Wikipedia:Reference desk/Archives/Mathematics/2010 November 6

= November 6 =

Showing that lim (1+1/n)^n = e as n –> infinity
$$\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\frac{x^4}{4}+...$$

Find a series for $$n\ln\left(1+\frac{1}{n}\right)$$:

$$n\ln\left(\frac{n+1}{n}\right)=n\left( \frac{n+1}{n} - \frac{(n+1)^2}{2n^2}+ \frac{(n+1)^3}{3n^3} - \frac{(n+1)^4}{4n^4}...\right) = n+1 - \frac{(n+1)^2}{2n}+ \frac{(n+1)^3}{3n^2} - \frac{(n+1)^4}{4n^3}...$$

Hence show that $$\left(1+\frac{1}{n}\right)^n = e$$ as n approaches infinity.

Since $$\ln\left(1+\frac{1}{n}\right)^n$$ is equal to the series $$n+1 - \frac{(n+1)^2}{2n}+ \frac{(n+1)^3}{3n^2} - \frac{(n+1)^4}{4n^3}...$$ if I can show that $$n+1 - \frac{(n+1)^2}{2n}+ \frac{(n+1)^3}{3n^2} - \frac{(n+1)^4}{4n^3}...$$ converges to 1 as n approaches infinity, I can conclude that $$\ln\left(1+\frac{1}{\infty}\right)^{\infty} = 1$$ and hence $$\left(1+\frac{1}{\infty}\right)^{\infty} = e$$, but how do I do this? --220.253.253.75 (talk) 00:58, 6 November 2010 (UTC)
 * Your series is wrong, firstly. We begin with
 * $$\ln(1 + x) = \sum^{\infin}_{k=1} (-1)^{k+1}\frac{x^k}k\text{ for }-1<x\le1.$$
 * Now substitute x = 1/n to get:
 * $$\ln \left( 1 + \frac{1}{n} \right ) = \sum^{\infin}_{k=1} (-1)^{k+1}\frac{(\frac{1}{n})^k}k = \frac{1}{n} - \frac{1}{2n^2} + \frac{1}{3n^3} - \frac{1}{4n^4} + \dots$$
 * Multiply through by n:
 * $$n \ln \left( 1 + \frac{1}{n} \right ) = \ln \left ( \left [ 1 + \frac{1}{n} \right ]^n \right ) = 1 - \left( \frac{1}{2n} - \frac{1}{3n^2} + \frac{1}{4n^3} - \dots \right ) \qquad (*).$$
 * Clearly as n goes to infinity, (*) goes to 1, so that the argument of the logarithm goes to e. — Anonymous Dissident  Talk 11:31, 6 November 2010 (UTC)

Correction to my previous Refdesk question
Hello,

I posted this a while back in August, and I thought I'd solved the problem then. But I've just had another think about it and I've realised that there's no guarantee K_m is going to be a subgroup of K[alpha]! Even if you take K_m=K_m[alpha] as K[alpha]/(X^m-c), it's not necessarily the same using the tower law on [K[alpha]:K_m[alpha]][K_m[alpha]:K] and then saying both m, n divide [K[alpha]:K], because there's no guarantee as far as I can see that [K_m[alpha]:K]=[K_m:K]: surely the fact we're introducing a new dependence relation on our powers of alpha is going to change things.

So could anyone tell me where I went wrong? At the time it seemed right but I've unconvinced myself now! Thankyou, 62.3.246.201 (talk) 01:14, 6 November 2010 (UTC)


 * I just looked at your previous question, and hopefully I'm following along correctly:
 * We assume that $$X^m - c$$ is irreducible, so $$K_m = K[X]/(X^m - c)$$ is a field. We wish to show that $$K_m \subset K[\alpha]$$, where $$\alpha$$ is any root of $$X^{mn} - c$$.  What do we know about $$\alpha^n$$?  Eric.  82.139.80.73 (talk) 13:28, 6 November 2010 (UTC)
 * That it's a root of $$X^m - c$$? In which case we know $$X^m - c$$ is a constant multiple of the min poly for $$\alpha^n$$, so $$K_m=K(\alpha^n)$$ which then is contained in $$K(\alpha)$$? Hope that's right! Also I think I meant to write K($$\alpha$$) rather than square brackets previously, sorry! 131.111.185.68 (talk) 14:27, 6 November 2010 (UTC)
 * I think I get it now anyway, thankyou! 62.3.246.201 (talk) 05:42, 7 November 2010 (UTC)

Series Summation
Hi. I have to sum the series $$ \sum_{n odd}^{\infty} \frac{r^n \sin n\theta}{n} $$ using the substitution $$ z=re^{i\theta}$$. I'v given it a go but I only get as far as $$ \sum_{n odd}^{\infty} \frac{r^n \sin n\theta}{n} = \sum_{n odd}^{\infty} Im{\frac{z^n}{n}}$$ and now don't see how to proceed. Originally, I was hoping to use the sum of a geometric series but clearly the n on the denominator stops this from being feasible. Can someone suggest what to do next? Thanks. asyndeton  talk  11:36, 6 November 2010 (UTC)


 * First, you can factor the Im out of the summation to get $$Im \sum_{n \text{ odd}}^\infty \frac {z^n}n$$ (because taking the imaginary part is linear). Second, a nice trick for summing things of the form $$\sum_{n = 0}^\infty n^kz^n$$ is to take the derivative or integral (depending on whether k is negative or positive) with respect to z and work from there.  Eric.  82.139.80.73 (talk) 13:17, 6 November 2010 (UTC)


 * A devilishly cunning trick. Cheers Eric. asyndeton   talk  13:56, 6 November 2010 (UTC)


 * Where, by linear, you only mean additive or R-linear, not C-linear. &#x2013; b_jonas 21:13, 7 November 2010 (UTC)
 * Right, I was thinking R-linear. "Additive" probably would have been clearer.  I just didn't want to leave it at "factor the Im out" so that someone reading along wouldn't think "Im" was a number.  Eric.  82.139.80.73 (talk) 18:09, 8 November 2010 (UTC)

Divisor
Hello, everybody! Suppose, that p is prime number in form $$p = 8k+1$$. Can we always find such natural number n so that $$n^4 + 1$$ is divisible by p? How we can prove this. Thank you! --RaitisMath (talk) 18:33, 6 November 2010 (UTC)
 * The multiplicative group mod p is cyclic of order divisible by 8, it therefore has an element n of order 8. If follows that n4+1 is 0 mod p.--RDBury (talk) 04:26, 7 November 2010 (UTC)