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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)

Alternate proof

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. 64.140.122.50 (talk) 08:14, 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. 64.140.122.50 (talk) 08:13, 12 February 2013 (UTC)

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)

What you probably want is (modular) addition, not multiplication. If A is a random variable in [0,1], and B is independent and uniformly distributed in [0,1], then $C=(A+B)\ \mod 1$ is uniformly distributed in [0,1], independent of A, and using B and C you can reconstruct A. This can be extended to (bounded) ranges which are not [0,1]. -- Meni Rosenfeld (talk) 13:44, 12 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).

Wouldn't the answer be (1 + x)(1 + x²)⋅⋅⋅(1 + xⁿ)? If you expand the product, the coefficient of x^k is a_k. 64.140.122.50 (talk) 04:56, 13 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)

Don't know of any, but there is a nice proof without words at [1] Dmcq (talk) 01:04, 13 February 2013 (UTC)

Not doubting the proof (though some more information as to which proof used by each of the discoverers might be nice. Part of the problem with the concept is that the cubical hyperpyramid has a single vertex which is unique so figuring out which of the 6 it lines up with.Naraht (talk) 19:01, 13 February 2013 (UTC)

February 12

Cosets and normal subgroup equality

Hello. I recently came across an exercise, and would be very grateful if someone could help.

Given a normal subgroup H in G, with g in G, if $gH=H$, then is $g\in H$?

Neuroxic (talk) 11:31, 12 February 2013 (UTC)

Any subgroup H contains the identity. Think what happens with that. Dmcq (talk) 12:11, 12 February 2013 (UTC)

That question is a special case of the following question, which may actually be easier (intuitively) to solve: if $gH=L$, then is $g\in L$? Basically you're just asking "Is $g\in gH$ for some subgroup $H$?" and it is a very well known fact that the answer is yes, and the proof is also very well known and straightforward: $1\in H\implies g\cdot 1\in gH\implies g\in gH$ --AnalysisAlgebra (talk) 19:21, 12 February 2013 (UTC)

There's no need for $H$ to be normal, by the way. --AnalysisAlgebra (talk) 19:23, 12 February 2013 (UTC)

Estimating parameters of a distribution from censored data

Suppose you hypothesize that your data are drawn from, say, a lognormal distribution or a gamma distribution. Your data are right-censored at the value x*. That is, you have exact data whenever x≤x*, but whenever x>x* all you know is that x>x*.

(1) How do you estimate the parameters of the hypothesized distribution? I know it would be by maximum likelihood, but how do you handle the censored data?

(2) Is there a command in, say, SAS that will do this?

(3) How do you test the hypothesis that the data did in fact come from that distribution? Duoduoduo (talk) 18:37, 12 February 2013 (UTC)

(1) If your data consists of uncensored points $x_{1},\ldots ,x_{m}$ and n censored points, and your pdf with parameters $\theta$ is $f(x|\theta )$, then the likelihood of $\theta$ is $\left(\int _{x^{*}}^{\infty }f(t|\theta )\ dt\right)^{n}\prod _{i=1}^{m}f(x_{i}|\theta )$. There's ostensibly an inconsistency in that you use a density for the contribution to likelihood of some points and a probability for others, but since likelihood functions are only meaningful up to a constant factor it doesn't matter. -- Meni Rosenfeld (talk) 19:49, 12 February 2013 (UTC)

Thanks, Meni! Duoduoduo (talk) 14:19, 14 February 2013 (UTC)

February 14

How do you solve ${\tbinom {n}{r}}=1-\epsilon$ for n?

How do you solve ${\tbinom {n}{r}}=1-\epsilon$ for $n$, given $r$ and $\epsilon$? The parameter values of interest are of the order $r\sim 10^{9}$ and $\epsilon \sim 10^{-5}$. Can someone give a good approximation formula that can be evaluated without gross loss of precision or overflows during computation? Thanks. --173.49.13.216 (talk) 03:56, 15 February 2013 (UTC)