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

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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 $n\ln (1+\frac{1}{n})$:

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

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

Since $\ln {(1+\frac{1}{n})}^n$ 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 $\ln {(1+\frac{1}{\mathrm{\infty }})}^{\mathrm{\infty }}=1$ and hence ${(1+\frac{1}{\mathrm{\infty }})}^{\mathrm{\infty }}=e$, 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

$\ln (1+x)=\sum _{k=1}^{\mathrm{\infty }}(-1{)}^{k+1}\frac{x^k}{k}\text{ for }-1<x\le 1.$

Now substitute x = 1/n to get:

$\ln (1+\frac{1}{n})=\sum _{k=1}^{\mathrm{\infty }}(-1{)}^{k+1}\frac{(\frac{1}{n}{)}^k}{k}=\frac{1}{n}-\frac{1}{2n^2}+\frac{1}{3n^3}-\frac{1}{4n^4}+\dots$

Multiply through by n:

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

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 [1] 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 $\sum _{nodd}^{\mathrm{\infty }}\frac{r^n\sin n\theta }{n}$ using the substitution $z=r{e}^{i\theta }$. I'v given it a go but I only get as far as $\sum _{nodd}^{\mathrm{\infty }}\frac{r^n\sin n\theta }{n}=\sum _{nodd}^{\mathrm{\infty }}Im\frac{z^n}{n}$ 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 $Im\sum _{n\text{ odd}}^{\mathrm{\infty }}\frac{z^n}{n}$ (because taking the imaginary part is linear). Second, a nice trick for summing things of the form $\sum _{n=0}^{\mathrm{\infty }}{n}^k{z}^n$ 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. – 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)