text stringlengths 1 2.12k | source dict |
|---|---|
Usually check if the number is divisible by 2, then 5, then 3, etc... (7, 11, 13,17, ....)
Here 441 is:
- clearly not divisible by 2 or 5 (not even or ending with a 5 or 0)
- divisible by 3 (sum of the digits is 4+4+1=9 divisible by 3)
441 = 3 * 100 + 141 = 3 * 100 + 120 + 21 = 3 *(100+40+7) = 3 * 147 (I would do the middle step in my head so to speak)
same thing with 147
147 = 120 + 27 = 3 * 49
and 49 = 7^2
so 441 = 3^2 * 7^2
Note that I found those links interesting:
http://www.f1gmat.com/data-sufficiency/ ... ility-rule (Not specific to GMAT)
Jeff Sackmann - specific to GMAT strategies:
http://www.gmathacks.com/mental-math/factor-faster.html
http://www.gmathacks.com/gmat-math/numb ... teger.html
http://www.gmathacks.com/main/mental-math.html (general tricks)
is 441 = 7 * 7 * 2 * 2...i think 441 cant be divided by 2
should it not be 7 * 7 * 3 * 3
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Re: How many different positive integers are factor of 441 ? [#permalink]
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14 Apr 2012, 02:13
harshavmrg wrote:
GreginChicago wrote:
sugu86 wrote:
How many different positive integers are factor of 441 ? (A) 4 (B) 6 (C) 7 (D) 9 (E) 11. OA is D. Can this be solved using the method of (p+1) * (q+1) * (r+1) when the given no is of the prime no of the form N=a^p b^q c^r where a,b,c are prime factors ? Can you please clarify whether my approach is correct?
Could you indicate where you found that method? I have not heard of it and I am curious b/c it may be faster than the way I usually approach prime factorization problems.... | {
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Usually check if the number is divisible by 2, then 5, then 3, etc... (7, 11, 13,17, ....)
Here 441 is:
- clearly not divisible by 2 or 5 (not even or ending with a 5 or 0)
- divisible by 3 (sum of the digits is 4+4+1=9 divisible by 3)
441 = 3 * 100 + 141 = 3 * 100 + 120 + 21 = 3 *(100+40+7) = 3 * 147 (I would do the middle step in my head so to speak)
same thing with 147
147 = 120 + 27 = 3 * 49
and 49 = 7^2
so 441 = 3^2 * 7^2
Note that I found those links interesting:
http://www.f1gmat.com/data-sufficiency/ ... ility-rule (Not specific to GMAT)
Jeff Sackmann - specific to GMAT strategies:
http://www.gmathacks.com/mental-math/factor-faster.html
http://www.gmathacks.com/gmat-math/numb ... teger.html
http://www.gmathacks.com/main/mental-math.html (general tricks)
is 441 = 7 * 7 * 2 * 2...i think 441 cant be divided by 2
should it not be 7 * 7 * 3 * 3
GreginChicago is saying exactly the same thing!
Check for a solution here: how-many-different-positive-integers-are-factor-of-130628.html#p1073364
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Re: How many different positive integers are factors of 441 [#permalink]
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27 Mar 2014, 21:02
$$441 = 21^2$$
$$= (7 * 3)^2$$
$$= 7^2 . 3^2$$
Prime factors = (2+1) * (2+1) = 9
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Re: How many different positive integers are factors of 441 [#permalink] 15 Aug 2016, 04:28
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# Machine X can complete a job in half the time it takes Machine Y to co
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Machine X can complete a job in half the time it takes Machine Y to co [#permalink]
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03 Feb 2015, 09:47
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Machine X can complete a job in half the time it takes Machine Y to complete the same job, and Machine Z takes 50% longer than Machine X to complete the job. If all three machines always work at their respective, constant rates, what is the ratio of the amount of time it will take Machines X and Z to complete the job to the ratio of the amount of time it will take Machines Y and Z to complete the job?
A. 5 to 1
B. 10 to 7
C. 1 to 5
D. 7 to 10
E. 9 to 10
Kudos for a correct solution.
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Re: Machine X can complete a job in half the time it takes Machine Y to co [#permalink]
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03 Feb 2015, 23:08
6
6
Since rates are additive, using rates is usually easier.
"Machine X can complete a job in half the time it takes Machine Y to complete the same job,"
So machine X's rate is twice machine Y's rate. Rate of X:Rate of Y = 2:1 | {
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} |
"Machine Z takes 50% longer than Machine X to complete the job"
Time taken by Z : Time taken by X = 3:2 so Rate of Z: Rate of X = 2:3
Rate of X : Rate of Y : Rate of Z = 6:3:4
Rate of X+Z : Rate of Y+Z = 10 : 7
Time taken by X+Z : Time taken by Y+Z = 7:10
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Re: Machine X can complete a job in half the time it takes Machine Y to co [#permalink]
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03 Feb 2015, 10:30
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Bunuel wrote:
Machine X can complete a job in half the time it takes Machine Y to complete the same job, and Machine Z takes 50% longer than Machine X to complete the job. If all three machines always work at their respective, constant rates, what is the ratio of the amount of time it will take Machines X and Z to complete the job to the ratio of the amount of time it will take Machines Y and Z to complete the job?
A. 5 to 1
B. 10 to 7
C. 1 to 5
D. 7 to 10
E. 9 to 10
Kudos for a correct solution.
Let time required by Y to complete the work = t
then time taken by X will be = $$\frac{t}{2}$$
and time taken by Z = $$t/2 + t/4$$ ----> $$\frac{t}{4}$$ as Z takes 50% longer then X
Now Rate for X = $$\frac{2}{t}$$
Rate for Y = $$\frac{1}{t}$$
Rate for Z = $$\frac{4}{3t}$$
amount of time it will take Machines X and Z to complete the job = $$1/[2/t + 4/3t]$$
i.e. $$\frac{3t}{10}$$
the amount of time it will take Machines Y and Z to complete the job = $$1/[1/t + 4/3t]$$
i.e. $$\frac{3t}{7}$$
Thus ratio = $$\frac{3t}{10}$$ / $$\frac{3t}{7}$$ = 7:10
[D]
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Re: Machine X can complete a job in half the time it takes Machine Y to co [#permalink]
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Updated on: 03 Feb 2015, 23:29
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Hi Bunuel, | {
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### Show Tags
Updated on: 03 Feb 2015, 23:29
1
2
Hi Bunuel,
Solutioning this exercise can be facilitated by using a R T W (rate time work) table:
We translate the exercise into the table:
R T W
X t/2 1
Y t 1
Z (t/2*3/2 =3t/4) 1
From this table we find the rates
Rx = 2/t
Ry = 1/t
Rz = 4/3t
The Q is what is the ratio of (Tx + Ty) / (Ty + Tz)
Rx + Ry = 2/t + 4/3t = 6/3t+4/3t = 10/3t
Ry+Rz = 1/t + 4/3t = 3/3t + 4/3t = 7/3t
The (10/3t)/(7/3t) = 10/7 then the work ratios is 10 to 7
Since Time Ratio is the inverse of work, the the answer is 7 to 10
Bunuel wrote:
Machine X can complete a job in half the time it takes Machine Y to complete the same job, and Machine Z takes 50% longer than Machine X to complete the job. If all three machines always work at their respective, constant rates, what is the ratio of the amount of time it will take Machines X and Z to complete the job to the ratio of the amount of time it will take Machines Y and Z to complete the job?
A. 5 to 1
B. 10 to 7
C. 1 to 5
D. 7 to 10
E. 9 to 10
Kudos for a correct solution.
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Originally posted by TudorM on 03 Feb 2015, 22:53.
Last edited by TudorM on 03 Feb 2015, 23:29, edited 2 times in total.
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Re: Machine X can complete a job in half the time it takes Machine Y to co [#permalink]
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04 Feb 2015, 04:14
1
x=0,5t
y=t
z=0,75t
t=4hours so we get:
x=2
y=4
z=3
x+z=1/2+3/4=5/6 -> they need 6/5 hours
y+z=1/4+1/3=7/12 -> they need 12/7 hours
ratio is 6/5 divided by 12/7, or multiplied by 7/12 -> we get 7/10
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Re: Machine X can complete a job in half the time it takes Machine Y to co [#permalink]
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### Show Tags
04 Feb 2015, 05:46
ans D..
let time taken by X =x,
so by Y =2x,
and by z = 1.5x...
time taken by X and Z = 1/( 1/x+1/1.5x)... = 1.5 X^2/2.5X =3X^2/5X
time taken by Y and Z = 1/( 1/2x+1/1.5x)... = 3 X^2/3.5X...= 6X^2/7X
ratio = (3X^2/5X)/ (6X^2/7X)= 7/10
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Re: Machine X can complete a job in half the time it takes Machine Y to co [#permalink]
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04 Feb 2015, 13:48
4
Hi All,
These types of rate questions can be solved in a few different ways. Since the prompt discusses how two machines will work together on a task, you can use the Work Formula. By TESTing VALUES, we can come up with a simple example that will answer the given question.
We're told that:
1) Machine X can do a job in HALF the time that it takes Machine Y
2) Machine Z takes 50% longer to the do the job than Machine X
Since Machine X appears in both 'facts', we want to TEST a VALUE for X that is easily doubled AND halved....
X = 2 hours to complete the job
With that value in place, we can figure out the other 2 values....
Y = 4 hours to complete the job
Z = 3 hours to complete the job
We're asked for the ratio of the time it takes (X and Z) working together vs. the time it takes (Y and Z) working together.
The Work Formula: (A)(B)/(A+B)
For Machines X and Z, we have: (2)(3)/(2+3) = 6/5 hours to complete the job
For Machines Y and Z, we have: (4)(3)/(4+3) = 12/7 hours to complete the job | {
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The ratio 6/5 : 12/7 needs to be "clean up" a bit. Since we're dealing with fractions, we can use common denominators....
6/5 : 12/7
42/35 : 60/35
Now we can eliminate the denominators...
42 : 60
And reduce the ratio....
7 : 10
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Re: Machine X can complete a job in half the time it takes Machine Y to co [#permalink]
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09 Feb 2015, 04:51
Bunuel wrote:
Machine X can complete a job in half the time it takes Machine Y to complete the same job, and Machine Z takes 50% longer than Machine X to complete the job. If all three machines always work at their respective, constant rates, what is the ratio of the amount of time it will take Machines X and Z to complete the job to the ratio of the amount of time it will take Machines Y and Z to complete the job?
A. 5 to 1
B. 10 to 7
C. 1 to 5
D. 7 to 10
E. 9 to 10
Kudos for a correct solution.
VERITAS PREP OFFICIAL SOLUTION:
In this ratio problem, it's helpful to begin with the ratio of each machine's RATES as opposed to times, as rates are additive when machines are working together. So if Machine X as a rate that's twice as fast as Machine Y, their rates have the ratio X:Y = 2:1. And the ratio of X to Z is 3:2, as X works 50% faster (and therefore would accomplish 3 jobs in the time that it takes for Z to complete 2). So to find a common three-way ratio, you can use X as the "anchor", and make the ratio of rates:
X : Y : Z = 6 : 3 : 4 | {
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X : Y : Z = 6 : 3 : 4
So X and Y working together would complete 10 jobs in the time that it would take Y and Z working together to complete 7 jobs, for a ratio of 10 : 7 in their respective rates. But since the question asks for times, not rates, you'll need to invert the ratio to 7 : 10, answer choice D.
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Machine X can complete a job in half the time it takes Machine Y to co [#permalink]
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20 Nov 2015, 18:32
let 2,4,3=respective times of X,Y,Z
1/2,1/4,1/3=respective rates of X,Y,Z
X+Z rate/Y+Z rate=(5/6)/(7/12)=10:7
inverting, X+Z time/Y+Z time=(6/5)/(12/7)=7:10
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Re: Machine X can complete a job in half the time it takes Machine Y to co [#permalink]
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28 Mar 2018, 10:39
Bunuel wrote:
Machine X can complete a job in half the time it takes Machine Y to complete the same job, and Machine Z takes 50% longer than Machine X to complete the job. If all three machines always work at their respective, constant rates, what is the ratio of the amount of time it will take Machines X and Z to complete the job to the ratio of the amount of time it will take Machines Y and Z to complete the job?
A. 5 to 1
B. 10 to 7
C. 1 to 5
D. 7 to 10
E. 9 to 10
We can let the time of machine Y to complete the job = y, so the time of machine X to complete the job is 0.5y = (½)y and the time of machine Z to complete the job = (1.5)(0.5y) = 0.75y = (3/4)y. .
Since rate is inverse of time, the rate of Y = 1/y, the rate of X = 1/[(½)y] = 2/y and the rate of Z = 1/[(3/4)y] = 4/(3y).
Thus, the amount of time it will take Machine X and Z to complete the job is
1/[2/y + 4/(3y)] = 1/[6/(3y) + 4/(3y)] = 1/[10/(3y)] = 3y/10
Similarly, the amount of time it will take Machine Y and Z to complete the job is
1/[1/y + 4/(3y)] = 1/[3/(3y) + 4/(3y)] = 1/[7/(3y)] = 3y/7 | {
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1/[1/y + 4/(3y)] = 1/[3/(3y) + 4/(3y)] = 1/[7/(3y)] = 3y/7
Therefore, the ratio is (3y/10)/(3y/7) = (1/10)/(1/7) = 7/10.
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Re: Machine X can complete a job in half the time it takes Machine Y to co &nbs [#permalink] 28 Mar 2018, 10:39
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# Finding the length of a side and comparing two areas from a given figure
The following figure came in two national-exam paper (different question in each paper).
First Question:
What is the length of $$AE$$?
Choices:
A) $$\frac{5}{4}$$ cm
B) $$\frac{\sqrt{41}}{5}$$ cm
C) $$\frac{7}{5}$$ cm
D) $$\frac{15}{8}$$ cm
Second Question
Compare:
First value: area of $$\triangle ABC$$
Second value: area of $$\triangle CDE$$
Choices:
A) First value > Second value
B) First value < Second value
C) First value = Second value
D) Given information is not enough
My Attempt:
Finding the equation of $$AC$$ and $$BD$$ in order to find the coordinates of E:
We can assume that $$C$$ is the origin $$(0,0)$$ and $$A$$ is $$(3,4)$$, then the equation of $$AC$$ is $$y=\frac{4}{3}x$$, and the equation of $$BD$$ is $$y=-\frac{4}{5}x+4$$.
Next, solve the system of equations for $$x$$ and $$y$$, we end up with the coordinates of $$E(\frac{15}{8},\frac{5}{2})$$.
To solve the first question, we can use the distance between two points formula,
therefore $$AE=\sqrt{(3-\frac{15}{8})^2+(4-\frac{5}{2})^2}=\frac{15}{8}$$ cm.
Hence D is the correct choice.
To solve the second question, we can use the area of triangle formula,
area of $$\triangle ABC=\frac{1}{2}\times 3\times 4=6$$ cm$$^2$$.
area of $$\triangle CDE=\frac{1}{2}\times 5\times \frac{5}{2}=6.25$$ cm$$^2$$.
Hence B is the correct choice.
In the exam, calculators are not allowed. It is simple to determine to coordinates of $$E$$ without a calculator, but it will take time!
For the first question, I think there is a good way to solve, maybe $$BD$$ divides $$AC$$ into two parts ($$AE$$ and $$CE$$) in a ratio, this ratio can be calculated somehow. | {
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For the second question, I think also a good way to think, like moving $$D$$ will make $$\triangle CDE$$ bigger or smaller, while $$\triangle ABC$$ will be unchanged. But this has a limitation since the areas $$6$$ and $$6.25$$ cm$$^2$$ are close, it will not be obvious to us.
If we have $$75$$ seconds (on average) to solve each of these questions, then how can we solve them? [remember that in some questions in the national exam, we do not need to be accurate $$100$$%].
Any help will be appreciated. Thanks.
• $AB\perp BC$? $AC\perp BD$? $AB||DC$? – Michael Rozenberg Jul 17 at 13:03
• @MichaelRozenberg Yes. – Hussain-Alqatari Jul 17 at 13:05
• If so, this quadrilateral does not exist because we need $4^2=3\cdot5,$ which is wrong. – Michael Rozenberg Jul 17 at 13:07
• @MichaelRozenberg $AB\perp BC$, $AB || DC$. But $AB$ is not $\perp BC$ – Hussain-Alqatari Jul 17 at 13:10
• @MichaelRozenberg Yes - No - Yes. – Hussain-Alqatari Jul 17 at 13:19
Triangles $$ABE$$ and $$CDE$$ are similar, with ratio $$3:5$$.
Hence $$AE=\frac{3}{8}AC=\frac{3}{8}\cdot 5=\frac{15}{8}$$
By the same similarity, if $$h$$ is the height of triangle $$CDE$$ from $$E$$ to $$CD$$, then $$h=\frac{5}{8}BC=\frac{5}{8}\cdot 4=\frac{5}{2}$$ hence the area of triangle $$CDE$$ is $$\frac{1}{2}\cdot CD\cdot h=\frac{1}{2}\cdot 5\cdot \frac{5}{2}=\frac{25}{4}=6.25$$ whereas the area of triangle $$ABC$$ is $$6$$.
• quasi your way looks very elegant. if I may ask how did you figure out $AE = \frac{3}{8}AC$ directly with out setting up a proportion ? Is there some nice trick here? – rsadhvika Jul 17 at 13:24
• @rsadhvika:$\;$Since $AE:EC=3:5$, it follows that $$AE:AC=AE:(AE+EC)=3:(3+5)=3:8$$ – quasi Jul 17 at 13:29
• Oh nice nice! With that in mind we can directly do: $AE = \frac{3}{8}AC$ and $EC = \frac{5}{8}AC$. I see now.. Thank you so much for explaining this you're awesome :) – rsadhvika Jul 17 at 13:31 | {
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For first question you may use similar triangles - $$\triangle AEB$$ and $$\triangle CED$$: $$\dfrac{3}{5} = \dfrac{x}{5-x}$$
For the first part use similarity of triangles to simplify computation.
You have similar triangles $$ABE$$ and $$CDE$$ with ratio of sides being $$3$$ to $$5$$
You also know $$AC=5$$ so you find $$AE=5\times \frac {3}{8}=\frac {15}{8}$$ which is choice $$D$$
Your have solved the second part correctly. | {
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# Why can we interchange summations?
Suppose we have the following
$$\sum_{i=1}^{\infty}\sum_{j=1}^{\infty}a_{ij}$$
where all the $a_{ij}$ are non-negative.
We know that we can interchange the order of summations here. My interpretation of why this is true is that both this iterated sums are rearrangements of the same series and hence converge to the same value, or diverge to infinity (as convergence and absolute convergence are same here and all the rearrangements of an absolutely convergent series converge to the same value as the series).
Is this interpretation correct. Or can some one offer some more insightful interpretation of this result?
Please note that I am not asking for a proof but interpretations, although an insightful proof would be appreciated.
• Another interpretation would be to use Tonnelli's theorem for counting measure. – TZakrevskiy Aug 13 '13 at 17:44
• The double sum is not a rearrangement of a single sum. At least, it's not of the form $\sum_{i=1}^\infty a_{\sigma(i)}$ where $\sigma$ is a permutation of the positive integers. However, let $I$ be your iterated sum and $J$ be the sum with the order interchanged. You can first show that $J\le I$. Then you can show that for any $\epsilon>0$, you have $J>I-\epsilon$. (This is mostly done by "lopping off small tails". Note each row and column is summable) – David Mitra Aug 13 '13 at 18:27
• We can show that it is rearrangement if you consider a bijection between $\mathbb{N}\times\mathbb{N}$ and $\mathbb{N}$. – Vishal Gupta Aug 14 '13 at 5:15
This isn't a proof, but perhaps can give you the insight you are looking for. Any nondecreasing sequence converges to its (possibly infinite) supremum. Thus a series of nonnegative terms converges to the supremum of its partial sums and interchanging the order of summation doesn't affect the value of the supremum: there is no accidental cancellation of terms of opposite sign. | {
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• Good thought. Then we only need to apply decoupling to interchange summations in this case. – Chival Sep 27 '14 at 20:13
• (+1)Consider the real valued function $f(a_{1},a_{2},...a_{n})$ then does the value of $\sum_{a_{1}} \sum_{a_{2}}...\sum_{a_{n}} f(a_{1},a_{2},...a_{n})$ depends upon the order of sum notions. In other words I want to know whether there is any theorem which talks about this interchanging of order of sum notions ? – NewBornMATH Apr 9 '19 at 15:11
The double sum you are asking about can be considered to be the sum of all terms of the infinite array of numbers $a_{ij}$:
$$\begin{pmatrix} a_{11} & a_{12} & \cdots & a_{1j} & \cdots\\ a_{21} & a_{22} & \cdots & a_{2j} & \cdots\\ \vdots& \vdots & &\vdots\\ a_{i1} & a_{i2} & \cdots & a_{ij} & \cdots\\ \vdots & \vdots & & \vdots \\ \end{pmatrix},$$
where the sum is taken with the particular order of first adding all elements of particular rows and then adding the resulting "row totals". The result you are citing claims that if all $a_{ij}$ is nonnegative, then it matters not in which order you add the entries of the infinite matrix, be that rows first and then row totals or columns first and then column totals.
