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I need to calculate value of $1000^{th}$ derivate of the following function at $0$:

$$ f(x) = \frac{x+1}{(x-1)(x-2)} $$

I've done similar problems before (e.g. $f(x)= \dfrac{x}{e^{x}}$) but the approach I've used would not work in this case and I believe I should expand this function into a power series. Could you please give me any hints on how to do it?

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3  
Partial fractions? –  David Mitra Jan 26 '13 at 19:29
    
"I believe I should expand this function into a power series" wouldn't work because you need the derivatives to get the Taylor expansion coefficients anyway. –  Fixed Point Jan 26 '13 at 19:47

3 Answers 3

up vote 6 down vote accepted

Hint: Note that $f(x)=\frac{3}{x-2}-\frac{2}{x-1}$ and if $g(x)=\frac{1}{x-a}$ then $g^n(x)=\frac{(-1)^n n! }{(x-a)^{n+1}}$ where $a$ is a constant and $g^n(x)$ is the $n$-th derivative of $g(x)$.

Here $$\begin{align} \frac{A}{x-2}+\frac{B}{x-1} &=\frac{x+1}{(x-2)(x-1)}\\ \implies A(x-1)+B(x-2) &= x+1 \\ \implies x(A+B)+(-A-2B) &= x+1\end{align}$$ From this we get $$\begin{align}A+B &=1 \\ -A-2B &=1 \end{align}$$

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Your formula for the $n$-th derivative of $g$ is not correct. You miss some factorial. –  1015 Jan 26 '13 at 19:35
    
@julien : you are right . I miss the factorial part. I will edit it. –  Argha Jan 26 '13 at 19:36

Starting with the partial fraction decomposition $$f(x)=\frac{3}{x-2}-\frac{2}{x-1}\;,$$ use the geometric series sum to write

$$\begin{align*} f(x)&=-\frac32\cdot\frac1{1-\frac{x}2}+2\frac1{1-x}\\\\ &=2\sum_{n\ge 0}x^n-\frac32\sum_{n\ge 0}\left(\frac{x}2\right)^n\\\\ &=\sum_{n\ge 0}\left(2-\frac32\left(\frac12\right)^n\right)x^n\\\\ &=\sum_{n\ge 0}\left(2-\frac3{2^{n+1}}\right)x^n\;. \end{align*}$$

Then equate this with the Maclaurin series

$$f(x)=\sum_{n\ge 0}\frac{f^{(n)}(0)}{n!}x^n$$

to get $$\frac{f^{(n)}(0)}{n!}=2-\frac3{2^{n+1}}$$ and then

$$f^{(n)}(0)=n!\left(2-\frac3{2^{n+1}}\right)\;.$$

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As suggested by Argha and David Mitra, begin with the decomposition into partial fractions: $$ f(x)=\frac{3}{x-2}-\frac{2}{x-1}. $$ Now if $g(x)=\frac{1}{x-a}$ for some constant $a$, then a proof by induction quickly shows that $$ g^{(n)}(x)=\frac{(-1)^nn!}{(x-a)^{n+1}}. $$ It follows that $$ f^{(1000)}(x)=\frac{3\cdot 1000!}{(x-2)^{1001}}-\frac{2\cdot 1000!}{(x-1)^{1001}}. $$ Now make $x=0$.

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