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February 5

A quetion in the real line topology

Hi! Here is my question: Given a countable union of countable sets $A_{i}$ such that $\cup A_{i}=R$. Does it implies that at list one of this sets is dense in some open segment of R? and if yes, how to prove it? Thanks! Topologia clalit (talk) 12:32, 5 February 2013 (UTC)

Are you sure you mean a countable union of countable sets? Because that requires rejecting the axiom of choice.--190.18.159.129 (talk) 12:49, 5 February 2013 (UTC)

Nevermind, our Baire category theorem article claims it can be proven for the reals without using choice. So the answer is yes, it follows from the Baire category theorem.--190.18.159.129 (talk) 13:08, 5 February 2013 (UTC)

Hi, As a matther of fact I have looked over this theorem. But I don't se how it's connected. The written states that the reals are a Baire space while: "A Baire space is a topological space with the following property: for each countable collection of open dense sets Un, their intersection ∩ Un is dense." But in my question there is a countable union of sets (not all countable) where $\cup A_{i}=R$. But it is not given that the sets are dense in R. and I want to show that at least one of them is dense in some open interval of R. I don't need to show that their intersection is dense.. I don't understand the connection.. Thanks Topologia clalit (talk) 13:58, 10 February 2013 (UTC)

The connection is via the complement. Let $B_{i}$ denote the closure of $A_{i}$. You wish to show that one of the $B_{i}$ contains an interval. Since $\bigcup _{i}B_{i}\supseteq \bigcup _{i}A_{i}=\mathbb {R}$, it follows that $\bigcap _{i}(\mathbb {R} -B_{i})=\emptyset$. But each $(\mathbb {R} -B_{i})$ is an open set, so by the contrapositive of the Baire category theorem, at least one of the $(\mathbb {R} -B_{i})$ is not dense. That means there is some open interval it does not intersect, and thus $B_{i}$ contains that open interval.--190.18.159.129 (talk) 03:49, 11 February 2013 (UTC)

February 6

Newton–Pepys problem

Reading about Samuel Pepys, I came across this interesting exercise in probabilities. I've read the article linked in the subject heading and struggled to understand the maths. Please go easy on me. I understand the simple maths required to work out the probability of proposition A ("Six fair dice are tossed independently and at least one “6” appears.") but am struggling with B/C.

It seems that B and C are both calculated using binomial distribution, but that article is gibberish to me. Is there a simple way of explaining how the calculation given for (say) proposition B was arrived at - and perhaps less likely, is there a simple way to explain why the calculation works? NB Newton's intuitive reasoning, given at the bottom of the page, makes perfect sense. A genius mathematician who can explain things to us mathematical plebs - a role model, eh? --Dweller (talk) 11:18, 6 February 2013 (UTC)

For B (two or more 6s in twelve throws) you first calculate the probability of throwing no 6s. Each of the twelve throws must be between 1 and 5, so this probability is (5/6)^12, which is approximately 11.2%. Then you calculate the probaqbility of throwing just one 6. This is 12 x (1/6) x (5/6)^11 - the factor of 12 is included because the one 6 can be in any one of twelve places in the sequence of throws - which is 26.9%. The probability of throwing two or more 6s is then 1 - prob of no 6s - prob of one 6 = 100% - 11.2% - 26.9% = 61.9%. Same method works for C, except you also need to work out the probability of exactly two 6s and subtract this too. Gandalf61 (talk) 11:37, 6 February 2013 (UTC)

Thanks, that's quite digestible. --Dweller (talk) 11:50, 6 February 2013 (UTC)

Identity

$\lim _{n\rightarrow \infty }{\frac {\int _{0}^{1}x^{n}(1-x)^{n}f(x)dx}{\int _{0}^{1}x^{n}(1-x)^{n}dx}}=f(1/2)$. I can understand the intuition of this, but is there a proof?

Someone suggested expanding $f$ as a Taylor series.

