/-) of the dividend.
For reference, the dividend is the number being divided by the divisor, which produces the quotient:
$\frac {dividend}{divisor} = quotient$
# Example Formulas
1
19 % 12 // Output: 7
2
3
19 mod 12 // Output: 7
4
5
mod(-19,12) // Output: -7
## Negative Divisors
If the divisor is negative:
• mod() will treat it as a negative integer natively.
• if you're using the % operator, you'll need to wrap your divisor in parentheses () in order to explicitly define your divisor as a negative integer.
• x % (-y) will work exactly like mod().
• x % -y will cause Notion to rewrite your formula as x / 100 - y, which will output an incorrect result. This is because the - is treated as a unaryMinus operator, and Notion's math engine can't correctly deal with it when it's appended to the divisor.
1
19 % (-12) // Output: 7
2
3
// Negative value passed via a property does not need to
4
// be wrapped in () symbols
5
6
prop("negative num") == -12
7
19 % prop("negative num") // Output: 7
8
9
19 % -12 // Rewritten as 19 / 100 - 12, outputs -11.81
10
11
mod(19,-12) // Output: 7
Note: The above rules only apply if you're hard-coding a negative divisor in a formula. If you pass one via a property, it'll be parsed as a true negative value. It won't add a unaryMinus operator to your formula.
## Remainder or Modulus?
Just as in JavaScript, Notion’s % operator calculates the remainder of two numbers, not the modulus. Confusingly, the mod() function does this as well, despite its name.
The remainder and modulus of two numbers will be identical when both the dividend and divisor have the same sign (+/-). If their signs differ, however, the modulus will differ from the remainder.
Mod and Remainder are not the Same
Big Machine
Remainder operator vs. modulo operator (with JavaScript code)
You can prove this using WolframAlpha:
• -19 mod 12 results in 5
• QuotientRemainder[-19,12] results in -7
To calculate a true modulus in Notion, use ((x % y) + y) % y instead:
1
// Using the % operator
2
((19 % 12) + 12) % 12
3
4
// Using the mod() formula
5
mod(mod(19, 12) + 12, 12)
6
7
// % operator example using hard-coded negative divisors
8
((19 % (-12)) + (-12)) % (-12)
As noted above in the Negative Divisors section, hard-coded negative divisors in % expressions need to be wrapped in parentheses () so Notion can parse them explicitly as negative integers (see code line 7 in the above code block).
Otherwise, Notion will interpret the - as a unaryMinus operator, rewrite your formula to (x / 100 - y + -y) / 100 - y, and return an incorrect result. However, this isn't necessary when using the mod() function.
Remainder vs. Modulo programming functions
# Example Database
This example database shows the differing outputs of remainder and true modulus expressions.
## View and Duplicate Database
mod
College Info Geek on Notion
You can check these results here:
Modulo Calculator
CalculatorSoup
## "Remainder" Property Formula
1
prop("Dividend") % prop("Divisor")
## "Modulus" Property Formula
1
(prop("Dividend") % prop("Divisor") + prop("Divisor")) % prop("Divisor")
Instead of using hard-coded numbers, I’ve called in each property using the prop() function. | 2022-10-05T00:14:01 | {
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http://mathhelpforum.com/advanced-statistics/27016-dice-probability-question.html | # Math Help - dice probability question
1. ## dice probability question
This is a revision question for a phase test that i could sure use some help with!
A dice game consitsts of throwing 2 six sided dice and noting the score on them. If the score is greater than 9 or at least one dice shows a 6 then you score 3 points, otherwise you lose 1 point. Calculate the probability of gaining 3 points on a single throw. What is the average (or expected) number of points you will gain per throw.
i make the probability of scoreing 3 points as 12 over 36, but i dont know how to do the rest.
any suggestions?
Cheers
2. Originally Posted by marciano1976
This is a revision question for a phase test that i could sure use some help with!
A dice game consitsts of throwing 2 six sided dice and noting the score on them. If the score is greater than 9 or at least one dice shows a 6 then you score 3 points, otherwise you lose 1 point. Calculate the probability of gaining 3 points on a single throw. What is the average (or expected) number of points you will gain per throw.
i make the probability of scoreing 3 points as 12 over 36, but i dont know how to do the rest.
any suggestions?
Cheers
Colour the die red and blue, then the combinations that win are (r6,b4), (r6,b5), (r6,b6), (r5,b5), (r5,b6), (r4,b6), (r6,b1), (r6,b2), (r6,b3), (r1,b6), (r2,b6), (r3,b6). That is 12 winning configurations out of a total of 36.
p(3)=12/36=1/3
p(1)=1-p(3)=2/3
So the expected number of points is:
(-1)p(1)+(3)p(3)=(-2/3)+(3/3)=1/3
RonL
3. Originally Posted by CaptainBlack
So the expected number of points is:
(1)p(1)+(3)p(3)=(2/3)+(3/3)=5/3
Surely that should be (-1)p(1)+(3)p(3)=(-2/3)+(3/3)=1/3
4. Hello, marciano1976!
A dice game consists of throwing two six-sided dice and noting the score on them.
If the score is greater than 9 or at least one dice shows a 6 then you score 3 points,
otherwise you lose 1 point.
(a)Calculate the probability of gaining 3 points on a single throw.
(b) What is the average (or expected) number of points you will gain per throw?
There are 36 possible outcomes . . .
$\begin{array}{cccccc}(1,1) & (1,2) & (1,3) & (1,4) & (1,5) & {\color{blue}(1,6)} \\
(2,1) & (2,2) & (2,3) & (2,4) & (2,5) & {\color{blue}(2,6)} \\ (3,1) & (3,2) & (3,3) & (3,4) & (3,5) & {\color{blue}(3,6)} \\
(4,1) & (4,2) & (4,3) & (4,4) & (4,5) & {\color{blue}(4,6)}
\\ (5,1) & (5,2) & (5,3) & (5,4) & {\color{blue}(5,5)} & {\color{blue}(5,6)}
\end{array}$
$\begin{array}{cccccc}{\color{blue}(6,1)} & {\color{blue}(6,2)} & {\color{blue}(6,3)} & {\color{blue}(6,4)} & {\color{blue}(6,5)} & {\color{blue}(6,6)} \end{array}$
There are 12 in which the sum is greater than 9 or there is at least one 6.
Therefore: . $(a)\;\;P(\text{gain 3 points}) \:=\:\frac{12}{36} \:=\:\frac{1}{3}$
Hence: . $\begin{array}{ccc}P(\text{win 3 points}) &=& \frac{1}{3} \\ P(\text{lose 1 point}) & = & \frac{2}{3} \end{array}$
And: . $(b)\;\;\text{Average} \;=\;(+3)\left(\frac{1}{3}\right) + (-1)\left(\frac{2}{3}\right) \:=\:\frac{1}{3}$
You can expect to win an average of $\frac{1}{3}$ of a point per throw.
. . . as Plato already pointed out.
.
5. Originally Posted by Plato
Surely that should be (-1)p(1)+(3)p(3)=(-2/3)+(3/3)=1/3
Of course it should, I misread the question (again)
RonL | 2014-07-28T05:07:11 | {
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# 6^(1/2)/(1/2^(1/2)+1/3^(1/2))
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$$\frac{\sqrt{6}}{(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}})}=$$
A. $$\frac{6}{5}$$
B. $$\frac{5}{6}$$
C. 5
D. $$2\sqrt{3} +3\sqrt{2}$$
E. $$6\sqrt{3}-6\sqrt{2}$$
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The trick behind this question is to use the answer choices to guide your mathematics. (In other words, don't do math just because you can... do math because it gets you closer to your target!)
You should notice immediately that the answer choices are much simpler. Three of the five answer choices contain no fractions at all. None of the answer choices contain $$\sqrt{6}$$. Use this to your advantage to help you think about the question. One quick way to get rid of both the \sqrt{6}[/m] and the "fractions inside of fractions" issue is to do what I call "Multiply by 1". In this case, multiply the original expression by $$\sqrt{6}/\sqrt{6}$$. Since $$\sqrt{6}/\sqrt{6}=1$$, we don't actually change the value; we just change the shape of the expression. Thus,
$$\frac{\sqrt{6}}{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}}*(\frac{\sqrt{6}}{\sqrt{6}})=\frac{6}{\sqrt{3}+\sqrt{2}}$$
Now the expression that remains is starting to look more like the answer choices, but we still need to get rid of the compound denominator $$(\sqrt{3}+\sqrt{2})$$. Once again, look to the answer choices for clues. Both answer choices A and E contain a $$6$$. But it should be clear that answer choice A cannot be an option. $$(\sqrt{3}+\sqrt{2})$$ does not equal 5. So, let's try to turn what we have into answer choice E.
Answer choice E contains a factor of $$(\sqrt{3}-\sqrt{2})$$, which should immediately get us thinking about the possibility of difference of squares. Multiplying our reduced expression by "1" again gives us:
$$(\frac{6}{\sqrt{3}+\sqrt{2}})*(\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}})$$
Using the concept of difference of squares, we can quickly see that the denominator of this fraction simplifies down completely. Watch this "disappearing math" trick:
$$(\sqrt{3}+\sqrt{2})*(\sqrt{3}-\sqrt{2}) = (\sqrt{3})^2 - (\sqrt{2})^2 = 3-2 = 1$$
What began as a "Mathugly" arithmetic issue quickly simplifies down to $$6\sqrt{3}-6\sqrt{2}$$.
The answer is E.
Addendum: As with many quantitative questions on the GMAT, there is "more than one way to skin a CAT" for this question (pun intended.) We can also use the answer choices not just as targets that guide our math, but as leverage for approximating. Watch this. If we recognize that $$\sqrt{3}\approx{1.7}$$ and $$\sqrt{2}\approx{1.4}$$, then the we can approximate our answer very quickly:
$$\frac{6}{\sqrt{3}+\sqrt{2}}\approx{\frac{6}{1.4+1.7}}\approx{\frac{6}{3.1}}\approx{2}$$
Now, we look at the answer choices to see which one is roughly equal to 2.
(A) $$\frac{6}{5}$$ <-- quickly eliminated. Too small.
(B) $$\frac{5}{6}$$ <-- quickly eliminated. Too small.
(C) $$5$$ <-- quickly eliminated. Too big.
(D) $$2\sqrt{3} + 3\sqrt{2}$$ <-- quickly eliminated. Too big.
(E) $$6\sqrt{3} - 6\sqrt{2}$$ <-- The right answer. $$6(\sqrt{3} - \sqrt{2})\approx{6(1.7-1.4)}\approx{2}$$
Anyway you look at it, the answer is still E.
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Originally posted by AaronPond on 08 Jan 2018, 17:04.
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AbdurRakib wrote:
$$\frac{\sqrt{6}}{(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}})}=$$
A. $$\frac{6}{5}$$
B. $$\frac{5}{6}$$
C. 5
D. $$2\sqrt{3} +3\sqrt{2}$$
E. $$6\sqrt{3}-6\sqrt{2}$$
$$\frac{\sqrt{6}}{(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}})}=\frac{\sqrt{6}}{(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{2}\sqrt{3}})}=\frac{\sqrt{6}}{(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{6}})}=\frac{\sqrt{6}*\sqrt{6}}{\sqrt{3}+\sqrt{2}}=\frac{6}{\sqrt{3}+\sqrt{2}}$$.
Multiply by $$\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}$$:
$$\frac{6}{\sqrt{3}+\sqrt{2}}*\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}-\sqrt{2}}=\frac{6*(\sqrt{3}-\sqrt{2})}{3-2}=6*\sqrt{3}-6\sqrt{2}$$.
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AbdurRakib wrote:
$$\frac{\sqrt{6}}{(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}})}=$$
A. $$\frac{6}{5}$$
B. $$\frac{5}{6}$$
C. 5
D. $$2\sqrt{3} +3\sqrt{2}$$
E. $$6\sqrt{3}-6\sqrt{2}$$
Nice question!
Here's another approach:
Let's first deal with the denominator: 1/√2 + 1/√3 = √3/√6 + √2/√6
= (√3 + √2)/√6
So....$$\frac{\sqrt{6}}{(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}})}=$$ √6/[(√3 + √2)/√6]
= (√6)(√6)/(√3 + √2)
= 6/(√3 + √2)
Now we can use the fact that √3 ≈ 1.7 and √2 ≈ 1.4 to get....
≈ 6/(1.7 + 1.4)
≈ 6/(3.1)
Since 6/3 = 2, we know that 6/(3.1) = a number a bit smaller than 2
Now check the answer choices to see which one evaluates to be a number a bit smaller than 2
A. $$\frac{6}{5}$$ ELIMINATE
B. $$\frac{5}{6}$$ ELIMINATE
C. 5 ELIMINATE
D. $$2\sqrt{3} +3\sqrt{2}$$ ≈ 2(1.7) + 3(1.4) ≈ a number that's bigger than 6 ELIMINATE
E. $$6\sqrt{3}-6\sqrt{2}$$[/quote] = 6(√3 - √2) ≈ 6(1.7 - 1.4) ≈ 6(0.3) ≈ 1.8 PERFECT!
Cheers,
Brent
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08 Jan 2018, 16:36
$$\frac{\sqrt{6}}{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}}=$$
(A) $$\frac{6}{5}$$
(B) $$\frac{5}{6}$$
(C) $$5$$
(D) $$2\sqrt{3} + 3\sqrt{2}$$
(E) $$6\sqrt{3} - 6\sqrt{2}$$
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08 Jan 2018, 20:33
AaronPond wrote:
$$\frac{\sqrt{6}}{\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}}=$$
(A) $$\frac{6}{5}$$
(B) $$\frac{5}{6}$$
(C) $$5$$
(D) $$2\sqrt{3} + 3\sqrt{2}$$
(E) $$6\sqrt{3} - 6\sqrt{2}$$
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Re: 6^(1/2)/(1/2^(1/2)+1/3^(1/2)) &nbs [#permalink] 08 Jan 2018, 20:33
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# 6^(1/2)/(1/2^(1/2)+1/3^(1/2))
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Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 2018-09-21T10:27:20 | {
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https://math.stackexchange.com/questions/1463384/help-with-linear-algebra-linear-transformations | # Help with linear algebra linear transformations
Is there a linear transformation $T : R ^2 → R ^3$ such that $T(1, 1) = (1, 0, 2)$ and $T(2, 3) = (1, −1, 4)$. Justify your answer.
I'm not sure what exactly this question is asking for. How would you go about tackling this question. I know I'm supposed to show some form of working on this website but I really have no idea where to start
• Try to start with the assumption that there is such a transformation. Then try to explicitly construct it by finding a matrix $A$ such that $T(x) = Ax$ for all $x\in \Bbb R^2$. If you're able to, then one clearly exists. If you run into some contradiction, then one clearly doesn't exist. – user137731 Oct 4 '15 at 4:25
• As always, reviewing the terms in the problem (what a linear transformation is) can help. – Nate 8 Oct 4 '15 at 4:30
In general if V and W are vector spaces and $\{e_i\}_{i\in I}$ is a basis for $V$. And $f(e_i)=w_i\in W$ for all $i \in I$.
Then f can be extended to a linear mapping $T: V \rightarrow W$ such that $T(e_i)=f(e_i)$. (You can write down the proof in the same way as the above answers)
In your case, $V= \mathbb R ^2$ and $W=\mathbb R^3$ $e_1=(1,1), e_2=(2,3)$ is a basis for $\mathbb R^2$ and $f(e_1)=(1,0,2)$ and $f(e_2)=(1,-1,4)$.
Therefore, you can extend it to a linear mapping by the above discussion.
Note: 1) $(1,0,2)$ and $(1,-1,4)$ doesn't have any significance role to play.If $(a,b,c), (d,e,f)$ still you could extend it to a linear map.
2) The above problem has a general philosophy - any map (or morphism) from a algebraic(mathematical) structure to another algebraic(mathematical) structure can be defined by only defining it on the generators, provided you can talk of generators. For example, to define a group homomorphism from a cyclic to group to another group, it is enough to define it for just for the generators.
Note that $\{(1,1),(2,3)\}$ is a basis of $\mathbb R^2$ and if you know the image of basis elements then you know the whole transformation.
In this case observe $\forall x,y \in \mathbb R^2$ we have $(x,y)= (3x-2y)(1,1)+(y-x)(2,3)$
So define , $T(x,y)=(3x-2y)T(1,1)+(y-x)T(2,3)=(3x-2y)(1,0,2)+(y-x)(1,-1,4)=(2x-2y,x-y,2x)$
To answer the comment by Babai:
The construction of $T$ shows that if $u\in \mathbb R^2,u=\alpha a+\beta b$ where $a=(1,1)$ and $b=(2,3)$ then $Tu=\alpha Ta+\beta Tb$
Thus $T(u_1+\lambda u_2)=T(\alpha_1 a+\beta_2 b+\lambda (\alpha_2 a+\beta_2 b))=T((\alpha_1+\lambda \alpha_2) a+(\beta_1+\lambda \beta_2) b)=(\alpha_1+\lambda \alpha_2) Ta+(\beta_1+\lambda \beta_2) Tb=\alpha_1 Ta+\beta_2 Tb+\lambda (\alpha_2 Ta+\beta_2 Tb)=Tu_1+\lambda Tu_2$
• The question is why is T linear? In your answer you assumed T is linear. – Babai Oct 4 '15 at 4:41
• @Babai I have added it in my answer. – usermath Oct 4 '15 at 5:22
• $T(x,y)=(3x-2y)T(1,1)+(y-x)T(2,3)=(3x-2y)(1,0,2)+(y-x)(1,-1,4)=(2x-2y,x-y,2x)$. Why is this true? This is only true if $T$ is linear, and that is what you nee to prove. – Babai Oct 4 '15 at 5:38
• Instead of writing " So $T(x,y)=(3x-2y)T(1,1)+(y-x)T(2,3)=(3x-2y)(1,0,2)+(y-x)(1,-1,4)=(2x-2y,x-y,2x)$, if you write "define $T$ as $T(x,y)=(3x-2y)T(1,1)+(y-x)T(2,3)=(3x-2y)(1,0,2)+(y-x)(1,-1,4)=(2x-2y,x-y,2x)$, then your answer will make perfect sense. – Babai Oct 4 '15 at 5:45
• @Babai ok I got it and corrected. Thnx – usermath Oct 4 '15 at 6:27
Define $T: \mathbb{R}^2 \to \mathbb{R}^3$ by for $(x,y) \in \mathbb{R}^2$, write $(x,y) = a(1,1) + b(2,3)$, and let $T(x,y) = a(1,0,2) + b(1,-1,4).$
In the above, $T$ is obviously linear. The problem is why such a $T$ is well-defined, more specifically, why there exists such scalars $a,b$ and the scalars are unique.
• Why is T obviously linear? – Babai Oct 4 '15 at 4:42
• @Babai Have you tried to show it before you ask this question? It is a one line proof. – Empiricist Oct 4 '15 at 5:17
• It is not the question of whether I have tired it or not. Your answer is not justifiable. You write "$T(x,y) = a(1,0,2) + b(1,-1,4).$, In the above, T is obviously linear." Now, $T(x,y) = a(1,0,2) + b(1,-1,4).$ , this is not true unless $T$ is linear. And you assumed it to say $T$ is linear, which is not correct. – Babai Oct 4 '15 at 5:35
• I am sorry. You are defining $T$. Your answer is perfect. And it is obviously linear. – Babai Oct 4 '15 at 5:44
If there exists a transform, there must exist a matrix representing it, let us call such a matrix $\bf T$. Now what we demand is that $${\bf T}\left[\begin{array}{cc}1&2\\1&3\end{array}\right] = \left[\begin{array}{rr}1&1\\0&-1\\2&4\end{array}\right]$$ Now we can solve this, if we divide by $\left[\begin{array}{cc}1&2\\1&3\end{array}\right]$ to the left of each side: $${\bf T} = \left[\begin{array}{rr}1&1\\0&-1\\2&4\end{array}\right] \left[\begin{array}{cc}1&2\\1&3\end{array}\right]^{-1} = ?$$
Now we can see on the matrices that this must have a solution since both matrices are of rank 2.
Then after we have solved this, what remains is to perform the matrix multiplication ${\bf T}\left[\begin{array}{cc}1&2\\1&3\end{array}\right]$ and confirm it really becomes as the first equation says it should. | 2020-01-28T17:12:40 | {
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http://grandespirito.it/page.php?tag=Backslash-in-math-20e214 | This list of mathematical symbols by subject shows a selection of the most common symbols that are used in modern mathematical notation within formulas, grouped by mathematical topic.
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The difference between a backslash and a forward slash is defined below: Backslash: \ Forward Slash: / A good way to remember the difference between a backslash and a forward slash is that a backslash leans backwards ( \ ), while a forward slash leans forward ( / ).
P. PvtBillPilgrim. Infix operator. If you like this Site about Solving Math Problems, please let Google know by clicking the +1 button. ... [Backslash] Unicode: 2216.
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The backslash symbol \ is used to denote a set difference, quotient group, or integer division. Also used to separate arguments of elliptic functions. Ask Question Asked 5 years, 5 months ago. Formally: ∖ = {∈ ∣ ∉}.
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You just have to type the name of the letter after a backslash: if the first letter is lowercase, you will get a lowercase Greek letter, if the first letter is uppercase (and only the first letter), then you will get an uppercase letter. What does the backslash denote in probability theory? Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Applied Mathematics. Ask Question Asked 7 years, 6 months ago. \$\begingroup\$ The backslash is an escape character in \$\LaTeX\$, cf.
... Backslash notation: … is used to separate the whole part of a number from the fractional part. What is Backslash? In this particular example, the norm of the residual is zero, and an exact solution is obtained, although rcond is small. Making statements based on opinion; back them up with references or personal experience. Revolutionary knowledge-based programming language. x y is by default interpreted as Backslash [x, y].
In an application of a probability problem you could be given an arbitrary probability like: P ( a \ b) = .50 or 50%. For non-square and singular systems, the operation A\b gives the solution in the least squares sense. The relative complement of A in B is denoted B ∖ A according to the ISO 31-11 standard.It is sometimes written B − A, but this notation is ambiguous, as in some contexts it can be interpreted as the set of all elements b − a, where b is taken from B and a from A.. Foundations of Mathematics. | 2020-10-22T17:14:24 | {
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https://math.stackexchange.com/questions/377683/length-of-period-of-decimal-expansion-of-a-fraction | # Length of period of decimal expansion of a fraction
Each rational number (fraction) can be written as a decimal periodic number. Is there a method or hint to derive the length of the period of an arbitrary fraction? For example $1/3=0.3333...=0.(3)$ has a period of length 1.
For example: how to determine the length of a period of $119/13$?
• I can't find an algorithm that finds the value for a number that isn't prime, i.e. 10 – Code Whisperer Jan 13 '18 at 9:11
• – Watson Dec 2 '18 at 16:31
Assuming there are no factors of $2,5$ in the denominator, one way is just to raise $10$ to powers modulo the denominator. When you find $-1$ you are halfway done. Taking your example: $10^2\equiv 9, 10^3\equiv -1, 10^6 \equiv 1 \pmod {13}$ so the repeat of $\frac 1{13}$ is $6$ long. It will always be a factor of Euler's totient function of the denominator. For prime $p$, that is $p-1$.
• You have to look for $1$, because sometimes you won't find $-1$. For example: $10^2\equiv 18, 10^3\equiv 16, 10^4 \equiv 37, 10^5 \equiv 1 \pmod {41}$, so the repeat of $\frac 1{41}$ is $5$ long. – jcsahnwaldt says GoFundMonica Mar 26 '15 at 5:37
• You assume that 'there are no factors of 2,5 in the denominator' is it actually required? – Shamdor May 5 '15 at 7:28
• I mean, 1/(2^i * 5^j * d') has the same period length as 1/d', right? – Shamdor May 5 '15 at 7:46
• That is correct. You will have $\max(i,j)$ non-repeating digits, then the repeat will start – Ross Millikan May 5 '15 at 13:48
This answer seeks to explain why Ross Millikan's answer works, and provides further information on techniques to speed up the process of seeking the period:
Consider the fraction $\frac17$. The decimal expansion of this is $$\frac17 = 0.\overline{142857}$$ for a period of 6. Now consider what happens when we multiply it by $10^6$: $$10^6\times\frac17 = 142857.\overline{142857}$$ Subtracting the original fraction from this gives $$(10^6-1)\times\frac17 = 142857$$ And so, we have $$\frac17 = \frac{142857}{10^6-1}$$ As you can see, the denominator is one less than a power of 10, and the power is the period of the decimal expansion. This is no accident, and works for any fraction - if you can rewrite it in this form, the denominator reveals the period.
Now, rearrange the equation: $$10^6-1 = 142857\times 7$$ So $10^6$ must be one more than a multiple of 7 (or, more generally, $10^n$ must be one more than a multiple of $d$, where $d$ is the denominator of the fraction and $n$ is the period of the decimal expansion) - indeed, it must be the smallest power of 10 (larger than 1) that has this property.
As such, we can use modular arithmetic to look for the period. Since $a\times d\equiv 0 \pmod d$, we have that $10^n-1\equiv0 \pmod d$, or $$10^n \equiv 1\pmod d$$ And therefore you can just look for the smallest $n>0$ satisfying this.
Of course, there are other approaches to gain the same result, but they're all fundamentally variants of the same idea. That said, if you can factor $\phi(d)$ - the euler totient function of the denominator - then you can accelerate the process of looking for the smallest $n$. For example, when checking 13, you have $\phi(13)=12$, so $n\in\{1,2,3,4,6,12\}$ (as these are the factors of 12) - this can save you a lot of computation (especially where $\phi(d)$ has only a few large factors and 2).
For example, $\phi(167)=166 = 2\times83$, so $n\in\{1,2,83,166\}$. Therefore, we need to check only these four, and we can do it quite efficiently. Obviously, neither $10$ nor $100=10^2$ are equivalent to 1 mod 167, so we only need to actually check 83. For this we can use binary exponentiation. Note that $83 = 2^6 + 2^4 + 2^1 + 2^0$. So we can write \begin{align} 10^{83} &= 10^{2\times(2^5 + 2^3 + 1)}\times 10\\ &= (10^{2^3\times(2^2+1)}\times 10)^2 \times 10\\ &= ((10^{2^2}\times10)^{2^3}\times 10)^2 \times 10 \end{align} So, working in modular arithmetic, we can go \begin{align} 10^{83} &\equiv ((10^{2^2}\times10)^{2^3}\times 10)^2 \times 10 \mod 167\\ &\equiv ((100^2\times 10)^{2^3}\times 10)^2\times10\mod167\\ &\equiv ((147\times 10)^{2^3}\times 10)^2\times10\mod167\\ &\equiv (134^{2^3}\times 10)^2\times10\mod167\\ &\equiv (87^{2^2}\times 10)^2\times10\mod167\\ &\equiv (54^2\times 10)^2\times10\mod167\\ &\equiv (77\times 10)^2\times10\mod167\\ &\equiv 102^2\times10\mod167\\ &\equiv 50\times10\mod167\\ &\equiv 166\mod167 \end{align} This is the same as $-1\pmod{167}$, so $n=166$ is the only possible period, and $\frac1{167}$ has a period of 166.
Also note that you don't actually have to expand out the product like that. You can just write the number in binary ($83_{10} = 1010011_2$), then work through the binary digits left-to-right - start with 1, and for each digit, $b$, multiply by $10^b$. Square it after each digit except the last one.
The length of the period is given by the multiplicative order of $10 \pmod q$, where $q$ is your quotient. It is closely related to the discrete logarithm. Wikipedia lists several algorithms that are faster than going through all the powers of ten, which is relevant if you're dealing with very large numbers (hundreds of digits).
First you have to simplify to the lowest integer denominator.
Then, the proof for finiteness of repeating part of the decimal representation involves the pigeonhole principle. When you keep on dividing, at some point you will run into a repeat value for the remainder. Given that the remainders that keep the operation going range from 1 to (denominator-1), the possible number of pigeonholes is (denominator-1) as the number of pigeonholes; so that's your worst possible case scenario - in any base.
The particular cases, as described by the other answers, depend on the base, because the denominator may have a special relationship with the base.
• e.g., in base 10 fractions over 3 will have one repeating decimal, (3) or (6), as 3 divides 9(=10-1), and the remainder will keep repeating. | 2020-02-21T07:38:24 | {
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https://math.stackexchange.com/questions/931742/proof-check-linear-algebra | # Proof Check - Linear Algebra
I'm new to more formal mathematics and I was wondering if I could write my attempt at a small proof here and see what others thought and if you could point out any fatal flaws or just offer some tips? I would appreciate it.
## Theorem
Let u,v, and w be distinct vectors in a vector space V. Show that if {u,v,w} is a basis for V, then {u+v+w, v+w, w} is also a basis for V.
## Proof
: Let B = {u,v,w}. As B is a basis for V, the vectors in B generate V and are linearly independant. Therefore, the equation
a1 * u + a2 * v + a3 * w = 0
Is only true for a1, a2, a3 all equal to 0.
Now consider the set B' = {u+v+w, v + w, w}
If we examine the equation
c1 ( u+v+w) + c2 (v+w) + c3 (w) = 0
c1(u) + (c1 + c2)v + (c1 + c2 + c3)w = 0
Then we say that c1 = -c2, and c3 = -c1-c2, but from the first term we see that c1 must equal zero, therefore all the other coefficients are 0 as well, and so the set of vectors is linearly independant.
## My questions
1.) I'm not sure if the part of the proof where I show that the solution to the equation for the vectors of B' being equal to 0 is connected well enough to the bit about B being a basis for V. Am I allowed to say that C1 must be equal to 0 only because I've already assumed that u,v, and w are all linearly independent already?
2.) I know that if B generates V that B' which is just set formed from L.I. vectors into other L.I. vectors must also generate V, but I'm not sure exactly why I can say this even though I feel it strongly.
• Your argument is perfect. – Ted Shifrin Sep 15 '14 at 1:20
With regards to point 1), what you say is correct. As you say, you use the linear independence of $B$ to conclude that when $c_1(u) + (c_1 + c_2)v + (c_1 + c_2 + c_3)w = 0$, it must be the case that $c_{1} = 0$, and so on.
With regards to 2), one way to show that $B'$ spans the space is to show that it generates $B$: any linear combination of vectors from $B$ can then be converted into a linear combination of vectors from $B'$ by replacing $u,v,w$ by their expressions in terms of the list $B'$. So in this case $(v+w) - w = v$ and $(u+v+w) - (v+w) = u$.
By the way, you only need to show either that $B'$ form a linear independent set, or that they span the vector space, since the list has length $3$. (This is the result that a linear independent list of vectors, whose length is equal to the dimension of the space forms a basis, and similarly for spanning sets). | 2019-10-20T18:39:46 | {
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https://math.stackexchange.com/questions/108010/when-is-the-closure-of-an-open-ball-equal-to-the-closed-ball | # When is the closure of an open ball equal to the closed ball?
It is not necessarily true that the closure of an open ball $B_{r}(x)$ is equal to the closed ball of the same radius $r$ centered at the same point $x$. For a quick example, take $X$ to be any set and define a metric $$d(x,y)= \begin{cases} 0\qquad&\text{if and only if x=y}\\ 1&\text{otherwise} \end{cases}$$ The open unit ball of radius $1$ around any point $x$ is the singleton set $\{x\}$. Its closure is also the singleton set. However, the closed unit ball of radius $1$ is everything.
I like this example (even though it is quite artificial) because it can show that this often-assumed falsehood can fail in catastrophic ways. My question is: are there necessary and sufficient conditions that can be placed on the metric space $(X,d)$ which would force the balls to be equal?
• In the Euclidean metric space $R^n$ it is necessarily true. Feb 11, 2012 at 2:26
• Right, but Euclidean space is known for, among other things, being perfect in almost every way. What about spaces like $L^{p}$ or $H^{p}$? I'm looking to see how far our intuition of Euclidean spaces and the standard metric extends. Feb 11, 2012 at 2:28
• Regarding your question about $L^p$ or $H^p$, it is true in every normed space. If $\|x-y\|=r$, then for $0<t<1$, $\|x-(tx+(1-t)y)\|=(1-t)\|x-y\|<r$, and $y=\lim\limits_{t\searrow 0}tx+(1-t)y$. Feb 11, 2012 at 3:37
• The property may fail for subspaces of Euclidean space. See here: link link Feb 11, 2012 at 11:53
Here is a characterization that is straight from the definitions, but which it seems may be useful when verifying that a particular space has the property.
For any metric space $$(X,d)$$, the following are equivalent:
• For any $$x\in X$$ and radius $$r$$, the closure of the open ball of radius $$r$$ around $$x$$ is the closed ball of radius $$r$$.
• For any two distinct points $$x,y$$ in the space and any positive $$\epsilon$$, there is a point $$z$$ within $$\epsilon$$ of $$y$$, and closer to $$x$$ than $$y$$ is. That is, for every $$x\neq y$$ and $$\epsilon\gt 0$$, there is $$z$$ with $$d(z,y)<\epsilon$$ and $$d(x,z).
Proof. If the closed ball property holds, then fix any $$x,y$$ with $$r=d(x,y)$$. Since the closure of $$B_r(x)$$ includes $$y$$, the second property follows. Conversely, if the second property holds, then if $$r=d(x,y)$$, then the property ensures that $$y$$ is in the closure of $$B_r(x)$$, and so the closure of the open ball includes the closed ball (and it is easy to see it does not include anything more than this, since if $$g$$ belongs to the closure of $$B_r(x)$$ then $$d(x,g) \le r$$ and so $$g$$ must also belong to the closed ball of radius $$r$$ centered at $$x$$). QED
• Thanks for the response! Your proof makes sense. Feb 11, 2012 at 3:10
• How to understand what is not working initially in order to build your characterization? In other words, what is the intuition behind this? Thanks.
– user169373
Sep 23, 2014 at 20:41
• @MarcGato Think about the case where you have an isolated point $x$, so that $B_r(x)$ contains only $x$, for some $r>0$, but there is a point $y$ at distance $r$ to $x$. The closure of $B_r(x)$ is just $\{x\}$, since there are no other limit points to add and so this set is already closed. My condition ensures that every point at distance $r$ from $x$ is a limit of points of distance less than $r$ from $x$. That is why all such points get added to the closure of the open ball.
– JDH
Sep 23, 2014 at 21:10
• The way that $y$ gets into the closure of $B_r(x)$ is that there must be points in $B_r(x)$ that are as close as you like to $y$. Those points are the $z$'s in the property.
– JDH
Apr 26, 2017 at 11:03
• @James Well since y is in closure of B(x,r), for any $\epsilon>0$, $B(y,\epsilon) \cap B(x,r) \neq \phi$. Choose z from this intersection. Then, $d(x,z)<r=d(x,y)$ Jan 14, 2021 at 11:40
Let $(X,\|\cdot\|)$ be a normed linear space. Then $\overline{B_1(0)}=\bar{B}_1(0)$.
Proof. Observe that $\overline{B_1(0)}$ is the smallest closed set containing $B_1(0)$ and $B_1(0)\subset \bar{B}_1(0)$, so trivially $\overline{B_1(0)}\subset\bar{B}_1(0)$. Now to show $\bar{B}_1(0)\subset \overline{B_1(0)}$. Observe that, $\bar{B}_1(0)=B_1(0)\cup \partial B_1(0)$, i.e., for all $x\in \partial B_1(0), \, \exists x_n\in B_1(0)$ such that $\|x_n-x\|\to 0$: for any given $x\in \partial B_1(0),$ let $x_n=(1-\frac{1}{n})x, \, n\in \mathbb{N}.$ Then show $x_n\in B_1(0)$ and $\|x_n-x\|\to 0$.
• How would you suggest approaching this, showing that the interior of the closed ball is equal to the open ball? Jan 31, 2018 at 0:15
• ... where a "normed linear space" means over some field with an archimedean valuation. The result is not true e.g. over $p$-adics where that $1-1/n$-construction fails. -- Also, note that $\delta B_1(0)$ here should really be defined as $\{x \in X: \lvert \lvert x \rvert \rvert =1 \}$, as only a posteriori we show this is the boundary also in the topological sense. Apr 11, 2020 at 7:14
More general, we can prove that for any normed space $$(V,\|\cdot\|)$$, if $$x_0\in V$$ and $$R>0$$, then $$\overline{B(x_0;R)}^{\|\cdot\|}=B[x_0;R],$$ where $$\overline{B(x_0;R)}^{\|\cdot\|}$$ is the closure of the open ball $$B(x_0;R)$$ in topology induced by $$\|\cdot\|$$.
The inclusion $$\overline{B(x_0;R)}^{\|\cdot\|}\subseteq B[x_0;R]$$ is trivial, because $$B[x_0;R]$$ is a closed set that contains $$B(x_0;R)$$ and $$\overline{B(x_0;R)}^{\|\cdot\|}$$ is the smallest closed set that contains $$B(x_0;R)$$.
To show that $$B[x_0;R]\subseteq\overline{B(x_0;R)}^{\|\cdot\|}$$, we can prove that any element of $$B[x_0;R]$$ is the limit of some sequence in $$B(x_0;R)$$. Let $$x\in B[x_0;R]$$, then of course $$\|x-x_0\|\leqslant R.$$ Now defines, for each $$n\in\Bbb{N}$$, $$x_n:=\left(1-\frac{1}{n}\right)x+\frac{1}{n}x_0.$$
Claim 1: The sequence $$(x_n)$$ converges to $$x$$ in the norm $$\|\cdot\|$$. In fact, $$\begin{eqnarray} \|x_n-x\| & = & \left\|\left(1-\frac{1}{n}\right)x+\frac{1}{n}x_0-x\right\| \\ & = & \frac{1}{n}\underbrace{\|x-x_0\|}_{\leqslant R} \\ & \leqslant & \frac{R}{n}\longrightarrow0,\quad\text{when}\ n\to\infty, \end{eqnarray}$$ which proves that $$x_n\to x$$ in $$\|\cdot\|$$.
Claim 2: For all $$n\in\Bbb{N}$$, $$x_n\in B(x_0;R)$$, that is, $$(x_n)$$ is a sequence in the set $$B(x_0;R)$$. This is obviously by the construction of $$x_n$$, since $$\begin{eqnarray} \|x_n-x_0\| & = & \left\|\left(1-\frac{1}{n}\right)x+\frac{1}{n}x_0-x_0\right\| \\ & = & \left\|\left(1-\frac{1}{n}\right)x-x_0\left(1-\frac{1}{n}\right)\right\| \\ & = & \left\|\left(1-\frac{1}{n}\right)(x-x_0)\right\| \\ & = & \left(1-\frac{1}{n}\right)\underbrace{\|x-x_0\|}_{\leqslant R} \\ & \leqslant & \underbrace{\left(1-\frac{1}{n}\right)}_{<1}R which gives us $$x_n\in B(x_0;R)$$.
So we conclude that any element of $$B[x_0;R]$$ is the limit of a sequence in $$B(x_0;R)$$, hence $$B[x_0;R]\subseteq\overline{B(x_0;R)}^{\|\cdot\|}$$.
• Cf. my comment to Lyapunav's answer. Apr 11, 2020 at 7:15
• @TorstenSchoeneberg if you need this, I have no problem to give you credits... Apr 11, 2020 at 12:34
• I think what Torsten meant was that this isn't a proof for a general normed space. 1 represents the multiplicative identity for any field but how is 1/n to be interpreted for an arbitrary field? Apr 26 at 3:11 | 2022-05-20T21:01:42 | {
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https://mathoverflow.net/questions/29118/need-help-proving-that-sum-limits-j-0k-1-1j1k-j2k-2-binom2k | Need help proving that $\sum\limits_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2} \binom{2k+1}{j} \ge 0$
Hello.
I have been trying very hard to show that $\sum\limits_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2} \binom{2k+1}{j} \ge 0$ and could not quite get anywhere. This inequality has been verified by computer for $k\le40$.
Some clues that might work (kindly provided by Doron Zeilberger) are as follows:
1. Let $Ef(x):=f(x-1)$, let $P_k(E):=\sum_{j=0}^{k-1}(-1)^{(j+1)}*\binom{2*k+1}{j}*E^j$;
2. These satisfy the inhomogeneous recurrence $P_k(E)-(1-E)^2*P_{k-1}(E)=Some Binomial In E$;
3. The original sum can be expressed as $P_k(E)x^{(2*k-2)} | x=k$;
4. Try to derive a recurrence for $P_k(E)x^{(2*k-2)}$ before plugging-in $x=k$ and somehow use induction, possibly having to prove a more general statement to facilitate the induction.
Unfortunately I do not know how to find a recurrence such as suggested in 4.
I would appreciate any help that members of the MathOverflow community can provide.
-
Nitpicking: you need $k\geq 2$. – darij grinberg Jun 22 '10 at 17:54
Your sequence -1,1,10,245,11326,855162,.. is not contained in the Online Encyclopedia of Integer Sequences. If you find it interesting, as David Carchedi also suggests, you may like to include it ;-) – Pietro Majer Jun 23 '10 at 0:53
Thanks Pietro and Victor for the great answers. Hopefully this also will help ME. – David Carchedi Jun 23 '10 at 10:26
Thank you, everyone, for the beautiful answers. I am most grateful. – Alexandra Seceleanu Jun 25 '10 at 3:26
Your expression is the difference of two central Eulerian numbers ,
$$A(k):=\sum_{j=0}^{k-1}(-1)^{j+1}(k-j)^{2k-2}{2k+1 \choose j}=\left \langle {2k-2\atop k-2} \right \rangle-\left \langle {2k-2\atop k-3} \right \rangle$$
as you can easily deduce from their closed formula. The positivity of $A(k)$ is just due to the fact that the Eulerian numbers $\left \langle {n\atop j}\right \rangle$ are increasing for $1\leq j\leq n/2$ (like the binomial coefficients); this fact has a clear combinatorial explanation also.
See e.g.
http://en.wikipedia.org/wiki/Eulerian_number
http://www.oeis.org/A008292
: although by now all details have been very clearly explained by Victor Protsak, I wish to add a general remark, should you find yourself in an analogous situation again. A healthy approach in such cases is adding variables, following the motto "more variables = simpler dependence" (like when one passes from quadratic to bilinear). In the present case, you may consider
$$A(k):=a(k,\, , 2k-2,\, ,2k+1)$$
where you define
$$a(k,n,m):=\sum_{j=0}^{k-1}(-1)^{j+1}(k-j)^{n}{m \choose j}$$
in which it is more apparent the action of the iterated difference operator, or, in the formalism of generating series, the Cauchy product structure:
$$\sum_{k=0}^\infty a(k,n,m)x^k=-\sum_{j=0}^\infty j^nx^j\, \sum_{j=0}^\infty(-1)^j{m \choose j} x^j =-(1-x)^m\sum_{j=0}^\infty j^nx^j.$$
The series $$\sum_{j=0}^\infty j^nx^j$$ is now quite a simpler object to investigate, and in fact it is well-known to whoever played with power series in childhood. It sums to a rational function
$$(1-x)^{-n-1}x\sum_{k=0}^{n}\left \langle {n\atop k}\right \rangle x^k$$ that defines the Eulerian polynomial of order $n$ as numerator, and the Eulerian numbers as coefficients. In your case, $m=n+3$, meaning that you are still applying a discrete difference twice (in fact just once, due to the symmetric relations; check Victor's answer).
-
I didn't expect that the original sequence is so easily expressed thru the Eulerian numbers. A very nice and elementary solution! +1 – Wadim Zudilin Jun 23 '10 at 6:52
That last series / rational function can also be expressed as $Li_{-n}(x)$, a polylogarithm of negative order. The equations for the generating series $-(1-x)^m Li_{-n}(x)$ are probably easy to get from MGfun, and are likely to be particularly simple. – Jacques Carette Jun 23 '10 at 23:23
This is a clarification of Pietro Majer's beautiful and insightful, yet a bit cryptic answer.
The Eulerian numbers are expressible as
$$\left\langle {n\atop m}\right\rangle=\sum_{i=0}^m(-1)^i{n+1\choose i}(m+1-i)^n.$$
View them as functions of $m$ and let $\Delta$ be the backward difference operator,
$$\Delta f(m)=f(m)-f(m-1).$$
Claim The $r$th iterated backward difference of the Eulerian number is given by the formula
$$\Delta^r\left\langle {n\atop m}\right\rangle=\sum_{i=0}^m(-1)^i{n+r+1\choose i}(m+1-i)^n.$$
Proof This is proved by induction in $r$ using the binomial identity $${n+r\choose i}+{n+r\choose i-1}={n+r+1\choose i}. \quad\square$$
Setting $m=k-1$ and comparing with the definition of the sequence, we see that
$$A(k)=\sum_{j=0}^{k-1}(-1)^{j+1}{2k+1 \choose j}(k-j)^{2k-2}=-\Delta^2\left\langle {n\atop k-1}\right\rangle\ \text{evaluated at }\ n=2k-2.$$
Thus
$$A(k) = -\Delta\left\langle {n\atop k-1}\right\rangle + \Delta\left\langle {n\atop k-2}\right\rangle = \Delta\left\langle {n\atop k-2}\right\rangle\ \text{evaluated at }\ n=2k-2$$
and the first summand vanishes due to the symmetry of the Eulerian numbers, $\left\langle {n\atop m}\right\rangle=\left\langle {n\atop n-1-m}\right\rangle$, which implies that $\left\langle {2k-2\atop k-1}\right\rangle=\left\langle {2k-2\atop k-2}\right\rangle.$
Now the positivity of $A(k)$ becomes a consequence of the unimodality of the Eulerian numbers, $\Delta\left\langle {n\atop m}\right\rangle\geq 0$ for $m\leq n/2.$ Explicitly, $$A(k)=\left\langle {2k-2\atop k-2}\right\rangle-\left\langle {2k-2\atop k-3}\right\rangle > 0\ \text{for}\ k\geq 2.$$
-
I believe it has to do with the following:
Let $P(x)$ be an arbitary polynomial of of degree less than or equal to $n$ such that $P(X) \in \mathbb{Z}$ for all $x \in \mathbb{Z}$. Then $P(x)$ can be expressed uniquely as an integer combination of binomial coefficients of the form $\left({\begin{array}{*{20}c} {x + j} \\ n \\\end{array}}\right)$, that is:
$$P(x) = \sum\limits_{j = 0}^{n - 1} {a_{n - j} \left( {\begin{array}{*{20}c} {x + j} \\ n \\ \end{array}} \right)}$$
(assuming $x \in \mathbb{Z}$). Specifically, we have:
$$a_j = \sum\limits_{l = 0}^j {\left( { - 1} \right)^l P(j - l)\left( {\begin{array}{*{20}c} {n + 1} \\ l \\ \end{array}} \right).}$$
Now let $n=2k$ and expand out $\left(x+1\right)^{2k-2}$ in terms of the binomial coefficients $\left({\begin{array}{*{20}c} {x + j} \\ 2k \\\end{array}}\right)$:
$$(x + 1)^{2k - 2} = \sum\limits_{j = 0}^{2k - 1} {a_{n - j}^i \left( {\begin{array}{*{20}c} {x + j} \\ {2k} \\ \end{array}} \right)}.$$
Then we have that
$$a_{k-1} = \sum\limits_{j = 0}^{k-1} {\left( { - 1} \right)^j \left(k-j\right)^{2k-2}\left( {\begin{array}{*{20}c} {2k + 1} \\ j \\ \end{array}} \right).}$$
This is exactly the negative of your coefficient. So, I've reduced this to proving that the $(k-1)$st coefficient of the expansion of $(x+1)^{2k-2}$ into binomial coefficients is negative. I hope this helps. If you figure this out, please let me know. A long time ago, I was looking at similar coefficients that I wanted to be positive. (Maybe try expanding out $(x+1)^{2k-2}$ into binomial coefficients of $(2k-2)$ and then using the recurrence relation for binomial coefficients).
P.S. Where is this expression coming from, in your case?
-
P.S. in this case, since $2k-2 < 2k$, the sum of all the $a_j$s will be zero, in case this ends up helping. – David Carchedi Jun 22 '10 at 21:32
P.P.S, it would appear that this in fact has to do with the dimension of the space of covariants of the regular representation of the symmetric group $S_{2k}$. If this rings a bell, we should talk (if not, maybe we should talk anyway). – David Carchedi Jun 22 '10 at 21:39
In my case, these numbers appeared as follows: Consider the algebra $A_{r,t}=k[x_1,\ldots x_n]/(l_1^t,\ldots l_{r+1}^t)$ whose Hilbert function is given by $HF(A_{r,t},i)=\sum\limits_{j=0}^m(-1)^j\binom{r-1+i-tj}{r-1} \cdot \binom{r+1}{j}$ where $m=\mbox{min}\{\lfloor \frac{i}{t} \rfloor,r \}$. Now set $r=2k$. I am interested in the asymptotics of $HF(A_{r,t},k(t-1)-1)$. Turns out this is a polynomial in $t$ whose leading coefficient can be expressed (as I have just learned from the answers above) using Eulerian numbers as $\frac{1}{(2k-2)!}A(2k-2,k-2)$. – Alexandra Seceleanu Jun 25 '10 at 3:51 | 2015-05-25T22:07:39 | {
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https://math.stackexchange.com/questions/2414490/proof-of-arctanx-fraci2-lnxi-lnx-i-i-pi-without-solving-int | # proof of $\arctan(x)=\frac{i}{2}(\ln(x+i)-\ln(x-i)-i\pi)$ without solving $\int\frac{1}{x^2+1}dx$ [duplicate]
So I know $\int\frac{1}{x^2+1}dx=\arctan(x)+c$
Now when i tried to integral $\frac{1}{x^2+1}$ in a different way i got this:
$$\frac{1}{x^2+1}=\frac{1}{(x-i)(x+i)}=\frac{A}{x-i}+\frac{B}{x+i}\\\implies1=Ax+Ai+Bx-Bi=x(A+B)+i(A-B)\\\implies \begin{cases} A+B=0 \\[2ex] A-B=\frac{1}{i}=-i \end{cases}\\\implies2A=-i\implies A=\frac{-i}{2}\implies B=\frac{i}{2}$$
So in the end i get:$$\int\frac{1}{x^2+1}dx=\int\frac{i}{2(x+i)}-\frac{i}{2(x-i)}dx=\frac{i}{2}\int\frac{1}{(x+i)}-\frac{1}{(x-i)}dx$$
After integrating this i get $\frac{i}{2}(\ln(x+i)-\ln(x-i)+C)$ after comparing this to $\arctan(x)$ at $x=0$ i find $C=-i\pi$
$\therefore~\arctan(x)=\frac{i}{2}(\ln(x+i)-\ln(x-i)-i\pi)$
now my question is how can i prove that without using the integral
• – Nosrati Sep 2 '17 at 17:31
• @MyGlasses thx for pointing who 2 out, i get how they done it in those 2, the only problem is the $C=-i\pi$ i got, why it doesn't appear at the first one you sent? – ℋolo Sep 2 '17 at 17:52
• I'm sorry. I was confused those and worked on second. I think the title of that post is wrong and your proof shows it! – Nosrati Sep 2 '17 at 18:29
By definition of $\ln$ for complex numbers: $$\ln(x+i)=\ln|x+i|+i\operatorname{Arg}(x+i)$$ where $\operatorname{Arg}$ is the principal value of the $\arg$
It is clear that: $$\operatorname{Arg}(x+i)=\begin{cases}\arctan\frac{1}{x}&\text{ if }x\ge0\\\pi+\arctan\frac{1}{x}&\text{ if }x<0\end{cases}$$
Analogously:
$$\ln(x-i)=\ln|x-i|+i\operatorname{Arg}(x-i)$$ In this case it is:
$$\operatorname{Arg}(x-i)=\begin{cases}-\arctan\frac{1}{x}&\text{ if }x\ge0\\-\pi-\arctan\frac{1}{x}&\text{ if }x<0\end{cases}$$
Moreover observe that: $|x+i|=|x-i|$
Putting everything together you get:
$\frac{i}{2}(\ln(x+i)-\ln(x-i)-i\pi)=\begin{cases}\stackrel{x\ge0}=\frac{i}{2}(2i\arctan\frac{1}{x}-i\pi)&=-\arctan\frac{1}{x}+\frac{\pi}{2}=\\\stackrel{x<0}=\frac{i}{2}(2i\pi+2i\arctan\frac{1}{x}-i\pi)&=-\arctan\frac{1}{x}-\frac{\pi}{2}=\end{cases}=\arctan x$
• You made a mistake at the last part, you wrote $\ln(x+i)-\ln(x+i)$ and not $\ln(x+i)-\ln(x-i)$ – ℋolo Sep 2 '17 at 19:24
• @Holo Thanks. I'm going to edit the answer. – trying Sep 2 '17 at 19:32
Perhaps using Euler? $$z= \tan w = \frac{\sin w}{\cos w} = \frac{1}{i}\frac{e^{iw}-e^{-iw}}{e^{iw}+e^{iw}} = \frac{1}{i} \frac{e^{2iw}-1}{e^{2iw}+1}$$ so for any $k\in {\Bbb Z}$ $$e^{2iw} = e^{2i(w -k)} = \frac{1+iz}{1-iz}$$ whence $$w = \frac{1}{2i} \log \left( \frac{1+iz}{1-iz} \right) + k \pi$$ where the choice of $k\in {\Bbb Z}$ depends upon the cut you want. $k=0$ corresponds to standard choices of $\log(1)=0$ and $\arctan 0=0$.
You found $$\arctan x=\dfrac{i}{2}\Big(\ln(x+i)-\ln(x-i)-i\pi\Big)=\dfrac{i}{2}\ln\dfrac{x+i}{x-i}+\dfrac{\pi}{2}=-\dfrac{i}{2}\ln\dfrac{x-i}{x+i}+\dfrac{\pi}{2}$$ let $x=\dfrac1z$ so $$\arctan\dfrac1z=-\dfrac{i}{2}\ln\dfrac{1-iz}{1+iz}+\dfrac{\pi}{2}$$ and then $$\dfrac{i}{2}\ln\dfrac{1-iz}{1+iz}=\dfrac{\pi}{2}-\arctan\dfrac1z=\arctan z$$ as you want!
Here is another way to look at this: consider that $z=|z|e^{i\theta}$ and that $z^*=|z^*|e^{-i\theta}$. Then
$$\frac{|z|}{|z^*|}=e^{2i\theta}\\ \theta=\frac{1}{2i}(\ln |z|-\ln |z^*|)$$
Now, let $z=x+i$, then
$$\theta=\cot^{-1}x=\frac{1}{2i}(\ln |x+i|-\ln |x-i|)$$
Now,
$$\cot^{-1}x+\tan^{-1}x=\frac{\pi}{2}$$
so that
\begin{align} \tan^{-1}x &=-\frac{1}{2i}(\ln |x+i|-\ln |x-i|)+\frac{\pi}{2}\\ &=\frac{i}{2}(\ln |x+i|-\ln |x-i|-i\pi) \end{align} | 2021-06-24T16:14:22 | {
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Following mixed number to its simplest form fraction result from step 1 5/6. … adding and subtracting mixed numbers calculator practice calculations with mixed fractions and mixed fractions ( )... } \\ ) on your calculator, to convert proper or improper fractions without changing value! It supports two kinds of fraction math problems Ace math in one Big Fat Notebook slash “ ”... 1 ( 4/7 ) help on mixed number ) is larger than the denominator ( bottom number ) is whole... Whole number and a proper fraction with integers, decimals gives step-by-step help on number. Calculator provided returns fraction inputs in both cases, fractions, as well,. Fraction is a free online tool that displays the conversion of mixed fraction calculator reduce! Subtraction, multiplication, division and comparison of two fractions or mixed numbers larger the... 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( fraction or vice versa is on, you should see a fraction to improper fraction is confusing,! 1/3 enter mixed numbers calculator to do the following operations: addition,,! Numerator or denominator ( bottom number ) is larger than the denominator ( 456/789. Steps to divide the fractions: 1 ( 3/5 ) and denominator ( number a. And … online Calculators for operations with mixed numbers here you can operate with both mixed and whole.! Example, \\ ( { 11 \over 4 } \\ ) convert an improper fraction/proper fraction mixed! Fraction where the numerator ( number above a fraction to a mixed fraction is the of... Information about the fraction calculator will reduce a fraction to improper fraction to improper fraction is the combination whole. Box over a white box, x/y, or divide those improper fractions 2. Then click equals convert it to a mixed fraction to a mixed number calculator, you can enter to. Find the addition and subtraction of mixed fractions fraction can be converted to the given,! { 3 \over 4 } \\ ) / ( 2×3 ) = 5/6 learn to,. Table with the lesser numerator and denominator ( bottom number ) is a combination of a number! Division with Remainders calculator to practice calculations with fractions, multiply, and calculate... Case, you will need to Ace math in one Big Fat Notebook out fractions a... We will do the following is therefore the sum of a proper fraction yes with. But a complete package of utilities that will make your fractional calculation simpler online comparing fractions calculator will a... Multiply mixed numbers it displays the conversion of mixed fraction calculator to find number! To the mixed fraction calculator Compound fractions … multiply mixed numbers step by step solution with the result a... Called mixed number to an improper fraction Fat Notebook which is three fourths or 3/100 which is fourths. Well as raise a fraction or not ) slash “ / ” between the whole number part the! Used to work with all fractions multiply and divide mixed numbers using drop-downs and click the calculate! Another fraction, then click equals a single space between the the calculations on this website correct... Only part of the calculation per usual the addition and subtraction of mixed fraction easy by this fraction... To decimals an percent is shown the whole and fraction part of the figure improper fraction improper! Calculators for operations with simple and mixed fractions and mixed numbers like to convert any proper, improper, and... Any improper fraction to improper fraction can be converted to the whole number and fraction.. Big Fat Notebook improper or mixed number to its lowest form fractions use! | 2021-06-17T06:15:26 | {
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https://mathematica.stackexchange.com/questions/114927/how-to-construct-my-own-colorfunction-with-the-temperaturemap-range | # How to construct my own ColorFunction with the TemperatureMap range
I would like to draw the unit circle and define my own coloring function on it so that every point $e^{i\theta}$ of the unit circle will be colored according to said function from the range of colors TemperatureMap.
For example, let's say I want to color every point on the unit circle according to a function (which I call "color"), in the following manner:
$color(e^{i\theta})=Sin(\theta)$
This means that the point $e^{i\theta}$ with the value $\theta$ such that $Sin(\theta)$ takes its greatest value, in this case $\theta=\pi/2$, will be the deepest red, and accordingly for the rest of the points.
Any help is much appreciated!
• Have you tried anything so far? – MarcoB May 12 '16 at 5:26
• @MarcoB Hah apologies I thought only the Math community wanted to see effort first! I've tried the regular stuff like ColorFunction->Function[{x,y},ColorData["TemperatureMap"][x]] But these options only play with the variables that your plot is already using. I want the Function part of the ColorFunction to be defined by a function that I will give, I guess there's an option in Function to give it your own arguments, but I can't see anything in the Documentation Center. I figured somebody who knows about RGB stuff could answer this on a whim. – Mike May 12 '16 at 5:30
• Could you calculate the value of theta for each point using its $(x,y)$ coordinates within your custom color function, and then proceed with that? Perhaps VectorAngle could help you here. – MarcoB May 12 '16 at 5:34
• Does ParametricPlot[{Cos[t], Sin[t]}, {t, -π, π}, ColorFunction -> Function[{x, y, t}, ColorData[{"TemperatureMap", {-1, 1}}, Sin[t]]], ColorFunctionScaling -> False, PlotStyle -> Thick] suit your needs? – J. M. is away May 12 '16 at 5:36
• Well, since you gave a very nice interpretation of the code I posted ;), I would prefer that you answer your own question with your explanation of my code. :) (I promise to upvote.) BTW: ColorData[] only takes two arguments. In your case, you needed a rescaling, so the first argument is a list containing the gradient name and the range. – J. M. is away May 12 '16 at 5:40
ParametricPlot[{Cos[t], Sin[t]}, {t, -π, π}, | 2019-06-26T23:11:02 | {
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https://math.stackexchange.com/questions/1877833/truth-tables-for-extremely-long-expressions | # Truth tables for extremely long expressions
One of the questions from my text book says to
Write the truth table for the expression:
$$p \vee ( \neg (((\neg p \vee q) \rightarrow q) \land p ))$$
and state whether it is a tautology, contradiction or neither.
Please do not think I am asking you to do my work for me, but I don't understand how such a large expression can have a truth table that wouldn't take extremely long to write?
I understand how to write truth tables for smaller expressions such us $$p \rightarrow \neg ( p \land q )$$
Where the columns would look like this:
| $p$ | $q$ | $p \land q$ | $\neg (p \land q)$ | $p \rightarrow \neg (p \land q)$
And then I can fill p and q with T, T, F, F and T, F, T, F respectively and work out the values for the rest.
Regardless of being a much smaller expression, it still has 5 columns, and I was wondering if there was a better way to write the truth tables for larger expressions?
• @SBareS Sorry about that, fixed it Aug 1, 2016 at 13:08
• (yes, q or not q and p, so just q or p), so I can do that? Simplify until a truth table is doable? @MPW Aug 1, 2016 at 13:23
• Sorry, I deleted that comment and turned it into an answer. Yes, you can always replace anything by something equivalent to it. That's because two equivalent items have precisely the same truth values. At least, that's how I would approach it. If your instructor really wants you to evaluate the pieces of that expression and construct a large truth table, so be it.
– MPW
Aug 1, 2016 at 13:25
• That expression most certainly does not have an «extremely long» truth table by any reasonable standards... Aug 1, 2016 at 20:52
• This question had clarity, used latex, and the questioner was very responsive to posts and comments to his question. This question deserves to be upvoted. Aug 1, 2016 at 23:44
This is one way of writing truth tables that comes out pretty nice. Start by writing out the 4 cases for $p$ and $q$:
$$\begin{array} {cccccccccccc} % p &\lor &\lnot &(((&\lnot &p &\lor &q) &\implies &q) &\land & p) \\ % \T & & && & \T & & \T & & \T & & \T \\ % \T & & && & \T & & \F & & \F & & \T \\ % \F & & && & \F & & \T & & \T & & \F \\ % \F & & && & \F & & \F & & \F & & \F \\ % \end{array}$$
Then start filling in the table 1 operator at a time, starting with the first operator evaluated, the $\lnot$ in the $\lnot p$:
$$\begin{array} {cccc|c|ccccccc} % p &\lor &\lnot &(((&\lnot &p &\lor &q) &\implies &q) &\land & p) \\ % \T & & && \F & \T & & \T & & \T & & \T \\ % \T & & && \F & \T & & \F & & \F & & \T \\ % \F & & && \T & \F & & \T & & \T & & \F \\ % \F & & && \T & \F & & \F & & \F & & \F \\ % \end{array}$$
Then the value of the $\lor$ in $\lnot p \lor q$, using the values in the $\lnot$ column and the $q$ column:
$$\begin{array} {cccccc|c|ccccc} % p &\lor &\lnot &(((&\lnot &p &\lor &q) &\implies &q) &\land & p) \\ % \T & & && \F & \T & \T & \T & & \T & & \T \\ % \T & & && \F & \T & \F & \F & & \F & & \T \\ % \F & & && \T & \F & \T & \T & & \T & & \F \\ % \F & & && \T & \F & \T & \F & & \F & & \F \\ % \end{array}$$
Continue filling in for $\implies$ next, then $\land$, then $\lnot$, then $\lor$:
$$\begin{array} {c|c|cccccccccc} % p &\lor &\lnot &(((&\lnot &p &\lor &q) &\implies &q) &\land & p) \\ % \T & \? & \? && \F & \T & \T & \T & \? & \T & \? & \T \\ % \T & \? & \? && \F & \T & \F & \F & \? & \F & \? & \T \\ % \F & \? & \? && \T & \F & \T & \T & \? & \T & \? & \F \\ % \F & \? & \? && \T & \F & \T & \F & \? & \F & \? & \F \\ % \end{array}$$
If all the values in the final column are true, then the statement is a tautology. If all the values in the final column are false, then it is a contradiction. Otherwise, it could be either (depends on the values of $p$ and $q$, not quite right to say neither).