I can't say your interpretation is right on the mark, because the issue is that even though one adds the entries of the same infinite array, the order of summation may result in summing the terms of different series in actuality. As the other answers point out this result holds essentially because we have no terms diminishing the total sum. Saying "both this iterated sums are rearrangements of the same series and hence converge to the same value, or diverge to infinity" really does not make any emphasis on this advantage, which I believe is sweeped under the phrase "same series". | {
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I'd like to mention yet another (measure theoretical) interpretation of this result, which I recently encountered in Rudin's Real and Complex Analysis (p. 23). In the book Rudin gives this result as a corollary of Lebesgue Monotone Convergence Theorem applied to series of functions.
Theorem: Let $X$ be a measure space. If $\forall n: f_n:X\to [0,\infty]$ is measurable, then
$$\int_X \sum_n f_n d\mu= \sum_n \int_X f_n d\mu (\ast).$$
Corollary: Let $X:=\{x_1,x_2,...,x_n,...\}$ be a countable set and $\mu:\mathcal{M}_X:=\mathcal{P}(X)\to[0,\infty]$ be the counting measure:
$$\mu(E):= \begin{cases} |E|&, \mbox{ if} |E|<\infty\\ \infty&, \mbox{ if} |E|=\infty \end{cases}.$$
If $\forall i,j:a_{ij}\geq0$, then
$$\sum_i \sum_j a_{ij}=\sum_j \sum_i a_{ij}.$$
Proof:
• Set $\forall j: f_j:X\to[0,\infty], f_j(x):=\sum_i a_{ij}\chi_{\{x_i\}}(x)$; and $\forall i: \bar{f_i}:X\to[0,\infty], \bar{f_i}(x):=\sum_j a_{ij}\chi_{\{x_i\}}(x).$ Then $\sum_j f_j=\sum_i \bar{f_i}$. Indeed, let $x_{i_0}\in X$. Then
$$\sum_j f_j(x_{i_0})= \sum_j \sum_i a_{ij} \chi_{\{x_i\}}(x_{i_0})= \sum_j a_{{i_0}j},$$
and
$$\sum_i \bar{f_i}(x_{i_0})=\bar{f_{i_0}}(x_{i_0})=\sum_j a_{{i_0}j}.$$
• Observe that the (inner) sum over $i$ is the integral of $f_j$ and the (inner) sum over $j$ is the integral of $\bar{f_i}$. Then we have:
$$\sum_j\sum_i a_{ij}= \sum_j \int_X f_j d\mu\stackrel{(\ast)}{=}\int_X \sum_j f_j d\mu=\int_X \sum_i \bar{f_i}d\mu\stackrel{(\ast)}{=}\sum_i\int_X\bar{f_i}d\mu =\sum_i \sum_j a_{ij}.$$
Now consider the interpretation induced by the above discourse, viz., the (countable) combinations of characteristic functions behave in such a way that one can concentrate nonnegative "weights" of individual points. In the above proof $f_j$ puts a single weight on each point, while $\bar{f_i}$ concentrates all the weights of the point $x_i$. | {
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Well, I think that interpretation is a bit circular, as the reason absolute convergence allows for rearrangements without changing the limit is precisely because of this point.
The gist of it is that no cancellations due to sign are possible. I think it is very illuminating to consider the counterexample for the rearrangement theorem when you do not have absolute convergence, but still have conditional convergence (Rearrangements can lead to any limiting value!).
You can add up as many negative terms as you want until you are happy (as it does not absolutely converge, you can continue adding until you are lower than any value in the real line you desire), and likewise for the positive terms (note, if adding the rest of the positive terms gives a finite value, then actually the whole sum didn't conditionally converge, and in fact will be $-\infty$). This procedure of adding negative terms until your sum is below $L$, and then adding positive terms until the sum is above $L$, and etc. will converge because conditional convergence requires the things you are summing to go to $0$ at the end of the day.
Okay, but to the non-negative double sum. Towards proving the double sum result (since you are only considering two orderings), I suppose one cute thing you could try is start with rows (arrange the numbers on the upper right quadrant for this picture). Let's assume the row sums converge first. Now, for each row sum, take out the first term of each, and set it aside, adding them upwards (the first column sum). Note that of course, the value of the sum didn't change, because you took out values one at a time, putting them into the column sum, and at no point did anything funny happen because only finite sums were involved (the full sum, the partial column sum, and the full sum minus the finitely many elements you took out). Now continue this procedure, extracting the second column sum, and etc. At the "end" you have what you want to prove. | {
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Two limiting procedures hidden: When you "finish" the first column sum, you have to take a limit procedure. When you "finish" all the columns, there's another limit there. The issue if you include negative terms is that the first column sum might already diverge. And even if all the column sums converge, summing the column sums might converge to an entirely different value! I think to really finish this proof idea, you should use the fact that the values you are working with vary monotonely at each step. | {
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# Counting the number of ways a test can be answered where at least one T/F and at least one multiple choice question must be answered
An exam has $15$ questions: eight true/false questions and seven multiple choice questions. You are asked to answer five of them, but your professor requires you to answer at least one true/false question and at least one multiple choice question. How many ways can you choose the questions you plan to answer.
step1. I chose $1$ question of true false type. $8$ ways of doing so.
step2. I chose $1$ question of MCQ type. $7$ ways of doing so.
step3. I chose $3$ questions of the remaining $13$ questions. $13 \cdot 12 \cdot 11$ ways of doing so.
step4. Hence, $8 \cdot 7 \cdot 13 \cdot 12 \cdot 11$ gives number of sequences of $5$ questions from a collection of $15$ questions such that there is at least one question from t/f type and mcq type each.
step5. Let the number of collections of $5$ questions from a collection of $15$ questions such that there is at least one question from t/f type and mcq type each be $N$.
step6. $N \cdot 5! = 8 \cdot 7 \cdot 11 \cdot 12 \cdot 13$
step7. $N = \dfrac{8 \cdot 7 \cdot 11 \cdot 12 \cdot 13}{5!}$
But this number above is not even a whole number. Where did I go wrong?
• when you chose one question you are also chosing four out of remaining questions you have to multiply them May 16 '18 at 14:27
• Please read this tutorial on how to typeset mathematics on this site. May 16 '18 at 14:41
• You choose tf first and then mc. You are dividing by all ways you can arrange 5. But you should only divide but the number of ways that tf comes first and then mc. May 16 '18 at 16:02
You don't have $5!$ ways of arranging the $5$ questions. You choose the questions specifically so that the first question was a TF and the second was MC. So the numbers of ways to arrange your five question must be that the first is TF and the second MC. | {
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That is, the reason you divide by $5!$ in the first place is that you need to account that picking $ABCDE$ is the same thing as picking $CEDAB$. But unless $C$ is TF and E is MFC picking $CEDAB$ was never an option.
The problem is the number of ways or arranging the $5$ questions depends on knowing how many TF and MC questions there are.
But to choose the number of way you can have $n$ TF question and $5-n$ MC questions first. You can have $1$ to $4$ TF questions and there are ${8 \choose n}$ ways of picking the $n$ TF question s an there are ${7\choose 5-n}$ ways of choosing the $5-n$ MC questions.
So the answer is $\sum_{n=1}^4 {8\choose n}{7\choose 5-n}=$
${8\choose 1}{7\choose 4}+{8\choose 2}{7\choose 3}+{8\choose 3}{7\choose 2}+{8\choose 4}{7\choose 1}=$
$8\frac {7!}{3!4!} + \frac {8!7!}{2!6!3!4!} + \frac {8!7!}{5!3!5!2!} + 7*\frac {8!}{4!4!}=$
$8*(5*7) + (4*7)(5*7) + (8*7)(7*3) + 7*(2*7*5)=$
$7*2(20 + 70 + 84 + 35)= 2926$
======
"The easier way is to count the number of ways you can choose any five questions, then subtract the ones that are all the same kind."
D'oh!
#any 5 - #all 5 TF - #all 5 mc =
${15 \choose 5} - {8\choose 5} - {7\choose 5}=$
$\frac {15!}{10!5!} - \frac {8!}{5!3!} - \frac {7!}{5!2!} =$
$3003 - 56- 21 = 2926$.
You overcount all of the cases. For a case where you answer two true/false questions and three multiple choice questions, there are two way to pick which one you count in step 1 and three in step 2, so this will get counted six times.
The easier way is to count the number of ways you can choose any five questions, then subtract the ones that are all the same kind. | {
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# fplot3
Plot 3-D parametric curve
## Syntax
fplot3(xt,yt,zt)
fplot3(xt,yt,zt,[tmin tmax])
fplot3(___,LineSpec)
fplot3(___,Name,Value)
fplot3(ax,___)
fp = fplot3(___)
## Description
example
fplot3(xt,yt,zt) plots the parametric curve xt = x(t), yt = y(t), and zt = z(t) over the default interval –5 < t < 5.
example
fplot3(xt,yt,zt,[tmin tmax]) plots xt = x(t), yt = y(t), and zt = z(t) over the interval tmin < t < tmax.
example
fplot3(___,LineSpec) uses LineSpec to set the line style, marker symbol, and line color.
example
fplot3(___,Name,Value) specifies line properties using one or more Name,Value pair arguments. Use this option with any of the input argument combinations in the previous syntaxes. Name,Value pair settings apply to all the lines plotted. To set options for individual lines, use the objects returned by fplot3.
fplot3(ax,___) plots into the axes object ax instead of the current axes gca.
example
fp = fplot3(___) returns a parameterized function line object. Use the object to query and modify properties of a specific parameterized line. For details, see ParameterizedFunctionLine Properties.
## Examples
### Plot 3-D Parametric Line
Plot the 3-D parametric line
$\begin{array}{c}x=\mathrm{sin}\left(t\right)\\ y=\mathrm{cos}\left(t\right)\\ z=t\end{array}$
over the default parameter range [-5 5].
syms t xt = sin(t); yt = cos(t); zt = t; fplot3(xt,yt,zt)
### Specify Parameter Range
Plot the parametric line
$\begin{array}{c}x={e}^{-t/10}\mathrm{sin}\left(5t\right)\\ y={e}^{-t/10}\mathrm{cos}\left(5t\right)\\ z=t\end{array}$
over the parameter range [-10 10] by specifying the fourth argument of fplot3.
syms t xt = exp(-t/10).*sin(5*t); yt = exp(-t/10).*cos(5*t); zt = t; fplot3(xt,yt,zt,[-10 10])
### Change Line Properties and Display Markers | {
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### Change Line Properties and Display Markers
Plot the same 3-D parametric curve three times over different intervals of the parameter. For the first curve, use a linewidth of 2. For the second, specify a dashed red line style with circle markers. For the third, specify a cyan, dash-dot line style with asterisk markers.
syms t fplot3(sin(t), cos(t), t, [0 2*pi], 'LineWidth', 2) hold on fplot3(sin(t), cos(t), t, [2*pi 4*pi], '--or') fplot3(sin(t), cos(t), t, [4*pi 6*pi], '-.*c')
### Plot 3-D Parametric Line Using Symbolic Functions
Plot the 3-D parametric line
$\begin{array}{c}x\left(t\right)=\mathrm{sin}\left(t\right)\\ y\left(t\right)=\mathrm{cos}\left(t\right)\\ z\left(t\right)=\mathrm{cos}\left(2t\right).\end{array}$
syms x(t) y(t) z(t) x(t) = sin(t); y(t) = cos(t); z(t) = cos(2*t); fplot3(x,y,z)
### Plot Multiple Lines on Same Figure
Plot multiple lines either by passing the inputs as a vector or by using hold on to successively plot on the same figure. If you specify LineSpec and Name-Value arguments, they apply to all lines. To set options for individual lines, use the function handles returned by fplot3.
Divide a figure into two subplots using subplot. On the first subplot, plot two parameterized lines using vector input. On the second subplot, plot the same lines using hold on.
syms t subplot(2,1,1) fplot3([t -t], t, [t -t]) title('Multiple Lines Using Vector Inputs') subplot(2,1,2) fplot3(t, t, t) hold on fplot3(-t, t, -t) title('Multiple Lines Using Hold On Command') hold off
### Modify 3-D Parametric Line After Creation
Plot the parametric line
$\begin{array}{l}x={e}^{-|t|/10}\mathrm{sin}\left(5|t|\right)\\ y={e}^{-|t|/10}\mathrm{cos}\left(5|t|\right)\\ z=t.\end{array}$
Provide an output to make fplot return the plot object.
syms t xt = exp(-abs(t)/10).*sin(5*abs(t)); yt = exp(-abs(t)/10).*cos(5*abs(t)); zt = t; fp = fplot3(xt,yt,zt) | {
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fp = ParameterizedFunctionLine with properties: XFunction: [1x1 sym] YFunction: [1x1 sym] ZFunction: [1x1 sym] Color: [0 0.4470 0.7410] LineStyle: '-' LineWidth: 0.5000 Show all properties
Change the range of parameter values to [-10 10] and the line color to red by using the TRange and Color properties of fp respectively.
fp.TRange = [-10 10]; fp.Color = 'r';
### Add Title and Axis Labels and Format Ticks
For $t$ values in the range $-2\pi$ to $2\pi$, plot the parametric line
$\begin{array}{c}x=t\\ y=t/2\\ z=\mathrm{sin}\left(6t\right).\end{array}$
Add a title and axis labels. Create the x-axis ticks by spanning the x-axis limits at intervals of pi/2. Display these ticks by using the XTick property. Create x-axis labels by using arrayfun to apply texlabel to S. Display these labels by using the XTickLabel property. Repeat these steps for the y-axis.
To use LaTeX in plots, see latex.
syms t xt = t; yt = t/2; zt = sin(6*t); fplot3(xt,yt,zt,[-2*pi 2*pi],'MeshDensity',30) view(52.5,30) xlabel('x') ylabel('y') title('x=t, y=t/2, z=sin(6t) for -2\pi < t < 2\pi') ax = gca; S = sym(ax.XLim(1):pi/2:ax.XLim(2)); ax.XTick = double(S); ax.XTickLabel = arrayfun(@texlabel, S, 'UniformOutput', false); S = sym(ax.YLim(1):pi/2:ax.YLim(2)); ax.YTick = double(S); ax.YTickLabel = arrayfun(@texlabel, S, 'UniformOutput', false);
### Create Animations
Create animations by changing the displayed expression using the XFunction, YFunction, and ZFunction properties and then by using drawnow to update the plot. To export to GIF, see imwrite.
By varying the variable i from 0 to 4π, animate the parametric curve
$\begin{array}{l}x=t+\mathrm{sin}\left(40t\right)\\ y=-t+\mathrm{cos}\left(40t\right)\\ z=\mathrm{sin}\left(t+i\right).\end{array}$
To play the animation, click the image.
syms t fp = fplot3(t+sin(40*t),-t+cos(40*t), sin(t)); for i=0:pi/10:4*pi fp.ZFunction = sin(t+i); drawnow end
## Input Arguments
collapse all | {
"domain": "mathworks.com",
"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9787126475856414,
"lm_q1q2_score": 0.8757417591632486,
"lm_q2_score": 0.894789457685656,
"openwebmath_perplexity": 8837.70278856056,
"openwebmath_score": 0.6026898622512817,
"tags": null,
"url": "https://de.mathworks.com/help/symbolic/fplot3.html"
} |
## Input Arguments
collapse all
Parametric input for x-axis, specified as a symbolic expression or function. fplot3 uses symvar to find the parameter.
Parametric input for y-axis, specified as a symbolic expression or function. fplot3 uses symvar to find the parameter.
Parametric input for z-axis, specified as a symbolic expression or function. fplot3 uses symvar to find the parameter.
Range of values of parameter, specified as a vector of two numbers. The default range is [-5 5].
Axes object. If you do not specify an axes object, then fplot3 uses the current axes.
Line style, marker, and color, specified as a character vector or string containing symbols. The symbols can appear in any order. You do not need to specify all three characteristics (line style, marker, and color). For example, if you omit the line style and specify the marker, then the plot shows only the marker and no line.
Example: '--or' is a red dashed line with circle markers
Line StyleDescription
-Solid line
--Dashed line
:Dotted line
-.Dash-dot line
MarkerDescription
'o'Circle
'+'Plus sign
'*'Asterisk
'.'Point
'x'Cross
'_'Horizontal line
'|'Vertical line
's'Square
'd'Diamond
'^'Upward-pointing triangle
'v'Downward-pointing triangle
'>'Right-pointing triangle
'<'Left-pointing triangle
'p'Pentagram
'h'Hexagram
ColorDescription
y
yellow
m
magenta
c
cyan
r
red
g
green
b
blue
w
white
k
black
### Name-Value Pair 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: 'Marker','o','MarkerFaceColor','red'
The properties listed here are only a subset. For a complete list, see ParameterizedFunctionLine Properties. | {
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"openwebmath_score": 0.6026898622512817,
"tags": null,
"url": "https://de.mathworks.com/help/symbolic/fplot3.html"
} |
Number of evaluation points, specified as a number. The default is 23. Because fplot3 uses adaptive evaluation, the actual number of evaluation points is greater.
Line color, specified as 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'
Here are the RGB triplets and hexadecimal color codes for the default colors MATLAB® uses in many types of plots.
[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'
Example: 'blue'
Example: [0 0 1]
Example: '#0000FF'
Line style, specified as one of the options listed in this table.
Line StyleDescriptionResulting Line
'-'Solid line
'--'Dashed line
':'Dotted line
'-.'Dash-dotted line
'none'No lineNo line | {
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"id": null,
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"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9787126475856414,
"lm_q1q2_score": 0.8757417591632486,
"lm_q2_score": 0.894789457685656,
"openwebmath_perplexity": 8837.70278856056,
"openwebmath_score": 0.6026898622512817,
"tags": null,
"url": "https://de.mathworks.com/help/symbolic/fplot3.html"
} |
'--'Dashed line
':'Dotted line
'-.'Dash-dotted line
'none'No lineNo line
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.
Marker symbol, specified as one of the values listed in this table. By default, the object does not display markers. Specifying a marker symbol adds markers at each data point or vertex.
ValueDescription
'o'Circle
'+'Plus sign
'*'Asterisk
'.'Point
'x'Cross
'_'Horizontal line
'|'Vertical line
'square' or 's'Square
'diamond' or 'd'Diamond
'^'Upward-pointing triangle
'v'Downward-pointing triangle
'>'Right-pointing triangle
'<'Left-pointing triangle
'pentagram' or 'p'Five-pointed star (pentagram)
'hexagram' or 'h'Six-pointed star (hexagram)
'none'No markers
Marker outline color, specified as 'auto', an RGB triplet, a hexadecimal color code, a color name, or a short name. The default value of 'auto' uses the same color as the Color property.
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. | {
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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.
[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'
Marker fill color, specified as 'auto', an RGB triplet, a hexadecimal color code, a color name, or a short name. The 'auto' value uses the same color as the MarkerEdgeColor property.
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 | {
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"id": null,
"lm_label": "1. YES\n2. YES",
"lm_name": "Qwen/Qwen-72B",
"lm_q1_score": 0.9787126475856414,
"lm_q1q2_score": 0.8757417591632486,
"lm_q2_score": 0.894789457685656,
"openwebmath_perplexity": 8837.70278856056,
"openwebmath_score": 0.6026898622512817,
"tags": null,
"url": "https://de.mathworks.com/help/symbolic/fplot3.html"
} |
'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.