$\lim _{n\rightarrow \infty }{\frac {\int _{0}^{1}x^{n}(1-x)^{n}\sum _{k=0}^{\infty }{\frac {(x-1/2)^{k}}{k!}}f^{(k)}(1/2)dx}{\int _{0}^{1}x^{n}(1-x)^{n}dx}}$

The integrand in the numerator is now a power series and can be easily integrated for known integer n (but remain a series), but I see no closed form for unknown n. What's the next step? --AnalysisAlgebra (talk) 19:30, 6 February 2013 (UTC)

Link to previous question

You didn't specify the conditions on f, I'll assume it's continuous at 1/2 and bounded.

Write the integral as the sum of an integral over a neighborhood around 1/2, and an integral over the rest of [0,1]. Bound the second integral using bounds on f and $x^{n}(1-x)^{n}$. In the neighborhood bound the distance between $f$ and $f(1/2)$ and use it to bound the distance between $\int _{1/2-\delta }^{1/2+\delta }x^{n}(1-x)^{n}f(x)dx$ and $\int _{1/2-\delta }^{1/2+\delta }x^{n}(1-x)^{n}f(1/2)dx$. Use these bounds and the triangle inequality to bound the distance between ${\frac {\int _{0}^{1}x^{n}(1-x)^{n}f(x)dx}{\int _{0}^{1}x^{n}(1-x)^{n}dx}}$ and $f(1/2)$. Tighten the bound as $n\to \infty$. -- Meni Rosenfeld (talk) 20:36, 6 February 2013 (UTC)

_______

Let $g(x;n):=x^{n}(1-x)^{n}$, and let $k_{n}$ = the integral of $g(x;n)$ from zero to one, and let $h(x;n):={\frac {g(x;n)}{k_{n}}}$. Hence the integral from zero to one of $h(x;n)$ equals one, so $h(x;n)$ (defined as zero outside the range zero to one) is a probability density function symmetric around 1/2. Using these, the OP's expression (upon cancelling out $k_{n}$ from numerator and denominator) becomes the limit of the integral from zero to one of $h(x;n)f(x)$, which is an expected value expression. My guess is that you can show that as n goes to infinity, $h(x;n)$ becomes zero everywhere except for a point mass of height one at x=1/2, which gives the desired result. (If my guess is wrong about the limit of h, then the desired result must be wrong.) Duoduoduo (talk) 23:02, 6 February 2013 (UTC)

Maclaurin's series, but where?

Dear Wikipedians:

I am confronted with the question "How many terms of the Maclaurin series for sin x are needed to guarantee an error less than 10^-5." But I feel that there is insufficient information given in the question for a single answer. Because depending on the specific value of x, either more or less terms are needed. As an example:

Maclaurin series for sin x is $\sum \limits _{n=0}^{\infty }{\frac {(-1)^{n}x^{2n+1}}{(2n+1)!}}$ where $x\in (-\infty ,\infty )$

If we truncate the series after the nth term, then the error E will be bounded by the size of the first omitted term $|E|\leq {\frac {1}{(2n+1)!}}x^{2n+1}$.

Since $x\in (-\infty ,\infty )$, if we let x = 1, then $|E|\leq {\frac {1}{(2n+1)!}}$ and we have

$|E|\leq {\frac {1}{7!}}={\frac {1}{5040}}=0.0001984>10^{-5}$, when n = 3;

$|E|\leq {\frac {1}{9!}}={\frac {1}{362800}}=0.000002755<10^{-5}$, when n = 4;

So we need at least 4 terms here.

But if we choose x = 2, then $|E|\leq {\frac {2^{2n+1}}{(2n+1)!}}$ and we have

$|E|\leq {\frac {2^{11}}{11!}}={\frac {8}{155925}}=0.00051306>10^{-5}$, when n = 5;

$|E|\leq {\frac {2^{13}}{13!}}={\frac {8}{6081075}}=0.000001315<10^{-5}$, when n = 6;

So we need at least 6 terms here.