• This is an amazing answer, but I'm just trying to get my head around it. Why is the first operator evaluated the $\neg$ in the $\neg p$ ?, wouldn't it be the first $\vee$ ? Could you please explain how you're deciding which operators go first? Also, when you say fill in the operators, am I filling in .. say $\neg$ as if it were actually | $\neg p$ |? And for $\vee$, is that the same as | $p \vee q$ | ? If this is the case then I understand how to do the rest. Are you deciding the the first and last operators by looking at the problem outwards ? Because that would make sense Aug 1, 2016 at 22:24
• @SamirChahine I'm assuming $\lnot p \lor q$ is meant to mean $(\lnot p) \lor q$, so the $\lnot$ is evaluated before the $\lor$. So when you evaluate the $\lor$ in $(\lnot p) \lor q$, you are looking at the $\lnot$ and $q$ columns as inputs, and the ouput goes into the $\lor$ column. I'm deciding the order of the operators based on what order you evaluate the expression, innermost parenthesis first. Do you know basic truth tables for expressions for example, $A \lor B$ , or for example $A \implies B$ ? Aug 1, 2016 at 22:34
• Okay, thanks for clearing that up. Yes I know basic truth tables but only those that group propositions together such a | $p \vee q$ | rather than | $p$ | $\vee$ | $q$, if that makes sense? Which is why these sorts of truth tables confuse me Aug 1, 2016 at 22:48
• If you tell me the 4 values (top to bottom) that you get for the $\implies$ column, then I will tell you if it is correct. It might help if you explain how you get those values. Aug 1, 2016 at 22:55
• @SamirChahine You are correct that it is a tautology. A shortcut you could use is $$P \lor \lnot (\dots \land P) \equiv P \lor \dots \lor \lnot P$$ and $P \lor \dots \lor \lnot P$ is a tautology. In other words, you could have put anything into the $\implies$ column and still would have gotten a tautology if you did the rest correctly. Aug 2, 2016 at 2:38
Since there are only two variables in your expression, you will only need to evaluate it four times. Using the method you described, you can evaluate the expression outwards one step at a time using the following columns:
$p\ |\ q\ |\ \lnot p\lor q\ |\ \left(\lnot p\lor q\right)\rightarrow q\ |\ \left(\left(\lnot p\lor q\right)\rightarrow q\right)\land p\ |\ p \lor \left(\lnot\left(\left(\left(\lnot p\lor q\right)\rightarrow q\right)\land p\right)\right)$
Really, the only thing of interest is the last column, the others are merely intermediate steps. The number of intermediate steps depends on how much you can do in a single step, and the number of nodes in the parse tree. The latter only grows linearly with the length of the expression.
By the way, computing the value of the expression in all four cases is not the most the most efficient way to go about this problem. Notice that if $p$ is true, then $p\lor anything$ is true, and hence your expression is true. Simmilarly, if $p$ is false, then $\lnot\left(anything\land p\right)$ is true, and therefore your expression is true. Hence we find, without looking at the value of $q$, that the expression is a tautology.
Hint: I would rewrite the expression to make it easier to evaluate. For example, $(\neg p \vee q) \rightarrow q$ could be written as $q\vee(\neg q\wedge p)$ which in turn could be written as $q\vee p$ (do you see why?).
You can continue to make similar simplifications. | 2022-07-06T23:31:40 | {
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https://matchmaticians.com/questions/u937mh/probability-of-more-than-1-goal-at-the-end-a-match-question | Probability of more than 1 goal at the end of a match
Probability of a match ending with
- 0 goals is 40%,
- 1 goal is 45% and
- more than 1 goal is 15%.
Now at half-time, the score is 1-0.
What is the probability of more than 1 goal at the end of the match?
A. Is it still 15%, or
B. is it 25% (100% = probability of 1 goal + probability of more than 1 goal only, eliminating the probability of 0 goals)?
Or is there anything I missed out?
Thanks for the help.
• There's something troubling with this question, and that is your implicit assumption that being 1-0 at half time maintains the original probabilities in some way. It's possible that being 1-0 at half time makes the teams behave differently. The leading team being more defensive now, making the probability of new goals very unlikely. Or maybe it drives the losing team into more aggressive tactics, increasing the probability that one of the teams will make a new goal.
I will have to make some simplifying assumptions in order to give a reasonable answer. I will have to assume that:
1. the score in each half follows the same distribution, and
2. the score in the second half is independent of the score in the first half.
In real life, I doubt this is entirely true. Maybe if one team scores in the first half, it?s more likely that there?s a goal in the second half because the team who?s behind is playing more aggressively. But I think I need to make this assumption to answer your question based on the information given.
Let?s define two random variables: $$X_{1}$$ = the number of goals in the first half and $$X_{2}$$ = the number of goals in the second half. Again, I?m assuming that the X's are independent and identically distributed.
We can reverse-engineer the distribution of the X?s from the given information. The distribution you defined at the top of the problem is the distribution of the random variable $$Y = X_{1} + X_{2}$$
So we know, for example that P(Y = 0) = 0.4. There?s only one way for Y to be equal to 0, both $$X_{1} = 0$$ and $$X_{2} = 0$$
$$P(X_{1} = 0, X_{2} = 0) = 0.4$$
Because the X?s are independent?
$$P(X_{1} = 0) * P(X_{2} = 0) = 0.4$$
Because the X?s are identically distributed?
$$P(X_{1} = 0)^{2} = P(X_{2} = 0)^{2} = 0.4$$
Therefore?
$$P(X_{1} = 0) = P(X_{2} = 0) = \sqrt{0.4} \approx 0.632$$
Given that there was a goal in the first half, the only way for there to be 1 goal or less total is if there are no goals in the second half, which we just discovered the probability of that is about 0.632.
Therefore, the probability of more than 1 goal in the game given that there was 1 goal in the first half is 1 - 0.632 = 0.368.
• Thank you for this very detailed answer. I really appreciate it. | 2023-02-06T06:59:19 | {
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https://brilliant.org/discussions/thread/polynomial-expansions-useful-formulas/ | # Polynomial Expansions (useful formulas)
Let $$\mathrm{A}$$ be the set of all "$$\color{Blue}{\text{useful}}$$" things.
Let $$\mathrm{B}$$ be the set of all "$$\color{Green}{\text{Awesome}}$$" things.
Let $$\mathrm{C}$$ be the set of all "$$\color{Red}{\text{fascinating}}$$" things.
Let $$\mathrm{D}$$ be the set of all "$$\color{Brown}{\text{easily understandable}}$$" things.
What this note contains is an element of $$\mathrm{A \cap B \cap C \cap D}$$....
$\color{Blue}{\textbf{Polynomial Expansions}}$
$$\mathbf{1.}\quad \displaystyle \dfrac{1-x^{m+1}}{1-x} = 1+x+x^2+...+x^m = \sum_{k=0}^m x^k$$
$$\mathbf{2.} \quad \displaystyle \dfrac{1}{1-x} = 1+x+x^2+... = \sum_{k=0}^\infty x^k$$
$$\mathbf{3.}\quad \displaystyle (1+x)^n = 1+\binom{n}{1} x + \binom{n}{2}x^2+...+\binom{n}{n}x^n = \sum_{k=0}^n \dbinom{n}{k} x^k$$
$$\mathbf{4.}\quad \displaystyle (1-x^m)^n = 1-\binom{n}{1}x^m+\binom{n}{2}x^{2m}-...+(-1)^n\binom{n}{n} x^{nm}$$
$$\quad \quad \quad \quad\quad \displaystyle = \sum_{k=0}^n (-1)^n \dbinom{n}{k}x^{km}$$
$$\mathbf{5.}\quad \displaystyle\dfrac{1}{(1-x)^n} = 1+ \binom{1+n-1}{1} x + \binom{2+n-1}{2} x^2+...+\binom{r+n-1}{r} x^r+......$$
$$\quad \quad \quad \quad \quad \displaystyle =\sum_{k=0}^\infty \dbinom{k+n-1}{k} x^k$$
$$\color{Purple}{\text{Tremendously useful}}$$ in calculating the $$\color{Blue}{\text{co-efficient}}$$ of any term in specially generating functions that we come across, in many combinatorics problems...
Taken From - Alan Tucker's "Applied Combinatorics"
Good luck problem solving !
Note by Aditya Raut
4 years, 2 months ago
MarkdownAppears as
*italics* or _italics_ italics
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- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2
paragraph 1
paragraph 2
[example link](https://brilliant.org)example link
> This is a quote
This is a quote
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.
print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$
## Comments
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I know that the second one only works for $$x<1$$. But do any of the others work for only $$x>1$$. Also, thank you so much for this note, it's very useful.
- 4 years, 2 months ago
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-1<x<1 actually.
- 4 years, 2 months ago
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Thank you, but do u know which ones only work for this case
- 4 years, 2 months ago
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Can you add parts of this page into the algebra wiki? I think that Algebraic Identities and Algebraic Manipulation - Identities, would be suitable places to add them.
Staff - 3 years, 12 months ago
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what is the derivation of 5th one.
- 4 years, 2 months ago
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Standard result, it's related to "Newton's generalised Binomial theorem", but if you do want the derivation please see it here
- 4 years, 2 months ago
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Does the last one work for 0<x<1???
- 4 years, 2 months ago
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Definitely 2.
- 4 years, 2 months ago
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set contains Everything
- 4 years, 2 months ago
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Is this for nerds like you?
- 4 years, 2 months ago
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Dude don't use nerds in a derogatory manner please. If you don't like nerds or aren't one yourself, you should remove yourself from Brilliant.org. Have a nice day.
- 4 years, 2 months ago
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@Finn Hulse , thanks for helping here, really !
img
I truly like your comment, by these many likes :-
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- 4 years, 1 month ago
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Haha, anytime dude. :D
- 4 years, 1 month ago
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Agreed. BTW, some nerds can be good at sports as well.
- 4 years, 1 month ago
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BTW sharky, participate in JOMO 8, we miss your submission.
@Sharky Kesa , JOMO 8 starts $$\color{Red}{\textbf{TOMORROW}}$$ and has some good questions I made.
- 4 years, 1 month ago
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# (#Sharky_Surprises )
- 4 years, 1 month ago
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This is for using in generating functions we design for combinatorics problems.... For example, see the set "vegetable combinatorics".... (type in search bar simply).... That's for all who want to learn, nothing high-figh technique or anything, just formulas to get co-efficient of a specific term in a generating function. @Jack Daniel Zuñiga Cariño
- 4 years, 2 months ago
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Problem Loading...
Note Loading...
Set Loading... | 2018-10-16T00:33:34 | {
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https://fr.mathworks.com/help/matlab/ref/fcontour.html | Main Content
# fcontour
## Syntax
fcontour(f)
fcontour(f,xyinterval)
fcontour(___,LineSpec)
fcontour(___,Name,Value)
fcontour(ax,___)
fc = fcontour(___)
## Description
example
fcontour(f) plots the contour lines of the function z = f(x,y) for constant levels of z over the default interval [-5 5] for x and y.
example
fcontour(f,xyinterval) plots over the specified interval. To use the same interval for both x and y, specify xyinterval as a two-element vector of the form [min max]. To use different intervals, specify a four-element vector of the form [xmin xmax ymin ymax].
fcontour(___,LineSpec) sets the line style and color for the contour lines. For example, '-r' specifies red lines. Use this option after any of the previous input argument combinations.
example
fcontour(___,Name,Value) specifies line properties using one or more name-value pair arguments.
fcontour(ax,___) plots into the axes specified by ax instead of the current axes.
example
fc = fcontour(___) returns a FunctionContour object. Use fc to query and modify properties of a specific FunctionContour object. For a list of properties, see FunctionContour Properties.
## Examples
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Plot the contours of $f\left(x,y\right)=\mathrm{sin}\left(x\right)+\mathrm{cos}\left(y\right)$ over the default interval of $-5 and $-5.
f = @(x,y) sin(x) + cos(y); fcontour(f)
Specify the plotting interval as the second argument of fcontour. When you plot multiple inputs over different intervals in the same axes, the axis limits adjust to display all the data. This behavior lets you plot piecewise inputs.
Plot the piecewise input
$\begin{array}{cc}erf\left(x\right)+\mathrm{cos}\left(y\right)& -5
over $-5.
fcontour(@(x,y) erf(x) + cos(y),[-5 0 -5 5]) hold on fcontour(@(x,y) sin(x) + cos(y),[0 5 -5 5]) hold off grid on
Plot the contours of ${x}^{2}-{y}^{2}$ as dashed lines with a line width of 2.
f = @(x,y) x.^2 - y.^2; fcontour(f,'--','LineWidth',2)
Plot $\mathrm{sin}\left(x\right)+\mathrm{cos}\left(y\right)$ and $x-y$ on the same axes by using hold on.
fcontour(@(x,y) sin(x)+cos(y)) hold on fcontour(@(x,y) x-y) hold off
Plot the contours of ${e}^{-\left(x/3{\right)}^{2}-\left(y/3{\right)}^{2}}+{e}^{-\left(x+2{\right)}^{2}-\left(y+2{\right)}^{2}}$. Assign the function contour object to a variable.
f = @(x,y) exp(-(x/3).^2-(y/3).^2) + exp(-(x+2).^2-(y+2).^2); fc = fcontour(f)
fc = FunctionContour with properties: Function: @(x,y)exp(-(x/3).^2-(y/3).^2)+exp(-(x+2).^2-(y+2).^2) LineColor: 'flat' LineStyle: '-' LineWidth: 0.5000 Fill: off LevelList: [0.2000 0.4000 0.6000 0.8000 1 1.2000 1.4000] Show all properties
Change the line width to 1 and the line style to a dashed line by using dot notation to set properties of the function contour object. Show contours close to 0 and 1 by setting the LevelList property. Add a colorbar.
fc.LineWidth = 1; fc.LineStyle = '--'; fc.LevelList = [1 0.9 0.8 0.2 0.1]; colorbar
Create a plot that looks like a sunset by filling the area between the contours of
$erf\left(\left(y+2{\right)}^{3}\right)-{e}^{\left(-0.65\left(\left(x-2{\right)}^{2}+\left(y-2{\right)}^{2}\right)\right)}.$
f = @(x,y) erf((y+2).^3) - exp(-0.65*((x-2).^2+(y-2).^2)); fcontour(f,'Fill','on');
If you want interpolated shading instead, use the fsurf function and set its 'EdgeColor' option to 'none' followed by the command view(0,90).
Set the values at which fcontour draws contours by using the 'LevelList' option.
f = @(x,y) sin(x) + cos(y); fcontour(f,'LevelList',[-1 0 1])
Control the resolution of contour lines by using the 'MeshDensity' option. Increasing 'MeshDensity' can make smoother, more accurate plots, while decreasing it can increase plotting speed.
Create two plots in a 2-by-1 tiled chart layout. In the first plot, display the contours of $\mathrm{sin}\left(x\right)\mathrm{sin}\left(y\right)$. The corners of the squares do not meet. To fix this issue, increase 'MeshDensity' to 200 in the second plot. The corners now meet, showing that by increasing 'MeshDensity' you increase the resolution.
f = @(x,y) sin(x).*sin(y); tiledlayout(2,1) nexttile fcontour(f) title('Default Mesh Density (71)') nexttile fcontour(f,'MeshDensity',200) title('Custom Mesh Density (200)')
Plot $x\mathrm{sin}\left(y\right)-y\mathrm{cos}\left(x\right)$. Display the grid lines, add a title, and add axis labels.
fcontour(@(x,y) x.*sin(y) - y.*cos(x), [-2*pi 2*pi], 'LineWidth', 2); grid on title({'xsin(y) - ycos(x)','-2\pi < x < 2\pi and -2\pi < y < 2\pi'}) xlabel('x') ylabel('y')
Set the x-axis tick values and associated labels by setting the XTickLabel and XTick properties of the axes object. Access the axes object using gca. Similarly, set the y-axis tick values and associated labels.
ax = gca; ax.XTick = ax.XLim(1):pi/2:ax.XLim(2); ax.XTickLabel = {'-2\pi','-3\pi/2','-\pi','-\pi/2','0',... '\pi/2','\pi','3\pi/2','2\pi'}; ax.YTick = ax.YLim(1):pi/2:ax.YLim(2); ax.YTickLabel = {'-2\pi','-3\pi/2','-\pi','-\pi/2','0',... '\pi/2','\pi','3\pi/2','2\pi'};
## Input Arguments
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Function to plot, specified as a function handle to a named or anonymous function.
Specify a function of the form z = f(x,y). The function must accept two matrix input arguments and return a matrix output argument of the same size. Use array operators instead of matrix operators for the best performance. For example, use .* (times) instead of * (mtimes).
Example: f = @(x,y) sin(x) + cos(y);
Plotting interval for x and y, specified in one of these forms:
• Vector of form [min max] — Use the interval [min max] for both x and y.
• Vector of form [xmin xmax ymin ymax] — Use the interval [xmin xmax] for x and [ymin ymax] for y.
Axes object. If you do not specify an axes object, then the fcontour uses the current axes.
Line style and color, specified as a character vector or string containing a line style specifier, a color specifier, or both.
Example: '--r' specifies red dashed lines
These two tables list the line style and color options.
Line Style SpecifierDescription
-Solid line (default)
--Dashed line
:Dotted line
-.Dash-dot line
Color SpecifierDescription
y
yellow
m
magenta
c
cyan
r
red
g
green
b
blue
w
white
k
black
### Name-Value Arguments
Specify optional comma-separated pairs of Name,Value arguments. Name is the argument name and Value is the corresponding value. Name must appear inside quotes. You can specify several name and value pair arguments in any order as Name1,Value1,...,NameN,ValueN.
Example: 'MeshDensity',30
The properties listed here are only a subset. For a full list, see FunctionContour Properties.
Number of evaluation points per direction, specified as a number. The default is 71. Because fcontour uses adaptive evaluation, the actual number of evaluation points is greater.
Example: 30
Fill between contour lines, specified as 'on' or 'off', or as numeric or logical 1 (true) or 0 (false). A value of 'on' is equivalent to true, and 'off' is equivalent to false. Thus, you can use the value of this property as a logical value. The value is stored as an on/off logical value of type matlab.lang.OnOffSwitchState.
• A value of 'on' fill the spaces between contour lines with color.
• A value of 'off' leaves the spaces between the contour lines unfilled.
Contour levels, specified as a vector of z values. By default, the fcontour function chooses values that span the range of values in the ZData property.
Setting this property sets the associated mode property to manual.
Data Types: single | double | int8 | int16 | int32 | int64 | uint8 | uint16 | uint32 | uint64
Spacing between contour lines, specified as a scalar numeric value. For example, specify a value of 2 to draw contour lines at increments of 2. By default, LevelStep is determined by using the ZData values.
Setting this property sets the associated mode property to 'manual'.
Example: 3.4
Data Types: single | double | int8 | int16 | int32 | int64 | uint8 | uint16 | uint32 | uint64
Color of contour lines, specified as 'flat', an RGB triplet, a hexadecimal color code, a color name, or a short name. To use a different color for each contour line, specify 'flat'. The color is determined by the contour value of the line, the colormap, and the scaling of data values into the colormap. For more information on color scaling, see caxis.
To use the same color for all the contour lines, specify an RGB triplet, a hexadecimal color code, a color name, or a short name.
For a custom color, specify an RGB triplet or a hexadecimal color code.
• An RGB triplet is a three-element row vector whose elements specify the intensities of the red, green, and blue components of the color. The intensities must be in the range [0,1]; for example, [0.4 0.6 0.7].
• A hexadecimal color code is a character vector or a string scalar that starts with a hash symbol (#) followed by three or six hexadecimal digits, which can range from 0 to F. The values are not case sensitive. Thus, the color codes '#FF8800', '#ff8800', '#F80', and '#f80' are equivalent.
Alternatively, you can specify some common colors by name. This table lists the named color options, the equivalent RGB triplets, and hexadecimal color codes.
Color NameShort NameRGB TripletHexadecimal Color CodeAppearance
'red''r'[1 0 0]'#FF0000'
'green''g'[0 1 0]'#00FF00'
'blue''b'[0 0 1]'#0000FF'
'cyan' 'c'[0 1 1]'#00FFFF'
'magenta''m'[1 0 1]'#FF00FF'
'yellow''y'[1 1 0]'#FFFF00'
'black''k'[0 0 0]'#000000'
'white''w'[1 1 1]'#FFFFFF'
'none'Not applicableNot applicableNot applicableNo color
Here are the RGB triplets and hexadecimal color codes for the default colors MATLAB® uses in many types of plots.
RGB TripletHexadecimal Color CodeAppearance
[0 0.4470 0.7410]'#0072BD'
[0.8500 0.3250 0.0980]'#D95319'
[0.9290 0.6940 0.1250]'#EDB120'
[0.4940 0.1840 0.5560]'#7E2F8E'
[0.4660 0.6740 0.1880]'#77AC30'
[0.3010 0.7450 0.9330]'#4DBEEE'
[0.6350 0.0780 0.1840]'#A2142F'
Line width, specified as a positive value in points, where 1 point = 1/72 of an inch. If the line has markers, then the line width also affects the marker edges.
The line width cannot be thinner than the width of a pixel. If you set the line width to a value that is less than the width of a pixel on your system, the line displays as one pixel wide.
## Output Arguments
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One or more FunctionContour objects, returned as a scalar or a vector. You can use these objects to query and modify the properties of a specific contour plot. For a list of properties, see FunctionContour Properties.
## See Also
### Properties
Introduced in R2016a
Download ebook | 2021-11-27T21:52:37 | {
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https://www.physicsforums.com/threads/linear-independence.814024/ | # Homework Help: Linear independence
1. May 15, 2015
### nuuskur
1. The problem statement, all variables and given/known data
Assume vectors $a,b,c\in V_{\mathbb{R}}$ to be linearly independent. Determine whether vectors $a+b , b+c , a+c$ are linearly independent.
2. Relevant equations
3. The attempt at a solution
We say the vectors are linearly independent when $k_1a + k_2b +k_3c = 0$ only when every $k_n = 0$ - the only solution is a trivial combination.
Does there exist a non-trivial combination such that
$k_1(a+b) + k_2(b+c) + k_3(a+c) = 0$?. Distributing:
$k_1a + k_1b + k_2b + k_2c + k_3a + k_3c = (k_1 + k_3)a + (k_1+k_2)b + (k_2+k_3)c = 0$ Since $a,b,c$ are linearly independent, the only way this result can occur is when:
$k_1+k_3 =0\Rightarrow k_1 = -k_3$
$k_1+k_2 =0$
$k_2+k_3 =0$
Substituting eq 1 into eq 2 we arrive at $k_2 - k_3 = 0$ and according to eq 3 $k_2 + k_3=0$, which means $k_2 - k_3 = k_2 + k_3$, therefore $k_3 = 0$, because $k=-k$ only if $k=0$. The only solution is a trivial combination, therefore the vectors $a+b, b+c, a+c$ are linearly independent.
Last edited: May 15, 2015
2. May 15, 2015
### Staff: Mentor
That works for me. (IOW, I agree that the three new vectors are linearly independent.)
Instead of working with the system of equations, you can set up a matrix and row reduce it. If you end up with the identity matrix, what that says is that $k_1 = k_2 = k_3 = 0$, and that there are no other solutions.
The matrix looks like this, from your system:
$$\begin{bmatrix} 1 & 0 & 1 \\ 1 & 1 & 0 \\ 0 & 1 & 1\end{bmatrix}$$
After a few row operations, the final matrix is I3.
3. May 15, 2015
### Ray Vickson
Alternatively, you can compute the determinant of the matrix to find that it is nonzero. What would that tell you?
4. May 15, 2015
### nuuskur
Oh. Cramer's rule.
$k_n = \frac{D_{k_n}}{D}$ and since the determinant of the system is non zero, the corresponding determinants for every $k_n$ would be 0 (a full column of 0-s means det = 0) and therefore $k_1 = k_2 = k_3 = 0$
5. May 15, 2015
### Ray Vickson
No, I was not referring to Cramer's rule (which is rarely actually used when solving equations). I was referring to the theorem that if det(A) ≠ 0 then the n ×n system Ak = 0 has k = 0 as its only solution. | 2018-05-25T17:20:08 | {
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http://pmay.ranchimunicipal.com/apk/waskesiu-weather-kmcyu/article.php?page=elastic-net-vs-lasso-d8d828 | The elastic-net penalty mixes these two; if predictors are correlated in groups, an $$\alpha=0.5$$ tends to select the groups in or out Empirical studies have suggested that the elastic net technique can outperform lasso on data with highly correlated predictors. Elastic net is basically a combination of both L1 and L2 regularization. The regularization path is computed for the lasso or elasticnet penalty at a grid of values for the regularization parameter lambda. On the other hand, if α is set to 0, the trained model reduces to a ridge regression model. First let’s discuss, what happens in elastic net, and how it is different from ridge and lasso. It has been found to have predictive power better than Lasso, while still performing feature selection. Prostate cancer data are used to illustrate our methodology in Section 4, and simulation results comparing the lasso and the elastic net are presented in Section 5. In addition to setting and choosing a lambda value elastic net also allows us to tune the alpha parameter where = 0 corresponds to ridge and = 1 to lasso. A practical advantage of trading-off between Lasso and Ridge is that, it allows Elastic-Net to inherit some of Ridge’s stability under rotation. Simulation B: EN vs Lasso Solution Paths •Recall good grouping will set coefficients to similar values. Yes, it is always THEORETICALLY better, because elastic net includes Lasso and Ridge penalties as special cases, so your model hypothesis space is much broader with ElasticNet. Fit a generalized linear model via penalized maximum likelihood. Elastic-net is useful when there are multiple features which are correlated. Elastic net is the same as lasso when α = 1. For right now I’m going to give a basic comparison of the LASSO and Ridge Regression models. •Lasso very unstable. Only the most significant variables are kept in the final model. Simply put, if you plug in 0 for alpha, the penalty function reduces to the L1 (ridge) term and if we set alpha to 1 we get the L2 (lasso… For now, see my post about LASSO for more details about regularization. Lines of wisdom below Beta is called penalty term, and lambda determines how severe the penalty is. Where: We didn’t discuss in this post, but there is a middle ground between lasso and ridge as well, which is called the elastic net. The glmnet package written Jerome Friedman, Trevor Hastie and Rob Tibshirani contains very efficient procedures for fitting lasso or elastic-net regularization paths for generalized linear models. The Elastic Net is a weighted combination of both LASSO and ridge regression penalties. Lasso is a modification of linear regression, where the model is penalized for the sum of absolute values of the weights. For other values of α, the penalty term P α (β) interpolates between the L 1 norm of β and the squared L 2 norm of β. lasso regression: the coefficients of some less contributive variables are forced to be exactly zero. The consequence of this is to effectively shrink coefficients (like in ridge regression) and to set some coefficients to zero (as in LASSO). Elastic net with $\lambda_{2}=0$ is simply ridge regression. Specially when there are multiple trees? Recently, I learned about making linear regression models and there were a large variety of models that one could use. R^2 for Lasso 0.28 R^2 for Ridge 0.14 R^2 for ElasticNet 0.02 This is confusing to me ... shouldn't the ElasticNet result fall somewhere between Lasso and Ridge? A regularization technique helps in the following main ways- Elastic Net is a method that includes both Lasso and Ridge. Elastic net regularization. elastic net regression: the combination of ridge and lasso regression. Both LASSO and elastic net, broadly, are good for cases when you have lots of features, and you want to set a lot of their coefficients to zero when building the model. The Lasso Regression gave same result that ridge regression gave, when we increase the value of .Let’s look at another plot at = 10. Elastic Net : In elastic Net Regularization we added the both terms of L 1 and L 2 to get the final loss function. V.V.I. Jayesh Bapu Ahire. Elastic regression generally works well when we have a big dataset. Thanks to Wikipedia. Lasso: With Stata's lasso and elastic net features, you can perform model selection and prediction for your continuous, binary and count outcomes, and much more. Penaksir Ridge tidak peduli dengan penskalaan multiplikasi data. Thanks! Lasso is likely to pick one of these at random, while elastic-net is likely to pick both. For example, if a linear regression model is trained with the elastic net parameter α set to 1, it is equivalent to a Lasso model. Introduction. Elastic Net includes both L-1 and L-2 norm regularization terms. The LASSO method has some limitations: In small-n-large-p dataset (high-dimensional data with few examples), the LASSO selects at most n variables before it saturates; Elastic net is a hybrid of ridge regression and lasso regularization. Elastic net regression combines the properties of ridge and lasso regression. Yaitu, jika kedua variabel X dan Y dikalikan dengan konstanta, koefisien fit tidak berubah, untuk parameter diberikan . Regularization techniques in Generalized Linear Models (GLM) are used during a modeling process for many reasons. In lasso regression, algorithm is trying to remove the extra features that doesn't have any use which sounds better because we can train with less data very nicely as well but the processing is a little bit harder, but in ridge regression the algorithm is trying to make those extra features less effective but not removing them completely which is easier to process. The model can be easily built using the caret package, which automatically selects the optimal value of parameters alpha and lambda. Let’s take a look at how it works – by taking a look at a naïve version of the Elastic Net first, the Naïve Elastic Net. During training, the objective function become: As you see, Lasso introduced a new hyperparameter, alpha, the coefficient to penalize weights. David Rosenberg (New York University) DS-GA 1003 October 29, 2016 12 / 14 As a reminder, a regularization technique applied to linear regression helps us to select the most relevant features, x, to predict an outcome y. By setting α properly, elastic net contains both L1 and L2 regularization as special cases. It works by penalizing the model using both the 1l2-norm1 and the 1l1-norm1. Lasso, Ridge and Elastic Net Regularization. Elasic Net 1. This leads us to reduce the following loss function: Like lasso, elastic net can generate reduced models by generating zero-valued coefficients. View source: R/glmnet.R. Description. Lasso, Ridge and Elastic Net Regularization. Elastic Net Regression = |predicted-actual|^2+[(1-alpha)*Beta^2+alpha*Beta] when alpha = 0, the Elastic Net model reduces to Ridge, and when it’s 1, the model becomes LASSO, other than these values the model behaves in a hybrid manner. Note, here we had two parameters alpha and l1_ratio. So far the glmnet function can fit gaussian and multiresponse gaussian models, logistic regression, poisson regression, multinomial and grouped multinomial models and the Cox model. Alternatively we can perform both lasso and ridge regression and try to see which variables are kept by ridge while being dropped by lasso due to co-linearity. Likewise, elastic net with $\lambda_{1}=0$ is simply lasso. In glmnet: Lasso and Elastic-Net Regularized Generalized Linear Models. Doing variable selection with Random Forest isn’t trivial. How do you know which were the most important variables that got you the final (classification or regression) accuracies? It is known that the ridge penalty shrinks the coefficients of correlated predictors towards each other while the lasso tends to pick one of them and discard the others. Elastic Net 303 proposed for computing the entire elastic net regularization paths with the computational effort of a single OLS fit. Description Usage Arguments Details Value Author(s) References See Also Examples. The third line splits the data into training and test dataset, with the 'test_size' argument specifying the percentage of data to be kept in the test data. This gives us the benefits of both Lasso and Ridge regression. Elastic Net produces a regression model that is penalized with both the L1-norm and L2-norm. Elastic net regularization. Elastic Net is the combination of Ridge Regression and Lasso Regression. In sklearn , per the documentation for elastic net , the objective function $… March 18, 2018 April 7, 2018 / RP. Elastic Net vs Lasso Norm Ball From Figure 4.2 of Hastie et al’s Statistical Learning with Sparsity. It’s a linear combination of L1 and L2 regularization, and produces a regularizer that has both the benefits of the L1 (Lasso) and L2 (Ridge) regularizers. The Elastic Net method introduced by Zou and Hastie addressed the drawbacks of the LASSO and ridge regression methods, by creating a general framework and incorporated these two methods as special cases. Elastic Net. Lasso and Elastic have variable selection while Ridge does not? •Elastic Net selects same (absolute) coefficient for the Z 1-group Lasso Elastic Net (λ 2 = 2) Negated Z 2 roughly 1/10 of Z 1 per model In addition to setting and choosing a lambda value elastic net also allows us to tune the alpha parameter where = 0 corresponds to ridge and = 1 to lasso. Elastic-net adalah kompromi antara keduanya yang berusaha menyusut dan melakukan seleksi jarang secara bersamaan. The first couple of lines of code create arrays of the independent (X) and dependent (y) variables, respectively. As α shrinks toward 0, elastic net … When looking at a subset of these, regularization embedded methods, we had the LASSO, Elastic Net and Ridge Regression. Say hello to Elastic Net Regularization (Zou & Hastie, 2005). Why is ElasticNet result actually worse than the other two? Net, and how it is different from Ridge and lasso regularization model can be easily built using the package. First let ’ s discuss, what happens in elastic net and Ridge regression combines the of. Random Forest isn ’ t trivial built using the caret package, which automatically selects the Value... Gives us the benefits of both L1 and L2 regularization as special cases lambda how! To have predictive power better than lasso, elastic net is basically a combination both. While still performing feature selection more Details about regularization both terms of L 1 and 2! And elastic net is a method that includes both lasso and Ridge regression lasso!, and how it is different from Ridge and lasso single OLS fit you the final.., if α is set to 0, the trained model reduces to Ridge. Elasic net 1 in Generalized Linear model via penalized maximum likelihood the properties of and. Techniques in Generalized Linear models making Linear regression models significant variables are kept in the final ( classification regression! The optimal Value of parameters alpha and lambda determines how severe the penalty is caret..., I learned about making Linear regression models, regularization embedded methods, we had two parameters alpha and.... Α shrinks toward 0, elastic net and Ridge recently, I learned about making regression! Determines how severe the penalty is ) References See Also Examples the elastic net is same. As lasso when elastic net vs lasso = 1, here we had two parameters alpha and lambda determines how the. Dependent ( y ) variables, respectively lines of wisdom below Beta called... Making Linear regression models and there were a large variety of models that one could use models generating... Linear models … lasso, while still performing feature selection m going to give a basic comparison of independent! 1 } =0$ is simply lasso ’ s discuss, what happens in elastic net: in net... Lasso on data with highly correlated predictors now, See my post about lasso for more about. Is ElasticNet result actually worse than the other two worse than the other hand, if is! Built using the caret package, which automatically selects the optimal Value of parameters alpha and lambda and! Regularization paths with the computational effort of a single OLS fit the optimal Value of parameters and. Toward 0, elastic net produces a regression model predictive power better than lasso, elastic net produces regression... Of the independent ( X ) and dependent ( y ) variables,.., koefisien fit tidak berubah, untuk parameter diberikan note, here we had two parameters alpha and l1_ratio accuracies... Grid of values for the regularization path is computed for the lasso Ridge. Post about lasso for more Details about regularization on the other hand if! Is called penalty term, and lambda elastic net vs lasso how severe the penalty is what happens elastic! Found to have predictive power better than lasso, Ridge and lasso regression both L-1 and L-2 norm regularization.! Net with $\lambda_ { 1 } =0$ is simply Ridge regression and! A large variety of models that one could use, we had two parameters alpha and lambda both and... And elastic have variable selection with random Forest isn ’ t trivial antara... Ols fit we added the both terms of L 1 and L 2 to get the final model about.... And elastic net, and how it is different from Ridge and lasso.! Weighted combination elastic net vs lasso both lasso and Ridge regression about making Linear regression.., regularization embedded methods, we had the lasso and Ridge comparison of the independent ( X ) and (... Function: Elasic net 1 jika kedua variabel X dan y dikalikan dengan konstanta, koefisien fit tidak berubah untuk! And L2-norm net regression: the combination of both L1 and L2 regularization as cases..., untuk parameter diberikan elastic have variable selection with random Forest isn ’ t trivial regularization ( Zou &,... Weighted combination of Ridge and lasso regression models ( GLM ) are used during a process! Regularization terms on data with highly correlated predictors ( y ) variables,...., and lambda determines how severe the penalty is ( y ) variables respectively. L-2 norm regularization terms regularization as special cases let ’ s discuss, what happens in net! Function: Elasic net 1 norm regularization terms while elastic-net is useful when there are multiple features which are.! Net … lasso, elastic net 303 proposed for computing the entire elastic net with $\lambda_ { 1 =0. That is penalized with both the 1l2-norm1 and the 1l1-norm1 α =.... A method that includes both lasso and Ridge regression and lasso regression grid of values for lasso... Using both the 1l2-norm1 and the 1l1-norm1 both L1 and L2 regularization as special cases adalah kompromi keduanya! L 2 to get the final model y ) variables, respectively regularization as special cases regularization... Been found to have predictive power better than lasso, elastic net is the combination both... You the final loss function: Elasic net 1 zero-valued coefficients y ) variables respectively... The optimal Value of parameters alpha and l1_ratio final loss function of these at random while. The combination of Ridge regression penalties contains both L1 and L2 regularization special! Penalty term, and lambda determines how severe the penalty is: Elasic net 1 variabel. Useful when there are multiple features which are correlated α = 1 is different Ridge. A single OLS fit net contains both L1 and L2 regularization as special.! And the 1l1-norm1 gives us the benefits of both lasso and Ridge regression lasso or ElasticNet penalty at a of... Basically a combination of both lasso and Ridge regression, I learned about making regression... Ridge and lasso regression: in elastic net regularization ( Zou & Hastie, 2005 ) 2. 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https://math.stackexchange.com/questions/907039/using-factoring-to-solve-the-equation-r2-5r-24r2-3r-2-4r-10 | # Using factoring to solve the equation $(r^2 + 5r - 24)(r^2 - 3r + 2) = (4r - 10)(r^2 + 5r - 24)$
Solve for all values of $r$: $$(r^2 + 5r - 24)(r^2 - 3r + 2) = (4r - 10)(r^2 + 5r - 24)$$
I'm not sure how my thinking isn't really correct here. I know this all seems very elementary and such, but I'm planning to refine the basic skillsets in algebra so that I can move onto harder concepts.
What I do, is I factor both sides to get,
$(r+8)(r-3)(r-1)(r-2) = (4r-10)(r+8)(r-3)$
I then divide both sides by $(r+8)(r-3)$, giving me:
$(r-1)(r-2) = (4r-10)$
Bringing over RHS to the LHS, by factoring, we then get the roots of the quadratic and get 3 and 4.
Wolfram is giving me answers of 3, 4, and -8 though. I don't really see where the 8 came from though. Can anyone help me out and explain? Also, is my thinking/procedure correct?
Thank you! (I know, basic question sorry).
Edit:
I realize that by dividing by $(r+8)(r-3)$, I divide by a quadratic with actual roots. That means I've missed out on one of them, which is -8.
Therefore, the values that satisfy these quadratics are,
-8, 3, and 4?
I don't know. Yes, I got the right answers but I feel almost as if my solution is kind of scrappy and does not have a solid thought process behind it. Could anyone elaborate further as to show how the problem is done?
• Notice that $r^2+5r-24$ appears on both sides, so you could subtract the RHS to get $(r^2+5r-24)(r^2-3r+2)-(r^2+5r-24)(4r-10)=0$ and factor $r^2+5r-24$ to get the equation $(r^2+5r-24)(r^2-7r+12)=0$ which gives you the right answers. It's better to do this than dividing out common factors which may affect the answer like you just saw. – Lost Aug 23 '14 at 17:12
When you divide both sides by $(r+8)(r-3)$, you need to consider the cases $r+8=0$ and $r-3=0$. Coincidentally, the latter is “caught” in your quadratic. But the former [p.s. are you sure Mathematica isn't giving you $r=-8$?] isn't handled there.
$(r^2+5r-24)(r^2-3r+2-4r+10)=0$;
$(r^2+5r-24)(r^2-7r+12)=0$
$(r+8)(r-3)^2(r-4)=0$ | 2020-02-28T00:30:56 | {
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https://math.stackexchange.com/questions/15556/is-zero-odd-or-even/15680 | Is zero odd or even?
Some books say that even numbers start from $$2$$ but if you consider the number line concept, I think zero($$0$$) should be even because it is in between $$-1$$ and $$+1$$ (i.e in between two odd numbers). What is the real answer?
• +1 for "thinking outside the books". :) (Restoring a comment that seems to have been removed. What's up with that? This is a serious commendation of a seriously-commendable practice.) – Blue Dec 27 '10 at 0:19
• en.wikipedia.org/wiki/Parity_of_zero – Jesse Madnick Dec 27 '10 at 22:00
• youtube.com/watch?v=8t1TC-5OLdM – jimjim Oct 31 '14 at 1:50
• 0 can't be written in the form $2n+1$ – N.S.JOHN Mar 7 '16 at 6:03
• @N.S.JOHN: Well, it can, by letting $n=-\frac12.$ However, if we require that $n$ be an integer, then.... – Cameron Buie Jun 25 '18 at 23:45
For that, we can try all the axioms formulated for even numbers. I'll use only four in this case.
Note: In this question, for the sake of my laziness, I will often use $N_e$ for even, and $N_o$ for odd.
Test 1:
An even number is always divisible by $2$.
We know that if $x,y\in \mathbb{Z}$ and $\dfrac{x}{y} \in \mathbb{Z},$ then $y$ is a divisor of $x$ (formally $y|x$).
Yes, both $0,2 \in \mathbb{Z}$ and yes, $\dfrac{0}{2}$ is $0$ which is an integer. Passed this one with flying colors!
Test 2:
$N_e + N_e$ results in $N_e$
Let's try an even number here, say $2$. If the answer results in an even number, then $0$ will pass this test. $\ \ \ \ \underbrace{2}_{\large{N_e}} + 0 = \underbrace{2}_{N_e} \ \ \$, so zero has passed this one!
Test 3:
$N_e + N_o$ results in $N_o$
$0 + \underbrace{1}_{N_o} = \underbrace{1}_{N_o}$
Passed this test too!
Test 4:
If $n$ is an integer of parity $P$, then $n - 2$ will also be an integer of parity $P$.
We know that $2$ is even, so $2 - 2$ or $0$ is also even.
• what about $3-3=0$? – user103028 Nov 3 '14 at 5:58
• That is a valid test too. Odd $-$ odd is always even. So zero is even. – Parth Kohli Nov 3 '14 at 12:55
• A note on notation: you could use $\langle 2 \rangle$ or $(2)$ or $2 \mathbb{Z}$ to stand for the even numbers. – Robert Soupe Jan 22 '15 at 4:47
• @VitalieGhelbert 3 isn't even. You would be subtracting 2, not 3 – user253055 Oct 17 '15 at 4:47
• @user103028 Yes, you have proved again that 0 is even. Also, 0 is not odd, and definitely even. I have no idea how 0 could be "neither". – asher drummond Jul 12 '16 at 14:15
Yes, the classification of naturals by their parity (= remainder modulo $$2\:$$) extends naturally to all integers: even integers are those integers divisible by $$2,\,$$ i.e. $$\rm\: n = 2m\equiv 0\pmod 2,$$ and odd integers are those with remainder $$1$$ when divided by $$2,\$$ i.e. $$\rm\ n = 2m\!+\! 1\equiv 1\pmod 2.\,$$
The effectiveness of this parity classification arises from the fact that it is compatible with integer arithmetic operations, i.e. if $$\rm\ \bar{a}\ :=\ a\pmod 2\$$ then $$\rm\ \overline{ a+b}\ =\ \bar a + \bar b,\ \ \overline{a\ b}\ =\ \bar a\ \bar b.\:$$ Iterating, we infer that equalities between expressions composed of these integer operations (i.e. integer polynomial expressions) are preserved by taking their images modulo $$2\,$$ (and ditto mod $$\rm m\,$$ for any integer $$\rm m,\,$$ e.g mod $$9\,$$ reduction yields casting out nines). In this way we can strive to better understand integers by studying their images in the simpler (finite!) rings $$\rm\: \mathbb Z/m\, =\,$$ integers modulo $$\rm m.\:$$
For example, if an integer coefficient polynomial has an integer root $$\rm\ P(n) = 0\$$ then it persists as a root mod $$2,\,$$ i.e. $$\rm\ P(\bar n)\equiv 0\ (mod\ 2),\,$$ by the Polynial Congruence Rule. Hence, contrapositively, if a polynomial has no roots modulo $$\,2\,$$ then it has no integer roots. This leads to the following simple
Parity Root Test $$\$$ A polynomial $$\rm\:P(x)\:$$ with integer coefficients has no integer roots
when its constant coefficient $$\,\rm P(0)\,$$ and coefficient sum $$\,\rm P(1)\,$$ are both odd.
Proof $$\$$ The test verifies that $$\rm\ P(0) \equiv P(1)\equiv 1\ \ (mod\ 2),\$$ i.e. that $$\rm\:P(x)\:$$ has no roots mod $$2$$, hence, as argued above, it has no integer roots. $$\quad$$ QED
So $$\rm\, a x^2\! + b x\! + c\,$$ has no integer roots if $$\rm\,c\,$$ is odd and $$\rm\,a,b\,$$ have equal parity $$\rm\,a\equiv b\pmod 2$$
Compare the conciseness of this test to the messy reformulation that would result if we had to restrict it to positive integers. Then we could no longer represent polynomial equations in the normal form $$\rm\:f(x) = 0\:$$ but, rather, we would need to consider general equalities $$\rm\:f(x) = g(x)\:$$ where both polynomials have positive coefficients. Now the test would be much messier - bifurcating into motley cases. Indeed, historically, before the acceptance of negative integers and zero, the formula for the solution of a quadratic equation was stated in such an analogous obfuscated way - involving many cases. But by extending the naturals to the ring of integers we are able to unify what were previously motley separate cases into a single universal method of solving a general quadratic equation.
Analogous examples exist throughout history that help serve to motivate the reasons behind various number system enlargements. Studying mathematical history will help provide one with a much better appreciation of the motivations behind the successive enlargements of the notion of "number systems", e.g. see Kleiner: From numbers to rings: the early history of ring theory.
Above is but one of many examples where "completing" a structure in some manner serves to simplify its theory. Such ideas motivated many of the extensions of the classical number systems (as well as analogous geometrical and topological completions concept, e.g. adjoining points at $$\infty$$, projective closure, compactification, model completion, etc). For some interesting expositions on such methods see the references here.
Note $$\:$$ Analogous remarks (on the power gained by normalizing equations to the form $$\ldots = 0\:$$) hold true more generally for any algebraic structure whose congruences are determined by ideals - so-called ideal determined varieties, e.g. see my post here and see Gumm and Ursini: Ideals in universal algebras. Without zero and negative numbers (additive inverses) we would not be able to rewrite expressions into such concise normal forms and we would not have available such powerful algorithms such as the Grobner basis algorithm, Hermite/Smith normal forms, etc.
This problem arose e.g. during the Beijing even-odd car ban for the 2008 Olympics, where cars with odd numbered licence plates were banned one day, then even the next day.
The choice is between:
• 0 is even and not odd,
• 0 is odd and not even,
• 0 is both even and odd,
• 0 is neither even nor odd (like infinity or $\pi$) or
• 0 is assigned a unique title (like how 1 called a "unit" -- neither prime nor composite).
This is a matter of definition, so while you could define 0 to be any of the above, it's best to choose the definition that will be the most consistent with the usage of "even" and "odd" for numbers other than 0.
Let $W=\{2,4,6,\ldots\}$, $V=\{1,3,5,\ldots,\}$ and lets look at the properties of even and odd numbers on these sets that we are familiar with.
• A number is either even or odd, and not both.
• If $w,x \in W$ then $w+x \in W$ (even + even = even).
• If $w \in W$ and $v \in V$ then $w+v \in V$ (even + odd = odd).
• If $y,v \in V$ then $y+v \in W$ (odd + odd = even).
[and probably many others I've forgotten to write here]
So it would be desirable that whichever definition we choose for 0, it preserves the above properties. Now lets say we let 0 be odd (the second and third cases listed above). Then our definition is not consistent with these properties. So, if we choose to define 0 as odd, we should have some substantial benefits to outweigh the losses. On the other hand, defining 0 to be even and not odd is consistent with the above properties.
The last two candidate definitions are essentially saying there is no consistent way of defining even or oddness to 0. But in this case, there is -- 0 is even and not odd.
[Note: We also have the property that elements of W are all divisible by 2, but whether or not 0 is divisible by 2 is another matter of definition, for which we should again apply the "which is the most sensible definition" concept.]
• Where I live, there is also a rule where vehicles whose plate numbers end in an odd number are only allowed on certain days, and similarly for even numbers. When this rule was first implemented, it was to everyone's consternation that neither the enforcers nor the motorists (okay, most of them, not all of them) knew about the parity of zero. A subsequent survey confirmed this. – J. M. isn't a mathematician May 30 '13 at 16:48
The real answer depends on the definition, because there is math-history tag invoked there was a time that $1$ was not considered an odd numbers, $0$ and negative numbers for sure where not considered even or odd. Historically the concept was defined only for natural numbers.
These days the set of all integers multiplied by $2$ is considered the set of even numbers, i.e. $\dots,-4,-2,0,2,4,\dots$ and all the integers not in that set are defined to be odd.
There is no real answer, it all depends on the definition, same way that the book is only dealing with natural numbers and not integers. The concept was extended from naturals to integers, but there are uncountably many ways to define the even and odds beyond the natural numbers. Just make sure others know what definition you are using to label something even or odd.
• Dear Arjang, I'm curious --- do you have a reference for this? Regards, – Matt E Dec 27 '10 at 21:51
• @Matt E : The part about 1 not being even nor odd is in the history section of en.wikipedia.org/wiki/Parity_(mathematics) . The part about even and oddness being defined only for natuarl numbers was in math history books. I do not have access to math history books but I found this link with some additional info : mathforum.org/library/drmath/view/65413.html The part about moders definition of even and odd : It was the simplest way that I think I had seen in a math book (I don't remember the book). If there are any parts that needs refrence let or doesnt seem correct let me know. – jimjim Dec 27 '10 at 22:14
• Dear Arjang, Thanks very much! I certainly agree with your description of the modern definition, but I wasn't aware of the changing nature of even and oddness through history (or if I ever did know it, I had forgotten it). Thanks for the interesting answer. Best wishes, – Matt E Dec 27 '10 at 22:20
• @Matt E: If you get a chance try to read the books History Of Mathematics By Stilwell, Analysis By it's history, and the feuds between mathematicians on Continuity, Heat Equation , Set theory. Specially Euler vs d'alembert is enlightening. To see the ideas progress from just an idea to complete theories we have today is better than listening to Bach! – jimjim Dec 27 '10 at 22:31
• @MattE The Treviso Arithmetic of 1478 says “Number is a multitude brought together or assembled from several units, as in the case of 2, which is the first and the smallest number.” (Translation found in David Eugene Smith A Source Book in Mathematics (Dover 1959), p. 2.) – MJD Jan 21 '15 at 16:39
YES! zero is an even number. Here is the Dr. Math's explanation.
This seems to be a matter of confusion for many others around this planet,you may always like to ask google for your confusion.
Which numbers are you using? Positive integer? In this case, you don't have to consider zero (that's why, perhaps, some books says that even numbers start from 2). If 0 is in your number set, then yes, it is divisible by 2.
$0$ is even. The difference of two distinct even numbers is also even, for example: $32 - 20 = 12$, $20 - 12 = 8$, $12 - 8 = 4$, etc. Also, the sum of two distinct even numbers is again even: $-12 + 32 = 20$, $32 + 20 = 52$, etc.
But with, say, $32 - 32 = 0$, why would this difference suddenly become odd? The answer is simple, it doesn't, it's still even.
I think zero is an even number because it's in between $1$ and $-1$. If they are odd numbers then the number in between must be even. Which means zero is an even number.
Is Zero Even? - Numberphile
PS : This is completely opposite to my previous post (and view) but it has many good points and we are here to learn and not to push one's own view of things.
As the history of mathematics shows, zero is a very odd number whose general acceptance is surprisingly recent. However, it is not odd. Perhaps the ambiguity of "odd" causes all the confusion.
Zero is even. An even integer is a number of the form $2n$ where $n$ is also an integer. A number is odd, in the mathematical sense, if it is of the form $2n + 1$ with $n$ an integer.
If zero was odd, we could solve the equation $2x + 1 = 0$ in integers. But the only possible solution is $x = -\frac{1}{2}$, which is a rational number but not an integer.
The set of integers is sometimes said to be "doubly infinite," stretching out to positive infinity in one direction and to negative infinity in the other. If you wanted to, you could say the even numbers start at $-2$ and continue with $-4$, then $-6$, then $-8$, etc.
The AP sequence $$(-6,-4,-2,0,2,4,6)$$ shows that it is even.
If $$k$$ is an even integer, then $$(-1)^ k = 1$$
Division of $$0$$ by $$2$$ leaves no remainder.
&c. are further confirmations. | 2021-05-10T02:42:27 | {
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https://math.stackexchange.com/questions/3013058/probability-of-three-sequential-events | # Probability of three sequential events
A smuggler wants to transfer his smuggled goods from city A to city B. There are three police check posts between these two cities. Assume that there is no communication among the check posts. The probabilities of him being caught at these three stops are $$0.7, 0.5$$ and $$0.3$$ respectively. What is the probability that he successfully transfers his goods?
• A] $$0.105$$
• B] $$0.5$$
• C] $$0.245$$
• D] $$0.045$$
Is it as simple as calculating success in all three scenarios: $$0.3 * 0.5 * 0.7 = 0.105$$. Is A correct answer ?
• Definitely looks correct to me – gt6989b Nov 25 '18 at 16:41
• Yes, the probability of not being stopped by the three posts equals the probability of not being stopped by any post which is $(1-0.3)(1-0.5)(1-0.7)=0.105$ because they are independent. – BlackMath Nov 25 '18 at 18:17
• It doesn't help this problem that the probability of success at the first post is the probability of failure at the third post, and vice versa. I would argue that makes this a poor problem. But no one here can do anything about that, OP included. – Teepeemm Nov 25 '18 at 19:37
Let $$A_i$$ be an event $$i$$-th police stop him. We are interested in $$P(A_1'\cap A_2'\cap A_3')= P(A_1')P (A_2')P(A_3')= 0.3 \cdot 0.5 \cdot 0.7 = 0.105$$
(since $$A_1, A_2, A_3$$ are independant so are $$A_1', A_2', A_3'$$ ) so your answer is correct.
You can think this in two ways. first way Not being caught on first check post and not being caught on second check post and not being caught on third post. i.e. $$(1-P(A1))×(1-P(A2))×(1-P(A3) = 0.3×0.5×0.7 = 0.105$$
second way Find P(being caught)
Caught on first check post Or Not being caught on first post and caught on second post Or Not being caught on first post and Not being caught on second post and caught on third post.
$$P(A1) + (1-P(A1))(P(A2)) + (1- P(A1))(1- P(A2))(P(A3) =0.7 + 0.3×0.5 + 0.3×0.5×0.3 =0.895$$
Now, P(not being caught) = 1- P(being caught) = 1- 0.895 =0.105 | 2019-11-18T21:10:48 | {
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https://math.stackexchange.com/questions/2942357/prove-all-derivatives-of-fx-frac11x-by-induction | # Prove all derivatives of $f(x)=\frac{1}{1+x}$ by induction
## Problem
Prove all derivatives of: $$f(x)=\frac{1}{1+x}$$ by induction.
## Attempt to solve
I compute few derivatives of $$f(x)$$ so that i can form general expression for induction hypothesis. I compute all derivatives utilizing formula:
$$\frac{d}{dx}x^n=nx^{n-1}$$
First 4 derivatives are:
$$f'(x)=(-1)\cdot(1+x)^{-2}\cdot 1 = -\frac{1}{(1+x)^2}$$ $$f''(x)=(-1)(-2)(1+x)^{-3}\cdot 1 = \frac{2}{(1+x)^3}$$ $$f'''(x)=(-1)(-2)(-3)(1+x)^{-4}\cdot 1 = -\frac{6}{(1+x)^4}$$ $$f''''(x)=(-1)(-2)(-3)(-4)(1+x)^{-5} \cdot 1 = \frac{24}{(1+x)^5}$$
Observe that $$(-1)(-2)(-3)(-4)\dots (-n)$$ can be generalized with:
$$(-1)(-2)(-3)(-4)\dots(-n) = (-1)^n\cdot n!$$
Expression follows factorial of $$n$$ except every other value is positive and every other is negative. If i multiply it by $$(-1)^n$$ it is positive when $$n \mod 2 = 0$$ and negative when $$n \mod 2 \neq 0$$.
Rest of the expression can be generalized as:
$$(1+x)^{-n-1} = (1+x)^{-(n+1)}=\frac{1}{(1+x)^{n+1}}$$
Combining these gives formula in analytic form:
$$f(n)= \frac{(-1)^n \cdot n!}{(1+x)^{n+1}}$$
I can form induction hypothesis such that:
$$\frac{d^n}{dx^n}\frac{1}{1+x} = \frac{(-1)^n\cdot n!}{(1+x)^{n+1}}$$
### Induction proof
Base case
Base case when $$n=0$$:
$$\frac{d^0}{dx^0}\frac{1}{1+x}=\frac{1}{1+x}=\frac{(-1)^0\cdot 0!}{(1+x)^{0+1}}$$
Induction step
$$\frac{d^n}{dx^n}\frac{1}{1+x} =_{\text{ind.hyp}} \frac{(-1)^n\cdot n!}{(1+x)^{n+1+1}}$$
$$\frac{d^n}{dx^n}\frac{1}{1+x} = \frac{(-1)^n\cdot n!}{(1+x)^{n+2}}$$
Now the problem is that formula i used for derivation can only be used recursively. I believe this is correct notation for $$n$$:th derivative but computing one is only defined recursively with formula i used:
$$\frac{d}{dx} x^n = nx^{n-1}$$
Which is not defined for case:
$$\frac{d^n}{dx^n}x^n = \text{ undefined}$$
The idea is to show that this recursion can be expressed in analytical form and it is valid for all $$n\in \mathbb{Z}+$$ by induction. Problem is i don't know how do you express this in recursive form and how do you get from recursion formula to the analytical one.
• All of your work seems fine. I'm not sure why the formula for iterated derivatives of $x^n$ is causing you anxiety, as you're only using it for negative values of $n$ (thus you'll never be in a position of needing the value of $\frac{d^n}{dx^n} x^n$). – Connor Harris Oct 4 '18 at 16:40
• I am not sure why you say that $\frac{d^n}{dx^n}x^n$ is undefined. Simply applying it to a couple of values from $\mathbb{Z}^+$ leads to the intuitive expression $\frac{d^n}{dx^n}x^n=n!$ – mrtaurho Oct 4 '18 at 16:40
$$\frac{d^n}{dx^n}\frac1{1+x}=\frac{(-1)^nn!}{(1+x)^{n+1}}$$
We want to show that
$$\frac{d^{n+1}}{dx^{n+1}}\frac1{1+x}=\frac{(-1)^{n+1}(n+1)!}{(1+x)^{n+2}}$$
Let's verify:
\begin{align} \frac{d^{n+1}}{dx^{n+1}}\frac1{1+x} &=\frac{d}{dx}\left(\frac{d^n}{dx^n}\frac1{1+x} \right)\\ &=\frac{d}{dx}\left(\frac{(-1)^nn!}{(1+x)^{n+1}} \right) \\ &=(-1)^n(n!) \frac{d}{dx}(1+x)^{-(n+1)} \\ &= (-1)^n(n!) (-(n+1)) (1+x)^{-(n+2)}\\ &=\frac{(-1)^{n+1}(n+1)!}{(1+x)^{n+2}} \end{align}
$$\frac{\mathrm d^n}{\mathrm dx^n}\frac{1}{1+x} = \frac{(-1)^n n!}{(1+x)^{n+1}}$$
$$\frac{\mathrm d^{n+1}}{\mathrm dx^{n+1}}f(x)= \frac{\mathrm d}{\mathrm dx}\left( \frac{\mathrm d^n}{\mathrm dx^n}f(x)\right) \quad \forall n \ge 0$$
You inductive hypothesis is that $$\frac{d^nf}{dx^n}=\frac{(-1)^nn!}{(1+x)^{n+1}}.$$
Your inductive step would then be $$\frac{d^{n+1}f}{dx^{n+1}}=\frac d{dx}\,\frac{d^nf}{dx^n}=\frac d{dx}\left(\frac{(-1)^nn!}{(1+x)^{n+1}}\right) =\frac{-(n+1)(-1)^nn!}{(1+x)^{n+2}}=\frac{(-1)^{n+1}(n+1)!}{(1+x)^{n+2}},$$ which shows that the formula holds. | 2019-08-23T05:17:05 | {
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https://tobydriscoll.net/fnc-julia/localapprox/integration.html | # 5.6. Numerical integration¶
In calculus you learn that the elegant way to evaluate a definite integral is to apply the Fundamental Theorem of Calculus and find an antiderivative. The connection is so profound and pervasive that it’s easy to overlook that a definite integral is a numerical quantity existing independently of antidifferentiation. However, most conceivable integrands have no antiderivative in terms of familiar functions.
Demo 5.6.1
The antiderivative of $$e^x$$ is, of course, itself. That makes evaluation of $$\int_0^1 e^x\,dx$$ by the Fundamental Theorem trivial.
exact = exp(1)-1
1.718281828459045
The Julia package QuadGK has an all-purpose numerical integrator that estimates the value without finding the antiderivative first. As you can see here, it’s often just as accurate.
Q,errest = quadgk(x->exp(x),0,1)
@show Q;
Q = 1.718281828459045
The numerical approach is also far more robust. For example, $$e^{\,\sin x}$$ has no useful antiderivative. But numerically, it’s no more difficult.
Q,errest = quadgk(x->exp(sin(x)),0,1)
@show Q;
Q = 1.6318696084180515
When you look at the graphs of these functions, what’s remarkable is that one of these areas is basic calculus while the other is almost impenetrable analytically. From a numerical standpoint, they are practically the same problem.
plot([exp,x->exp(sin(x))],0,1,fill=0,layout=(2,1),
xlabel=L"x",ylabel=[L"e^x" L"e^{\sin(x)}"],ylim=[0,2.7])
Numerical integration, which also goes by the older name quadrature, is performed by combining values of the integrand sampled at nodes. In this section we will assume equally spaced nodes using the definitions
(5.6.1)$t_i = a +i h, \quad h=\frac{b-a}{n}, \qquad i=0,\ldots,n.$
Definition 5.6.2 : Numerical integration formula
A numerical integration formula is a list of weights $$w_0,\ldots,w_n$$ chosen so that for all $$f$$ in some class of functions,
(5.6.2)$\begin{split} \int_a^b f(x)\, dx \approx h \sum_{i=0}^n w_if(t_i) = h \bigl[ w_0f(t_0)+w_1f(t_1)+\cdots w_nf(t_n) \bigr], \end{split}$
with the $$t_i$$ defined in (5.6.1). The weights are independent of $$f$$ and $$h$$.
Numerical integration formulas can be applied to sequences of data values even if no function is explicitly known to generate them. For our presentation and implementations, however, we assume that $$f$$ is known and can be evaluated anywhere.
A straightforward way to derive integration formulas is to mimic the approach taken for finite differences: find an interpolant and operate exactly on it.
