Thread: Taylor expansion - find limit value

1. Taylor expansion - find limit value

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

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

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

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

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...

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

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

3. Originally Posted by liquidFuzz
$\displaystyle -e^1 \displaystyle \sum_{n=0}^{\infty} \frac{(z-1)^n (z-1)^n } {n! (z-1)^n}$ = $\displaystyle -e^1 \displaystyle \sum_{n=0}^{\infty} \frac{(z-1)^n } {n! }$
I found a table with formulas of Taylor expansions. It's stated that $\displaystyle e^x \rightarrow 1+x+\frac{x^2}{2!} + ... + \frac{x^n}{n!}$

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

4. Originally Posted by liquidFuzz
..., right?
No, it isn't, $\displaystyle f(z)$ has a pole at $\displaystyle z=1$ , so there is no Taylor expasion around $\displaystyle z=1$. Look Opalg's post.

Fernando Revilla

5. What if I said Laurent expansion..?

6. Originally Posted by liquidFuzz
What if I said Laurent expansion..?
You can say just what you want.

Fernando Revilla

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

I shuffled around a bit and used the geometric series and the expansion of $\displaystyle e^z$ and got this. $\displaystyle -e^1 \displaystyle \frac {\sum_{n=0}^{\infty} \frac{(z-1)^n} {n!} \sum_{n=0}^{\infty} (z-1)^n}{(z-1)^n}$
The fatal flaw with this approach (quite apart from any other mistakes) is that you are using the same letter $\displaystyle n$ for three different purposes. You are given the denominator in the form $\displaystyle (z-1)^n$, but then you have used the same letter $\displaystyle n$ (twice!) as a summation index in the numerator. That makes nonsense of the subsequent calculation.

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

8. 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!