[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'
Example: [0.3 0.2 0.1]
Example: 'green'
Example: '#D2F9A7'
Marker size, specified as a positive value in points, where 1 point = 1/72 of an inch.
## Output Arguments
collapse all
One or more parameterized line objects, returned as a scalar or a vector. You can use these objects to query and modify properties of a specific parameterized line. For details, see ParameterizedFunctionLine Properties.
### Topics
Introduced in R2016a
## Support
#### Mathematical Modeling with Symbolic Math Toolbox
Get examples and videos | {
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"openwebmath_score": 0.6026898622512817,
"tags": null,
"url": "https://de.mathworks.com/help/symbolic/fplot3.html"
} |
# How many possible passwords of a four digit length contain at least one uppercase character and at least one number?
How many possible passwords of a four digit length contain at least one uppercase character and at least one number?
• 95 total ACII Symbols
• 26 uppercase letters
• 26 lowercase letters
• 10 numbers
• 33 special characters
So I have calculated that there are $26$ possibilities for the uppercase letter digit and $10$ possibilities for the number digit and also a variation of all the other possibilites, thus:
$26 \cdot 10 \cdot |var_2([1,95])| = 26 \cdot 10 \cdot 95^2 =2 346 500$
Unfortunately this is the wrong solution. It should be $18700240$.
Question: Were is my mistake?
• Your mistake is that you counted the number of passwords where the first character is upper case, the second character is a number, and your remaining two characters are anything. You neglected to count the password "3aBc" as well as neglected to count the password "133T" etc... For a correct solution, apply inclusion-exclusion. There are $95^4$ total passwords. $(95-26)^4$ of which contain no capital letters so we remove those from the count. $(95-10)^4$ of which contain no numbers so we remove those too, but in doing so the passwords with neither got removed twice so we correct it by... – JMoravitz Mar 30 '17 at 8:44
Your mistake is that you counted the number of passwords where the first character is upper case, the second character is a number, and your remaining two characters are anything. You neglected to count the password "3aBc" as well as neglected to count the password "133T" etc...
For a correct solution, apply inclusion-exclusion. There are $95^4$ total passwords. $(95-26)^4$ of which contain no capital letters so we remove those from the count. $(95-10)^4$ of which contain no numbers so we remove those too, but in doing so the passwords with neither got removed twice so we correct it by adding that amount back in. | {
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"lm_q1q2_score": 0.8757417525282729,
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"openwebmath_perplexity": 365.107649185972,
"openwebmath_score": 0.710651695728302,
"tags": null,
"url": "https://math.stackexchange.com/questions/2209916/how-many-possible-passwords-of-a-four-digit-length-contain-at-least-one-uppercas"
} |
In more detail, let $\Omega$ be the set of passwords of length $4$ using these $96$ characters with no other restrictions. Let $A$ be the set of passwords with no capital letters. Let $B$ be the set of passwords with no numbers.
We are tasked with counting $|A^c\cap B^c|$, i.e. the set of passwords where there is at least one capital letter and there is at least one number.
By demorgan's, $A^c\cap B^c = (A\cup B)^c = \Omega\setminus(A\cup B)$
We have then by inclusion-exclusion $$|A^c\cap B^c|=|\Omega\setminus(A\cup B)|=|\Omega|-|A|-|B|+|A\cap B|$$
We find $|\Omega|=95^4$, $|A|=(95-26)^4$, $|B|=(95-10)^4$, and $|A\cap B|=(95-26-10)^4$ each by straightforward application of multiplication principle.
The final total is then
$$95^4-69^4-85^4+59^4$$
• What does the $c$ mean in $|A^c\cap B^c|$? – jublikon Mar 30 '17 at 9:38
• @jublikon the complementary set. Given a universal set $\Omega$ one has $A^c:= \{x\in\Omega~:~x\notin A\}=\Omega\setminus A$ – JMoravitz Mar 30 '17 at 9:48
For one, this does not account for the possible variations of order.
Here's a short way to compute this: Let $\Omega$ denote the set of all passwords, $U$ denote the set of passwords that include no uppercase letter, and $N$ the set of passwords that include no digit. Then, the set of passwords that include an uppercase letter and a number is $$\Omega - (N \cup C),$$ so the number of passwords satisfying the criteria is $$|\Omega - (N \cup C)| = |\Omega| - |N \cup C| .$$ Now, use the Inclusion-Exclusion principle to evaluate $|N \cup C|$.
You have worked out the number of passwords that have, say, the first characters an uppercase letter and the second character a number. This does not allow for cases where, say, our first two spaces are lowercase characters and/or special characters and our uppercase letter and number fill the last two spaces of the password. | {
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"openwebmath_perplexity": 365.107649185972,
"openwebmath_score": 0.710651695728302,
"tags": null,
"url": "https://math.stackexchange.com/questions/2209916/how-many-possible-passwords-of-a-four-digit-length-contain-at-least-one-uppercas"
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Because the question says "at least $1$" I would immediately be drawn to considering the the compliment case i.e. The passwords with no uppercase letters, no numbers or neither uppercase letters nor numbers.
$$\text{total unrestriced passwords}=95^4$$ $$\text{passwords with no uppercase letters}=(95-26)^4=69^4$$ $$\text{passwords with no numbers}=(95-10)^4=85^4$$ $$\text{passwords with no uppercase letters or numbers}=(95-26-10)^4=59^4$$
then by the principle of inclusion-exclusion we subtract the two complement cases from the total and add back their intersection:
$$\text{desired count}=95^4-(69^4+85^4)+59^4=18\,700\,240\tag{Answer}$$ | {
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# Convergent sequences and accumulation points
I am very confused about the relationship between convergent sequences and accumulation points.
Basically, my question boils down to: if $(a_n)_{n\in N}$ is a convergent sequence in $\mathbb{R}$ then,
(1) Does the set $\{a_n\}$ have exactly one accumulation point?
(2) If so, does $(a_n)_{n\in N}$ necessarily converge to the said accumulation point?
I'm tempted to say no to (1), but I'm afraid that I'm missing something. My counter-example to (1) is $\{a_n\} = \{ 4, 3, 2, 1, 0,0,0,...\}$ (i.e. inserting $0$s after the 4th element). Then the set has no accumulation point and it converges to 0. Is that correct?
If I'm right about (1), let's assume that $\{a_n\}$ has exactly one accumulation point. Then does $(a_n)_{n\in N}$ necessarily converge to the said accumulation point?
Lastly, is it possible that $\{a_n\}$ could have more than one accumulation point?
Update:
I define accumulation point as: Let $a$ be an accumulation point of $A$. Then $\forall \ \epsilon >0$, $B_{\epsilon}(a) \setminus \{a\}$ contains an element of $A$.
• What is your definition of accumulation point? – wjm Dec 4 '17 at 5:44
• @Raptor updated. – user1691278 Dec 4 '17 at 5:57
• The situation is quite simple. If a sequence $(a_n)_{n\in\Bbb N}$ is convergent, then every accumulation point of the set $\{\, a_n\mid n\in\Bbb N\,\}$ is equal to the limit of the sequence (easy proof using an extracted sequence). This implies there is at most one accumulation point, but not that such an accumulation point exists. – Marc van Leeuwen Dec 4 '17 at 9:52
I am calling $x$ an accumulation point of the set $A$ iff $(B(x,\epsilon) \cap A)\setminus \{x\} \neq \emptyset$ for all $\epsilon>0$.
Suppose $a_n \to a$.
The set $\{a_n\}_n$ can have at most one accumulation point which would have to be $a$. If $b \neq a$, then there is some $\epsilon>0$ such that $B(b,\epsilon)$ contains a finite number of points hence $b$ cannot be an accumulation point. | {
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Note that the sequence $a_n = 1$ has $\{a_n\}_n = \{1\}$ which has no accumulation points.
In general, the set $\{a_n\}_n$ will have $a$ as an accumulation point iff for all $N$ there is some $n \ge N$ such that $a_n \neq a$.
If $\{a_{n}\}$ has an accumulation point, say, $a$, and $(a_{n})$ is convergent. Then choose some $n_{1}$ such that $a_{n_{1}}\in B_{1}(a)-\{a\}$. Then choose some $n_{2}$ such that $B_{1/2}(a)-\{a,a_{1},...,a_{n_{1}}\}$, proceed in this way we have $a_{n_{k}}\rightarrow a$. Since $(a_{n})$ is convergent, one has $a_{n}\rightarrow a$.
Here I use the following definition:
$a$ is an accumulation point for $A$ if for every $\delta>0$, $(B_{\delta}(a)-\{a\})\cap A\ne\emptyset$.
And note that in the topology of ${\bf{R}}$, being such an accumulation point also implies that $(B_{\delta}(a)-\{a\})\cap A$ contains infinitely many points.
• If $\{a_{n}\}$ has two distinct accumulation points, say, $a$, $b$, one can find subsequences $(a_{n_{k}})$, $(a_{n_{l}})$ such that $a_{n_{k}}\rightarrow a$ and $a_{n_{l}}\rightarrow b$, a contradiction. – user284331 Dec 4 '17 at 5:54
• Your counterexample $(a_n)$ is not a convergent sequence though... – user1691278 Dec 4 '17 at 6:04
• I have edited, let's see if there is any mistake. – user284331 Dec 4 '17 at 6:13
The usual definition of an accumulation point (for a subset of $\mathbb{R}$) is as follows:
Let $A \subseteq \mathbb{R}$. We say that $a$ is an accumulation point of $A$ if for all $r > 0$ the set $B(a,r) \cap A \setminus \{a\}$ is nonempty. That is, every ball centered at $a$ contains a point of $A$ other than $a$ itself. | {
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Even if the sequence $(a_n)$ converges, the set $\{ a_n \}$ needn't have any accumulation points. For example, any constant set $\{a, a, a, \dotsc, \} = \{a\}$ does not have any accumulation points (as no ball centered at $a$ contains any point of the set other than $a$, but the sequence $(a,a,a,\dotsc)$ converges to $a$. On the other hand, if $\{a_n\}$ has an accumulation point, and $(a_n)$ is convergent, then the accumulation point is necessarily the limit.
• Thank you. Does that mean it cannot have more than one accumulation point since limits are unique? – user1691278 Dec 4 '17 at 6:05
• In $\mathbb{R}$ with the usual topology, a convergent sequence can have at most one accumulation point, yes. As noted above, it needn't have even that. – Xander Henderson Dec 4 '17 at 14:38
Here's a somewhat more general proof of this:
Let $X$ be a Hausdorff topological space, and let $\left\{x_n\right\}_{n\in\mathbb N}$ be a sequence in $X$ that converges to $x\in X$. Suppose that $y\in X$ is an accumulation point of $\left\{x_n\right\}_{n\in\mathbb N}$ such that $x\neq y$. Then, since $X$ is Hausdorff, there are disjoint open neighborhoods $U$ and $V$ of $x$ and $y$, respectively. Since $\left\{x_n\right\}_{n\in\mathbb N}$ converges to $x$, there is an $N\in\mathbb N$ such that $x_n\in U$ whenever $n\geq N$. However, this implies that there are at most $N-1$ elements of $\left\{x_n\right\}_{n\in\mathbb N}$ in $V$ different from $y$. Denote this set of finite elements by $\left\{y_n\right\}_{n=1}^m$. Again, since $X$ is Hausdorff, there are disjoint open neighborhoods $U_n$ and $V_n$ of $y_n$ and $y$, respectively. Then $\bigcap_{n=1}^m V_n$ is an open neighborhood of $y$ containing no elements of $\left\{x_n\right\}_{n\in\mathbb N}$, which is a contradiction. | {
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• In the context of this question, it is rather confusing to use braces to designate the sequence, especially when you then immediately use the very same notation for the set. OP rightly uses parentheses for the sequence, and braces for the set. – Marc van Leeuwen Dec 4 '17 at 9:56
Yes, you are right. If $x_n$ is a convergent sequence, then
• $\{x_n\}$ has an accumulations point if and only if $\{x_n\}$ is infinite.
• $\{x_n\}$ has at most one accumulation point and if it has one, then it coincides with the limit.
None of these statements is true for non-convergent squences. | {
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Find a multiplicative inverse…
Find a multiplicative inverse of $a=11$ modulo $m=13$.
What is this saying?
This seems like such a simple question, I just don't understand what it is asking for.
An additional question related to this I have is:
If $a$ has a multiplicative inverse modulo $m$, explain why $\gcd(a,m)=1$.
Is this also saying $ab \equiv 1 (mod \space m)$? Do $a$ and $m$ have to be prime since the $\gcd(a,m)=1$?
Thanks again.
-
It's asking to find $b$ such that
$$ab \equiv 1 \pmod m$$
Hint: Euclidean algorithm
Edit:
Suppose that $a$ has a multiplicative inverse mod $m$. That is, there is a number $b$ such that $ab = 1 + km$ (for some $k \in \mathbb{Z}$). Hence, $ab - km = 1$.
If $a$ and $m$ have a common divisor $d$, then $d$ is a divisor of $ba − km$ and hence of 1.
Therefore $d=1$, which means that $\gcd(a,m)=1$.
And, no $a$ and $m$ don't have to be prime! Just co-prime.
-
So, for $11b \equiv 1 (mod 13)$ we get $b=6$ This doesn't seem like the multiplicative inverse. Where do I go from here? Thanks. – student.llama Oct 16 '12 at 16:40
@student.llama No, that's it! The modular multiplicative inverse of 11 modulo 13 is 6. – Alex Oct 16 '12 at 16:42
That is the multiplicative inverse since $$11\cdot 6=66=1+5\cdot 13=1\pmod {13}$$ – DonAntonio Oct 16 '12 at 16:43
Oh, ok, thanks! Could you take a look at what I added to the OP, if you haven't already? – student.llama Oct 16 '12 at 16:47
@student.llama See my edit :) – Alex Oct 16 '12 at 16:59
$11^2=121\equiv 4\pmod{13}$
$11^3\equiv4\cdot 11\equiv 5$
$11^4=(11^2)^2\equiv 4^2=16\equiv 3$
$11^5\equiv3\cdot11=33\equiv 7$
$11^6\equiv7\cdot 11= 77=-1$
$\implies 11\cdot 11^5\equiv-1\implies (11)^{-1}\equiv -11^5\equiv -7\equiv 6\pmod{13}$
Alternatively, using convergent property of continued fraction,
$\frac{13}{11}=1+\frac 2{11}=1+\frac1{\frac{11}2}=1+\frac1{5+\frac12}$
So, the last but one convergent is $1+\frac15=\frac 6 5$ | {
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So, the last but one convergent is $1+\frac15=\frac 6 5$
So, $11\cdot 6-13\cdot 5=66-65=1\implies 11\cdot 6\equiv 1\pmod{13}$ $\implies (11)^{-1}\equiv 6\pmod{13}$
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# Need help with the general formula for the Taylor series of $e^x+\sin(x)$
A while ago, I asked about ways to derive the sigma notation for the infinite series of $$e^x+sin(x)$$.
$$e^x+\sin(x) = 1+2x+\dfrac{x^2}{2!}+\dfrac{x^4}{4!}+\dfrac{2x^5}{5!}+\dfrac{x^6}{6!}+\dfrac{x^8}{8!}+\dfrac{2x^9}{9!}+...$$
Clive Newstead brilliantly gave me two formulas.
There, he said that the pattern of the series, regarding the coefficient of the numerators, is as followed: 1,2,1,0,1,2,1,0
He said that the pattern is "is periodic with period 4"
This comment still puzzles me. I have tried to validate his formula and it is true, they are all correct, but I don't know why he can come up with succinct representation of the pattern:
$$\sum_{k=0}^{\infty} \left( \dfrac{x^{4k}}{(4k)!} + \dfrac{2x^{4k+1}}{(4k+1)!} + \dfrac{x^{4k+2}}{(4k+2)!} \right)$$
How does one realize that the power is related to $$4k, 4k+1$$ and $$4k+2$$ and $$4k+3$$?
This seems to relate to finding the general formula for a sequence or series of number, for example, trivially
$$1, 3, 5, 7, ...$$ can be represented as $$2k+1$$
$$0, 2, 4, 6, ...$$ can be represented as $$2k$$
with $$k$$ as the position of the term in the sequence.
So my thought is you to count the terms that have the numerator's coefficient as $$1$$, $$2$$ and $$0$$
The ones that have $$1$$ is $$1^{st}$$, $$3^{sd}$$, $$5^{th}$$, $$7^{th}$$, etc.
The ones that have $$2$$ is $$2^{sd}$$, $$6^{th}$$, $$10^{th}$$, $$14^{th}$$, etc.
The ones that have $$0$$ is $$4^{th}$$, $$8^{th}$$, $$12^{th}$$, $$16^{th}$$, etc.
So the difference between the positions of the terms that contain the coefficient $$1$$ is $$2$$
So the difference between the positions of the terms that contain the coefficient $$2$$ is $$4$$
So the difference between the positions of the terms that contain the coefficient $$0$$ is $$4$$
I find that the general term for terms that contain $$1$$ is $$2k+1$$
I find that the general term for terms that contain $$2$$ is $$4k-2$$ | {
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I find that the general term for terms that contain $$2$$ is $$4k-2$$
I find that the general term for terms that contain $$0$$ is $$4k$$
But I don't know how to derive these into $$4k$$, $$4k+1$$, $$4k+2$$, $$4k+3$$
I realize that this is similar to finding patterns of a sequence on an IQ test, but I don't know how to derive it. One my friend said I should look into Lagrange's interpolation, but I don't why this technique has anything to do with writing down the general terms for sequences and series.
Could you help me on this?
Methinks, this is a simple question. I am not good at writing down general term of a sequence yet.
• note that the coefficients of the Maclaurin series for $\sin x$ are periodic with period $4$ – J. W. Tanner Jan 24 at 22:33
• The general formula for $\sin(x)$ is $$\sum_{k=0}^{\infty}\dfrac{x^{2n+1}}{(2n+1)!}$$, the coefficients of the numerators are all ones, so I don't quite understand what are you trying to say? Sorry, could you clarify more? – James Warthington Jan 24 at 23:17
• I realize that he breaks terms that contain $1$ in the numerators into 2, $4k$ and $4k+2$, quite clever. – James Warthington Jan 24 at 23:23
• @JamesWarthington: That is the formula for $\sinh x$, not $\sin x$. – TonyK Jan 24 at 23:39
• Yeah, thanks, I forgot to add $(-1)^{n}$ – James Warthington Jan 24 at 23:44
$$e^x = 1 + x + \frac {x^2}{2} + \frac {x^3}{3!} + \cdots$$
It seem pretty natural to write this as
$$e^x = \sum_\limits{n=0}^\infty \frac {x^n}{n!}$$
But we could look at pairs of terms.
$$e^x = \sum_\limits{n=0}^\infty \left(\frac {x^{2n}}{2n!}+\frac {x^{2n+1}}{(2n+1)!}\right)$$
Or, even triples of terms.
$$e^x = \sum_\limits{n=0}^\infty \left(\frac {x^{3n}}{3n!}+\frac {x^{3n+1}}{(3n+1)!}+\frac {x^{3n+2}}{(3n+3)!}\right)$$
And do the same thing with $$\sin x$$
$$\sin x = \sum_\limits{n=0}^\infty (-1)^n \frac {x^{2n+1}}{(2n+1)!} = \sum_\limits{n=0}^\infty \left(\frac {x^{4n+1}}{(4n+1)!} - \frac {x^{4n+3}}{(4n+3)!}\right)$$ | {
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Now find a representation of $$e^x$$ that plays nicely with our representation of $$\sin x$$ add them together and you have what you show above.
• Thanks, very instructive.I've understood! – James Warthington Jan 24 at 23:34
Remember that for the series centered around the point b, the coefficient $$a_{n} = \frac{f^{(n)}(b)}{n!}$$. So if $$f^{(4)}(x) = f(x)$$ then $$f^{(n)}(b)$$ is periodic with not necessarily prime period 4.
This is why you can be sure that the $$f^{(n)}(b)$$ is periodic with period 4 in your case.
Let $$f(x) = e^{x}$$, then $$f'(x) = e^{x}$$, and so continuing to take derivatives $$f^{(4)}(x)=e^{x}$$.