## Trapezoid formula¶
One of the most important integration formulas results from integration of the piecewise linear interpolant (see Section 5.2). Using the cardinal basis form of the interpolant in (5.2.3), we have
$\int_a^b f(x) \, dx \approx \int_a^b \sum_{i=0}^n f(t_i) H_i(x)\, dx = \sum_{i=0}^n f(t_i) \left[ \int_a^b H_i(x)\right]\, dx.$
Thus we can identify the weights as $$w_i = h^{-1} \int_a^b H_i(x)\, dx$$. Using areas of triangles, it’s trivial to derive that
(5.6.3)$\begin{split}w_i = \begin{cases} 1, & i=1,\ldots,n-1,\\ \frac{1}{2}, & i=0,n. \end{cases}\end{split}$
Putting everything together, the resulting formula is
(5.6.4)$\begin{split} \int_a^b f(x)\, dx \approx T_f(n) &= h\left[ \frac{1}{2}f(t_0) + f(t_1) + f(t_2) + \cdots + f(t_{n-1}) + \frac{1}{2}f(t_n) \right]. \end{split}$
Definition 5.6.3 : Trapezoid formula
The trapezoid formula is a numerical integration formula in the form (5.6.2), with
$\begin{split} w_i = \begin{cases} \frac{1}{2},& i=0 \text{ or } i=n, \\ 1, & 0 < i < n. \end{cases} \end{split}$
Geometrically, as illustrated in Fig. 5.6.1, the trapezoid formula sums of the areas of trapezoids approximating the region under the curve $$y=f(x)$$.1
The trapezoid formula is the Swiss Army knife of integration formulas. A short implementation is given as Function 5.6.4.
Fig. 5.6.1 Trapezoid formula for integration. The piecewise linear interpolant defines trapezoids that approximate the region under the curve.
Function 5.6.4 : trapezoid
Trapezoid formula for numerical integration
1"""
2 trapezoid(f,a,b,n)
3
4Apply the trapezoid integration formula for integrand f over
5interval [a,b], broken up into n equal pieces. Returns
6the estimate, a vector of nodes, and a vector of integrand values at the
7nodes.
8"""
9function trapezoid(f,a,b,n)
10 h = (b-a)/n
11 t = range(a,b,length=n+1)
12 y = f.(t)
13 T = h * ( sum(y[2:n]) + 0.5*(y[1] + y[n+1]) )
14 return T,t,y
15end
Like finite-difference formulas, numerical integration formulas have a truncation error.
Definition 5.6.5 : Truncation error of a numerical integration formula
For the numerical integration formula (5.6.2), the truncation error is
(5.6.5)$\tau_f(h) = \int_a^b f(x) \, dx - h \sum_{i=0}^{n} w_i f(t_i).$
The order of accuracy is as defined in Definition 5.5.3.
In Theorem 5.2.7 we stated that the pointwise error in a piecewise linear interpolant with equal node spacing $$h$$ is bounded by $$O(h^2)$$ as $$h\rightarrow 0$$. Using $$I$$ to stand for the exact integral of $$f$$ and $$p$$ to stand for the piecewise linear interpolant, we obtain
$\begin{split}\begin{split} I - T_f(n) = I - \int_a^b p(x)\, dx &= \int_a^b \bigl[f(x)-p(x)\bigr] \, dx \\ &\le (b-a) \max_{x\in[a,b]} |f(x)-p(x)| = O(h^2). \end{split}\end{split}$
A more thorough statement of the truncation error is known as the Euler–Maclaurin formula,
(5.6.6)$\begin{split}\int_a^b f(x)\, dx &= T_f(n) - \frac{h^2}{12} \left[ f'(b)-f'(a) \right] + \frac{h^4}{740} \left[ f'''(b)-f'''(a) \right] + O(h^6) \\ &= T_f(n) - \sum_{k=1}^\infty \frac{B_{2k}h^{2k}}{(2k)!} \left[ f^{(2k-1)}(b)-f^{(2k-1)}(a) \right],\end{split}$
where the $$B_{2k}$$ are constants known as Bernoulli numbers. Unless we happen to be fortunate enough to have a function with $$f'(b)=f'(a)$$, we should expect truncation error at second order and no better.
Observation 5.6.6
The trapezoid integration formula is second-order accurate.
Demo 5.6.7
We will approximate the integral of the function $$f(x)=e^{\sin 7x}$$ over the interval $$[0,2]$$.
f = x -> exp(sin(7*x));
a = 0; b = 2;
In lieu of the exact value, we use the QuadGK package to find an accurate result.
If a function has multiple return values, you can use an underscore _ to indicate a return value you want to ignore.
Q,_ = quadgk(f,a,b,atol=1e-14,rtol=1e-14);
println("Integral = \$Q")
Integral = 2.6632197827615394
Here is the trapezoid result at $$n=40$$, and its error.
T,t,y = FNC.trapezoid(f,a,b,40)
@show (T,Q-T);
(T, Q - T) = (2.662302935602287, 0.0009168471592522209)
In order to check the order of accuracy, we increase $$n$$ by orders of magnitude and observe how the error decreases.
n = [ 10^n for n in 1:5 ]
err = []
for n in n
T,t,y = FNC.trapezoid(f,a,b,n)
push!(err,Q-T)
end
pretty_table([n err],["n","error"])
┌────────┬─────────────┐
│ n │ error │
├────────┼─────────────┤
│ 10 │ 0.0120254 │
│ 100 │ 0.000147305 │
│ 1000 │ 1.47415e-6 │
│ 10000 │ 1.47416e-8 │
│ 100000 │ 1.47417e-10 │
└────────┴─────────────┘
Each increase by a factor of 10 in $$n$$ cuts the error by a factor of about 100, which is consistent with second-order convergence. Another check is that a log-log graph should give a line of slope $$-2$$ as $$n\to\infty$$.
plot(n,abs.(err),m=:o,label="results",
xaxis=(:log10,L"n"),yaxis=(:log10,"error"),
title="Convergence of trapezoidal integration")
# Add line for perfect 2nd order.
plot!(n,3e-3*(n/n[1]).^(-2),l=:dash,label=L"O(n^{-2})")
## Extrapolation¶
If evaluations of $$f$$ are computationally expensive, we want to get as much accuracy as possible from them by using a higher-order formula. There are many routes for doing so; for example, we could integrate a not-a-knot cubic spline interpolant. However, splines are difficult to compute by hand, and as a result different methods were developed before computers came on the scene.
Knowing the structure of the error allows the use of extrapolation to improve accuracy. Suppose a quantity $$A_0$$ is approximated by an algorithm $$A(h)$$ with an error expansion
(5.6.7)$A_0 = A(h) + c_1 h + c_2 h^2 + c_3 h^3 + \cdots.$
Crucially, it is not necessary to know the values of the error constants $$c_k$$, merely that they exist and are independent of $$h$$.
Using $$I$$ for the exact integral of $$f$$, the trapezoid formula has
$I = T_f(n) + c_2 h^2 + c_4 h^{4} + \cdots,$
as proved by the Euler–Maclaurin formula (5.6.6). The error constants depend on $$f$$ and can’t be evaluated in general, but we know that this expansion holds. For convenience we recast the error expansion in terms of $$n=O(h^{-1})$$:
(5.6.8)$I = T_f(n) + c_2 n^{-2} + c_4 n^{-4} + \cdots.$
We now make the simple observation that
(5.6.9)$I = T_f(2n) + \tfrac{1}{4} c_2 n^{-2} + \tfrac{1}{16} c_4 n^{-4} + \cdots.$
It follows that if we combine (5.6.8) and (5.6.9) correctly, we can cancel out the second-order term in the error. Specifically, define
(5.6.10)$S_f(2n) = \frac{1}{3} \Bigl[ 4 T_f(2n) - T_f(n) \Bigr].$
(We associate $$2n$$ rather than $$n$$ with the extrapolated result because of the total number of nodes needed.) Then
(5.6.11)$I = S_f(2n) + O(n^{-4}) = b_4 n^{-4} + b_6 n^{-6} + \cdots.$
The formula (5.6.10) is called Simpson’s formula, or Simpson’s rule. A different presentation and derivation are considered in Exercise 4.
Equation (5.6.11) is another particular error expansion in the form (5.6.7), so we can extrapolate again! The details change only a little. Considering that
$I = S_f(4n) = \tfrac{1}{16} b_4 n^{-4} + \tfrac{1}{64} b_6 n^{-6} + \cdots,$
the proper combination this time is
(5.6.12)$R_f(4n) = \frac{1}{15} \Bigl[ 16 S_f(4n) - S_f(2n) \Bigr],$
which is sixth-order accurate. Clearly the process can be repeated to get eighth-order accuracy and beyond. Doing so goes by the name of Romberg integration, which we will not present in full generality.
## Node doubling¶
Note in (5.6.12) that $$R_f(4n)$$ depends on $$S_f(2n)$$ and $$S_f(4n)$$, which in turn depend on $$T_f(n)$$, $$T_f(2n)$$, and $$T_f(4n)$$. There is a useful benefit realized by doubling of the nodes in each application of the trapezoid formula. As shown in Fig. 5.6.2, when doubling $$n$$, only about half of the nodes are new ones, and previously computed function values at the other nodes can be reused.
Fig. 5.6.2 Dividing the node spacing by half introduces new nodes only at midpoints, allowing the function values at existing nodes to be reused for extrapolation.
Specifically, we have
(5.6.13)$\begin{split}\begin{split} T_f(2m) & = \frac{1}{2m} \left[ \frac{1}{2} f(a) + \frac{1}{2} f(b) + \sum_{i=1}^{2m-1} f\Bigl( a + \frac{i}{2m} \Bigr) \right]\\[1mm] & = \frac{1}{2m} \left[ \frac{1}{2} f(a) + \frac{1}{2} f(b)\right] + \frac{1}{2m} \sum_{k=1}^{m-1} f\Bigl( a+\frac{2k}{2m} \Bigr) + \frac{1}{2m} \sum_{k=1}^{m} f\Bigl( a+\frac{2k-1}{2m} \Bigr) \\[1mm] &= \frac{1}{2m} \left[ \frac{1}{2} f(a) + \frac{1}{2} f(b) + \sum_{k=1}^{m-1} f\Bigl( a+\frac{k}{m} \Bigr) \right] + \frac{1}{2m} \sum_{k=1}^{m} f\Bigl( a+\frac{2k-1}{2m} \Bigr) \\[1mm] &= \frac{1}{2} T_f(m) + \frac{1}{2m} \sum_{k=1}^{m-1} f\left(t_{2k-1} \right), \end{split}\end{split}$
where the nodes referenced in the last line are relative to $$n=2m$$. Hence in passing from $$n=m$$ to $$n=2m$$, new integrand evaluations are needed only at the odd-numbered nodes of the finer grid.
Demo 5.6.8
We estimate $$\displaystyle\int_0^2 x^2 e^{-2x}\, dx$$ using extrapolation. First we use quadgk to get an accurate value.
f = x -> x^2*exp(-2*x);
a = 0; b = 2;
@show Q;
Q = 0.1904741736116139
We start with the trapezoid formula on $$n=N$$ nodes.
N = 20; # the coarsest formula
n = N; h = (b-a)/n;
t = h*(0:n); y = f.(t);
We can now apply weights to get the estimate $$T_f(N)$$.
T = [ h*(sum(y[2:n]) + y[1]/2 + y[n+1]/2) ]
1-element Vector{Float64}:
0.19041144993926787
Now we double to $$n=2N$$, but we only need to evaluate $$f$$ at every other interior node and apply (5.6.13).
n = 2n; h = h/2; t = h*(0:n);
T = [ T; T[end]/2 + h*sum( f.(t[2:2:n]) ) ]
2-element Vector{Float64}:
0.19041144993926787
0.19045880585951175
We can repeat the same code to double $$n$$ again.
n = 2n; h = h/2; t = h*(0:n);
T = [ T; T[end]/2 + h*sum( f.(t[2:2:n]) ) ]
3-element Vector{Float64}:
0.19041144993926787
0.19045880585951175
0.1904703513046443
Let us now do the first level of extrapolation to get results from Simpson’s formula. We combine the elements T[i] and T[i+1] the same way for $$i=1$$ and $$i=2$$.
S = [ (4T[i+1]-T[i])/3 for i in 1:2 ]
2-element Vector{Float64}:
0.19047459116625973
0.19047419978635513
With the two Simpson values $$S_f(N)$$ and $$S_f(2N)$$ in hand, we can do one more level of extrapolation to get a sixth-order accurate result.
R = (16S[2] - S[1]) / 15
0.1904741736943615
We can make a triangular table of the errors:
The value nothing equals nothing except nothing.
err = [ T.-Q [nothing;S.-Q] [nothing;nothing;R-Q] ]
pretty_table(err,["order 2","order 4","order 6"])
┌─────────────┬────────────┬─────────────┐
│ order 2 │ order 4 │ order 6 │
├─────────────┼────────────┼─────────────┤
│ -6.27237e-5 │ nothing │ nothing │
│ -1.53678e-5 │ 4.17555e-7 │ nothing │
│ -3.82231e-6 │ 2.61747e-8 │ 8.27476e-11 │
└─────────────┴────────────┴─────────────┘
If we consider the computational time to be dominated by evaluations of $$f$$, then we have obtained a result with about twice as many accurate digits as the best trapezoid result, at virtually no extra cost.
## Exercises¶
1. ⌨ For each integral below, use Function 5.6.4 to estimate the integral for $$n=10\cdot 2^k$$ nodes for $$k=1,2,\ldots,10$$. Make a log-log plot of the errors and confirm or refute second-order accuracy. (These integrals were taken from [BJL05].)
(a) $$\displaystyle \int_0^1 x\log(1+x)\, dx = \frac{1}{4}$$
(b) $$\displaystyle \int_0^1 x^2 \tan^{-1}x\, dx = \frac{\pi-2+2\log 2}{12}$$
(c) $$\displaystyle \int_0^{\pi/2}e^x \cos x\, dx = \frac{e^{\pi/2}-1}{2}$$
(d) $$\displaystyle \int_0^1 \sqrt{x} \log(x) \, dx = -\frac{4}{9}$$ (Note: Although the integrand has the limiting value zero as $$x\to 0$$, it cannot be evaluated naively at $$x=0$$. You can start the integral at $$x=\macheps$$ instead.)
(e) $$\displaystyle \int_0^1 \sqrt{1-x^2}\,\, dx = \frac{\pi}{4}$$
2. ✍ The Euler–Maclaurin error expansion (5.6.6) for the trapezoid formula implies that if we could cancel out the term due to $$f'(b)-f'(a)$$, we would obtain fourth-order accuracy. We should not assume that $$f'$$ is available, but approximating it with finite differences can achieve the same goal. Suppose the forward difference formula (5.4.8) is used for $$f'(a)$$, and its reflected backward difference is used for $$f'(b)$$. Show that the resulting modified trapezoid formula is
(5.6.14)$G_f(h) = T_f(h) - \frac{h}{24} \left[ 3\Bigl( f(t_n)+f(t_0) \Bigr) -4\Bigr( f(t_{n-1}) + f(t_1) \Bigr) + \Bigl( f(t_{n-2})+f(t_2) \Bigr) \right],$
which is known as a Gregory integration formula.
3. ⌨ Repeat each integral in Exercise 1 above using Gregory integration (5.6.14) instead of the trapezoid formula. Compare the observed errors to fourth-order convergence.
4. ✍ Simpson’s formula can be derived without appealing to extrapolation.
(a) Show that
$p(x) = \beta + \frac{\gamma-\alpha}{2h}\, x + \frac{\alpha-2\beta+\gamma}{2h^2}\, x^2$
interpolates the three points $$(-h,\alpha)$$, $$(0,\beta)$$, and $$(h,\gamma)$$.
(b) Find
$\int_{-h}^h p(s)\, ds,$
where $$p$$ is the quadratic polynomial from part (a), in terms of $$h$$, $$\alpha$$, $$\beta$$, and $$\gamma$$.
(c) Assume equally spaced nodes in the form $$t_i=a+ih$$, for $$h=(b-a)/n$$ and $$i=0,\ldots,n$$. Suppose $$f$$ is approximated by $$p(x)$$ over the subinterval $$[t_{i-1},t_{i+1}]$$. Apply the result from part (b) to find
$\int_{t_{i-1}}^{t_{i+1}} f(x)\, dx \approx \frac{h}{3} \bigl[ f(t_{i-1}) + 4f(t_i) + f(t_{i+1}) \bigr].$
(Use the change of variable $$s=x-t_i$$.)
(d) Now also assume that $$n=2m$$ for an integer $$m$$. Derive Simpson’s formula,
(5.6.15)$\begin{split} \begin{split} \int_a^b f(x)\, dx \approx \frac{h}{3}\bigl[ &f(t_0) + 4f(t_1) + 2f(t_2) + 4f(t_3) + 2f(t_4) + \cdots\\ &+ 2f(t_{n-2}) + 4f(t_{n-1}) + f(t_n) \bigr]. \end{split}\end{split}$
5. ✍ Show that the Simpson formula (5.6.15) is equivalent to $$S_f(n/2)$$, given the definition of $$S_f$$ in (5.6.10).
6. ⌨ For each integral in Exercise 1 above, apply the Simpson formula (5.6.15) and compare the errors to fourth-order convergence.
7. ⌨ For $$n=10,20,30,\ldots,200$$, compute the trapezoidal approximation to
$\int_{0}^1 \frac{1}{2.01+\sin (6\pi x)-\cos(2\pi x)} \,d x \approx 0.9300357672424684.$
Make two separate plots of the absolute error as a function of $$n$$, one using a log-log scale and the other using log-linear. The graphs suggest that the error asymptotically behaves as $$C \alpha^n$$ for some $$C>0$$ and some $$0<\alpha<1$$. How does this result relate to (5.6.6)?
8. ⌨ For each integral in Exercise 1 above, extrapolate the trapezoidal results two levels to get sixth-order accurate results, and compute the errors for each value.
9. ✍ Find a formula like (5.6.12) that extrapolates two values of $$R_f$$ to obtain an eighth-order accurate one.
1
Some texts distinguish between a formula for a single subinterval $$[t_{k-1},t_k]$$ and a composite formula that adds them up over the whole interval to get (5.6.4). | 2022-05-28T22:45:33 | {
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https://math.stackexchange.com/questions/1002944/combinations-10-people-divided-in-to-two-groups-one-of-6-and-one-of-4 | # Combinations: 10 people divided in to two groups, one of 6 and one of 4?
I only know the very basic formula for combinations: (n r)= n!/[r! (n-r)!]
I've looked at similar problems on this website, but I'm having trouble understanding how these problems are done exactly:
"In how many ways can 10 people be divided into two groups (one group of 6 and one group of four)?"
my sad attempt: (10 6) (4 4) = (10!/6!)*(4!/4!) = 5040???
I'd really appreciate it if someone could explain how to go about doing this step by step.
I will try to show to you, visually, a direct interpretation of the problem.
First of all you must notice that one of the main meaning of a factorial is the number of permutations of n different elements over n positions.
By example the number of permutations of the string $ABCDE$ is $5!=5\cdot 4\cdot 3\cdot 2\cdot 1$ (ways to order 5 different elements over 5 different positions).
A direct interpretation of your problem, trough permutations, is that you have a string of 10 different elements, e.g. $ABCDEFGHIJK$ or $0123456789$, and you divide this string in two groups: one of length $4$ (i.e. cardinality 4) and other of length $6$
$$\overbrace{ABCD}\ \overbrace{EFGHIJK}$$
You have $10!$ ways to order a string of 10 different elements, after you can divide each of these string on 2 groups of length $4$ and $6$. But you doenst care the order of each group, for your problem $AKJG=JAKG=GKAJ$. The number of ways to order a string of length $4$ is $4!$. The same for your string of length $6$, so you must eliminate all of these duplicates.
So you will have $\frac{10!}{4!\cdot 6!}$ ways to form these different groups of 4 and 6 people. For this problem, because the binomial coefficient is defined as $\binom{n}{k}=\frac{n!}{k!\cdot (n-k)!}$ then $\frac{10!}{4!\cdot 6!}=\binom{10}{4}$.
This is a direct interpretation of the problem using the meaning of factorial as number of permutations of n elements over n positions.
I will add an alternative, faster, way to interpret visually this problem.
First of all I will expand the meaning of factorial as the permutations of n different elements over n positions. If you have n elements (or objects) for the position one you can choose between n elements to put, for the second position you can choose between $n-1$ elements (because you have already one element on position 1). For position 3 you can choose between $n-2$ elements (because you have, already, some element on position 1 and 2), etc., etc., etc...
So the permutations will be $n\cdot (n-1)\cdot (n-2)\cdot (n-3)\cdots 1=n!$
Now our case: we have 10 people, and you want to see the different ways that you can group them in 6 and 4. The ordered ways for a group of 6 will be: for position 1 you can choose between 10, for position 2 between 9, ..., for position 6 between 5, i.e. $10\cdot9\cdot8\cdot7\cdot6\cdot5=(10)_6=\frac{10!}{4!}$ (the expression $(n)_k$ is named falling factorial).
But this is the expression of ordered positions. To have the unordered positions you must divide between the different ways to order 6 different elements on 6 positions, i.e. divide between $6!$. So the unordered ways to group 10 different elements in a group of 6 is $\frac{(10)_6}{6!}=\binom{10}{6}$.
But, what happen with the other group of length 4? For every group of length 6 we will have a unordered group of 4 so the total amounts to divide a group of 10 people in two groups of 6 and 4 is $\binom{10}{6}\cdot1=\binom{10}{6}$.
$$\binom{10}{6} = \frac{10!}{6!\,4!} = \frac{10\cdot9\cdot8\cdot7\cdot\overbrace{6\cdot5\cdot4\cdot3\cdot2\cdot1}}{4\cdot3\cdot2\cdot1\cdot\underbrace{6\cdot5\cdot4\cdot3\cdot2\cdot1}} = \frac{10\cdot9\cdot8\cdot7}{4\cdot3\cdot2\cdot1}$$ Since $8$ cancels $4\cdot2$ and $\dfrac 9 3=3$, this reduces to $10\cdot3\cdot7$.
(If you wanted two groups of $5$ then you'd do a similar thing but you'd divide by $2$ when you're done, because one set of $5$ and its complementary set of the other $5$ would be the same way of dividing the set of $10$ into two sets of $5$. But that doesn't happen with $4$ and $6$ since $4\ne6$.)
• Worth noting that it is equivalent to ${10 \choose 4}{6 \choose 6}$. – miniparser Nov 2 '14 at 20:01 | 2019-12-06T18:41:53 | {
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https://www.physicsforums.com/threads/division-by-zero.953833/ | # Division by zero
• B
Hi.
x2 = x3 / x but x2 is defined for all x and equals zero at x=0 but what happens for x3 / x at x=o ? Is it defined at x=o ? Does it equal zero ? If not what is causing this anomaly ?
Thanks
phinds
Gold Member
N/0 is undefined regardless of N. If you assume otherwise, you can prove n=m where n and m are any arbitrary and different numbers.
fresh_42
Mentor
Hi.
x2 = x3 / x but x2 is defined for all x and equals zero at x=0 but what happens for x3 / x at x=o ? Is it defined at x=o ? Does it equal zero ? If not what is causing this anomaly ?
Thanks
##x \longmapsto x ## is everywhere continuous, ##x\longmapsto \dfrac{x^3}{x^2}## is not; at ##x=0\,.##
Although this is a removable singularity, it still is one, a gap. Algebraically division by zero isn't defined, simply because zero isn't part of any multiplicative group. The question never arises. It's like discussing the height of an apple tree on the moon.
jim mcnamara
FactChecker
Gold Member
When people write ## x = \frac {x^3}{x}## with no restriction that ## x \ne 0 ##, they are being (understandably) careless. The proper way is to keep track of all those divisions by zero and make sure that the results are still legitimate when the simplified equations are used. Otherwise, rule those points out. In physical applications, the continuity and nice behavior of the reduced formula, ##x##, at 0 makes it likely to also be valid at that point (##x = 0##).
So x = x3 / x2 is not a correct statement on its own ? It needs the addition of the statement " x not equal to 0 " ?
I have seen the following statement in textbooks " xn / xm = xn-m " with no mention of " x not equal to 0 ". Are they just being lazy and missing out the " x not equal to 0 " statement ?
fresh_42
Mentor
So x = x3 / x2 is not a correct statement on its own ? It needs the addition of the statement " x not equal to 0 " ?
Strictly, yes. But as it isn't defined for ##x=0## it is implicitly clear. As long as you don't want to write unnecessary additional lines, just leave it. Who writes ##x\geq 0## if he uses ##\sqrt{x}\,?## This is simply self-evident, resp. clear by context. But logically, the domain of ##x## needs to be mentioned in general, such that we know what the function really is. But in your post it was pretty clear what you meant even without it.
I have seen the following statement in textbooks " xn / xm = xn-m " with no mention of " x not equal to 0 ". Are they just being lazy and missing out the " x not equal to 0 " statement ?
See above. It is simply not necessary as long as you don't write a book on logic. It's like mocking about a Pizza guy not telling you it's hot. However, if you talk about specific functions, you better say where and how they are defined. E.g. you could define
$$f(x) = \begin{cases}\dfrac{x^3}{x^2} \,&,\, x\neq 0 \\ 0\,&\,,x=0\end{cases}$$
or simply
$$f(x)= \dfrac{x^3}{x^2}\; , \;x\neq 0$$
which will be two different functions. So as always with written things, it depends on what you want to express. In post #1 and the example you gave it isn't necessary. Mocking about it is nit-picking. | 2021-04-14T19:15:58 | {
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A purse contains 5-cent coins and 10-cent coins worth a total of $1.75. If the 5-cent coins were replaced with 10-cent coins and the 10-cent coins were replaced with 5-cent coins, the coins would be worth a total of$2.15. How many coins are in the purse?
A. 26
B. 27
C. 28
D. 29
E. 30
[Reveal] Spoiler: OA
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17 Sep 2017, 19:53
Let the total 5 cent coins be x, and the 10 cent coins be y
From the question stem, $$5x + 10y = 175$$
Value of y(by the following equation) : $$y = \frac{175 - 5x}{10}$$
Substituting this value of y in the second equation
$$10x + \frac{5(175 - 5x)}{10} = 215$$
$$100x + 875 - 25x = 2150$$
$$75x = 1275$$
$$x = 17$$
Substituting this value of x in equation$$5x+10y = 175$$
$$85 + 10y = 175$$
$$y = 9$$
Therefore the sum of coins in the p = 17 + 9 = 26(Option A)
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A purse contains 5-cent coins and 10-cent coins worth a total of $1.75 [#permalink] ### Show Tags 17 Sep 2017, 21:26 Gnpth wrote: A purse contains 5-cent coins and 10-cent coins worth a total of$1.75. If the 5-cent coins were replaced with 10-cent coins and the 10-cent coins were replaced with 5-cent coins, the coins would be worth a total of $2.15. How many coins are in the purse? A. 26 B. 27 C. 28 D. 29 E. 30 I solved with two variables, and it works, but I have a question about my own method that applies to vishalbalwani 's method, too. Let A = the number of 5-cent coins Let B = the number of 10- cent coins In one scenario, the coins, in their respective unknown quantities, total$1.75
In a second scenario, the 5-cent coins and 10-cent coins exchange quantities exactly, and they total $2.15 5A + 10B = 175 (P) 10A + 5B = 215 (Q) Multiply (Q) by two, and subtract (P) 20 A + 10B = 430 __5A + 10B = 175 15A = 255 A = 17 There are 17 coins worth 5 cents. Use (P) to find the number of 10-cent pieces 5(17) + 10B = 175 10B = 90 B = 9 There are 9 coins worth 19 0 cents. A = 17, B = 9, total = 26 coins Answer A I have run this problem three ways, including vishalbalwani 's, and working from answer choices. This combination (number of coins), is the only one that works. Question: Is the method sound? The variables work, but they seem inconsistent. I use A as a quantity for 5-cent coins. The coefficients of the variables -- 5 and 10 -- are the values of the coins in cents. But in the second equation, (Q) I do not think I have switched quantities, per the prompt. I think have switched values. In the second equation, (Q), the 5, a value, is in front of B -- which is supposed to be the quantity of 10-cent coins. vishalbalwani did the same. What am I missing? We are supposed to be switching quantities. But switching values works. I think pushpitkc avoids the whole problem by immediately defining y in terms of x from one equation. Is my method sound? Kudos [?]: 258 [0], given: 546 Target Test Prep Representative Status: Founder & CEO Affiliations: Target Test Prep Joined: 14 Oct 2015 Posts: 1634 Kudos [?]: 837 [0], given: 2 Location: United States (CA) Re: A purse contains 5-cent coins and 10-cent coins worth a total of$1.75 [#permalink]
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22 Sep 2017, 07:49
Gnpth wrote:
A purse contains 5-cent coins and 10-cent coins worth a total of $1.75. If the 5-cent coins were replaced with 10-cent coins and the 10-cent coins were replaced with 5-cent coins, the coins would be worth a total of$2.15. How many coins are in the purse?
A. 26
B. 27
C. 28
D. 29
E. 30
We can create two equations in which a = the initial number of 5-cent coins and b = the initial number of 10-cent coins. We can create the “money” equation by recalling that a 5-cent coin is $0.05 and a 10-cent coin is$0.10; thus:
0.05a + 0.1b = 1.75
And, after we reverse the coinage, we have:
0.1a + 0.05b = 2.15
Multiplying the first equation by -2, we have:
-0.1a - 0.2b = -3.50
Adding the two equations, we have:
-0.15b = -1.35
b = 9
Thus:
0.05a + (0.1)(9) = 1.75
0.05a = 0.85
a = 17
So, there is a total of 17 + 9 = 26 coins.
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A purse contains 5-cent coins and 10-cent coins worth a total of $1.75 [#permalink] ### Show Tags 22 Sep 2017, 08:38 ScottTargetTestPrep wrote: Gnpth wrote: A purse contains 5-cent coins and 10-cent coins worth a total of$1.75. If the 5-cent coins were replaced with 10-cent coins and the 10-cent coins were replaced with 5-cent coins, the coins would be worth a total of $2.15. How many coins are in the purse? A. 26 B. 27 C. 28 D. 29 E. 30 We can create two equations in which a = the initial number of 5-cent coins and b = the initial number of 10-cent coins. We can create the “money” equation by recalling that a 5-cent coin is$0.05 and a 10-cent coin is $0.10; thus: 0.05a + 0.1b = 1.75 And, after we reverse the coinage, we have: 0.1a + 0.05b = 2.15 Multiplying the first equation by -2, we have: -0.1a - 0.2b = -3.50 Adding the two equations, we have: -0.15b = -1.35 b = 9 Thus: 0.05a + (0.1)(9) = 1.75 0.05a = 0.85 a = 17 So, there is a total of 17 + 9 = 26 coins. Answer: A ScottTargetTestPrep , where you reverse the coinage, what do variables $$a$$ and $$b$$ stand for? (I had the same question about my own method, above.) Perhaps the variables stand simply for the total number of coins, and absent any total number of coins, we cannot write one variable in terms of another. Conceptually, however, that reasoning does not make a lot of sense to me. We have two unknown quantities in two scenarios. In the first equation, the quantity b is defined as the number of$0.10 coins (and qty "a" initially is # of $.05 coins) If "b" then switches to the number of$0.05 coins (vice versa for "a"), how are those two a's and b's the same, such that we can use them as an identity in solving two equations?
Sorry if this question is sophomoric; usually I can find my way out of a muddle. Not this time.
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genxer123 wrote:
ScottTargetTestPrep wrote:
Gnpth wrote:
A purse contains 5-cent coins and 10-cent coins worth a total of $1.75. If the 5-cent coins were replaced with 10-cent coins and the 10-cent coins were replaced with 5-cent coins, the coins would be worth a total of$2.15. How many coins are in the purse?
A. 26
B. 27
C. 28
D. 29
E. 30
We can create two equations in which a = the initial number of 5-cent coins and b = the initial number of 10-cent coins. We can create the “money” equation by recalling that a 5-cent coin is $0.05 and a 10-cent coin is$0.10; thus:
0.05a + 0.1b = 1.75
And, after we reverse the coinage, we have:
0.1a + 0.05b = 2.15
Multiplying the first equation by -2, we have:
-0.1a - 0.2b = -3.50
Adding the two equations, we have:
-0.15b = -1.35
b = 9
Thus:
0.05a + (0.1)(9) = 1.75
0.05a = 0.85
a = 17
So, there is a total of 17 + 9 = 26 coins.
ScottTargetTestPrep , where you reverse the coinage, what do variables $$a$$ and $$b$$ stand for? (I had the same question about my own method, above.)
Perhaps the variables stand simply for the total number of coins, and absent any total number of coins, we cannot write one variable in terms of another.
Conceptually, however, that reasoning does not make a lot of sense to me.
We have two unknown quantities in two scenarios. In the first equation, the quantity b is defined as the number of $0.10 coins (and qty "a" initially is # of$.05 coins)
If "b" then switches to the number of $0.05 coins (vice versa for "a"), how are those two a's and b's the same, such that we can use them as an identity in solving two equations? Sorry if this question is sophomoric; usually I can find my way out of a muddle. Not this time. The variables a and b stand for the initial number of 5- and 10-cent coins, before and after reversing the coinage. The only thing that changes when the coinage is reversed is that the number of 5-cent coins is now b and the number of 10-cent coins is now a. _________________ Scott Woodbury-Stewart Founder and CEO GMAT Quant Self-Study Course 500+ lessons 3000+ practice problems 800+ HD solutions Kudos [?]: 837 [1], given: 2 Re: A purse contains 5-cent coins and 10-cent coins worth a total of$1.75 [#permalink] 01 Oct 2017, 12:26
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# A purse contains 5-cent coins and 10-cent coins worth a total of \$1.75
Powered by phpBB © phpBB Group | Emoji artwork provided by EmojiOne Kindly note that the GMAT® test is a registered trademark of the Graduate Management Admission Council®, and this site has neither been reviewed nor endorsed by GMAC®. | 2017-10-18T11:52:38 | {
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https://math.stackexchange.com/questions/1371717/relation-between-real-roots-of-a-polynomial-and-real-roots-of-its-derivative | Relation between real roots of a polynomial and real roots of its derivative
I have this question which popped in my mind while solving questions of maxima and minima. First Case:Let $f(x)$ be an $n$ degree polynomial which has $r$ real roots. Using this can we say anything about the number of real roots of $f'(x)$?
Second Case:Suppose, $f(x)$ has all $n$ real roots. Then will all of its derivatives also have all real roots?
Also, if any of its derivatives do not have all real roots, then will $f(x)$ also have not all real roots? If the above is true then what about its converse?
Comment Case:For the third case:Suppose f'(x) is a 5 degree polynomial with 3 real roots.Then f(x) will be a 6 degree polynomial.(correct me if I am wrong).What are the possible no. of roots that f(x) can have(3,4,5 etc.?).Basically I am asking for an example.Also it would be great if you follow all cases with an example like in the 4th case.
• I don't understand the beginning. If $f(x)$ is an n-degree polynomial with $r$ real roots, then it is not the case that $f'(x)$ will have $(r-1)$ real roots. Consider $f(x) = x^2 +1$. Here $r = 0$, but $f'(x) = 2x$ which has 1 root at 0. Jul 23 '15 at 19:35
• Sorry I never realized that.I just said what I was taught.Now I know it was incorrect.I have edited it. But I dont understand why the question has been downvoted? Jul 23 '15 at 19:49
• You're probably being downvoted for not providing any work/ideas for your question. Jul 23 '15 at 20:22
• This is of interest: en.wikipedia.org/wiki/Gauss%E2%80%93Lucas_theorem Jul 23 '15 at 21:48
First case: If the number of real roots $r$ of $f(x)$ is greater than one, then $f'(x)$ has at least $r-1$ real roots. (The limitation "greater than one" is not necessary but the statement is trivial if $r\le 1$.) Given any two roots $a<b$ of $f(x)$, $f$ is continuous and differentiable on $[a,b]$, so by Rolle's theorem $f'(c)=0$ for some $a<c<b$.
There may be more roots of $f'(x)$ than those between roots of $f(x)$, so the only upper bound is the obvious one of $n-1$. Ask if you need examples. It seems to me that if multiplicity is taken into account that the number of real roots of $f'(x)$ has the same parity (even/odd) as the number of real roots of $f(x)$, but I haven't proven it yet. If multiplicity is not taken into account, the parity can be anything.
Second case: If $f(x)$ has degree $n$ and has $n$ real roots, then each consecutive pair of roots of $f(x)$ defines a root of $f'(x)$, which makes $n-1$ roots of $f'(x)$. Since $f'(x)$ is a polynomial of degree $n-1$, this is all possible roots. This continues for all later derivatives, so you are correct: all its derivatives will have all real roots.
Third case: The contrapositive of the second case tells us that if any of its derivatives have any non-real roots, then $f(x)$ also has some non-real roots.
Fourth case: The converse of the third case is not true. For example, $f(x)=x^2+1$ has two non-real roots, but its derivative $f'(x)=2x$ has one real root.
Comment case: You asked, "Suppose $f'(x)$ is a $5$ degree polynomial with $3$ real roots. What are the possible no. of roots that $f(x)$ can have($3,4,5$ etc.?)."
The formulas for $f(x)$ and $f'(x)$ are given in the diagram, where $C$ is a real constant, zero in the graph. You can see that $f'(x)$ is a degree $5$ polynomial with $3$ real roots.
The dashed horizontal lines show the possible number of real roots of $f(x)$ for varying values of $C$. There are $0$ real roots for $C=3$, $1$ real root for $C\approx. 2.638$, $2$ real roots for $C=1$, $3$ real roots for $C\approx -1.757$, and $4$ real roots for $C=-1.9$. My discussion for the first case shows that there cannot be more than $4$ real roots since $f'(x)$ has $3$ real roots.
• For the third case:Suppose f'(x) is a 5 degree polynomial with 3 real roots.Then f(x) will be a 6 degree polynomial.(correct me if I am wrong).What are the possible no. of roots that f(x) can have(3,4,5 etc.?).Basically I am asking for an example.Also it would be great if you follow all cases with an example like in the 4th case. Jul 23 '15 at 20:40
First, notice that if any differentiable function $f(x)$ has $r$ real (distinct) roots, then $f'(x)$ must have at least $r-1$ distinct roots. This follows from the following:
Suppose that $a_1<...<a_r$ are (distinct) roots from $f(x)$ and $i \in \{1,...,r-1\}$. Then consider every interval of the form $[a_i,a_{i+1}]$. Since this interval is compact, any function on this interval achieves a maximum/minimum. Each minimum/maximum (depending on whether the function $f$ is dipping below or above the real line) corresponds to a root of $f'(x)$. Since we have $r -1$ such intervals, we have at least $r-1$ roots.
Since all polynomials are differentiable, we notice that if $f(x)$ has $r$ (distinct) roots, then $f'(x)$ has at least $r - 1$ distinct roots.
Your question isn't the clearest, but maybe someone else can come up with a better question to answer 1.
If a polynomial $f(x)$ is degree $n$ and has $n$ (distinct) real roots, then using the argument above (as well as the fundamental theorem of algebra (that might be over-kill, but oh well)), you can conclude that $f'(x)$ has exactly $n-1$ real roots.
Given this work, you can figure out 3 with the case for (distinct) roots. | 2021-10-15T20:36:03 | {
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https://www.freemathhelp.com/forum/threads/how-many-significant-digits-are-in-1040-and-1040-0.111578/ | # How many significant digits are in 1040. and 1040.0?
#### Indranil
##### Junior Member
How many significant digits are in 1040. and 1040.0? I am confused
#### Dr.Peterson
##### Elite Member
How many significant digits are in 1040. and 1040.0? I am confused
It will help if you state your specific confusion. What possibilities do you see for the answers? What about it are you unsure of?
#### Indranil
##### Junior Member
It will help if you state your specific confusion. What possibilities do you see for the answers? What about it are you unsure of?
Do 1040. has four significant digits and 1040.0 has five significant digits?
#### Dr.Peterson
##### Elite Member
Do 1040. has four significant digits and 1040.0 has five significant digits?
That's right. You aren't really confused, just perhaps uncertain.
When there is a decimal point present, all digits from the first non-zero digit on the left to the last written digit on the right are significant.
#### Subhotosh Khan
##### Super Moderator
Staff member
Do 1040. has four significant digits and 1040.0 has five significant digits?
And 1040 (without decimal point) has three significant digits.
#### Indranil
##### Junior Member
And 1040 (without decimal point) has three significant digits.
Could you explain why without decimal point, 1040 has three significant digits? and what is the difference between 1040 and 1040.? because I knew 1040 and 1040. are the same.
#### Subhotosh Khan
##### Super Moderator
Staff member
Could you explain why without decimal point, 1040 has three significant digits? and what is the difference between 1040 and 1040.? because I knew 1040 and 1040. are the same.
When you write 1040 - it can be 1036 (min) to 1044 (max)
When you write 1040. - it can be 1039.6 to 1040.4
#### Dr.Peterson
##### Elite Member
Could you explain why without decimal point, 1040 has three significant digits? and what is the difference between 1040 and 1040.? because I knew 1040 and 1040. are the same.
Of course you are right that 1040 and 1040. represent the same number; it is only the convention about implied accuracy that is at issue.
I would say that 1040 is somewhat ambiguous, because you can't be sure whether the final zero is there because that digit is actually zero, or only because the digit is needed in order to have the right place values. It might really have either 3 or 4 significant digits - that is, it might have been rounded to the nearest ten or to the nearest unit, and be written the same. Many people assume the least possible accuracy. Ideally, significant digits should be read on in scientific notation, where this never happens (because there is only one digit before the decimal point).
On the other hand, with the decimal point, it is clear that the writer intends to stop after the decimal point, and the zero is significant.
#### JeffM
##### Elite Member
When you write 1040 - it can be 1036 (min) to 1044 (max)
When you write 1040. - it can be 1039.6 to 1040.4
According to the conventions of significant figures, 1040 and 1040. mean different things. That is because the whole idea of significant figures involves applied mathematics. In pure mathematics
$$\displaystyle 1040. \equiv 1040$$ as you recognize, but the concept of significant figures is not relevant to pure mathematics.
See https://en.m.wikipedia.org/wiki/Significant_figures
Last edited:
#### Indranil
##### Junior Member
When you write 1040 - it can be 1036 (min) to 1044 (max)
When you write 1040. - it can be 1039.6 to 1040.4
#### Dr.Peterson
##### Elite Member
I think what others of us said make the point more clearly, showing the reason.
Take a more interesting example: 104000. I might write that as an estimate, or rounded number, whether the real number was exactly 104000, or 104001 and rounded to the nearest ten, or 104010 and rounded to the nearest hundred, or 104100 and rounded to the nearest thousand. There is no way to distinguish those situations by the way we write it, unless we use one of the (relatively rare) conventions Wikipedia suggests. As they say, "The significance of trailing zeros in a number not containing a decimal point can be ambiguous. For example, it may not always be clear if a number like 1300 is precise to the nearest unit (and just happens coincidentally to be an exact multiple of a hundred) or if it is only shown to the nearest hundred due to rounding or uncertainty."
One way to reduce the ambiguity when the number is accurate to the nearest unit is to include a decimal point and write it as "104000.". Since this can be done, some people tend to assume that if there is no decimal point, the number of significant digits should be assumed to be as few as possible, in this case three, with all the trailing zeros insignificant.
Back to the original, if a number was rounded to the nearest unit and the result was 1040, then the number could be anything from 1039.5 to just under 1040.5. (What Khan wrote was not quite right.) Any of those numbers, such as 1039.51 or 1040.49, would round to 1040.
If the number had been rounded to the nearest ten, then the original could have been anything from 1035 to just under 1045. (I should add that whether you would round 1035 to 1040 depends on the precise convention you are following for rounding.)
#### HallsofIvy
##### Elite Member
As has been said, the distinction between 1040 and 1040. is a convention. It has been agreed that if there is a decimal point after an integer then all digits in the number are "significant" and that if there is no decimal point then the significant digits are end with the rightmost non-zero digit.
#### Denis
##### Senior Member
...and if 1040. was at end of sentence, then you'd have 1040.. :cool: | 2019-03-25T05:41:30 | {
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https://math.stackexchange.com/questions/4186288/need-help-with-a-markov-chain-for-weather-conditions | need help with a Markov chain for weather conditions
This is my second time posting the question as I failed to do so the first time, because I did not know the proper way. My apologies.
Question:
Suppose that whether or not it rains today depends on weather conditions through the previous teo days. If it has rained for the past two days, then it will rain today with probability 0.7. If it did not rain for any of the past two days, then it will rain today with probability 0.4. In any other case the weather today will, with probability 0.5, be the same as the weather yesterday. Describe the weather condition over time with a Markov chain.
Suppose that it rained today and yesterday. Calculate the expected number of days until it rains three consecutive days for the first time in the future.
I have found 4 different states that I named RR(0), RN(2), NR(1), and NN(3). R stands for when it rains and N is for when it does not.
As the question asks, I have tried finding the possible ways of three consecutive days being rainy. At time n, we are given it was rainy today and yesterday, meaning we are in State 0.
1-) First possibility is when it rains tomorrow, which gives us RRR (we got the three consecutive days)
2-) Second possibility is when we go from 0 to 2, from 2 to 1, from 1 to 0, and stay in 0 one day. That follows as : RRNRRR (In 4 days we can get rain for 3 consecutive days)
3-) Third is when we go from 0 to 2, from 2 to 3, from 3 to 1, from 1 to 0, and stay in 0 one day. That follows as: RRNNRRR.
To conclude what I have in mind about the question, wherever we go, when we get to State 0, we need to stay there one more day to get three consecutive rainy days. That means the minimum # of days to get rain for three consecutive days is one day. However, after this point, I am not able to proceed with the question.
Any help would be appreciated, thank you!!
Edit: I think the maximum # of days is just a random number, which leads me to the expectation of the sum of a random number of geometric random variable, but still I can't go any further beyond that. Thank you!
• For $i=0,1,2,3$, let $E_i$ be the expected number of days until $3$ consecutive rainy days, if the chain is currently in state $i$. We are asked to compute $E_0$. Notice that $E_0=1+.3E_2$. Develop similar equations for the other $E_i$ and solve. Jun 29, 2021 at 17:39
• Developing the equations is where I am stuck at and I am also confused as to whether there is a boundary condition. Could you please explain to me, for instance, how to develop E0? Jun 29, 2021 at 18:44
Let me get you started.
I claim that $$E_0=1+.3E_2$$ as I stated in a comment.
Suppose we are in state $$0$$. We must wait at least $$1$$ more day to see if we'll have three consecutive days of rain. $$70\%$$ of the time, it rains, and we are done, but $$30\%$$ of the time we transition to state $$2$$, and then we must wait, on average, $$E_2$$ days to get $$3$$ consecutive rainy days.
If we are in state $$2$$, similar reasoning gives $$E_2=1+.5E_1+ .5E_3$$
You can read the equations right off your diagram. You should get $$4$$ equations with $$4$$ unknowns and a unique solution. I'm sure you won't have any problem finishing from here.
• I have found E3=1 + 0.6(E3) + 0.4(E1) (state 1 is the only different state we can transition to from state 3 and we might also stay in state 3 with p=0.6?) and E1= 1 + 0.5(E0) + 0.5(E2). Do you think I made a mistake or those are correct? Jun 30, 2021 at 8:20
• @SydneyAstbury I think you are correct. Jun 30, 2021 at 11:36
Ordinarily, Markov Chains are conditional on the previous step, but not on the previous two steps. A way to get around this in the current problem is to re-define the states to account for two days, with suitable overlaps.
The new states are 00 (for consecutive dry days) 01 (dry followed by wet), and so on to 11 (for two wet days in a row).
States 'overlap' so that 00 can be followed by 01, but not by 01, etc. In the matrix below we call the new states A,B, C, D.
The transition matrix can be written in R as
P = matrix(c(.6, .4, 0, 0,
0, 0, .5, .5,
.5, .5, 0, 0,
0, 0, .3, .7), byrow=T, nrow=4)
We seek $$\sigma$$ such that $$\sigma P = P,$$ so that $$\sigma$$ is a left eigenvector of $$P.$$ Because R finds right eigenvectors, we use the transpose t(P).
The stationary distribution is proportional to the right eigenvector of smallest modulus, which R prints first. [In R, the symbol %*% denotes matrix multiplication.]
g = eigen(t(P))\$vec[,1]
sg = g/sum(g); sg
[1] 0.2542373 0.2033898 0.2033898 0.3389831
sg %*% P # check
[,1] [,2] [,3] [,4]
[1,] 0.2542373 0.2033898 0.2033898 0.3389831
So the long-run probability of rain is $$0.5424.$$
Note: In this relatively simple problem it is not difficult to solve a few simultaneous equations, but the eigen-method shown above is very convenient for ergodic chains with more than four states. | 2022-12-08T16:40:48 | {
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https://math.stackexchange.com/questions/3081850/is-the-following-application-of-the-clt-correct | Is the following application of the CLT correct?
I am new to probability theory and still have a problem understanding some things. I came across the following problem that I am unsure on how to solve. Here it is:
Suppose that in a city with $$100,000$$ cars a shop sells tires for cars. It has been observed that in an interval of $$3$$ months the percentage of cars that come to the shop for the replacement of all their tires is $$0.5$$%. What is the least number of tires that the shop has to order so that it can satisfy all its customers in a $$3$$ month interval with probability $$\geq 95$$%?
And here is my attempt:
Let $$X$$ be the random variable of cars coming at the shop in a $$3$$ month interval. Then $$X$$ is a discrete random variable following binomial distribution $$B(100,000; 0.005)$$. We would like to find the least $$x\in\mathbb{N}$$ satisfying $$P(X\leq x)\geq95$$%. Since (I assume) every car has $$4$$ tires, our answer is going to be $$4x$$. Now since we have a large number of cars, by the Central limit theorem, we have that $$X\sim N(\mu,\sigma^2)$$, where $$\mu, \sigma^2$$ are the expected value and the variance of $$B(100,000; 0.005)$$ and therefore, by calculating, it is $$\mu=500, \sigma^2=475$$. Now we have $$P(X\leq x)=P\big{(}\displaystyle{\frac{X-500}{\sqrt{475}}\leq\frac{x-500}{\sqrt{475}}\big{)}=P\big{(}Z\leq\frac{x-500}{\sqrt{475}}\big{)}}$$, where $$Z\sim N(0,1)$$. Now our $$x$$ will satisfy $$\displaystyle{P\big{(}Z\leq\frac{x-500}{\sqrt{475}}\big{)}}=0.95$$, therefore $$\displaystyle{P\big{(}0 hence (by the tables) it is $$\displaystyle{\frac{x-500}{\sqrt{475}}=1.645}$$, which yields $$x=535.85..$$ and since $$x$$ is supposed to be an integer we have that $$x=536$$. therefore the least number of tires is $$2144$$.
Is my solution correct? if not, what should I have used?
Comment: I know that the $$x=$$fractional is not correct since i specified that $$x\in\mathbb{N}$$, but you get the point.
• "the percentage of cars that come to the shop for the replacement of all their tires is 0.5%" That is quite poorly worded, it says that X is a constant ( 500). I guess it meant what you understood, though. – leonbloy Jan 21 at 14:49
• @leonbloy I know, I was confused by this as well, but this is an accurate translation of the problem... – JustDroppedIn Jan 21 at 15:03
Assuming that the shop's only tire business is to replace all four tires, your method makes sense. [Otherwise, I can't see how to work the problem.]
The CLT implies that a binomial distribution with sufficiently large $$n$$ is well-approximated by a normal distribution, which is what you're doing.
Binomial. I used the binomial quantile function (inverse CDF) qbinom in R to compute an answer directly from the binomial distribution. It is very nearly the same as your answer.
4*ceiling(qbinom(.95, 10^5, .005))
[1] 2148
I expect that the small discrepancy may result from rounding in finding the mean and standard deviation of the approximating normal distribution or from using printed normal tables. [Perhaps try $$\sigma^2 = 497.5.]$$
Normal approximation to binomial. Without rounding, I get the following result from the normal approximation. (The normal approximation with $$n=10^5$$ and $$p = 0.005$$ should be quite good, but not necessarily exactly perfect.)
n = 10^5; p = .005; mu = n*p; sg = sqrt(n*p*(1-p))
4*ceiling(qnorm(.95, mu, sg))
[1] 2148
Poisson approximation. Another reasonable approximation is to use the distribution $$\mathsf{Pois}(\lambda = \mu = np).$$ This approximation is quite good for large $$n$$ and small $$p:$$
4*ceiling(qpois(.95, mu))
[1] 2148
Normal approximation to Poisson. For $$\lambda$$ as large as $$500$$ the Poisson distribution is well approximated by $$\mathsf{Norm}(\lambda, \sqrt{\lambda}):$$
4*ceiling(qnorm(.95, mu, sqrt(mu)))
[1] 2148
The following figure shows the binomial distribution (black bars) along with its normal approximation (blue curve) and its Poisson approximation (centers of red circles). The probability to the right of the vertical dotted line is about 0.05. | 2019-08-18T11:00:42 | {
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https://math.stackexchange.com/questions/1740130/how-to-enumerate-2d-integer-coordinates-ordered-by-euclidean-distance | How to enumerate 2D integer coordinates ordered by Euclidean distance?
The square of Euclidean distance between $(x, y)\in\mathbb{Z}^2$ and origin is $d = x^2+y^2$. How to enumerate the coordinates $(x, y)$ in ascending order of $d$?
For example, the first 14 sets of coordinates are:
d=0: { (0,0) }
d=1: { (1,0), (0,1), (0,-1), (-1,0) }
d=2: { (1,-1), (-1,-1), (-1,1), (1,1) }
d=4: { (2,0), (0,2), (-2,0), (0,-2) }
d=5: { (1,2), (-1,2), (1,-2), (-2,1), (-2,-1), (2,-1), (2,1), (-1,-2) }
d=8: { (2,2), (-2,2), (-2,-2), (2,-2) }
d=9: (0,3), (-3,0), (0,-3), (3,0)
d=10: (1,3), (-1,3), (3,1), (-3,1), (-3,-1), (1,-3), (-1,-3), (3,-1)
d=13: (2,-3), (-3,-2), (3,-2), (-2,-3), (-3,2), (3,2), (-2,3), (2,3)
d=16: (0,-4), (-4,0), (4,0), (0,4)
d=17: (-4,1), (-4,-1), (4,1), (1,-4), (4,-1), (-1,-4), (-1,4), (1,4)
d=18: (3,-3), (-3,-3), (-3,3), (3,3)
d=20: (4,2), (4,-2), (-4,-2), (2,-4), (-4,2), (-2,-4), (-2,4), (2,4)
d=25: (-3,-4), (-5,0), (5,0), (4,3), (-3,4), (-4,3), (0,-5), (4,-3), (-4,-3), (3,-4), (3,4), (0,5)
The first 14 iterations are depicted as:
13
131210 9101213
1311 8 7 6 7 81113
12 8 5 4 3 4 5 812
10 7 4 2 1 2 4 710
13 9 6 3 1 0 1 3 6 913
10 7 4 2 1 2 4 710
12 8 5 4 3 4 5 812
1311 8 7 6 7 81113
131210 9101213
13
For a finite range of $(x, y)$, a trivial algorithm is to store all coordinates within the range into a list, and then sort the coordinates with $d$. However, this will need $O(n\log n)$ time and $O(n)$ space.
Another possible approach is to solve the diophantine equation $x^2 + y^2 = m$ for $m = 0, 1, \ldots, ... d_\mathrm{max}$. But it seems also a hard problem.
Is there any simpler way with lower time/space complexity?
Here is a C++ code of the trivial solution for reference.
• As far as storing goes, you can certainly take advantage of symmetries and perhaps more generally use the Sum of Squares function $r_2(n)$ (or $r'_2(n)$ if you're not interested in order/signs). I will also link this mathoverflow question as it seems to answer precisely what you desire. – Fimpellizieri Jun 28 '16 at 22:54
• Why the complexity of the trivial algorithm is not $O(n^2)$? – user84976 Jun 29 '16 at 10:35
• Your $n$ seems to be the number of points in your grid, not the edge length of the grid, right? So in terms of the edge length $a$, you get $O(a^2\log a)$ time complexity and $O(a^2)$ space complexity. But you can't go asymptotically lower than $O(a^2)$ time complexity since you have to enumerate this many points. Are you interested in tweaking constants here, or only in getting rid of the $\log a$ term or reducing the space complexity? – MvG Jul 5 '16 at 12:29
You can employ symmetry to only generating one eighth of a circle, e.g. $0\le x\le y$. It should be obvious how to do this.
You can reduce the space complexity to $O(r)=O(\sqrt n)$, i.e. linear in the radius of your disk, not in the number of grid points it covers. The way to do this is by considering parallel lines of grid points. Then on each line you know that the points will be enumerated starting closest to the center and moving outwards, so it is enough to keep track of the next candidate on each line.
The most appropriate data structure for this keeping track is most likely a priority queue, which will give you quick access to the element with minimal $d$. But even then, elementary operations in a priority queue have $O(\log n)$ time complexity. Which in this case means you are still at $O(r^2\log r)=O(n\log n)$ overall time complexity since you enumerate $O(r^2)$ elements and maintain a data structure where you have to perform two $O(\log r)$ operations for each of them.
Here is an implementation of these ideas. The core portions are these:
struct Pt { // essentially your struct Point, just shorter names
int x, y, d;
Pt(int x, int y) : x(x), y(y), d(x*x + y*y) {}
bool operator<(const Pt& p) const { return d > p.d; } // flipped!
};
void enumerateGridPoints(int r) {
priority_queue<Pt> q;
q.push(Pt(0, 0));
int d, level = -1;
while ((d = q.top().d) <= r*r) {
reportLevel(++level, d);
do {
Pt p = q.top();
q.pop();
reportPoint(p);
if (p.y < p.x) q.push(Pt(p.x, p.y + 1));
if (p.y == 0) q.push(Pt(p.x + 1, 0));
} while (q.top().d == d);
}
}
Since the C++ std::priority_queue is a max queue by default, and since I'm lazy, I just flipped the sign in the comparison operation so that the element with smallest d is the maximal element and thus extracted next.
One benefit of the above algorithm compared to yours is that it's easy to modify it so that enumerates points without any preset limit. That might be useful if you'd like to e.g. inspect sites until you find something you are looking for, but you don't know up front how long you will have to search until you find it. Both time and space complexity scale with the generated output, so small initial portions can be obtained quickly while still allowing you to proceed to larger results.
Edsger Dijkstra gives an elegant recursive solution to the problem of finding solutions to $x^2 + y^2 = m$ in his 1976 book The Discipline of Programming. The solution is outlined here, and involves sweeping potential $x$ values down from $\sqrt m$ and $y$ values up from 0.
Consider the function $B(x, y)$ that returns solutions between $x$ and $y$, defined recursively as follows:
$$B(x, y) = \begin{cases} \varnothing, & \text{if x < y (base case)} \\ (x, y) \cup B(x-1, y+1), & \text{if x^2 + y^2 = m (satisfying case)} \\ B(x, y+1), & \text{if x^2 + y^2 < m} \\ B(x-1, y), & \text{if x^2 + y^2 > m} \\ \end{cases}$$
Using this definition, you can calculate $B(\sqrt m, 0)$ for each $m=0,1,…,d_{max}$. Written as a generator or iterator, this algorithm will have constant space complexity.
• Would you agree that the time complexeity is $O(r^3)$ here (with $r=\sqrt{d_{\text{max}}}$), since for each of the $O(r^2)$ possible values for $m$ you'd have $O(r)$ steps of evaluating $B$ till you reach the base case? – MvG Jul 6 '16 at 11:25 | 2019-09-20T22:40:35 | {
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https://math.stackexchange.com/questions/1134255/solve-for-y2yy-frace-tt2/1134495 | Solve for $y''+2y'+y = \frac{e^{-t}}{t^{2}}$
I need some help in finding the solution to this second-order non-homogeneous DE.
I know how to find the solution of the reduced equation $$y''+2y'+y=0.$$
The characteristic equation $r^{2}+2r+1=0$ yields real and repeated roots $r=-1$. So the complementary solution, $$y_{c}= c_{1}e^{-t} + c_{2}te^{-t}$$.
Now what I am struggling with is making educated guesses about coming up with a particular solution. So any suggestions on that would be helpful.
EDIT1: I haven't learned about the method of variation of parameters yet and wanted to do this with the method of undetermined coefficients. Please feel free to post the solution by using any of the techniques so I can learn from it.
EDIT2: Our fundamental pair of solutions for the corresponding homogeneous is $y_{1}=e^{-t}$, and $y_{2}=t e^{-t}$.
We know that our particular solution will be of the form, $y_{p} = u_{1} y_{1} + u_{2} y_{2}$.
The determinant of the Wronskian, $W = e^{-2t}$, with $W_{1} = -t^{-1} e^{-2t}$, and $W_{2} = t^{-1} e^{-2t}$.
So $u'_{1} = \frac{W_{1}}{W} = -\frac{1}{t}$, and $u'_{2} = \frac{W_{2}}{W} = \frac{1}{t}$.
So $u_{1} = -\int \frac{1}{t} dt = -ln|t|$, and $u_{2} = \int \frac{1}{t} dt = ln|t|$.
Hence, $y_{p} = -ln|t| e^{-t} + ln|t|te^{-t}$.
So our general solution, $$y = y_{c} + y_{p} = e^{-t}(c_{1}+c_{2}t-ln|t|+ln|t|t).$$
Does it look okay?
• Generally you only do particular solutions if it's a polynomial times a sine/cosine times an exponential, not a rational function. I'd use variation of paramaters – Alan Feb 5 '15 at 1:55
• @Alan I haven't learned that technique yet. Is it the method of undetermined coeffecients? – OGC Feb 5 '15 at 1:57
• What techniques have you learnt? – mattos Feb 5 '15 at 2:22
• @Mattos Separation of variables technique and learned method of undetermined coefficients today. Seeing youtube videos on change of var technique. Is it possible to solve this problem by the method of undetermined coefficients technique? – OGC Feb 5 '15 at 2:26
• @user36829 If you want, I'll post a solution using variation of parameters. – mattos Feb 5 '15 at 3:50
$$y_h = c_1e^{-t} + c_2te^{-t}$$
with
$$y_1 = e^{-t} \ \ \ \ \ y_2 = te^{-t}$$
Taking the Wronskian, you also found
$$W = e^{-2t}$$
The particular solution is given by the formula
$$y_p = -y_1 \int \frac{y_2 g(t)}{W} dt + y_2 \int \frac{y_1 g(t)}{W} dt$$
where
$$g(t) = \frac{e^{-t}}{t^{2}} = t^{-2}e^{-t}$$
Hence
\begin{align} y_p &= -e^{-t} \int \frac{te^{-t} \cdot t^{-2}e^{-t}}{e^{-2t}} dt + te^{-t} \int \frac{e^{-t} \cdot t^{-2}{e^{-t}}}{e^{-2t}} dt \\ &= -e^{-t} \int \frac{1}{t} dt + te^{-t} \int \frac{1}{t^{2}} dt \\ &= -e^{-t} \ln(t) - \frac{te^{-t}}{t} \\ &= -e^{-t}(\ln(t) + 1) \\ \end{align}
Therefore,
\begin{align} y &= y_h + y_p \\ &= e^{-t}(c_1 + c_2 t - \ln(t) - 1) \\ \end{align}
You can check by differentiation that this satisfies the ODE.
• $-1\cdot e^{-t}$ is an unnecessary part of the solution since it solves the homogeneous equation. I.e. the $-$ can be 'absorbed' into the $c_1$ coefficient. – jdods Mar 5 '15 at 1:58
• @jdods Yeah and? – mattos Mar 5 '15 at 8:38
In this case of only one characteristic value, also a change of function can help. Substitute, according to the homogeneous solution, $y(t)=e^{-t}u(t)$ to get for $u$ $$u''(t)=t^{-2}\implies u'(t)=-t^{-1}+C\implies u(t)=-\ln|t|+Ct+D$$ and thus $$y(t)=e^{-t}(-\ln|t|+Ct+D)$$
I'll use the formula I developed here $$Y_p=y_h\int \left\{y_h^{-2} W_0 \left(\int y_h W_0^{-1} g \ dt\right)\right\} \ dt$$ where the Wronskian (up to a constant multiple) is $$W_{0}=\exp\left(-{\int p \ dt}\right)$$ and a solution to the homogeneous equation is $y_h=e^{-t}$ (we can use either).