And I would imagine that for higher values of x even more terms are needed for the Maclaurin series to converge as any polynomial would start to deviate more from the x axis around which sin x oscillates given sufficiently large x.

Therefore I feel that the question is not well-formed in the way it is being asked here and that x needs to be fixed around a particular value in order for the question to have a particular answer. Am I right with this line of reasoning?

Thanks,

76.75.148.30 (talk) 22:15, 6 February 2013 (UTC)

Maybe the question is intended to limit x to the range from 0 to 2$\pi$? If so, then work it out for the worst case, which is x=2$\pi$. Or maybe the intended range is $-\pi$ to $\pi$. Duoduoduo (talk) 23:36, 6 February 2013 (UTC)

Yes, the larger the argument, the more terms you will need. For any of the trig functions, the first step in numerical evaluation (whatever method is used (see for example CORDIC)), is to reduce the argument to the first quadrant [0,π/2], by using the periodicity and other symmetries of the function. So the question ought to specify 0 ≤ x ≤ π/2. --catslash (talk) 23:54, 6 February 2013 (UTC)

Thank you all for your response. That is what I have suspected all along. 69.158.77.73 (talk) 22:35, 7 February 2013 (UTC)

February 7

3 in each column and row of 6 by 6 grid?

How many ways are there to put 18 items into a 6 by 6 grid so that 1 item goes in each grid square and there are a total of 3 items in each row and each column. For example, you could have rows 1-3 filled in for columns 1-3 and rows 4-6 filled in for columns 4-6 or conversely row n could be filled in from column n+2 to n+4 mod 6. Secondly, how many of them are unique in that they can not be changed into each other by any combination of row switches or column switches? I'm hoping for formulae since what I'm actually looking for is a 20 by 20 grid with 200 items. :) Naraht (talk) 20:18, 7 February 2013 (UTC)

You are looking for the number X of 6x6 bitmatrices A_ij satisfying the column sum condition Σ_i A_ij = 3 and the row sum condition Σ_j A_ij = 3. You have initially that X ≤ 2³⁶ = 68719476736. The row condition Σ_j A_ij = 3 says that there are only ${\binom {6}{3}}$ = 20 possible rows, and so X ≤ 20⁶ = 64000000. The column sum condition tells you that once you have the first 5 rows in place the 6'th row is fixed, so X ≤ 20⁵ = 3200000. Switching rows reduces the possibilities further. You may demand that the rows are in nondescending order. Number the 20 possible rows from 1 through 20. How many nondescending 5-tupples of numbers from 1 through 20 are there? Bo Jacoby (talk) 11:05, 8 February 2013 (UTC).

But even with that number of nondescending 5-tuples, each nondescending 5-tuple must be tested to see if actually has 3 in 3 of the columns and 2 in the other three (so as to give the ability to have a fixed sixth row. For example, a 5-tuple that contains 4 of the first 10 possible rows isn't allowed since that would have 4 in column 1, but keeping track of which rows would give 4 in column 3 isn't as easy to remember.Naraht (talk) 01:19, 9 February 2013 (UTC)

Yes, surely, but it could further improve the upper boundary. Bo Jacoby (talk) 12:35, 10 February 2013 (UTC).

I ran a computer simulation, and got 297,200 possibilities for the 6×6 case. I don't know how many of those are unique, though. I don't think I can brute force the 20×20 case, unfortunately. StuRat (talk) 02:24, 9 February 2013 (UTC)

Surely you mean "distinct" (or something like "essentially distinct"), not "unique". If there are 297,200 of them, they aren't unique. :-) —Bkell (talk) 03:26, 11 February 2013 (UTC)

I'm using "unique" in the way the OP defined it: "unique in that they can not be changed into each other by any combination of row switches or column switches". And, I don't believe they are all unique, in this sense. StuRat (talk) 04:33, 11 February 2013 (UTC)

