Taylor series and 100th derivative on this function

Question

I have this real function:

$f(x) = \frac{1}{(x^2-2x+3)^2}$ and I need to find Taylor series at $x = 1$ and find 100th derivative at $f^{(100)}(1)$.

Can anybody help me???

Answer 1

Notice $$f(x) = \frac{1}{((x-1)^2+2)^2} = \frac14 \left(1 + \frac{(x-1)^2}{2}\right)^{-2} \stackrel{\color{blue}{[1]}}{=} \frac14\sum_{n=0}^\infty \frac{(-1)^n (n+1)}{2^n} (x-1)^{2n} $$ The coefficient for $(x-1)^{100}$ in the Taylor expansion of $f(x)$ around $x = 1$ is $\frac{51}{2^{52}}$ and hence

$$f^{(100)}(1) = \frac{51\times 100!}{2^{52}}$$

Notes

• $\color{blue}{[1]}$ Please note that for general $\alpha$, we have $$\frac{1}{(1-z)^\alpha} = 1 + \alpha z + \frac{\alpha(\alpha+1)}{2} z^2 + \cdots + \frac{\prod\limits_{k=0}^{n-1} (\alpha+k)}{n!} z^n + \cdots$$ When $\alpha = 2$, this reduces to $$\frac{1}{(1-z)^2} = 1 + 2 z + \frac{3!}{2!} z^2 + \cdots + \frac{(n+1)!}{n!} z^n = \sum_{n=0}^\infty (n+1)z^n$$ Another way to look at this is start from the power series: $$\frac{1}{1-z} = 1 + z + z^2 + \cdots = \sum_{n=0}^\infty z^n$$ If one differentiate both sides once, you get $$\frac{1}{(1-z)^2} = 1 + 2z + 3z^2 \cdots = \sum_{n=0}^\infty(n+1)z^n$$ Expansion of $\frac{1}{(1-z)^\alpha}$ for other integral $\alpha > 0$ can be derived in similar manner. You can really view this sort of expansion as as an extension of ordinary binomial theorem to negative integral power.