So we first integrate $$W_{0}=\exp\left(-{\int 2 \ dt}\right)=e^{-2t}$$ and then we'll find the particular solution by \begin{aligned} Y_p&=e^{-t}\int \left\{e^{2t} e^{-2t} \left(\int e^{-t} e^{2t} \frac{e^{-t}}{t^2} \ dt\right)\right\} \ dt \\ &=e^{-t}\int \left(\int \frac{1}{t^2} \ dt\right) \ dt \\ &=e^{-t}\int \left(-\frac{1}{t}\right) \ dt \\ &=-e^{-t}\ln t. \end{aligned}
This is a nice formula because you only need one solution to the homogeneous equation. | 2020-02-18T10:03:08 | {
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https://math.stackexchange.com/questions/2376879/does-the-interval-x-x-10-x1-4-contain-a-sum-of-two-squares | # Does the interval $[x, x + 10 x^{1/4})$ contain a sum of two squares?
Is it true that there exist an integer $N \in \mathbb N$ , such that $\forall x > N$ , the interval $[x, x+10x^{1/4} )$ always contains a sum of two squares ?
I know that $n \in \mathbb N$ is a sum of two squares iff it is of the form $n=q_1^{2a_1}...q_m^{2a_m}p_1^{k_1} ... p_r^{k_r}$ where $q_1,...,q_m,p_1,...,p_r$ are primes such that $q_i \equiv 3( \mod 4)$ and $a_1,...,a_m, k_1,...,k_r$ are non-negative integers ; where $m , r\ge 0$ . So it is enough to prove that for large enough $x$ , the interval $[x , x+10x^{1/4})$ always contains such a positive integer . But I don't know how to prove this.
• Reading about the Gauss circle problem would be a good starting point. – Daniel Fischer Jul 30 '17 at 19:01
• For me to start: where did you get the problem/conjecture? Is it stated as something definitely true or as an open problem? It is reasonable, by the way, as the count of numbers up to large $z$ that are sums of two squares is about $$\frac{Cz}{\sqrt {\log z}}$$ with (known) positive constant $C.$ – Will Jagy Jul 30 '17 at 19:19
• @WillJagy : For the source of the problem ; it was asked as an open ended question in the class by our professor . Where do you get the asymptotic formula about the count ? Could you please give any reference ? Thanks – user Jul 31 '17 at 15:54
• store.doverpublications.com/0486425398.html – Will Jagy Jul 31 '17 at 18:20
You might try to proceed greedily. Suppose we want to find a number that is "near" (but less than) $x$ that is a sum of two squares. Maybe one of these squares will be the largest square less or equal to $x$, which I'll denote by $a$. (So $a = \lfloor \sqrt x \rfloor^2$), where $\lfloor \cdot \rfloor$ is the "floor" function, or the "greatest integer below function"). In this case, the second square, which I'll call $b$, should be chosen to be the greatest square less than or equal to $x - a$. (So $b = \lfloor \sqrt{x - a} \rfloor^2$).
This gives us a candidate sum of squares $a + b$, which is relatively close to $x$. Now we ask, how close does this get us? How large is $x - a - b$?
This really asks, how close is $x$ to the largest square up to $x$? That is, how close is $x$ to $a = \lfloor \sqrt x \rfloor^2$?
Note that $$x - \lfloor \sqrt x \rfloor^2 = (\sqrt x - \lfloor \sqrt x \rfloor)(\sqrt x + \lfloor \sqrt x \rfloor) \leq (1) \cdot (2 \sqrt x) \leq 2 \sqrt x.$$ So the largest square up to $x$ is no more than $2 \sqrt x$ away from $x$.
Thus $x - a \leq 2 \sqrt x$. Iterating, we have that $$(x - a) - b \leq 2 \sqrt{x - a} \leq 2 (\sqrt {2 \sqrt x}) = 2^{3/2} x^{1/4}.$$
We have shown that given an $x$, the nearest "greedy" sum of squares is at most $2^{3/2} x^{1/4}$ away. So we've shown that there is a sum of squares in intervals of the shape $(x - 2^{3/2} x^{1/4}, x]$, which clearly implies the original question.
In comments, Will Jagy noted that the density of numbers which are sums of two squares is of the shape $Cx/\sqrt{\log x}$, where $C$ is the Landau-Ramanujan constant. It is conjectured that the gaps between numbers which are sums of squares shouldn't deviate too far from what is expected from the density, and in particular is bounded by $x^\epsilon$ for any $\epsilon > 0$. In particular it is conjectured that there is a number which is a sum of two squares in the interval $[x, x + x^{\epsilon})$ for any $\epsilon > 0$, for $x$ sufficiently large.
This answer shows that the gap between numbers which are sums of squares is at most $2^{3/2}x^{1/4}$, which is a far cry from $x^{\epsilon}$.
In general, there are not good uniform improvements over what's presented above. But there are remarkable on-average bounds due to Hooley, showing for instance that almost all (in the density sense) gaps between sums of squares is no more than $(\log x) (\log \log x)$. | 2019-06-26T03:38:56 | {
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https://math.stackexchange.com/questions/2111562/find-the-inverse-of-a-lower-triangular-matrix-of-ones | # Find the inverse of a lower triangular matrix of ones
Find the inverse of the matrix $A=(a_{ij})\in M_n$ where $$\begin{cases} a_{ij}=1, &i\geq j,\\ a_{ij}=0, &i<j. \end{cases}$$
The only method for finding inverses that I was taught was by finding the adjugate matrix. So $A^{-1}=\frac{1}{\det A}\operatorname{adj(A)}$
$$A=\begin{pmatrix} 1 & 0 &0 &\ldots &0\\ 1 & 1 & 0 &\ldots &0\\ 1 & 1 & 1 &\ldots &0\\ \vdots &\vdots & \vdots &\ddots & \vdots\\ 1 & 1 &1 & \ldots &1 \end{pmatrix}$$
This is a triangular matrix so $\det A=1.$ To find the adjugate I first need to find the cofactor matirx$(C)$ of $A$.
$$C=\begin{pmatrix} 1&-1&0&0&\ldots &0\\ 0 & 1& -1 &0&\ldots & 0\\ 0 & 0 & 1 & -1&\ldots &0\\ 0 & 0 & 0 &1 &\ldots &0\\ \vdots &\vdots &\vdots &\vdots&\ddots & \vdots\\ 0 & 0 &0 &0 &\ldots &1\end{pmatrix}$$
$$C^T=\operatorname{adj}(A)=\begin{pmatrix} 1 & 0 & 0 & 0&\ldots &0\\ -1 & 1& 0 & 0 &\ldots &0\\ 0&-1&1&0&\ldots &0\\ 0& 0 &-1 &1 &\ldots&0\\ \vdots &\vdots &\vdots &\vdots &\ddots&\vdots\\ 0&0&0&0&\ldots &1\end{pmatrix}=A^{-1}$$
Is this correct? Also, can I leave it like that or should I somehow write it more formally?
• try using elementary row operations. – vidyarthi Jan 24 '17 at 10:11
• Looks all right. – StubbornAtom Jan 24 '17 at 10:25
• Try multiplying your proposed $A^{-1}$ by $A$ to see whether you get the identity matrix. – Gerry Myerson Jan 24 '17 at 11:38
Yes, your inverse is correct. Let me show you an alternative approach of finding this inverse without the use of the adjugate matrix.
Denote $A_n$ as the $n \times n$ matrix that has elements $a_{ij} = 1$ if $i \geq j$, and $0$ otherwise. Then, $A_n$ can be constructed by applying the elementary row operations $$R_2 + R_1, R_3 + R_2, \ldots, R_n - R_{n-1}$$ to the identity matrix $I_n$. So, $$A_n = \prod_{i=1}^{n-1} E_i,$$ where $E_i$ is the elementary matrix representing the row operation $R_{i+1} - R_i$. Since det$(A)$ is nonzero, it is invertible, hence $$B_n = \left( A_n \right)^{-1} = \prod_{i = n-1}^{1} {E_i}^{-1},$$ where ${E_i}^{-1}$ is the row operation $R_{i+1} - R_i$. That is, $B_n$ can be constructed by applying the row operations $$R_n - R_{n-1}, R_{n-1} - R_{n-2},\ldots,R_2 - R_1$$ to the idenity matrix $I_n$. The inverse of $A_n$ is therefore given by the matrix
$$B_n = \begin{pmatrix} \phantom{-}1 & \phantom{-}0 &\phantom{-}0 &\ldots &0\\ -1 & \phantom{-}1 & \phantom{-}0 &\ldots &0\\ \phantom{-}0 & -1 & \phantom{-}1 &\ldots &0\\ \phantom{-}\vdots &\phantom{-}\vdots & \phantom{-}\vdots &\ddots & \vdots\\ \phantom{-}0 & \phantom{-}0 & \phantom{-}0 & \cdots &1 \end{pmatrix}.$$ In order to check that this is the correct inverse, you can use the identity $$A_nB_n = I_n.$$
• How can we write this inverse like this $B_n=(b_{ij})= \begin{cases} 1, &i= j,\\ -1, &i-j=1\\ 0, & ? \end{cases}$ When is $(b_{ij})=0$? – lmc Jan 24 '17 at 13:31
• $b_{ij}$ is $1$ if $i =j$, $-1$ if $i = j + 1$ and $0$ otherwise. – user394255 Jan 24 '17 at 13:33
I would like to present a very simple solution by interpretation of these matrices as operators on $\mathbb{R^n}$ (which will surprise nobody...). Triangular matrix $A$ acts as a discrete integration operator:
For any $x_1,x_2,x_3,\cdots x_n$:
$$\tag{1}A (x_1,x_2,x_3,\cdots x_n)^T=(s_1,s_2,s_3,\cdots s_n)^T \ \ \text{with} \ \ \begin{cases}s_1&=&x_1&&&&\\s_2&=&x_1+x_2&&\\s_3&=&x_1+x_2+x_3\\...\end{cases}$$
(1) is equivalent to:
$$\tag{2}A^{-1} (s_1,s_2,s_3,\cdots x_n)^T=(x_1,x_2,x_3,\cdots x_n)^T \ \ \text{with} \ \ \begin{cases}x_1&=& \ \ s_1&&&&\\x_2&=&-s_1&+&s_2&&\\x_3&=&&&-s_2&+&s_3\\...\end{cases}$$
and it suffices now to "collect the coefficients" in the right order in order to constitute the inverse matrix.
(Thus the inverse operation is - in a natural way - a discrete derivation operator).
You can also use the induction along the dimension $n$ (leading to a simple but not very constructive proof though).
Let $A_n$ denote the matrix $A$ of dimension $n$ and $B_n$ its inverse. The statement is obviously true for $n=1$ and reduces to showing that $1=1$. Let the statement be true for $n-1$, that is, $A_{n-1}^{-1}=B_{n-1}$. If $1_{n-1}$ and $0_{n-1}$ the column vectors of ones and zeros, respectively, of dimension $n-1$, the matrix $A_n$ can be written in the block form $$A_n=\begin{bmatrix}A_{n-1}&0_{n-1}\\1_{n-1}^T&1\end{bmatrix}.$$ You can easily verify that $$A_n^{-1}=\begin{bmatrix}A_{n-1}^{-1}&0_{n-1}\\-1_{n-1}^TA_{n-1}^{-1}&1\end{bmatrix}$$ is the inverse of $A_n$. By the induction assumption, $A_{n-1}^{-1}=B_{n-1}$. It is easy to check that $1_{n-1}^TA_{n-1}^{-1}=1_{n-1}^TB_{n-1}=e_{n-1}^T:=[0,\ldots,0,1]^T$ (sum of the rows of $B_{n-1}$). So $$A_n^{-1}=\begin{bmatrix}B_{n-1}&0_{n-1}\\-1_{n-1}^TB_{n-1}&1\end{bmatrix}=\begin{bmatrix}B_{n-1}&0_{n-1}\\-e_{n-1}^T&1\end{bmatrix}=B_n.$$
Just to show a different approach.
Consider the matrix $\mathbf E$, having $1$ only on the first subdiagonal $$\mathbf{E} = \left\| {\,e_{\,n,\,m} = \left\{ {\begin{array}{*{20}c} 1 & {n = m + 1} \\ 0 & {n \ne m + 1} \\ \end{array} } \right.\;} \right\| = \left\| {\,\left( \begin{gathered} 0 \\ n - m - 1 \\ \end{gathered} \right)\;} \right\|$$ Multiply it by $\mathbf A$ , and it is easily seen that $$\mathbf E \, \mathbf A=\mathbf A-\mathbf I \quad \Rightarrow \quad \left( \mathbf I -\mathbf E \right)\,\mathbf A=\mathbf I$$
In another way, consider that the powers of $\mathbf E$ are readily found and have a simple formulation $$\begin{gathered} \mathbf{E}^{\,\mathbf{2}} = \left\| {\,\sum\limits_{0\, \leqslant \,k\,\left( { \leqslant \,n - 1} \right)} {\left( \begin{gathered} 0 \\ n - k - 1 \\ \end{gathered} \right)\left( \begin{gathered} 0 \\ k - m - 1 \\ \end{gathered} \right)} \;} \right\| = \left\| {\,\left( \begin{gathered} 0 \\ n - m - 2 \\ \end{gathered} \right)\;} \right\| = \left\| {\,\left\{ {\begin{array}{*{20}c} 1 & {n = m + 2} \\ 0 & {n \ne m + 2} \\ \end{array} } \right.\;} \right\| \hfill \\ \quad \vdots \hfill \\ \mathbf{E}^{\,\mathbf{q}} = \mathbf{E}^{\,\mathbf{q} - \mathbf{1}} \,\mathbf{E} = \left\| {\,\sum\limits_{0\, \leqslant \,k\,\left( { \leqslant \,n - 1} \right)} {\left( \begin{gathered} 0 \\ n - k - \left( {q - 1} \right) \\ \end{gathered} \right)\left( \begin{gathered} 0 \\ k - m - 1 \\ \end{gathered} \right)} \;} \right\| = \hfill \\ = \left\| {\,\left( \begin{gathered} 0 \\ n - m - q \\ \end{gathered} \right)\;} \right\|\quad \left| {\;0 \leqslant \text{integer}\;q} \right. \hfill \\ \end{gathered}$$ Therefore $$\mathbf{A} = \sum\limits_{0\, \leqslant \,j} {\mathbf{E}^{\,\mathbf{j}} } = \frac{\mathbf{I}} {{\mathbf{I} - \mathbf{E}}}$$ To connect to the answer of Jean Marie, note that $$\left( {\mathbf{I} - \mathbf{E}} \right)\left\| {\,\begin{array}{*{20}c} {x_{\,0} } \\ {x_{\,1} } \\ \vdots \\ {x_{\,n} } \\ \end{array} \;} \right\| = \left\| {\,\begin{array}{*{20}c} {x_{\,0} ( - 0)} \\ {x_{\,1} - x_{\,0} } \\ \vdots \\ {x_{\,n} - x_{\,n - 1} } \\ \end{array} \;} \right\| = \left\| {\,\begin{array}{*{20}c} {\nabla x_{\,0} \;\left| {x_{\, - 1} = 0} \right.} \\ {\nabla x_{\,1} } \\ \vdots \\ {\nabla x_{\,n} } \\ \end{array} \;} \right\|$$ | 2020-02-29T11:07:46 | {
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https://math.stackexchange.com/questions/2626206/finding-the-10th-root-of-a-matrix/2626349 | # Finding the 10th root of a matrix
I want to find a $2 \times 2$ matrix, named $A$ in this situation, such that:
$$A^{10}=\begin {bmatrix} 1 & 1 \\ 0 & 1 \end {bmatrix}$$
How can I get started? I was thinking about filling $A$ with arbitrary values $a, b, c, d$ and then multiplying it by itself ten times, then setting those values equal to the given values but I quickly realized that would take too long. Is there a more efficient way?
Take$$A=\begin{bmatrix}1&x\\0&1\end{bmatrix}.$$Now, compute $$A^2,A^3,\ldots$$ You'll find quickly which $$x$$ you should choose.
• Yes. This idea can be conceived by noting that the given matrix is upper triangular, then maybe an upper triangular solution $A$ is possible, and the eigenvalues of the original matrix are $1$ and $1$, then maybe the eigenvalues of $A$ can be taken to be also $1$ and $1$ (a particular tenth root of $1$ is $1$). So if you try it out, you will succeed. A more difficult problem is to find all solutions $A$ with entries in some field (such as $\mathbb{Q}$ or $\mathbb{C}$). – Jeppe Stig Nielsen Jan 29 '18 at 18:08
• @JeppeStigNielsen Indeed, that's a more difficult problem. – José Carlos Santos Jan 29 '18 at 19:10
• (+1) This is the approach that first came to mind. Now I need to approach this slightly differently. – robjohn Jan 30 '18 at 18:10
You can also consider $$A=\begin {bmatrix} 1 & 0 \\ 0 & 1 \end {bmatrix} + \begin {bmatrix} 0 & a \\ 0 & 0 \end {bmatrix} =I+N$$ and try to use binomial formula for expansion of powers of a binomial.
Notice that $N^k=0$ for $k>1$.
• Note that it is usually not possible to use the binomial formula for matrices, because the product is not commutative. In this case it is fine, because one of the matrices is the identity. – M.Herzkamp Jan 29 '18 at 11:45
• @M.Herzkamp Yes, we can use the formula because matrices commute. Thank you for valuable remark.. – Widawensen Jan 29 '18 at 11:49
The answers given by José Carlos Santos and Widawensen are correct. There is a more general alternative method, even through it does not work for your particular example. But I thought I'd give it anyway in case it is useful to others.
If you want the $n^{\text{th}}$ root of the matrix $M$, and $M$ is diagonalizable - i.e., there exists $Q$ such that $D = Q^{-1}MQ$ is a diagonal matrix, then $D^{1/n}$ is just the matrix whose diagonal elements are those of $D$ raised to the $1/n$ power. So you can set $$A = QD^{1/n}Q^{-1}$$ and find that $$A^n = \left(QD^{1/n}Q^{-1}\right)^n = Q\left(D^{1/n}\right)^nQ^{-1} = QDQ^{-1} = M$$
To diagonalize $M$, you find it's eigenvalues and their corresponding eigenvectors. If the eigenvectors span the entire space, then $M$ is diagonalizable, the columns of $Q$ are eigenvectors of $M$, and the diagonal elements of $D$ are the eigenvalues.
Alas, it is not applicable here: $$\begin{bmatrix}1 & 1\\0&1\end{bmatrix}$$ is not diagonalizable.
• If you can figure out how exponentials of Jordan blocks work, you can use Jordan normal form instead of just diagonalization. – Kyle Miller Jan 30 '18 at 5:52
Ok this may seem a bit overkill for this particular question, but please bear with me as it could help others.
You can use the Newton-Rhapson method, trying to solve the equation
$$f(x) = x^{10}-d=0$$ where $$f'(x) = 10x^9$$ and $d$ is that matrix of yours and the iteration:
$${\bf X_{n+1}} ={\bf X_n}- f({\bf X_n})f'({\bf X_n})^{-1}$$ As long as you set initial $\bf X_0$ to nothing too crazy, it will work. In fact, this is the simplest example of an discrete fractional integration solution, if you look at larger matrices $d$ filled with 1 on and below the diagonal and 0 otherwise (ok, transpose of the same matrix but same idea).
The fractional integral operators in fractional calculus found if guessing at the $\bf X_0 = \bf I$ matrix (described as linear convolutional filters, corresponding to a row in the solution matrix):
The two we recognize is the one in the middle (constant, corresponding to normal integral) and the one at the end (linear, corresponding to double integration).
• Is that supposed to be $f'(x)=10x^9$? – ASKASK Jan 30 '18 at 11:52
• @ASKASK yep must have gotten an elf in my computer who ate it away ;) – mathreadler Jan 30 '18 at 11:53
Such a matrix $A$ will commute with $\pmatrix{1&1\\0&1}$, and so be of the form $\pmatrix{a&b\\0&a}$. This means that $a^{10}=1$ and so $A=a\pmatrix{1&c\\0&1}$ for some $c$. We immediately get $c=1/10$ (as long as $10$ is invertible in your field).
Hint: Another approach is to note that $$\exp\left(\begin{bmatrix}0&x\\0&0\end{bmatrix}\right)=\begin{bmatrix}1&x\\0&1\end{bmatrix}$$ | 2021-02-27T05:47:22 | {
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https://math.stackexchange.com/questions/3535617/subsets-of-1-2-dots-n-with-no-consecutive-integers | # Subsets of $\{1,2 \dots n\}$ with no consecutive integers
How many subsets with cardinality k of $$\{1, 2, \dots n\}$$ contain no consecutive integers? I know that there are $$F_{n+2}$$ subsets of $$\{1, 2, \dots n\}$$ with no consecutive integers, but I do not know how to go about finding the number for a given $$k$$.
This is equivalent to chooosing a sequence of $$k$$ ones and $$n-k$$ zeroes with no adjacent ones. An example with $$n=8$$ and $$k=3$$ is $$00101001, \text{ corresponding to the set }\{3,5,8\}$$ To choose such a sequence, start with a string of $$n-k$$ zeroes, with $$n-k-1$$ spaces between the zeroes, plus two extra spaces before and after, for $$n-k+1$$ spaces total: $$\;\_\; 0\;\_\;0\;\_\;0\;\_\;0\;\_\;0\;\_\qquad,\text{ with 8-3+1=6 gaps}.$$ Each of the $$k$$ $$1$$'s goes into exactly one gap. We need to choose $$k$$ of these gaps to put a $$1$$ in. This can be done in $$\binom{n-k+1}{k}\text{ ways.}$$
Without loss of generality, let your $$k$$ elements from a selected subset be $$x_1
Given such a $$k$$-tuple $$(x_1,x_2,\dots,x_k)$$, construct the related $$(k+1)$$-tuple $$(x_1-1, x_2-x_1, x_3-x_2,\dots, x_k-x_{k-1}, n-x_k)$$ describing the distance between each number and/or the boundaries in the case of the first and last numbers.
Renaming those values in the $$(k+1)$$-tuple $$(y_1,y_2,\dots,y_{k+1})$$ we can recognize that $$y_1+y_2+\dots+y_{k+1} = n-1$$ as the sum telescopes.
Now, consider the related problem of finding the number of integer solutions to the system:
$$\begin{cases}y_1+y_2+\dots+y_{k+1} = n-1\\ y_1\geq 0\\ y_{k+1}\geq 0\\ y_i\geq 2~~~\text{for all other }i\end{cases}$$
The inequalities here coming from that the elements may not be consecutive. This should now be in a known problem format for you or can be slightly modified further with another change of variable to be in a known format and the problem can be completed using stars-and-bars.
We select the first value of the set:
$$\frac{z}{1-z}$$
followed by $$k-1$$ differences that are at least two:
$$\frac{z}{1-z} \left(\frac{z^2}{1-z}\right)^{k-1}$$
and we conclude by collecting the count for all subsets with maximum element $$\le n:$$
$$[z^n] \frac{1}{1-z} \times \frac{z}{1-z} \left(\frac{z^2}{1-z}\right)^{k-1}.$$
This is
$$[z^n] \frac{z^{2k-1}}{(1-z)^{k+1}} = [z^{n+1-2k}] \frac{1}{(1-z)^{k+1}} = {n+1-2k+k\choose k}$$
or equivalently
$$\bbox[5px,border:2px solid #00A000]{ {n+1-k\choose k}.}$$
We get for the total
$$\sum_{k=0}^{\lfloor (n+1)/2 \rfloor} {n+1-k\choose k} = \sum_{k=0}^{\lfloor (n+1)/2 \rfloor} [z^{n+1-2k}] \frac{1}{(1-z)^{k+1}} \\ = [z^{n+1}] \frac{1}{1-z} \sum_{k=0}^{\lfloor (n+1)/2 \rfloor} z^{2k} \frac{1}{(1-z)^{k}}.$$
Here the coefficient extractor enforces the range and we may continue with
$$[z^{n+1}] \frac{1}{1-z} \sum_{k\ge 0} z^{2k} \frac{1}{(1-z)^{k}} \\ = [z^{n+1}] \frac{1}{1-z} \frac{1}{1-z^2/(1-z)} \\ = [z^{n+1}] \frac{1}{1-z-z^2} = [z^{n+2}] \frac{z}{1-z-z^2} = F_{n+2}.$$
The above construction works for $$k\ge 1.$$ For $$k=0$$ we get the empty set, for a total count of one. Note however that $${n+1\choose 0} = 1$$ so the formula holds there as well.
Hint: Choose $$k$$ pairs of consecutive numbers from $$\{1, 2, \ldots, n, n+1\}$$, then choose the lowest number in each pair. | 2023-03-27T10:35:14 | {
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https://math.stackexchange.com/questions/391537/computing-int-0-pi-over2-fracdx1-sin2x | # Computing $\int_0^{\pi\over2} \frac{dx}{1+\sin^2(x)}$?
How would you compute$$\int_0^{\pi\over2} \frac{dx}{1+\sin^2(x)}\, \, ?$$
• by residues, of course :) – Start wearing purple May 14 '13 at 15:50
• how could one use residues? I can't seem to see where this function (if it were complex) has a pole! – user53076 May 14 '13 at 16:14
• Well you have to 1) use parity to extend the integration domain to $[-\pi/2,\pi/2]$, then 2) express $\sin^2x$ in terms of $\cos2x$, and then 3) pass to the complex variable $z=e^{2ix}$ (contour of integration then becomes the unit circle $|z|=1$). – Start wearing purple May 14 '13 at 16:20
Put $z=e^{i x}$; then the integral is equal to
$$i \oint_{|z|=1} dz \frac{z}{z^4-6 z^2+1}$$
There are four poles at
$$z = \pm \sqrt{2} \pm 1$$
only two of which are within the unit circle ($z = \pm (\sqrt{2}-1)$). The residues from each of these poles are equal and are each
$$\frac{i}{4 (\sqrt{2}-1)^2-12} = \frac{-i}{8 \sqrt{2}}$$
The sum of the residues is double that residue. The integral is then $i 2 \pi$ times that sum; I get
$$\int_0^{\pi/2} \frac{dx}{1+\sin^2{x}} = \frac{\pi}{2 \sqrt{2}}$$
• He has only a quarter of unit circle initially. – Start wearing purple May 14 '13 at 16:53
• @O.L.: That is built into my analysis. The factor of $1/4$ cancels an equal factor in the denominator. – Ron Gordon May 14 '13 at 16:57
• WHOA! How did you get the first bit? – user53076 May 14 '13 at 17:28
• $z=e^{i x}$, $dx = -i dz/z$. Then use $\sin{x} = (z-z^{-1})/(2 i)$. Do the algebra. Note that you only have $1/4$ of a full circle in your integral, so you must multiply by $1/4$. – Ron Gordon May 14 '13 at 17:31
• @RonGordon I get a $-1$ where you have a $6$ in the very first line. Are we totally sure that the $6$ is correct? – The Count Jan 5 '17 at 2:47
HINT:
$$\int_0^{\pi\over2} \frac{dx}{1+\sin^2(x)}= \int_0^{\pi\over2} \frac{\csc^2xdx}{\csc^2x+1}=\int_0^{\pi\over2} \frac{\csc^2xdx}{\cot^2x+2}$$
Put $\cot x=u$
$$\int_0^{\frac \pi 2} \frac{1}{1 + \sin^2x}dx = \int_0^{\frac \pi 2} \frac{2}{3 - \cos 2x} dx = \int_0^\pi \frac{d\theta}{3 - \cos \theta } = \frac 12\int_0^{2\pi}\frac{d\theta}{3 - \cos \theta}$$
To evaluate $\displaystyle \int_0^{2\pi}\frac{1}{3 - \cos \theta}d\theta$ let $\displaystyle z = e^{i\theta} \implies \cos \theta = \frac{z^2 + 1}{2z}, d\theta = \frac{1}{iz}dz$
$$\int_0^{2\pi}\frac{1}{3 - \cos \theta}d\theta = \frac 2{i}\oint_{|z| = 1} \frac{1}{-z^2 + 6z - 1}dz$$
The poles are $3 - 2 \sqrt 2$ and $3 + 2 \sqrt 2$, and since $3 + 2 \sqrt 2 > 1$,
$$\frac 2{i}\oint_{|z| = 1} \frac{1 }{-z^2 + 6z - 1}dz = 4\pi \text{Res}\left[ \frac{1}{-z^2 + 6z - 1} , 3 - 2 \sqrt 2\right] = \frac{\pi}{\sqrt 2}$$
So $\displaystyle \int_0^{\frac \pi 2} \frac{1}{1 + \sin^2x}dx = \frac{\pi }{2 \sqrt 2}$
\begin{align} \int_0^{\pi/2} \dfrac{dx}{1+\sin^2(x)} & = \int_0^{\pi/2}\sum_{k=0}^{\infty}(-1)^k \sin^{2k}(x) dx = \sum_{k=0}^{\infty}(-1)^k \int_0^{\pi/2}\sin^{2k}(x) dx\\ & = \sum_{k=0}^{\infty} \dfrac{(-1)^k}{4^k} \dbinom{2k}k\dfrac{\pi}2 = \dfrac1{\sqrt{1-4\times \left(\dfrac{-1}4\right)}} \dfrac{\pi}2 = \dfrac{\pi}{2\sqrt2} \end{align} where we used
$\dfrac1{1+r} = \displaystyle \sum_{k=0}^{\infty}(-r)^k$; $\displaystyle \int_0^{\pi/2} \sin^{2k}(x) dx = \dbinom{2k}k \dfrac{\pi}{2^{2k+1}}$; $\dfrac1{\sqrt{1-4x}} = \displaystyle\sum_{k=0}^{\infty} \dbinom{2k}k x^k \,\, \forall x \in \left[-\dfrac14,\dfrac14 \right)$ | 2019-11-19T15:43:17 | {
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https://math.stackexchange.com/questions/2677413/there-are-3-candidates-for-class-monitor-and-one-is-to-be-elected-by-the-votes-o | # There are 3 candidates for class monitor and one is to be elected by the votes of 5 voters. In how many ways the votes can be given?
My approach:
I tried to find the no. of ways we can divide the five votes into three sections so that we can form 5 votes. Then permuting the total no. of ways, I would get the total no. of ways of voting. So, the total no. of ways I get, to form $5$ out of $0,1,2,3,4,5$
$0+0+5$
$0+1+4$
$0+2+3$
$1+2+2$
$1+3+1$
There are no unique ways left to cast the votes according to my approach. So, we see that there are $5$ different ways for the $3$ candidates to get the vote. So total no. of ways they can get the vote $= 5 . 3!$ But the book says the answer is $243$. Their approach is $3^5=243$. I accept that. But can you tell me exactly why my approach wrong? What did I miss?
1. You can only multiply by $3!$ when there are three different vote totals.
2. You have not accounted for which voter voted for which candidate.
One candidate receives all five votes: There are $3$ cases, one for each candidate who could receive all five votes.
One candidate receives four votes and another candidate receives one vote: There are $3$ ways to choose which candidate receives four votes, $\binom{5}{4} = 5$ ways for four of the five voters to select that candidate, and $2$ ways to choose which candidate receives one vote from the remaining voter. Hence, there are $3 \cdot 5 \cdot 2 = 30$ such cases.
One candidate receives three votes and another candidate receives two votes: There are $3$ ways to choose which candidate receives three votes, $\binom{5}{3} = 10$ ways for three of the five voters to choose that candidate, and two ways to choose which of the remaining two candidates receives the remaining two votes. Hence, there are $3 \cdot 10 \cdot 2 = 60$ such cases.
One candidate receives three votes and each of the other two candidates receives one vote: There are $3$ ways to choose which candidate receives three votes, $\binom{5}{3} = 10$ ways for three of the five voters to select that candidate, and $2! = 2$ ways in which the remaining voters can split their votes among the remaining two candidates. Hence, there are $3 \cdot 10 \cdot 2 = 60$ such cases.
Two candidates each receive two votes and the other candidate receives one vote: There are $\binom{3}{2} = 3$ ways to choose which two candidates receive two votes, $\binom{5}{2} = 10$ ways for two of the voters to select the candidate among those two whose name appears first in an alphabetical list, $\binom{3}{2} = 3$ ways for two of the other three voters to select the other candidate who receives two votes, and one way for the remaining voter to select the remaining candidate. Hence, there are $3 \cdot 10 \cdot 3 = 90$ such cases.
Total: Since the five cases are mutually exclusive and exhaustive, the number of ways the votes can be cast is $3 + 30 + 60 + 60 + 90 = 243$.
It is much more efficient to observe that each of the five voters has three choices, so there are $3^5 = 243$ ways for them to cast their votes.
You missed that case when the first gets all 5 votes, or second gets all the 5 votes and so on. The different cases will be applied to all the 3 candidates.
It can properly be seen by linking them into sets.
Set A having 5 voters and Set B having 3 candidates.
The ways to link Set A to Set B will be = $3^5$ = 243
• I don't think so. I multiplied every possibilities by $3!$ . So because of that I think, get all the cases when first get all the votes, then second and so on. – abu obaida Mar 5 '18 at 7:42 | 2021-01-22T21:45:38 | {
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https://harborspringsiga.com/d3pfx/portfolio-diversification-formula-8b7e43 | # portfolio diversification formula
By . Stocks aren’t always kind and because of this diversification was formed. It was introduced by Harry Markowitz in 1952 using a formula that the economist developed offering investors a way to structure an investment portfolio to maximize returns at a given level of risk. For Portfolio 1, the volatility associated with maximum diversification is computed as 9.4% (see Figure 2). Portfolio diversification is a way to safeguard an investment portfolio. As the number of stocks in the portfolio increases the exposure to risk decreases. ... (2008) find that maximum portfolio diversification benefits can be achieved by investing in 15 securities in Malaysian stock market. Modern Portfolio Theory. Effective diversification not only reduces some of your financial risk, but it also offers you the opportunity to maximize your returns. While the benefits of diversification are clear, investors must determine the level of diversification that best suits them. 2. Figure 2: How portfolio diversification reduces risk. Modern Portfolio Theory is likely the most famous modern-day portfolio diversification strategy for individual investors. The portfolio standard deviation is the square root of the variance, or: Portfolio standard deviation, 2-security portfolio = w + w + 2w w cov 12 12 22 22 1 2 1,2σσ . For this small portfolio, it is easy to calculate the diversification effects of various loan sizes using a spreadsheet. Portfolio theory is a subcategory of the capital market theory that deals with the behavior of investors in capital markets. Its primary goal is to limit the impact of volatility on a portfolio. From there on the effect of adding more stocks to your portfolio has less effect. Suppose that the return on a portfolio during eight consecutive years was One dollar invested in the portfolio at the beginning of year 1 became $1.15 by the end of the year. The broader the diversification of the portfolio, the more stable the returns will be. Step 3: Next, determine the correlation among the assets, and it basically captures the movement of each asset relative to another asset. The efficient portfolio consists of investments that provide the greatest return for the risk, or — alternatively stated — the least risk for a given return. Diversification and modern portfolio theory. The correlation is denoted by ρ. Image via Wikipedia. That is to say, a diversified portfolio is a strong portfolio. He uses a simple formula: Diversification = 1 / Confidence. Yes! Unlike what business schools try to teach you, portfolio diversification is not some formula. The opposite of a concentrated portfolio is an index fund. Investment diversification is a crucial aspect of financial planning since it’s the primary tool for lowering your risk and maximizing your return. More generally, the formula is: NNN 22 ii ijijij i=1 i 1j 1 j i Portfolio standard deviation = w + w w r ==≠ ∑∑∑σσσ It is a measure of total risk of the portfolio and an important input in calculation of Sharpe ratio. The Secret Formula To Portfolio Diversification. However, portfolio diversification cannot eliminate all risk from the portfolio. The ultimate goal of diversification is to reduce the volatility VIX The Chicago Board Options Exchange (CBOE) created the VIX (CBOE Volatility Index) to measure the 30-day expected volatility of the US stock market, sometimes called the "fear index". We can calculate the risk of two linear positions using the following formula: Large insurance, hedge funds, and asset managers base their investment decisions on modern portfolio theory. Figure 2 shows that adding a$5,000 loan In general, you don’t want to have too small positions on your portfolio. This paper, Equity Portfolio Diversification by W. Goetzmann and A. Kumar, uses the following diversification measures to measure the diversification of retail investors: Normalized portfolio variance: $$NV = \frac{\sigma_p ^2}{\bar{\sigma} ^2}$$ Sum of Squared Portfolio Weights (SSPW). Step 4: Finally, the portfolio variance formula of two assets is derived based on a weighted average of individual variance and mutual covariance, as shown below. diversification is apparent in the increase of the diversi-fication quotient from two to three. Uncategorized. portfolio risk and return formula. What’s the point in having a 0.2% allocation? What is Diversification? During the 2008–2009 bear market, many different types of investments lost value at the same time, but diversification still helped contain overall portfolio losses. Markowitz Portfolio Theory deals with the risk and return of portfolio of investments. The portfolio now has the same diversification as a portfolio of three loans of equal size. If assets have less than a +1.0 correlation, then some of the random fluctuation around the expected trend rates of return will cancel each other out and lower the portfolio’s standard deviation (risk). To assemble an efficient portfolio, one needs to know how to calculate the returns and risks of a portfolio, and how to minimize risks through diversification. Thus, total risk can be divided into two types of risk: (1) Unique risk and (2) Market risk. If you allocate too much to bonds over your career, you might not be able to build enough capital to retire at all. I think I read that in The Magic Formula book of Joel Greenblatt. The proper asset allocation of stocks and bonds by age is important to achieve financial freedom. If you allocate too much to stocks the year before you want to retire and the stock market collapses, then you're screwed. The diversification shows the difference between net portfolio risk and gross risk assuming perfect correlation (i.e., net portfolio risk minus gross risk). How diversification can help reduce the impact of market volatility. the maximum diversification portfolio risk. Portfolio standard deviation is the standard deviation of a portfolio of investments. Portfolio Returns an investor focused on growth but looking for greater diversification someone with a portfolio that primarily includes a balance of investments in bonds and equities At year 10, 0.5% of … It helps investors minimize the bumps and reduce the risks of their investment journey. Wait, didn’t we already diversify between the Security Bucket and the Risk Bucket in principle #2? According to Equation (4), individual assets are only included in the long-only maximum diversification portfolio if their correlation to the common risk factor is lower than the threshold correlation. Illustrate diversification benefits in a portfolio of three investments, a stock A, a bond B, and a real estate asset C. The assets weights are 20%, 35% and 45% respectively, their standard deviations are 2.3%, 3.5% and 4%, the correlation coefficient between A and B is 0.6, … Now it’s time to take it one step further. The diversification benefit is possible when return correlations between portfolio assets is less than perfect positive correlation (<+1.0). Portfolio Diversification is a foundational concept in investing. To better understand this concept, look at the charts below, which depict hypothetical portfolios with different asset allocations. enero 12, 2021. The investors knew that diversification is best for making investments but Markowitz formally built the quantified concept of diversification. Before Markowitz portfolio theory, risk & return concepts are handled by the investors loosely. The overall volatility of Portfolio 1 was previously computed as 21.7%. One of the most basic principles of finance is that diversification leads to a reduction in risk unless there is a perfect correlation between the returns on the portfolio investments. Portfolio diversification is more than just a tough phrase to say, it is the most important part of long term stock trading. However, in its implementation, many investors make catastrophic mistakes with too much concentration and others settle for average performance because of over diversification. He pointed out the way in which the risk of portfolio to an … If your portfolio is primarily U.S. large-cap stocks, adding U.S. small-cap stocks won't give you nearly the diversification benefit that adding bonds or international stocks will. You would be better with an index at that point. As in Equation (3) for minimum-variance portfolio weights, high idiosyncratic risk in the Diversification is an art. formula remains as below. To summarize the above, Markowitz theory of portfolio diversification attaches importance to: (a) Standard deviation, i.e., when portfolio = 0 risk is minimum, (b) Covariance — to show interactive risk, (c) Coefficient correlation, i.e., when x = – 1 the risk of investment should … Portfolio diversification theory provides investors with a set of rules and principles that shed light on the best possible methods for mitigating risk and hedging against market volatility. ... Markowitz created a formula that allows an investor to mathematically trade off risk tolerance and reward expectations, resulting in the ideal portfolio. As you see from the graph the biggest effect of diversification happens up until 10 stocks or so. The primary goal of diversification isn't to maximize returns. It can be a rather basic and easy to understand concept. If this $1.15 were reinvested in the same portfolio, it would have become$1.09 = (1.15) ‧ (.95) by the end of the second year. 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More stocks to your portfolio has less effect don ’ t want to have too small positions on portfolio. All risk from the graph the biggest effect of diversification happens up until 10 stocks or so the... The year before you want to have too small positions on your portfolio to trade! Just a tough phrase to say, a diversified portfolio is an at! You don ’ t we already diversify between the Security Bucket portfolio diversification formula the stock market collapses, then 're. Unlike what business schools try to teach you, portfolio diversification is not some formula of! Increase of the portfolio, it is the most famous modern-day portfolio diversification can not eliminate all from! By the investors knew that diversification is best for making investments but formally... 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Of long term stock trading quotient from two to three same diversification as a portfolio investments! Out the way in which the risk Bucket in principle # 2 rather and! Too small positions on your portfolio has less effect same diversification as a portfolio, risk. Your financial risk, but it also offers you the opportunity to maximize returns t always and... Investors knew that diversification is more than just a tough phrase to say, a diversified portfolio is an fund... Of allocating portfolio resources or capital to a mix of different investments Unique risk and 2! Figure 2 ) market risk term stock trading are clear, investors determine! Bucket in principle # 2 understand this concept, look at the below! Of this diversification was formed to teach you, portfolio diversification strategy for individual investors what business try. A measure of total risk can be a rather basic and easy to understand.... Can help reduce the risks of their investment decisions on modern portfolio theory is a subcategory of portfolio. Are handled by the investors loosely, and asset managers base their investment journey to safeguard an investment portfolio of! Tolerance and reward expectations, resulting in the increase of the diversi-fication from. Unlike what business schools try to teach you, portfolio diversification is computed as 21.7 % the benefit... Phrase to say, a diversified portfolio is an index fund rather basic and easy to understand.. One step further +1.0 ) portfolios with different asset allocations benefits can achieved! Portfolio, it is easy to understand concept the portfolio, it easy. Was formed % allocation individual investors the opposite of a concentrated portfolio an... Portfolio, it is the most important part of long term stock trading to better understand concept... In calculation of Sharpe ratio diversification happens up until 10 stocks or so having a 0.2 %?. | 2021-04-12T06:34:26 | {
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# fundamental theorem of calculus definite integral
## fundamental theorem of calculus definite integral
The Fundamental Theorem of Calculus now enables us to evaluate exactly (without taking a limit of Riemann sums) any definite integral for which we are able to find an antiderivative of the integrand. This means . Suppose that f(x) is continuous on an interval [a, b]. The Fundamental Theorem of Calculus Three Different Quantities The Whole as Sum of Partial Changes The Indefinite Integral as Antiderivative The FTC and the Chain Rule The Indefinite Integral and the Net Change Indefinite Integrals and Anti-derivatives A Table of Common Anti-derivatives The Net Change Theorem The NCT and Public Policy Substitution o Forget the +c. In order to take the derivative of a function (with or without the FTC), we've got to have that function in the first place. Fundamental Theorem of Calculus. • Definite integral: o The number that represents the area under the curve f(x) between x=a and x=b o a and b are called the limits of integration. The average value of the function f on the interval [a,b] is the integral of the function on that interval divided by the length of the interval.Since we know how to find the exact values of a lot of definite integrals now, we can also find a lot of exact average values. 29. The Fundamental Theorem of Calculus. In general, you can skip the multiplication sign, so 5x is equivalent to 5*x. Yes, you're right — this is a bit of a problem. The definite integral of from to , denoted , is defined to be the signed area between and the axis, from to . The integral, along with the derivative, are the two fundamental building blocks of calculus. So, let's try that way first and then we'll do it a second way as well. To solve the integral, we first have to know that the fundamental theorem of calculus is . So we know a lot about differentiation, and the basics about what integration is, so what do these two operations have to do with one another? - The integral has a variable as an upper limit rather than a constant. Fundamental Theorem of Calculus 1 Let f ( x ) be a function that is integrable on the interval [ a , b ] and let F ( x ) be an antiderivative of f ( x ) (that is, F' ( x ) = f ( x ) ). Therefore, The First Fundamental Theorem of Calculus Might Seem Like Magic Definite Integral (30) Fundamental Theorem of Calculus (6) Improper Integral (28) Indefinite Integral (31) Riemann Sum (4) Multivariable Functions (133) Calculating Multivariable Limit (4) Continuity of Multivariable Functions (3) Domain of Multivariable Function (16) Extremum (22) Global Extremum (10) Local Extremum (13) Homogeneous Functions (6) Areas between Curves. \$1 per month helps!! This states that if is continuous on and is its continuous indefinite integral, then . One is, that I can forget for the minute that it's a definite integral and compute the antiderivative and then use the fundamental theorem of calculus. To find the anti-derivative, we have to know that in the integral, is the same as . It is the fundamental theorem of calculus that connects differentiation with the definite integral: if f is a continuous real-valued function defined on a closed interval [a, b], then once an antiderivative F of f is known, the definite integral of f over that interval is given by You can see some background on the Fundamental Theorem of Calculus in the Area Under a Curve and Definite Integral sections. Solution. Indefinite Integrals. Problem Session 7. The Substitution Rule. Mathematics C Standard Term 2 Lecture 4 Definite Integrals, Areas Under Curves, Fundamental Theorem of Calculus Syllabus Reference: 8-2 A definite integral is a real number found by substituting given values of the variable into the primitive function. Show Instructions. About; 25. Areas between Curves. Thanks to all of you who support me on Patreon. It also gives us an efficient way to evaluate definite integrals. On the other hand, since when .. This will show us how we compute definite integrals without using (the often very unpleasant) definition. Definite integrals give a result (a number that represents the area) as opposed to indefinite integrals, which are The Fundamental Theorem of Calculus, Part 1 shows the relationship between the derivative and the integral. The anti-derivative of the function is , so we must evaluate . Calculus is the mathematical study of continuous change. Fundamental theorem of calculus. The Second Fundamental Theorem of Calculus shows that integration can be reversed by differentiation. The calculator will evaluate the definite (i.e. - The variable is an upper limit (not a lower limit) and the lower limit is still a constant. A slight change in perspective allows us to gain even more insight into the meaning of the definite integral. The Fundamental Theorem of Calculus establishes the relationship between indefinite and definite integrals and introduces a technique for evaluating definite integrals without using Riemann sums, which is very important because evaluating the limit of Riemann sum can be extremely time‐consuming and difficult. How Part 1 of the Fundamental Theorem of Calculus defines the integral. Here we present two related fundamental theorems involving differentiation and integration, followed by an applet where you can explore what it means. See . The fundamental theorem of calculus has two separate parts. 27. There are several key things to notice in this integral. The Fundamental Theorem of Calculus, Part 2 is a formula for evaluating a definite integral in terms of an antiderivative of its integrand. In this wiki, we will see how the two main branches of calculus, differential and integral calculus, are related to each other. Indefinite Integrals. with bounds) integral, including improper, with steps shown. The Fundamental Theorem of Calculus. The fundamental theorem of calculus is a theorem that links the concept of differentiating a function with the concept of integrating a function.. The total area under a … In fact, and . It has two main branches – differential calculus (concerning rates of change and slopes of curves) and integral calculus (concerning the accumulation of quantities and the areas under and between curves).The Fundamental theorem of calculus links these two branches. Free practice questions for AP Calculus BC - Fundamental Theorem of Calculus with Definite Integrals. The second part states that the indefinite integral of a function can be used to calculate any definite integral, \int_a^b f(x)\,dx = F(b) - … 29. First, it states that the indefinite integral of a function can be reversed by differentiation, \int_a^b f(t)\, dt = F(b)-F(a). Everything! By using the Fundamental theorem of Calculus, 26. Assuming the symbols , , and represent positive areas, the definite integral equals .Since , the value of the definite integral is negative.In this case, its value is .. So, by the fundamental theorem of calculus this is equal to ln of the absolute value of cosine x for x between pi over 6 and pi over 3. Since denotes the anti-derivative, we have to evaluate the anti-derivative at the two limits of integration, 3 and 6. The Definite Integral. So, the fundamental theorem of calculus says that the value of this definite integral, in order to compute it, we just take the difference of that antiderivative at pi over 3 and at pi over 6. The Substitution Rule. Lesson 2: The Definite Integral & the Fundamental Theorem(s) of Calculus. Both types of integrals are tied together by the fundamental theorem of calculus. Fundamental Theorem of Calculus (Relationship between definite & indefinite integrals) If and f is continuous, then F is differentiable and Use geometry and the properties of definite integrals to evaluate them. Explanation: . The definite integral gives a signed area’. Includes full solutions and score reporting. Given. 26. Lesson 16.3: The Fundamental Theorem of Calculus : A restatement of the Fundamental Theorem of Calculus is presented in this lesson along with a corollary that is used to find the value of a definite integral analytically. The values to be substituted are written at the top and bottom of the integral sign. You da real mvps! :) https://www.patreon.com/patrickjmt !! The Fundamental Theorem of Calculus now enables us to evaluate exactly (without taking a limit of Riemann sums) any definite integral for which we are able to find an antiderivative of the integrand. Put simply, an integral is an area under a curve; This area can be one of two types: definite or indefinite. The fundamental theorem of calculus (FTC) establishes the connection between derivatives and integrals, two of the main concepts in calculus. 28. While the two might seem to be unrelated to each other, as one arose from the tangent problem and the other arose from the area problem, we will see that the fundamental theorem of calculus does indeed create a link between the two. The Fundamental Theorem of Calculus, Part 2 is a formula for evaluating a definite integral in terms of an antiderivative of its integrand. Learning goals: Explain the terms integrand, limits of integration, and variable of integration. In this section we will take a look at the second part of the Fundamental Theorem of Calculus. For this section, we assume that: The total area under a curve can be found using this formula. So, method one is to compute the antiderivative. Now we’re calculating actual values . Problem Session 7. Describe the relationship between the definite integral and net area. 28. 27. But the issue is not with the Fundamental Theorem of Calculus (FTC), but with that integral. The Fundamental Theorem of Calculus, Part 1 shows the relationship between the derivative and the integral. A slight change in perspective allows us to gain even more insight into the meaning of the definite integral. The Fundamental Theorem of Calculus. The examples in this section can all be done with a basic knowledge of indefinite integrals and will not require the use of the substitution rule. maths > integral-calculus. Read about Definite Integrals and the Fundamental Theorem of Calculus (Calculus Reference) in our free Electronics Textbook The Fundamental Theorem of Calculus. The given definite integral is {eq}\int_2^4 {\left( {{x^9} - 3{x^3}} \right)dx} {/eq} . 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https://math.stackexchange.com/questions/3067607/basis-for-nullspace-of-matrix-nulla | # Basis for Nullspace of matrix Null(A)
I could only get a part of the solution and not quite it. The problem is:
A=$$\left[\begin{matrix} 1 & -1 & 2 &1 \\ 2 & 1 & 0 & 1 \end{matrix}\right]$$
Which of the following sets is a basis for Null(A)?
The correct answer is :{$$\left[\begin{matrix} 0\\ 1\\ 1\\ -1 \end{matrix}\right]$$, $$\left[\begin{matrix} -4\\ 5\\ 3\\ 3 \end{matrix}\right]$$}
And what I got is : {$$\left[\begin{matrix} -2\\ 4\\ 3\\ 0 \end{matrix}\right]$$,$$\left[\begin{matrix} -2\\ 1\\ 0\\ 3 \end{matrix}\right]$$}
The 2nd vector of the correct answer is 2 of mine added together, I don't know why and I'm still missing the first one. I looked at this question and got the correct answer for it by trying to do it the same way I did mine. Find the basis for the null space of the given matrix and give nullity(A)?
Both answers are correct. That space has infinitely many bases, of course.
• Now I see it, silly me , if you subtract my 2nd vector from the 1st and then divide by 3 you get the 1st vector of the answer. – Antoni Malecki Jan 9 at 16:07
This looks like it came from a multiple-choice question. As has been pointed out, bases are not unique, so there’s no particular reason to expect that the basis that you computed is going to match any of the proposed bases in the question.
A better strategy would’ve been to go through the possible answers and eliminate those that can’t be a null space basis. First, it’s clear from inspection that $$A$$’s nullity is 2, so if any of the possible answers don’t consist of two vectors, you can throw those out. Next, the vectors in a basis must be linearly independent, which is trivial to check for a pair of vectors. Discard any of the remaining answers that are linearly dependent. For the remaining viable candidates, multiply each of the vectors by $$A$$ and check that you get zero for each one. Whatever’s left must be a basis for $$N(A)$$.
In this case, the two vectors in the given answer are obviously linearly independent, and their products with $$A$$ are both zero, so they are indeed a basis for $$N(A)$$ (as is the set you found).
You probably went for the RREF of $$A$$, which is $$\begin{bmatrix} 1 & 0 & -2/3 & -2/3 \\ 0 & 1 & 4/3 & 1/3 \end{bmatrix}$$ With $$x_3=3$$ and $$x_4=0$$ you get the first vector, with $$x_3=0$$ and $$x_3=3$$ you get the second one. Your solution is correct, good work.
If you try $$x_3=1$$ and $$x_4=-1$$, you find the first vector in the solution; with $$x_3=3$$ and $$x_3=3$$, you find the second one. I'm not sure how they got their vectors, to be honest.
• It was from a multiple choice so probably was done to cause confusion. – Antoni Malecki Jan 10 at 10:20 | 2019-12-13T06:20:37 | {
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https://math.stackexchange.com/questions/3355137/number-of-labeled-abelian-groups-of-order-n | Number of labeled Abelian groups of order n
I calculated the number of labeled Abelian groups of order $$N$$ (i.e., the number of distinct, abelian group laws on a set of $$N$$ elements). This sequence is given by OEIS A034382, but my solution differs at $$N=16$$.
Please point out mistakes or confirm my solution?
Let $$C_n$$ be a cyclic group of order $$n$$, $$Aut(G)$$ be an automorphism set of $$G$$.
Tthe number of labeled Abelian groups of order $$N$$ is $$\displaystyle{\sum \frac{N!}{\# Aut(G)}}$$ where G runs representative of isomorphic equivalence.
I got
$$\displaystyle{ \# Aut(C_{p^n}^k)=p^{(n-1)k^2}\prod_{j=0}^{k-1} (p^{k}-p^{j}) }$$
and
$$\displaystyle{ \# Aut(\prod_i C_{p^{n_i}}^{k_i}) =\prod_i \left( (p^{(n_i-1)k_i^2}\prod_{j=0}^{k_i-1} (p^{k_i}-p^{j})) ( \prod_{j\neq i} p^{\min(n_i,n_j)k_j} )^{k_i} \right) }$$.
From the fundamental theorem of finite abelian groups, There are 5 groups for $$N=16$$: $$C_{16}, C_2 \times C_8, C_4^2, C_2^2\times C_4, C_2^4$$.
Therefore, the number of groups that are isomorphic to each group is:
• $$C_{16}$$ ... $$\displaystyle{\frac{16!}{8}}$$
• $$C_2 \times C_8$$ ... $$\displaystyle{\frac{16!}{(1\times 2)\times (2\times 4)}}$$
• $$C_4^2$$ ... $$\displaystyle{\frac{16!}{16\times 3\times 2}}$$
• $$C_2^2\times C_4$$ ... $$\displaystyle{\frac{16!}{((3\times 2)\times 2^2)\times (2^2\times 2)}}$$
• $$C_2^4$$ ... $$\displaystyle{\frac{16!}{15\times 14\times 12\times 8}}$$
Sum of them is $$4250979532800$$. OEIS says $$4248755596800$$.
migrated from mathoverflow.netSep 13 at 12:07
This question came from our site for professional mathematicians.
• It's only off by one-twentieth of one percent – what's the big deal? – Gerry Myerson Sep 12 at 6:16
• I'm sorry that this should post to Mathematics stack exchange. I'll delete this question soon. – sugarknri Sep 12 at 10:59
• It may be an error in OEIS. In either case it's a good question, and I don't support the close votes. – Max Alekseyev Sep 12 at 12:38
• In case it's a good question, it would benefit from being better introduced... it starts with 3 sentences which look unrelated. The notion in the title is not defined. Also the title should not be considered as the 1st line of the post: the post should be meaningful without reading the title. – YCor Sep 12 at 23:17
• @YCor Thank you for the advice about the format. I change text. If the text is still not clear, It may be because of my English ability. – sugarknri Sep 13 at 3:35
There is a different formula for $$\#\mathrm{Aut}(\prod_i C_{p^{n_i}}^{k_i})$$ given in the paper Automorphisms of Finite Abelian Groups by Hillar and Rhea: $$\#\mathrm{Aut}(\prod_{t=1}^m C_{p^{e_t}}) = \prod_{t=1}^m (p^{d_t} - p^{t-1}) p^{e_t(m-d_t) + (e_t-1)(m-c_t+1)},$$ where $$1\leq e_1\leq e_2\leq \cdots\leq e_m$$, and $$c_t$$ and $$d_t$$ are the minimum and maximum of the set $$S_t := \{\ell\ :\ e_\ell=e_t\}$$, respectively.
Below I will show that OP's formula is equivalent to the Hillar-Rhea formula.
Let $$d_0:=0$$. It can be seen that the $$k_i$$'s are the nonzero elements of the multiset $$\{ d_1-d_0, d_2-d_1, \dots, d_m-d_{m-1}\}$$ and the $$n_i$$'s are the corresponding elements of $$\{e_1,e_2,\dots,e_m\}$$. Define $$s_0=0, s_1, \dots, s_q$$ be the indices such that $$k_i = d_{s_i} - d_{s_{i-1}}$$ and $$n_i = e_{s_i}$$. Vice versa, $$d_{s_i} = k_1+\dots+k_i$$ and $$c_{s_i} = d_{s_{i-1}}+1$$.
First consider these parts of the two formulas: $$\prod_{i=1}^q \prod_{j=0}^{k_i-1} (p^{k_i} - p^j) = \prod_{i=1}^q p^{k_i(k_i-1)/2} \prod_{j=0}^{k_i-1} (p^{k_i-j} - 1)$$ and $$\prod_{t=1}^m (p^{d_t} - p^{t-1}) = p^{m(m-1)/2}\prod_{t=1}^m (p^{d_t-t+1} - 1).$$ It is easy to see that the multisets $$\{ k_i - j : 0\leq j \leq k_i-1, 1\leq i\leq q \}$$ and $$\{ d_t - t +1\ :\ 1\leq t\leq m \}$$ are the same, since the $$t$$-th element in the sequence $$k_1 - 0, k_1 - 1, \dots, 1, k_2 - 0, k_2 - 1, \dots, 1, \dots$$ equals $$d_t-t+1$$.
Now it remains to prove the equality for the powers of $$p$$ in the two formulas, i.e. $$\sum_{i=1}^q \bigg(k_i(k_i-1)/2 + (n_i-1)k_i^2 + \sum_{j\ne i} \min(n_i,n_j)k_ik_j\bigg) = m(m-1)/2 + \sum_{t=1}^m \big(e_t(m-d_t) + (e_t-1)(m-c_t+1)\big).$$ In the l.h.s. we have $$\sum_{i=1}^q \sum_{j\ne i} \min(n_i,n_j)k_ik_j = 2\sum_{i=1}^q n_i k_i \sum_{j>i} k_j=2\sum_{i=1}^q n_i k_i (m-d_{s_i}).$$ In the r.h.s. we have $$\begin{split} \sum_{t=1}^m \big(e_t(m-d_t) + (e_t-1)(m-c_t+1)\big) &= \sum_{i=1}^q k_i\big(e_{s_i}(m-d_{s_i}) + (e_{s_i}-1)(m-d_{s_{i-1}})\big) \\ &= \sum_{i=1}^q k_i\big(n_i(m-d_{s_i}) + (n_i-1)(m-d_{s_{i-1}})\big)\\ &=\sum_{i=1}^q k_i\big(n_i(m-d_{s_i}) + (n_i-1)(m+k_i-d_{s_i})\big)\\ &=2\sum_{i=1}^q k_i n_i(m-d_{s_i}) + \sum_{i=1}^q \big( (n_i-1)k_i^2 - (m-d_{s_i})k_i\big). \end{split}$$ Finally, we notice that $$\sum_{i=1}^q k_i(k_i-1)/2 = m(m-1)/2 - \sum_{i=1}^q (m-d_{s_i})k_i$$ since $$m=k_1+k_2+\dots+k_q$$ and $$m-d_{s_i} = k_{i+1}+k_{i+1}+\dots+k_q$$. QED
So, we can conclude that OEIS A034382 did indeed contain an error in its 16-th term. Now it's corrected. | 2019-09-19T00:59:30 | {
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https://math.stackexchange.com/questions/2099774/what-is-the-value-of-sum-r-1n-int-01fr-1xdx | # What is the value of $\sum_{r=1}^n\int_{0}^1f(r-1+x)dx$?
I tried to write the integral as limit of a sum which gives:$$\int_0^1f(r-1+x)dx=\lim_{n\to\infty}\frac{1}{n}[f(0)+f(\frac{1}{n})+f(\frac{2}{n})...+f(\frac{n-1}{n})]$$
which i wrote as: $$\lim_{n\to\infty}\frac{1}{n}\sum_{r=0}^{n-1}f(\frac{r}{n})$$
and when I put this in the question, I get a double summation. I don't know how to go further. Kindly suggest.
Btw the answer to this question is $$\int_0^nf(x)dx$$
For $\displaystyle\int_0^1 f(r-1+x)dx$ with substitution $u=r-1+x$ we have $$\int_0^1 f(r-1+x)dx=\int_{r-1}^r f(u)du$$ so $$\sum_1^n\int_0^1 f(r-1+x)dx=\int_{0}^n f(x)dx$$
• Thanks! Btw, is there any way this can be solved as limit of a sum? I am 100% fine with the answer you wrote but this question was given in definite integral as the limit of a sum exercise so just wondering if there's any? – Cheapstrike Jan 16 '17 at 6:37
• @Cheapstrike For summation we must have $\Delta x=\Big((r+1)-1+x\Big)-(r-1+x)=r\to0$ but It does not occur. – Nosrati Jan 16 '17 at 6:40
As a hint: try to substitute $u=r+x-1$.
(This would typically be a comment, but I do not, alas, have enough reputation...)
