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

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

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

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

Now substitute x = 1/n to get:

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

Multiply through by n:

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

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 [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 $\sum _{nodd}^{\infty }{\frac {r^{n}\sin n\theta }{n}}$ using the substitution $z=re^{i\theta }$. I'v given it a go but I only get as far as $\sum _{nodd}^{\infty }{\frac {r^{n}\sin n\theta }{n}}=\sum _{nodd}^{\infty }Im{\frac {z^{n}}{n}}$ 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 $\sum _{n=0}^{\infty }n^{k}z^{n}$ 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)