February 9

significance of a series

what is the significance of the series 1, 1, 1, 1, 1, 1, etc. Does it have any special properties? This is A000012 in the online integers database. — Preceding unsigned comment added by 178.48.114.143 (talk) 06:17, 9 February 2013 (UTC)

The OEIS entry for A000012 lists many properties of this sequence. —Bkell (talk) 18:45, 9 February 2013 (UTC)

For my sandbox

Can somebody calculate the midway point between 500 and 2,200,000,000 please? Please show the calculation as well. This is for a wiki-essay to calculate the largest and smallest religious populations, 2.2 billion being Christians. Thanks Pass a Method talk 20:34, 9 February 2013 (UTC)

What you want is the arithmetic mean, in your case

${500+2,200,000,000 \over 2}$

See also http://www.freemathhelp.com/arithmetic-mean.html for some illustrative examples. You might also want to take a look at midpoint. The midpoint is just a special case of an arithmetic mean where you have two terms. -- Toshio Yamaguchi 20:57, 9 February 2013 (UTC)

For something like this I think the geometric mean is more appropriate. For these values you get

${\sqrt {500*2200000000}}=1048809$

In other words, about a million. Looie496 (talk) 00:30, 10 February 2013 (UTC)

February 10

How to prove

How to prove that tan20⋅tan30=tan10⋅tan50 ? 117.227.40.234 (talk) 03:32, 10 February 2013 (UTC)

Are these in degrees ? If so, why not just multiply them out ? StuRat (talk) 03:34, 10 February 2013 (UTC)

Ya, they are in degrees. How I can I do it using tanA = cot(90-A)? 117.227.145.29 (talk) 03:47, 10 February 2013 (UTC)

Or some any other basic identities. 117.227.18.193 (talk) 06:06, 10 February 2013 (UTC)

((( My browser doesn't show the symbol used between the tans here, can someone say what it is please (* / + - ???) -- SGBailey (talk) 09:49, 10 February 2013 (UTC) )))

Hello, I think the symbol is meant to show multiplication, like latex's '\cdot' (in a hex editor it comes up as 0xB7 which is Unicode (UTF-16)'s 'MIDDLE DOT') so it's

$\tan(20)\cdot \tan(30)=\tan(10)\cdot \tan(50)$.77.86.3.26 (talk) 10:44, 10 February 2013 (UTC)

There may be simpler ways, but here's an outline of mine. Multiply both sides by cos 20 cos 30 cos 10 cos 50 so that the entire identity deals only with sin and cos. Now using the product-to-sum formulas at Trigonometric_identity#Product-to-sum_and_sum-to-product_identities, you can get rid of the 20's and 30's entirely and only be left with angles of 10 and 50. Using the double-angle formulas, you can then replace some of the 50's with 100's. The 100's can in turn be replaced with 10's using the formulas for shifting by 90. Now it's not too hard to prove the identity. You'll still need an angle addition formula, though. 96.46.195.35 (talk) 13:01, 10 February 2013 (UTC)

The identity $\tan(10)=\tan(20)\cdot \tan(30)\cdot \tan(40)$ is mentioned at Trigonometric identities#Identities without variables (though not with a clear explanation, unfortunately) and this identity is identical to the one above by recognising that $\tan(40)=\cot(90-40)=\cot(50)={\frac {1}{\tan(50)}}$. So, the identity the IP mentions is certainly true, provided that WP page is correct. EdChem (talk) 13:03, 10 February 2013 (UTC)

Yes, I just noticed that. See also Morrie's law, which relates directly to the OP's identity in the same way. 96.46.195.35 (talk) 13:08, 10 February 2013 (UTC)

How about this: we can write $\tan(30^{\circ })=1/{\sqrt {(}}3)$, then we have:

$\tan(20)/{\sqrt {(}}3)=\tan(10)\cdot \tan(50)$.

Using your "tanA = cot(90-A)" we have

$\tan(20)\cdot \tan(40)/{\sqrt {(}}3)=\tan(10)$.