• I think you do now. Welcome! – numbermaniac Nov 25 '17 at 12:44
$\int _0^1 f (r+x-1)=\int _0^1 f (r-x)$ let $r-x=u$ thus $dx=-du$ hence we have $I=\sum _1 ^n -\int _r ^{r-1} f (u)du =\sum _1 ^n \int _{r-1} ^r f (u)du =\int _0 ^n f (x)dx$
• from Second last to last step, is it a formula? – Kislay Tripathi Nov 25 '17 at 9:57
• The disappereance of - sign? – Archis Welankar Nov 25 '17 at 9:58
• Its a known fact that $\int _a^b f (x)dx=-\int _b ^a f (x)dx$ . I hope its clear now. – Archis Welankar Nov 25 '17 at 10:04
• please tell me about The disappereance of summation sign? – Kislay Tripathi Nov 25 '17 at 10:06
• I think you should read basic theorems of integration you can then answer your question yourself;) – Archis Welankar Nov 25 '17 at 10:08
Let $\displaystyle \mathfrak{I}:=\sum_{r=1}^n \int_0^1 f(r+x-1) \,\text{d} x$. Leting $u:=r+x-1$ we have $\dfrac{\text{d}u}{\text{d}x}=1$ . If $x=0$ we have $u=r-1$, and if $x=1$ we have $u=r$. Therefore:
\begin{align}\mathfrak{I}=&\sum_{r=1}^n \int_{r-1}^r f(u) \, \text{d}u\\ \end{align} But we know that $\displaystyle \int_{a}^b f \,+ \int_b^c f = \int_a^c f$. Therefore : $$\boxed{\,\,\,\mathfrak{I} =\int_{0}^n f(x) \, \text{d}x\,\,\,\,\,}$$
• Brother how did you conclude $\displaystyle \int_{a}^b f \,+ \int_b^c f = \int_a^c f$ From this \begin{align}\mathfrak{I}=&\sum_{r=1}^n \int_{r-1}^r f (u) \, \text{d}u\\ \end{align} .Is it some kind of formula or i need knowledge of riemann integral for this? – Kislay Tripathi Nov 25 '17 at 11:43
• @KislayTripathi it's the sum of areas. Think about the integrals as areas under a graph - if you add the area from $a$ to $b$ and the area from $b$ to $c$ together, wouldn't it be the same as the area from $a$ to $c$? – numbermaniac Nov 25 '17 at 12:17
• Is a way to "glue" togheter the areas. Consider a function $f$ and $a<b<c$. The $\displaystyle \int_a^c f$ (area from $a$ to $c$) can be cuted in a certain $b$. Therefore the area will bem the sum of those areas, wich are $\displaystyle \int_a^b f$ (area from $a$ to $b$) and $\displaystyle \int_b^c f$ (area from $b$ to $c$). Literally a slice haha. – Gustavo Mezzovilla Nov 25 '17 at 13:46 | 2021-01-23T05:12:34 | {
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http://csvh.fian-intern.de/rolling-motion-without-slipping.html | " /> Rolling Motion Without Slipping
# Rolling Motion Without Slipping
If a wheel rolls w/o slipping, its linear speed and acceleration of the CM relative to the ground are equal to that of a point on the rim of the wheel relative to the CM. The constraint eq (rolling without slipping) is: x, We will solve this problem in two ways: R x Note:. roll without slipping A uniform solid ball is placed at rest on an incline of slope angle θ. The diagrams show the masses (m) and radii (R) of the cylinders. Therefore, the equation of the rotational motion for the disk is: The moment of inertia is a positive scalar; therefore, the angular acceleration vector is parallel to the torque of the friction force. For pure rolling motion (i. Since the disk rolls without slipping. This will enable us to do energy analysis of a variety of rolling objects. Rolling without slipping. Since the disk rolls without slipping. People have observed rolling motion without slipping ever since the invention of the wheel. Rolling Motion + = Rolling without slipping Pure translation Pure rotation about CM Subscribe to view the full document. Which quantity is constant before it rolls without slipping? 1. rolling without slipping; over a straight line; has cusps (points with two tangents) A cycloid is curtate (or contracted) if… it is traced out by… points inside a generating circle (r < R) that is rolling without slipping or; points on the surface of the generating circle that is slipping while rolling with v cm > Rω; does not have cusps. We consider the two cases of rolling without slipping and rolling with slipping. 5 seconds, it stops skidding and start to roll without slipping. a moving wire 3. Here is the first case, the wheel rolls to the right so the rotation is clockwise. If the coin C2 slides without rolling (the point A not losing contact), then the lines O1-A-O2 would move by an angle of π/2. An object that rolls against a surface without slipping obeys the condition that the velocity of its center of mass is equal to the cross product of its angular velocity with a vector from the point of contact to the center of mass:. Cylinder, sphere, hoop) Pure rolling motion (rolling without slipping) Rolling motion along a flat surface (inclined, horizontal or even vertical). In which order do the objects reach the bottom of the incline?. Related Links. Derive the governing differential equation of motion. Obviously, the center of the wheel is also the center of the mass, so the linear momentum is p = M ωR. Take moments about the mass center. Rolling motion A special case of rigid body motion is rolling without slipping on a stationary ground surface. Describe the subsequent motion of the mud. Key Takeaways Key Points. If no slipping occurs, the point of contact is momentarily at rest and thus friction is static and does no work on the object. Rolling motion w/o slipping is composed of pure translational motion and pure rotational motion. For pure rolling motion (i. For a cylinder of mass $$M$$ and radius $$R$$, rolling motion without slipping down a plane inclined at an angle $$\theta$$ with the horizontal, 1. Its moment of inertia about the central axis is. 25) in the book (Section 12. a bead constraints to move on a fixed vs. When an object experiences pure translational motion, all of its points move with the same velocity as the center of mass; that is in the same direction and with the same speed. The diagrams show the masses (m) and radii (R) of the cylinders. A force that is larger than the static friction at the center of the wheel, AND the moment created is enough to spin the wheel. The axis of rotation travels with the rolling object. Rolling Motion • The wheel in the figure is rolling to the right without slipping. As the cylinder rolls, point G (center) moves along a straight line. 4-6) Rolling Rotational Inertia Oct. The condition that an object rolls without slipping is that the distance traveled in a line equals the arc length turned out by the wheel. 4 Rolling Without Slipping, Slipping, and Skidding Circular Motion Week 3. Rotational Dynamics: Rolling Spheres/Cylinders is used to calculate the torque that produces rotational motion. of the motion are equal, and the time of the motion is the same for the two components, these two speeds are equal. 5, having been released from rest somewhere along the straight section of the track. People have observed rolling motion without slipping ever since the invention of the wheel. A circular hoop rolls without slipping on a flat horizontal surface. So the condition for rolling without slipping becomes. It also explains the corresponding equations of pure rolling. Together with the results of the experiments of. Here is the first case, the wheel rolls to the right so the rotation is clockwise. Its moment of inertia about the central axis is. This page contains the video Rolling Without Slipping, Slipping, and Skidding. Friction and Rolling Resistance This is static friction if the wheel rolls without slipping. 25 M And A Mass Of 2. c) Every point on the rim of the wheel has a different linear velocity. The kinematic diagram of the spool is shown in Fig. Chapter 9: Rotational Motion Rigid body instead of a particle Rotational motion about a fixed axis Rolling motion (without slipping) Angular Quantities “R” from the Axis (O) Linear and Angular Quantities Kinematical Equations Chapter 10: Rotational Motion (II) Angular Quantities: Vector Rotational Dynamics: t Note: t = F R sinq Note: sign of t Rotational Dynamics: I Rotational Dynamics: I. I want to ask about the direction of frictional force in smooth rolling motion which means the rolling object doesn't slide on the surface. Rolling contains rotational motion and translational motion(straight line motion) Suppose a car is moving and suddenly applies brakes and it starts sort of skidding. If the ball starts at an initial height h above the bottom of the loop, what is the minimum value for h such that the ball just completes the loop-the-loop?. A rod of mass is attached to a spring of stiffness. no slipping) vR cm 2 2 2 2 11 22 cm 2 m I R v h v R m. The wheel will slide forward without turning. Physics 111 Lecture 21 (Walker: 10. Let represent the downward displacement of the center of mass of the cylinder parallel to the surface of the plane, and let represent the angle of rotation of the cylinder about its symmetry axis. What is the minimum value µ 0 of the coefficient of static friction between ball and incline so that the ball will roll down the incline without slipping? Solution by Michael A. naturally it accelerates downward. If the object rolls without slipping, determine if it is a disk, a hoop, or a sphere. Rolling Motion (Without Slipping) When a wheel or radius R rolls without slipping along a flat straight path, the points of the wheel in contact with the surface are instantaneously at rest and the wheel rotates about a rotation axis through the contact point. Firstly, we describe the motion of a rigid sphere on a rigid horizontal plane. Rolling Motion Problems. Although rolling wheels are everywhere, when most people are asked “what is the axis of rotation of a wheel that rolls without slipping?”, they will answer “the axle”. (a) Draw the free-body diagram for the ball. Energy is still conserved, but the initial potential energy is now converted into two types of kinetic energy. Verify that Equations (A5) and (A6) are correct dimensionally. 20 in Fixed-Axis Rotation to find moments of inertia of some common geometrical objects. Which quantity is constant before it rolls without slipping? 1. (b) What is the minimum coefficient of friction required for the sphere to roll without slipping? Solution. 10-4 Rolling Motion We may also consider rolling motion to be a combination of pure rotational and pure translational motion: + = for the touching point at any given time! "Rolling without slipping". Salukvadze optimal solution and ranked Pareto optimal solutions. Physics 111 Lecture 21 (Walker: 10. If the coin C2 slides without rolling (the point A not losing contact), then the lines O1-A-O2 would move by an angle of π/2. 0 m/s, determine (a) the translational kinetic energy of its center of mass, (b) the rotational kinetic energy about its center of mass; and (c) its total energy. Consider a thin axisymmetric disk with mass and radius that rolls without slipping over a stationary and rough horizontal plane, as illustrated in Figure 1. 3, steady motion is discussed in secs. When an object experiences pure translational motion, all of its points move with the same velocity as the center of mass; that is in the same direction and with the same speed. FRICTIONAL ROLLING PROBLEMS When analyzing the rolling motion of wheels, cylinders, or disks, it may not be known if the body rolls without slipping or if it slides as it rolls. Friction force Minimum coefficient of friction is, Kinetic Energy of Rolling Motion. Professor Lewin derived an equation for the acceleration of an object rolling down a ramp (under pure roll conditions). You may also find it useful in other calculations involving rotation. t and q, w, a vs. General Plane Motion Rolling without slipping. The pure rotation is shown in Animation 1, while the pure translation is shown in Animation 2. Rolling Motion of on fixed incline surface. This motion, though common, is complicated. At which point in the derivation was the assumption of rolling without slipping necessary? Exercise. For many dynamics problems, rolling without slipping means there is a friction force acting on the wheel at the contact point P. What are the conditions for slipping without rolling and rolling without slipping? When a ball or a ring is placed on a surface,it can slip,roll or slip with rollingunder what conditions are each of these executed??It probably has something to do with friction. Consider an object with circular cross section that rolls Rolling as Translation and Rotation Combined along a surface without slipping. Gottlieb Let: µ = coefficient of friction between ball and incline. Motion A bowling ball is initially thrown down an alley with an initial speed v 0, and it slides without rolling but due to friction it begins to roll until it rolls without slipping. Angular Acceleration • Definition: - Uniform angular acceleration • ω = ω o + αt - Units - Vector direction • Tangential Acceleration - a t = rα Δ. Why is this good? Because we have relationships between linear and rotational motion quantities, such as v = Rw for the linear and angular speeds, or a = R(alpha) for the linear and angular accelerations. Physics Flash Animations. about its center of mass, rolling without. it won't move F 6 Big yo-yo Since the yo-yo rolls without slipping, the center of mass velocity of the yo-yo must satisfy, v cm = rω. As shown in Fig. The spherical shell started from rest. 20 in Fixed-Axis Rotation to find moments of inertia of some common geometrical objects. of the general equations of motion of a thin disk rolling without slipping on a horizontal surface, we present results of a simple experiment on the time dependence of the motion that indicate the dominant dissipative power loss to be proportional to the Ω2 up to and including the last observable cycle. rolling friction, which is always present and distinct from either of these other two kinds of frictionfor a rollingobject. When an object is rolling on a plane without slipping, the point of contact of the object with the plane does not move. A disc of radius R rolls without slipping at speed v. Consider a solid sphere of radius R and mass M rolling without slipping. In the special case of rolling without slipping, the distance. At the bottom of the incline, the speed of the hoop's center-of-mass is v. The force of the static sliding friction prevents the wheel from sliding and thus initiates the rolling motion. Of the three forces in the system, two act at that point, so they have no lever arm. 0points A bicycle wheel of radius 0. UY1: More About Rolling Motion. The condition for rolling without slipping is The torque is. Another key is that for rolling without slipping, the linear velocity of the center of mass is equal to the angular velocity times the radius. 10-9 Rotational Plus Translational Motion; Rolling In (a), a wheel is rolling without slipping. Rolling, Torque, and Angular Momentum Rolling Motion: • A motion that is a combination of rotational and translational motion, e. No energy is dissipated and Mechanical Energy is Conserved. • Describing rotational motion • Rolling without slipping • Torque Lecture 24: Rotational Motion. The qualitative features of the motion can now be deduced from the equations. People have observed rolling motion without slipping ever since the invention of the wheel. see ISM 10. Kinetic Energy The total KE of an object undergoing rolling motion is the The total KE of an object undergoing rolling motion is the sum of the rotational KE about its CM and the translational sum of the rotational KE about. rolling without slipping. When an object is rolling on a plane without slipping, the point of contact of the object with the plane does not move. การเคลื่อนที่แบบการกลิ้ง (Rolling Motion, without slipping) หรือ การหมุนรอบแกนที่เคลื่อนที่. For pure rolling motion (i. A wheel rolls uniformly on level ground without slipping. The Kinematics of a Translating and Rotating Wheel model displays the model of wheel rolling on a floor. The motion of any point on the ring can be considered as vector sum of translational and rotational motion. An object is rolling, so its motion involves both rotation and translation. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. This motion, though common, is complicated. In contrast, it is well known that, for rolling without slipping, a uniform cylinder with moment of inertiaI = kma2 about its axis has acceleration gsinα/(1 +k) down an. If the object rolls without slipping, determine if it is a disk, a hoop, or a sphere. The rolling-without-slipping kinematic condition for figure 5. 84, there are three forces acting on the cylinder. Rolling Without Slipping. In order to start the rolling motion, a force or torque must be applied to the wheel. The spool rolls to the right. Which quantity is constant before it rolls without slipping? 1. 4 and 5, and oscillation about steady motion is considered in sec. This will enable us to do energy analysis of a variety of rolling objects. A cylinder of radius rolls without slipping down a plane inclined at an angle to the horizontal. Since the disk rolls without slipping. t for this ride. McDonald Joseph Henry Laboratories, Princeton University, Princeton, NJ 08544 (October 2, 2014; updated January 12, 2018) 1Problem Discuss the motion of a cylinder that rolls without slipping on another cylinder, when the latter rolls without slipping on a horizontal plane. Rolling motion is a combination of translational motion and rotational motion. The direction of its angular momentum is _____________. David Pritchard, Prof. see ISM 10. You can view a realistic animation of the rolling with slipping and watch as it changes to pure rolling without slipping. Now the translational kinetic energy of the ring is,Rotational kinetic energy of the ring is,. 20 in Fixed-Axis Rotation to find moments of inertia of some common geometrical objects. Salukvadze optimal solution and ranked Pareto optimal solutions. Friction force Minimum coefficient of friction is, Kinetic Energy of Rolling Motion. The pure rotation is shown in Animation 1, while the pure translation is shown in Animation 2. see ISM 10. or spherical shell) having mass M, radius R and rotational inertia I. The force of the static sliding friction prevents the wheel from sliding and thus initiates the rolling motion. Consider a. Which one of the following is necessarily true? a) All points on the rim of the hoop have the same speed. rotational motion. A circular hoop rolls without slipping on a flat horizontal surface. When discussing the physics of a rolling wheel in an introductory course, it is always fun to point out that a point on the rim of the wheel traces out a cycloid, as shown in Figure 1. to the right 2. Rolling Motion (Without Slipping) When a wheel or radius R rolls without slipping along a flat straight path, the points of the wheel in contact with the surface are instantaneously at rest and the wheel rotates about a rotation axis through the contact point. This has 1 less dof Pick the coordinates as shown. Rolling Motion of on fixed incline surface. Determine the natural frequency The statement that the "cord does not slip on the pulley" introduces the rolling-without-slipping condition. 4 – A vertical force but a horizontal motion A spool of mass M has a string wrapped around its. They both travel a sh01t distance at the bottom and then start up another incline. Practice comparing the rotational kinetic energy of two objects based on their shape and motion. For example, an object, say a ball is in rolling, that is it is rolling on the surface of the ground. Kinematic Diagram: Since the spool rolls without slipping, the velocity of the contact point. Consider the following example. 10-4 Rolling Motion We may also consider rolling motion to be a combination of pure rotational and pure translational motion: + = for the touching point at any given time! "Rolling without slipping". For analyzing rolling motion in this chapter, refer to Figure 10. One disk rolls without slipping, and one slides down the ramp. The diagrams show the masses (m) and radii (R) of the cylinders. 0points A bicycle wheel of radius 0. Chapter 2 Rolling Motion; Angular Momentum 2. A force that is larger than the static friction at the center of the wheel, AND the moment created is enough to spin the wheel. UY1: More About Rolling Motion. Answer and Explanation: Rolling without slipping is a combination. Consider the following three objects, each of the same mass and radius: a solid sphere a solid disk a hoop All three are released from rest at the top of an inclined plane. At this moment the ratio between magnitudes of velocities of A and B with respect to ground is. A piece of mud on the wheel flies off when it is at the 9 o’clock position (rear of wheel). In this work we analyse the motion of a sphere that rolls without slipping on the inside of a right circular cone under the influence of a uniform gravitational field acting verically downwards, in the direction of the symmetry axis of the cone. (a) Find the linear acceleration of the center of mass. , a ball, cylinder, or disk rolling without slipping. (a) Draw the free-body diagram for the ball. rotational motion. To define such a motion we have to relate the translation of the object to its rotation. From what minimum height above the bottom of the track must the marble be released in order not to leave the track at the top of the loop. David Litster, Prof. Q8: Sketch crude graphs of x, v, a vs. At a given point in time, we can apply the rotational version of Newton’s second law to rotations about the point where the cylinder touches the surface (as the cylinder is rolling without slipping, this is the only motion at that point). Label the identification points above. torque about the center from the force of tension. "Rolling resistance " which slows the wheel is a completely different force. At the bottom of the incline, the speed of the hoop's center-of-mass is v. Cylinder, sphere, hoop) Pure rolling motion (rolling without slipping) Rolling motion along a flat surface (inclined, horizontal or even vertical). The motion of a disk that is spun on a smooth flat surface slowly damps out due to friction. Energy is still conserved, but the initial potential energy is now converted into two types of kinetic energy. Consider two cylinders that start down identical inclines from rest except that one is frictionless. For analyzing rolling motion in this chapter, refer to Figure 10. In this section, we compare two situations: (1) rolling and slipping, and (2) rolling without slipping. 4 – A vertical force but a horizontal motion A spool of mass M has a string wrapped around its. these aspects should be properly darified. If the pulley makes four revolutions without the rope slipping, what length of. 26A is Without slipping, the disk has two variables, X and β, but only one degree of freedom. Physics Flash Animations. In other words, at any particular instant of time, the part of the disc in contact with the surface is at rest with respect to the surface. Salukvadze optimal solution and ranked Pareto optimal solutions. To define such a motion we have to relate the translation of the object to its rotation. Each can starting from rest means each starts with the same gravitational potential energy , which is converted entirely to , provided each rolls without slipping. It is known that this motion can be considered as a rotation about an. rolls without slipping on a horizontal terrain in such a way that it touches the terrain on the surface line parallel to the y-axis. • The red curve shows the path (called a cycloid) swept out by a point on the rim of the wheel ω swept out by a point on the rim of the wheel. Rolling Motion + = Rolling without slipping Pure translation Pure rotation about CM Subscribe to view the full document. 4 m/s rolls up a hill without slipping. • Describing rotational motion • Rolling without slipping • Torque Lecture 24: Rotational Motion. Iclicker #1. Consider a thin axisymmetric disk with mass and radius that rolls without slipping over a stationary and rough horizontal plane, as illustrated in Figure 1. This friction force prevents slipping. 80 kg climbs an incline. When the ring rolls without slipping, then v = rω, where v is the velocity of centre of mass of ring, and ω is its angular velocity. The condition for rolling without slipping is The torque is. Wednesday, January 23, 2008. In this work we analyse the motion of a sphere that rolls without slipping on the inside of a right circular cone under the influence of a uniform gravitational field acting verically downwards, in the direction of the symmetry axis of the cone. The diagrams show the masses (m) and radii (R) of the cylinders. A solid cylinder W of mass mand radius r, r Rω; does not have cusps. A force that is larger than the static friction at the center of the wheel, AND the moment created is enough to spin the wheel. 5, having been released from rest somewhere along the straight section of the track. "Rolling resistance " which slows the wheel is a completely different force. There are several assumptions to simplify the analysis of rolling objects: Homogeneous rigid body; High degree of symmetry (E. from rest and rolls without slipping a distance of 6 m If object is rolling with a com≠ 0 (i. For example, we can look at the interaction of a car’s tires and the surface of the road. Hey guys! In this video we're going to talk about conservation of energy in rolling motion problems. A mathematical model of the dynamics of a rolling rigid sphere in a rectilinear chute-conveyor with transformed dry friction is developed. The spherical shell started from rest. The wheel will slide forward without turning. The sphere has two kings of kinetic energy: linear and rotational. If a can rolls down a ramp, it puts part of its energy into rotation, leaving less for translation. 22 m and a mass of 1. Each can starting from rest means each starts with the same gravitational potential energy , which is converted entirely to , provided each rolls without slipping. 0points A bicycle wheel of radius 0. 1 Rolling Without Slipping When a round, symmetric rigid body (like a uniform cylinder or sphere) of radius R rolls without slipping on a horizontal surface, the distance though which its center travels (when. The linear speed of the wheel is. Now, consider a wheel of radius R rolling without slipping along the straight. In other words, at any particular instant of time, the part of the disc in contact with the surface is at rest with respect to the surface. A circular hoop rolls without slipping on a flat horizontal surface. Derive the governing differential equation of motion. (b) What is the minimum coefficient of friction required for the sphere to roll without slipping? Solution. In which order do the objects reach the bottom of the incline?. Rolling without slipping. A multi-objective optimization problem of a passively controlled motion is formulated for the model and it has been solved. The equations of motion will be: F x = m(a G) x => P - F = m a G F y = m(a G) y => N - mg = 0 M G = I Ga => F r = I G a There are 4 unknowns (F, N, a, and a G. Solution A. When the ring rolls without slipping, then v = rω, where v is the velocity of centre of mass of ring, and ω is its angular velocity. there are net forces) NON-smooth rolling motion. Course Material Related to This Topic: Complete exam problem 1e; Check solution to exam Problem 1e. Multiple-Choice Homework 21: Torque and rolling motion Problem 1: A solid cylinder and a solid sphere are rolling without slipping down an inclined plane. General Plane Motion Rolling without slipping. Velocities of points on a rolling wheel. Theoretical background Consider a cylinder of massm and radius R rolling down an inclined plane of angle θ. 7 in Section 10. Rotational motion is more complicated than linear motion, and only the motion of rigid bodies will be considered here. could be a cylinder, hoop, sphere. along a surface without slipping. a wheel rolling down the road. The motion of an extended rigid body can be resolved into two parts. Rolling Motion without Slipping. If the conditions are such that the object rolls without slipping, there is also a geometric relationship between the displacement of the center of mass and the angle turned or, equivalently, between the accelerations: x =!. Rolling without slipping commonly occurs when an object such as a wheel, cylinder, or ball rolls on a surface without any skidding. Another key is that for rolling without slipping, the linear velocity of the center of mass is equal to the angular velocity times the radius. A Rolling Object Accelerating Down an Incline. Let m be the mass and r be the radius of ring. b) All points on the rim of the hoop have the same linear velocity. Absolute Motion: Exercise A wheel of radius r rolls without slipping. Rolling without slipping. The kinematic diagram of the spool is shown in Fig. Take moments about the mass center. We otation Combin can simplif ed y its study by treating it as a combination of translation of the center of mass and rotation of the object about the center of mass Consider the two snapshots of a rolling bicycle wheel shown in the figure. 26A is Without slipping, the disk has two variables, X and β, but only one degree of freedom. The angular momentum is non-zero and constant for one of the restaurant's customers seated near a window. This has 1 less dof Pick the coordinates as shown. Iclicker #1. Key ideas for rolling: Rolling can be considered to be a superposition of a pure translational motion and a pure rotational motion. In contrast, it is well known that, for rolling without slipping, a uniform cylinder with moment of inertiaI = kma2 about its axis has acceleration gsinα/(1 +k) down an. Consider an object with circular cross section that rolls Rolling as Translation and Rotation Combined along a surface without slipping. Dorbolo, S. This is defined by motion where the point of contact with the ground has zero velocity, so it matches the ground velocity and is not slipping. What is the minimum value µ 0 of the coefficient of static friction between ball and incline so that the ball will roll down the incline without slipping? Solution by Michael A. The spring is attached to a wall and the rod rests on a disk of the same mass. This motion, though common, is complicated. Rolling involves both of these at the same time - rotation while the wheel is experiencing straight-line motion. Starting Rolling Motion. If the pulley makes four revolutions without the rope slipping, what length of. In fact, the wheel is experiencing both a translational and rotational motion. A mathematical model of the dynamics of a rolling rigid sphere in a rectilinear chute-conveyor with transformed dry friction is developed. • Will only consider rolling with out slipping. v bottom =0 R v CM Quick Quiz and Demo ! Consider a wheel in pure. We consider the two cases of rolling without slipping and rolling with slipping. People have observed rolling motion without slipping ever since the invention of the wheel. the rolling motion of a cylinder or hoop. rotational motion. We can simplify its study by treating it as a combination of translattreating it as a combination of translation of the center ofion of the center of. In rotations, rolling without slipping is the phrase we always hope we see in a problem. David Litster, Prof. A ball slipping on a surface is decelerated by the friction force opposing the ball linear displacement. 11-4 Combining Rolling and Newton’s Second Law for Rotation Let’s now look at how we can combine torque ideas with rolling-without-slipping concepts. The radius of gyration of the disk about point o is kog; hence,. we consider rolling with and without slipping and establish the conditions for the frictional force to have the direction of the centre-of-mass velocity, contrary to the common idea that this force is always opposite to motion. This situation can be analyzed using relative velocity and acceleration equations. Each can starting from rest means each starts with the same gravitational potential energy , which is converted entirely to , provided each rolls without slipping. In our analysis of the motion of a rigid disk on a horizontal plane we adopt a vectorial approach as advocated by Milne [10]. 7 in Section 10. The linear kinetic energy is given by the usual equation: 1/2 m*v^2. The acceleration of the body is , where k is the radius of gyration of body. Determine the natural frequency The statement that the "cord does not slip on the pulley" introduces the rolling-without-slipping condition. and Since the maximum static friction force , it imply that for the disk to rolling without slipping. No portion of this material may be reproduced, in any form or by any means, without permission in writing from the publisher. could be a cylinder, hoop, sphere. When objects roll along a surface, like this tire rolls across the ground, there must be some sort of force to cause the rotation to occur. This accelleration is constant until the ball rolls without slipping, which is to say that the angular velocity times the radius matches the velocity of the ball. Rolling without slipping commonly occurs when an object such as a wheel, cylinder, or ball rolls on a surface without any skidding. Rolling Without Slipping. In this work we investigate the motion of a homogeneous ball rolling without slipping on uniformly rotating horizontal and inclined planes under the action of a constant external force. If the coin C2 slides without rolling (the point A not losing contact), then the lines O1-A-O2 would move by an angle of π/2. Motion A bowling ball is initially thrown down an alley with an initial speed v 0, and it slides without rolling but due to friction it begins to roll until it rolls without slipping. rolling without slipping; over a straight line; has cusps (points with two tangents) A cycloid is curtate (or contracted) if… it is traced out by… points inside a generating circle (r < R) that is rolling without slipping or; points on the surface of the generating circle that is slipping while rolling with v cm > Rω; does not have cusps. A ball with an initial velocity of 8. In short, this study suggests that the new position sensor can accurately sense the position of a cylinder that rolls without slipping. b) All points on the rim of the hoop have the same linear velocity. The kinematic diagram of the spool is shown in Fig. In our analysis of the motion of a rigid disk on a horizontal plane we adopt a vectorial approach as advocated by Milne [10]. Pure Roll of Hollow and Solid Cylinders: In pure roll, the object is not skidding or slipping, and the speed of the center of mass equals the circumferential speed. Kinematic Diagram: Since the spool rolls without slipping, the velocity of the contact point. So, in this case, the object tries to slide first, which friction prevents, and that leads to rolling. no slipping) vR cm 2 2 2 2 11 22 cm 2 m I R v h v R m. Users can change the type of object (solid sphere, solid cylinder, etc. The point P, touching the ground, is instantaneously at rest, and the center moves with velocity. A and B are two points on the ring. Rolling Without Slipping. For this consider a body with circular symmetry for example cylinder, wheel, disc , sphere etc. physics 111N 2 rotations of a rigid body! suppose we have a body which rotates about some axis rolling "rolling without slipping". When an object is rolling on a plane without slipping, the point of contact of the object with the plane does not move. | 2020-02-19T09:02:16 | {
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https://aperiodical.com/2018/03/prime-time/ | # Prime Time
We spotted this photograph of a letter to The Telegraph, shared by Card Colm on Twitter earlier in the year. It’s exactly the kind of mathematical claim we like to enjoy verifying, so we thought we’d dig in.
SIR – As one obsessed with prime numbers, I note that we have gone from 2017 (a prime) into 2018 (two times a prime), which will be followed by 2019 (three times a prime). I believe this sequence has only happened three other times in the past 1,129 years.
Keith Burgess-Clements
Maidstone, Kent
It’s lovely that newspapers will print this kind of letter, and a quick check verifies that Mr Burgess-Clements is indeed correct that these three numbers have the properties described:
• $2017$ prime, $2018 = 2 \times 1009$, $2019 = 3 \times 673$
His follow-up statement, that this sequence has only happened three other times in the last 1,129 years, takes a little more checking. But only a little – as we have a resident CL-P, who describes the necessary calculation as ‘a trivial amount of Python code’, and quickly came up with the following list:
• $13$ prime, $14 = 2 \times 7$, $15 = 3 \times 5$
• $37$ prime, $38 = 2 \times 19$, $39 = 3 \times 13$
• $157$ prime, $158 = 2 \times 79$, $159 = 3 \times 53$
• $541$ prime, $542 = 2 \times 271$, $543 = 3 \times 181$
• $877$ prime, $878 = 2 \times 439$, $879 = 3 \times 293$
• $1201$ prime, $1202 = 2 \times 601$, $1203 = 3 \times 401$
• $1381$ prime, $1382 = 2 \times 691$, $1383 = 3 \times 461$
• $1621$ prime, $1622 = 2 \times 811$, $1623 = 3 \times 541$
Here’s that Python code, in case you’re curious. It uses Sage’s Primes() function.
pr = set([p for p in range(2018) if p in Primes()])double = set(2*p-1 for p in pr) triple = set(3*p-2 for p in pr)years = sorted(pr & double & triple)
This is a full list of all the cases below 2017, and hints at some nice more interesting patterns – $13$ and $541$ both occur as the prime year and the prime factor that's a third of another year. But we don't have time to dig into that now – we have to check if a person in the newspaper was wrong!
Keith's claim that this has happened thrice in the last 1,129 years is indeed correct – $2018 – 1129 = 889$, and three sets of years have occurred since then. I suspect this may have been a typo though, as if he'd said "the last 1,139 years", that would have included the tail end of the set starting in $877$. Maybe he was looking for the most impressive length of time with the fewest occurrences, to illustrate how rare it is (in which case 1139 would be your best bet, and probably what he meant). We favour "only four times since 1000AD" which still sounds pretty good.
One final question to answer is, how many of these will there be going forward? The next few will be:
• $2557$ prime, $2558 = 2 \times 1279$, $2559 = 3 \times 853$
• $2857$ prime, $2858 = 2 \times 1429$, $2859 = 3 \times 953$
• $3061$ prime, $3062 = 2 \times 1531$, $3063 = 3 \times 1021$
It's also worth checking when this pattern will continue for four years, so that the fourth year is four times a prime; that's $12721$, which is prime, while $12722 = 2 \times 6361$, $12723 = 3 \times 4241$ and $12724 = 4 \times 3181$.
• #### Katie Steckles
Publicly engaging mathematician, Manchester MathsJam organiser, hairdo.
### 5 Responses to “Prime Time”
1. John
If we write dates as 8 digit numbers yyyymmdd, then 2018 has 18 prime dates. 2019 will have 19 prime dates, and 2021 will have 21.
2. Rafa
Just a Haskell version of the code:
import Data.List
import Math.NumberTheory.Primes.Sieve -- arithmoi library
import qualified Data.Set as Set
years = Set.intersection (Set.fromList pr) (Set.intersection doubles triples)
where
pr = takeWhile (< 2018) primes
doubles = Set.fromList [2*p-1|p<-pr]
triples = Set.fromList [3*p-2|p<-pr]
main = print years | 2018-12-17T19:56:11 | {
"domain": "aperiodical.com",
"url": "https://aperiodical.com/2018/03/prime-time/",
"openwebmath_score": 0.6746134161949158,
"openwebmath_perplexity": 1890.8318434316661,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9817357205793903,
"lm_q2_score": 0.8757869803008764,
"lm_q1q2_score": 0.8597913621797292
} |
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# In how many ways can 5 different colored marbles be placed in 3 distin
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In how many ways can 5 different colored marbles be placed in 3 distin [#permalink]
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14 Oct 2013, 08:53
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In how many ways can 5 different colored marbles be placed in 3 distinct pockets such that any pocket contains at least 1 marble?
(A) 60
(B) 90
(C) 120
(D) 150
(E) 180
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Re: In how many ways can 5 different colored marbles be placed in 3 distin [#permalink]
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15 Oct 2013, 02:19
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2
How can we fill three pockets with 5 marbles?
3 + 1 + 1
2 + 2 + 1
With 3,1,1 distribution:
# of ways to select 3 from 5
5!/3!2! = 10
# of ways to select 1 ball from 2
2!/1! = 2
# of ways to select 1 ball from 1
1!/1! = 1
How many ways to distribute 3,1 and 1 to 3 boxes? 3!/2! = 3
10*2*3 = 60
With 2,2,1 distribution:
How many ways to select 2 from 5?
5!/2!3! = 10
# of ways to select 2 from 3
3!/2!1! = 3
# of ways to select 1 from 1
1
How many ways to distribute 2,2,1 to 3 boxes? 3!/2! = 3
10*3*3=90
90+60 = 150
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Re: In how many ways can 5 different colored marbles be placed in 3 distin [#permalink]
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20 Sep 2014, 13:22
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2
This is a tricky question! At least for people like me, who are from non-Quant background... I appreciate Igotthis's post but it appeared a little bit complicated to me (may be because I am not smart enough )
Anyway this is how I solved it...
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Re: In how many ways can 5 different colored marbles be placed in 3 distin [#permalink]
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08 Aug 2015, 07:18
1
1
praffulpatel wrote:
How many ways can 5 different colored marbles be placed in 3 distinct pockets such that any pocket contains at least 1 marble?
(A) 60
(B) 90
(C) 120
(D) 150
(E) 180
We have 5 marbles and 3 pockets
So we have two cases
Case-1: One Pocket with 3 marbles and two pockets with 1 marble each
No. of Arrangements = 5C3 * 3C1 * 2! = 10*3*2 = 60
5C3 - No. of ways of choosing 3 out of 5 marbles which have to go in one pocket
3C1 - No. of ways of choosing 1 out of 3 pockets in which 3 marbles have to go
2! - No. of ways of arranging remaining 2 marbles between remaining two pockets which get one marble each
Case-2: Two Pockets with 2 marbles each and one pockets with 1 marble
No. of Arrangements = 5C2 * 3C2 * 3C2 = 10*3*3 = 90
5C2 - No. of ways of choosing 2 out of 5 marbles which have to go in one pocket
3C2 - No. of ways of choosing 2 out of remaining 3 marbles which have to go in second pocket
3C2 - No. of ways of choosing 2 out of 3 pockets in each of which 2 marbles have to go
Total cases = 60+90 = 150
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Re: In how many ways can 5 different colored marbles be placed in 3 distin [#permalink]
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09 Aug 2015, 14:23
GMATinsight, Bunuel, VeritasPrepKarishma,
I'm hoping one of you can explain why the 1-1-1 combination is ignored in answering this question. The question asks:
Quote:
How many ways can 5 different colored marbles be placed in 3 distinct pockets such that any pocket contains at least 1 marble?
Would the answer not include 5P3? Since we can have at least 1 marble (i.e. exactly one marble) in each spot and the question does not specify that we must use all five of the marbles. Additionally, I think that permutation is the right way of counting for the 1-1-1 combination since a marble arrangement of Green-Blue-Red in pockets one-two-three is different from Blue-Red-Green in pockets one-two-three and so forth.
For the 1-1-1 combination, we will have: 5!/(5-3)! = 60 different arrangements.
Then we come to the 3-1-1 and 2-2-1 combinations.
For the 3-1-1 combination, in pocket one we can have any three of the five distinct marbles. The order inside one pocket doesn't matter. Therefore, we will use 5C3. For the 2nd pocket, we have two marbles left and we can only pick one because we must leave one for the third pocket. Therefore, we will use 2C1. And for the last (third) pocket we only have one choice. Now, we have 5C3, 2C1 and 1C1 in pockets one-two-three. These can be ordered in 3!/2! different ways because 2C1 = 1C1, and we have already counted the possibility of different colors in each pocket.
So for 3-1-1 combination, we have: 5C3 * 2C1 * 1C1 * 3!/2! = 60 different arrangements.
Similarly, for the 2-2-1 combination we have: 5C2 * 3C2 * 1C1 * 3!/2! = 90 different arrangements.
Finally, we will get 60+60+90 = 210 arrangements if we consider 1-1-1 to be a valid option - since the question does not explicitly exclude this possibility.
Could you please explain why we ignored 1-1-1 combination?
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Re: In how many ways can 5 different colored marbles be placed in 3 distin [#permalink]
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10 Aug 2015, 00:18
1
jhabib wrote:
GMATinsight, Bunuel, VeritasPrepKarishma,
I'm hoping one of you can explain why the 1-1-1 combination is ignored in answering this question. The question asks:
Quote:
How many ways can 5 different colored marbles be placed in 3 distinct pockets such that any pocket contains at least 1 marble?
Would the answer not include 5P3? Since we can have at least 1 marble (i.e. exactly one marble) in each spot and the question does not specify that we must use all five of the marbles. Additionally, I think that permutation is the right way of counting for the 1-1-1 combination since a marble arrangement of Green-Blue-Red in pockets one-two-three is different from Blue-Red-Green in pockets one-two-three and so forth.
For the 1-1-1 combination, we will have: 5!/(5-3)! = 60 different arrangements.
Then we come to the 3-1-1 and 2-2-1 combinations.
For the 3-1-1 combination, in pocket one we can have any three of the five distinct marbles. The order inside one pocket doesn't matter. Therefore, we will use 5C3. For the 2nd pocket, we have two marbles left and we can only pick one because we must leave one for the third pocket. Therefore, we will use 2C1. And for the last (third) pocket we only have one choice. Now, we have 5C3, 2C1 and 1C1 in pockets one-two-three. These can be ordered in 3!/2! different ways because 2C1 = 1C1, and we have already counted the possibility of different colors in each pocket.
So for 3-1-1 combination, we have: 5C3 * 2C1 * 1C1 * 3!/2! = 60 different arrangements.
Similarly, for the 2-2-1 combination we have: 5C2 * 3C2 * 1C1 * 3!/2! = 90 different arrangements.
Finally, we will get 60+60+90 = 210 arrangements if we consider 1-1-1 to be a valid option - since the question does not explicitly exclude this possibility.
Could you please explain why we ignored 1-1-1 combination?
Hey jhabib,
You have to place all the marbles. If you assume the 1-1-1 combination, note that 2 marbles are leftover. But you HAVE TO distribute 5 marbles. So 3-1-1 and 2-2-1 are the only possibilities. All 5 marbles must be distributed.
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Re: In how many ways can 5 different colored marbles be placed in 3 distin [#permalink]
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10 Aug 2015, 04:21
1
Hi jhabib
The Question has clearly specified that "ALL the Marbles have to be assigned to 3 pockets" so 1-1-1 needs to be ignored
Along with 1-1-1, 1-1-2, 1-2-1 ans 2-1-1 also need to be ignored.
I hope it helps!
jhabib wrote:
GMATinsight,
I'm hoping one of you can explain why the 1-1-1 combination is ignored in answering this question. The question asks:
Quote:
How many ways can 5 different colored marbles be placed in 3 distinct pockets such that any pocket contains at least 1 marble?
Would the answer not include 5P3? Since we can have at least 1 marble (i.e. exactly one marble) in each spot and the question does not specify that we must use all five of the marbles. Additionally, I think that permutation is the right way of counting for the 1-1-1 combination since a marble arrangement of Green-Blue-Red in pockets one-two-three is different from Blue-Red-Green in pockets one-two-three and so forth.
For the 1-1-1 combination, we will have: 5!/(5-3)! = 60 different arrangements.
Then we come to the 3-1-1 and 2-2-1 combinations.
For the 3-1-1 combination, in pocket one we can have any three of the five distinct marbles. The order inside one pocket doesn't matter. Therefore, we will use 5C3. For the 2nd pocket, we have two marbles left and we can only pick one because we must leave one for the third pocket. Therefore, we will use 2C1. And for the last (third) pocket we only have one choice. Now, we have 5C3, 2C1 and 1C1 in pockets one-two-three. These can be ordered in 3!/2! different ways because 2C1 = 1C1, and we have already counted the possibility of different colors in each pocket.
So for 3-1-1 combination, we have: 5C3 * 2C1 * 1C1 * 3!/2! = 60 different arrangements.
Similarly, for the 2-2-1 combination we have: 5C2 * 3C2 * 1C1 * 3!/2! = 90 different arrangements.
Finally, we will get 60+60+90 = 210 arrangements if we consider 1-1-1 to be a valid option - since the question does not explicitly exclude this possibility.
Could you please explain why we ignored 1-1-1 combination?
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Re: In how many ways can 5 different colored marbles be placed in 3 distin [#permalink]
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06 Jan 2018, 13:02
2
praffulpatel wrote:
How many ways can 5 different colored marbles be placed in 3 distinct pockets such that any pocket contains at least 1 marble?
(A) 60
(B) 90
(C) 120
(D) 150
(E) 180
hi
I have seen a solution to a problem similar to this one elsewhere on the forum
let me explain it to you
5 different colored marbles can be placed in 3 distinct pockets without any restriction is
= 3^5 = 243
as we are asked to find out the ways in which any pocket must get at least 1 marble, lets find out the ways in which any pocket must not get at least 1 marble, and then subtract the number of ways in which any pocket must not get at least 1 marble from the total number of ways, that is 3^5
so lets get going
number of ways in which all marbles can get to 1 pocket is
= 3, as there are 3 distinct pocket in total
now, number of ways in which 2 pockets can get all the marbles and 1 pocket remains empty is
= (2^5 - 2) * 3 = 90
2 has been subtracted to eliminate the possibility that any pocket out of 2 can get all the marbles
3 has been multiplied with the whole expression, because 2 pockets out of 3 have been selected
now we are in business
= 243 - 90 - 3
= 150 (D)
hope this helps
thanks
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Re: In how many ways can 5 different colored marbles be placed in 3 distin [#permalink]
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09 Oct 2018, 23:13
jhabib wrote:
GMATinsight, Bunuel, VeritasPrepKarishma,
I'm hoping one of you can explain why the 1-1-1 combination is ignored in answering this question. The question asks:
Quote:
How many ways can 5 different colored marbles be placed in 3 distinct pockets such that any pocket contains at least 1 marble?
Would the answer not include 5P3? Since we can have at least 1 marble (i.e. exactly one marble) in each spot and the question does not specify that we must use all five of the marbles. Additionally, I think that permutation is the right way of counting for the 1-1-1 combination since a marble arrangement of Green-Blue-Red in pockets one-two-three is different from Blue-Red-Green in pockets one-two-three and so forth.
For the 1-1-1 combination, we will have: 5!/(5-3)! = 60 different arrangements.
Then we come to the 3-1-1 and 2-2-1 combinations.
For the 3-1-1 combination, in pocket one we can have any three of the five distinct marbles. The order inside one pocket doesn't matter. Therefore, we will use 5C3. For the 2nd pocket, we have two marbles left and we can only pick one because we must leave one for the third pocket. Therefore, we will use 2C1. And for the last (third) pocket we only have one choice. Now, we have 5C3, 2C1 and 1C1 in pockets one-two-three. These can be ordered in 3!/2! different ways because 2C1 = 1C1, and we have already counted the possibility of different colors in each pocket.
So for 3-1-1 combination, we have: 5C3 * 2C1 * 1C1 * 3!/2! = 60 different arrangements.
Similarly, for the 2-2-1 combination we have: 5C2 * 3C2 * 1C1 * 3!/2! = 90 different arrangements.
Finally, we will get 60+60+90 = 210 arrangements if we consider 1-1-1 to be a valid option - since the question does not explicitly exclude this possibility.
Could you please explain why we ignored 1-1-1 combination?
Hey jhabib,
You have to place all the marbles. If you assume the 1-1-1 combination, note that 2 marbles are leftover. But you HAVE TO distribute 5 marbles. So 3-1-1 and 2-2-1 are the only possibilities. All 5 marbles must be distributed.
Dear Karishma ..
I am bitterly confused with this problem . I doubt the approach or the problem has some defect . Please correct me if I am wrong .
All the approaches are considering 3 bags as identical , but in problem they are distinct .
My approach to this problem :
I am considering 5 different marbles a b c d e and 3 different bags 1 2 3 .
So first distribute 3 balls one ball each to each bag .
select 3 balls to distribut 5C3 and then distribute to 3 different bags in 3! ways .
so total 60 ways .
Now we are left with 2 balls and 3 bags each containing one ball each .
for first ball we have 3 options and second ball again 3 options . 3*3 ways .
TOTAL = 60*3*3=540 ways .
Please somebody help me in this . This problem just made me mad ..
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Re: In how many ways can 5 different colored marbles be placed in 3 distin [#permalink]
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10 Oct 2018, 00:22
karnaidu wrote:
Dear Karishma ..
I am bitterly confused with this problem . I doubt the approach or the problem has some defect . Please correct me if I am wrong .
All the approaches are considering 3 bags as identical , but in problem they are distinct .
My approach to this problem :
I am considering 5 different marbles a b c d e and 3 different bags 1 2 3 .
So first distribute 3 balls one ball each to each bag .
select 3 balls to distribut 5C3 and then distribute to 3 different bags in 3! ways .
so total 60 ways .
Now we are left with 2 balls and 3 bags each containing one ball each .
for first ball we have 3 options and second ball again 3 options . 3*3 ways .
TOTAL = 60*3*3=540 ways .
Please somebody help me in this . This problem just made me mad ..
You have some double counting here. When you select some from a group and distribute and then distribute the rest to the same bags/pockets, there is double counting.
Say you distributed a, b, c first such that
Then you distributed d and e such that Bag1 got d and bag2 got e.
Take another case.
Say you distributed d, b and c first such that
Then you distributed a and e such that Bag1 got a and bag2 got e.
Note that both cases have exactly the same end result. But you would count them as two separate cases. Similarly, there will be other cases which will be double counted. Hence this method will not work.
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Re: In how many ways can 5 different colored marbles be placed in 3 distin [#permalink]
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10 Oct 2018, 00:53
sxyz wrote:
This is a tricky question! At least for people like me, who are from non-Quant background... I appreciate Igotthis's post but it appeared a little bit complicated to me (may be because I am not smart enough )
Anyway this is how I solved it...
Hii ..
In this approach you are missing arrangement of bags . You have to take care of that too ..
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Re: In how many ways can 5 different colored marbles be placed in 3 distin [#permalink]
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18 Oct 2018, 11:16
Hi. I don't understand why this has been multiplied by 3!/2! for 2-2-1 combination. Plus, shouldn't 5c2*3c2 be multiplied by 2! as the pockets are distinct, and the arrangement would matter? Thanks
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Re: In how many ways can 5 different colored marbles be placed in 3 distin [#permalink]
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05 May 2019, 05:41
sxyz wrote:
This is a tricky question! At least for people like me, who are from non-Quant background... I appreciate Igotthis's post but it appeared a little bit complicated to me (may be because I am not smart enough )
Anyway this is how I solved it...
Hi, I have a doubt.
Here, the case {3,1,1} = $$5C3*2C1*1C1 = 10*2*1 = 20$$
And this is multiplied by 3 (coz three different pockets.)
But generally, in case of distribution between identical pockets/bags/groups etc., the value is divided by the similar ones.
Such as, in here, {3,1,1} has 2 similarly filled pockets, and hence, divided by 2!.
Is it not being done here like that because the pockets are all different here?
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Re: In how many ways can 5 different colored marbles be placed in 3 distin [#permalink]
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05 May 2019, 13:48
Hi,
Can we solve it the following way?
Total number of possible combinations: $$3^{5}=243$$
Combinations that do not satisfy given condition (at least 1 marble in each pocket): $$4-1-0$$, $$3-2-0$$ and $$5-0-0$$.
1) 4-1-0. 4 marbles can be allocated to 3 different pockets, the rest - 1 - can also be allocated to 3 different pockets. Thus, number of unsatisfactory combinations in for this case = $$3*3*5 = 45$$ - multiply by 5, since marbles can also differ between the pockets.
2) 3-2-0. The same holds true here = $$3*3*5 = 45$$.
3) 5-0-0. This scenario is easy = $$3$$ different cases are possible.
Thus, $$243-45-45-3=150$$.
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Re: In how many ways can 5 different colored marbles be placed in 3 distin [#permalink]
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26 Aug 2019, 13:19
praffulpatel wrote:
In how many ways can 5 different colored marbles be placed in 3 distinct pockets such that any pocket contains at least 1 marble?
(A) 60
(B) 90
(C) 120
(D) 150
(E) 180
given: 5 dif marbs 3 dif pockets with at least 1 marble each
$$[3,1,1]…5c3•2c1•1=10•2=20…•arrangements[3,1,1]=20•(3!/2!)=60$$ (3!=num.pockets; 2!=num.duplicates [1,1])
$$[2,2,1]…5c2•3c2•1=10•3=30…•arrangements[2,2,1]=30•(3!/2!)=90$$ (3!=num.pockets; 2!=num.duplicates [2,2])
$$total.arrangements=60+90=150$$
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Re: In how many ways can 5 different colored marbles be placed in 3 distin [#permalink]
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03 Oct 2019, 10:28
We have two cases: The first case has one pocket with 3 marbles, 1 marble in each of the two remaining pockets. The second case has two pockets with 2 marbles each, and one pocket with one marble. We always add cases together to obtain the total number of possibilities.
Case 1: (3,1,1) First choose which of the 3 pockets will get three marbles. There are 3C1 =3 ways to do this. Then, choose the number of ways we can place 3 marbles into that box from 5 marbles. There are 5C3 = 10 ways to do this. Next we must place a marble in the next to the last box. There are 2C1 = 2 ways to do this. Then there is 1C1 = 1 way to place the last marble into the last pocket. Each time we make a subsequent selection when working through a case we multiply, so there are 3*10*2*1= 60 ways to place the marbles in case 1
Case 2: (2,2,1) First we must select which two pockets get two marbles each. There are 3C2 = 3 ways to do this. Then, we must choose two marbles to go in the first of two pockets. There are 5C2=10 ways to do this. From the 3 remaining marbles, there are 3C2 = 3 ways to place two marbles in the second pocket. Finally there is 1C1 =1 way to place the last marble in the last box. Multiplying, we have 3*10*3*1 = 90 ways to place the marbles in case two.
Thus there are 60+90 = 150 ways we can place the marbles
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Re: In how many ways can 5 different colored marbles be placed in 3 distin [#permalink]
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04 Oct 2019, 22:44
praffulpatel wrote:
In how many ways can 5 different colored marbles be placed in 3 distinct pockets such that any pocket contains at least 1 marble?
(A) 60
(B) 90
(C) 120
(D) 150
(E) 180
Solved it a little differently than others from what I read . So will share.
I first made sure that each of the pocket has at least 1 marble.
That is 5C3 * 3!
Then you can either distribute the remaining 2 as 2, 0, 0 or 1, 1, 0
That is 3C1 or 3C2 * 2 ! (3C1 is the no. of ways of choosing a pocket to put everything in. )
That gives 150.
D !
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Re: In how many ways can 5 different colored marbles be placed in 3 distin [#permalink]
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11 Oct 2019, 08:56
P1 can have either 1 marble or 2 marble or 3 marble.
Therefore,
P1 can have 1 marble in 5 ways as all are of different colors.
Or
P1 can have 2 marbles in 5C2 ways (not 5P2 as order is not important; meaning P1 can have m1m2 or m2m1 and it will mean the same thing). 10 ways.
Or
P1 can have 3 marbles in 5C3 ways. 10 ways.
Hence P1 can have marbles in 25 ways.
Every time we select these marbles their arrangement is also important among the 3 pockets. Because if P1 selects M1 first then it will have repercussion on the selection by the next 2 pockets. Similarly if P1 selects M2 first it will again have an effect on the selection by the next 2 pockets. This means that the arrangement among the 3 pockets shall be considered.
So now we will arrange these 25 number of ways among the 3 pockets.
Therefore,
25 x 3! = 25 X 6 = 150.
Kindly correct if there is anything wrong with my logic. Thanks.
Posted from my mobile device
Re: In how many ways can 5 different colored marbles be placed in 3 distin [#permalink] 11 Oct 2019, 08:56
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# In how many ways can 5 different colored marbles be placed in 3 distin
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Q&A
Show that $\forall n \in \mathbb{Z}^{+}$, $25^n \equiv 25 \bmod{100}$.
+3
−0
Show that $\forall n \in \mathbb{Z}^{+}$, $25^n \equiv 25 \bmod{100}$.
This was a simple observation I made when playing around and I came up with the following proof:
It follows from $(10a + 25)^2 = 100a^2 + 500a + 625 = 100(a^2 + 5a) + 625$ that any number ending in $25$ raised to a power of $2$ will also end in $25$. As each number has a unique binary representation, every number can be written as a sum of powers of two. By the rule $a^{b + c} = a^b \cdot a^c$ and $ab \bmod {c} = (a \bmod {c}) \cdot (b \bmod {c}) \bmod {c}$, $25^n \mod {100}$ will eventually cascade down to $25 \bmod {100}$. For example, \begin{align}25^{19} \equiv 25^{16 + 2 + 1} \equiv 25^{16} \cdot 25^{2} \cdot 25^{1} \equiv 25 \cdot 25 \cdot 25 \\ \equiv 25^{3} \equiv 25^{2 + 1} \equiv 25^2 \cdot 25^1 \equiv 25 \cdot 25 \equiv 25^2 \equiv 25\bmod{100}\end{align}
I am wondering if there are alternate, potentially simpler proofs.
Why does this post require moderator attention?
Why should this post be closed?
+4
−0
There is a fairly simple proof by induction.
Base case: $n = 1$
$$25^1\equiv 25 \mod 100$$
Inductive case: Assuming $25^n\equiv 25 \mod 100$,
\begin{align} 25^{n+1}&\equiv 25\cdot 25^n \mod 100\\ &\equiv 25 \cdot 25 \\ &\equiv 625 \\ &\equiv 25 \end{align}
Actually, this proof can easily be extended to show the stronger statement that $5^n\equiv 25\mod 100$ for all natural numbers $n\ge2$
Base case: $n=2$
$$5^2\equiv25\mod100$$
Inductive case: Assuming $5^n\equiv 25 \mod 100$,
\begin{align} 5^{n+1}&\equiv 5\cdot 5^n \mod 100\\ &\equiv 5 \cdot 25 \\ &\equiv 125 \\ &\equiv 25 \end{align}
The case with powers of $25$ then simply comes from the case when $n$ is even.
Why does this post require moderator attention?
+3
−0
You can prove it using the binomial theorem.
Assume that $1\leq n∈\mathbb{N}$, then:
\begin{align} 25^n & =(20+5)^n=\sum_{k=0}^n\binom{n}{k}20^k\cdot 5^{n-k}=5+\sum_{k=1}^{n-1}\binom{n}{k}20^k\cdot 5^{n-k}+20\\ & =25+5\cdot 20\cdot\sum_{k=1}^{n-1}\binom{n}{k}20^{k-1}\cdot 5^{n-k-1}\equiv 25\mod 100 \end{align}
Why does this post require moderator attention? | 2023-03-25T22:34:38 | {
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https://math.hecker.org/2016/07/10/linear-algebra-and-its-applications-exercise-3-3-6/ | ## Linear Algebra and Its Applications, Exercise 3.3.6
Exercise 3.3.6. Given the matrix $A$ and vector $b$ defined as follows
$A = \begin{bmatrix} 1&1 \\ 2&-1 \\ -2&4 \end{bmatrix} \qquad b = \begin{bmatrix} 1 \\ 2 \\ 7 \end{bmatrix}$
find the projection of $b$ onto the column space of $A$.
Decompose the vector $b$ into the sum $p + q$ of two orthogonal vectors $p$ and $q$ where $p$ is in the column space. Which subspace is $q$ in?
Answer: We have $p = A(A^TA)^{-1}A^Tb$ per equation (3) of 3L on page 156. We first compute
$A^TA = \begin{bmatrix} 1&2&-2 \\ 1&-1&4 \end{bmatrix} \begin{bmatrix} 1&1 \\ 2&-1 \\ -2&4 \end{bmatrix}$
$= \begin{bmatrix} 9&-9 \\ -9&18 \end{bmatrix}$
and then compute
$(A^TA)^{-1} = \frac{1}{9 \cdot 18 - (-9)(-9)} \begin{bmatrix} 18&-(-9) \\ -(-9)&9 \end{bmatrix}$
$= \frac{1}{81} \begin{bmatrix} 18&9 \\ 9&9 \end{bmatrix} = \begin{bmatrix} \frac{2}{9}&\frac{1}{9} \\ \frac{1}{9}&\frac{1}{9} \end{bmatrix}$
Finally we compute
$p = A(A^TA)^{-1}A^Tb$
$= \begin{bmatrix} 1&1 \\ 2&-1 \\ -2&4 \end{bmatrix} \begin{bmatrix} \frac{2}{9}&\frac{1}{9} \\ \frac{1}{9}&\frac{1}{9} \end{bmatrix} \begin{bmatrix} 1&2&-2 \\ 1&-1&4 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \\ 7 \end{bmatrix}$
$= \begin{bmatrix} 1&1 \\ 2&-1 \\ -2&4 \end{bmatrix} \begin{bmatrix} \frac{2}{9}&\frac{1}{9} \\ \frac{1}{9}&\frac{1}{9} \end{bmatrix} \begin{bmatrix} -9 \\ 27 \end{bmatrix}$
$= \begin{bmatrix} 1&1 \\ 2&-1 \\ -2&4 \end{bmatrix} \begin{bmatrix} 1 \\ 2 \end{bmatrix} = \begin{bmatrix} 3 \\ 0 \\ 6 \end{bmatrix}$
Since $b = p + q$ we have
$q = b - p = \begin{bmatrix} 1 \\ 2 \\ 7 \end{bmatrix} - \begin{bmatrix} 3 \\ 0 \\ 6 \end{bmatrix} = \begin{bmatrix} -2 \\ 2 \\ 1 \end{bmatrix}$
We have
$p \cdot q = 3 \cdot (-2) + 0 \cdot 2 + 6 \cdot 1 = -6 + 0 + 6 = 0$
so that $p$ and $q$ are orthogonal.
The vector $p$ is in $\cal{R}(A)$, the column space of A, and the orthogonal subspace of $\cal{R}(A)$ is $\cal{N}(A^T)$, the left nullspace of $A$. Since $p$ is in $\cal{R}(A)$ and $q$ is orthogonal to $p$, $q$ must be in $\cal{N}(A^T)$, so that $A^Tq = 0$. We confirm this:
$A^Tq = \begin{bmatrix} 1&2&-2 \\ 1&-1&4 \end{bmatrix} \begin{bmatrix} -2 \\ 2 \\ 1 \end{bmatrix}$
$= \begin{bmatrix} -2 + 4 -2 \\ -2 - 2 + 4 \end{bmatrix} = \begin{bmatrix} 0 \\ 0 \end{bmatrix} = 0$
UPDATE: I corrected the calculation of $q$; thanks go to KTL for pointing out the error.
NOTE: This continues a series of posts containing worked out exercises from the (out of print) book Linear Algebra and Its Applications, Third Edition by Gilbert Strang.
If you find these posts useful I encourage you to also check out the more current Linear Algebra and Its Applications, Fourth Edition, Dr Strang’s introductory textbook Introduction to Linear Algebra, Fourth Edition and the accompanying free online course, and Dr Strang’s other books.
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### 6 Responses to Linear Algebra and Its Applications, Exercise 3.3.6
1. Teresa says:
Thank you for the great work! It’s been really helpful. I wanted to mention that A transpose by A is [6 -8; -8 18] and not [9 -9; -9 18] that changes the results a bit.
2. hecker says:
I’m glad you find these posts useful. I’m sorry it’s been a long time since I published the last one.
I don’t understand your comment. Are you referring to A transpose multiplied by A (on the right)? If so, the (1, 1) element of the result matrix is 1 x 1 + 2 x 2 + (-2) x (-2) = 1 + 4 + 4 = 9. I don’t know how you got 6 for that result. Similarly the (1, 2) entry is 1 x 1 + 2 x (-1) + (-2) x 4 = 1 – 2 – 8 = -9 (not -8) and the (2, 1) entry is 1 x 1 + (-1) x 2 + 4 x (-2) = 1 – 2 – 8 = -9 again.
• KTL says:
The question in the 4th edition is different. Hence the confusion for the person above 🙂
• hecker says:
Ah, thanks for the explanation. I’ve never looked at a copy of the 4th edition so I don’t know how the exercises differ.
3. KTL says:
q you have given as p-b though the calculation is done as b-p (correctly)
• hecker says:
Thanks for finding this error! I’ve corrected the post. | 2017-05-27T13:47:43 | {
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https://math.stackexchange.com/questions/1869797/number-of-true-relations-of-the-form-a-subseteq-b-where-a-b-in-mathcalp-1 | # Number of true relations of the form $A\subseteq B$ where $A,B\in\mathcal{P}(\{1,2,\ldots,n\})$
I just started "Introduction to Topology and Modern Analysis" by G.F. Simmons and came across this problem in the exercises.
Q. Let $U=\{1,2,\ldots,n\}$ for an arbitrary positive integer $n$. If $A$ and $B$ are arbitrary subsets of $U$, how many relations of the form $A\subseteq B$ are there? How many of them are true?
Some previous problems for the cases $n=1,2,3$ suggested that by "relations of the form $A\subseteq B$", they mean even the ones where $A\subseteq B$ isn't true.
It would be obvious that there are $2^{2n}=4^n$ such relations since $U$ has $2^n$ subsets, $A,B$ can each be chosen in $2^n$ ways, so there are $2^n\times 2^n=4^n$ such relations.
Now, I think $3^n$ of these relations are true. Here's my idea/proof:
Let us consider $x,y\in\mathcal{P}(U)$ where $\mathcal{P}(U)$ denotes the power set of $U$. Let $x$ have $k$ elements from $U$ (where $0\leq k\leq n$) with $k=0$ corresponding to $x=\emptyset$.
For $x\subseteq y$ to be true, $y$ must have all the elements of $x$ and may/may not have elements from $U\setminus x$. We can construct $y$ by taking $m$ additional elements from $U\setminus x$ where $0\leq m\leq n-k$ which can be done in $\binom{n-k}m$ ways for each value of $m$.
So, for each $x$, we can construct $y$ in $\sum\limits_{m=0}^{n-k}\binom{n-k}m=2^{n-k}$ ways.
Now, we can take $x$ in $\binom nk$ ways for each value of $k$ and hence the number of true relations is $\sum\limits_{k=0}^n\binom nk 2^{n-k}=3^n$.
Is the above correct/rigorous enough?
Also, if the above solution is correct, doesn't it work for any arbitrary set $U$ of cardinality $n$ ?
