Question

If f (n)(0) = (n + 1)! for n = 0, 1, 2, , find the taylor series at a=0 for f.

Answer 1

Given that $f^{(n)}(0)=(n+1)!$, we have for $f(x)$ the Taylor series expansion about 0 as

$f(x)=\displaystyle\sum_{n=0}^\infty\frac{(n+1)!}{n!}x^n=\sum_{n=0}^\infty(n+1)x^n$

Replace $n+1$ with $n$, so that the series is equivalent to

$f(x)=\displaystyle\sum_{n=1}^\infty nx^{n-1}$

and notice that

$\displaystyle\frac{\mathrm d}{\mathrm dx}\sum_{n=0}^\infty x^n=\sum_{n=1}^\infty nx^{n-1}$

Recall that for $|x|$, we have

$\displaystyle\sum_{n=0}^\infty x^n=\frac1{1-x}$

which means

$f(x)=\displaystyle\sum_{n=1}^\infty nx^{n-1}=\frac{\mathrm d}{\mathrm dx}\frac1{1-x}$
$\implies f(x)=\dfrac1{(1-x)^2}$