From List_of_trigonometric_identities#Double-angle.2C_triple-angle.2C_and_half-angle_formulae we have:

$\tan(2A)={\frac {2\tan(A)}{1-\tan ^{2}(A)}}$, so we have:

$\tan(20)\cdot {\frac {\tan(20)}{1-\tan ^{2}(20)}}/{\sqrt {(}}3)=\tan(10)$.

Using the half-angle formula:

$\tan(A/2)={\frac {\tan(A)}{1+{\sqrt {1+\tan ^{2}(A)}}}}$ we have

$\tan(20)\cdot {\frac {\tan(20)}{1-\tan ^{2}(20)}}/{\sqrt {(}}3)={\frac {\tan(20)}{1+{\sqrt {1+\tan ^{2}(20)}}}}$.

What do people think?77.86.3.26 (talk) 13:15, 10 February 2013 (UTC)

You dropped a factor of 2 along the way. Also, although if you restore the 2 the calculations are correct, the argument is still incomplete. It's something special about the number tan(20) that makes the last equation true - for tan(21) it would be false.96.46.195.35 (talk) 14:42, 10 February 2013 (UTC)

Thanks for spotting my mistake. One possible way to go is by expanding the above into an equation for $\tan(20^{\circ })$ and noting that:

$\tan(3x)={\frac {3\tan x-\tan ^{3}x}{1-3\tan ^{2}x}}$, so

$\tan(60^{\circ })={\frac {3\tan(20^{\circ })-\tan ^{3}(20^{\circ })}{1-3\tan ^{2}(20^{\circ })}}={\sqrt {3}}$

and comparing? 77.86.3.26 (talk) 16:28, 10 February 2013 (UTC)

Sorry that's no help, one equation is quartic and the other cubic. 77.86.3.26 (talk) 16:35, 10 February 2013 (UTC)

Complete proof

Recalling the double angle formula for the sine function and rearranging, we can state that, for any angle α:

$\cos(\alpha )={\frac {\sin(2\alpha )}{2\sin(\alpha )}}$     and similarly that     $\cos(2\alpha )={\frac {\sin(4\alpha )}{2\sin(2\alpha )}}$     and     $\cos(4\alpha )={\frac {\sin(8\alpha )}{2\sin(4\alpha )}}$

On multiplying these together, we get that

$\cos(\alpha )\cdot \cos(2\alpha )\cdot \cos(4\alpha )={\frac {\sin(2\alpha )}{2\sin(\alpha )}}\cdot {\frac {\sin(4\alpha )}{2\sin(2\alpha )}}\cdot {\frac {\sin(8\alpha )}{2\sin(4\alpha )}}={\frac {1}{2\sin(\alpha )}}\cdot {\frac {1}{2}}\cdot {\frac {\sin(8\alpha )}{2}}={\frac {\sin(8\alpha )}{8\sin(\alpha )}}$

Setting α = 20°, we get

$\cos(20^{\circ })\cdot \cos(40^{\circ })\cdot \cos(80^{\circ })={\frac {\sin(160^{\circ })}{8\sin(20^{\circ })}}$

and since $\sin(20^{\circ })=\sin(180^{\circ }-20^{\circ })=\sin(160^{\circ })$

$\cos(20^{\circ })\cdot \cos(40^{\circ })\cdot \cos(80^{\circ })={\frac {1}{8}}$     . .     . .     . .     (1)

This is the so-called Morrie's law.