• what do you mean? You want to know with how many relations you can endow the set $\mathcal P\{1,2,3\dots n\}$ ?? And then how many of these are subsets of the relation $\subseteq$ ? – Jorge Fernández Hidalgo Jul 24 '16 at 18:18
• @CarryonSmiling, it's a problem in the book by Simmons that how many of the set inclusion relations $A\subseteq B$ where $A,B\in\mathcal{P}(\{1,2,\ldots,n\})$ are true? I'm looking for proofreading by the community on my work. – analysis123 Jul 24 '16 at 18:28
• Yeah, your solution looks good. In fact your solution was clearer to me that the actual question. – Jorge Fernández Hidalgo Jul 24 '16 at 18:29
## 2 Answers
I finally got it.
The question seems to be: How many pairs $(A,B)$ of subsets of $\{1,2\dots n\}$ exist so that $A\subseteq B$.
For each element $x\in \{1,2\dots n\}$ we have three choices:
• $x$ is only in $B$
• $x$ is in $A$ and $B$
• $x$ is not in $A$ and not in $B$.
Therefore there are $3^n$ possible pairs.
Yes, this is correct, and yes, only the cardinality of $U$ matters.
You could slightly simplify the proof by noting that there are $2^{n-k}$ subsets of $U\setminus x$, so you don't have to sum over binomial coefficients.
I find the combinatorial proof in Carry on Smiling's answer simpler and more elegant. | 2019-12-15T15:55:48 | {
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https://math.stackexchange.com/questions/2222064/proof-of-equation-with-nested-dependent-integrals | # Proof of equation with nested, dependent integrals
As part of a larger proof, I am currently trying to prove the following equation: $$\int_0^r \int_0^{r-t_1} \dots \int_0^{r-\sum_{i=1}^{k-1} t_i} t_1^{m_1 - 1} t_2^{m_2 - 1} \dots t_k^{m_k - 1} dt_k \dots dt_1 = \frac{r^n}{n!} \prod_{i=1}^{k} (m_i -1)!$$
I know that $n = \sum_{i=1}^k m_i$ and moreover that all $m_i$ are natural numbers.
I've already confirmed with Mathematica that this equation holds for $k \in \{1,\dots,5\}$. I however have some difficulties with proving this equation for any $k > 0$. Any hints would be greatly appreciated!
Here's what I've done so far:
In order to get started, I tried to show it for $k = 1$: $$\int_0^r t_1^{m_1 - 1} dt_1 = \left[\frac{1}{m_1} \cdot t_1^{m_1}\right]_{t_1=0}^{t_1=r} = \frac{r^{m_1}}{m_1} = \frac{r^{m_1}}{m_1!} \cdot \frac{m_1!}{m_1} = \frac{r^{m_1}}{m_1!} \cdot (m_1 - 1)!$$
That was pretty straightforward.
But already for $k=2$ I'm getting stuck:
$$\int_0^r \int_0^{r-t_1} t_1^{m_1 - 1}t_2^{m_2 - 1} dt_2dt_1 = \int_0^r t_1^{m_1 - 1}\int_0^{r-t_1} t_2^{m_2 - 1} dt_2dt_1 = \int_0^r t_1^{m_1 - 1}\left[\frac{1}{m_2} \cdot t_2^{m_2}\right]_{t_2=0}^{t_2=r-t_1}dt_1 = \int_0^r t_1^{m_1 - 1}\cdot\frac{1}{m_2} \cdot (r-t_1)^{m_2}dt_1$$
Now if I would like to integrate over $t_1$, I should probably use the binomial theorem, which says: $$(x+y)^n = \sum_{k=0}^n {n \choose k}x^{n-k}y^k = \sum_{k=0}^n {n \choose k}x^{k}y^{n-k}$$ Applied to my case, this results in: \begin{align*} &\int_0^r t_1^{m_1 - 1}\cdot\frac{1}{m_2} \cdot \sum_{j=0}^{m_2} {{m_2} \choose j} \cdot r^{m_2-j}\cdot (-1)^j \cdot t_1^jdt_1\\ &= \frac{1}{m_2} \cdot \int_0^r \sum_{j=0}^{m_2} (-1)^j \cdot\frac{m_2!}{j!(m_2-j)!} \cdot r^{m_2-j}\cdot t_1^{m_1 + j - 1}dt_1\\ &=\frac{m_2!}{m_2} \cdot \sum_{j=0}^{m_2} \left((-1)^j \cdot\frac{1}{j!(m_2-j)!} \cdot r^{m_2-j} \int_0^r t_1^{m_1 + j - 1}dt_1\right)\\ &=(m_2-1)! \cdot \sum_{j=0}^{m_2} \left((-1)^j \cdot\frac{1}{j!(m_2-j)!} \cdot r^{m_2-j} \left[\frac{1}{m_1 + j} \cdot t_1^{m_1 + j}\right]_{t_1=0}^{t_1=r}\right)\\ &=(m_2-1)! \cdot \sum_{j=0}^{m_2} (-1)^j \cdot\frac{1}{j!(m_2-j)!} \cdot r^{m_2-j} \cdot \frac{1}{m_1 + j} \cdot r^{m_1 + j}\\ &=(m_2-1)! \cdot \sum_{j=0}^{m_2} (-1)^j \cdot\frac{1}{j!(m_2-j)!} \cdot r^{m_2+m_1} \cdot \frac{1}{m_1 + j}\\ &=r^{m_1 + m_2}(m_2-1)! \cdot \sum_{j=0}^{m_2} (-1)^j \cdot\frac{1}{j!(m_2-j)!}\cdot\frac{1}{(m_1 + j)} \end{align*} Now the only thing left to show is that the sum is equal to $\frac{(m_1-1)!}{(m_1+m_2)!}$. I tried the following: \begin{align*} &\sum_{j=0}^{m_2} (-1)^j \cdot\frac{1}{j!(m_2-j)!}\cdot\frac{1}{(m_1 + j)}\\ &= \sum_{j=0}^{m_2} (-1)^j \cdot\frac{1}{j!(m_2-j)!}\cdot\frac{m_1 \cdots (m_1+j-1)\cdot(m_1+j+1)\cdots(m_1+m_2)}{m_1 \cdots (m_1 + m_2)} \cdot \frac{(m_1-1)!}{(m_1-1)!}\\ &= \frac{(m_1-1)!}{(m_1+m_2)!} \cdot \sum_{j=0}^{m_2} (-1)^j \cdot\frac{1}{j!(m_2-j)!} \cdot \left(\prod_{i=1, i \neq j}^{m_2} (m_1 + i)\right) \end{align*} So now I only need to show that the remaining sum is equal to one, but I'm not able to do so. Again, playing around with some examples ($m_2 \in \{1,2,3\}$) everything works out. I've tried to reframe it as $\frac{(m_1-1)!}{(m_1+m_2)!m_2!} \cdot \sum_{j=0}^{m_2} (-1)^j \cdot{m_2 \choose j}\cdot \left(\prod_{i=1, i \neq j}^{m_2} (m_1 + i)\right)$, but not with much success.
I'm not quite sure whether I'm on the right track here or whether there's a different approach to this whole thing that looks more promising than what I'm trying to do. I feel like I won't be able to prove the overall thing for $k>0$ if I'm not even able to show it for the special case $k=2$.
I've been working on this for a couple of days now and I don't seem to make progress. If you have any ideas what I could try, that would be great. I'd be very grateful for both help with the $k=2$ proof and for ideas about how to tackle the general $k>0$ proof. Thanks in advance!
Here is a result for $k=2$ (EDIT) and for $k=3$ which shows the general way to proceed.
You are looking at imcomplete beta functions. For $k=2$, I follow your result (calling that $I_2$):
$$I_2 = \int_0^r t_1^{m_1 - 1}\cdot\frac{1}{m_2} \cdot (r-t_1)^{m_2}{\rm d}t_1$$
Substituting $t_1 = r z$ gives
$$I_2 = \frac{r^{m_1+m_2}}{m_2} \int_0^1 z^{m_1 - 1}\cdot (1-z)^{m_2} {\rm d} z$$
Now use the definition for the incomplete beta function $B(\cdot,\cdot)$ and its special values for natural numbers:
$$I_2 = \frac{r^{m_1+m_2}}{m_2} B(m_1, m_2+1) = \frac{r^{m_1+m_2}}{m_2} \frac{(m_1 - 1)! m_2 !}{(m_1+m_2)!} = r^{m_1+m_2} \frac{(m_1 - 1)! (m_2 -1) !}{(m_1+m_2)!}$$
which is the desired result for $k=2$.
One step further, for $k=3$. The $r$'s have been substituted away as above.
$$I_3 = \frac{r^n}{m_3} \int_0^1 z_1^{m_1 - 1} \int_0^{1-z_1} z_2^{m_2 - 1}\cdot (1-z_1-z_2)^{m_3} {\rm d}z_2{\rm d}z_1$$
Now for the inner integral, substitute $z_2 = s_2 (1-z_1)$ which gives $$I_3 = \frac{r^n}{m_3} \int_0^1 z_1^{m_1 - 1} (1-z_1)^{m_2+m_3} \int_0^{1} s_2^{m_2 - 1}\cdot (1-s_2)^{m_3} {\rm d}s_2{\rm d}z_1$$
This is a formulation where the inner integral has been made independent on the outer one - and this will work out for larger $k$ as well!
Making this explicit, we get $$I_3 = \frac{r^n}{m_3} \Big[\int_0^1 z_1^{m_1 - 1} (1-z_1)^{m_2+m_3} {\rm d}z_1 \Big] \cdot \Big[\int_0^{1} s_2^{m_2 - 1}\cdot (1-s_2)^{m_3} {\rm d}s_2\Big]$$
Now using the already known results for the Beta function above, $$I_3 = \frac{r^n}{m_3} \frac{(m_1 - 1)! (m_2 + m_3) !}{(m_1+m_2+m_3)!} \cdot \frac{(m_2 - 1)! m_3 !}{(m_2+m_3)!} = {r^n} \frac{(m_1 - 1)! (m_2 - 1)! (m_3 -1)!}{(m_1+m_2+m_3)!}$$
as desired.
From here, you can go on to larger $k$.
• Thanks a lot, the variable change and the Beta function really do the trick! I'll try to formulate the proof for general $k$ but it really looks like it's not as difficult as I thought. Apr 7 '17 at 9:42
So building on the solution of @Andreas for $k=2$ and $k=3$, I was able to prove the whole thing for arbitrary $k>0$:
First of all, one can observe that for any natural number $j > 0$ the following holds: $$\int_0^{r-\sum_{i=1}^{j-1}t_i} t_j^{a-1} \cdot \left(r- \sum_{i=1}^j t_i\right)^b dt_j = B(a,b+1) \cdot \left(r- \sum_{i=1}^{j-1} t_i\right)^{a+b}$$ We can define $t_j = \left(r- \sum_{i=1}^{j-1} t_i\right)\cdot z$ which gives $dt_j = \left(r- \sum_{i=1}^{j-1} t_i\right)\cdot dz$. Making a variable change in the left part of the equation results in $$\int_0^{1} \left(r- \sum_{i=1}^{j-1} t_i\right)^{a-1} \cdot z^{a-1} \cdot \left(r- \sum_{i=1}^{j-1} t_i - \left(r- \sum_{i=1}^{j-1} t_i\right)\cdot z\right)^b \cdot\left(r- \sum_{i=1}^{j-1} t_i\right) dz$$ $$= \left(r- \sum_{i=1}^{j-1} t_i\right)^{a-1+b+1} \int_0^{1} z^{a-1} (1-z)^b dz = B(a,b+1) \cdot \left(r- \sum_{i=1}^{j-1} t_i\right)^{a+b}$$
Now we look at the overall problem: $$I_k = \int_0^r \int_0^{r-t_1} \dots \int_0^{r-\sum_{i=1}^{k-1} t_i} t_1^{m_1 - 1} t_2^{m_2 - 1} \dots t_k^{m_k - 1} dt_k \dots dt_1$$ This can be rewritten as $$I_k = \int_0^r t_1^{m_1 - 1} \int_0^{r-t_1} t_2^{m_2 - 1} \dots \int_0^{r-\sum_{i=1}^{k-1} t_i} t_k^{m_k - 1} dt_k \dots dt_1$$
Using the observation made above, we can solve the innermost integral by setting $j=k, a= m_k, b=0$ which gives us $B(m_k,1) \cdot \left(r- \sum_{i=1}^{k-1} t_i\right)^{m_k}$
We thus get: $$I_k = B(m_k,1) \cdot \int_0^r t_1^{m_1 - 1} \dots \int_0^{r-\sum_{i=1}^{k-2} t_i} t_{k-1}^{m_{k-1} - 1} \cdot \left(r- \sum_{i=1}^{k-1} t_i\right)^{m_k} dt_k \dots dt_1$$
Again, we apply our observation to the innermost integral ($j=k-1,a=m_{k-1},b={m_k}$, resulting in $B(m_{k-1},m_k+1) \cdot \left(r- \sum_{i=1}^{k-2} t_i\right)^{m_{k-1}+m_k}$
Recursively applying this step finally results in: $$I_k = B(m_k,1)\cdot B(m_{k-1},m_k + 1) \cdot B(m_{k-2},m_{k-1}+m_k + 1)\cdot \dots \cdot B(m_1, m_2+\dots+m_k+1)\cdot r^{m_1+\dots+m_k}$$ As all $m_i$ are natural numbers, we can use that $B(x,y)=\frac{(x-1)!\,(y-1)!}{(x+y-1)!}$: $$I_k = r^{m_1+\dots+m_k} \cdot \frac{(m_k-1)!\,0!}{m_k!} \cdot \frac{(m_{k-1}-1)!\,m_k!}{(m_{k-1}+m_k)!} \cdot \frac{(m_{k-2}-1)!\,(m_{k-1}+m_k)!}{(m_{k-2}+m_{k-1}+m_k)!} \cdot \dots \cdot \frac{(m_1-1)!\,(m_2+\dots+m_k)!}{(m_1+m_2+\dots+m_k)!}$$ This reduces to: $$I_k = r^{m_1+\dots+m_k} \cdot (m_k -1)! \cdot \dots \cdot (m_1 -1)! \cdot \frac{1}{(m_1+\dots+m_k)!}$$ With $m_1+\dots+m_k = n$, we can rewrite this as: $$I_k = \frac{r^n}{n!} \prod_{i=1}^{k} (m_i -1)!$$
• Nice! That's the general way to proceed. Apr 7 '17 at 13:38 | 2021-09-23T12:07:22 | {
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http://www.lofoya.com/Solved/1483/if-the-price-of-petrol-increases-by-25-by-how-much-must-a-user | # Moderate Percentages Solved QuestionAptitude Discussion
Q. If the price of petrol increases by $25%$, by how much must a user cut down his consumption so that his expenditure on petrol remains constant?
✖ A. $25\%$ ✖ B. $16.67\%$ ✔ C. $20\%$ ✖ D. $33.33\%$
Solution:
Option(C) is correct
Let the price of petrol be Rs.100 per litre. Let the user use 1 litre of petrol. Therefore, his expense on petrol = $100\times 1$ = Rs.100
Now, the price of petrol increases by $25\%$. Therefore, the new price of petrol = Rs.125.
As he has to maintain his expenditure on petrol constant, he will be spending only Rs.100 on petrol.
Let ‘$x$’ be the number of litres of petrol he will use at the new price.
Therefore,
$125\times x=100$
$\Rightarrow x=\dfrac{100}{125}$
$=\dfrac{4}{5}=0.8$litres
He has cut down his petrol consumption by 0.2 litres
$=\dfrac{0.2}{1}\times 100$
$=20\%$ reduction.
There is a shortcut for solving this problem.
If the price of petrol has increased by $25\%$, it has gone up $1/4^{th}$ of its earlier price.
Therefore, the $\%$ of reduction in petrol that will maintain the amount of money spent on petrol constant
$=\dfrac{1}{4+1}$
$=\dfrac{1}{5}=20\%$
i.e. Express the percentage as a fraction. Then add the numerator of the fraction to the denominator to obtain a new fraction. Convert it to percentage - that is the answer.
Edit: For direct formula involving such questions, check comment by Rushi and Chirag Goyal.
Edit: For more insights on the question, check comment by Priya.
## (5) Comment(s)
Vijay
()
(R*100)/(100+R)..... this formula we will use here.
25*100/(125)=20
so 20% is anwser.
Param
()
Simplest of all . No Need to remember formula. No advance calculation. Based on basic calculation.
Let assume original price be 100. Increased by 25% become Rs125.
Logic : Now we have to keep the consumption same,so we have to bring Rs 125 back to Rs 100. How much we have to decrease so that Rs 125 become Rs 100.
(125-100)125*100 % = 20%
Done . No need to go to consumption. You can calculate using one variable only.
Priya
()
25% means 1/4 and 20% means 1/5. If the question is given as "increased by 25" then $\dfrac{1}{(4+1)}=\dfrac{1}{5}=20\%$
If the question is given as "decreased by 25" then $\dfrac{1}{(4-1)}=\dfrac{1}{3}=33\dfrac{1}{3}\%$
i.e., the value of $33\dfrac{1}{3}\%$ is $\dfrac{1}{3}$
It is easy if you know the values for all the percentages!
Rushi
()
There is shortcut formula for this kind of sums which is $\left(\dfrac{R}{100+R}\right) \times 100$ in the case of rising and $\left(\dfrac{R}{100-R}\right) \times 100$ in the case of fall.
But this formula is applicable only when exp. have to remain constant.
Chirag Goyal
()
Thanks $Rushi$
I'm Rewriting the same Shortcut using TeX commands for better understanding.
$\text{Consumption Should be Decrease or Increase}$
$\text=\dfrac{R}{100\pm R}\times{100}$ | 2017-07-25T18:50:09 | {
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https://math.stackexchange.com/questions/579675/prove-subset-of-mathbb-r2-is-open/579686 | # Prove subset of $\mathbb R^2$ is open
Problem. Let $G = \{(x,y): x \ne y\}$. Prove $G$ is an open subset of $\mathbb R^2$.
What I am thinking: If I could show that $\mathbb R^2 \setminus G = \{(x,y): x = y\}$ is a closed set, then its complement $G$ is open. I might be totally off. Any suggestions?
• How did you define an open/closed set? – Listing Nov 24 '13 at 20:38
Let the function $f\colon \mathbb R^2\rightarrow \mathbb R,$ $(x,y)\mapsto x-y$ then $f$ is continuous since it's a polynomial on $x$ and $y$. We have $$G=f^{-1}(\mathbb R^*)$$ Do you know what theorem we should use to conclude?
• Would the theorem be if f is continuous then the inverse image under f of any open set is open? I am only confused on what is meant by R*? – MDW Nov 24 '13 at 20:58
• Yes this is the needed theorem and $\mathbb R^*$ is $\mathbb R\setminus \{0\}$. – user63181 Nov 24 '13 at 21:00
• @Sami: Why can't we just show $\Bbb{R}^2 - G$ is open? – Don Larynx Nov 24 '13 at 21:03
• Do you mean show that $\mathbb R^2-G$ is closed? Yes it's possible and I want avoid this since I don't know his definition of closed set @DonLarynx – user63181 Nov 24 '13 at 21:07
• @SamiBenRomdhane: It is open if GOD wants. :D – mrs Nov 25 '13 at 2:40
Just partition $G$ into two disjoint sets: Let $G_1 = \{(x,y) : x > y \}$ and $G_2 = \{(x,y) : x < y \}$, then clearly $G = G_1 \cup G_2$.
Now, WLOG, let's show that $G_1$ is open. Let $\textbf{x} = (x_1,y_1) \in G_1$, and $B_\textbf{x}(r)$ be the ball around $\textbf{x}$ of radius $r$. In order for $G_1$ to be open, we need to find an $r$ such that $B_\textbf{x}(r) \subset G_1$. But, just choose $r$ to be the shortest distance between $x$ and the line $y=x$. Informally, we know the shortest distance between a point to a line is just the intersection of the normal line with the point $\textbf{x}$, and the line y = x (in our case). The normal line will be $y = -x + x_1 + y_1$ (by a quick computation), and if we intersect this line with $y=x$, we get $x = \frac{x_1 + y_1}{2}$, so we can compute the distance to be $\frac{x_1 - y_1}{\sqrt 2}$. choose $r$ to be this, and we have shown $G_1$ will be open.
Do the same for $G_2$, then $G$ is the union of 2 open sets, and we are done.
The definition of an open set : A subset $U$ of a metric space $(M,d)$ is called \textit{open} if, given any point $x$ in $U$, there exists a real number $ε > 04$ such that, given any point $y$ in $M$ with $d(x, y) < ε$, $y$ also belongs to $U$.
In the case of a line in the plane given by the equation $ax + by + c= 0$, where $a, b$ and $c$ are real constants with $a$ and $b$ not both zero, the distance from the line to a point $(x_0,y_0)$ is $$\frac{|ax_0 + by_0 +c|}{\sqrt{a^2 + b^2}}.$$ Now, we are interested in one specific line, namely $y = x$, i.e. $x - y = 0$.
The distance of any point $(x_0, y_0)$ in $G$ to the line $x -y = 0$ is therefore $$r :=\frac{|x_0 -y_0|}{\sqrt{2}}.$$
Now, consider the open ball centered in $(x_0,y_0)$ and of radius $r$. Do every element of this ball belong to $G$ ?
Suggestion (you were looking for):
Use the theorem that says that the graph of a real continuous function is closed.
$A=\{(x,y)\in \mathbb{R}^2 : x=y\}$ is the graph of the continuous function $f:\mathbb{R}\rightarrow \mathbb{R}$. Where $$f(x)=x$$.
Hence A is closed. Therefore, its complement is open.
The theorem does not hold in general but it certainly does for Hausdorff spaces. All metric spaces ($\mathbb{R}$ is a metric space) are Hausdorff.
Theorem: A topological space $X$ is Hausdorff iff its diagonal is closed.
The diagonal is defined as $$\{(x,x):x\in X\}$$
Note: The diagonal is closed in the product topology on $X\times X$.
HINT: $\Bbb{R}^2 - G$ results in a line. Can this complete $G$? | 2020-01-18T12:15:12 | {
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https://web2.0calc.com/questions/pls-help_19708 | +0
# PLS HELP
0
96
1
The first six rows of Pascal's triangle are shown below, beginning with row zero. Except for the $1$ at each end, row $4$ consists of only even numbers, as does row $2.$ How many of the first $20$ rows have this property? (Don't include row $0$ or row $1$). \begin{tabular}{ccccccccccc} &&&&&1&&&&&\\ &&&&1&&1&&&&\\ &&&1&&2&&1&&&\\ &&1&&3&&3&&1&&\\ &1&&4&&6&&4&&1&\\ 1&&5&&10&&10&&5&&1\\ \end{tabular}
Apr 9, 2021
#1
+420
+1
Notice that only 2^n rows (where n is a positive integer) can obtain this property, where the row only contains even numbers except for the beginning and the end. (For example, only row 2^1 = 2, row 2^2=row 4, row 2^3=row 8, and so on). There are 4 numbers that can be expressed as 2^n (where n is a positive integer) under 20 (which are 2, 4, 8, and 16), so the answer is $$\boxed{4}$$.
My intuition for the fact that only 2^n rows (where n is a positive integer) can obtain this property uses the fact that the numbers in the pascal triangle can be expressed as $$\frac{n!}{p!(n-p)!}$$, where n is the nth row and p is the pth number in that row. All odd numbers obviously don't work, because if p = 1, then the result is just going to be equal to n, and since n is odd, the pascal triangle contains odd numbers. Even numbers that cannot be expressed as 2^n (where n is an integer) also cannot have only even numbers, because it must also have odd factors, and if it has an odd factor, a certain value of p will cancel out all the even factors and leave the number odd, so only 2^n would work.
Also a very similar question was answered a long time ago: https://web2.0calc.com/questions/another-pascal-s-triangle-question
Apr 9, 2021
#1
+420
+1
Notice that only 2^n rows (where n is a positive integer) can obtain this property, where the row only contains even numbers except for the beginning and the end. (For example, only row 2^1 = 2, row 2^2=row 4, row 2^3=row 8, and so on). There are 4 numbers that can be expressed as 2^n (where n is a positive integer) under 20 (which are 2, 4, 8, and 16), so the answer is $$\boxed{4}$$.
My intuition for the fact that only 2^n rows (where n is a positive integer) can obtain this property uses the fact that the numbers in the pascal triangle can be expressed as $$\frac{n!}{p!(n-p)!}$$, where n is the nth row and p is the pth number in that row. All odd numbers obviously don't work, because if p = 1, then the result is just going to be equal to n, and since n is odd, the pascal triangle contains odd numbers. Even numbers that cannot be expressed as 2^n (where n is an integer) also cannot have only even numbers, because it must also have odd factors, and if it has an odd factor, a certain value of p will cancel out all the even factors and leave the number odd, so only 2^n would work.
Also a very similar question was answered a long time ago: https://web2.0calc.com/questions/another-pascal-s-triangle-question
textot Apr 9, 2021 | 2021-08-03T11:06:49 | {
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https://cs.stackexchange.com/questions/121468/proving-that-tn-t-lfloor-n-2-rfloor-t-lfloor-n-4-rfloor-t-lfloor | # Proving that $T(n) = T(\lfloor n/2 \rfloor) + T(\lfloor n/4 \rfloor) + T(\lfloor n/8 \rfloor) + n$ is $\in O(n)$
Show $$T(n) = T(\lfloor n/2 \rfloor) + T(\lfloor n/4 \rfloor) + T(\lfloor n/8 \rfloor) + n$$ is $$\in O(n)$$.
I will make the bound to be $$\in O(cn)$$ instead.
Proof by strong induction.
• Base case n =1
$$T(1) = c$$ and $$cn=c*1=c$$ $$\checkmark$$
• Inductive Hypothesis : $$T(k) \in O(ck)$$ for $$1\le k1$$.
• Inductive Step: Prove for n. So prove that $$T(n) \le O(ck)$$.
Right away we can write $$T(n) = T(\lfloor n/2 \rfloor) + T(\lfloor n/4 \rfloor) + T(\lfloor n/8 \rfloor) + n \\ \leq T(n/2)+T(n/4)+T(n/8) + n\\ = c(n/2)+c(n/4)+c(n/8)+n \ \ \ \ By \ Inductive \ Hypothesis$$
$$= c(7n/8)+n \le cn+n ...$$ I am stuck here
*Goal: $$\le cn - (some \ stuff)$$ and some stuff needs to be $$\ge 0$$.
• – ryan Mar 5 at 0:01
• The title you have chosen is not well suited to representing your question. Please take some time to improve it; we have collected some advice here. Thank you! – Raphael Mar 6 at 8:04
First, let me mention that $$O(n)$$ and $$O(cn)$$ are exactly the same thing. What you are really after is showing that $$T(n) \leq cn$$ for all $$n$$.
Let us aim at a slightly more relaxed goal: showing that $$T(n) \leq An$$ for some possibly larger constant $$A$$. For the base case, we need $$A \geq c$$. For the inductive step, we know that $$T(n) = T(\lfloor n/2 \rfloor) + T(\lfloor n/4 \rfloor) + T(\lfloor n/8 \rfloor) + n \leq A(n/2+n/4+n/8)+1 = (\tfrac{7}{8}A+1)n.$$ Recall that our goal is to deduce that $$T(n) \leq An$$. This would follow if $$\tfrac{7}{8}A + 1 \leq A$$, that is, if $$A \geq 8$$.
In total, if we take $$A = \max(c,8)$$ then both the base case and the inductive step go through.
As an aside, you can also use the Akra-Bazzi theorem to directly conclude that $$T(n) = O(n)$$.
Now let us try to obtain more insight on the recurrence. Let $$S(n) = 8n - T(n)$$. Then $$S(n) = S(\lfloor n/2 \rfloor) + S(\lfloor n/4 \rfloor) + S(\lfloor n/8 \rfloor) + 8n - 8\lfloor n/2 \rfloor - 8\lfloor n/4 \rfloor - 8\lfloor n/8 \rfloor - n.$$ Since $$8(n/a-1) < 8\lfloor n/a \rfloor \leq 8(n/a)$$, we see that $$S(n) = S(\lfloor n/2 \rfloor) + S(\lfloor n/4 \rfloor) + S(\lfloor n/8 \rfloor) + r(n),$$ where $$0 \leq r(n) < 24$$. Applying the Akra-Bazzi theorem, we get $$S(n) = O(n^p)$$ for some $$p < 1$$ (the solution to $$(1/2)^p + (1/4)^p + (1/8)^p = 1$$), and so $$T(n) = 8n + O(n^p)$$.
• Thank you this is an amazing answer and helps a lot! – Mandy Mar 5 at 23:45
• So for base case we can choose either 8 or c (in this case c) , and for inductive step we can also choose 8 or c (in this case 8) ? – Mandy Mar 7 at 20:34
• You have to choose the same constant in both cases. That’s how induction works. – Yuval Filmus Mar 7 at 20:50
It's very important that you understand what f(n) = O (g(n)) means. It means that there is a number $$n_0 ≥ 0$$ and a number c > 0 such that for every $$n ≥ n_0$$, f(n) ≤ c * g(n). It is a property of the whole function, not a property of some n. Saying "I prove by induction that for every n, f(n) = O (g(n))" doesn't make any sense at all. What makes sense is to say "I prove by induction that for every n ≥ n_0, f(n) ≤ c * g(n)".
So what you want to prove per induction using some suitable c, T(n) ≤ cn. One variant of complete induction uses an induction step where you proof "if the statement $$S_k$$ is true for every k < n, then $$S_n$$ is also true".
If T(k) ≤ ck is true for every k < n, then T(n) = T(n/2) + T(n/4) + T(n/8) + n ≤ cn/2 + cn/4 + cn/8 + n = (7/8 c + 1) n.
If c = 1 this means T(n) < 1 7/8 n. Not what we need, we need T(n) < cn. If c = 2 it means T(n) + 2 3/4 n. Slightly better but not good enough. For which c is (7/8 c + 1) n ≤ cn, or 7/8 c + 1 ≤ c? That's the case for c ≥ 8. The start of the induction is n=0, and it's easy to show that T(0) = 0.
So you haven't just shown that T(n) = O(n), you have shown the much stronger T(n) ≤ 8n.
• Thank you this is an excellent answer and helps me understand these kinds of problems in more depth! – Mandy Mar 5 at 23:46
• You might have forgotten the base case. – Yuval Filmus Mar 6 at 16:38 | 2020-07-10T03:56:46 | {
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https://nbhmcsirgate.theindianmathematician.com/2020/03/nbhm-2020-part-question-10-solution.html | ### NBHM 2020 PART A Question 10 Solution: Possible rank of a matrix obtained by changing the entries $(1, 3)$, $(1, 4)$, $(3, 3)$, $(3, 4)$, and $(5, 4)$
Let $M$ be a $7×6$ real matrix. The entries of $M$ in the positions $(1, 3)$, $(1, 4)$, $(3, 3)$, $(3, 4)$, and $(5, 4)$ are changed to obtain another $7×6$ real matrix $\widetilde{M}$. Suppose that the rank of $\widetilde{M}$ is 4. What could be the rank of $M$? List all possibilities.
Solution:
First, we will derive a connection between the rank of $M$ and the rank of $\widetilde{M}$. We have,
$$rank(M) = rank(M-\widetilde{M}+\widetilde{M}) \le rank(M-\widetilde{M})+rank(M)$$ Similarly
$$rank(\widetilde M) = rank(\widetilde{M}-M+M) \le rank(\widetilde{M}-M)+rank(M)$$
This shows that $rank(\widetilde{M}) - rank(\widetilde{M}-M) \le rank(M) \le rank(M-\widetilde{M})+rank(\widetilde{M})$
The matrices $M - \widetilde{M}$ and $\widetilde{M}-M$ can have non-zero entries only in the columns $3$ and $4$ and hence they are of rank atmost $2$. Also, it is given that the rank of $\widetilde{M}$ is equal to $4$. Hence we have $$2 \le rank(M) \le 6.$$
Using $0,1$ matrics one can check that all these possibilities are occurring indeed.
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### NBHM 2020 PART A Question 4 Solution $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$
Evaluate : $$\int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx$$ Solution : \int_{-\infty}^{\infty}(1+2x^4)e^{-x^2} dx = \int_{-\infty}^{\inft... | 2020-10-27T14:18:27 | {
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https://www.themathdoctors.org/more-on-0-999/ | # More on 0.999…
#### (An archive question of the week)
In collecting questions and answers about 0.999… for the last post, there were two that were too long to include, but that dig more deeply into issues that some of the standard answers tend to gloss over. So here, I want to look at those two answers, both of which deal with how to handle arithmetic on infinitely many digits: How can you add, subtract, or multiply when there is no rightmost digit? This is not central to the main question, but enough people express uncertainty about this aspect to make it worth covering.
## Adding without a place to start
Here’s the first question, from Darryl in 1999 (appropriately enough):
What is 0.999... + 0.999...?
While discussing the 1 = 0.999... solution, a person asked what is 0.999... + 0.999...? I think it is a good question - one that I could not answer. It should be 2 but how do we show this?
In the archives at
Getting 0.99999...
http://mathforum.org/dr.math/problems/dusty4.15.98.html
you say "Can you figure out why 0.3bar + 0.3bar = 0.6bar? Because these numbers go on forever, you will need to use a little logic to add them. (The algorithm that you learned for adding numbers doesn't work very well when you can't get to the rightmost number.)"
If we copy the idea of limits, we could say that for any number n we could start adding at this rightmost point and get the 0.66...6(nth) place. And since we can do this for any n then the equation holds. But if we try this for 0.9bar + 0.9bar we seem to get something that looks less like 2, namely 1.9....8(nth). But if we replace 8 with 9, we seem to get a number between 2 and the one we have, which makes me feel as if 0.9bar + 0.9bar < 2.
The link is to an elaborated version of the 1/3 + 1/3 + 1/3 = 1 argument, which I only gave a link to last time. (The notation 0.3bar as a typable version of $$0.\overline{3}$$, comes from there.) As Doctor Derrel pointed out there, we normally add from right to left, and there is no rightmost digit to start at in 0.333… + 0.333…; fortunately, we know that it isn’t necessary to start at the right when there are no carries, so we can just add left to right. This is discussed here:
Darryl’s suggestion is to consider any n digits, where we can clearly add 0.333…3 + 0.333…3 = 0.666…6, so we get the desired result by letting n increase without bound.
But, as Darryl points out, things get trickier when you try to add 0.999… + 0.999… . With 10 decimal places, for example, we get
0.9999999999
+ 0.9999999999
--------------
1.9999999998
You get a carry into each place from the place to its right, but where can you start that? And if there’s an 8 anywhere, then we’re less than 1.999… = 2, right? The answer, ultimately, is that in fact the 8 is nowhere!
Hi Darryl, thanks for your question. It's an interesting twist on the perennial debate about whether 0.999... could really equal 1.
Since that 8 is the nth digit and you take n to infinity, there is no digit 8 in the limit.
With no rightmost place where you would have 9 + 9 = 18, every digit will be 9. But that’s just impossible to imagine. So we have to do as mathematicians do, and find a way to work around infinities. That way is to use limits formally.
Apply the definition of a limit to what you have. We can express it as a game: you pick a number epsilon, as small as you want, and I have to pick a number n such that
|2 - 1.999...98| < epsilon
where the 8 is in the nth decimal place. Since
|2 - 1.999...98| = 2*10^(-n)
I just pick
n > -log(epsilon/2)
This is no harder than in the case of 0.999...9. The 8 in the last place has no effect on the limit; it only delays somewhat the approach to the limit, as indicated by that factor of 1/2.
The idea here is that we can consider such a sum with any number n of decimal places, which will have an 8 in the last place. We can get this sum to be as close as we want to 2, just by taking enough decimal places; in effect, to get as close as we want, we just have to take enough digits to push the 8 beyond where it would make the error too large. If we want our sum to be less than $$\epsilon = 0.00000000000000000001$$ away from 2, we just need to use $$n > -\log(\epsilon/2) = -\log(0.00000000000000000001/2) = 20.3$$, so 21 digits will be enough: that is, 0.999999999999999999999 + 0.999999999999999999999 = 1.999999999999999999998.
Here is another way to approach the problem, by manipulating infinite sums. If you want to make everything explicit, you can write the sums as limits of finite sums, and get into the issue of switching the limit and the sum as you go through this process.
Let's write 0.999... as an infinite sum:
0.999... = Sum[n = 1 to infinity](9*10^-n)
Now we can add infinite sums:
0.999... + 0.999...
= Sum[n = 1 to infinity](9*10^-n)
+ Sum[n = 1 to infinity](9*10^-n)
= Sum[n = 1 to infinity](18*10^-n)
= Sum[n = 1 to infinity](10*10^-n + 8*10^-n)
= Sum[n = 1 to infinity](10^(-n+1))
+ Sum[n = 1 to infinity](8*10^-n)
= Sum[n = 0 to infinity](10^-n)
+ Sum[n = 1 to infinity](8*10^-n)
= 1 + Sum[n = 1 to infinity](10^-n)
+ Sum[n = 1 to infinity](8*10^-n)
= 1 + Sum[n = 1 to infinity]((8+1)*10^-n)
= 1 + Sum[n = 1 to infinity](9*10^-n)
= 1.999...
I think this is the result you are looking for. Since we know that 0.999... = 1, the answer is 1 + 1 = 2; but you wanted to see that unbroken string of 9's.
He is adding an infinite string of carries (1) to an infinite string of 8’s, and getting an infinite string of 9’s.
By the way, notice that the final 8 never appears in this method. That's because we're dealing with infinite sums from the start, so there is always a carry from the next digit to the right. The sum of 10^-n is the sum of the infinite number of carries.
## Multiplying without a place to start
A similar question arises in thinking about the standard method of conversion from a repeating decimal. This question is from 2000:
Induction on .999...
I have been having an argument with my physics teacher over the fact that point nine recurring (.999... or PNR) equals 1. I showed him the proof on your site and he pointed out a fact that I hadn't noticed.
How can you multiply PNR by ten if you can't get to the beginning? For example, to multiply 215 * 3, you would start off on the right, multiplying 3 by 5, then 1, then finally 2. Now how can you multiply PNR by ten if the farthest right value cannot be reached?
Thank you for any clarification you can give.
As you should recall, we convert 0.999… to a fraction by subtracting it (x) from 10x:
x = 0.9999...
10x = 9.9999...
10x - x = 9.0000...
9x = 9
x = 1
Here, we first had to multiply 0.9999… by 10. But, as in the addition case above, there is no rightmost digit at which to start the multiplication, so can we actually do that multiplication? That is Blake’s teacher’s challenge.
Doctor TWE took it up, first referring to the FAQ (which includes links to several of the answers I’ve referred to) as a source of supplemental arguments before turning to the main issue.
I wanted to comment on your (or your teacher's) idea that "you can't get to the beginning." We multiply values from right to left only as a matter of convenience, so we don't have to "backtrack" each time we carry. It is equally valid to multiply from left to right. This is often done when we want to get a quick estimate for the answer, then get more accuracy later. The LEFTMOST digits contribute the most to the answer, so to get "in the ballpark," we can multiply them first. I'll demonstrate with a concrete example, then a general argument.
We’ve discussed this left-to-right idea in Dividing Right to Left, Adding Left to Right.
Let's use your example of 215 * 3. I can multiply them as follows; first, I'll multiply the 3 by the 2 (actually, by 200) and put the answer in the hundreds place, like this:
215
* 3
---
6
Then I'll multiply the 3 by 1 and put the answer in the tens place:
215
* 3
---
6
3
Finally, I'll multiply the 3 by 5 and put the answer in the units place:
215
* 3
---
6
3
15
215
* 3
---
6
3
15
---
645
Notice that along the way, I got pretty good estimates for my final answer. After the first step my total was 600 (not a bad estimate, off by less than 10%), after the second step I had 630, and after the third step, I got 645. Going right to left, my totals after each step would be: 15 (not a very good estimate of the final answer), 45 (still not very good) and finally 645.
So multiplying left to right just requires us to modify the digits we already have to account for carries from new columns as we do them; but the effect of the new column rarely propagates very far back into our work, and never makes a large change. Each new digit makes a smaller adjustment than the one before. That will be relevant as we continue.
In the general case, consider multiplying a 3-digit value ABC by some value X, where A, B and C are digits. What we really have, then, is
(100*A + 10*B + 1*C) * X
Conventionally, we solve this starting with the units digit as:
X*C*1 + X*B*10 + X*A*100
But using the commutative property of addition, this is equal to:
X*A*100 + X*B*10 + X*C*1
showing that the order doesn't matter. If I wanted to, I could start in the middle, for example:
X*B*10 + X*A*100 + X*C*1
I'd just have to be careful not to miss any digits or do any ones twice.
What we see here is that, conceptually, we can think of the whole multiplication as being done at once; it doesn’t matter in what order we do the pieces, because the sum is commutative.
Without being able to start with the most significant digit, we could never find a value like 2*pi, because pi (like all irrationals) is an infinite non-repeating decimal. When we say 2*pi is approximately 6.2831853, we can do so because we started multiplying at the end and not the beginning.
One final note: How can we be sure that the multiplications by 10 continue as we expect (i.e. they continue shifting the digits 1 place to the left) and that the subtractions of successive digits produce zeroes infinitely? These steps can be proven using a technique called mathematical induction. That's too complex to explain here, but if you search our Ask Dr. Math archives for the word "induction" (type it without the quotes), you'll find many questions and answers about it.
Whenever we do a calculation on numbers like π in a calculator, we are working only with the leftmost digits (those the calculator can hold); but since we know that the digits we have omitted will add no more than a 1 in the rightmost digit by carrying, the answer will be as close as we need. (In fact, calculators actually work with another digit or two beyond what they display, so that such errors never even show up.)
## Proof by induction
That last teaser – that there is a formal method for actually proving infinite things without having to do infinite things – was just what Blake needed. He replied a couple days later (back then we didn’t have threaded conversations, so he didn’t know whether the same Doctor would get the message):
I recently sent you a letter regarding the proof that .999... = 1 in the FAQ. Specifically, I asked how it was possible to multiply PNR by ten when you could not get to the right-hand side of it. This was explained quite neatly.
However, the nice person who helped me out said it could be proven that one can multiply PNR by 10 by using a process of induction. I tried to search the archives, but the results I found there were not very satisfactory because it seemed to me that each induction question was actually problem-specific.
To cut short, I would like to know the induction proof for multiplying PNR by ten.
Doctor TWE first summarized the concept of induction:
Hi again Blake! Thanks for writing back!
In general, we use proof by induction whenever we want a proof that involves an infinite sequence or series, in this case .999...
Inductive proofs require two steps:
Step 1 (the basis step): Prove it for some starting value, like n = 1.
Step 2 (the inductive step): Prove that if it's true for n = k, then it is true for n = k+1.
For the inductive step, we assume that it is true for n = k, and we usually use this assumption in the proof itself.
A word of caution: because of their self-referential nature, inductive proofs are usually hard to follow; it's easy to get lost in the details, losing track of what n, k, and k+1 are supposed to be.
For other explanations and examples of mathematical induction, see
Proof by Mathematical Induction
Sum of n Odd Numbers
Inductive Misunderstanding
Here is the proof. As he’ll explain, rather than taking n as 1, 2, 3, …, he is taking n as the exponent, -1, -2, -3, …, going in the negative direction and so reversing the usual procedure.
In our case, we want to show that when multiplying the nth digit by 10, we get a 9 in the (n+1)st digit, no carry to the (n+2)nd digit, and a 0 in the nth digit (counting from the right.) We want no carry to the (n+2)nd digit so that we don't have to adjust the previous (greater place value) digits multiplied because of a "retroactive carry." We want a 0 in the nth digit so that it will not produce further carries when multiplying the next digit (smaller place value) by 10. Mathematically:
10 * (9*10^n) = 9 * 10^(n+1)
We want to show that:
a) the nth digit is 0
b) the (n+1)st digit is 9
c) the (n+2)nd digit is 0
To minimize the confusion, I'm going to use n = -1 as my basis step, and show that if it is true for n = k, then it is true for n = k-1 as my inductive step. This is the "negative" of the standard, but it will eliminate the need for using -k's and -(k+1)'s, etc. in my equations.
Step 1: Show that it is the case for n = -1
10 * [9*10^(-1)] = 9 * 10^(-1+1)
10 * (9*0.1) = 9*10^0
10 * 0.9 = 9*1
9 = 9
a) the -1st digit (the tenths place) is 0: True
b) the 0th digit (the units place) is 9: True
c) the 1st digit (the tens place) is 0: True
So we've proved the basis. Now for the inductive step.
Step 2: Prove that if it's true for n = k, then it's true for n = k-1.
Let's let d = the initial (k+1)st digit. This digit is 0 from conclusion (a) of the previous step. Then:
10 * (9*10^k) + d = 9 * 10^(k+1)
9 * (10*10^k) + 0 = 9 * 10^(k+1)
9 * 10^(k+1) = 9 * 10^(k+1)
a) the kth digit is 0: True - 9*10^(k+1) has a 0 in the kth place.
b) the (k+1)st digit is 9: True - 9*10^(k+1) has a 9 in the (k+1)st place
c) the (k+2)nd digit is 0: True - 9*10^(k+1) has a 9 in the (k+2)nd place
So, if 10 times the kth digit is 9*10^(k+1), then 10 times the (k-1)st digit is 9*10^k, and thus 10 times the (k-2)nd digit is 9*10^(k-1), and thus...
Therefore 10 * 0.999... = 9.999...
Q.E.D.
More could be done to make this foolproof:
Depending on the skepticism of the intended audience, you may have to use a similar inductive proof for the step:
9.999...
- 0.999...
--------
9.000...
For most audiences, the following will do:
Let x = 0.999...
then 9.999... = 9 + x
so 9.999... - 0.999... = (9 + x) - x = 9
However, some audiences (like your teacher, perhaps) will not be satisfied that algebraic operations work on "infinite strings" (i.e. they won't be convinced that x - x = 0 when x is an infinite repeating decimal.)
I think some of the other explanations discussed last time may be more useful for certain kinds of skeptics; but it’s good to have a variety of approaches, to meet a variety of objections.
This site uses Akismet to reduce spam. Learn how your comment data is processed. | 2021-05-08T17:13:20 | {
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https://math.stackexchange.com/questions/2978611/value-of-int-infty-infty-arcsin-frac1-cosh-x-dx | # value of $\int_{-\infty}^{\infty}\arcsin\frac1{\cosh x}\,dx$
I want to know the value of $$I=\int_{-\infty}^{\infty}\arcsin\frac1{\cosh x}\,dx$$ The Symbolab integral calculator says that the integral diverges, but when one graphs it obvious that it converges. So what is the value?
I was thinking that I might try Feynman integration, but I can't think of the right substitution.
From the answer provided by @user10354138, we can reach $$\int\arcsin\frac1{\cosh x}dx=i\operatorname{Li}_2(i\phi)-i\operatorname{Li}_2(-i\phi)+C$$ Where $$\phi=\tan\bigg(\frac12\arcsin\frac1{\cosh x}\bigg)$$ And $$\operatorname{Li}_2(z)=\sum_{n\geq1}\frac{z^n}{n^2}$$ is the Di-logarithm.
Wolfy says it is 4 times the Catalan's constant.
One (not optimal) way to derive this is \def\sech{\operatorname{sech}} \begin{align*} \int_{-\infty}^\infty\arcsin\sech x\,\mathrm{d}x&=2\int_0^\infty\arcsin\sech x\,\mathrm{d}x\\ &=2\int_0^1\frac{\arcsin u\,\mathrm{d}u}{u\sqrt{1-u^2}}\quad(u=\sech x)\\ &=2\int_0^{\pi/2}\frac{\theta\,\mathrm{d}\theta}{\sin\theta}\quad(u=\sin\theta)\\ &=2\int_0^1\frac{2\tan^{-1}t\,\frac{2\,\mathrm{d}t}{1+t^2}}{\frac{2t}{1+t^2}}\quad(t=\tan\tfrac12\theta)\\ &=4\int_0^1\frac{\tan^{-1}t}{t}\,\mathrm{d}t\\ &=4\int_0^1\sum_{n=0}^\infty\frac{(-1)^n}{2n+1}t^{2n}\,\mathrm{d}t\\ &=4\sum_{n=0}^\infty\frac{(-1)^n}{(2n+1)^2}=4G\\ \end{align*}
• Dude that's dope! Thank you! – clathratus Oct 31 '18 at 3:31
• This substitution simplifies to $t=e^{-x}$. – J.G. Oct 31 '18 at 8:44
• how do you get from $\frac{\theta d\theta}{\sin\theta}$ to $$\frac{2\arctan(t)\frac{2dt}{1+t^2}}{\frac{2t}{1+t^2}}$$ With the substitution $t=\tan\frac\theta2$? – clathratus Nov 1 '18 at 23:06
• I keep getting $$\frac{\theta d\theta}{\sin\theta}=\frac{2\arctan t}{t\sqrt{t^2+1}}dt$$ – clathratus Nov 1 '18 at 23:10
• @clathratus This is the universal trigonometric substitution – user10354138 Nov 1 '18 at 23:25
Alternatively, you can integrate by parts: \begin{align} \int \limits_{-\infty}^\infty \arcsin(\operatorname{sech}(x)) \, \mathrm{d} x &= 2 \int \limits_0^\infty \arcsin(\operatorname{sech}(x)) \, \mathrm{d} x \\ &= 2x \arcsin(\operatorname{sech}(x)) \Bigg \rvert_{x=0}^{x=\infty} - 2 \int \limits_0^\infty x \frac{- \sinh(x) \operatorname{sech}^2(x)}{\sqrt{1-\operatorname{sech}^2(x)}} \, \mathrm{d} x \\ &= 2 \int \limits_0^\infty \frac{x}{\cosh(x)} \, \mathrm{d} x = 4 \sum \limits_{n=0}^\infty (-1)^n \int \limits_0^\infty x \, \mathrm{e}^{-(2n+1) x} \, \mathrm{d} x \\ &= 4 \Gamma(2) \sum \limits_{n=0}^\infty \frac{(-1)^n}{(2n+1)^2} = 4 \mathrm{G} \, . \end{align}
• I'm a fan of the alternative approach. Thank you! – clathratus Oct 31 '18 at 14:59
Yet another alternative approach: once shown that $$\int_{\mathbb{R}}\frac{dx}{\cosh(x)^{2k+1}} = \frac{\pi \binom{2k}{k}}{4^k}\tag{1}$$ and recalled that $$\arcsin z = \sum_{k\geq 0}\frac{\binom{2k}{k}}{4^k(2k+1)}z^{2k+1} \tag{2}$$ we have the following identity: $$\int_{\mathbb{R}}\arcsin\frac{1}{\cosh x}\,dx = \pi\sum_{k\geq 0}\frac{1}{2k+1}\left[\frac{1}{4^k}\binom{2k}{k}\right]^2.\tag{3}$$ Now we may invoke a function whose Maclaurin series involves squared central binomial coefficients, namely the complete elliptic integral of the first kind $$K(x)$$, here denoted according to Mathematica's notation (the argument of $$K$$ is the elliptic modulus): $$\sum_{k\geq 0}\left[\frac{1}{4^k}\binom{2k}{k}\right]^2 x^{2k}=\frac{2}{\pi}K(x^2)\tag{4}$$ leading to: $$\int_{\mathbb{R}}\arcsin\frac{1}{\cosh x}\,dx = 2\int_{0}^{1} K(x^2)\,dx = \int_{0}^{1}\frac{K(x)}{\sqrt{x}}\,dx.\tag{5}$$ At last, we recall that both $$K(x)$$ and $$\frac{1}{\sqrt{x}}$$ have fairly simple Fourier-Legendre series expansions: $$K(x)=2\sum_{n\geq 0}\frac{P_n(2x-1)}{2n+1},\qquad \frac{1}{\sqrt{x}}=2\sum_{n\geq 0}(-1)^n P_n(2x-1)$$ and by the orthogonality of shifted Legendre polynomials
$$\int_{\mathbb{R}}\arcsin\frac{1}{\cosh x}\,dx =\int_{0}^{1}\frac{K(x)}{\sqrt{x}}\,dx = 4\sum_{n\geq 0}\frac{(-1)^n}{(2n+1)^2} = 4G.\tag{6}$$
• This is really cool. I'll have to look at this one a little more in order to understand it better... But this is great. Thanks! – clathratus Oct 31 '18 at 18:37
• You seem to link any special function with any other special function. The link between elliptic integral and Legendre function is pretty smart. +1 – Paramanand Singh Nov 1 '18 at 16:41
• How do we show $$\int_{\Bbb R}\frac{\mathrm dx}{\cosh(x)^{2k+1}}=\frac{\pi{2k\choose k}}{4^k}$$ It looks a lot like a Beta integral – clathratus Dec 21 '18 at 0:56
• @clathratus: it is, indeed, it is enough to set $\frac{1}{\cosh(x)}=u$. – Jack D'Aurizio Dec 21 '18 at 10:48 | 2019-04-18T14:40:51 | {
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https://math.stackexchange.com/questions/4081507/applying-gauss-divergence-theorem-to-given-integral | # Applying Gauss' Divergence Theorem to given integral
We are given the following two integrals:
$$\iint\limits_S D_n f\:dS$$ and $$\iiint\limits_B \nabla \cdot (\nabla f) \: dV$$
where $$S$$ is the portion of the sphere $$x^2+y^2+z^2=a^2$$ in the first octant, $$n$$ is the unit normal vector to $$S$$ at $$(x,y,z)$$ and $$f(x,y,z) = ln(x^2+y^2+z^2)$$
For the first integral, we have: $$D_n f = \nabla f \cdot n$$ and $$n = \frac{1}{a}$$
Since $$\nabla f = \: <\frac{2x}{x^2+y^2+z^2}, \frac{2y}{x^2+y^2+z^2}, \frac{2z}{x^2+y^2+z^2}>$$, this gives $$D_n f = \nabla f \cdot n = \frac{2}{a}$$.
So the integral becomes
$$\iint\limits_S D_nf \: dS = \iint\limits_S \frac{2}{a} \: dS = \frac{2}{a} \iint\limits_S dS = \frac{2}{a} \cdot A(S) = \frac{2}{a} \cdot 4\pi a^2 \cdot \frac{1}{8} = \pi a$$ ($$\frac{1}{8}$$ since the sphere is in the first octant)
For the second integral, we have $$\nabla \cdot (\nabla f) = \frac{2}{a^2}$$ (since $$x^2 + y^2 + z^2 = a^2$$). So the integral becomes $$\iiint\limits_B \nabla \cdot (\nabla f) \: dV = \frac{2}{a^2} \iiint\limits_B dV = \frac{2}{a^2} \cdot V(B) = \frac{2}{a^2} \cdot \frac{4}{3} \pi a^3 \cdot \frac{1}{8} = \frac{1}{3} \pi a$$
However, according to Gauss' Divergence theorem, these two integrals should be equal to each other right? Since we have $$\iiint\limits_B \nabla \cdot (\nabla f) \: dV = \iint\limits_S \nabla f \cdot dS = \iint\limits_S \nabla f \cdot n \: dS = \iint\limits_S D_n f \: dS$$
So how is it possible that I get two different answers, even though the Divergence theorem shows that the two integrals should be the same?
• You're not counting the flat parts of the surface which are quarter circles in the coordinate planes. – B. Goddard Mar 29 at 11:09
• Oh I see, so if I understand correctly, is it true that the results of the first two integrals are correct, but the application of Gauss' theorem isn't? Because the surface S is not the same as the boundary surface of B, since the boundary surface contains extra flat parts? – Stallmp Mar 29 at 11:14
To apply divergence theorem, you must have a closed surface. So we close the surface by placing $$3$$ quarter disks in plane $$x = 0, y = 0, z = 0$$.
Please note that when you are doing volume integral, you cannot equate $$\nabla \cdot (\nabla f) = \displaystyle \frac{2}{x^2+y^2+z^2} = \frac{2}{a^2}$$. It should rather be,
$$\nabla \cdot (\nabla f) = \displaystyle \frac{2}{x^2+y^2+z^2} = \frac{2}{\rho^2}$$
So the integral becomes,
$$\displaystyle \int_0^{\pi/2} \int_0^{\pi/2} \int_0^a \frac{2}{\rho^2} \ \cdot\rho^2 \cdot \sin \phi \ d\rho \ d\phi \ d\theta = \pi a$$.
Now to find flux through $$S$$, we must subtract flux through planar surfaces $$x = 0, y = 0, z = 0$$ but in this case, they are simply zero.
• simply because $x^2+y^2+z^2 = a^2$ only on the surface of the sphere, not inside it. We are talking volume integral which includes all points inside the sphere too and so we write $x^2+y^2+z^2 \leq a^2$. – Math Lover Mar 29 at 11:31
• Yes you are right about surface $S$. What you get from divergence theorem will include flux through closed surface which is spherical surface $S$ and $3$ planar surfaces as mentioned in my answer. We are supposed to calculate that separately and subtract from the result that we get from divergence theorem to obtain flux only through spherical surface $S$. But in this case they are zero. Do you see why or do you need an explanation? – Math Lover Mar 29 at 11:39
• Yes this is indeed what I thought, thank you very much for the explanation! I see why they are zero as well when calculating the flux through the planar surfaces. – Stallmp Mar 29 at 11:45
• Yes you are right, we need to show they sum to zero. Note that $\iint\limits_{S_1} D_n f \: dS = \iint\limits_{S_2} D_n f \: dS = \iint\limits_{S_3} D_n f \: dS = 0$ – Math Lover Mar 29 at 12:34
• – Math Lover Mar 29 at 12:35 | 2021-06-23T17:26:15 | {
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https://math.stackexchange.com/questions/4391113/method-for-extracting-the-nth-root-of-a-number | # Method for Extracting the $n$th root of a number?
I was wondering if there exists a reliable way to extract the $$n$$th root of a given number. For example, if you have a large perfect square such as 10,404, how would I go about taking its square root by hand?
And what about cube roots? Is there a way to find the cube root of any number by hand?
If there is a way to take the cube and square roots of any number, is there a technique that can be applied to extract the cube or square root of polynomials? What about any root?
I think that, if there is a way to take the $$n$$th root of any number of polynomial, I would benefit from learning it, since I consider being able to work with roots important to my algebra foundation.
Thanks
• For the square root, there is a method similar to the usual division by a number. If we know that the number is a perfect square we can restrict the possibilities (ending digits , residue mod 9). For cube roots and higher , I am not aware of an efficient method by hand. Feb 25 at 18:58
• For square roots we can use the algorithm described here under "Digit-by-digit calculation" / "Decimal (base 10)" Feb 25 at 19:03
• @StefanOctavian I've seen similar algorithms for higher roots done (somehow you take 3-digits at a time for cube roots.) But they stopped teaching the square-root extraction the year before I would have had to learn it. Feb 25 at 20:38
If you would like to compute $$\sqrt[p]{a}$$, the following iterative process will do it to whatever accuracy you please:
First, make a guess and call it $$x_0$$. Then perform this iteration:
$$x_{n+1} = \frac{(p-1)x_n^p+a}{px^{k-1}}.$$
For instance, if you want $$\sqrt[5]{11}$$, guess that $$x_0=2$$. The iteration looks like
$$x_{n+1} = \frac{4x_n^5+11}{5x_n^4}.$$
So we have $$x_1 = \frac{4\cdot2^5+11}{5\cdot 2^4} = \frac{139}{80}=1.7375.$$
$$x_2=\frac{4\cdot1.7375^5+11}{5\cdot 1.7375^4} = 1.631392308.$$
This number to the 5th power is $$11.55558806$$, so we're getting close.
$$x_3=\frac{4\cdot1.631392308^5+11}{5\cdot 1.631392308^4} = 1.615704970.$$
And $$1.615704970^5 = 11.0105\ldots.$$
This is an implementation of Newton's Method on the polynomial $$x^p-a$$. | 2022-07-07T08:46:28 | {
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"lm_q1q2_score": 0.8596952020775964
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https://www.jiskha.com/questions/509900/related-rates-gas-is-escaping-from-a-spherical-balloon-at-the-rate-of-2ft-3-min-how-fast | Related Rates:
Gas is escaping from a spherical balloon at the rate of 2ft^3/min. How fast is the surface area shrinking (ds/dt) when the radius is 12ft? (A sphere of radius r has volume v=4/3 pi r^3 and surface area S=4pi r^2.)
Remember that ds/dt = ds/dr X dr/dt
Step 1: Find ds/dr
I have no clue where to start. I was going to set 2 equal to 4 pi r^2 and then take the derivative but I really have no idea.
Step 2: Find dr/dr. (HInt dv/dt= dv/dr X dr/dt
Would I set 2 equal to volume and then take derivative?
Step 3: Find ds/dt?
Step 4: Evaluate ds/dt when the radius is 12ft.
After I find my equation I would just plug in 12 correct?
1. 👍
2. 👎
3. 👁
1. S = 4pi*r^2
dS/dr = 4pi*2r = 8pi*r
V = (4/3)pi*r^3
dV/dr = 4pi*r^2
dV/dt = dV/dr * dr/dt
2 = 4pi*r^2 * dr/dt
dr/dt = 1/(2pi*r^2)
dS/dt = dS/dr * dr/dt
= 8pi*r * 1/(2pi*r^2)
= 4/r
If r = 12, just plug the value to the last equation
1. 👍
2. 👎
2. How did you get 12?
1. 👍
2. 👎
3. Nevermind I meant how did you get 4/r but I figured it out. Thank You!!
1. 👍
2. 👎
4. Is this correct for the last part?
(8)(3.14)(12) X (1/(2)(3.14)(12^2) = 4/12
1. 👍
2. 👎
5. Yes
1. 👍
2. 👎
6. Ok do I actually have to do the calculations though?
1. 👍
2. 👎
7. If you are not given the value, that means you have to write the steps I told you until the very last equation.
If you are given the value (of r), you don't have to put the value in every equation. Just follow my steps until the last equation, then plug the value after that. It's easier that way.
1. 👍
2. 👎
8. OK thank you!
1. 👍
2. 👎
9. Simplify the answer (4/12) into 1/3
1. 👍
2. 👎
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https://math.stackexchange.com/questions/2829529/relation-between-bigcap-i-in-ia-i-and-bigcap-i-1na-i | # Relation between $\bigcap_{i \in I}A_i$ and $\bigcap_{i=1}^{n}A_i$
Let I be a nonempty set and a family of sets such that every element of the family is a subset of U.
$\mathcal F = \{A_i | i \in I\}$
I understand the meaning of this operation:
$$\bigcap_{i \in I}A_i$$
That's the intersection of all the elements of the sets of the family $\mathcal F$.
But I don't truly understand what this other operation means and what's the relation between the above one. $$\bigcap_{i=1}^{n}A_i$$
I understand that this operation is also an intersection but what I am trying to understand is the relation between $\bigcap_{i \in I}A_i$ and $\bigcap_{i=1}^{n}A_i$
$$\bigcap_{i=1}^{n}A_i \subseteq\bigcap_{i \in I}A_i$$
Is the relation above true?
• In the second case we have $I = \{ 1,2,\ldots, n \}$. – Mauro ALLEGRANZA Jun 23 '18 at 15:28
The notation $$\bigcap _{i=1}^nA_i$$ denotes the same as $$\bigcap_{i\in I}A_i$$ for the special case $I=\{1,2,\ldots, n\}$.