Let $$g(x) = sin(x)$$ then $$g''(x)=-sin(x)$$ and so $$g^{(4)}(x) = sin(x)$$
So for $$h(x)=f(x) + g(x)$$ we have $$h^{(4)}(x) = (f(x)+g(x))'''' = f^{(4)}(x) + g^{(4)}(x) = f(x) + g(x) = h(x)$$.
In this way you can see that the $$f^{(n)}(b)$$ has the period 4 structure.
• What if, by continuously taking the derivatives, the function we are dealing with do not have periodic derivatives? For example $\tan(x)$? – James Warthington Jan 25 at 0:06
• Insightful comment, thank you, BTW! :) – James Warthington Jan 25 at 0:07
• @JamesWarthington Then the coefficients will likely not have this periodic form, Of course what they do will depend on the function you are working with, they could still have some sort of pattern. – open problem Jan 25 at 0:09
• So I guess it is not always possible to write a general formula for all power series? Am I correct? Or is it otherwise? – James Warthington Jan 25 at 0:11
• To do so you will need a formula for the evaluation of the nth derivative of the function at the point you have centered your power series around. – open problem Jan 25 at 0:14
The mod $$4$$ periodicity can perhaps be most easily understood using the formula $$e^{ix}=\cos x+i\sin x$$, so that $$\sin x={e^{ix}-e^{-ix}\over2i}$$, which gives
$$e^x+\sin x={1\over2i}\sum_{n=0}^\infty{(2i+i^n-(-i)^n)x^n\over n!}$$ | {
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$$e^x+\sin x={1\over2i}\sum_{n=0}^\infty{(2i+i^n-(-i)^n)x^n\over n!}$$
Since $$i^4=1$$, the coefficients $$2i+i^n-(-i)^n$$ cycle through the values
\begin{align} 2i+i^0-(-i)^0&=2i+1-1=2i\\ 2i+i^1-(-i)^1&=2i+i-(-i)=4i\\ 2i+i^2-(-i)^2&=2i-1-(-1)=2i\\ 2i+i^3-(-i)^3&=2i-i-i=0 \end{align} | {
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# How to Handle Stronger Induction Hypothesis - Strong Induction
I'm having trouble understanding strong induction proofs
I understand how to do ordinary induction proofs and I understand that strong induction proofs are the same as ordinary with the exception that you have to assume that the theorem holds for all numbers up to and including some $n$ (starting at the base case) then we try and show: theorem holds for $n+1$.
How do you show this exactly.
Here is a proof by induction:
Thm: $n≥1, 1+6+11+16+\dots+(5n-4) = (n(5n-3))/2$
Proof (by induction)
Basis step: for $n=1: 5-4=(5-3)/2 \Rightarrow 1=1$. The basis step holds
Induction Step: Suppose that for some integer $k≥1$, $$1+6+11+16+...+(5k-4) = \frac{k(5k-3)}{2} \qquad\text{(inductive hypothesis)}$$
Want to show: $$1+6+11+16+...+(5k-4)+(5(k+1)-4) = \frac{(k+1)(5(k+1)-3)}{2}$$ so $$\frac{k(5k-3)}{2} + (5(k+1)-4) = \frac{(k+1)(5(k+1)-3)}{2}$$
then you just show that they are equal.
So how can I do the same proof using strong induction? What are the things I need to add/change in order for this proof to be a strong induction proof?
You wrote:
i understand how to do ordinary induction proofs and i understand that strong induction proofs are the same as ordinary with the exception that you have to show that the theorem holds for all numbers up to and including some n (starting at the base case) then we try and show: theorem holds for $n+1$
No, not at all: in strong induction you assume as your induction hypothesis that the theorem holds for all numbers from the base case up through some $n$ and try to show that it holds for $n+1$; you don’t try to prove the induction hypothesis. | {
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In your example the simple induction hypothesis that the result is true for $n$ is already enough to let you prove that it’s true for $n+1$, so there’s neither need nor reason to use a stronger induction hypothesis. The proof by ordinary induction can be seen as a proof by strong induction in which you simply didn’t use most of the induction hypothesis.
I suggest that you read this question and my answer to it and see whether that clears up some of your confusion; at worst it may help you to pinpoint exactly where you’re having trouble.
Added: Here’s an example of an argument that really does want strong induction. Consider the following solitaire ‘game’. You start with a stack of $n$ pennies. At each move you pick a stack that has at least two pennies in it and split it into two non-empty stacks; your score for that move is the product of the numbers of pennies in the two stacks. Thus, if you split a stack of $10$ pennies into a stack of $3$ and a stack of $7$, you get $3\cdot7=21$ points. The game is over when you have $n$ stacks of one penny each.
Claim: No matter how you play, your total score at the end of the game will be $\frac12n(n-1)$.
If $n=1$, you can’t make any move at all, so your final score is $0=\frac12\cdot1\cdot0$, so the theorem is certainly true for $n=1$. Now suppose that $n>1$ and the theorem is true for all positive integers $m<n$. (This is the strong induction hypothesis.) You make your first move; say that you divide the pile into a pile of $m$ pennies and another pile of $n-m$ pennies, scoring $m(n-m)$ points. You can now think of the rest of the game as splitting into a pair of subgames, one starting with $m$ pennies, the other with $n-m$. | {
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Since $m<n$, by the induction hypothesis you’ll get $\frac12m(m-1)$ points from the first subgame. Similarly, $n-m<n$, so by the induction hypothesis you’ll get $\frac12(n-m)(n-m-1)$ points from the second subgame. (Note that the two subgames really do proceed independently: the piles that you create in one have no influence on what you can do in the other.)
Your total score is therefore going to be
$$m(n-m)+\frac12m(m-1)+\frac12(n-m)(n-m-1)\;,$$
which (after a bit of algebra) simplifies to $\frac12n(n-1)$, as desired, and the result follows by (strong) induction. | {
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• A question related to "No, not at all: in strong induction you ... from the base case up through some n ...". Agree with this statement, but I usually see proofs describing it without considering the base case, for example: "Assume $P(x)$ holds $\forall x, x\le i$", so strictly speaking this kind of statement is incorrect? (The correct one should change the range to $b\le x\le i$, where $b$ is the base?) Jul 23, 2019 at 8:19
• Nice strong induction example! But BTW here is (IMHO) an even nicer non-induction solution. Model the coins as the vertices of a graph, two vertices adjacent iff the coins are in the same stack. Initially all coins are in one stack so we have the complete graph on $n$ vertices. Splitting into stacks of say $a$ and $b$ coins means removing all the edges between the remaining graphs $K_a$ and $K_b$; the number of edges removed is $ab$, which is the score for this step. Completing the process means removing all $n(n-1)/2$ original edges, so this is the final score. Apr 28 at 7:43
• @David: That is indeed an elegant solution. And a student who looks closely can even see the structure of the strong induction proof in a sense ‘hiding’ in it: it could be recast as essentially the same induction, but in this form we can actually see the inevitable end result without having to go through the induction. Apr 28 at 17:23
• Thanks @Brian! The graph theory solution also makes it clear why the score for a move (in a sense) has to be the product in order for this problem to work. Apr 28 at 22:57 | {
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# How do I solve $x^4 -2x^3-6x^2-2x+1=0$? [closed]
$$x^4 -2x^3-6x^2-2x+1=0$$ $\left[\text{Hint let } v = x + \frac{1}{x}\right]$
I am stumped, and have no idea how to proceed. I have tried solving it, but have had no success.
P.S: This question is meant to be solved, using only techniques for solving quadratic equations.
• Do you mean "find the roots" or something like that? Have you searched for rational roots? Note: there is an obvious rational root. – lulu Jul 22 '17 at 15:56
• – lab bhattacharjee Jul 22 '17 at 15:57
• Ok, now there is a question. In which case my hint is valid, and there is no need for the substitution. – lulu Jul 22 '17 at 15:59
• Doesn't "solve" mean "find the roots"? A difference in language? – Tobi Alafin Jul 22 '17 at 16:38
• A simplistic way to find the roots is to graph the polynomial, look to see if there are any obvious rational roots, then divide the polynomial by the factors corresponding to rational roots ($x+1$ in this case). Repeat as many times as you can. – Χpẘ Jul 22 '17 at 19:07
I think you meant for $$x^4 -2x^3-6x^2-2x+1=0$$
Obviously $x=0$ is not a solution. So we divide by $x^2$ and get
$$x^2 -2x-6-2x^{-1}+x^{-2}=0$$ $$(x^2+2+x^{-2}) - 2(x+x^{-1}) -8=0$$ $$v^2-2v-8=0$$ $$(v-4)(v+2)=0$$
• Thanks for the pointer. :) – Tobi Alafin Jul 22 '17 at 16:39
• @TobiAlafin you're welcome :) – Yujie Zha Jul 22 '17 at 17:15
• I'm just curious: how did you know to divide by $x^2$? Obviously the $v$ contained a $x^{-1}$ term, but one could make that pattern appear in a variety of ways… – gen-ℤ ready to perish Jul 24 '17 at 7:36
• @ChaseRyanTaylor Just observed that equation has $x$ interns with power of 4,3,2,1,0, and thus it would make it symmetrical to divide by $x^2$ and make it to be $x$ items with power of 2,1,0,-1,-2.. – Yujie Zha Jul 24 '17 at 11:05
• Care explaining the downvote? – Yujie Zha Jul 24 '17 at 12:45 | {
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since $x=0$ is not a solution we can divide by $x^2$ $$x^2+\frac{1}{x^2}-2\left(x+\frac{1}{x}\right)-6=0$$ and no set $$t=x+\frac{1}{x}$$ then you will get $$t^2=x^2+\frac{1}{x^2}+2$$ and $$t^2-2=x^2+\frac{1}{x^2}$$ and you have to solve $$t^2-2t-8=0$$
I would not bother with the substitution. The only possible rational roots are $\pm 1$ and it is easy to check that $-1$ works. A quick calculation shows that your polynomial is $$(x + 1)^2 \,(x^2 - 4 x + 1)$$ and we are done.
• Aren't solutions generally defined on $\mathbb{R}$, and not on $\mathbb{Q}$? – Tobi Alafin Jul 22 '17 at 17:29
• Not following. I imagined you were looking for complex solutions, no? In any case, searching for rational solutions is a very easy and natural first step...as that solves the problem entirely (well, up to a trivial quadratic) then I don't see any point in looking for a more elaborate solution. – lulu Jul 22 '17 at 17:35
• To be clear: there was no a priori reason why searching for rational roots should have solved the problem. But then, there was no a priori reason why this particular substitution should have solved it. For what it's worth: In general, there is a closed formula for the roots of quartics (granted, it isn't very nice). – lulu Jul 22 '17 at 17:42
$x^4 -2x^3-6x^2-2x+1=0$
Divide all terms by $x^2$
$x^2-2x-6-\dfrac{2}{x}+\dfrac{1}{x^2}=0$
Rearrange in this way
$\left(x^2+\dfrac{1}{x^2}\right)-2\left(x+\dfrac{1}{x}\right)-6=0$
Now set $v=x+\dfrac{1}{x}$
squaring you get
$v^2=x^2+\dfrac{1}{x^2}+2\to x^2+\dfrac{1}{x^2}=v^2-2$
Plug in the equation
$v^2-2 -2v -6=0\to v^2-2v-8=0\to v_1=4;\;v_2=-2$
As we want to solve for $x$ two more steps
For $v=4\to x+\dfrac{1}{x}=4 \to x^2-4x+1=0 \to x= 2\pm\sqrt{3}$
for $v=-2\to x+\dfrac{1}{x}=-2\to x^2+2x+1\to x=-1$
Sometimes you can add and subtract the right thing and land in Pascal's triangle. The 4th row of Pascal's triangle is $1 4 6 4 1$, so add and subtract $6x^2+12x^2+6x$ to get | {
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$$x^4+4x^3+6x^2+4x+1 - 6x^3-12x^2-6x = 0$$
or
$$(x+1)^4 - 6x(x+1)^2=0.$$
Factor out $(x+1)^2$ to get
$$(x+1)^2((x+1)^2 - 6x) = 0$$
or $$(x+1)^2(x^2-4x+1)=0.$$ | {
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# how many unique patterns exist for a $N\times N$ grid
I'm trying to figure out if there is a way to determine how many unique patterns exist for a given $N\times N$ grid if you choose N points on the grid. For example, for a $2\times 2$ grid we can get two unique patterns from the six possible combinations. The rest are just rotations and mirrors of the two unique patterns below
[x] [x]
[ ] [ ]
and
[x] [ ]
[ ] [x]
Is there a mathematical way of determining a unique number of patterns for a NxN grid where N=3,4,5,6,7,8?
I figured for a 3x3, there are 14 unique patterns for picking 3 random points on the grid, but it gets tedious after that.
N: N^2 : N^2 Choose N Unique pattern
2 4 6 2
3 9 84 14
4 16 1820 ????
5 25 53130 ????
6 36 1947792 ????
7 49 85900584 ????
• one must choose $N$ points on a binary grid? – nbubis Nov 17 '13 at 4:46
• yes, forgot to add that in – user109380 Nov 17 '13 at 4:52
• Are you familiar with group theory? Polya enumeration could be the way to go here. – Gerry Myerson Nov 17 '13 at 5:26
We can actually do a bit more and compute the cycle index $Z(G_N)$ for general $N$, where we have to distinguish between $N$ even and $N$ odd. This will permit lookup in the OEIS, which in turn leads to more material about this interesting problem.
We proceed to enumerate the permutations of $G_N$ by their cycle structure. For $N$ even, we get the identity, which contributes $$a_1^{N^2}.$$ There is a vertical reflection, which contributes $$a_2^{N^2/2},$$ the same for a horizontal reflection, i.e. $$a_2^{N^2/2}.$$ The reflection in the rising diagonal contributes $$a_1^N a_2^{(N^2-N)/2},$$ the same for the other diagonal, i.e. $$a_1^N a_2^{(N^2-N)/2}.$$ | {
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What remains are the rotations. Two of these contribute (recall that $N$ is even) $$2\times a_4^{N^2/4}$$ and one of them, $$a_2^{N^2/2}.$$ This gives for even $N$ the cycle index $$Z(G_N) = \frac{1}{8} \left( a_1^{N^2} + 3 a_2^{N^2/2} + 2 a_1^N a_2^{(N^2-N)/2} + 2 a_4^{N^2/4}\right).$$ For $N$ odd, we get the identity, which is $$a_1^{N^2}.$$ The two reflections now contribute $$2\times a_1^N a_2^{(N^2-N)/2}.$$ The reflection in the two diagonals are unchanged and contribute $$2\times a_1^N a_2^{(N^2-N)/2}$$ What remains is the rotations, two of which have cycle structure $$2\times a_1 a_4^{(N^2-1)/4}$$ and the last one has cycle structure $$a_1 a_2^{(N^2-1)/2}.$$
This gives for odd $N$ the cycle index $$Z(H_N) = \frac{1}{8} \left( a_1^{N^2} + 4 a_1^N a_2^{(N^2-N)/2} + 2 a_1 a_4^{(N^2-1)/4}+a_1 a_2^{(N^2-1)/2}\right).$$ Evidently for $N$ marks being placed on the grid we seek to compute $$[z^N] Z(G_N)(1+z) \quad\text{and}\quad [z^N] Z(H_N)(1+z),$$ alternating between the two for $N$ even and $N$ odd.
This produces the sequence $$1, 2, 16, 252, 6814, 244344, 10746377, 553319048, 32611596056, 2163792255680,\ldots$$ which is A019318 from the OEIS.
Here is the Maple program that was used to compute these cycle indices.
with(numtheory);
with(group):
with(combinat):
pet_varinto_cind :=
proc(poly, ind)
local subs1, subs2, polyvars, indvars, v, pot, res;
res := ind;
polyvars := indets(poly);
indvars := indets(ind);
for v in indvars do
pot := op(1, v);
subs1 :=
[seq(polyvars[k]=polyvars[k]^pot,
k=1..nops(polyvars))];
subs2 := [v=subs(subs1, poly)];
res := subs(subs2, res);
od;
res;
end;
G :=
proc(N)
if type(N,odd) then return FAIL; fi;
1/8*(a[1]^(N^2)+3*a[2]^(N^2/2)+
2*a[1]^N*a[2]^((N^2-N)/2) + 2*a[4]^(N^2/4));
end;
H :=
proc(N)
if type(N,even) then return FAIL; fi;
1/8*(a[1]^(N^2)+4*a[1]^N*a[2]^((N^2-N)/2)+
a[1]*a[2]^((N^2-1)/2) + 2*a[1]*a[4]^((N^2-1)/4));
end;
v :=
proc(N)
option remember;
local p, k, gf; | {
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v :=
proc(N)
option remember;
local p, k, gf;
if type(N, even) then
gf := expand(pet_varinto_cind(1+z, G(N)));
else
gf := expand(pet_varinto_cind(1+z, H(N)));
fi;
coeff(gf, z, N);
end;
Here is another interesting MSE cycle index computation I. This MSE cycle index computation II is relevant also.
• I want to ask about the notation $Z(G_N)(1+z).$ I assume this means that $a_1$ gets replaced by $1+z,$ $a_2$ gets replaced by $1+z^2,$ and $a_4$ gets replaced by $1+z^4,$ but I wanted to double check. – Will Orrick Jan 1 '14 at 15:27
• That is correct. Wikipedia has useful data on this: Cycle Index and Polya Enumeration Theorem. – Marko Riedel Jan 1 '14 at 15:46
You're looking for the number of orbits of the dihedral group $D_N$ acting on the $N\times N$ grid with $N$ chosen cells. (Caution: This group is often expressed as $D_{2N}$ as it has $2N$ elements.)
You want to use the Cauchy-Frobenius-Burnside Lemma (aka Burnside Lemma, aka "Not Burnside Lemma") to compute this. That is, you have a group action of $D_N$ on the set $\Omega$ of all such patterns ($N\times N$ grids with $N$ marked cells), and the number of orbits in this action (which is what you want) is then given by $$\frac{1}{2N}\sum_{g\in G} |\Omega^g|$$ where $|\Omega^g|$ is the number of elements in $\Omega$ fixed by $g$.
• and I agree with Gerry (above) that Polya enumeration is an efficient way of doing this. (See also, cycle index.) – Doc Nov 17 '13 at 6:03
• For the uninitiated (and probably the OP as well), would you like to present the final formula? – nbubis Nov 17 '13 at 6:34
• @nbubis, by "formula" I'm interpreting that you mean a closed form for the answer. I doubt this is even possible. However, your comment inspired me to cook up an example (below) that hopefully clarifies most of the procedure. – Doc Nov 17 '13 at 16:50 | {
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Consider the action of the group $G$ on a set $\Omega$ of cardinality $n$. The cycle index of $(G,\Omega)$ is the polynomial defined by $$Z_G(X_1,X_2,\dots, X_n)=\frac{1}{|G|} \sum_{g\in G} X_1^{c_1(g)}X_2^{c_2(g)}\cdots X_n^{c_n(g)}$$ where $c_i(g)$ denotes the number of $i$-cycles in the cycle decomposition of the element $g$ acting on $\Omega$.
For example, if we consider the action of $D_4$ on the square, we would first label the vertices of the square by $\{1,2,3,4\}$ in some order. Let's do this in cyclic clockwise order, starting from the upper left corner. Then we get the following table. \begin{array}{c|c|c} g & \mbox{permutation} & \mbox{cycle index term}\\ \hline \mbox{identity} & (1)(2)(3)(4) & X_1^4 \\ 90^o \mbox{rotation} & (1,2,3,4) & X_4^1 \\ 180^o \mbox{rotation} & (1,3)(2,4) & X_2^2 \\ 270^o \mbox{rotation} & (1,4,3,2) & X_4^1 \\ \mbox{horizontal reflection} & (1,4)(2,3) & X_2^2\\ \mbox{vertical reflection} & (1,2)(3,4) & X_2^2\\ \mbox{diagonal reflection} & (1,3)(2)(4) & X_1^2X_2\\ \mbox{opposite diag reflection} & (2,4)(1)(3)& X_1^2X_2 \end{array}
Thus we obtain as cycle index $$Z_G(X_1,X_2,\dots, X_n)=\frac{1}{8}(X_1^4 + 2X_1^2X_2+3X_2^2+2X_4).$$
Now consider $$Z_G(b+1,b^2+1,b^3+1,b^4+1).$$ The coefficient of $b^i$ will give the number (up to isomorphism!) of positions in which there are precisely $i$ marked squares. Returning to the grid problem, we are only interested in the coefficient of the $b^2$ term. This turns out to be $\frac{1}{8}(16) = 2$.