Now, recalling the sums-to-products formula for the sine function we can state that, for any angles α and β:

$\sin(\alpha )\cdot \sin(\beta )={\frac {\cos(\alpha -\beta )-\cos(\alpha +\beta )}{2}}$

Taking α = 80° and β = 40°:

$\sin(80^{\circ })\cdot \sin(40^{\circ })={\frac {\cos(80^{\circ }-40^{\circ })-\cos(80^{\circ }+40^{\circ })}{2}}={\frac {\cos(40^{\circ })-\cos(120^{\circ })}{2}}$

Then, nothing that   $\cos(120^{\circ })=-\cos(180^{\circ }-120^{\circ })=-\cos(60^{\circ })={\frac {-1}{2}}$:

$\sin(80^{\circ })\cdot \sin(40^{\circ })={\frac {\cos(40^{\circ })+{\frac {1}{2}}}{2}}$

On multiplying by sin(20°), we get:

$\sin(20^{\circ })\cdot \sin(40^{\circ })\cdot \sin(80^{\circ })=(\cos(40^{\circ })+{\frac {1}{2}})\cdot {\frac {\sin(20^{\circ })}{2}}={\frac {\cos(40^{\circ })\cdot \sin(20^{\circ })}{2}}+{\frac {\sin(20^{\circ })}{4}}$

Now, invoking the sums-to-products formula for the product of a sine and a cosine function, for any angles α and β:

$\sin(\alpha )\cdot \cos(\beta )={\frac {\sin(\alpha +\beta )+\sin(\alpha -\beta )}{2}}$

with α = 20° and β = 40°:

$\sin(20^{\circ })\cdot \cos(40^{\circ })={\frac {\sin(20^{\circ }+40^{\circ })+\sin(20^{\circ }-40^{\circ })}{2}}={\frac {\sin(60^{\circ })+\sin(-20^{\circ })}{2}}$

Now, as $\sin(60^{\circ })={\frac {\sqrt {3}}{2}}$     and     $\sin(-20^{\circ })=-\sin(20^{\circ })$

$\sin(20^{\circ })\cdot \cos(40^{\circ })={\frac {{\frac {\sqrt {3}}{2}}-\sin(20^{\circ })}{2}}$

and it follows that:

$\sin(20^{\circ })\cdot \sin(40^{\circ })\cdot \sin(80^{\circ })={\frac {\frac {{\frac {\sqrt {3}}{2}}-\sin(20^{\circ })}{2}}{2}}+{\frac {\sin(20^{\circ })}{4}}={\frac {\sqrt {3}}{8}}-{\frac {\sin(20^{\circ })}{4}}+{\frac {\sin(20^{\circ })}{4}}={\frac {\sqrt {3}}{8}}$     . .     . .     . .     (2)

Now, divind equation (2) by equation (1), we get:

${\begin{aligned}{\frac {\sin(20^{\circ })\cdot \sin(40^{\circ })\cdot \sin(80^{\circ })}{\cos(20^{\circ })\cdot \cos(40^{\circ })\cdot \cos(80^{\circ })}}={\frac {\frac {\sqrt {3}}{8}}{\frac {1}{8}}}\\\tan(20^{\circ })\cdot \tan(40^{\circ })\cdot \tan(80^{\circ })={\sqrt {3}}\\\end{aligned}}$

Recalling that one of the properties of the tangent function is that:

$\tan(90^{\circ }-\theta )=\cot \theta$     and applying it at θ = 50° and θ = 10° and recalling that     $\cot(30^{\circ })={\sqrt {3}}$     and substituting, we find:

$\tan(20^{\circ })\cdot \tan(90^{\circ }-50^{\circ })\cdot \tan(90^{\circ }-10^{\circ })=\cot(30^{\circ })$

$\tan(20^{\circ })\cdot \cot(50^{\circ })\cdot \cot(10^{\circ })=\cot(30^{\circ })$

$\tan(20^{\circ })\cdot {\frac {1}{\tan(50^{\circ })}}\cdot {\frac {1}{\tan(10^{\circ })}}={\frac {1}{\tan(30^{\circ })}}$

$\tan(20^{\circ })\cdot \tan(30^{\circ })=\tan(10^{\circ })\cdot \tan(50^{\circ })$

as required. EdChem (talk) 06:22, 11 February 2013 (UTC)

Here is an alternate proof, loosely modelled on the outline above. We compute cos² 50 + tan 10 sin 50 cos 50 = (1 + cos 100)/2 + tan 10 (sin 100 /2 ) = (1 - sin 10)/2 + tan 10 (cos 10 / 2) = 1/2 - (sin 10)/2 + (sin 10)/2 = 1/2 = cos 60 = cos 10 cos 50 - sin 10 sin 50.