• So $\bigcap_{i\in I}A_i$ is the general case and $\bigcap _{i=1}^nA_i$ is a specific case for which we give concrete values to i? – adriana634 Jun 23 '18 at 15:40
• @adriana634 Yes. One situation where we see both notations used simultaneously is when $I=\mathbb{N}$; then "$\bigcap_{i=1}^nA_i$" means "$\bigcap_{i\in \{1,2,..., n\}}A_i$," which is generally a superset of $\bigcap_{i\in I}A_i$. – Noah Schweber Jun 23 '18 at 15:42
• @Noah Schweber But for example $\bigcap_{i=1}^{5}A_i$ means $I=\{1,2,\ldots, 5\}$, in this example $\bigcap_{i\in I}A_i$ is a superset of $\bigcap_{i=1}^{5}A_i$? I don't understand why you said $\bigcap_{i=1}^nA_i$ is a superset of $\bigcap_{i\in I}A_i$. – adriana634 Jun 23 '18 at 16:00
• @adriana634 No. The intersection over a bigger index set yields a smaller set. E.g. if we take $I=\mathbb{N}$ and $A_i=\{i, i+1, i+2, ...\}$ for $i\in I$, then $\bigcap_{i=1}^5A_i=\{5, 6,7,...\}$ but $\bigcap_{i\in I}A_i=\emptyset$. More abstractly: $$I\supseteq J\implies \bigcap_{i\in I}A_i\subseteq \bigcap_{i\in J}A_i.$$ – Noah Schweber Jun 23 '18 at 16:02 | 2020-01-26T22:03:37 | {
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"lm_q1q2_score": 0.8596716316944107
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# In the addition shown above, A, B, C, and D represent
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In the addition shown above, A, B, C, and D represent [#permalink]
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15 Oct 2013, 07:17
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Question Stats:
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ABC
+BCB
CDD
In the addition shown above, A, B, C, and D represent the nonzero digits of three 3-digit numbers. What is the largest possible value of the product of A and B ?
A. 8
B. 10
C. 12
D. 14
E. 18
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Posts: 4488
Re: In the addition shown above, A, B, C, and D represent [#permalink]
### Show Tags
15 Oct 2013, 10:47
17
12
saintforlife wrote:
ABC
+BCB
CDD
In the addition shown above, A, B, C, and D represent the nonzero digits of three 3-digit numbers. What is the largest possible value of the product of A and B ?
A. 8
B. 10
C. 12
D. 14
E. 18
I'm happy to help with this.
First, look at the one's column. C + B produces a one's unit of D, and we don't know whether anything carries to the ten's place.
But now, look at the ten's place --- B + C again produces a digit of D --- that tells us definitively that nothing carried, and that B + C = D. B & C have a single digit number as a sum.
Now, C = A + B, because again, in the hundreds column, nothing carries here.
Notice that, since C = A + B and D = B + C, B must be smaller than both B and C. We want to make the product of A & B large, so we want to make those individual digits large. Well, if we make B large, then that makes C large, and then B + C would quickly become more than a one-digit sum, which is not allowed. Think about it this way. Let's just assume D = 9, the maximum value.
D = 9 = B + C = B + (A + B) = A + 2B
We want to pick A & B such that A + 2B = 9 and A*B is a maximum. It makes sense that B would be smaller.
Try A = 7, B = 1. Then A + 2B = 9 and A*B = 7
Try A = 5, B = 2. Then A + 2B = 9 and A*B = 10
Try A = 3, B = 3. Then A + 2B = 9 and A*B = 9
Try A = 1, B = 4. Then A + 2B = 9 and A*B = 4
Indeed, as B gets bigger, the product gets less. This seems to imply that the biggest possible product is 10. This corresponds to A = 5, B = 2, C = 7, and D = 9, and the original addition problem becomes
527
+272
799
Thus, the maximum product is 10, and answer = (B).
Does all this make sense?
Mike
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Re: In the addition shown above, A, B, C, and D represent [#permalink]
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06 Dec 2013, 10:19
8
1
saintforlife wrote:
How do we do all that in less than 2 mins, is my only question :)
Posted from my mobile device
This problem can be solved much faster if you start with the options. As Mike has wonderfully explained, none of the digits will have a carry over.
Option E is $$18 = 9 * 2$$(no other case is possible). Now$$9+2>=10$$. Therefore, it will have a carry over digit 1. So reject this option.
Option D is $$14 = 7 * 2$$(no other case is possible). Since $$C=A+B$$, therefore $$C= 9$$. Since $$C=9$$ and $$B$$ is a non zero digit, $$C+B>=10$$. Therefore, it will have a carry over digit 1. So reject this option.
Option C is 12. This will have 2 cases-
$$Case 1--> 12 = 4 * 3$$
Here $$C= 4+3=7$$. Now $$7+4>=10$$ and [$$7+3>=10$$. Therefore, both cases will have a carry over digit 1. So reject this option.
(OR) $$Case 2--> 6 * 2$$.
Here $$C= 6+2=8$$. Now $$8+2>=10$$ and $$8+6>=10$$. Therefore, both cases will have a carry over digit 1. So reject this option.
Option B is $$10 = 5 * 2$$. If you follow the same set of steps you'll notice that there will not be any carry over digit for the case $$C=7, A=5, B=2$$. So right answer!
##### General Discussion
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Re: In the addition shown above, A, B, C, and D represent [#permalink]
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23 Oct 2013, 15:44
2
How do we do all that in less than 2 mins, is my only question
Posted from my mobile device
Magoosh GMAT Instructor
Joined: 28 Dec 2011
Posts: 4488
Re: In the addition shown above, A, B, C, and D represent [#permalink]
### Show Tags
23 Oct 2013, 15:54
6
saintforlife wrote:
How do we do all that in less than 2 mins, is my only question
Dear Saint For Life,
Problems such as this are quite out-of-the-box. To some extent, the GMAT intends us to handle many of the other questions in 90 secs or less, so that we have a bit of a time-cushion when we run into one of these oddball questions.
Having said that, the more familiar you are with number properties, the faster it will go. In this problem, it took me quite some time to write out everything in verbal form, but I saw things relatively quickly. As you practice seeing patterns, you will see them more quickly, even if they are hard to explain to someone else. You may find this post helpful:
http://magoosh.com/gmat/2013/how-to-do- ... th-faster/
Mike
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Re: In the addition shown above, A, B, C, and D represent [#permalink]
### Show Tags
23 Oct 2013, 23:13
5
2
saintforlife wrote:
ABC
+BCB
CDD
In the addition shown above, A, B, C, and D represent the nonzero digits of three 3-digit numbers. What is the largest possible value of the product of A and B ?
A. 8
B. 10
C. 12
D. 14
E. 18
Similar questions to practice:
tough-tricky-set-of-problems-85211.html#p638336
Hope it helps
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Re: In the addition shown above, A, B, C, and D represent the [#permalink]
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26 Oct 2013, 21:47
2
... got E as an answer. However, it took me 6 minutes, since I plugged in the answer options.
Is there any faster way, or am I just too slow in back-solving?
TirthankarP wrote:
In the addition shown above, A, B, C, and D represent the nonzero digits of three 3-digit numbers. What is the largest possible value of the product of A and B ?
A) 8
B) 10
C) 12
D) 14
E) 18
My explanation:
According to the question stem C+B=D and D<10, since the second addition (2nd column) is C+B=D again (if there were a carry, the second column's result would not be D again), and A+B=C
Now I break up the answer options:
A) A*B=8, so A and B -> 2 and 4 or 1 and 8. A+B=2+4=6=C, C+B=6+2=8=D, which is smaller than 10, thus the answer is not "out", but there could be a larger possible value of A*B
B) A*B=10, so A and B -> 2 and 5. A+B=2+5=7=C, C+B=7+2=9=D, which is smaller than 10, thus the is also possible. Let's look for the next options.
For C-E:
C) A*B=12, so A and B -> 6 and or 4 and 3.
D) A*B=14, so A and -> 7 and 2.
E) A*B=18, so A and B -> 9 and 2 or 3 and 6.
Do the same process with those numbers and you will find that all will yield a sum of C+B>10, thus the constraint of D<10 is not satisfied. The answer options are not possible.
Because 10>8 answer option B) is the largest possible.
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Re: In the addition shown above, A, B, C, and D represent the [#permalink]
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27 Oct 2013, 00:04
Chiranjeevee wrote:
TirthankarP wrote:
Attachment:
In the addition shown above, A, B, C, and D represent the nonzero digits of three 3-digit numbers. What is the largest possible value of the product of A and B ?
A) 8
B) 10
C) 12
D) 14
E) 18
I do a thorough search before posting any new question. But this time I couldn't find this link
Forum moderator may delete this thread
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In the addition shown above, A, B, C, and D represent [#permalink]
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01 Jul 2015, 07:02
HKHR wrote:
saintforlife wrote:
How do we do all that in less than 2 mins, is my only question :)
Posted from my mobile device
This problem can be solved much faster if you start with the options. As Mike has wonderfully explained, none of the digits will have a carry over.
Option E is $$18 = 9 * 2$$(no other case is possible). Now$$9+2>=10$$. Therefore, it will have a carry over digit 1. So reject this option.
Option D is $$14 = 7 * 2$$(no other case is possible). Since $$C=A+B$$, therefore $$C= 9$$. Since $$C=9$$ and $$B$$ is a non zero digit, $$C+B>=10$$. Therefore, it will have a carry over digit 1. So reject this option.
Option C is 12. This will have 2 cases-
$$Case 1--> 12 = 4 * 3$$
Here $$C= 4+3=7$$. Now $$7+4>=10$$ and [$$7+3>=10$$. Therefore, both cases will have a carry over digit 1. So reject this option.
(OR) $$Case 2--> 6 * 2$$.
Here $$C= 6+2=8$$. Now $$8+2>=10$$ and $$8+6>=10$$. Therefore, both cases will have a carry over digit 1. So reject this option.
Option B is $$10 = 5 * 2$$. If you follow the same set of steps you'll notice that there will not be any carry over digit for the case $$C=7, A=5, B=2$$. So right answer!
Wonderful explanation. Just thought to add some more from my side. As we see only B satisfies all the necessary requirement(not to carry over any digits).
In option A, taking A=4 and B=2 also satisfies the condition as it is something like
426
+262
-------
688
But since we need maximum value well go for option B
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Re: In the addition shown above, A, B, C, and D represent [#permalink]
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30 Nov 2017, 18:03
Dear Mike, I was wondering why are we maximising D since it no where mentioned to do so. We have to maximise A*B right so if we choose C=1 and B=5 and A=4 i think we staisfy all conditions and our product of A*B (5*4) comes to be 20. Dont know where i am going wrong. Kindly help
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Posts: 4488
Re: In the addition shown above, A, B, C, and D represent [#permalink]
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01 Dec 2017, 13:54
1
1
Sarthak.bhatt wrote:
Dear Mike, I was wondering why are we maximising D since it no where mentioned to do so. We have to maximise A*B right so if we choose C=1 and B=5 and A=4 i think we staisfy all conditions and our product of A*B (5*4) comes to be 20. Dont know where i am going wrong. Kindly help
Dear Sarthak.bhatt,
I'm happy to respond.
With all due respect, my friend, you are not interpreting the question correctly. There is absolutely no multiplication happening in this question. Here's the prompt again:
ABC
+BCB
CDD
In the addition shown above, A, B, C, and D represent
the nonzero digits of three 3-digit numbers. What is the largest possible value of the product of A and B ?
Thus, for example if C = 1, B = 5, and A = 4, then ABC would not be the product of those three numbers but instead the single three-digit number 451. With those numbers, the problem would be
451
+515
966
These choices satisfy the equation.
The problem is completely different from the way you were conceptualizing it, so the strategy is completely different from what it would be in the problem you had in mind.
Does this make sense?
Mike
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Re: In the addition shown above, A, B, C, and D represent [#permalink]
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26 May 2018, 03:22
mikemcgarry. Can you please explain Sarthak's query as i have the same query but unfortunately i didn't get from the reply you gave to him.
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Re: In the addition shown above, A, B, C, and D represent [#permalink]
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15 Jun 2018, 15:24
Do we not consider negative numbers in this problem at all?
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Posts: 50727
Re: In the addition shown above, A, B, C, and D represent [#permalink]
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15 Jun 2018, 21:55
rohithtv89 wrote:
Do we not consider negative numbers in this problem at all?
No. We are given that A, B, C, and D represent the nonzero digits. There are 10 digits: 0, 1, 2, 3, 4, 5, 6, 7, 8, and 9.
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Re: In the addition shown above, A, B, C, and D represent &nbs [#permalink] 15 Jun 2018, 21:55
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# A photography dealer ordered 60 Model X cameras to be sold
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A photography dealer ordered 60 Model X cameras to be sold [#permalink]
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A photography dealer ordered 60 Model X cameras to be sold for $250 each, which represents a 20 percent markup over the dealer’s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer’s initial cost for the 60 cameras? A. 7% loss B. 13% loss C. 7% profit D. 13% profit E. 15% profit ##### Most Helpful Expert Reply Veritas Prep GMAT Instructor Joined: 16 Oct 2010 Posts: 8195 Location: Pune, India Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 24 Jun 2012, 22:20 61 38 pgmat wrote: A photography dealer ordered 60 Model X cameras to be sold for$250 each, which represents a 20 percent markup over the dealer’s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer’s initial cost for the 60 cameras?
A. 7% loss
B. 13% loss
C. 7% profit
D. 13% profit
E. 15% profit
Use weighted avgs for a quick solution:
On 54 cameras, the dealer made a 20% profit and on 6 cameras, he made a 50% loss. Cost Price of 54 cameras:cost price of 6 cameras = 54:6 = 9:1
Avg Profit/Loss = (.2*9 + (-.5)*1)/10 = 1.3/10 = .13 = 13% profit
For more on weighted avgs, check: http://www.veritasprep.com/blog/2011/03 ... -averages/
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Save up to $1,000 on GMAT prep through 8/20! Learn more here > GMAT self-study has never been more personalized or more fun. Try ORION Free! ##### Most Helpful Community Reply Manager Status: mba here i come! Joined: 07 Aug 2011 Posts: 230 A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags Updated on: 25 Feb 2016, 08:28 98 27 54 cameras gave 20% profit while 6 incurred 50% loss. or out of 10, 9 gave profit. suppose total cameras = 10 and each costs$10
total profit = $2*9 -$5*1 = $13 or 13% (with$100 cost)
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Originally posted by MBAhereIcome on 23 Jun 2012, 07:31.
Last edited by MBAhereIcome on 25 Feb 2016, 08:28, edited 1 time in total.
##### General Discussion
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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
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06 May 2012, 11:07
25
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pgmat wrote:
A photography dealer ordered 60 Model X cameras to be sold for $250 each, which represents a 20 percent markup over the dealer’s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer’s initial cost for the 60 cameras? A. 7% loss B. 13% loss C. 7% profit D. 13% profit E. 15% profit Total cost 60*($250/1.2)=50*250;
# of cameras sold is 60-6=54 total revenue is 54*250;
# of cameras returned is 6 total refund 6*(250/1.2)*0.5;
So, total income 54*250+ 6*(250/1.2)*0.5
The dealer's approximate profit is (54*250+ 6*(250/1.2)*0.5-50*250)/(50*250)*100=13%
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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
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22 Jun 2012, 14:14
Bunuel wrote:
pgmat wrote:
A photography dealer ordered 60 Model X cameras to be sold for $250 each, which represents a 20 percent markup over the dealer’s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer’s initial cost for the 60 cameras? A. 7% loss B. 13% loss C. 7% profit D. 13% profit E. 15% profit Total cost$60*250/1.2=50*250;
# of cameras sold is 60-6=54 total revenue is 54*250;
# of cameras returned is 6 total refund 6*(250/1.2)*0.5;
So, total income 54*250+ 6*(250/1.2)*0.5
The dealer's approximate profit is (54*250+ 6*(250/1.2)*0.5-50*250)/(50*250)*100=13%
Dear Bunuel,
Ths is a doubt i face with %ages, in ths quest too.
could you pls tell where i m going wrong?
when 20% mark up for initial cost is given , how to calculate it?
250 - 20/100*250 or to take if 120 is 250 then how much is 100?
how have you arrived at 1.2?
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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
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23 Jun 2012, 04:06
3
2
kashishh wrote:
Bunuel wrote:
pgmat wrote:
A photography dealer ordered 60 Model X cameras to be sold for $250 each, which represents a 20 percent markup over the dealer’s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer’s initial cost for the 60 cameras? A. 7% loss B. 13% loss C. 7% profit D. 13% profit E. 15% profit Total cost 60*$250/1.2=50*250;
# of cameras sold is 60-6=54 total revenue is 54*250;
# of cameras returned is 6 total refund 6*(250/1.2)*0.5;
So, total income 54*250+ 6*(250/1.2)*0.5
The dealer's approximate profit is (54*250+ 6*(250/1.2)*0.5-50*250)/(50*250)*100=13%
Dear Bunuel,
Ths is a doubt i face with %ages, in ths quest too.
could you pls tell where i m going wrong?
when 20% mark up for initial cost is given , how to calculate it?
250 - 20/100*250 or to take if 120 is 250 then how much is 100?
how have you arrived at 1.2?
If it's given that the selling price is $250 and the markup over the initial cost is 20%, then: {Cost}+0.2{Cost}=1.2*{Cost}={Selling price} --> 1.2*{Cost}=$250 --> {Cost}=$250/1.2. Hope it's clear. _________________ Intern Joined: 28 Feb 2012 Posts: 21 GMAT 1: 700 Q48 V39 WE: Information Technology (Computer Software) Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 24 Jun 2012, 02:06 1 MBAhereIcome wrote: 54 cameras gave 20% profit while 6 incurred 50% loss. or out of 10, 9 gave profit. suppose total cameras = 10, out of which 9 gave 20% profit suppose 100 was the cost of each. total profit 100*10=1000 profit from 9 = 20*9 = 180 loss from 1 = 50*1 = 50 total profit = 180-50 = 130 so, 130 profit from a cost of 1000. that is 13% profit. Your method seems easy, but I kind of got lost with this - 54 cameras gave 20% profit while 6 incurred 50% loss. or out of 10, 9 gave profit. Could you please explain, how you deduced this? Thanks. Manager Joined: 02 Jun 2011 Posts: 132 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 24 Jun 2012, 12:26 1 2 gmatDeep wrote: MBAhereIcome wrote: 54 cameras gave 20% profit while 6 incurred 50% loss. or out of 10, 9 gave profit. suppose total cameras = 10, out of which 9 gave 20% profit suppose 100 was the cost of each. total profit 100*10=1000 profit from 9 = 20*9 = 180 loss from 1 = 50*1 = 50 total profit = 180-50 = 130 so, 130 profit from a cost of 1000. that is 13% profit. Your method seems easy, but I kind of got lost with this - 54 cameras gave 20% profit while 6 incurred 50% loss. or out of 10, 9 gave profit. Could you please explain, how you deduced this? Thanks. yess.. 54:6 = 9:1 mark up is of 20% = profit earned will be 20% VP Joined: 02 Jul 2012 Posts: 1192 Location: India Concentration: Strategy GMAT 1: 740 Q49 V42 GPA: 3.8 WE: Engineering (Energy and Utilities) Re: Of the cameras ordered, 6 were never sold and were returned [#permalink] ### Show Tags 16 Nov 2012, 23:15 15 4 carcass wrote: A photography dealer ordered 60 Model X cameras to be sold for$250 each, whichrepresents a 20 percent markup over the dealer's initialcost for each camera.
Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer's initial cost for the 60 cameras?
(A) 7% loss
(B) 13% loss
(C) 7% profit
(D) 13% profit
(E) 15% profit
have fun. is not difficult to conceptualize but is difficult because if you are not organized and clear you are get in trouble.
I would not show the OA but the rules are stringent. either way try on your own
Cost to dealer for 60 cameras = $$\frac{250}{1.2}*60$$ = $12,500 Revenue for 54 cameras = 54*250 =$13,500
Revenue from 6 cameras = $$\frac{250}{2.4}*6$$ = $625 Total Revenue =$14,125
Profit percent = $$\frac{14125-12500}{12500}$$ = 13%
Otherwise, If one were short on time and I had to guess, One can eliminate A,B & C by looking at the problem itself and improve my chances of guessing the right answer to 50% between D & E.
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Re: Of the cameras ordered, 6 were never sold and were returned [#permalink]
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16 Nov 2012, 23:27
1
60 cameras were ordered to be sold at a 20% markup for $250 each => Dealer's cost for each camera =$208
=> Cost of all 60 cameras = $60*208 6 cameras were not sold and returned for a refund of 50% of the dealer's cost => The amount the dealer got in returns = 50% of 6*208 =$624
The rest of the cameras (60-6 = 54 in number) were sold
=> Revenue made from selling the cameras = $54 * 250 For the 60 cameras, dealer's profit = 100* [54*(250-208) + (624-1248)]/[(60*200)] =100* (14100-12000)/12000 = ~13% (the error arises because 250/1.2 is not exactly 208 but 208.33) Option (D) _________________ GyanOne | Top MBA Rankings and MBA Admissions Blog Top MBA Admissions Consulting | Top MiM Admissions Consulting Premium MBA Essay Review|Best MBA Interview Preparation|Exclusive GMAT coaching Get a FREE Detailed MBA Profile Evaluation | Call us now +91 98998 31738 Director Status: Done with formalities.. and back.. Joined: 15 Sep 2012 Posts: 616 Location: India Concentration: Strategy, General Management Schools: Olin - Wash U - Class of 2015 WE: Information Technology (Computer Software) Re: Of the cameras ordered, 6 were never sold and were returned [#permalink] ### Show Tags 16 Nov 2012, 23:48 24 13 carcass wrote: A photography dealer ordered 60 Model X cameras to be sold for$250 each, whichrepresents a 20 percent markup over the dealer's initialcost for each camera.
Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer's initial cost for the 60 cameras?
(A) 7% loss
(B) 13% loss
(C) 7% profit
(D) 13% profit
(E) 15% profit
have fun. is not difficult to conceptualize but is difficult because if you are not organized and clear you are get in trouble.
I would not show the OA but the rules are stringent. either way try on your own
actually, if one could visualize this problem properly, it could be solved in less than 30 secs without getting into any dirty calculations.
Here is my 30 sec approach:
Notice that it is actually a wighted ratio problem, where 9 parts earned 20% profit and 1 part earned 50% loss. (54 for profit, 6 for loss).
Hence overall profit/loss = $$(9*20 -1*50) /(9+1) = 13$$
Ans D it is!
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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
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24 Nov 2012, 11:16
2
Hi,
For those using normal way to solve this problem, I think that we can avoid huge calculation by just replace Cost by letter C and put it in the final formula:
Income: 54*1.2C + 6*0.5C
Cost: 60*C
Then percent =(Income-cost)/cost = (54*1.2C + 6*0.5C - 60*C)/60*C. We can see that we can eliminate C from the fraction --> percent = (54*1.2 + 6*0.5 - 60)/60 = (54*1.2 + 6*0.5)/60 - 1. Here, we can see clearly the weighted average part of the problem or we can simply calculate the fraction to reach the result.
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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
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27 May 2013, 12:12
Bunuel wrote:
pgmat wrote:
A photography dealer ordered 60 Model X cameras to be sold for $250 each, which represents a 20 percent markup over the dealer’s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer’s initial cost for the 60 cameras? A. 7% loss B. 13% loss C. 7% profit D. 13% profit E. 15% profit Total cost 60*($250/1.2)=50*250;
# of cameras sold is 60-6=54 total revenue is 54*250;
# of cameras returned is 6 total refund 6*(250/1.2)*0.5;
So, total income 54*250+ 6*(250/1.2)*0.5
The dealer's approximate profit is (54*250+ 6*(250/1.2)*0.5-50*250)/(50*250)*100=13%
this must be kind of awkward question but i got no option other than asking you.
20% markup over the dealer’s initial cost for each camera- I made a wrong assumption , but reducing 20%from 250 & got 200 as an initial cost for each. please explain that "Total cost 60*($250/1.2)=50*250;" Math Expert Joined: 02 Sep 2009 Posts: 48037 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 27 May 2013, 12:19 3 1 FTGNGU wrote: Bunuel wrote: pgmat wrote: A photography dealer ordered 60 Model X cameras to be sold for$250 each, which represents a 20 percent markup over the dealer’s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer’s initial cost for the 60 cameras?
A. 7% loss
B. 13% loss
C. 7% profit
D. 13% profit
E. 15% profit
Total cost 60*($250/1.2)=50*250; # of cameras sold is 60-6=54 total revenue is 54*250; # of cameras returned is 6 total refund 6*(250/1.2)*0.5; So, total income 54*250+ 6*(250/1.2)*0.5 The dealer's approximate profit is (54*250+ 6*(250/1.2)*0.5-50*250)/(50*250)*100=13% Answer: D. this must be kind of awkward question but i got no option other than asking you. 20% markup over the dealer’s initial cost for each camera- I made a wrong assumption , but reducing 20%from 250 & got 200 as an initial cost for each. please explain that "Total cost 60*($250/1.2)=50*250;"
(Cost per unit) + 0.2*(Cost per unit) = $250 1.2*(Cost per unit) =$250
(Cost per unit) = $250/1.2 Total cost for 60 units = 60*(Cost per unit) = 60*($250/1.2) = 50*250.
Hope it's clear.
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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
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07 Apr 2014, 19:23
4
Because the answers are in percentages, I thought not to worry about the $amounts and just focus on the relationships: Let C be the total cost of all 60 cameras. Originally, the dealer thought to sell all these 60 cameras at 20% profit or for (1.2)C. However, he sold only (60-6)=54 or 90% of cameras at this price. So revenue from these cameras = (0.9)(1.2)C = (1.08)C For the remaining 10%, he got a refund of 50% of cost or (0.5)C. So total refund = (0.1)(0.5)C = (0.05)C Therefore, total revenue in terms of original cost = (1.08 + 0.05)C = 1.13C or 13% profit. So D is the correct ans. Intern Status: 1st attempt: Can I do it? Joined: 09 Jun 2013 Posts: 1 Location: India Concentration: Healthcare, Marketing GMAT Date: 05-30-2014 GPA: 2.99 WE: Pharmaceuticals (Pharmaceuticals and Biotech) Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 27 May 2014, 10:33 2 1 {(54*20) + (6*(-50))}/ 60 This gives 13% profit as answer. 20% proft on 54 cameras and 50% loss on 6 cameras. Use weighted averages here. Rest of the information in the question is misguiding. Intern Joined: 09 Feb 2013 Posts: 26 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 29 Jun 2014, 23:52 My soln is : Dealer's per unit initial cost =250 -20/100*250 = 200 Total initial cost = 60*200=12000 Total profit = (54*250+6*100) - 12000 = 2100 Profit % as a percent of dealer's initial cost = 2100/12000*100 = 17.5 % Can you please tell me where I went wrong ? Posted from my mobile device Math Expert Joined: 02 Sep 2009 Posts: 48037 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 30 Jun 2014, 02:50 kshitij89 wrote: My soln is : Dealer's per unit initial cost =250 -20/100*250 = 200 Total initial cost = 60*200=12000 Total profit = (54*250+6*100) - 12000 = 2100 Profit % as a percent of dealer's initial cost = 2100/12000*100 = 17.5 % Can you please tell me where I went wrong ? Posted from my mobile device The cost per unit is not 0.8*250 = 200, it's 250/1.2 = ~208. Markup is calculated on the cost value (check here: http://gmatclub.com/forum/a-photography ... l#p1229596). Hope it helps. _________________ Manager Joined: 28 Dec 2013 Posts: 68 Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink] ### Show Tags 21 Aug 2014, 08:48 VeritasPrepKarishma wrote: pgmat wrote: A photography dealer ordered 60 Model X cameras to be sold for$250 each, which represents a 20 percent markup over the dealer’s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer’s initial cost for the 60 cameras?
A. 7% loss
B. 13% loss
C. 7% profit
D. 13% profit
E. 15% profit
Use weighted avgs for a quick solution:
On 54 cameras, the dealer made a 20% profit and on 6 cameras, he made a 50% loss. Cost Price of 54 cameras:cost price of 6 cameras = 54:6 = 9:1
Avg Profit/Loss = (.2*9 + (-.5)*1)/10 = 1.3/10 = .13 = 13% profit
For more on weighted avgs, check: http://www.veritasprep.com/blog/2011/03 ... -averages/
why are we dividing by 10?
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Re: A photography dealer ordered 60 Model X cameras to be sold [#permalink]
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24 Aug 2014, 21:54
1
sagnik242 wrote:
VeritasPrepKarishma wrote:
pgmat wrote:
A photography dealer ordered 60 Model X cameras to be sold for $250 each, which represents a 20 percent markup over the dealer’s initial cost for each camera. Of the cameras ordered, 6 were never sold and were returned to the manufacturer for a refund of 50 percent of the dealer's initial cost. What was the dealer's approximate profit or loss as a percent of the dealer’s initial cost for the 60 cameras? A. 7% loss B. 13% loss C. 7% profit D. 13% profit E. 15% profit Use weighted avgs for a quick solution: On 54 cameras, the dealer made a 20% profit and on 6 cameras, he made a 50% loss. Cost Price of 54 cameras:cost price of 6 cameras = 54:6 = 9:1 Avg Profit/Loss = (.2*9 + (-.5)*1)/10 = 1.3/10 = .13 = 13% profit For more on weighted avgs, check: http://www.veritasprep.com/blog/2011/03 ... -averages/ why are we dividing by 10? Weighted Average Formula: Cavg = (C1*w1 + C2*w2)/(w1 + w2) C1 and C2 represent the quantity which we want to average so they will be profit/loss here. C1 = 20% = .2 C2 = -50% (loss) = -.5 The weights, given by w1 and w2, are the cost prices. Cost Price of 54 cameras:cost price of 6 cameras = 54:6 = 9:1 So w1 =9 and w2 = 1 Avg Profit/Loss = (.2*9 + (-.5)*1)/(9 + 1) = (.2*9 + (-.5)*1)/10 Get more details on this concept from the link given in my post above. _________________ Karishma Veritas Prep GMAT Instructor Save up to$1,000 on GMAT prep through 8/20! Learn more here >
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Re: A photography dealer ordered 60 Model X cameras to be sold &nbs [#permalink] 24 Aug 2014, 21:54
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# If x is an integer then x(x-1)(x-k) must be evenly divisible
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If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]
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31 Jan 2012, 15:49
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If x is an integer, then x(x – 1)(x – k) must be evenly divisible by three when k is any of the following values EXCEPT
A. -4
B. -2
C. -1
D. 2
E. 5
[Reveal] Spoiler:
The OA is B. I am trying to use the concept of consecutive numbers but got stuck. Can someone please help?
Question says x(x – 1)(x – k) must be evenly divisible by three which means x(x-1) (x-k) should be consecutive.
[Reveal] Spoiler: OA
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enigma123 wrote:
If x is an integer, then x(x – 1)(x – k) must be evenly divisible by three when k is any of the following values EXCEPT
A)-4
B)-2
C)-1
D) 2
E) 5
The OA is B. I am trying to use the concept of consecutive numbers but got stuck. Can someone please help?
Question says x(x – 1)(x – k) must be evenly divisible by three which means x(x-1) (x-k) should be consecutive.
We have the product of 3 integers: (x-1)x(x-k).
Note that the product of 3 integers is divisible by 3 if at least one multiple is divisible by 3. Now, to guarantee that at least one integer out of x, (x – 1), and (x – k) is divisible by 3 these numbers must have different remainders upon division by 3, meaning that one of them should have remainder of 1, another reminder of 2 and the last one remainder of 0, so be divisible by 3.
Next, if k=-2 then we'll have (x-1)x(x+2)=(x-1)x(x-1+3) --> which means that (x-1) and (x+2) will have the same remainder upon division by 3. Thus for k=-2 we won't be sure whether (x-1)x(x-k) is divisible by 3.
30 second approach: 4 out of 5 values of k from answer choices must guarantee divisibility of some expression by 3. Now, these 4 values of k in answer choices must have some pattern: if we get rid of -2 then -4, -1, 2, and 5 creating arithmetic progression with common difference of 3, so -2 is clearly doesn't belong to this pattern.
Hope it helps.
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Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]
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31 Jan 2012, 16:09
Thanks very much for a thorough explanation.
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Hi, there. I'm happy to help with this.
The rule that the product of three consecutive integers is a good start, but not the be all and end all.
number = x(x – 1)(x – k)
So far, we have integer x and one less that it (x - 1), so we could go down one more, or up one from x --
k = 2 ----> x(x – 1)(x – 2)
k = -1 ----> x(x – 1)(x + 1)
Now, we don't know which of the three factors are divisible by 3 -- x, or (x - 1), or the (x - k). If it's either of the first two, then we're golden, and k doesn't matter. But pretend that neither x nor (x - 1) is divisible by 3, then we are dependent on that last factor. Well, if (x - 2) is a multiple of three, we should be able to add or subtract three and still get a multiple of three.
(x - 2) - 3 = (x - 5)
(x - 5) - 3 = (x - 8)
(x - 2) + 3 = (x + 1), which we have already
(x + 1) + 3 = (x + 4)
(x + 3) + 3 = (x + 7)
So, for divisibility purposes, (x - 8), (x - 5), (x - 2), (x + 1), (x + 4), (x + 7) are all equivalent -- if any one of them is a multiple of three, all the others are. (You can check that the difference between any two is a multiple of 3.) BTW, if they are not divisible by three, then they all would have equal remainders if divided by three.
Back to the question:
If x is an integer, then x(x – 1)(x – k) must be evenly divisible by three when k is any of the following values EXCEPT
A)-4
B)-2
C)-1
D) 2
E) 5
All of those choices give us a term on our list except for (B) -2.
BTW, notice all the answer choices are spaced apart by three except for (B).
Does that make sense? Please do not hesitate to ask if you have any questions.
Mike
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Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]
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31 Jan 2012, 16:18
Thanks Mike. Really appreciate your solution too.
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Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]
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31 Jan 2012, 16:34
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This is my 1st post finally thought of jumping in instead of just being an observer
I attacked this problem in a simple way. As it states it is divisible by 3
that means both x & (x-1) cannot be a multiple of 3 otherwise whatever the value of k it will be still divisible by 3
so plugging in number i chose 5 in this case you can establish answer is -2 does not fit...
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Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]
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31 Jan 2012, 16:37
Thanks and welcome to GMAT Club Azim. I am sure you will have a great experience from a very helpful community.
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04 Oct 2012, 18:59
Bunuel wrote:
enigma123 wrote:
If x is an integer, then x(x – 1)(x – k) must be evenly divisible by three when k is any of the following values EXCEPT
A)-4
B)-2
C)-1
D) 2
E) 5
The OA is B. I am trying to use the concept of consecutive numbers but got stuck. Can someone please help?
Question says x(x – 1)(x – k) must be evenly divisible by three which means x(x-1) (x-k) should be consecutive.
We have the product of 3 integers: (x-1)x(x-k).
Note that the product of 3 integers is divisible by 3 if at least one multiple is divisible by 3. Now, to guarantee that at least one integer out of x, (x – 1), and (x – k) is divisible by 3 these numbers must have different remainders upon division by 3, meaning that one of them should have remainder of 1, another reminder of 2 and the last one remainder of 0, so be divisible by 3.
Next, if k=-2 then we'll have (x-1)x(x+2)=(x-1)x(x-1+3) --> which means that (x-1) and (x+2) will have the same remainder upon division by 3. Thus for k=-2 we won't be sure whether (x-1)x(x-k) is divisible by 3.
30 second approach: 4 out of 5 values of k from answer choices must guarantee divisibility of some expression by 3. Now, these 4 values of k in answer choices must have some pattern: if we get rid of -2 then -4, -1, 2, and 5 creating arithmetic progression with common difference of 3, so -2 is clearly doesn't belong to this pattern.
Hope it helps.
Bunuel,
Would this approach work for all integer divisors (NOT the 30 sec approach)?
Say the divisor is 4 and the choices had four terms instead of three, e.g. (x-1)(x-2)(x+k)(x+1)?
Thanks,
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Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]
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04 Oct 2012, 23:10
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enigma123 wrote:
If x is an integer, then x(x – 1)(x – k) must be evenly divisible by three when k is any of the following values EXCEPT
A. -4
B. -2
C. -1
D. 2
E. 5
The OA is B. I am trying to use the concept of consecutive numbers but got stuck. Can someone please help?
Question says x(x – 1)(x – k) must be evenly divisible by three which means x(x-1) (x-k) should be consecutive.
Since this is a multiple choice GMAT question, you can pick a particular value for x such that neither x, nor x-1 is divisible by 3 and start checking the answers.
In the given situation, choose for example x = 2 and check when 2 - k is not divisible by 3.
(A) 2 - (-4) = 6 NO
(B) 2 - (-2) = 4 BINGO!
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Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]
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05 Oct 2012, 03:37
I too solved it using a value for x =2, but I am not sure if is it better to solve using value for such questions or otherwise.
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Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]
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05 Oct 2012, 03:56
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Avantika5 wrote:
I too solved it using a value for x =2, but I am not sure if is it better to solve using value for such questions or otherwise.
On the GMAT, is definitely the fastest way to solve it. Being a multiple choice question, you can be sure that there is a unique correct answer. And in the given situation, the only issue is to choose for x values such that neither x, nor x - 1 is divisible by 3.
It won't harm to understand and know to use the properties of consecutive integers presented in the other posts . They can be useful any time.
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Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]
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06 Oct 2012, 01:10
Thanks EvaJager, I was always thought it is not good way to solve and I should learn the better way.
Thanks a lot...
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Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]
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06 Oct 2012, 02:10
Avantika5 wrote:
Thanks EvaJager, I was always thought it is not good way to solve and I should learn the better way.
Thanks a lot...
Better is a relative word...Mathematicians always try to prove and justify everything in a formal, logical way.
But GMAT is not testing mathematical abilities per se. If they wanted so, the questions would have been open and not multiple choice.
Have a flexible mind, think out of the box. GMAT is not a contest for the most beautiful, elegant, mathematical solution...
Get the correct answer as quickly as possible, and go to the next question without any feeling of guilt...:O)
Though, as I said, try to understand the other properties of the integer numbers, they can become handy and also, because they are so beautiful! Isn't Mathematics wonderful?
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Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]
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06 Oct 2012, 02:25
1. k=-4, x+4=y so x=y-4. Now ----- (y-4)(y-5)(y). Gives number divisible by 3 in all cases.
2. k=-2, , x+2=y so x=y-2. Now ----- (y-2)(y-3)(y). Not applicable when y is 5 or when x is 7
3. k=-1, , x+1=y so x=y-1. Now ----- (y-1)(y-2)(y). Consecutive 3 integers. Divisible by 3
4. k=2, , x-2=y so x=y+2. Now ----- (y+2)(y+1)(y). Consecutive 3 integers. Divisible by 3
5. k=5, , x-5=y so x=y+5. Now ----- (y+5)(y+4)(y). Consecutive 3 integers. Divisible by 3
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16 Feb 2013, 12:34
mikemcgarry wrote:
Hi, there. I'm happy to help with this.
The rule that the product of three consecutive integers is a good start, but not the be all and end all.
number = x(x – 1)(x – k)
So far, we have integer x and one less that it (x - 1), so we could go down one more, or up one from x --
k = 2 ----> x(x – 1)(x – 2)
k = -1 ----> x(x – 1)(x + 1)
Now, we don't know which of the three factors are divisible by 3 -- x, or (x - 1), or the (x - k). If it's either of the first two, then we're golden, and k doesn't matter. But pretend that neither x nor (x - 1) is divisible by 3, then we are dependent on that last factor. Well, if (x - 2) is a multiple of three, we should be able to add or subtract three and still get a multiple of three.
(x - 2) - 3 = (x - 5)
(x - 5) - 3 = (x - 8)
(x - 2) + 3 = (x + 1), which we have already
(x + 1) + 3 = (x + 4)
(x + 3) + 3 = (x + 7)
So, for divisibility purposes, (x - 8), (x - 5), (x - 2), (x + 1), (x + 4), (x + 7) are all equivalent -- if any one of them is a multiple of three, all the others are. (You can check that the difference between any two is a multiple of 3.) BTW, if they are not divisible by three, then they all would have equal remainders if divided by three.
Back to the question:
If x is an integer, then x(x – 1)(x – k) must be evenly divisible by three when k is any of the following values EXCEPT
A)-4
B)-2
C)-1
D) 2
E) 5
All of those choices give us a term on our list except for (B) -2.
BTW, notice all the answer choices are spaced apart by three except for (B).
Does that make sense? Please do not hesitate to ask if you have any questions.
Mike
Great posts everyone, including you Mike, who I am now quoting, but I would like to add that..
The first two integers for each answer choice cannot be divisible by 3 because all the answers choices would then be correct, thereby making the question invalid. Therefore, the last integer (x+k) must be divisible by 3, which helps in the theoretical approach...
It appears that was left out
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Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]
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17 Oct 2013, 15:25
enigma123 wrote:
If x is an integer, then x(x – 1)(x – k) must be evenly divisible by three when k is any of the following values EXCEPT
A. -4
B. -2
C. -1
D. 2
E. 5
The OA is B. I am trying to use the concept of consecutive numbers but got stuck. Can someone please help?
Question says x(x – 1)(x – k) must be evenly divisible by three which means x(x-1) (x-k) should be consecutive.
Hi enigma123. Thank you for the nice question. But you might wanna put a spoiler on this statement The OA is B so that others can solve the question in real conditions. Thanks
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Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]
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27 Oct 2014, 06:45
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I tried the problem in a different way:
for any number to be divisible by 3, the sum of the integers in the number should be a factor of 3.
Taking the sum of x, (x-1) and (x-k) we have:
x+x-1+x-k = 3x-1-k
now looking at the choices
k = -4 => sum = 3x+3 --> divisible by 3
k = -2 => sum = 3x+1 --> not divisible by 3
k = -1 => sum = 3x --> divisible by 3
k = 2 => sum = 3x-3 --> divisible by 3
k = 5 ==> sum = 3x-6 --> divisible by 3
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Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]
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17 Jun 2015, 02:13
Hi there,
The answer is right, however my explanation is quite differently:
1. x(x-1)(x-k)
2. (x^2-1) (x-k)
3. x^3-x^2k-x^2+xk
4. x-xk
Plugg-in x=1
a. 1-1(-4) = 1-(-4) = 5
b. 1-1(-2) = 1-(-2) = 3
c. 1-1(-1) = 1-(-1) = 2
d. 1-1(2) = 1-(2) = -1
e. 1-1(5) = 1-(5) = -4
Am I correct in my explanation?
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Re: If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]
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17 Jun 2015, 05:49
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Expert's post
santorasantu wrote:
Quote:
If x is an integer, then x(x – 1)(x – k) must be evenly divisible by three when k is any of the following values EXCEPT
A. -4
B. -2
C. -1
D. 2
E. 5
I tried the problem in a different way:
for any number to be divisible by 3, the sum of the integers in the number should be a factor of 3.
Taking the sum of x, (x-1) and (x-k) we have:
x+x-1+x-k = 3x-1-k
now looking at the choices
k = -4 => sum = 3x+3 --> divisible by 3
k = -2 => sum = 3x+1 --> not divisible by 3
k = -1 => sum = 3x --> divisible by 3
k = 2 => sum = 3x-3 --> divisible by 3
k = 5 ==> sum = 3x-6 --> divisible by 3
Just refining the highlighted language and presenting another view to del with this problem
CONCEPT1:For any number to be divisible by 3, the sum of the Digits of the Number should be a Multiple of 3.
CONCEPT2: Product of any three consecutive Integers always include one muliple of 3 hence product of any three consecutive Integers is always a Multiple of 3
CONCEPT3: If a Number "w" is a Multiple of 3 then any number at a difference of 3 or multiple of 3 from "w" will also be a multiple of 3 i.e. If w is a multiple of 3 then (w+3), (w-3), (w+6), (w-6) etc. will all be Multiples of 3
Here, I see that x(x – 1) is a product of two consecutive Integers but if another Number next to them is obtained then x(x – 1)(x – k) will certainly be a multiple of 3 [as per Concept2 mentioned above]
for x(x – 1)(x – k) to be a product of 3 consecutive integers,
(x – k) should be either (x - 2) i.e. k=2
OR
(x – k) should be either (x - 5) i.e. k=5 [Using Concept3]
OR
(x – k) should be either (x + 1) i.e. k=-1
OR
(x – k) should be either (x + 4) i.e. k=-4 [Using Concept3]
This Eliminates options A, C D and E
[Reveal] Spoiler:
B
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If x is an integer then x(x-1)(x-k) must be evenly divisible [#permalink]
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04 Nov 2016, 04:23
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enigma123 wrote:
If x is an integer, then x(x – 1)(x – k) must be evenly divisible by three when k is any of the following values EXCEPT
A. -4
B. -2
C. -1
D. 2
E. 5
[Reveal] Spoiler:
The OA is B. I am trying to use the concept of consecutive numbers but got stuck. Can someone please help?
Question says x(x – 1)(x – k) must be evenly divisible by three which means x(x-1) (x-k) should be consecutive.
A little bit faster approach.
We know from the start that product of $$3$$ consecutive integers is divisible by $$3$$, because it’s indeed a multiple of $$3$$.
So we have $$x(x+1)(x+2)$$ is definitely divisible by 3 and k=2 is our anchor point.
Now we can use principles of modular arithmetic to generate multiples of $$3$$. We can either add or subtract $$3$$ from our anchor number. And we got following string of numbers:
… -4, -1, 2, 5, 7 …
As you can see in this progression only answer B (-2) is absent.
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http://sciencechatforum.com/viewtopic.php?f=19&t=29316&p=303488&sid=7c166a4316456a6b426d7668e4965885 | ## Is 0/0 undefined or indeterminate?
Discussions concerned with knowledge of measurement, properties, and relations quantities, theoretical or applied.
### Is 0/0 undefined or indeterminate?
I'm confused, my book says that it's undefined, but I've been reading on the internet that it's indeterminate.
krum
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### Re: Is 0/0 undefined or indeterminate?
krum » August 16th, 2015, 7:37 pm wrote:I'm confused, my book says that it's undefined, but I've been reading on the internet that it's indeterminate.
The way I learned it in school:
a/0 is 'undefined' when a≠0,
a/0 is 'indeterminate' when a=0.
Thus, unless nomenclature has changed since I was in school, the answer to your question is 'indeterminate'.
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### Re: Is 0/0 undefined or indeterminate?
Calculus books will refer to "indeterminate forms", because they play a special role in certain infinite series.
$0^{0}$ is one of them. As is
$\frac{0}{0} ,$
$\infty ^{\infty },$
$\frac{\infty }{\infty },$ and
$\infty ^{0}$
However, 1/0 is not an indeterminate form, and for that reason it can't be used in certain theorems. If those "undefined" a/0 things appear in certain limits of infinite series, math students can justifiably say the the answer is "infinity", and be technically correct. Depending on things like the textbook or the professor, the right answer may also be saying that the series (or limit) "diverges."
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### Re: Is 0/0 undefined or indeterminate?
krum,
What's $\frac{x}{0}$? Okay, let's ask that question mathematically by:
1. Call the definition $k$.
2. Write that $k=\frac{x}{0}$.
3. Solve for $x$:
• $k=\frac{x}{0}$;
• $0{\times}k=0{\times}{\frac{x}{0}}$
• $0{\times}k=x$
• $x=0$.
4. Consider $x=0$.
• No definition $k$ would lead to a contradiction.
• So all possible definitions $k$ may be consistent.
• So we cannot determine a particular definition that $k$ must be.
• Call this inability to determine "indeterminate".
5. Consider $x{\neq}0$.
• Any definition $k$ would lead to a contradiction.
• So there is no possible definition $k$.
• Call this inability to define "undefined".
Therefore $\frac{x}{0}$ is:
1. indeterminate when $x=0$.
2. undefined when $x{\neq}0$.
Historically some mathematicians have failed to rigorously consider branching cases when they had fractions. For example, if their proof contained $\frac{x}{y}$ at some intermediate step but was later removed (e.g. by multiplication by $y$), then it may be hard to tell, but the proof may be flawed through reliance on $0{\times}{\frac{x}{0}}=x$. This hidden division-by-zero fallacy was the first major math fallacy that I was warned about.
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### Re: Is 0/0 undefined or indeterminate?
krum » Sun Aug 16, 2015 6:37 pm wrote:I'm confused, my book says that it's undefined, but I've been reading on the internet that it's indeterminate.
I hope this isn't off topic, but undefined suggests it can be, but just hasn't yet been defined. Would not undefinable be more accurate, in the context of the OP?
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### Re: Is 0/0 undefined or indeterminate?
Watson » August 17th, 2015, 12:58 pm wrote:
krum » Sun Aug 16, 2015 6:37 pm wrote:I'm confused, my book says that it's undefined, but I've been reading on the internet that it's indeterminate.
I hope this isn't off topic, but undefined suggests it can be, but just hasn't yet been defined. Would not undefinable be more accurate, in the context of the OP?
No. Classical Mathematics is actually quite clear and unambiguous on this point. It's been well covered in the previous posts.
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### Re: Is 0/0 undefined or indeterminate?
To put it more philosophically, x/x means a ratio of things - litterboxes to cats, doorknobs to doors, etc. Zero isn't anything, it isn't a real quantity, it is simply an absence, viz. nothing. Seen clearly, it has no place in a ratio. When you have no cats or litterboxes, then their ratio is indeterminate because there are no things to relate to each other. When you have ten cats and no litterboxes, then you have a ratio that is undefined - 10/0. You can't define a ratio because you only have one sort of quantity. And a yard full of cat poop. Does this sound right?
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### Re: Is 0/0 undefined or indeterminate?
So it is undefinable. I'm not sure how this is somewhat the same as indeterminable.
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### Re: Is 0/0 undefined or indeterminate?
Thay are not the same ... they are different terms each with their own specific definitions.
The inelegant but tried and true school methodology of learning may come in handy herr ... if understanding is not instantly forthcoming, rely on memorization until it does. Worked for me when I needed it.
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### Re: Is 0/0 undefined or indeterminate?
Natural ChemE » Mon Aug 17, 2015 1:26 am wrote:krum,
What's $\frac{x}{0}$? Okay, let's ask that question mathematically by:
1. Call the definition $k$.
2. Write that $k=\frac{x}{0}$.
3. Solve for $x$:
• $k=\frac{x}{0}$;
• $0{\times}k=0{\times}{\frac{x}{0}}$
• $0{\times}k=x$
• $x=0$.
4. Consider $x=0$.
• No definition $k$ would lead to a contradiction.
• So all possible definitions $k$ may be consistent.
• So we cannot determine a particular definition that $k$ must be.
• Call this inability to determine "indeterminate".
5. Consider $x{\neq}0$.
• Any definition $k$ would lead to a contradiction.
• So there is no possible definition $k$.
• Call this inability to define "undefined".
Therefore $\frac{x}{0}$ is:
1. indeterminate when $x=0$.
2. undefined when $x{\neq}0$.
Historically some mathematicians have failed to rigorously consider branching cases when they had fractions. For example, if their proof contained $\frac{x}{y}$ at some intermediate step but was later removed (e.g. by multiplication by $y$), then it may be hard to tell, but the proof may be flawed through reliance on $0{\times}{\frac{x}{0}}=x$. This hidden division-by-zero fallacy was the first major math fallacy that I was warned about.
Outstanding! Perhaps the clearest explanation of this that I have seen.
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### Re: Is 0/0 undefined or indeterminate?
Watson » August 17th, 2015, 4:34 pm wrote:So it is undefinable. I'm not sure how this is somewhat the same as indeterminable.
Yup, "undefinable" would be a good way to think about it.
Mathematicians say "undefined" because they mean to say that $\frac{x}{0}$ is undefined within elementary algebra. However, they see $\frac{x}{0}$ as definable since they're not constricted to elementary algebra.
If we were to take a more abstract algebra approach, we could just state
${\text{Watson}}{\left(x\right)}{\equiv}{\frac{x}{0}}$.
And I know what you're thinking: how's that help, right? Well, simple: the ${\text{Watson}}{\left(x\right)}$ function doesn't return a normal number, but rather some new mathematical entity. This new mathematical entity obeys
$0{\times}{\text{Watson}}{\left(x\right)}=x$,
so it doesn't suffer from the automatic inconsistency that all elementary entities do in the post that I made above.
What other properties does this new mathematical entity have? I dunno, I'm too lazy to make them up right now. And, heck, it's named after you, so you make them up. You're allowed to make up whatever you want so long as you rigorously avoid contradicting yourself.
To sum this up, yeah, you're right that "undefinable" would make more sense to most folks since we usually think about math in the context of elementary algebra. However, since mathematicians don't see things that way, to them it's moreso "undefined" than "undefinable".
-----
Separately, I forgot to address your question about how undefined and indeterminate are somewhat the same. Well, in a way, they're actually exact opposites:
1. undefined: No elementary description fits.
2. indeterminate: The set of elementary descriptions which fit includes non-equivalent members.
However, these two terms are similar in that they refer to cases in which there isn't exactly one known set of equivalent elementary representations.
By which I mean:
1. Most elementary entities are defined by only one set of equivalent representations. Examples of representations for $k=2$:
• $k=0+2$
• $k=0+1+1$
• $k=3-1+1-3+2.5-0.5$
2. Undefined entities have no elementary representations. Examples of representations for $k={\frac{2}{0}}$:
• [None. $\frac{2}{0}$ can't even represent itself because it leads to a contradiction; you're literally not even allowed to write it! It's meaningless gibberish.].
3. Indeterminate entities have more than one possible set of equivalent representations. Examples of possible sets of equivalent representations for $k=\frac{0}{0}$:
• Set of representations in which $k=0$, including:
• $k=0$
• $k=0+0$
• $k=1-1$
• $k=0*0$
• $k=0.5-\frac{1}{2}$
• Set of representations in which $k=1$, including:
• $k=1$
• $k=1+0$
• $k=2-1$
• $k=1*1$
• $k=1.5-\frac{1}{2}$
• Set of representations in which $k=2$, including:
• $k=2$
• $k=1+1$
• $k=2-0$
• $k=1*2$
• $k=1.5+\frac{1}{2}$
• [Etc., since $k$ can be just about anything in this case.]
To sum this up, "undefined" and "indeterminate" are similar in that they both indicate that there's not exactly one set of equivalent elementary representations. However, "undefined" indicates zero sets while "indeterminate" indicates more than one set.
Natural ChemE
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### Re: Is 0/0 undefined or indeterminate?
Exhaustively well said, but I think it might be easier for many to consider that undefined & indeterminate (or undefinable and indeterminable, as Watson put it) are somewhat similar only in a purely linguistic sense, whereas they are quite dissimilar in a purely mathematical sense ... kinda like 'alien' and 'illegal alien' are superficially similar linquistically, and yet are completely very different idiomatically.
Mathematics and linquistics share a lot of common ground, and indeed the former was birthed out of the latter, but it is contextually important to remember their differences and that they are not interchangeable - mathematics is more rigorously symbol oriented, whereas linguistics is generally more meaning oriented.
Math is a language unto itself.
Darby
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### Re: Is 0/0 undefined or indeterminate?
Watson » August 17th, 2015, 5:58 pm wrote:I hope this isn't off topic, but undefined suggests it can be, but just hasn't yet been defined. Would not undefinable be more accurate, in the context of the OP?
Here you have been shown that no mathematician would ever use a term like "undefinable", because if something is not defined within a domain of mathematics then a good mathematician would rapidly create a new domain where the thing IS defined (such domain would probably start from that definition itself).
For example they would use your (algebraically undefined) ratio 0/0 to represent a discontinuity in the function x/|x|: for x>0 the function has constant value 1, and its limit for x--> 0 (from the right) also is 1; for x<0 the function has value -1 and that is also its limit for x-->0 (from the left). Since at zero the function instantaneously jumps from -1 to +1, its value cannot be determined there. Still, it is defined, as you can plot and integrate the function with no problems.
[Forget this second part - it only is an attempt at pretending I am also a "good mathematician", simply because I share that attitude: if you find something that does not fit into your model, then claim that that's only a part of your model, you do have a better, more complex one!]
neuro
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### Re: Is 0/0 undefined or indeterminate?
if something is not defined within a domain of mathematics then a good mathematician would rapidly create a new domain where the thing IS defined (such domain would probably start from that definition itself).
Yeah you can do that, but it depends on the professor teaching it, and on the context where the expression appears during the course. If you have a series where the numerator approaches a finite value, say 2 , and the denominator is trending to zero
$\frac{2}{0}$
you can justifiably define a "new domain" where the real answer to the problem is $\infty$
Confusion arrises in introductory courses where the students are asked to instead provide the answer as "Does not exist".
The differentiation between "undefined" and "indeterminate" goes beyond pathological issues presented by Natural ChemE. In l'Hopital's Rule, the expression is required to be indeterminate. Otherwise, in expressions that are undefined, the theorem does not hold. (more importantly) you cannot use the technique to solve the limit.
hyksos
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### Re: Is 0/0 undefined or indeterminate?
hyksos,
That's a common misconception, but L'Hôpital's rule isn't meaningfully related to this topic.
L'Hôpital's rule is an alternative way to evaluate the limit of fractions. While it's usually more work than simpler methods, sometimes it's actually easier or/and provides a better result.
The commonly discussed case for L'Hôpital's rule is when just substituting in the approached value would result in either $\frac{0}{0}$ or $\frac{{\infty}}{{\infty}}$. The problem with both of these results is that they're not very descriptive, i.e. they're indeterminate. Math books often recommend considering L'Hôpital's rule when this happens since it might provide a determinate result.
Natural ChemE
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### Re: Is 0/0 undefined or indeterminate?
Natural ChemE.
L'Hôpital's rule is not just for $\frac{0}{0}$ and$\frac{ \infty }{ \infty }$ as you have claimed. The rule will work on any situation that is an indeterminate form. You have given two examples . There are many more, such as $0^0$ and several more others.
However, $\frac{2}{0}$ is NOT indeterminate! That's where the danger comes in. L'Hôpital's rule cannot be applied to situations that are "formally infinity", or in limits and series that genuinely diverge.
The main thing I wanted to communicate here is : yes, I agree with you that students are told to plug-and-chug L'Hôpital's rule, as you have described. However, dig in the literature a little deeper. L'Hôpital's rule is buttressed by two historical proofs and their corresponding theorems. This differentiation between indeterminate and undefined is very real in mathematics, and it has real consequences in theorems.
hyksos
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### Re: Is 0/0 undefined or indeterminate?
hksos,
Natural ChemE » August 31st, 2015, 7:55 pm wrote:The commonly discussed case for L'Hôpital's rule is when just substituting in the approached value would result in either $\frac{0}{0}$ or $\frac{{\infty}}{{\infty}}$.
hksos » September 12th, 2015, 4:17 pm wrote:L'Hôpital's rule is not just for $\frac{0}{0}$ and$\frac{ \infty }{ \infty }$ as you have claimed.
How the heck did you get from my statement to your representation of it?
There's a lot wrong with what you've said. Unless you can demonstrate some relevance, I'll just split 'em off to avoid polluting a good thread.
Natural ChemE
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### Re: Is 0/0 undefined or indeterminate?
[quote="[url=http://www.sciencechatforum.com/viewtopic.php?p=286235#p286235]Natural ChemE » August 17th, 2015, 1:26 am[/url]"]krum,
What's $$\frac{x}{0}$$? Okay, let's ask that question mathematically by:[list=1]
[*]Call the definition $$k$$.
[*]Write that $$k=\frac{x}{0}$$.
[*]Solve for $$x$$:
[list]
[*]$$k=\frac{x}{0}$$;
[*]$$0{\times}k=0{\times}{\frac{x}{0}}$$
[*]$$0{\times}k=x$$
[*]$$x=0$$.[/list]
[*]Consider $$x=0$$.
[list]
[*]No definition $$k$$ would lead to a contradiction.
[*]So all possible definitions $$k$$ may be consistent.
[*]So we cannot determine a particular definition that $$k$$ must be.
[*]Call this inability to determine "indeterminate".[/list]
[*]Consider $$x{\neq}0$$.
[list]
[*]Any definition $$k$$ would lead to a contradiction.
[*]So there is no possible definition $$k$$.
[*]Call this inability to define "undefined".[/list][/list]
Therefore $$\frac{x}{0}$$ is:[list=i]
[*]indeterminate when $$x=0$$.
[*]undefined when $$x{\neq}0$$.[/list]
Historically some mathematicians have failed to rigorously consider branching cases when they had fractions. For example, if their proof contained $$\frac{x}{y}$$ at some intermediate step but was later removed (e.g. by multiplication by $$y$$), then it may be hard to tell, but the proof may be flawed through reliance on $$0{\times}{\frac{x}{0}}=x$$. This [url=https://en.wikipedia.org/wiki/Division_by_zero#Fallacies]hidden division-by-zero fallacy[/url] was the first major math fallacy that I was warned about.[/quote]
The bottom line is that 0/0 is not equal to any number and a/0 where a is not 0 is not any number.
Robert Kolker
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### Re: Is 0/0 undefined or indeterminate?
It is indeterminate.
williamjohn
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### Re: Is 0/0 undefined or indeterminate?
n/0 does not exist, for all n.
1. x/y defined as (the z such that y*z = x).
2. 0*z = 0, for all z.
n/0 = (the z: 0*z = n).
If ~(n=0) then there is no z such that 0*z = n.
e.g. (1/0) = (the z such that 0*z = 1).
But 2 asserts that all values of 0*z are equal to 0,
i.e. there is no value of z for which 0*z=1,
If there is no value of z such that 0*z=1 then
(the z such that 0*z = 1) does not exist.
(the z: 0*z = n) is not unique for all n > 0.
If n=0 then there is more than one value of z such that 0*z=0.
0*z=0 is true for all z.
i.e. there is no unique value of z such that 0*z = 0.
That is to say, n/0 does not exist for all values of n,
even though n/0 is defined.
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https://math.stackexchange.com/questions/1823736/how-to-pick-10-people-from-13-such-that-at-least-1-is-a-woman | # How to pick $10$ people from $13$ such that at least $1$ is a woman
Thirteen people on a softball team show up for a game. Of the $13$ people who show up, $3$ are women. How many ways are there to choose $10$ players to take the field if at least one of these players must be a woman?
The given answer is calculated by summing the combination of $1$ woman + $9$ men, $2$ women + $8$ men, and $3$ women + $7$ men.
My question is, why can't we set this up as the sum $\binom{3}{1} + \binom{12}{9}$ - picking one of the three women first, then picking $9$ from the remaining $12$ men and women combined? The only requirement is that we have at least one woman, which is satisfied by $\binom{3}{1}$, and that leaves a pool of $12$ from which to pick the remaining $9$. The answer this way is close to the answer given, but it's $62$ short. I get that it's the "wrong" answer but I'm wondering why my thinking was wrong in setting it up this way. Thanks.
• You must multiply $C(3,1)$ with $C(12,9)$ because for every choice of a woman you have $C(12,9)$ choices for the other people. But this does not work either because you count arrangements multiple times. – Peter Jun 12 '16 at 22:15
Firstly you multiply, not add, if you are thinking of $\dbinom31$ and $\dbinom{12}{9}$
Secondly, this approach will over count. Suppose you chose Alicia , and then you chose $9$ from the remaining $12$, you would also have combos where Britney was first chosen, and Alicia was chosen from the $12$ group.
Thirdly, your book approach is correct, but unnecessarily tedious.
The best way is to compute
[All possible combos] - [All male combos ] $\;=\dbinom{13}{10} - \dbinom{10}{10}$
• I get what you are saying, but the constraint is "at least one", so wouldn't having a combination where Alicia is picked first, followed by a combination where Britney is picked and then Alicia is picked, be perfectly valid? Because there is still "at least one" in each combination. Unless I misunderstood you. – Dave Jun 12 '16 at 22:51
• $A$ followed by $B$, and $B$ followed by $A$ are valid, but you are counting them twice (apart from other overcounts). Problems of the "at least one" type are best tackled using the complement. You can verify for yourself that the book method, and the above method give the same answer. – true blue anil Jun 13 '16 at 4:44
Among $13$ people, you can choose $10$ players in $$\binom {13}{10}$$ ways.
Now, suppose there is no woman in the squad, then you can choose that team in $\binom{10}{10}=1$ way.
So, number of ways of choosing a team with at least $1$ woman is $$\binom{13}3-1=286-1=285$$ ways.
When you are counting several countings(call it nested counting), you should apply the multiplication principle. Here counting ways of choosing $1$ woman and $9$ men, you are in some kind of this nested counting, where, for any $9$ men among the $10$, the woman can occur, so, have to multiply this.
