Taylor expansion

**liquidFuzz** · October 30th 2010, 12:46 PM

Find the limet value of the residue $R_n$ when $n \rightarrow \infty$ in $z=1$ to the function $\displaystyle \frac{e^z}{(z+2)(z-1)^n}$ .

I shuffled around a bit and used the geometric series and the expansion of $e^z$ and got this. $-e^1 \displaystyle \frac {\sum_{n=0}^{\infty} \frac{(z-1)^n} {n!} \sum_{n=0}^{\infty} (z-1)^n}{(z-1)^n}$

Now, I don't quite know how to simplify this expression... Ah well, here goes nothing!

$-e^1 \displaystyle \sum_{n=0}^{\infty} \frac{(z-1)^n (z-1)^n } {n! (z-1)^n}$ = $-e^1 \displaystyle \sum_{n=0}^{\infty} \frac{(z-1)^n } {n! }$

So, I'm not sure of how to simplify the expression with al the sums. And finally, I'm not sure of how to get further now...

**Opalg** · October 31st 2010, 09:50 AM

Let $w = z-1$ . Then you want the residue at $w=0$ of the function $f(w) = \dfrac{e^w}{(w+3)w^n}$ . Let $g(w) = \dfrac{e^w}{(w+3)}$ and let $a_0+a_1w+a_2w^2+\ldots$ be the Taylor series for $g(w)$ , which converges in the region $|w|<3$ . For the series to converge, it is necessary that $a_n\to0$ as $n\to\infty$ .

The residue at 0 for $f(w)$ is the coefficient of $w^{-1}$ in its Laurent expansion, which will be the coefficient of $w^{n-1}$ in the Taylor expansion of $g(w)$ , namely $a_{n-1}$ . But that goes to 0. So the limiting value of $R_n$ is 0.

**liquidFuzz** · December 31st 2010, 12:25 AM

Originally Posted by liquidFuzz

$-e^1 \displaystyle \sum_{n=0}^{\infty} \frac{(z-1)^n (z-1)^n } {n! (z-1)^n}$ = $-e^1 \displaystyle \sum_{n=0}^{\infty} \frac{(z-1)^n } {n! }$

I found a table with formulas of Taylor expansions. It's stated that $e^x \rightarrow 1+x+\frac{x^2}{2!} + ... + \frac{x^n}{n!}$

So all in all I get $-e^2$ and this would be my limit value when $n \rightarrow \infty$ , right?

**FernandoRevilla** · December 31st 2010, 01:04 AM

Originally Posted by liquidFuzz

..., right?

No, it isn't, $f(z)$ has a pole at $z=1$ , so there is no Taylor expasion around $z=1$ . Look Opalg's post.

Fernando Revilla

**liquidFuzz** · December 31st 2010, 01:38 AM

What if I said Laurent expansion..?

**FernandoRevilla** · December 31st 2010, 02:07 AM

Originally Posted by liquidFuzz

What if I said Laurent expansion..?

You can say just what you want.

Fernando Revilla

**Opalg** · December 31st 2010, 03:54 AM

Originally Posted by liquidFuzz

Find the limet value of the residue $R_n$ when $n \rightarrow \infty$ in $z=1$ to the function $\displaystyle \frac{e^z}{(z+2)(z-1)^n}$ .

I shuffled around a bit and used the geometric series and the expansion of $e^z$ and got this. $-e^1 \displaystyle \frac {\sum_{n=0}^{\infty} \frac{(z-1)^n} {n!} \sum_{n=0}^{\infty} (z-1)^n}{(z-1)^n}$

The fatal flaw with this approach (quite apart from any other mistakes) is that you are using the same letter $n$ for three different purposes. You are given the denominator in the form $(z-1)^n$ , but then you have used the same letter $n$ (twice!) as a summation index in the numerator. That makes nonsense of the subsequent calculation.

Also, the original function had a term $z+2$ in the denominator, which seems to have disappeared after that.

**liquidFuzz** · December 31st 2010, 04:49 AM

You're actually right, it seems as if I confused (z-2) with (z+2) when I tried to find the expansion/series. Even if I use different letters for the indexes isn't it ok to use the same in the last expression? Oh and thanks for the cheerful answer!

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