Dividing the result by cos 50, we find cos 50 + tan 10 sin 50 = cos 10 - sin 10 tan 50.

We rearrange to obtain cos 10 - cos 50 = tan 10 sin 50 + sin 10 tan 50 = tan 10 tan 50 cos 10 + tan 10 cos 10 tan 50 = tan 10 tan 50 (cos 10 + cos 50).

Now we divide by cos 10 + cos 50 and use the identity for a product of tangents to find tan 10 tan 50 = (cos 10 - cos 50)/(cos 10 + cos 50) = tan 20 tan 30. — Preceding unsigned comment added by 64.140.122.50 (talk) 08:12, 12 February 2013 (UTC)

Extension

Above, an IP noted that the result is equivalent (after re-adding a lost 2) to the statement that:

$\tan(20)\cdot {\frac {2\tan(20)}{1-\tan ^{2}(20)}}/{\sqrt {3}}={\frac {\tan(20)}{1+{\sqrt {1+\tan ^{2}(20)}}}}$

which is the same as the polynominalin t:

${\frac {2t}{1-t^{2}}}={\frac {\sqrt {3}}{1+{\sqrt {1+t^{2}}}}}$

Solving this for t in the range 0 < t < 1 will give the exact value for tan 20°. EdChem (talk) 06:22, 11 February 2013 (UTC)

Letting t = tan 20, we have sqrt(3) = tan 60 = tan 3t = (t^3 - 3t)/(3t^2 - 1), hence t^3 -3sqrt(3)t^2 - 3t + sqrt(3) = 0.

Multiplying this equation by t^3 + 3sqrt(3)t^2 - 3t -sqrt(3), we find that t^6 - 33t^4 + 27t^2 - 3 = 0.

By Eisenstein's criterion, the polynomial on the left is irreducible over the rational numbers, so it is not possible to find a polynomial equation of lower degree with rational coefficients of which t is a root.

Multiplying by a uniformly-distributed random variable

If I multiply random variable A by random variable B, and random variable B has a uniform distribution, will random variable A*B then have the same type of distribution as what random variable A had? Thorstein90 (talk) 22:29, 10 February 2013 (UTC)

No. I don't think it will ever be the same, and it can be very different, as for example if you multiply a narrow Gaussian r.v. by a broad uniform r.v. Looie496 (talk) 00:40, 11 February 2013 (UTC)

But what I mean is if you multiply a narrow Gaussian by a broad uniform, would it still be (maybe a broader) Gaussian? Thorstein90 (talk) 04:51, 11 February 2013 (UTC)

Clearly not, consider the Gaussian N(1,0). (That's the constant 1.) McKay (talk) 05:54, 11 February 2013 (UTC)

A zero-variance Gaussian? Thorstein90 (talk) 06:48, 11 February 2013 (UTC)

Yes, a zero-variance Gaussian is just a limiting case of a Gaussian as the variance goes to zero (it has all its probability mass at the mean). If something (such as preservation of the class of distributions) were true for all members of the class, it would be true in the limiting case as well. But here, with a unit-mean, zero-variance Gaussian (or any other distribution of unit mean and zero variance), the product with a uniform distribution is simply the same uniform distribution. Duoduoduo (talk) 15:20, 11 February 2013 (UTC)

See Normal distribution#Zero-variance limit. Duoduoduo (talk) 15:50, 11 February 2013 (UTC)

So how do you get the answer to be "yes, it will be random", the same as xoring by a random one time pad? You have a good source of randomness so is there some way to "xor" by it in the same way, to get an equally random result as the "OTP" what you're xoring by? 91.120.48.242 (talk) 07:42, 11 February 2013 (UTC)