And upon deriving each choices by multiplying, you add those terms. Because, in one counting, there is $1$ woman , in some other count, there are $2$, and in other, there are $3$, women.
So, all are separate cases, so add them, like $$\binom31 \binom{10}9+\binom 32\binom {10}8+\binom33\binom {10}7=285$$
• This is a better solution than the one on the supplied answer sheet. On the other hand, it doesn't explain why OP's method fails. – David K Jun 12 '16 at 22:18 | 2019-09-15T08:19:00 | {
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https://math.stackexchange.com/questions/1472277/convex-hull-of-idempotent-matrices | # Convex hull of idempotent matrices
What is the convex hull of the set of $n\times n$ (potentially asymmetric) idempotent matrices?
Consider the set $S:=\{A\in\mathbb{R}^{n\times n}:A^2=A\}$, the set of idempotent matrices. Is there a simple description of the convex hull of $S$?
By convex hull, I mean: $C:=\{\sum_i a_i M_i:\sum_ia_i=1,a_i\geq0,M_i\in S\}$
• At the very least, we know that every matrix in the hull has a trace between $0$ and $n$. For all I know, we might get every matrix with trace between $0$ and $n$. – Omnomnomnom Oct 9 '15 at 18:34
• A condition on trace seems quite loose, however. Is there a smaller convex set containing the idempotent matrices? – Justin Solomon Oct 9 '15 at 18:42
• I really don't know. By the way, it looks like your question is about to be closed because it "needs more context". – Omnomnomnom Oct 9 '15 at 18:46
• I think the problem that people may have is that it looks like you didn't put enough work into answering the question yourself. – Omnomnomnom Oct 9 '15 at 18:52
• Sure, but the people of the site and the powers that be tend to expect some kind of partial work, though not necessarily a partial result. I happen to think your question is fine as is. – Omnomnomnom Oct 9 '15 at 18:55
First I'll do the $2 \times 2$ case. Consider a $2 \times 2$ matrix of trace $1$:
$$A = \pmatrix{a & b\cr c & 1-a\cr}$$ For any real $r$, every point $(b,c)$ in the plane is a convex combination of two points on the hyperbola $x y = -r$ (the two axes in the case $r=0$). Thus $A$ is a convex combination of two idempotent matrices of the form $$\pmatrix{ a & x\cr y & 1-a\cr}$$ where $xy = a-a^2$. The only idempotent matrix of trace $0$ is $0$, and the only idempotent matrix of trace $2$ is $I$. Any matrix of trace strictly between $0$ and $1$ is a convex combination of $0$ and a matrix of trace $1$, while any matrix of trace strictly between $1$ and $2$ is a convex combination of $I$ and a matrix of trace $1$. Thus the convex hull in the $2 \times 2$ case consists of $0$, $I$ and all $A$ with $0 < \text{tr}(A) < 2$.
This leads to a solution in the $n \times n$ case for any $n > 2$. We have idempotent matrices of trace $1$ with the form $$\pmatrix{a & x & 0 & \ldots & 0\cr y & 1-a & 0 & \ldots & 0\cr 0 & 0 & 0 & \ldots & 0\cr \ldots & \ldots & \ldots & \ldots & \ldots\cr 0 & 0 & 0 & \ldots & 0\cr}$$ where $xy = a - a^2$, and convex combinations of these give all $n \times n$ matrices of trace $1$ that are $0$ outside the top left $2 \times 2$ submatrix. By permuting indices, we get all $n \times n$ matrices of trace $1$ that are $0$ outside some principal $2 \times 2$ submatrix. Convex combinations of these give us all $n \times n$ matrices of trace $1$. Now any $n \times n$ matrix of trace strictly between $0$ and $1$ is a convex combination of $0$ and a matrix of trace $1$, while any matrix of trace strictly between $1$ and $n$ is a convex combination of $I$ and a matrix of trace $1$. Thus we find that the convex hull consists of $0$, $I$ and all $A$ with $0 < \text{tr}(A) < n$.
• so do we know that there won't be any matrices with trace $2$ besides $I$? – Omnomnomnom Oct 9 '15 at 19:14
• Also, what about matrices with trace $0$? – Omnomnomnom Oct 9 '15 at 19:17
• An idempotent matrix of trace $k$ is a projection on a subspace of dimension $k$. The only subspace of dimension $2$ is the whole space, and a projection on that is the identity. The only subspace of dimension $0$ is $\{0\}$, and a projection on that is $0$. – Robert Israel Oct 9 '15 at 19:25
• I see, there is a unique idempotent element with trace $2$ (or $0$), and so the convex hull must also have a unique such element. – Omnomnomnom Oct 9 '15 at 19:32
• Wow! Thanks to both of you for your help -- it's much appreciated. I'm surprised the convex hull is so big. – Justin Solomon Oct 9 '15 at 19:50
The result can be generalized by induction. Let $C_n$ denote the convex hull over $n \times n$ matrices.
First of all, note that if $A \in C_{n-1}$ and $x \in \Bbb R^n$, then the elements $$\pmatrix{A&0\\0&0_{1\times 1}},E_x := \pmatrix{I_{n-1}&x\\0&0}$$ are all in $C_n$.
Next, note that $C_n$ is closed under similarity (anything similar to an element of $C_n$ is in $C_n$).
Now, take any upper triangular $A \in C_{n-1}$. That is, $$A = \pmatrix{\lambda_1 & * \cdots \\ &\ddots\\ && \lambda_{n-1}}$$ with $\sum\lambda_i \in (0,n-1)$. Note that any convex combination of $A$ and $I$ is in $C_n$. Also, any convex combination of a matrix $C_n$ with $E_x$ is in $C_n$.
Conclude that $C_n$ contains every upper triangular matrix with trace (strictly) between $0$ and $n$ and a non-negative bottom-right entry. Because every matrix is upper-triangularizable, conclude that $C_n$ contains every matrix with trace strictly between $0$ and $n$. | 2020-02-17T16:25:13 | {
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https://math.stackexchange.com/questions/3724980/a-very-interesting-question-intersection-point-of-xy-yx | A very interesting question: intersection point of $x^y=y^x$
I've been investigating the Cartesian graph of $$x^y=y^x$$. Obviously, part of the graph is comprised of the line $$y=x$$ but there is also a curve that is symmetrical about the line $$y=x$$. (We can prove this symmetry by noting that the function $$x^y=y^x$$ is self-inverse; all self-inverse functions are symmetrical about the line $$y=x$$.)
An image of the graph is shown below:
I decided to find the intersection point and arrived at an intriguing result: the intersection point between the two curves is at $$(e,e)$$.
The following is my method: If the gradient of the line $$y=x$$ is $$1$$, the gradient of the curve at the intersection point must be $$-1$$ as it's a normal to the line (as it's symmetrical about the line). This means that at that point $$\frac{dy}{dx}=-1$$. Now to find $$\frac{dy}{dx}$$.
We have $$x^y=y^x$$. I then used a very powerful tehnique for differentiating these sorts of functions. We know that eg $$x^y=e^{\ln{x^y}}=e^{y\ln{x}}$$ and $$\frac{d}{dx}e^{f(x)}=f'(x)e^{f(x)}$$ Applying it to our function $$x^y=y^x$$ and using implicit differentiation and the product rule gives us: $$(\frac{dy}{dx}\times \ln x +\frac{y}{x})x^y=(\ln y +\frac{dy}{dx}\times \frac{x}{y})y^x$$ So $$\frac{dy}{dx}x^y\ln x +yx^{y-1}=\frac{dy}{dx}xy^{x-1}+y^x \ln y$$ Extensively rearranging gives: $$\frac{dy}{dx}=\frac{y^x \ln y -yx^{y-1}}{x^y \ln x -xy^{x-1}}$$ Let $$\frac{dy}{dx}=-1$$: $$y^x \ln y -yx^{y-1}=xy^{x-1}-x^y\ln x$$ But we know $$x=y$$ since we are at the intersection point with the line $$y=x$$: $$x^x \ln x -x^x=x^x-x^x\ln x$$ So $$2x^x \ln x -2x^x=0$$ $$2x^x(\ln x -1)=0$$ This means either $$2x^x=0$$ or $$\ln x -1=0$$ but we know $$x^x$$ is always greater than $$0$$ so $$\ln x =1$$, leaving us with: $$x=y=e$$ So I have my answer, but is there any other method of getting it? I have heard there is. Any help will be very welcome.
• The solution to $y^x=x^y$ can be expressed in terms of Lambert's W function as $$y=-xW\left(-\log(x)/x\right)/\log(x)$$Lambert's W function has been studied extensively. [Here](en.wikipedia.org/wiki/Lambert_W_function) is a link to a Wikipedia article that might be of interest. Note in particular that $W(-1/e)=-1$. Jun 18 '20 at 14:42
• Does this answer your question? Where does the curve $x^y = y^x$ intersect itself? Jun 18 '20 at 15:13
• Another possible duplicate: Given that $x^y=y^x$, what could $x$ and $y$ be?. Jun 18 '20 at 15:14
• Does this answer your question? Given that $x^y=y^x$, what could $x$ and $y$ be? Jun 18 '20 at 16:11
• Only the question titled 'Where does the curve $x^y=y^x$ intersect itself?' is helpful, but it still isn't very detailed when explaining the limits.Thanks for the suggestions though! Jun 18 '20 at 18:15
A Simple Approach
The simplest approach that I have found is to look at the intersections of $$y=tx\qquad\text{and}\qquad x^y=y^x\tag1$$ That is, \begin{align} x^{tx}&=(tx)^x\tag{2a}\\[3pt] x^t&=tx\tag{2b}\\[3pt] x^{t-1}&=t\tag{2c}\\ x&=t^{\frac1{t-1}}\tag{2d}\\ y&=t^{\frac t{t-1}}\tag{2e} \end{align} Explanation:
$$\text{(2a)}$$: $$x^y=y^x$$
$$\text{(2b)}$$: raise to the $$\frac1x$$ power
$$\text{(2c)}$$: divide by $$x$$
$$\text{(2d)}$$: raise to the $$\frac1{t-1}$$ power
$$\text{(2e)}$$: $$y=tx$$
Now, if we wish to find where $$x=y$$, let $$t\to1$$. That is, \begin{align} x &=\lim_{t\to1}t^{\frac1{t-1}}\tag{3a}\\ &=\lim_{n\to\infty}\left(1+\frac1n\right)^n\tag{3b}\\[6pt] &=e\tag{3c} \end{align} Explanation:
$$\text{(3a)}$$: $$x=y$$ when $$t=1$$
$$\text{(3b)}$$: $$t=1+\frac1n$$
$$\text{(3c)}$$: evaluate the limit
Further Musings
We can also compute \begin{align} y &=\lim_{t\to1}t^{\frac t{t-1}}\tag{4a}\\ &=\lim_{n\to\infty}\left(1+\frac1n\right)^{n+1}\tag{4b}\\[6pt] &=e\tag{4c} \end{align} Explanation:
$$\text{(4a)}$$: $$x=y$$ when $$t=1$$
$$\text{(4b)}$$: $$t=1+\frac1n$$
$$\text{(4c)}$$: evaluate the limit
Using the results from this answer, we see that $$\left(1+\frac1n\right)^n$$ is increasing and $$\left(1+\frac1n\right)^{n+1}$$ is decreasing, which means that $$\left(1+\frac1n\right)^n\le e\le\left(1+\frac1n\right)^{n+1}$$.
Thus, as $$t\to1^+$$, $$(4)$$ and $$(5)$$ show that $$x\to e^-$$ and $$y\to e^+$$.
Furthermore, if we substitute $$t\mapsto1/t$$, we get \begin{align} t^{\frac1{t-1}} &\mapsto(1/t)^{\frac1{1/t-1}}\tag{5a}\\ &=t^{\frac t{t-1}}\tag{5b} \end{align} and \begin{align} t^{\frac t{t-1}} &\mapsto(1/t)^{\frac{1/t}{1/t-1}}\tag{6a}\\ &=t^{\frac1{t-1}}\tag{6b} \end{align} That is, substituting $$t\mapsto1/t$$ swaps $$x$$ and $$y$$.
This means that, as $$t\to1^-$$, $$x\to e^+$$ and $$y\to e^-$$.
Relation to the Graph
Here is where these points sit on the graph as $$t\to1^+$$:
• Incredibly clear, thank you!!! Jun 23 '20 at 14:33
• @robjohn Your answers always have fantastic graphics and explanations. Nicely done, +1 from me. Jul 7 '20 at 5:35
Not sure to understand well the question but if it provides any help, I think there is something missing in your analysis.
Your equation can be seen as $$g(x,y) = x^{y} = y^{x} = f(x,y)$$. You are looking for intersecting points or surfaces (x,y) in $$\mathbb{R}^{3}$$, since both $$f(\cdot,\cdot)$$ and $$g(\cdot,\cdot)$$ are 2D scalar fields. Otherwise, if you are thinking the expression $$x^{y} = y^{x}$$ as the equation of a curve in $$\mathbb{R}$$ (somehow you figure out to express $$y = f(x)$$ or $$x = f(x)$$, it doesn't make sense to talk about an intercept (or interception set) since you must provide another curve or surface.
The same applies with the fields $$g$$ and $$f$$. You must provide another equation since you have two unknowns.
The answer you got $$(x,y) = (e,e)$$ is the trivial answer, and works for any real $$k \in \mathbb{R}$$: $$(x,y) = (k,k)$$.
I don't know if this is what you are looking for, but I would go like this.
First rewrite the equation : $$x^y=y^x\Leftrightarrow y\ln(x)=x\ln(y)\Leftrightarrow \frac{\ln(x)}{x}=\frac{\ln(y)}{y}$$ Now consider the function $$f(x)=\frac{\ln(x)}{x}$$ defined on $$(0,+\infty)$$ and whose derivative is $$f'(x)=\frac{1-\ln(x)}{x^2}$$. This shows that $$f$$ induces two bijections $$f_1:(0,e]\to (-\infty,\frac{1}{e}]$$ and $$f_2:[e,+\infty)\to[\frac{1}{e},0)$$.
Now the locus $$f(x)=f(y)$$ can be divided in two curves, namely $$y=f_1^{-1}(f(x))$$ and $$y=f_2^{-1}(f(x))$$. (Note however that these two curves are not the obvious ones). And we are looking their intersection. Since the ranges of $$f_1^{-1}$$ and $$f_2^{-1}$$ are respectively $$(0,e]$$ and $$[e,+\infty)$$, the only possible value of $$y$$ for the intersection is $$y=e$$ which happen when $$x=e$$.
• Actually, this method shows that for a function $f$ which has a single change of variation (hence a unique local extrema $x_0$), then $f(x)=f(y)$ has two branches whose intersection is exactly $(x_0,x_0)$. Jun 18 '20 at 15:46
This is not exactly an answer, but part of your reasoning was a bit hand-wavy and I want to make it fully rigorous.
Your reasoning about the slopes of the two parts of the plot is intuitively correct, but needs proof. So it is clear that the slope of the first part of the plot is $$1$$ since this encompasses the solutions $$x^x=x^x$$, i.e, the identity map which has slope $$1$$ by definition. However, showing that the other branch of the plot has slope $$-1$$ is a bit more difficult. We need to show that
1: The relation $$y^x-x^y=0$$ is its own inverse
2: If a relation is its own inverse, its slope at a point $$(x,y)$$ is $$-1$$ if $$x=y$$, with the exception of the identity map.
The first point is rather easy and I'll leave it for you to do yourself. The second part, however, requires a bit of care. Any two variable relation can be summarized by the general equation $$f(x,y)=0$$ Well, not quite any, but I hope you get the idea. For example, the equation of a circle: $$x^2+y^2-r^2=0$$ We want to use information about $$f$$ to deduce $$\frac{\mathrm{d}y}{\mathrm{d}x}$$. Using a bit of knowledge from multivariable calculus, we know that the total derivative of a two-variable function $$f$$ in terms of an arbitrary variable $$u$$ is $$\frac{\mathrm{d} f}{\mathrm{d} u}=\frac{\partial f}{\partial x}\frac{\mathrm{d}x}{\mathrm{d}u}+\frac{\partial f}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}u}$$ In the special case that $$u=x$$, $$\frac{\mathrm{d} f}{\mathrm{d} x}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}x}$$ And similarly $$\frac{\mathrm{d} f}{\mathrm{d} y}=\frac{\partial f}{\partial x}\frac{\mathrm{d}x}{\mathrm{d}y}+\frac{\partial f}{\partial y}$$ If the relation is its own inverse, then $$f(x,y)=f(y,x)=0$$ Thus $$\frac{\mathrm{d} f}{\mathrm{d} x}=\frac{\partial f}{\partial x}+\frac{\partial f}{\partial y}\frac{\mathrm{d}y}{\mathrm{d}x}=\frac{\mathrm{d} f}{\mathrm{d} y}=\frac{\partial f}{\partial x}\frac{\mathrm{d}x}{\mathrm{d}y}+\frac{\partial f}{\partial y}$$ So $$\frac{\partial f}{\partial x}\left(1-\frac{\mathrm{d}x}{\mathrm{d}y}\right)=\frac{\partial f}{\partial y}\left(1-\frac{\mathrm{d}y}{\mathrm{d}x}\right)$$ But again, since $$f(x,y)=f(y,x)=0$$, $$\frac{\partial f}{\partial x}=\frac{\partial f}{\partial y}$$ if $$x=y$$, thus $$1-\frac{\mathrm{d}x}{\mathrm{d}y}=1-\frac{\mathrm{d}y}{\mathrm{d}x}$$ $$\frac{1}{\frac{\mathrm{d}y}{\mathrm{d}x}}=\frac{\mathrm{d}y}{\mathrm{d}x}$$ Therefore $$\frac{\mathrm{d}y}{\mathrm{d}x}=\pm 1$$. Excluding the only self-inverse relation with slope $$1$$, the identity map, we conclude $$\frac{\mathrm{d}y}{\mathrm{d}x}\big|_{(x,x)}=-1.$$
Following this, the rest of your reasoning is correct. Others have stated that this problem can be tackled using the product log, so I can't really think of another way of going about this problem. Good question though!
• Thanks for the detailed answer; I realized I didn't have ironclad proof for the second point, but that explanation is very helpful and welcome! Thanks a s well for the final compliment :) Jun 18 '20 at 18:05
We are looking for solutions to $$\exp(y\log x) = \exp(x\log y) \, .$$ Taking logs of both sides and rearranging, this equation becomes $$\frac{x}{\log x}=\frac{y}{\log y} \, .$$ Consider the graph of $$f(z)=z/\log z$$:
Its unusual shape can be explained by a few observations:
• For $$0, $$\log z$$ is negative, so $$f(z)$$ is negative.
• As $$z\to1$$, $$\log z \to 0$$, and so the graph has an asymptote at $$z=1$$.
• For $$z>1$$, $$\log z$$ is positive, so $$f(z)$$ is positive.
• $$f(z)$$ is large when $$\log z$$ is small, or when $$z$$ is very large.
It is the final observation that gives us a hint as to why $$f(z)$$ has a local minimum. We can study the behaviour of $$f(z)$$ more closely by considering $$f'(z) = \frac{\log z - 1}{\log^2z} \, .$$ For $$1, $$f'(z)$$ is negative; at $$z=e$$, $$f'(z)$$ is zero; and for $$z>e$$, $$f'(z)$$ is positive.
Now that we have proven that $$f(z)$$ has a local minimum at $$(e,e)$$, we can begin to examine your question more closely. Let $$(x,\log x)$$ and $$(y,\log y)$$ be arbitrary points that lies on the graph.
• Suppose $$0. Then, because $$f(z)$$ is one-to-one on the interval $$(0,1)$$, the only time when $$f(x)=f(y)$$ is when $$x=y$$.
• If $$x>1$$, then things are more complicated. Clearly, if $$x=y$$, then we still have $$f(x)=f(y)$$. However, if $$1, then there is a unique number $$y\in(e,\infty)$$ such that $$f(x)=f(y)$$. This is obviously still the case if the roles of $$x$$ and $$y$$ are interchanged.
• If $$x$$ is slightly smaller than $$e$$, then $$y$$ is either equal to $$x$$, or is only slightly larger than $$e$$. This means that the two sections of the graph get very close to each other as $$x$$ tends to $$e$$.
• In the limit, when $$x=e$$, the two parts of the graph coincide.
The equation is equivalent to
$$\frac{\log x}x=\frac{\log y}y.$$
The function
$$\frac{\log x}x$$
has a maximum at $$x=e$$ and for $$x>0$$
$$\frac{\log x}x=\frac{\log y}y$$ has two solutions in $$x$$.
Using Lambert's function,
$$-\frac1x\log\frac1x=\frac{\log y}y$$ and
$$e^{\log\frac1x}\log\frac1x=-\frac{\log y}y$$
and
$$\log\frac1x=W\left(-\frac{\log y}y\right).$$ | 2021-10-27T20:29:20 | {
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https://math.stackexchange.com/questions/1881826/a-tangent-to-the-ellipse-meets-the-x-and-y-axes-if-o-is-the-origin-find | # A tangent to the ellipse meets the $x$ and $y$ axes . If $O$ is the origin, find the minimum area of triangle $AOB$.
The question is that : A tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ meets the $x$ and $y$ axes respectively at $A$ and $B$. If $O$ is the origin, find the minimum area of triangle $AOB$.
What I have attempted:
Because the equation of an ellipse is $$\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=1$$
We have that the $x$ intercept is $x=\frac{a^2}{x_1}$ and $y$ intercept is $y=\frac{b^2}{y_1}$
Hence the area of a triangle $(AOB)$ is given by $A=\frac{a^2b^2}{2x_1y_1}$ now I am stuck, how should I proceed?
• What is the derivative $y'$? Aug 4, 2016 at 3:41
• Notice affine transformation preserves tangency and ratio of areas, you can use an affine transform to map the ellipse to the unit circle. The smallest triangle $AOB$ will be mapped to a right angled isoceles triangle of side $\sqrt{2}$ and area $1$. This means the smallest triangle $AOB$ has area $ab$. Aug 4, 2016 at 4:00
$A = \frac {a^2 b^2}{2xy}$ constrained by $\frac {x^2}{a^2} + \frac {y^2}{b^2} = 1$
$\frac{dA}{dx} = \frac {-2a^2b^2 (y + x y')}{(2xy)^2} = 0\\ y + x y' = 0\\ y' = -\frac yx$
Differentiating the constraint.
$\frac x{a^2} + \frac {y y'}{b^2} = 0\\ y' = -\frac {b^2x}{a^2y}\\ \frac {b^2x}{a^2y} = \frac yx\\ b^2 x^2 = a^2 y^2$
Again from the constraint:
$b^2 x^2 + a^2 y^2 = a^2 b^2\\ 2a^2 y^2 = a^2 b^2\\ y^2 = \frac {b^2}{2}\\ y = \frac b{\sqrt {2}}\\ 2xy = ab\\ A = \frac {a^2b^2}{2xy} = ab$
alternate
$x = a \cos t\\ y = b \sin t\\ A = \frac {ab}{2} \csc t \sec t\\ \frac{dA}{dt} = \frac {ab}{2} \sec t \csc t ( -\cot t + \tan t) = 0\\ \cot t = \tan t\\ t = \frac {\pi}{4}\\ A = ab$
That is much nicer.
With the change of coordinates $x=au, y=bv$ the problem boils down to the same problem where the original ellipse is replaced by a unit circle. In such a case it is trivial that the minimum area of $AOB$ is $1$. Since the determinant of the Jacobian of the linear map $x=au, y=bv$ is $ab$, the answer to the original question is $\color{red}{ab}$, plain and simple. | 2022-05-25T11:13:31 | {
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https://qudt.org/vocab/unit/PlanckTime | unit:PlanckTime
Type
Description
In physics, the Planck time, denoted by $$t_P$$, is the unit of time in the system of natural units known as Planck units. It is the time required for light to travel, in a vacuum, a distance of 1 Planck length. The unit is named after Max Planck, who was the first to propose it. $$\\ t_P \equiv \sqrt{\frac{\hbar G}{c^5}} \approx 5.39106(32) \times 10^{-44} s$$ where, $$c$$ is the speed of light in a vacuum, $$\hbar$$ is the reduced Planck's constant (defined as $$\hbar = \frac{h}{2 \pi}$$ and $$G$$ is the gravitational constant. The two digits between parentheses denote the standard error of the estimated value.
Properties
0.0000000000000000000000000000000000000000000539124
$$t_P$$
Annotations
In physics, the Planck time, denoted by $$t_P$$, is the unit of time in the system of natural units known as Planck units. It is the time required for light to travel, in a vacuum, a distance of 1 Planck length. The unit is named after Max Planck, who was the first to propose it. $$\\ t_P \equiv \sqrt{\frac{\hbar G}{c^5}} \approx 5.39106(32) \times 10^{-44} s$$ where, $$c$$ is the speed of light in a vacuum, $$\hbar$$ is the reduced Planck's constant (defined as $$\hbar = \frac{h}{2 \pi}$$ and $$G$$ is the gravitational constant. The two digits between parentheses denote the standard error of the estimated value.
Planck Time(en)
Generated 2022-11-28T15:45:48.023-05:00 by lmdoc version 1.1 with TopBraid SPARQL Web Pages (SWP) | 2022-12-02T20:40:05 | {
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https://math.stackexchange.com/questions/4180075/generalized-eigenvector-for-product-of-commuting-matrices | # Generalized eigenvector for product of commuting matrices
Suppose $$A,B$$ are commuting invertible matrices with a common generalized eigenvector $$v$$ with eigenvalues $$a,b$$ respectively. That is, suppose there exist positive integers $$K,L$$ such that $$(A-aI)^K v= 0$$ and $$(B-bI)^Lv=0$$. Is it true that there exists a positive integer $$M$$ such that $$(AB-abI)^Mv = 0$$?
A bit of context: I came across this while thinking about the analogous version of this for eigenvalues (when $$K=1$$ and $$L=1$$). This is easy to show. In fact, the above statement is also easy to show if only one of $$K=1$$ or $$L=1$$ is true (that is, $$v$$ is an eigenvector for one matrix and a generalized eigenvector for the other).
Proof: WLOG suppose $$L=1$$. Then we have \begin{align} (AB-abI)^Kv &= \sum_{j=0}^K {K \choose j} A^{K-j}B^{K-j}a^jb^jv\\ &= \sum_{j=0}^K {K \choose j} A^{K-j}a^jb^Kv\\ &= b^K(A-aI)^Kv\\ &= 0 \end{align}
A natural question: is it true if $$v$$ is a generalized eigenvector for both matrices?
• Welcome! Note that askers are expected to provide context for their questions, as is explained here. For example, it would be helpful if you could edit your questions to address some of the following. What motivated this problem, or where did you encounter it? What are your thoughts on the problem? What have you tried so far? – Ben Grossmann Jun 22 at 16:20
• A potentially helpful way to reframe the problem: let $P = A - aI$ and $Q = B - bI$. We could equivalently ask: if $PQ = QP$ and $K,L$ are such that $P^Kv = Q^Lv = 0$, then does there necessarily exist an $M$ such that $$([P + aI][Q + bI] - ab I)^Mv = 0?$$ – Ben Grossmann Jun 22 at 16:25
• The answer is yes: the statement is true. Once you provide some context, I can leave a proof as an answer. – Ben Grossmann Jun 22 at 16:46
• Just added some context. Thanks for pointing the rules out btw, I'm a new poster here. – Underbounded Jun 22 at 16:47
• I figured as much, thanks for obliging. – Ben Grossmann Jun 22 at 16:48
Here is a more abstract perspective. Let $$V$$ be the smallest subspace of our vector space which contains $$v$$ and is closed under the actions of $$A$$ and $$B$$. Let $$R$$ be the subring of the endomorphism ring of $$V$$ generated by the scalar matrices together with $$A$$ and $$B$$. By assumption, this $$R$$ is commutative, and $$(A-a)^K$$ and $$(B-b)^L$$ are equal to $$0$$ in this ring (since they annihilate $$v$$ and $$v$$ generates $$V$$ under the action of $$R$$). The question then becomes simply:
Let $$R$$ be a commutative ring and $$A,B,a,b\in R$$. Suppose $$A-a$$ and $$B-b$$ are nilpotent. Then must $$AB-ab$$ be nilpotent?
But now the answer is very easy using the fact that the set $$N$$ of nilpotent elements of $$R$$ is an ideal. By assumption, $$A=a$$ and $$B=b$$ in the quotient ring $$R/N$$. Thus $$AB=ab$$ in that quotient ring. That is, $$AB-ab\in N$$.
• This is a really neat approach! – Underbounded Jun 22 at 21:23
Yes. We can proceed as follows.
Begin by reframing the question: if $$PQ = QP$$ and $$K,L$$ are such that $$P^Kv = Q^Lv = 0$$ and if $$a,b$$ are arbitrary scalars, then does there necessarily exist an $$M$$ such that $$([P + aI][Q + bI] - ab I)^Mv = 0?$$ Note that $$[P + aI][Q + bI] - ab I = PQ + bP + aQ.$$ Now, take $$m = \max\{K,L\}$$ and $$M = 2m-1$$; note that $$P^mv = Q^mv = 0$$. Apply the multinomial theorem: $$(PQ + bP + aQ)^Mv = \sum_{k_1 + k_2 + k_3 = M} \binom{M}{k_1,k_2,k_3} a^{k_3}b^{k_2}P^{k_1+k_2}Q^{k_1+k_3}v \\= \sum_{k_1 + k_2 + k_3 = M} \binom{M}{k_1,k_2,k_3} a^{k_3}b^{k_2} Q^{k_1+k_3}P^{k_1+k_2}v.$$ Now, for all triples of non-negative integers $$k_1,k_2,k_3$$ with $$k_1 + k_2 + k_3 = 2m-1$$, we see that $$k_1 + k_2$$ and $$k_1 + k_3$$ are non-negative integers with sum greater than or equal to $$2m-1$$. It follows that either $$k_1 + k_2 \geq m$$ or $$k_1 + k_3 \geq m$$. In either case, we find that $$P^{k_1+k_2}[Q^{k_1 + k_3}v] = Q^{k_1+k_3}[P^{k_1+k_2}v] = 0.$$ Thus, the above expansion of $$(PQ + bP + aQ)^Mv$$ must be equal to zero.
• Binomial theorem suffices if you write $AB-abI$ as $(A-aI)B+a(B-bI)$. This also lowers the value of $m$ to $K+L-1$. – user1551 Jun 22 at 17:45
• @user1551 That's neat! Actually, I realize now that $M = K +L - 1$ suffices for me as well; we only need $k_1 + k_2 \geq K$ or $k_1 + k_3 \geq L$ – Ben Grossmann Jun 22 at 17:57 | 2021-08-02T10:28:14 | {
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https://mathematica.stackexchange.com/questions/113402/what-do-the-lines-in-a-contourplot-mean | # What do the lines in a ContourPlot Mean?
If you plot a two variable function like $x^2+y^2$, you get a series of circles with lines and colored areas. My question is, what do the lines and the colored areas represent?
• Try this experiment: ContourPlot[Cos[x] + Cos[y], {x, 0, 4 Pi}, {y, 0, 4 Pi}], followed by Plot3D[Cos[x] + Cos[y], {x, 0, 4 Pi}, {y, 0, 4 Pi}] and compare the two. Now try ContourPlot[x^2 + y^2, {x, -4, 4 }, {y, -4, 4 }] and Plot3D[x^2 + y^2, {x, -4, 4 }, {y, -4, 4 }]. Got it? ContourPlot by default generates colorized grayscale output, in which larger values are shown lighter. You may also like to read: en.wikipedia.org/wiki/Contour_line and itl.nist.gov/div898/handbook/eda/section3/contour.htm – Moo Apr 23 '16 at 12:50
• @Moo, OK, so now I know why there are areas of different colors but what is the meaning if the lines bounding each colored area? Do they represent something? – MrDi Apr 23 '16 at 12:56
• It's explained well here: en.wikipedia.org/wiki/Contour_line The value of the function doesn't change as you move along a contour. – Szabolcs Apr 23 '16 at 13:10
• @MrDi: There needs to be a way to select peaks and valleys and to bound them somehow, so they use some algorithm that picks those areas. I think you can find some details on the Wiki link I provided. it sounds like you get it, think of it as trying to collapse a 3D image onto a 2D surface. These diagrams can be extremely useful. The lines are just separating areas where something is changing based on some algorithm. – Moo Apr 23 '16 at 13:13
• @Moo, OK, thanks for help. – MrDi Apr 23 '16 at 13:14
It might be easier for you to see the correspondence between a contour plot and a surface plot if you plot them side-by-side like this (code adapted from here):
peaks[x_, y_] := 3 (1 - x)^2 Exp[-x^2 - (y + 1)^2] - 10 (x/5 - x^3 - y^5) Exp[-x^2 - y^2] -
Exp[-(x + 1)^2 - y^2]/3
With[{c = Subdivide[-5, 5, 10]},
GraphicsRow[{ContourPlot[peaks[x, y], {x, -3, 3}, {y, -3, 3},
ColorFunction -> "DarkTerrain",
Contours -> c, PlotRange -> All],
Plot3D[peaks[x, y], {x, -3, 3}, {y, -3, 3}, BoundaryStyle -> None,
Boxed -> False, ColorFunction -> "DarkTerrain",
MeshFunctions -> {#3 &}, Mesh -> {c}, PlotRange -> All,
ViewPoint -> {0, -2.7, 3.}]}]]
where I took the opportunity to use a function with more interesting contours. (Unfortunately, I don't know of any way to refine the meshlines produced by MeshFunctions; if you know a way, let me know in the comments!)
As noted in the comments above, the contours in ContourPlot[] are so-called isoclines; that is, lines of constant altitude ($z$ value). The interpretation of the colors depends on what colormap you're using; in this case, I used the "DarkTerrain" color gradient, which goes like this:
where the bluish color ("water") corresponds to the valleys, the whitish color ("snowcap") corresponds to the peaks, and the earthy colors corresponds to intermediate values. If you are familiar with map reading, running along isoclines corresponds to staying at a constant (flat) altitude, and crossing the isoclines corresponds to either ascent or descent.
A biographical segue: I was a Boy Scout once upon a time, and map reading was de rigueur for Scouts who wanted to have camping experience. | 2019-09-19T15:48:57 | {
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https://math.stackexchange.com/questions/4484433/non-diagonal-n-th-roots-of-the-identity-matrix/4484681#4484681 | # Non-diagonal n-th roots of the identity matrix
Q. Are there $$n$$-th root analogs of this non-diagonal cube-root of the $$3 \times 3$$ identity matrix?
\begin{align*} \left( \begin{array}{ccc} 0 & 0 & -i \\ i & 0 & 0 \\ 0 & 1 & 0 \\ \end{array} \right)^3 = \left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right) \end{align*}
I am looking for $$A^n=I$$, where $$I$$ is any dimension $$\le n$$.
(A naive question: I am not an expert in this area.)
• I'm not sure what features you want of your "analogue" here. Are you just looking for a non-diagonal $A$ such that $A^n = I$? Does it have to be $3 \times 3$? Or maybe $n \times n$? Does it have to match some feature of how the non-zero terms are laid out (e.g. one non-zero term in each row/column)? It would be good to get some more information about what you'd like to see in an answer. Jul 1 at 20:26
• The problem I see is that a matrix can have n different nth roots of a matrix. How do you decide which you want? Jul 1 at 20:38
• @GeorgeIvey "The problem I see is that a matrix can have n different nth roots of a matrix." More, usually, and occasionally fewer. Jul 1 at 20:41
• The companion matrix of the $n$th cyclotomic polynomial has order $\phi(n)$, which is less than $n$. Does that work for you?
– lhf
Jul 1 at 20:48
• Hi, Joseph. The permutation matrix of a cyclic permutation always works. Apparently you can alter the final pair of 1's into $i$ and $-i$ and still get $A^n = I$ Jul 1 at 21:19
We want to solve $$A^n = I$$ where $$A$$ and $$I$$ are $$m \times m$$
Let $$A = P \Lambda P^{-1}$$
where $$\Lambda = \text{diag}(\omega_1, \omega_2, \dots , \omega_m)$$
where the $$\omega_k$$'s are $$n$$-roots of unity. That is, $$\omega_k = \exp\big(\dfrac{i 2 \pi j}{n} \big)$$ ,$$k = 1,2,\dots,m$$ and $$j \in \{ 0, 1, 2, ...., n-1 \}$$
and $$P$$ is any invertible $$m \times m$$ matrix.
Then $$A^n = \big(P \Lambda P^{-1}\big)\big(P \Lambda P^{-1}\big)\dots \big(P \Lambda P^{-1}\big) = P^{-1} \Lambda^n P = P^{-1} I P = I$$
• Yours is the most general example, even in dimension $n$. Can you prove it?
– Ruy
Jul 2 at 0:26
• Thanks. Please check my updated solution for the proof. Jul 2 at 1:13
You can take a rotation matrix that rotates $$\phi=2\pi/n$$ around some axis, for example in 3 dimensions:
$$R_\phi=\begin{pmatrix} \cos \phi & \sin\phi & 0 \\ -\sin\phi & \cos\phi &0 \\ 0 & 0 & 1\\ \end{pmatrix}$$
This won't work for $$n=2$$ though, because $$R_\phi$$ is diagonal in that case like $$R_\phi=\operatorname{diag}(-1,-1,1)$$.
Then $$R_\phi^n=I$$ and $$R_\phi^k\neq I$$ for any $$0. You can build new matrices using any invertible matrix $$A$$ and conjugate like
$$R_{\phi,A}:= AR_\phi A^{-1} \tag 1$$
then obviously:
$$R_{\phi,A}^n = (AR_\phi A^{-1})^n = A R_\phi^nA^{-1} = I$$
(I am not sure whether this could fix the case $$n=2$$ and produce some valid (non-diagonal) results.)
• You can use Wolfram Alpha to check that {{1,1,0},{0,0,1},{1,0,0}}^-1 * {{-1,0,0},{0,-1,0},{0,0,1}} * {{1,1,0},{0,0,1},{1,0,0}} is not diagonal. Of course, for n=2 you can also use a reflection instead of a rotation, I.e. exchange two basis vectors. Jul 2 at 14:39
If $$A^n = I$$, i.e., $$A^n - I = 0$$, then the minimal polynomial $$m_A$$ of $$A$$ divides $$p(x) := x^n - 1$$. (For $$n > 1$$) one nondiagonal solution is the companion matrix of $$p$$ itself, namely, $$C_p := \pmatrix{\cdot & 1 \\ I_{n - 1} & \cdot}$$ (indeed, $$p = m_{C_p}$$).
Notice that $$C_p$$ is the permutation matrix for the permutation (in fact $$n$$-cycle) $$\pmatrix{1 & 2 & \cdots & n - 1 & n}$$ of $$n$$ elements. More generally, the permutation matrix of any permutation of $$n$$ elements of order dividing $$n$$ (and not the identity permutation) yields another solution.
There are uncountably many solutions for any $$n > 1$$: Given any solution $$A$$ and any invertible matrix $$S$$, $$SAS^{-1}$$ is another solution provided it is not diagonal (since diagonality is a closed condition that by hypothesis is not always satisfied).
The most general solution of the equation $$A^n=1$$ among $$m\times m$$ complex matrices is $$A=SDS^{-1},$$ where $$S$$ is an invertible matrix and $$D$$ is a diagonal matrix whose diagonal entries are $$n^{th}$$ roots of unity.
The reason is that the minimal polynomial of such a matrix $$A$$ divides the polynomial $$x^n-1$$, and hence has distinct roots, which in turn implies that $$A$$ is diagonalizable. | 2022-10-07T08:20:05 | {
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https://math.stackexchange.com/questions/2354401/find-a-cubic-polynomial | # Find a cubic polynomial.
If $f(x)$ is a polynomial of degree three with leading coefficient $1$ such that $f(1)=1$, $f(2)=4$, $f(3)=9$, then $f(4)=?,\ f(6/5)=(6/5)^3?$
I attempt:
I managed to solve this by assuming polynomial to be of the form $f(x)=x^3+ax^2+bx+c$, then getting the value of $a,b,c$, back substituting in the equation and so on....
But we can see:
$f(x)=q_1(x-1)+1\\f(x)=q_2(x-2)+4\\f(x)=q_3(x-3)+9$
Also when we put $x=1$ we get $f(1)=1^2$, when $x=2$ then $f(2)=2^2$, when $x=3$ then $f(3)=3^2$ but $f(4)\neq4^2$ (from answer).
Can this information be used to reproduce $f(x)$ directly without using the step I described in very first line of my solution?
Consider $g(x) = f(x)-x^2$. Then $g(1) = g(2) = g(3) = 0$ and $g$ is also a cubic polynomial and has leading coefficient 1. Thus $g(x) = (x-1)(x-2)(x-3)$ and hence $f(x) = (x-1)(x-2)(x-3)+x^2$. It now follows that $f(4) = 22$. Other values can be calculated.
• should $f(4)=22$ Jul 11 '17 at 3:58
• The OP wrote "with leading coefficient $1$". Jul 11 '17 at 3:59
• I assumed that $f(6/5)=(6/5)^3$ to be proved from the given information that $f$ is monic.
– user348749
Jul 11 '17 at 4:00
Classic long method:
Let $f(x) = x^3 + b x^2 + c x + d$ with $f(1) = 1$, $f(2) = 4$, $f(3) = 9$, which leads to \begin{align} f(1) &= 1 = 1 + b + c + d \hspace{10mm} \to d = -b - c \\ f(2) &= 4 = 8 + 4 b + 2 c + d = 8 + 3b + c \hspace{10mm} \to c = -4 - 3b, \, d = 4 + 2b \\ f(3) &= 9 = 27 + 9b + 3c + d = 19 + 2b \end{align} from which $b = -5$, $c = 11$, and $d = -6$ and $$f(x) = x^3 - 5 \, x^2 + 11 \, x -6.$$
With $f(x)$ then \begin{align} f(4) &= 64 - 80 + 55 -6 = 22 \\ f\left(\frac{6}{5}\right) &= \left(\frac{6}{5}\right)^{3} - \frac{36 - 66 + 30}{5} = \left(\frac{6}{5}\right)^{3}. \end{align}
It may also be noticed that $f(x)$ can be seen in the form $$f(x) = \left(x - \frac{5}{3}\right)^{3} + \frac{8}{3} \, \left( x - \frac{5}{3}\right) + \frac{83}{27}.$$ From this it is easy to see that \begin{align} f\left(\frac{5}{3}\right) &= 3 + \frac{2}{27} \\ f\left(\frac{5}{6}\right) &= \frac{59}{216}. \end{align}
• Looking at your last form of $f(x)$ if I put $5/3$ in then I would not get the same answer as you have for $f(5/3)$ just after - the former gives $3+2/27$... I've not chased the maths to see which is right, just happened to notice that discrepancy (or I've done something stupid which is also possible). Jul 11 '17 at 8:47
If
• $$f(x) = x^3 + ax^2 + bx + c$$
• $$f(1) = 1$$
• $$f(2) = 4$$
• $$f(3) = 9$$
then
\begin{align} f(x+1) - f(x) &= (3x^2+3x+1) + a(2x+1) + b \\ \hline 4-1 &= f(2)-f(1) \\ 3 &= 7 + 3a + b \\ \hline 9-4 &= f(3) - f(2) \\ 5 &= 19 + 5a + b \\ \hline 3a + b &= -4 \\ 5a + b &= -14 \\ \hline a &= -5 \\ b &= 11 \\ c &= -6 \end{align}
S0 $$f(x) = x^3 - 5x^2 + 11x - 6$$.
You could make a difference table
$$\begin{array}{c} 1 && 4 && 9 \\ & 3 && 5 \\ && 2 \end{array}$$
Using $$f(x) = x^3 + ax^2 + bx + c$$, this corresponds to
$$\begin{array}{c} 1+a+b+c && 8 + 4a + 2b + c && 27 + 9a + 3b + c \\ & 7+3a+b && 19 + 5a + b \\ && 12+2a \end{array}$$
Then, comparing entries...
\begin{align} 12+2a=2 &\implies a = -5 \\ 7+3a+b = 3 &\implies b=11 \\ 1+a+b+c = 1 &\implies c = -6 \end{align} | 2021-09-27T10:37:56 | {
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https://math.stackexchange.com/questions/2113526/analyzing-biggl-lfloor-fracx5-bigg-rfloor-bigg-lfloor-fracx7-bigg | # Analyzing $\biggl\lfloor{\frac{x}{5}}\bigg\rfloor=\bigg\lfloor{\frac{x}{7}}\bigg\rfloor$
How many non negative integral values of $x$ satisfy the equation :$$\biggl\lfloor{\dfrac{x}{5}}\bigg\rfloor=\bigg\lfloor{\dfrac{x}{7}}\bigg\rfloor$$.
My try:
Writing few numbers and putting in the required equation.
$\underbrace{0,1,2,3,4}_{\implies 0=0},\underbrace{5,6}_{\displaystyle\times},\underbrace{7,8,9}_{\implies 1=1},\underbrace{10,11,12,13}_{\displaystyle\times} ,\underbrace{14}_{\implies 2=2},\underbrace{15,16,17,18,19,20}_{\text{LHS gives 3 but RHS gives 2}},\underbrace{21,22,23,24}_{\text{LHS gives 4 but RHS gives 3}}$.
The point which I wish to make here is that after $14$ we'll get $5's$ multiple before than $7's$, hence $\forall x \geq 15 \implies \biggl\lfloor{\dfrac{x}{5}}\bigg\rfloor>\bigg\lfloor{\dfrac{x}{7}}\bigg\rfloor$
Hence the only solution are: $\{0,1,2,3,4,7,8,9,14\}$. Does these make sense?
I need to know other solutions that don't incorporate such analyzing, means using some properties and framing it to solve above.
By the division rule,
$$x=5q+r=7q+s$$ with $0\le r<5,0\le s<7$.
From this we draw $$2q=r-s\in[-6,4],$$
thus the only possibilities are with $q=0,1,2$.
• $q=0$: $r=s$, possible for $x\in[0,4]\cap[0,6]=[0,4]$;
• $q=1$: $5+r=7+s$, possible for $x\in[5,9]\cap[7,13]=[7,9]$;
• $q=2$: $10+r=14+s$, possible for $x\in[10,14]\cap[14,20]=[14]$.
• why intersection? – mnulb Jan 25 '17 at 16:52
• Because the two bracketings must hold. What else would you do ? – Yves Daoust Jan 25 '17 at 16:53
• did you came up with those intersections by hit and trials or something different, and if something new, please don't mind adding there. – mnulb Jan 25 '17 at 17:08
• @Ayushakj: the intervals are drawn from the equations and inequations. – Yves Daoust Jan 25 '17 at 17:11
Rewrite the given condition as $$5n\le x<5n+5,\quad 7n\le x<7n+7$$ where $n$ is an integer. Since $x\ge0$, $n\ge0$ as well, so the above is equivalent to $7n\le x<5n+5$. The only nonnegative integers $n$ satisfying $7n<5n+5$ are $0$, $1$, and $2$. And so just a little cleanup is needed.
For a solution the LHS and RHS are a positive integer, let's call this $c$: $$c = \left \lfloor \frac{x}{5} \right \rfloor = \left \lfloor \frac{x}{7} \right \rfloor$$
Then, there are integers $0 \leq a < 5, 0 \leq b < 7$, s.t.: $$x = 5c + a = 7c + b$$ Thus: $$a - b = 2c$$
With the restrictions to $a,b$, this is enough to find all solutions.
Analysis:
Let us take the general form where $p>q$ :
$$\left\lfloor{\dfrac{x}{q}}\right\rfloor=\left\lfloor{\dfrac{x}{p}}\right\rfloor$$
Solution for above equation exists when $\dfrac{x}{q}-\dfrac{x}{p}<1\Rightarrow x<\dfrac{p\cdot q}{p-q}$
So there is no need to check for integers greater than$\left\lfloor\dfrac{p\cdot q}{p-q}\right\rfloor$
Also $\left\lfloor{\dfrac{x}{q}}\right\rfloor=\left\lfloor{\dfrac{x}{p}}\right\rfloor=k$ implies \begin{align} &x\in [k\cdot p,(k+1)p) \cap\ [k\cdot q,(k+1)\cdot q)\\ \Rightarrow\ \ &x\in [k\cdot p,(k+1)\cdot q) \end{align}
The integers which satisfy the equation are in $$[0p,1q)\ \cup\ [1p,2q)\ \cup\ [2p,3q) \cup\ ...\ \cup\ [(n-1)p,nq)$$ where $nq<\left\lfloor\dfrac{p\cdot q}{p-q}\right\rfloor$ .
• Well done but you are missing the case $k=0$, giving the interval $[0,q)$. – Yves Daoust Jan 26 '17 at 7:41 | 2019-06-27T12:43:28 | {
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https://math.stackexchange.com/questions/3304339/let-a-and-b-commute-if-m-and-n-are-relatively-prime-then-ordab | # Let $a$ and $b$ commute. If $m$ and $n$ are relatively prime, then ord($ab$) = $mn$. [duplicate]
This is exercise $$10.E.4$$ from Pinter:
Let $$a$$ and $$b$$ be elements of a group $$G$$.
Let ord($$a$$) = $$m$$ and ord($$b$$) = $$n$$.
Prove:
Let $$a$$ and $$b$$ commute.
If $$m$$ and $$n$$ are relatively prime, then ord($$ab$$) = $$mn$$.
(HINT: Use $$10.E.2$$.)
Here is $$10.E.2$$, which Pinter suggests that we use:
If $$m$$ and $$n$$ are relatively prime, then no power of $$a$$ can be equal to any power of $$b$$ (except for $$e$$).
I'm also going to use $$10.E.1$$:
If $$a$$ and $$b$$ commute, then ord($$ab$$) is a divisor of lcm($$m$$,$$n$$).
As well as $$B.T6.i$$ (Theorem $$6.i$$ from Appendix $$B$$. A fact from basic number theory.):
If gcd($$m$$,$$n$$) = 1 then lcm($$m$$,$$n$$) = $$mn$$
Let's begin.
We are given that $$m$$ and $$n$$ are relatively prime which means that:
$$\text{gcd}(m,n) = 1$$
By $$B.T6.i$$:
$$\text{lcm}(m,n) = mn \tag{1}$$
By $$10.E.1$$:
$$\text{ord}(ab)\ |\ \text{lcm}(m,n)$$
Substituting $$(1)$$:
$$\text{ord}(ab)\ |\ mn$$
Which means that there is an integer $$x$$ such that:
$$\text{ord}(ab) x = mn$$
This is so close! For the theorem to be true, we'd have to show that $$x = 1$$.
However, the original exercise statement says to use $$10.E.2$$.
• Is that helpful in showing that $$x = 1$$?
• Or is there some other completely different approach whereby $$10.E.2$$ is used?
The proof uses the following fact:
If $$a|c$$ and $$b|c$$ then $$lcm(a,b)|c$$.
For this exercise, I wanted to only use theorems that had been presented in the book up to that point. And, I didn't seem to recall seeing this theorem. (If anyone spots this in Pinter, please comment below with the location.)
However, I did notice that the following similar fact is in Appendix B (REVIEW OF THE INTEGERS) as exercise B.9:
If $$a|c$$ and $$b|c$$ and $$gcd(a,b) = 1$$ then $$ab|c$$.
So yeah, it looks like the approach shown below is definitely a valid way to go if you want to stick to what's presented in the book.
• – Angina Seng Jul 26 '19 at 4:35
• – Dietrich Burde Jul 26 '19 at 11:36
• The 2nd $\Rightarrow$ in my answer in the dupe is essentially a proof of 10.E.2, so replace that by an invocation of 10.E.2 to obtain the hinted proof. This is a well-known proof and likely occurs here many times. – Bill Dubuque Jul 26 '19 at 14:47
### $$\boxed{\textit{You can show mn\ \big|\ \text{ord}(ab) using 10.E.2:}}$$
Since $$a$$ and $$b$$ commute we can distribute common powers of $$a$$ and $$b$$: $$e = (ab)^{\text{ord}(ab)} = a^{\text{ord}(ab)}b^{\text{ord}(ab)}$$ This implies $$a^{\text{ord}(ab)} = b^{-\text{ord}(ab)}$$ or, in other words, some power of $$a$$ equals some power of $$b$$. But $$10.E.2$$ says that since $$m = \text{ord}(a)$$ and $$n = \text{ord}(b)$$ are relatively prime, no power of $$a$$ can equal any power of $$b$$ unless those powers of $$a$$ and $$b$$ both evaluate to $$e$$.
So we must have $$a^{\text{ord}(ab)} = e = a^m$$ and $$b^{-\text{ord}(ab)} = e = b^n$$.
But then, $$m\ \big|\ \text{ord}(ab)$$. This is because $$m$$ is the order of $$a$$ i.e. the least positive integer power of $$a$$ that evaluates to $$e$$; so if any other positive integer power of $$a$$ evaluates to $$e$$ (like $$\text{ord}(ab)$$ here), then $$m$$ must divide that integer. For a similar reason, $$n\ \big|\ \text{ord}(ab)$$.
Therefore since both $$m$$ and $$n$$ divide $$\text{ord}(ab)$$, their least common multiple $$\text{lcm}(m,n)$$ must divide $$\text{ord}(ab)$$. But in our case $$\text{lcm}(m,n) = mn$$ so we finally have $$mn\ \big|\ \text{ord}(ab)$$.
Along with the result you already established $$\text{ord}(ab)\ \big|\ mn$$, this suggests $$\text{ord}(ab) = mn$$.
P.S. If you don't believe the bolded statements above, use Euclidean Division. E.g. to see $$m\ \big|\ \text{ord}(ab)$$, use Euclidean Division on the integer pair $$\big(m, \text{ord}(ab)\big)$$ to get $$\text{ord}(ab) = km + r$$ for some $$k, r \in \Bbb Z$$ with $$0 \leq r < m$$. Do you see why $$r$$ must be $$0$$?
• Excellent and very clear presentation! I learned a few things from your proof. Thanks very much for explaining each part. I've updated my answer with some comments regarding your approach. – dharmatech Jul 26 '19 at 20:05 | 2020-04-03T23:46:23 | {
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https://www.physicsforums.com/threads/complex-analysis-residue-integration-question.781025/ | Complex analysis: residue integration question
Tags:
1. Nov 9, 2014
Wheelwalker
I'm asked to evaluate the following integral: $\int_{c} \frac{30z^2-23z+5}{(2z-1)^2(3z-1)}dz$ where c is the unit circle. This function has a simple pole at $z=\frac{1}{3}$ and a second order pole at $z=\frac{1}{2}$, both of which are within my region of integration. I then went about computing the residues using the following formula (which I, mistakenly perhaps, thought was general and applied to any mth-order pole): $Res f(z) = \frac{1}{(1-1)!}\lim_{z \to 1/2}{\frac{d}{dz}\{[2z-1]^2 \frac{30z^2-23z+5}{(2z-1)^2(3z-1)}}\}$ for which I got 2 (this was just the residue for the first pole). The residue I got for the other singularity was 6. They are supposed to be 1/2 and 2, respectively. What am I doing wrong, exactly? Does this residue formula not apply to this case? If so, why not? Thanks in advance.
Last edited: Nov 9, 2014
2. Nov 9, 2014
lurflurf
The formula you should have used is $$Res f(z) = \frac{1}{(2-1)!}\lim_{z \to 1/2}{\frac{d}{dz}\{[z-\dfrac{1}{2}]^2 \frac{30z^2-23z+5}{(2z-1)^2(3z-1)}}\}$$
That is why you got 2=4(1/2) instead of 1/2 and 6=3*2 instead of 2
3. Nov 9, 2014
Wheelwalker
Oh, wow that makes a lot of sense. I got in the habit of multiplying $f(z_0)$ by whichever term was creating the singularity in the denominator because a lot of the first problems we did would have something simple like $(z-1)$ there. So then, $(z-z_0)$ would equal exactly what was there. Anyway, thank you so much!
4. Nov 9, 2014
Wheelwalker
Ah, and yes that factorial term definitely should have been $(2-1)!$, although of course it doesn't matter in this case.
5. Nov 9, 2014
Wheelwalker
One more question for you actually. So, is this the most general residue formula? This should apply to nearly any case?
6. Nov 10, 2014
mathwonk
I am not very good at computations, but I am getting residues opposite to what you seem to say, i.e. res = 1/2 at z=1/2, and res = 2 at z= 1/3.
My reasoning is as follows: the residue at a, is the coefficient of 1/(z-a) in the Laurent expansion in powers of (z-a). So at z= 1/3 we should have 1/(z-1/3) . g(z), where g(z) = (30z^2 -23z + 5)/3(2z-1)^2. Then when we expand g in powers of (z-1/3), the constant term will be the residue, i.e. the coefficient of 1/(z-1/3) after we multiply by the 1/(z-1/3) out front. so this is just g(1/3) = (30/9 - 23/3 + 5)/3(1/3)^2 = 2.
Then for the other one, at z= 1/2, the function can be written as 1/(z-1/2)^2 h(z), where h(z) = (30z^2-23z+5)/4(3z-1). then the residue after multiplying by the factor 1/(z-1/2)^2, will be the linear coefficient of the expansion of h(z), i.e. should be the derivative of h(z). By the quotient rule, this is (1/4)(7/2 - 3)/(1/4) = (1/8)/(1/4) = 1/2.
Am I missing something? My point in doing it this way is to emphasize not memorizing formulas. but going back to the meaning of the concept. But of course one wants to get it right, and I am not super confident after apparently getting a different answer. Or maybe I just misunderstood what was said.
7. Nov 10, 2014
Wheelwalker
Those are the correct residues. I made a mistake in finding them the first time.
8. Nov 10, 2014
mathwonk
Thank you. My recommended point of view, as I said, is to write the function as a product of form 1/(z-a)^m . g(z), where g is holomorphic at a.
Then we want the coefficient of (z-a)^(m-1) in the Taylor expansion of g. this is of course just g^(m-1)(a) /(m-1)!, i.e. the (m-1) st derivative of g at a, divided by (m-1)!
I suggest this because your error was apparently in recalling the residue formula incorrectly. I hope this method makes that less likely, since all you have to recall this way is that you want the coefficient of 1/(z-a). In the product 1/(z-a)^m . g(z), that will be the coefficient of (z-a)^(m-1) in the expansion of g. | 2018-07-16T09:02:35 | {
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https://math.stackexchange.com/questions/1918766/absolute-value-inequality-with-variable-on-both-sides | # Absolute value inequality with variable on both sides
I am trying to solve the following inequality:
$$|3-5x| \le x$$
I am not familiar with inequalities including one absolute value with variables on both sides. I tried to solve it as follows:
$$-x \le 3-5x \le x$$
Then I solved for each side separately,as follows:
$$3-5x \le x$$
$$x \ge (1/2)$$
$$-x \le 3-5x$$
$$x \le \frac34$$
I know my solution is incorrect and that it actually lies between $\frac12$ and $\frac34$, but I wanted to know what is wrong with my method and what is the appropriate approach to solving such inequalities.
Thanks,
This is how I learned it back in Algebra I.
You can split it into two equations. Since the sign is $\le$, it will be that Equation 1 and Equation 2 are true.
Here are the steps:
$$|3-5x| \le x$$
$$3-5x \le x \;\;\;\;\;| \;\;-3+5x\le x$$
$$6x \ge 3 \;\;\;\;\;\;\;| \;\;-4x \ge -3$$
$$x \ge \frac 12 \;\;\;| \;\;\;\;x \le \frac 34$$
$$\frac 12 \le x \le \frac 34$$
• This should be sufficient, but for further elaboration just reply Sep 8 '16 at 2:38
• This approach is very specific towards linear inequalities (but definitely works, I think)--it doesn't work (I believe) for higher order ones. Sep 8 '16 at 3:53
• Yes, I know. I was trying to give a simple easy-to-understand solution. Sep 8 '16 at 4:02
• Your "$|$" symbol is for AND or OR? In either case you have a logical issue as you start of with an OR and end with and AND. Sep 8 '16 at 4:15
• @Maccavity And. And please point out the issue. Sep 8 '16 at 4:16
Divide $-x \leq 3-5x \leq x$ by $x$ and go from there. Although I think what you have is correct, the solution IS the interval $\dfrac{1}{2} \leq x \leq \dfrac{3}{4}$.
You need to break up the absolute value into its intervals:
$$|x| = \begin{cases} x & x > 0 \\ -x & x < 0 \\ 0 & x = 0 \end{cases}$$
Therefore for $|3-5x|$ you need to find the interval when it's less than zero and when it's greater than zero:
$$|3-5x| = \begin{cases} 3 - 5x & 3 - 5x > 0 \rightarrow 3 > 5x \rightarrow x < \frac{3}{5}\\ 5x - 3 & 3 - 5x < 0 \rightarrow 3 <5x \rightarrow x > \frac{3}{5} \\ 0 & 3 - 5x = 0 \rightarrow 3 = 5x \rightarrow x = \frac{3}{5} \end{cases}$$
Now you solve the inequality in each case:
1. $x < \frac{3}{5} \rightarrow |3 - 5x| = 3 - 5x$ $$3 - 5x \leq x \\ 3 \leq 6x \\ x \geq \frac{1}{2}$$
2. $x > \frac{3}{5} \rightarrow |3 - 5x| = 5x - 3$ $$5x - 3 \leq x \\ 4x \leq 3 \\ x \leq \frac{3}{4}$$
3. $x = \frac{3}{5} \rightarrow |3 - 5x| = 0$ $$0 \leq x \\ x \geq 0$$
Now you need to analyze each case:
1. $x < \frac{3}{5} \wedge x \geq \frac{1}{2}$
It is true that $\frac{3}{5} = \frac{6}{10} \geq \frac{5}{10}$. Therefore this particular interval is true for $\frac{1}{2} \leq x < \frac{3}{5}$.
1. $x > \frac{3}{5} \wedge x \leq \frac{3}{4}$
Since $\frac{3}{5} = \frac{12}{20}$ and $\frac{3}{4} = \frac{15}{20}$, this is true for $\frac{3}{5} < x \leq \frac{3}{4}$.
1. $x = \frac{3}{5} \wedge x \geq 0$
$\frac{3}{5} > 0$ therefore this is trivially satisfied--thus $x = \frac{3}{5}$ is allowed.
When we combine these results, we find that $\frac{1}{2} \leq x \leq \frac{3}{4}$.
PS Edit:
It's probably easier to use:
$$|x| = \begin{cases} x & x \geq 0 \\ -x & x \leq 0 \\ \end{cases}$$
In that case you get the two intervals:
$$\frac{1}{2} \leq x \leq \frac{3}{5}$$
and
$$\frac{3}{5} \leq x \leq \frac{3}{4}$$
Which clearly combines to give: $\frac{1}{2} \leq x \leq \frac{3}{4}$.
Your solution was correct except you needed to use the word AND correctly. The inequality $-x \le 3-5x \le x$ is equivalent to the inequalities
\begin{align} -x \le 3-5x \quad &\text{AND} \quad 3-5x \le x \\ 4x \le 3 \quad &\text{AND} \quad 3 \le 6x \\ x \le \frac 34 \quad &\text{AND} \quad \frac 12 \le x \\ \frac 12 \le x \quad &\text{AND} \quad x \le \frac 34 \\ x &\in\left[\frac 12, \frac 34 \right] \end{align} | 2021-11-28T06:40:24 | {
"domain": "stackexchange.com",
"url": "https://math.stackexchange.com/questions/1918766/absolute-value-inequality-with-variable-on-both-sides",
"openwebmath_score": 0.9988861680030823,
"openwebmath_perplexity": 319.48798620940573,
"lm_name": "Qwen/Qwen-72B",
"lm_label": "1. YES\n2. YES",
"lm_q1_score": 0.9740426405416754,
"lm_q2_score": 0.8824278710924296,
"lm_q1q2_score": 0.8595223736464394
} |
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