Here's an answer that leads to explicit formulas whose most complicated term is a sum over a product of binomial coefficients. I haven't yet made any attempt to connect it to the excellent solutions in the answers given earlier. | {
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First some notation. The dihedral group of the square, $D_4,$ has elements $\{e,\rho,\rho^2,\rho^3,h,v,d_1,d_2\},$ where $e$ is the identity, $\rho$ is the $90^\circ$ rotation, $h$ is the reflection about the horizontal axis, $v$ is the reflection about the vertical axis, $d_1$ is the reflection about the forward diagonal, and $d_2$ is the reflection about the backward diagonal.
(Added: Having never learned Pólya enumeration properly, I made my answer overly complicated. I leave my previous answer, but add a more direct route to equation (**) which appears below. In this problem we have $D_4$ acting on the set $X,$ consisting of the $\binom{n^2}{n}$ ways of coloring $n$ small squares in the $n\times n$ grid black, leaving the remaining squares white. The orbit counting theorem leads immediately to (**) since $$K(n)=\lvert\mathcal{O}\rvert=\frac{1}{8}\sum_{g\in D_4}C(n,n,\langle g\rangle)$$ is identical to (**) when one realizes that $\langle\rho\rangle=\langle\rho^3\rangle=C_4.$ The orbit counting theorem is described in many places. For convenience I've appended some background information about it to the end of this post.)
Since the order of $D_4$ is $8,$ all subgroups are of order $1,$ $2,$ $4,$ or $8.$ Either of the two elements of order $4,$ $\rho$ and $\rho_3,$ generates $C_4,$ the rotation group of the square. Each of the five elements of order $2$, $\rho^2,$ $h,$ $v,$ $d_1,$ and $d_2,$ generates a subgroup of order $2.$ There are two additional subgroups of order $4,$ isomorphic to $C_2\times C_2.$ They are $\{e,\rho^2,h,v\}$ and $\{e,\rho^2,d_1,d_2\}.$ Any other group generated by two elements of order $2$ is all of $D_4.$
There are therefore $10$ subgroups with inclusions given by the following Hasse diagram. | {
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There are therefore $10$ subgroups with inclusions given by the following Hasse diagram.
Define $C(n,b,G)$ to be the set of configurations of the $n\times n$ grid with $b$ marked squares that are invariant under the group $G.$ We are ultimately interested in $b=n,$ but will start out with slightly greater generality. Define $\overline{C}(n,b,G)$ to be the subset of $C(n,b,G)$ consisting of elements that are not invariant under any group that properly contains $G.$ The number we wish to compute, which we denote $K(n),$ is \begin{aligned} (*)\quad K(n)=&\frac{\lvert\overline{C}(n,n,D_4)\rvert}{1}+\frac{\lvert\overline{C}(n,n,\{e,\rho^2,h,v\})\rvert}{2}+\frac{\lvert\overline{C}(n,n,C_4)\rvert}{2}+\frac{\lvert\overline{C}(n,n,\{e,\rho^2,d_1,d_2\})\rvert}{2}\\ &+\frac{\lvert\overline{C}(n,n,\{e,h\})\rvert}{4}+\frac{\lvert\overline{C}(n,n,\{e,v\})\rvert}{4}+\frac{\lvert\overline{C}(n,n,\{e,\rho^2\})\rvert}{4}+\frac{\lvert\overline{C}(n,n,\{e,d_1\})\rvert}{4}\\ &+\frac{\lvert\overline{C}(n,n,\{e,d_2\})\rvert}{4}+\frac{\lvert\overline{C}(n,n,\{e\})\rvert}{8}. \end{aligned} | {
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(Added: The set $C(n,n,G)$ is the set of colorings whose stabilizer contains $G,$ while $\overline{C}(n,n,G)$ is the set of colorings whose stabilizer equals $G.$ The formula (*) for $K(n),$ the number of orbits, comes from counting each coloring with a weight equal to the reciprocal of the size of its orbit, the weight having been computed using the orbit-stabilizer theorem. (See the end of the post.) For example, if $x$ has stabilizer $D_4,$ then the weight is $1/1=8/8;$ if $x$ has stabilizer $\{e,\rho^2,h,v\},$ then the weight is $1/2=4/8.$ Of course, as noted above, the use of (*) and the inclusion-exclusion argument below is rendered superfluous by the use of the orbit counting theorem. It is interesting that although (*) involves all ten subgroups of $D_4,$ the final expression (**) involves only the seven cyclic subgroups. My naive derivation does not explain this, but the orbit counting theorem makes it manifest that that must be the case.) | {
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From the diagram, we obtain expressions for the $\overline{C}(n,b,G)$ in terms of the $C(n,b,G):$ \begin{aligned} \lvert\overline{C}(n,b,D_4)\rvert&=\lvert C(n,b,D_4)\rvert\\ \vert\overline{C}(n,b,\{e,\rho^2,h,v\})\rvert&=\lvert C(n,b,\{e,\rho^2,h,v\})\rvert- \lvert C(n,b,D_4)\rvert\\ \vert\overline{C}(n,b,C_4)\rvert&=\lvert C(n,b,C_4)\rvert- \lvert C(n,b,D_4)\rvert\\ \vert\overline{C}(n,b,\{e,\rho^2,d_1,d_2\})\rvert&=\lvert C(n,b,\{e,\rho^2,d_1,d_2\})\rvert- \lvert C(n,b,D_4)\rvert\\ \vert\overline{C}(n,b,\{e,h\})\rvert&=\lvert C(n,b,\{e,h\})\rvert- \lvert C(n,b,\{e,\rho^2,h,v\})\rvert\\ \vert\overline{C}(n,b,\{e,v\})\rvert&=\lvert C(n,b,\{e,v\})\rvert- \lvert C(n,b,\{e,\rho^2,h,v\})\rvert\\ \vert\overline{C}(n,b,\{e,d_1\})\rvert&=\lvert C(n,b,\{e,d_1\})\rvert- \lvert C(n,b,\{e,\rho^2,d_1,d_2\})\rvert\\ \vert\overline{C}(n,b,\{e,d_2\})\rvert&=\lvert C(n,b,\{e,d_2\})\rvert- \lvert C(n,b,\{e,\rho^2,d_1,d_2\})\rvert\\ \vert\overline{C}(n,b,\{e,\rho^2\})\rvert&=\lvert C(n,b,\{e,\rho^2\})\rvert- \lvert \overline{C}(n,b,\{e,\rho^2,h,v\})\rvert- \lvert \overline{C}(n,b,C_4)\rvert\\ &\quad- \lvert \overline{C}(n,b,\{e,\rho^2,d_1,d_2\})\rvert- \lvert \overline{C}(n,b,D_4)\rvert\\ &=\lvert C(n,b,\{e,\rho^2\})\rvert- \lvert C(n,b,\{e,\rho^2,h,v\})\rvert- \lvert C(n,b,C_4)\rvert\\ &\quad- \lvert C(n,b,\{e,\rho^2,d_1,d_2\})\rvert+2 \lvert C(n,b,D_4)\rvert\\ \vert\overline{C}(n,b,\{e\})\rvert&=\lvert C(n,b,\{e\})\rvert- \lvert \overline{C}(n,b,\{e,h\})\rvert- \lvert \overline{C}(n,b,\{e,v\})\rvert- \lvert \overline{C}(n,b,\{e,\rho^2\})\rvert\\ &\quad- \lvert \overline{C}(n,b,\{e,d_1\})\rvert- \lvert \overline{C}(n,b,\{e,d_2\})\rvert- \lvert \overline{C}(n,b,\{e,\rho^2,h,v\})\rvert\\ &\quad- \lvert \overline{C}(n,b,C_4)\rvert- \lvert \overline{C}(n,b,\{e,\rho^2,d_1,d_2\})\rvert- \lvert \overline{C}(n,b,D_4)\rvert\\ &=\lvert C(n,b,\{e\})\rvert- \lvert C(n,b,\{e,h\})\rvert- \lvert C(n,b,\{e,v\})\rvert- \lvert C(n,b,\{e,\rho^2\})\rvert\\ &\quad- \lvert C(n,b,\{e,d_1\})\rvert- \lvert | {
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\lvert C(n,b,\{e,\rho^2\})\rvert\\ &\quad- \lvert C(n,b,\{e,d_1\})\rvert- \lvert C(n,b,\{e,d_2\})\rvert+2\lvert C(n,b,\{e,\rho^2,h,v\})\rvert\\ &\quad+2\lvert C(n,b,\{e,\rho^2,d_1,d_2\})\rvert \end{aligned} These imply that \begin{aligned} (**)\quad K(n)=&\frac{\lvert C(n,n,C_4)\rvert}{4}+\frac{\lvert C(n,n,\{e,h\})\rvert}{8}+\frac{\lvert C(n,n,\{e,v\})\rvert}{8}+\frac{\lvert C(n,n,\{e,\rho^2\})\rvert}{8}\\ &+\frac{\lvert C(n,n,\{e,d_1\})\rvert}{8}+\frac{\lvert C(n,n,\{e,d_2\})\rvert}{8}+\frac{\lvert C(n,n,\{e\})\rvert}{8}. \end{aligned} | {
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It remains to compute the quantities $C(n,b,G)$ for the seven groups that appear in the expression above. (Added: It is usual in Pólya enumeration at this point to use the cycle index and generating function techniques. My more bare-handed approach yields the final expressions in a fairly direct fashion.) We have $$C(n,b,\{e\})=\binom{n^2}{b}.$$ In general, the counting depends on whether $n$ is even or odd. If $n$ is odd, then the action of $C_4$ fixes the center square; every other square has an orbit of size $4.$ Therefore, a configuration that is invariant under the action of $C_4$ is specified by choosing the markings of the center square and of one square from each of the $\frac{n^2-1}{4}$ orbits of size $4.$ (The other three squares in an orbit of size $4$ must be marked the same way as the first square.) Since the total number of marked squares must be $b,$ we have $$C(n,b,C_4)=\begin{cases}\binom{(n^2-1)/4}{b/4} & \text{if b\equiv0\pmod{4}}\\ \\ \binom{(n^2-1)/4}{(b-1)/4} & \text{if b\equiv1\pmod{4}}\\ \\ 0 & \text{otherwise.}\end{cases}$$ The center square is unmarked when $b\equiv0\pmod{4}$ and marked when $b\equiv1\pmod{4}.$ If $n$ is even then, under the action of $C_4,$ every square has an orbit of size $4.$ Therefore $$C(n,b,C_4)=\begin{cases}\binom{n^2/4}{b/4} & \text{if b\equiv0\pmod{4}}\\ \\ \binom{n^2/4}{(b-1)/4} & \text{if b\equiv1\pmod{4}}\\ \\ 0 & \text{otherwise.}\end{cases}$$ Specializing to $b=n,$ we get $$C(n,n,C_4)=\begin{cases}\binom{n^2/4}{n/4} & \text{if n\equiv0\pmod{4}}\\ \\ \binom{(n^2-1)/4}{(n-1)/4} & \text{if n\equiv1\pmod{4}}\\ \\ 0 & \text{otherwise.}\end{cases}$$
The analysis for $G=\{e,\rho^2\}$ is similar. The orbit sizes are now $2$ instead of $4,$ with the exception of the center square in the $n$ odd case, which has orbit size $1.$ The result, for $b=n,$ is $$C(n,n,\{e,\rho^2\})=\begin{cases}\binom{n^2/2}{n/2} & \text{if n even}\\ \\ \binom{(n^2-1)/2}{(n-1)/2} & \text{if n odd.}\end{cases}$$ | {
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When $G=\{e,h\}$ and $n$ is odd, the action of $G$ fixes the $n$ squares on the horizontal centerline. All other squares have orbit size $2.$ A configuration invariant under the action of $\{e,h\}$ is specified by marking $j$ of these orbits and $b-2j$ squares on the centerline. When $n$ is even, all squares have orbit size $2,$ so we need only mark $b/2$ of these orbits. The analysis when $G=\{e,v\}$ is similar. In the $b=n$ case, this gives $$C(n,n,\{e,h\})=C(n,n,\{e,v\})=\begin{cases}\binom{n^2/2}{n/2} & \text{if n even}\\ \\ \sum_j\binom{(n^2-n)/2}{j}\binom{n}{n-2j} & \text{if n odd.}\end{cases}$$ | {
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When $G=\{e,d_1\}$ or $\{e,d_2\},$ it is a diagonal that is fixed by the action of $G.$ So the analysis is the same as for $G=\{e,h\}$ with $n$ odd. We have $$C(n,n,\{e,d_1\})=C(n,n,\{e,d_2\})= \sum_j\binom{(n^2-n)/2}{j}\binom{n}{n-2j}.$$ Mathematica recognizes this sum, which we denote by $S,$ as a generalized hypergeometric function with argument $-1:$ $$S=\sum_j\binom{(n^2-n)/2}{j}\binom{n}{n-2j} = \ _3F_2\left[\begin{matrix}1/2-n/2 & -n/2 & (n-n^2)/2\\1/2 & 1 & \end{matrix};-1\right].$$ Indeed, \begin{aligned} S=&\sum_j\binom{(n^2-n)/2}{j}\binom{n}{n-2j} =\sum_j\binom{(n^2-n)/2}{j}\binom{n}{2j}\\ =&\sum_j\frac{\frac{n^2-n}{2}(\frac{n^2-n}{2}-1)\ldots(\frac{n^2-n}{2}-j+1)}{j!}\frac{n(n-1)\ldots(n-2j+1)}{2j(2j-1)\ldots1}\\ =&\sum_j\frac{n-n^2}{2}\left(\frac{n-n^2}{2}+1\right)\ldots\left(\frac{n-n^2}{2}+j-1\right)\frac{\frac{-n}{2}\frac{-n+1}{2}\ldots\frac{-n+2j-1}{2}}{j(j-\frac{1}{2})\ldots\frac{2}{2}\frac{1}{2}}\frac{(-1)^j}{j!}\\ =&\sum_j\frac{n-n^2}{2}\left(\frac{n-n^2}{2}+1\right)\ldots\left(\frac{n-n^2}{2}+j-1\right)\\ &\times\frac{\frac{-n}{2}\left(\frac{-n}{2}+1\right)\ldots\left(\frac{-n}{2}+j-1\right)\frac{1-n}{2}\left(\frac{1-n}{2}+1\right)\ldots\left(\frac{1-n}{2}+j-1\right)}{\frac{1}{2}\frac{3}{2}\ldots\left(\frac{1}{2}+j-1\right)\cdot1\cdot2\ldots\cdot j}\frac{(-1)^j}{j!}, \end{aligned} which, by definition, is the stated generalized hypergeometric function. | {
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Combining these results, we have $$K(n)=\begin{cases} \frac{1}{8}\binom{n^2}{n} + \frac{1}{4}\binom{n^2/4}{n/4} + \frac{3}{8}\binom{n^2/2}{n/2} + \frac{1}{4}S & \text{for n\equiv0\pmod{4}}\\ \frac{1}{8}\binom{n^2}{n} + \frac{1}{4}\binom{(n^2-1)/4}{(n-1)/4} + \frac{1}{8}\binom{(n^2-1)/2}{(n-1)/2} + \frac{1}{2}S & \text{for n\equiv1\pmod{4}}\\ \frac{1}{8}\binom{n^2}{n} + \frac{3}{8}\binom{n^2/2}{n/2} + \frac{1}{4}S & \text{for n\equiv2\pmod{4}}\\ \frac{1}{8}\binom{n^2}{n} + \frac{1}{8}\binom{(n^2-1)/2}{(n-1)/2} + \frac{1}{2}S & \text{for n\equiv3\pmod{4}.} \end{cases}$$ This answer agrees with that of Marko Riedel and the OEIS sequence he cites. (Added: These formulas can be obtained without much difficulty by extracting the relevant coefficient from the final generating function in Marko Riedel's answer.)
Added: (Some background about the orbit-stabilizer theorem and the orbit counting theorem.) Let $G$ be a group acting on a set $X.$. The stabilizer of $x\in X$ is the set of elements of $G$ that fix $x,$ that is, the set of $g\in G$ such that $gx=x.$ The orbit of $x\in X$ is the set of elements of $X$ to which $x$ is sent by some element of $G.$
The orbit-stabilizer theorem states that, for any $x\in X,$ $$\lvert G\rvert=\lvert\text{orbit}(x)\rvert\cdot\lvert\text{stabilizier}(x)\rvert.$$ It is a consequence of the fact that $\text{stabilizer}(x)$ is a subgroup of $G$ and the fact that the elements of $\text{orbit}(x)$ are in one-to-one correspondence with the left cosets of $\text{stabilizer}(x).$ | {
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The action of $G$ on $X$ partitions $X$ into disjoint orbits. Let $\mathcal{O}$ be the set of orbits. The fixed set of $g\in G$ is the set of elements of $X$ fixed by $g,$ that is, the set of $x\in X$ such that $gx=x.$ The orbit counting theorem, which is often called Burnside's lemma, states that the number of orbits equals the average (over all $g\in G$) of $\lvert\text{fixed}(g)\rvert.$ It is proved by considering the set $P$ of pairs $(g,x)$ satisfying $gx=x.$ We may count the elements of $P$ either by running over $G$ or by running over $X:$ \begin{aligned} \lvert P\rvert=\sum_{g\in G}\lvert\text{fixed}(g)\rvert&=\sum_{x\in X}\lvert\text{stabilizer}(x)\rvert\\ &=\sum_{x\in X}\frac{\lvert G\rvert}{\lvert\text{orbit}(x)\rvert}\\ &=\lvert G\rvert\sum_{O\in\mathcal{O}}\sum_{x\in O}\frac{1}{\lvert\text{orbit}(x)\rvert}\\ &=\lvert G\rvert\sum_{O\in\mathcal{O}}1\\ &=\lvert G\rvert\cdot\lvert\mathcal{O}\rvert, \end{aligned} where the orbit-stabilizer theorem was used to obtain the second line. This implies that $$\lvert\mathcal{O}\rvert=\frac{1}{\lvert G\rvert}\sum_{g\in G}\lvert\text{fixed}(g)\rvert,$$ which is the desired result.
http://puzzlemaker.discoveryeducation.com/MathSquareForm.asp
Above is a link to a puzzlemaker referred to as Math Square.
It allows you to create matrice puzzle.
For example, you can create a 6x6 grid and the rows and columns are maths equation having multiplication, division, addition and subtraction. The answers to the equation are available on the extreme right and the extreme bottom of the grid. However, no number in the grid is provided. Can such a puzzle be logically or mathematically solved using any of the above methods mentioned in this thread. | {
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# Math Help - number of integers bet 90 and 990
1. ## number of integers bet 90 and 990
How many integers bet 90 and 990 are divisible by 7?
is there any short and simple way of finding the answer?
thanks
2. ## Re: number of integers bet 90 and 990
Originally Posted by rcs
How many integers bet 90 and 990 are divisible by 7?
is there any short and simple way of finding the answer?
$\left\lfloor {\frac{{990}}{7}} \right\rfloor - \left\lfloor {\frac{{90}}{7}} \right\rfloor=~?$
3. ## Re: number of integers bet 90 and 990
Originally Posted by rcs
How many integers bet 90 and 990 are divisible by 7?
is there any short and simple way of finding the answer?
thanks
Hi rcs!
The answer is 990 / 7 - (90 - 1) / 7, where "/" denotes a division that is rounded down.
I am assuming that you would count 990 and 90 if they would have been divisible by 7 (which they are not anyway).
To explain:
(90 - 1) / 7 is the number of integers divisible by 7 below and excluding 90.
990 / 7 is the number of integers divisible by 7 below and including 990.
The difference is the total number.
Suppose you are in the middle of a road with poles alongside the road.
And suppose you want to know the number of poles from where you are to the end of the road.
To find out you count the number that you can see from where you stand to the beginning of the road.
And if you are next to a pole you do not count that one.
Subtract from the total number of poles and there you are.
