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I'm working on expressing the function $f(x)=\frac{6}{x}$ as a taylor series about $-4$. I've got the general idea, but I'm not quite there yet. I've come up with the equation $$f^{(n)}(-4)=\frac{(-1)^n6n!}{(-4)^{n+1}}$$ which appears to be correct. I've tried substitution this into the formula $$f(x)=\sum_{n=0}^\infty\frac{f^{(n)}(a)}{n!}(x-a)^n$$ to get $$f(x)=\sum_{n=0}^\infty\frac{(-1)^n6}{(-4)^{n+1}}(x+4)^n$$ But this does not appear to be correct, as far as I can tell. Any help would be appreciated.

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Please don't write $-1^n$ if you mean $(-1)^n$. Otherwise it looks fine. –  mrf Feb 28 '13 at 21:18
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Looks basically OK. do use parentheses, and note that most of those minus signs disappear, we get simply $-\frac{6}{4^{n+1}}(x+4)^n$. –  André Nicolas Feb 28 '13 at 21:25
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Don't you mean $(-4)^{n+1}$? Try simplifying the expression before you plug things into it, it helps you keep track of the calculations.

How do you know it's not correct?

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