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Math Help - Taylor expansion - find limit value

  1. #1
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    Taylor expansion - find limit value

    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...
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  2. #2
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    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.
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  3. #3
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    Quote Originally Posted by liquidFuzz View Post
    -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?
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  4. #4
    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by liquidFuzz View Post
    ..., 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
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  5. #5
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    What if I said Laurent expansion..?
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    MHF Contributor FernandoRevilla's Avatar
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    Quote Originally Posted by liquidFuzz View Post
    What if I said Laurent expansion..?
    You can say just what you want.

    Fernando Revilla
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  7. #7
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    Quote Originally Posted by liquidFuzz View Post
    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.
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  8. #8
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    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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