4. ## Re: number of integers bet 90 and 990
thanks Prof . Plato
5. ## Re: number of integers bet 90 and 990
Originally Posted by Plato
$\left\lfloor {\frac{{990}}{7}} \right\rfloor - \left\lfloor {\frac{{90}}{7}} \right\rfloor=~?$
the answer is 128 4/7 is this correct sir?
i wonder why the book the answer they posted is 56... however they do not post the solution
6. ## Re: number of integers bet 90 and 990 | {
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6. ## Re: number of integers bet 90 and 990
Originally Posted by rcs
the answer is 128 4/7 is this correct sir?
i wonder why the book the answer they posted is 56... however they do not post the solution
First that is the floor function. It returns an integer.
There are $\left\lfloor {\frac{{990}}{7}} \right\rfloor=141$ integers from 1 to 990 that are by seven.
There are $\left\lfloor {\frac{{90}}{7}} \right\rfloor=12$ integers from 1 to 90 that are by seven.
Note that neither 990 nor 90 is divisible by seven.
So you book is not correct. The answer is 129.
7. ## Re: number of integers bet 90 and 990
Originally Posted by Plato
First that is the floor function. It returns an integer.
There are $\left\lfloor {\frac{{990}}{7}} \right\rfloor=141$ integers from 1 to 990 that are by seven.
There are $\left\lfloor {\frac{{90}}{7}} \right\rfloor=12$ integers from 1 to 90 that are by seven.
Note that neither 990 nor 90 is divisible by seven.
So you book is correct. The answer is 129.
you mean to say that the book i've read whose answer is 56 is correct and your answer 129 is also correct... therefore, there are two correct answers now... 56 and 129
8. ## Re: number of integers bet 90 and 990
Sorry that was a typo. It should be NOT CORRECT
9. ## Re: number of integers bet 90 and 990
ahhh i see... well then thank you again Sir. God Bless
10. ## Re: number of integers bet 90 and 990
Suggest we find the multiple of 7 just greater than 90 and just less than 990.
We will get 91 and 987. Now all the multiples of 7 between these two numbers will form an AP with first term 91, common difference 7 and last term 987.
Question now simply is to find n such that the nth term is 987.
If a is the first term, common difference d then the nth term is given by
an = a + ( n-1) d
Thus we have 91 + ( n – 1 ) x 7 = 987
OR 91 + 7n – 7 = 987
7n = 987 – 91 + 7 = 903
n = 129 Hence there are 129 integers between 90 and 990 which are divisible by 7 | {
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# Orthogonality of Legendre polynomials using specific properties
I'm having significant issues with a problem and would appreciate any help at all with it. It is regarding proving the orthogonality of Legendre polynomials using a specific recursion formula and property of Legendre polynomials.
I have been given the following relation $$x P'_n(x) -P'_{n-1}(x) = n P_n (x)$$ And that $$\int^{1}_{-1} f(x) P_n (x) dx = 0$$ for any polynomial $$f(x)$$ of degree less than $$n$$.
I must then show that for $$n \geq 1$$
$$\int^{1}_{-1} P_n (x)^2 dx = \frac{1}{2n} \int^{1}_{-1} x \frac{d}{dx} (P_n (x)^2)dx$$
And then determine the value of the integral, that being $$\frac{2}{2n+1}$$. I must use the two stated facts, I'd be grateful for any help with this, thank you guys.
• Have you tried integration by parts yet? – Somos Nov 14 '20 at 0:31
• My first thought too but I decided its a no-go. Id like to see your approach, @Somos. – CogitoErgoCogitoSum Nov 14 '20 at 1:31
$$\int\limits_{-1}^{1} P_n(x)^2 \;dx =$$ $$\int\limits_{-1}^{1} P_n(x)P_n(x) \;dx =$$ $$\frac{1}{2n}\int\limits_{-1}^{1} 2n P_n(x)P_n(x) \;dx =$$
Substituting your first relation in: $$\frac{1}{2n}\int\limits_{-1}^{1} 2 \left[x P'_n(x) - P'_{n-1}(x)\right] P_n(x)\;dx =$$
$$\frac{1}{2n}\int\limits_{-1}^{1} 2 x P'_n(x) P_n(x) \; dx - \frac{1}{2n}\int\limits_{-1}^{1} 2 P'_{n-1}(x) P_n(x)\;dx =$$
Given your second property defining orthogonality (since after all $$P'$$ is of lesser degree than $$P$$) the entire second integral drops out.
$$\frac{1}{2n}\int\limits_{-1}^{1} x 2 P_n(x) P'_n(x) \; dx =$$
Reverse of the chain rule:
$$\frac{1}{2n}\int\limits_{-1}^{1} x \frac{d}{dx} P^2_n(x) \; dx$$
QED
And for anyone who asks/wonders, or is inclined to criticize... I answered this question because it was asked, and this is a Q&A forum. I shouldnt have to justify answering questions on a Q&A forum though, but this is the culture here. | {
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• Did you read the part in the question stating: "And then determine the value of the integral, that being 2/(2n+1)."? You seem to have not done that part yet. – Somos Nov 14 '20 at 21:55
• Youre right, I didnt. I dont feel the need to do all 100% of the work when OP is hindered only by the first 10%. Dont know what your point is. I dont believe in "handing out answers". OP should be able to do some of it, dont you think? – CogitoErgoCogitoSum Nov 14 '20 at 23:20
I immediately thought of integration by parts. Here is my work:
By the product rule of derivative we have $$\frac{d}{dx} x\, P_n(x)^2 = P_n(x)^2 + x\, \frac{d}{dx}(P_n(x)^2) \tag{1}$$
Note that Legendre polynomials have the property $$P_n(1) = 1,\quad \text{ and } \quad P_n(-1)=(-1)^n. \tag{2}$$
By integrating equation $$(1)$$ and using equation $$(2)$$ we get $$2 = 1 - (-1) = \int_{-1}^1 P_n(x)^2 dx + \int_{-1}^1 x\, \frac{d}{dx}(P_n(x)^2) dx. \tag{3}$$ | {
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Define $$\, a_n := \int_{-1}^1 P_n(x)^2 dx.\,$$ Equation $$(3)$$ implies that $$2 - a_n = \int_{-1}^1 x\, \frac{d}{dx}(P_n(x)^2)\,dx. \tag{4}$$ We are given the property that $$x P'_n(x) -P'_{n-1}(x) = n P_n (x). \tag{5}$$ Multiply this equation $$(5)$$ by $$\,P_n(x)\,$$ and integrate to get $$\int_{-1}^1\!P_n(x)(xP'_n(x)\!-\!P'_{n-1}(x))dx \!=\!\int_{-1}^1\!P_n(x)nP_n(x)dx. \tag{6}$$ We are also given the property that $$\int_{-1}^1 P_n(x)\,f(x)\,dx = 0 \tag{7}$$ for any polynomial $$\,f(x)\,$$ of degree less than $$\,n.\,$$ Now notice that $$x\, \frac{d}{dx}(P_n(x)^2) = 2\,x\,P_n(x)P'_n(x). \tag{8}$$ Combining equations $$(6)$$, $$(7)$$, and $$(8)$$ we get that $$\frac12(2-a_n) = n\,a_n. \tag{9}$$ Dividing this equation $$(9)$$ through by $$\,n\,$$ and using the definition of $$\,a_n\,$$ and equation $$(4)$$ gives the result we wanted: $$\int_{-1}^1 P_n(x)^2 dx = \frac{2-a_n}{2n} = \frac1{2n} \int_{-1}^1 x\, \frac{d}{dx}(P_n(x)^2)\,dx. \tag{10}$$ Also, solving for $$\,a_n\,$$ in equation $$(9)$$ gives $$a_n = \frac2{2n+1} \tag{11}$$ which was also requested.
NOTE: The property in equation $$(2)$$ was not given to us. It would have to be proven using equations $$(5)$$ and $$(7)$$ which are given to us. It can be done, and would take some effort, but I have not given the steps required. | {
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• A few points/criticisms, if you dont mind. For one thing, it isnt obvious how your suggestion to use "integration by parts" is being utilized. Frankly I dont see I by P in your work. Secondly, your work doesnt flow all that well... you make a few leaps. The logic is sound but effort must be used to comprehend the somewhat convoluted steps. Its just a bit scattered and disorderly. Thirdly, property (2), although true, was not one of the properties OP provided; Im not sure its appropriate to use it. Just some things to consider... – CogitoErgoCogitoSum Nov 14 '20 at 20:27
• In the Wikipedia article on Integration by parts is stated "The rule can be thought of as an integral version of the product rule of differentiation." That is what I used. You say "leaps". I say "steps". I don't see anything convoluted. In my NOTE at the end I admit the status of property (2), but the original quesition did not explicitly state than no other properties of Legendre polynomials can be used. If you have a better proof, you can answer the question yourself, which, of course, you have done. Thanks for commenting. – Somos Nov 14 '20 at 20:45 | {
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# Show that there exists $x_0\in(a,b)$ such that $f(x_0)=\frac{1}{n}(f(x_1)+f(x_2)+\cdots+f(x_n)).$
Question: Suppose that $$f:[a,b]\to\mathbb{R}$$ is continuous. Let $$x_1,x_2,\cdots, x_n$$ be any $$n$$ points in $$(a,b).$$ Show that there exists $$x_0\in(a,b)$$ such that $$f(x_0)=\frac{1}{n}(f(x_1)+f(x_2)+\cdots+f(x_n)).$$
Solution: Let $$g:[a,b]\to\mathbb{R}$$ be such that $$g(x)=nf(x)-\sum_{k=1}^nf(x_k), \forall x\in[a,b].$$ Observe that to prove the statement of the problem it is enough to show that $$g(x_0)=0$$ for some $$x_0\in(a,b)$$.
Now note that by the 3rd form of the Pigeon Hole principle we can conclude that there exists $$1\le i,j\le n$$ such that $$f(x_i)\le \frac{1}{n}\sum_{k=1}^nf(x_k)\le f(x_j)\\\implies nf(x_i)\le \sum_{k=1}^nf(x_k)\le nf(x_j).$$ Thus, $$g(x_i)=nf(x_i)-\sum_{k=1}^nf(x_k)\le 0$$ and $$g(x_j)=nf(x_j)-\sum_{k=1}^nf(x_k)\ge 0.$$ Now if $$g(x_i)=0$$ or $$g(x_j)=0$$, then we are done. Thus, let us assume that $$g(x_i)<0$$ and $$g(x_j)>0$$. Now since $$f$$ is continuous on $$[a,b]$$, implies that $$g$$ is continuous on $$[a,b]$$. Therefore, by IVT we can conclude that there exists $$x_0\in(x_i,x_j)$$ or $$x_0\in(x_j,x_i)$$ such that $$g(x_0)=0$$. This completes the proof.
Is this solution correct and rigorous enough and is there any other way to solve the problem?
• What is the PHP? Aug 26, 2020 at 19:22
• @MartinR, it's the "Pigeon Hole Principle". Sorry, I will expand that term in a bit. Aug 26, 2020 at 19:24
Your proof looks fine to me. There is no need however to introduce the function $$g$$. You know that $$f(x_i)\le \frac{1}{n}\sum_{k=1}^nf(x_k)\le f(x_j)$$ for some indices $$i, j$$, so you can just apply the intermediate value theorem to $$f$$ on the interval $$I = [\min(x_i, x_j), \max(x_i, x_j)]$$ and conclude that $$\frac{1}{n}\sum_{k=1}^nf(x_k) = f(x)$$ for some $$x \in I$$. | {
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Instead of using the pigeon hole principle you can also apply the mean value theorem to $$f$$ on the interval $$J= [\min_k x_k, \max_k x_k] \subset (a, b)$$ because $$m\le \frac{1}{n}\sum_{k=1}^nf(x_k)\le M$$ with $$m = \min_J f(x)$$ and $$M = \max_J f(x)$$.
Given a continuous $$f(x)$$, an iterated application of the Intermediate value Theorem gives \eqalign{ & \exists x_{1,2} \in \left[ {x_1 ,x_2 } \right]:f(x_{1,2} ) = t\;f(x_1 ) + \left( {1 - t} \right)f(x_2 )\quad \left| {\,0 \le t \le 1} \right. \cr & \exists x_{2,3} \in \left[ {x_2 ,x_3 } \right]:f(x_{2,3} ) = u\;f(x_2 ) + \left( {1 - u} \right)f(x_3 )\quad \left| {\,0 \le u \le 1} \right. \cr} which express the possibility of finding a point corresponding to the weighted mean within each interval.
Putting $$t=2/3, \, u=1/3$$, we can write \eqalign{ & \exists x_{1,2} \in \left[ {x_1 ,x_2 } \right]:f(x_{1,2} ) = {2 \over 3}\;f(x_1 ) + {1 \over 3}f(x_2 )\quad \left| {\,0 \le t \le 1} \right. \cr & \exists x_{2,3} \in \left[ {x_2 ,x_3 } \right]:f(x_{2,3} ) = {1 \over 3}\;f(x_2 ) + {2 \over 3}f(x_3 )\quad \left| {\,0 \le u \le 1} \right. \cr & \exists x_{1,3} \in \left[ {x_1 ,x_2 } \right] \cup \left[ {x_2 ,x_3 } \right]:f(x_{1,3} ) = {1 \over 2}\,f(x_{1,2} ) + {1 \over 2}f(x_{2,3} ) = \cr & = {{f(x_1 ) + f(x_2 ) + f(x_3 )} \over 3} \cr} and the extension to n points is clear.
Pick $$i$$ with \begin{align} f(x_i) &\le f(x_k) & \text{for all k = 1, \ldots, n.} \tag{1} \end{align}
Pick $$j$$ with \begin{align} f(x_j) &\ge f(x_k) & \text{for all k = 1, \ldots, n.} \tag{2} \end{align}
If $$i = j$$, then all $$x_k$$ are equal, and $$x_0 = x_i$$ solves the problem.
Consider the case $$i < j$$; the $$i > j$$ case is almost identical. But equation $$1$$, we have $$n f(x_i) \le \sum_k f(x_k)$$ By equation 2, similarly $$n f(x_j) \ge \sum_k f(x_k)$$.
Then by the Intermediate value theorem, there's an $$x_0 \in [x_i, x_j]$$ such that $$f(x_0) = \frac{1}{n} \sum_k f(x_k).$$ | {
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is 30 an integer 1. For example, 2 , 5 , 0 , − 12 , 244 , − 15 and 8 are all integers. Integer is a math term for a number that is a whole number. The answer to the question: "Is 30 an integer?" Walk through the difference between whole numbers & integers. > How do you check if a value in Java is of an integer type? MATLAB ® has four signed and four unsigned integer classes. Integers are the set of whole numbers and their opposites. He has been teaching from the past 9 years. He provides courses for … These numbers are used to perform various arithmetic operations, like addition, subtraction, multiplication and division.The examples of integers are, 1, 2, 5,8, -9, -12, etc. Is it an integer? An integer.An integer.An integer.An integer. An integer, also called a "round number" or “whole number,” is any positive or negative number that does not include decimal parts or fractions. Quantity A is greater. How to use integer in a sentence. Signed types enable you to work with negative integers as well as positive, but cannot represent as wide a range of numbers as the unsigned types because one bit is used to … $\begingroup$ "For this ratio to not be an integer m and n must be distinct which implies that x is not an integer quantity." 5- (-4)=1 please help i have a test tommorrow and i just dont get what an integer is. Because you can't "count" zero. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. My only temporary fix was to change the JavaScript that handles the fields, as it was using regex to check if it was and Integer so i changed it to convert the string to int and check if it was NaN Comment Share Find out if 24 is an Integer here! Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. Yes. An even number is any integer (a whole number) that can be divided by 2. You could say that 30/6 is just another name for 5, and then it is an integer. 30 is | {
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divided by 2. You could say that 30/6 is just another name for 5, and then it is an integer. 30 is the integer in this case. Why don't libraries smell like bookstores? integer - any of the natural numbers (positive or negative) or zero; "an integer is a number that is not a fraction" whole number. (30 points) An integer n is prime iff n >= 2 and n's only factors (divisors) are 1 and itself. And 30 is an integer and not a fraction. Suppose x and y are even. A comprehensive database of more than 56 integer quizzes online, test your knowledge with integer quiz questions. What are the three consecutive odd integers with sum of -30? It is a special set of whole numbers comprised of zero, positive numbers and negative numbers and denoted by the letter Z. My only temporary fix was to change the JavaScript that handles the fields, as it was using regex to check if it was and Integer so i changed it to convert the string to int and check if … go, not everyone agrees on a simple thing! Integer Question: Is 24 an Integer? We are simply taking "5 lots of −3", like this: It should be noted that an integer can never be a fraction, a decimal or a per cent. When I as doing my graduation one of my professors Mr. Jaggi told us “Maths student must be polite” thats why I used here ‘may’. sound intelligent. You might be wondering, what is an opposite? python 2.7.4 error: exactly the same as integers. In fact I can't think of any proof of this sentence which doesn't prove the whole theorem along the way. Integer Question: Is 20 an Integer? The opposite of 24 is -24. The number line goes on forever in both directions. Solution for 1. Is a none repeating decimal an integer? Thanks for your help! The number 0 is also considered an integer even though it has neither a positive or negative value. What details make Lochinvar an attractive and romantic figure? Question 256872: The sum of an integer and its square is 30. 30 is the smallest sphenic number, and the smallest of the form 2 × 3 × r, | {
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and its square is 30. 30 is the smallest sphenic number, and the smallest of the form 2 × 3 × r, where r is a prime greater than 3. What is the conflict of the story of sinigang? What is an Integer? So they are 1, 2, 3, 4, 5, ... (and so on). What are the steps to the solution? If x is even, then x y is even. An integer is what is more commonly known as a whole number.It may be positive, negative, or the number zero, but it must be whole. In some cases, the definition of whole number will exclude the number zero, or even the set of negative numbers, but this is not as common as the more inclusive use of the term. n is an integer, and k is not an integer. De tekst is beschikbaar onder de licentie Creative Commons Naamsvermelding/Gelijk delen, er kunnen aanvullende voorwaarden van toepassing zijn. The symbol of integers is “ Z “. See more. Examples– -2.4, 3/4, 90.6. Every integer is a rational number. Quantity B is greater. "Natural Numbers" can mean either "Counting Numbers" {1, 2, 3, ...}, or "Whole Numbers" {0, 1, 2, 3, ...}, depending on the subject. The set of integers is infinite and has no smallest element and no largest element. > is.integer(66L) [1] TRUE > is.integer(66) [1] FALSE Also functions like round will return a declared integer, which is what you are doing with x==round(x). This is shown below. And some people say that zero is NOT a whole number. An integer is a positive or negative whole number. The … PHP does not support unsigned integers. The basic rules of math. For 20 to be an integer, 20 has to be a whole number. How is 30 an even number? Our online integer trivia quizzes can be adapted to suit your requirements for taking some of the top integer quizzes. Furthermore, for 20 to be an integer, you should be able to write 20 without a fractional or decimal component. Definitions. You could try this. (use indirect… It is impossible to answer your question without … 0 0 1 ... An integer is a number that is not a fraction, and decimal numbers are | {
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question without … 0 0 1 ... An integer is a number that is not a fraction, and decimal numbers are decimal fractions. There are three Proper… Integers The integers are the set of whole numbers and their opposites. Find out if 20 is an Integer here! The farther the number is to the right, the larger the value is. Base Representation; 2: … This is one of those tricky definition questions. Even and Odd Numbers (use indirect… Find the number. It cannot be done. Hello, We have a drafter who's Map 3D 2013 Drafting Settings dialog box is giving him a pop up that says "Please enter an integer from 1 to 30". Solution for 1. Here are some examples: We can compare integers the same way that we compare positive whole numbers. The number 30 as a Roman Numeral XXX The number 30 in various Numeral Systems. Fractions and decimals are not included in the set of integers. please give me an example or two of an integer. I want to know how to test whether an input value is an integer or not. Please note this from the Integer datatype page: "The size of an integer is platform-dependent, although a maximum value of about two billion is the usual value (that's 32 bits signed). Some examples of integers include 1, 3, 4, 8, 99, 108, -43, … Examples of Integers – 1, 6, 15. I have tried using the function isinteger, but I obtain, for example, isinteger(3) = 0.Apparently, any constant is double-precision by Matlab default, and is therefore not recognized as an integer. In the Registry Editor window, navigate to the folder path below depending on your Photoshop version.. Photoshop CC 2018: Navigate to HKEY_CURRENT_USER\SOFTWARE\Adobe\Photoshop\120.0 Photoshop CC 2017: Navigate to HKEY_CURRENT_USER\SOFTWARE\Adobe\Photoshop\110.0 Photoshop CC 2015.5: Navigate to HKEY_CURRENT_USER\SOFTWARE\Adobe\Photoshop\100.0 … For 24 to be an integer, 24 has to be a whole number. A modular type is an integer type with all arithmetic modulo a specified positive modulus; such a type corresponds to an unsigned | {
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type with all arithmetic modulo a specified positive modulus; such a type corresponds to an unsigned type with wrap-around semantics. 30 is a whole number that can be written without a fractional component, thus 30 is an integer. It is not only accurate, it makes you In Maths, integers are the numbers which can be positive, negative or zero, but cannot be a fraction. An integer is any number including 0, positive numbers, and negative numbers. But every rational number is not an integer, since for b not equal to 1, if b does not divide a then w don't get an integer. Integer definition, one of the positive or negative numbers 1, 2, 3, etc., or zero. Examples of Integer Multiplication (a) Here is an integer times an integer: 5 xx -3 = -15 Why is this so? But you for sure can say it is a rational number, which is any number that can be expressed as a … Choose from 500 different sets of 30 math integers flashcards on Quizlet. How will understanding of attitudes and predisposition enhance teaching? In the equation 2 + 1/2, the number 2 is the integer and 1/2 is the fraction. Find the integer The sum of an integer and its square is - Answered by a verified Math Tutor or Teacher Whole Numbers are simply the numbers 0, 1, 2, 3, 4, 5, ... (and so on), (But numbers like ½, 1.1 and 3.5 are not whole numbers.). The number line is used to represent integers. Start by using a number line. Compare -9 and -2. 0 0 1 ... 176 + 125 = 301 Therefore, the lowest positive integer you can add is 125. When did organ music become associated with baseball? An integer (pronounced IN-tuh-jer) is a whole number (not a fractional number) that can be positive, negative, or zero. An integer (from the Latin integer meaning "whole") is colloquially defined as a number that can be written without a fractional component.For example, 21, 4, 0, and −2048 are integers, while 9.75, 5 + 1 / 2, and √ 2 are not. Suppose x and y are even. I want to know how to test whether an input value is an integer or not. | {
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Suppose x and y are even. I want to know how to test whether an input value is an integer or not. Where is telephone area code 035? 4. Consider a function Factor_In_Range (k , n) to return true iff some integer in the range k to n … And everyone agrees on the definition of an integer, so when in doubt say "integer". This is indicated by the arrows. All Rights Reserved. If a a a and b b b are integers such that a ⋅ b = 0 a \cdot b = 0 a ⋅ b = 0, then a = 0 a=0 a = 0 or b = 0 b=0 b = 0. Rational number is a number which can be expressed as a/b where a,b are integers and b is not 0. a) -3 or 10 b) 3 or -10 c) -5 or 6 d) 5 or -6 Help! For 30 to be an integer, 30 has to be a whole number. Ex 6.1, 2 Solve –12x > 30, when x is a natural number –12x > 30 –x > 3012 –x > 2.5 Since x is negative, we multiply both sides by -1 & change the signs (–x) × (-1) < 2.5 × (-1) x < − 2.5 (i) Since x is a natural number, it can be only positive num The base range of a signed integer type includes at least the values of the specified range. is.integer tests if the number is declared as an integer. B. An integer_type_definition defines an integer type; it defines either a signed integer type, or a modular integer type. You need to prove this, as nothing you have said makes it obvious this is true. Learn 30 math integers with free interactive flashcards. But you for sure can say it is a rational number, which is any number that can be expressed as a fraction of two integers.. An octal integer is an integer represented by a string of digits from the set $\{0,1,2,3,4,5,6,7\}$, each representing (in that order) the usual first eight non-negative integers. Integers are like whole numbers, but they also include negative numbers ... but still no fractions allowed! This is one of those tricky definition questions. If a is an odd integer, then a2 +3a+5 is odd. These are all integers (click to mark), and they continue left and right infinitely: Some people (not me) say that whole numbers can also | {