My interpretation if 91.120's question is this: A one-time pad (OTP) is uniformly distributed over some fixed range of integers. Is there any random variable A, somehow distributed, that we could multiply by the OTP distribution to get another distribution that is also uniformly distributed? This is a different question from the original one, which wanted the product to be distributed like A rather than like the uniform. This question is also different because one-time pads are discretely distributed. My own answer is that I doubt it, but I'm not sure. Duoduoduo (talk) 15:32, 11 February 2013 (UTC)

February 11

Variance of the product of two normals

I know that for independent random variables, $Var(XY)=Var(X)Var(Y)+Var(X)[E(Y)]^{2}+Var(Y)[E(X)]^{2}$.

Does this mean that the variance of the product of $X\sim N(0,\sigma _{X}^{2})$ and $Y\sim N(0,\sigma _{Y}^{2})$ is simply $\sigma _{X}^{2}\sigma _{Y}^{2}$?

Thorstein90 (talk) 01:22, 11 February 2013 (UTC)

With all respect, a person who has the sophistication to ask that question should have the sophistication to answer it, since the product of two zero-mean Gaussians is a zero-mean Gaussian. Looie496 (talk) 04:36, 11 February 2013 (UTC)

Oh really? Then what is this: http://mathworld.wolfram.com/NormalProductDistribution.html Thorstein90 (talk) 04:47, 11 February 2013 (UTC)

At a glance that looks okay to me. In equation 2 of your wolfram reference, the expression $(|u|)/(\sigma _{x}\sigma _{y})$, looks to me like the ratio of the centered random variable to the standard deviation, so the variance would be $\sigma _{x}^{2}\sigma _{y}^{2}$. Duoduoduo (talk) 14:54, 11 February 2013 (UTC)

However, the product of two zero mean Gaussians is not Gaussian. That's true for sums not products. Duoduoduo (talk) 14:59, 11 February 2013 (UTC)

The answer to your (original) question is "yes". McKay (talk) 05:52, 11 February 2013 (UTC)

The number of subsets of {1,2,...,n} with a total sum k

What's the generation function $f_{n}$ of the series $a_{k}=\left|\left\{A\subseteq \{1,2,\dots ,n\}:\sum _{i\in A}i=k\right\}\right|$? Thanks, 79.179.188.147 (talk) 11:56, 11 February 2013 (UTC)

See Restricted partition generating functions in partition (number theory). Gandalf61 (talk) 13:19, 11 February 2013 (UTC)

Still can't solve it. The function $q(k)$ is similar but doesn't restrict the subsets $A$ to be contained in $\{1,2,\dots ,n\}$. 79.179.188.147 (talk) 14:17, 11 February 2013 (UTC)

The restriction to subsets of $\{1,2,\dots ,n\}$ means that the upper bound of the product is not infinity but is ... what ? Gandalf61 (talk) 14:31, 11 February 2013 (UTC)

Oh, thanks! 79.179.188.147 (talk) 17:11, 11 February 2013 (UTC)

The number of ways to color N balls with I colors is. ${\binom {N+I-1}{I-1}}$. Bo Jacoby (talk) 02:51, 12 February 2013 (UTC).

Squared triangular number representative polychora.

Not sure if this belongs here or on WP:MATH. In the article Squared triangular number, is the following text:

These numbers can be viewed as figurate numbers, a four-dimensional hyperpyramidal generalization of the triangular numbers and square pyramidal numbers.

. Now this makes it sound like there is a single 4 dimensional figure that if sliced differently into two dimensional pieces would make it hypergeometically obvious. While the sum of cubes is obviously a cubical hyperpyramid (shape made of a cube and 6 square pyramids), it is unclear to me how the hyper pyramid would be cut to give the square of the triangular number. Can someone please help with explaining this to me here, which might help in changing the explanation in the article?Naraht (talk) 15:08, 11 February 2013 (UTC)