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and they continue left and right infinitely: Some people (not me) say that whole numbers can also be negative, which makes them So they are 1, 2, 3, 4, 5, ... (and so on). What is an integer? Therefore, these numbers can never be integers: fractions; decimals; percents; Adding and Subtracting Integers. Given an integer number n, return the difference between the product of its digits and the sum of its digits.. He provides courses for Maths and Science at Teachoo. ok i dont get.... is it just considered an integer if the dominate is 1? Counting Numbers are Whole Numbers, but without the zero. Examples of integers are: … If you divide by 2 and there is a remainder left, then the number is odd. The set of integers consists of zero (), the positive natural numbers (1, 2, 3, ...), also called whole numbers or … For example, 3, -10, and 1,025 are all integers, but 2.76 (decimal), 1.5 … Looking at a number line can help you when you need to add or subtract integers. Integers are positive whole numbers and their opposites, including zero. characteristic - the integer part (positive or negative) of the representation of a logarithm; in the expression log 643 = 2.808 the characteristic is 2 Furthermore, for 30 to be an integer, you should be able to write 30 without a fractional or decimal component. If a is an odd integer, then a2 +3a+5 is odd. For any integer a a a, the additive inverse − a-a − a is an integer. (use direct proof) 2. (ii) x is an integer Now, x < 2.5 Since x is an integer, it can be negative but cannot be a decimal So, x can be any negative number less than 3 Hence, the values of x which will satisfy the inequality are = { , 6, 5, 4, 3} Show More The word integer originated from the Latin word “Integer” which means whole. Integers are whole numbers and their negative opposites. Copyright © 2020 Multiply Media, LLC. Integers Integer Classes. Is 30 an integer? When multiplying integers, we can think of multiplying "blocks" of negative numbers. expressed in | {
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When multiplying integers, we can think of multiplying "blocks" of negative numbers. expressed in rational form as 30/1. 2020-09-30 17:50:29 2020-09-30 17:50:29. First of all we should know that “Every whole number is an integer, but every integer may not be a whole number.”. An integer (from the Latin integer meaning "whole") is colloquially defined as a number that can be written without a fractional component.For example, 21, 4, 0, and −2048 are integers, while 9.75, 5 + 1 / 2, and √ 2 are not. Intager. What is the contribution of candido bartolome to gymnastics? You could say that 30/6 is just another name for 5, and then it is an integer. Like this (note: zero isn't positive or negative): For an interesting look at other types of numbers read The Evolution of Numbers, { ... −4, −3, −2, −1, 0, 1, 2, 3, 4, ... }, Adding and Subtracting Positive and Negative Numbers, Negative Integers = { ..., −4, −3, −2, −1 }, Non-Negative Integers = { 0, 1, 2, 3, 4, ... }. Essay On Nature And Environment, How To Get Rid Of Spotted Knapweed, Projection Matrix Songho, Thermomix Recipes Chinese Food, Cherry Ambrosia Salad, Burton Michigan Chamber Of Commerce, Fundamentals Of General, Organic, And Biological Chemistry Pdf, Solar Energy Activities For Elementary Students, " /> | {
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The Distinctive Designer Roof | {
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# is 30 an integer | {
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Answer by drk(1908) (Show Source): So if we consider b=1 then we get the set of integers. If x is even, then x y is even. Example 1: Input: n = 234 Output: 15 Explanation: Product of digits = 2 * 3 * 4 = 24 Sum of digits = 2 + 3 + 4 = 9 Result = 24 - 9 = 15 Example 2: Input: n = 4421 Output: 21 Explanation: Product of digits = 4 * 4 * 2 * 1 = 32 Sum of digits = 4 + 4 + 2 + 1 = 11 Result = 32 - 11 = 21 Integer definition is - any of the natural numbers, the negatives of these numbers, or zero. Whats wrong with this code: n = 10 ((n/3)).is_integer() I do not understand why I cannot set n = any number and check if it is an integer or not. So, integers can be negative {−1, −2,−3, −4, ... }, positive {1, 2, 3, 4, ... }, or zero {0}, Integers = { ..., −4, −3, −2, −1, 0, 1, 2, 3, 4, ... }, (But numbers like ½, 1.1 and 3.5 are not integers). (use direct proof) 2. (n)(k+1) A. Integers vs. decimals and fractions An integer, also called a "round number" or “whole number,” is any positive or negative number that does not include decimal parts or fractions. I have tried using the function isinteger, but I obtain, for example, isinteger(3) = 0.Apparently, any constant is double-precision by Matlab default, and is therefore not recognized as an integer. Because you can't \"count\" zero. First of all we should know that “Every whole number is an integer, but every integer may not be a whole number.”. Counting Numbers are Whole Numbers, but without the zero. The opposite of 2 is -2. Integer size can be determined from PHP_INT_SIZE, maximum value from PHP_INT_MAX since PHP 4.4.0 and PHP 5.0.5." Integer Multiplication. when you only want positive integers, say "positive integers". No. When I as doing my graduation one of my professors Mr. Jaggi told us “Maths student must be polite” thats why I used here ‘may’. So there you Fractions, decimals, and percents are out of this basket. Integers are helpful when modeling real life situations. and which one of these would be an integer? Who | {
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are helpful when modeling real life situations. and which one of these would be an integer? Who is the longest reigning WWE Champion of all time? In most programming languages, you can convert a number into an integer … If you can cleanly divide a number by 2, the number is even. The integer 30 is a Composite number. Integer kan verwijzen naar: Integriteit (persoon), ... Deze pagina is voor het laatst bewerkt op 30 jul 2017 om 14:55. Does pumpkin pie need to be refrigerated? Examples of negative integers are -1, -2, -3, and -4. 0 k n k + 2. Properties of the number 30 The integer 30 is an even number. Thus, when x is a natural number, there is no solution. What are the factors of 602? 42 is greater than 30, so 30 is an abundant number. -5 + 5=0 2. 30 has an aliquot sum of 42; the second sphenic number and all sphenic numbers of this form have an aliquot sum 12 greater than themselves. You can declare an integer by putting a L after it. Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. For example, 3, -10, and 1,025 are all integers, but 2.76 (decimal), 1.5 (decimal), and 3 ½ (fraction) are not. For example, is the number -8 a whole number? Integer Rules: A Video Watch this video to better understand the correct procedure for adding, subtracting, multiplying, and dividing positive and negative whole numbers. is YES. Furthermore, for 24 to be an integer, you should be able to write 24 without a fractional or decimal component. 2012-06-30 12:44:11 2012-06-30 12:44:11. The problem with this approach is what you consider to be practically an integer. C. The two quantities are equal. In the Registry Editor window, navigate to the folder path below depending on your Photoshop version.. Photoshop CC 2018: Navigate to HKEY_CURRENT_USER\SOFTWARE\Adobe\Photoshop\120.0 Photoshop CC 2017: Navigate to HKEY_CURRENT_USER\SOFTWARE\Adobe\Photoshop\110.0 … The opposite of -13 is 13. The square of an integer is 30 more than the integer. Examples of positive | {
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opposite of -13 is 13. The square of an integer is 30 more than the integer. Examples of positive integers are 1, 2, 3, and 4. He has been teaching from the past 9 years. -----> 1. For example, 2 , 5 , 0 , − 12 , 244 , − 15 and 8 are all integers. Integer is a math term for a number that is a whole number. The answer to the question: "Is 30 an integer?" Walk through the difference between whole numbers & integers. > How do you check if a value in Java is of an integer type? MATLAB ® has four signed and four unsigned integer classes. Integers are the set of whole numbers and their opposites. He has been teaching from the past 9 years. He provides courses for … These numbers are used to perform various arithmetic operations, like addition, subtraction, multiplication and division.The examples of integers are, 1, 2, 5,8, -9, -12, etc. Is it an integer? An integer.An integer.An integer.An integer. An integer, also called a "round number" or “whole number,” is any positive or negative number that does not include decimal parts or fractions. Quantity A is greater. How to use integer in a sentence. Signed types enable you to work with negative integers as well as positive, but cannot represent as wide a range of numbers as the unsigned types because one bit is used to … $\begingroup$ "For this ratio to not be an integer m and n must be distinct which implies that x is not an integer quantity." 5- (-4)=1 please help i have a test tommorrow and i just dont get what an integer is. Because you can't "count" zero. The material on this site can not be reproduced, distributed, transmitted, cached or otherwise used, except with prior written permission of Multiply. My only temporary fix was to change the JavaScript that handles the fields, as it was using regex to check if it was and Integer so i changed it to convert the string to int and check if it was NaN Comment Share Find out if 24 is an Integer here! Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. | {
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if 24 is an Integer here! Davneet Singh is a graduate from Indian Institute of Technology, Kanpur. Yes. An even number is any integer (a whole number) that can be divided by 2. You could say that 30/6 is just another name for 5, and then it is an integer. 30 is the integer in this case. Why don't libraries smell like bookstores? integer - any of the natural numbers (positive or negative) or zero; "an integer is a number that is not a fraction" whole number. (30 points) An integer n is prime iff n >= 2 and n's only factors (divisors) are 1 and itself. And 30 is an integer and not a fraction. Suppose x and y are even. A comprehensive database of more than 56 integer quizzes online, test your knowledge with integer quiz questions. What are the three consecutive odd integers with sum of -30? It is a special set of whole numbers comprised of zero, positive numbers and negative numbers and denoted by the letter Z. My only temporary fix was to change the JavaScript that handles the fields, as it was using regex to check if it was and Integer so i changed it to convert the string to int and check if … go, not everyone agrees on a simple thing! Integer Question: Is 24 an Integer? We are simply taking "5 lots of −3", like this: It should be noted that an integer can never be a fraction, a decimal or a per cent. When I as doing my graduation one of my professors Mr. Jaggi told us “Maths student must be polite” thats why I used here ‘may’. sound intelligent. You might be wondering, what is an opposite? python 2.7.4 error: exactly the same as integers. In fact I can't think of any proof of this sentence which doesn't prove the whole theorem along the way. Integer Question: Is 20 an Integer? The opposite of 24 is -24. The number line goes on forever in both directions. Solution for 1. Is a none repeating decimal an integer? Thanks for your help! The number 0 is also considered an integer even though it has neither a positive or negative value. What details make Lochinvar an | {
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an integer even though it has neither a positive or negative value. What details make Lochinvar an attractive and romantic figure? Question 256872: The sum of an integer and its square is 30. 30 is the smallest sphenic number, and the smallest of the form 2 × 3 × r, where r is a prime greater than 3. What is the conflict of the story of sinigang? What is an Integer? So they are 1, 2, 3, 4, 5, ... (and so on). What are the steps to the solution? If x is even, then x y is even. An integer is what is more commonly known as a whole number.It may be positive, negative, or the number zero, but it must be whole. In some cases, the definition of whole number will exclude the number zero, or even the set of negative numbers, but this is not as common as the more inclusive use of the term. n is an integer, and k is not an integer. De tekst is beschikbaar onder de licentie Creative Commons Naamsvermelding/Gelijk delen, er kunnen aanvullende voorwaarden van toepassing zijn. The symbol of integers is “ Z “. See more. Examples– -2.4, 3/4, 90.6. Every integer is a rational number. Quantity B is greater. "Natural Numbers" can mean either "Counting Numbers" {1, 2, 3, ...}, or "Whole Numbers" {0, 1, 2, 3, ...}, depending on the subject. The set of integers is infinite and has no smallest element and no largest element. > is.integer(66L) [1] TRUE > is.integer(66) [1] FALSE Also functions like round will return a declared integer, which is what you are doing with x==round(x). This is shown below. And some people say that zero is NOT a whole number. An integer is a positive or negative whole number. The … PHP does not support unsigned integers. The basic rules of math. For 20 to be an integer, 20 has to be a whole number. How is 30 an even number? Our online integer trivia quizzes can be adapted to suit your requirements for taking some of the top integer quizzes. Furthermore, for 20 to be an integer, you should be able to write 20 without a fractional or decimal component. Definitions. | {
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an integer, you should be able to write 20 without a fractional or decimal component. Definitions. You could try this. (use indirect… It is impossible to answer your question without … 0 0 1 ... An integer is a number that is not a fraction, and decimal numbers are decimal fractions. There are three Proper… Integers The integers are the set of whole numbers and their opposites. Find out if 20 is an Integer here! The farther the number is to the right, the larger the value is. Base Representation; 2: … This is one of those tricky definition questions. Even and Odd Numbers (use indirect… Find the number. It cannot be done. Hello, We have a drafter who's Map 3D 2013 Drafting Settings dialog box is giving him a pop up that says "Please enter an integer from 1 to 30". Solution for 1. Here are some examples: We can compare integers the same way that we compare positive whole numbers. The number 30 as a Roman Numeral XXX The number 30 in various Numeral Systems. Fractions and decimals are not included in the set of integers. please give me an example or two of an integer. I want to know how to test whether an input value is an integer or not. Please note this from the Integer datatype page: "The size of an integer is platform-dependent, although a maximum value of about two billion is the usual value (that's 32 bits signed). Some examples of integers include 1, 3, 4, 8, 99, 108, -43, … Examples of Integers – 1, 6, 15. I have tried using the function isinteger, but I obtain, for example, isinteger(3) = 0.Apparently, any constant is double-precision by Matlab default, and is therefore not recognized as an integer. In the Registry Editor window, navigate to the folder path below depending on your Photoshop version.. Photoshop CC 2018: Navigate to HKEY_CURRENT_USER\SOFTWARE\Adobe\Photoshop\120.0 Photoshop CC 2017: Navigate to HKEY_CURRENT_USER\SOFTWARE\Adobe\Photoshop\110.0 Photoshop CC 2015.5: Navigate to HKEY_CURRENT_USER\SOFTWARE\Adobe\Photoshop\100.0 … For 24 to be an | {
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Photoshop CC 2015.5: Navigate to HKEY_CURRENT_USER\SOFTWARE\Adobe\Photoshop\100.0 … For 24 to be an integer, 24 has to be a whole number. A modular type is an integer type with all arithmetic modulo a specified positive modulus; such a type corresponds to an unsigned type with wrap-around semantics. 30 is a whole number that can be written without a fractional component, thus 30 is an integer. It is not only accurate, it makes you In Maths, integers are the numbers which can be positive, negative or zero, but cannot be a fraction. An integer is any number including 0, positive numbers, and negative numbers. But every rational number is not an integer, since for b not equal to 1, if b does not divide a then w don't get an integer. Integer definition, one of the positive or negative numbers 1, 2, 3, etc., or zero. Examples of Integer Multiplication (a) Here is an integer times an integer: 5 xx -3 = -15 Why is this so? But you for sure can say it is a rational number, which is any number that can be expressed as a … Choose from 500 different sets of 30 math integers flashcards on Quizlet. How will understanding of attitudes and predisposition enhance teaching? In the equation 2 + 1/2, the number 2 is the integer and 1/2 is the fraction. Find the integer The sum of an integer and its square is - Answered by a verified Math Tutor or Teacher Whole Numbers are simply the numbers 0, 1, 2, 3, 4, 5, ... (and so on), (But numbers like ½, 1.1 and 3.5 are not whole numbers.). The number line is used to represent integers. Start by using a number line. Compare -9 and -2. 0 0 1 ... 176 + 125 = 301 Therefore, the lowest positive integer you can add is 125. When did organ music become associated with baseball? An integer (pronounced IN-tuh-jer) is a whole number (not a fractional number) that can be positive, negative, or zero. An integer (from the Latin integer meaning "whole") is colloquially defined as a number that can be written without a fractional component.For example, 21, | {
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colloquially defined as a number that can be written without a fractional component.For example, 21, 4, 0, and −2048 are integers, while 9.75, 5 + 1 / 2, and √ 2 are not. Suppose x and y are even. I want to know how to test whether an input value is an integer or not. Where is telephone area code 035? 4. Consider a function Factor_In_Range (k , n) to return true iff some integer in the range k to n … And everyone agrees on the definition of an integer, so when in doubt say "integer". This is indicated by the arrows. All Rights Reserved. If a a a and b b b are integers such that a ⋅ b = 0 a \cdot b = 0 a ⋅ b = 0, then a = 0 a=0 a = 0 or b = 0 b=0 b = 0. Rational number is a number which can be expressed as a/b where a,b are integers and b is not 0. a) -3 or 10 b) 3 or -10 c) -5 or 6 d) 5 or -6 Help! For 30 to be an integer, 30 has to be a whole number. Ex 6.1, 2 Solve –12x > 30, when x is a natural number –12x > 30 –x > 3012 –x > 2.5 Since x is negative, we multiply both sides by -1 & change the signs (–x) × (-1) < 2.5 × (-1) x < − 2.5 (i) Since x is a natural number, it can be only positive num The base range of a signed integer type includes at least the values of the specified range. is.integer tests if the number is declared as an integer. B. An integer_type_definition defines an integer type; it defines either a signed integer type, or a modular integer type. You need to prove this, as nothing you have said makes it obvious this is true. Learn 30 math integers with free interactive flashcards. But you for sure can say it is a rational number, which is any number that can be expressed as a fraction of two integers.. An octal integer is an integer represented by a string of digits from the set $\{0,1,2,3,4,5,6,7\}$, each representing (in that order) the usual first eight non-negative integers. Integers are like whole numbers, but they also include negative numbers ... but still no fractions allowed! This is one of those tricky definition questions. If a is an | {
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... but still no fractions allowed! This is one of those tricky definition questions. If a is an odd integer, then a2 +3a+5 is odd. These are all integers (click to mark), and they continue left and right infinitely: Some people (not me) say that whole numbers can also be negative, which makes them So they are 1, 2, 3, 4, 5, ... (and so on). What is an integer? Therefore, these numbers can never be integers: fractions; decimals; percents; Adding and Subtracting Integers. Given an integer number n, return the difference between the product of its digits and the sum of its digits.. He provides courses for Maths and Science at Teachoo. ok i dont get.... is it just considered an integer if the dominate is 1? Counting Numbers are Whole Numbers, but without the zero. Examples of integers are: … If you divide by 2 and there is a remainder left, then the number is odd. The set of integers consists of zero (), the positive natural numbers (1, 2, 3, ...), also called whole numbers or … For example, 3, -10, and 1,025 are all integers, but 2.76 (decimal), 1.5 … Looking at a number line can help you when you need to add or subtract integers. Integers are positive whole numbers and their opposites, including zero. characteristic - the integer part (positive or negative) of the representation of a logarithm; in the expression log 643 = 2.808 the characteristic is 2 Furthermore, for 30 to be an integer, you should be able to write 30 without a fractional or decimal component. If a is an odd integer, then a2 +3a+5 is odd. For any integer a a a, the additive inverse − a-a − a is an integer. (use direct proof) 2. (ii) x is an integer Now, x < 2.5 Since x is an integer, it can be negative but cannot be a decimal So, x can be any negative number less than 3 Hence, the values of x which will satisfy the inequality are = { , 6, 5, 4, 3} Show More The word integer originated from the Latin word “Integer” which means whole. Integers are whole numbers and their negative opposites. Copyright | {
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word “Integer” which means whole. Integers are whole numbers and their negative opposites. Copyright © 2020 Multiply Media, LLC. Integers Integer Classes. Is 30 an integer? When multiplying integers, we can think of multiplying "blocks" of negative numbers. expressed in rational form as 30/1. 2020-09-30 17:50:29 2020-09-30 17:50:29. First of all we should know that “Every whole number is an integer, but every integer may not be a whole number.”. An integer (from the Latin integer meaning "whole") is colloquially defined as a number that can be written without a fractional component.For example, 21, 4, 0, and −2048 are integers, while 9.75, 5 + 1 / 2, and √ 2 are not. Intager. What is the contribution of candido bartolome to gymnastics? You could say that 30/6 is just another name for 5, and then it is an integer. Like this (note: zero isn't positive or negative): For an interesting look at other types of numbers read The Evolution of Numbers, { ... −4, −3, −2, −1, 0, 1, 2, 3, 4, ... }, Adding and Subtracting Positive and Negative Numbers, Negative Integers = { ..., −4, −3, −2, −1 }, Non-Negative Integers = { 0, 1, 2, 3, 4, ... }. | {
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# Thread: When theta must be positive and when negative
1. ## When theta must be positive and when negative
I'm grappling with just not seeing when to use negative angles and when to use positive angles when solving trig functions.
If I pick a point on a graph with coordinates (5,-5) the angle formed between the x-axis and the hypotenuse can be either $-45^\circ$ or $315^\circ$ (along with all multiples of $360^\circ$). Yes?
If this is true, then I would think that $\sin(-45^\circ) = \sin(315^\circ) = \dfrac{-\sqrt{2}}{2}$
If I plot the sine function I see that $\sin(-45^\circ) = \dfrac{-\sqrt{2}}{2}$ but my calculator gives me an error when entering $\sin(-45^\circ)$ or $\sin(\dfrac{-\pi}{4})$
Likewise, my current lesson on complex numbers requires that I use the sin and cos functions of $\dfrac{7\pi}{4}$ rather than $\dfrac{-\pi}{4}$
I would appreciate an explanation of when to use a negative reference angle and when to use the positive angle $< 360^\circ$
2. ## Re: When theta must be positive and when negative
Originally Posted by B9766
I'm grappling with just not seeing when to use negative angles and when to use positive angles when solving trig functions. | {
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If I pick a point on a graph with coordinates (5,-5) the angle formed between the x-axis and the hypotenuse can be either $-45^\circ$ or $315^\circ$ (along with all multiples of $360^\circ$). Yes?
If this is true, then I would think that $\sin(-45^\circ) = \sin(315^\circ) = \dfrac{-\sqrt{2}}{2}$
If I plot the sine function I see that $\sin(-45^\circ) = \dfrac{-\sqrt{2}}{2}$ but my calculator gives me an error when entering $\sin(-45^\circ)$ or $\sin(\dfrac{-\pi}{4})$
Likewise, my current lesson on complex numbers requires that I use the sin and cos functions of $\dfrac{7\pi}{4}$ rather than $\dfrac{-\pi}{4}$
I would appreciate an explanation of when to use a negative reference angle and when to use the positive angle $< 360^\circ$
Some older texts & authors still use degrees as well as reference angles which acute. If the terminal side is in quadrant II the reference angle is $180^{\circ}-\theta$;
terminal side is in quadrant III the reference angle is $180^{\circ}+\theta$;terminal side is in quadrant IV the reference angle is $360^{\circ}-\theta$.
In II the $\sin$ is positive, in III the $\tan$ is positive, & in IV the $cos$ is positive.
Look Here. The basic parts of wolframalpha are free to use. The whole site is inexpensive for students & the is a version for small screen devices.
3. ## Re: When theta must be positive and when negative
I don't know why your calculator can't determine sin(-pi/4)). Mine is ok with it, as well as sin(7 pi/4) | {
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Whether to measure angles as clockwise from the x-axis or counterclockwise depends on the application, and may be influenced by the particulars of the problem you are working. As an example, to calculate the y-coordinate of a point on a wheel that is rotating about the origin you would use y=R sin(theta) where theta is positive if the wheel rotates counterclockwise and theta is negative if it's rotating clockwise. This allows for a nice smooth plot of theta and y versus time. As for complex analysis - you are correct that by convention angles are measured clockwise from the real axis, and of course they are given in radians.
4. ## Re: When theta must be positive and when negative
Really, thanks ChipB.
As for the calculator, it was user error. Much to my chagrin, I keep forgetting the TI-84 requires the use of the (-) key rather than the minus key for negative values.
My Trig and Pre-Calc course is a self-study program. I worry sometimes when I don't come up with the same answers as the text.
As an example:
Find Trigonometric form of $5 - 5i$. I came up with $z = 5\sqrt{2} (\cos\dfrac{-\pi}{4}+i \sin\dfrac{-\pi}{4})$
But the text answer was $z = 5\sqrt{2} (\cos\dfrac{7\pi}{4}+i \sin\dfrac{7\pi}{4})$
From your answer it seems either should be accepted as correct. Is that true? Would there normally be a preference?
5. ## Re: When theta must be positive and when negative | {
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5. ## Re: When theta must be positive and when negative
Originally Posted by B9766
Really, thanks ChipB.
As for the calculator, it was user error. Much to my chagrin, I keep forgetting the TI-84 requires the use of the (-) key rather than the minus key for negative values.
My Trig and Pre-Calc course is a self-study program. I worry sometimes when I don't come up with the same answers as the text.
As an example:
Find Trigonometric form of $5 - 5i$. I came up with $z = 5\sqrt{2} (\cos\dfrac{-\pi}{4}+i \sin\dfrac{-\pi}{4})$
But the text answer was $z = 5\sqrt{2} (\cos\dfrac{7\pi}{4}+i \sin\dfrac{7\pi}{4})$
From your answer it seems either should be accepted as correct. Is that true? Would there normally be a preference?
The "angle" associated with a complex is known as its argument. So $\displaystyle \arg (5 - 5\bf{i}) = - \frac{\pi }{4}$
Note that I used the negative value. In resent times that has become the preferred notation. However older texts often used the convention that
$\displaystyle 0\le \arg (z) <2\pi$ that seems to be the case with your textbook.
Just note that $2\pi -\frac{\pi}{4}=\frac{7\pi}{4}$ That means that the two are equivalent.
6. ## Re: When theta must be positive and when negative
Thanks for the confirmation, Plato. | {
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# Are polynomials dense in Gaussian Sobolev space?
Let $\mu$ be standard Gaussian measure on $\mathbb{R}^n$, i.e. $d\mu = (2\pi)^{-n/2} e^{-|x|^2/2} dx$, and define the Gaussian Sobolev space $H^1(\mu)$ to be the completion of $C_c^\infty(\mathbb{R}^n)$ under the inner product $$\langle f,g \rangle_{H^1(\mu)} := \int f g\, d\mu + \int \nabla f \cdot \nabla g\, d\mu.$$
It is easy to see that polynomials are in $H^1(\mu)$. Do they form a dense set?
I am quite sure the answer must be yes, but can't find or construct a proof in general. I do have a proof for $n=1$, which I can post if anyone wants. It may be useful to know that the polynomials are dense in $L^2(\mu)$.
Edit: Here is a proof for $n=1$.
It is sufficient to show that any $f \in C^\infty_c(\mathbb{R})$ can be approximated by polynomials. We know polynomials are dense in $L^2(\mu)$, so choose a sequence of polynomials $q_n \to f'$ in $L^2(\mu)$. Set $p_n(x) = \int_0^x q_n(y)\,dy + f(0)$; $p_n$ is also a polynomial. By construction we have $p_n' \to f'$ in $L^2(\mu)$; it remains to show $p_n \to f$ in $L^2(\mu)$. Now we have \begin{align*} \int_0^\infty |p_n(x) - f(x)|^2 e^{-x^2/2} dx &= \int_0^\infty \left(\int_0^x (q_n(y) - f'(y)) dy \right)^2 e^{-x^2/2} dx \\ &\le \int_0^\infty \int_0^x (q_n(y) - f'(y))^2\,dy \,x e^{-x^2/2} dx \\ &= \int_0^\infty (q_n(x) - f'(x))^2 e^{-x^2/2} dx \to 0 \end{align*} where we used Cauchy-Schwarz in the second line and integration by parts in the third. The $\int_{-\infty}^0$ term can be handled the same with appropriate minus signs.
The problem with $n > 1$ is I don't see how to use the fundamental theorem of calculus in the same way.
-
Nate, I once needed this result, so I proved it in Dirichlet forms with polynomial domain (Math. Japonica 37 (1992) 1015-1024). There may be better proofs out there, but you could start with this paper. | {
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-
Hi Byron, I'm interested as well. I couldn't not find the paper using MathSciNet. Can I get a copy? – Jonas Teuwen Nov 2 '11 at 21:34
Thanks Byron! I think my library has it, I'll head there right now. The MathSciNet link is ams.org/mathscinet-getitem?mr=1196376, for reference. – Nate Eldredge Nov 2 '11 at 21:51
@Jonas I have scanned the paper and added it to my "Publications" page. You want Proposition 1.3. I hope my argument holds up to scrutiny! It is not very profound, just a multi-dimensional version of Nate's argument. – Byron Schmuland Nov 2 '11 at 22:16
@Byron: Thanks. I've got it. – Jonas Teuwen Nov 2 '11 at 22:16
Thanks a lot for that! – t.b. Nov 2 '11 at 22:28
Byron's paper, which he linked in his (accepted) answer, has a proof in a more general setting (where the Gaussian measure can be replaced by any measure with exponentially decaying tails). Here is a specialization of it to the Gaussian case, which I wrote up to include in some lecture notes. I guess I was on the right track, the trick was to differentiate more times. | {
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$\newcommand{\R}{\mathbb{R}}$ For continuous $\psi : \R^k \to \R$, let \begin{equation*} I_i \psi(x_1, \dots, x_k) = \int_0^{x_i} \psi(x_1, \dots, x_{i-1}, y, x_{i+1}, \dots, x_k)dy. \end{equation*} By Fubini's theorem, all operators $I_i, 1 \le i \le k$ commute. If $\psi \in L^2(\mu)$ is continuous, then $I_i \psi$ is also continuous, and $\partial_i I_i \psi = \psi$. Moreover, \begin{align*} \int_{0}^\infty |I_i \psi (x_1, \dots, x_k)|^2 e^{-x_i^2/2} dx_i &= \int_{0}^\infty \big\lvert\int_0^{x_i} \psi(\dots, y,\dots)\,dy\big\rvert^2 e^{-x_i^2/2} dx_i \\ &\le \int_0^\infty \int_0^{x_i} |\psi(\dots, y, \dots)|^2 \,dy x_i e^{-x_i^2/2}\,dx_i && \text{Cauchy--Schwarz} \\ &= \int_0^\infty |\psi(\dots, x_i, \dots)|^2 e^{-x_i^2/2} dx_i \end{align*} where in the last line we integrated by parts. We can make the same argument for the integral from $-\infty$ to $0$, adjusting signs as needed, so we have \begin{equation*} \int_\R |I_i \psi(x)|^2 e^{-x_i^2/2} dx_i \le \int_\R |\psi_i(x)|^2 e^{-x_i^2/2} dx_i. \end{equation*} Integrating out the remaining $x_j$ with respect to $e^{-x_j^2/2}$ shows $$||{I_i \psi}||_{L^2(\mu)}^2 \le ||{\psi}||_{L^2(\mu)}^2,$$ i.e. $I_i$ is a contraction on $L^2(\mu)$.
Now for $\phi \in C_c^\infty(\R^k)$, we can approximate $\partial_1 \dots \partial_k \phi$ in $L^2(\mu)$ norm by polynomials $q_n$. If we let $p_n = I_1 \dots I_k q_n$, then $p_n$ is again a polynomial, and $p_n \to I_1 \dots I_k \partial_1 \dots \partial_k \phi = \phi$ in $L^2(\mu)$. Moreover, $\partial_i p_n = I_1 \dots I_{i-1} I_{i+1} \dots I_k q_n \to I_1 \dots I_{i-1} I_{i+1} \dots I_k \partial_1 \dots \partial_k \phi = \partial_i \phi$ in $L^2(\mu)$ also.
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I think I have a different proof. Let $\gamma$ be the Gaussian measure, that is, $\gamma$ is given by the Radon-Nikodym density, $$\mathrm{d}\gamma(x) = \frac{\mathrm{e}^{-x^2}}{\sqrt{\pi}} \mathrm{d}x.$$ Also, consider the Ornstein-Uhlenbeck operator given as, $$L := -\frac12 \Delta + x \cdot \nabla.$$ We can verify that for $u$ and $v$ in $C_{\mathrm{c}}^\infty(\mathbf R^d)$ we have the symmetricity $$\int_{\mathbf{R}^d} u L v \, \gamma(\mathrm{d}x) = \int_{\mathbf{R}^d} \nabla u \cdot \nabla v \, \gamma(\textrm{d}x).$$ The nice thing about this is that the Hermite polynomials are an orthogonal basis for the Gaussian Hilbert space, that is, $L^2(\gamma)$. These can be define using the Rodrigues' formula, that is, $$H_n(x) = (-1)^n \mathrm{e}^{x^2} \partial_x^n \mathrm{e}^{-x^2}.$$ Furthermore, we have, $$L H_n = n H_n.$$ Also, we have, $$H_n' = 2n H_{n - 1}.$$ Proving that the polynomials form a basis for $L^2(\gamma)$ is not hard, just consider the entire function $$F(z) = \int_{-\infty}^\infty \mathrm e^{zx - x^2} \, \frac{\mathrm{d}x}{\sqrt\pi}.$$ Hence, so are the Hermite polynomials as they are linear combinations of the monomials $(x \mapsto x^n)_n$. Higher-order Hermite polynomials are the canonical tensor extensions. So, given an $f$ in $L^2$ we have that $f$ is given in the form $$f = \lim_N f_N = \lim_N \sum_{n = 0}^N a_n \frac{a_n}{\sqrt{n! 2^n}}.$$ And $f_N$ gives the limit of functions that converges to $f$.
Also, recall the orthogonality (after scaling), $$\int_{\mathbf{R}^d} h_n h_m \, \gamma(\mathrm{d}x) = \delta_{nm}.$$ where $$h_n = \frac{H_n}{\sqrt{n! 2^n}}.$$ | {
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We can rewrite the inner product on the Sobolev space as $$\langle u, v \rangle_{H^1} = \langle u, (L + 1) v \rangle_{L^2(\gamma)}.$$ We only have to care about the bilinear form $$\mathcal E(u, v) = \int_{\mathbf{R}^d} u L v \, \gamma(\mathrm{d}x).$$ Or so, picking $u = v = f_N - f$ we have after noting that $$g_N := f - f_N = \sum_{n = N + 1}^\infty a_n \frac{H_n}{\sqrt{n! 2^n}},$$ \begin{align} \mathcal E(u, v) &= \int_{\mathbf{R}^d} f_N L f_N \, \gamma(\mathrm{d}x)\\ &\quad + \int_{\mathbf{R}^d} f L f \, \gamma(\mathrm{d}x)\\ &\quad - 2\int_{\mathbf{R}^d} f_N L f \, \gamma(\mathrm{d}x)\\ &= \sum_{n = 0}^\infty n \frac{|a_n|^2}{n! 2^n} - \sum_{n = 0}^N n \frac{|a_n|^2}{n! 2^n}\\ &= \sum_{n = N + 1}^\infty n \frac{|a_n|^2}{n! 2^n}. \end{align}. And as $\sum |a_n|^2$ converges due to the $L^2$ density, so should this.
I hope I did not make any mistakes, just occurred to me while I was biking...
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Kronecker Product (G Dataflow)
Calculates the Kronecker product of two input matrices.
matrix A
The first input matrix.
This input accepts a 2D array of double-precision, floating point numbers or 2D array of complex double-precision, floating point numbers.
Default: Empty array
matrix B
The second input matrix.
This input accepts a 2D array of double-precision, floating point numbers or 2D array of complex double-precision, floating point numbers.
Default: Empty array
error in
Error conditions that occur before this node runs. The node responds to this input according to standard error behavior.
Default: No error
kronecker product
Matrix containing the Kronecker product of the first and second input matrices.
The number of rows in kronecker product is the product of the number of rows in the first and second input matrices. The number of columns in kronecker product is the product of the number of columns in the first and second input matrices.
error out
Error information. The node produces this output according to standard error behavior.
Algorithm for Calculating the Kronecker Product
If A is an n-by-m matrix and B is a k-by-l matrix, the Kronecker product of A and B, C = AB, results in a matrix C with dimensions nk-by-ml. This node calculates the Kronecker product using the following equation.
$C={\left[\begin{array}{cccc}{a}_{11}B& {a}_{12}B& \dots & {a}_{1m}B\\ {a}_{21}B& {a}_{22}B& \dots & {a}_{2m}B\\ ⋮& ⋮& \ddots & ⋮\\ {a}_{n1}B& {a}_{n2}B& \dots & {a}_{nm}B\end{array}\right]}_{nk×ml}$
For example, if
$\begin{array}{cc}A=\left[\begin{array}{cc}1& 2\\ 3& 4\end{array}\right]& B=\left[\begin{array}{cc}5& 6\\ 7& 8\end{array}\right]\end{array}$
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