find $f^{(15)}(0)$ if $f = \frac{10x^2+12x+4}{(x+2) (x^2 +1)}$

Question

$$f = \frac{10x^2+12x+4}{(x+2)(x^2 +1)}=\frac{4}{x+2}+\frac{6x}{x^2+1}=\frac{2}{1-\left(-\frac x 2 \right)}+\frac{6x}{1-(-x^2)}$$

So we end up with $$2 \sum_{n=0}^{15} \left(-\frac x 2 \right)^n + 6x \sum_{n=0}^{15} (-x^2)^n$$ but because we are looking for $f^{(15)}(0)$ we get the result of $2$?

Answer 1

If you have a taylor polynomial for $f(x)$ i.e. $f(x) = \sum a_n x^n$ then $f^{(n)}(0) = n! a_n$

Normally we think about this the other way around. We want to find the coefficients and we say $a_n = \frac {f^{(n)}(0)}{n!}$

$f(x) = 2\sum (-\frac 12)^n x^n + 6x\sum (-1)^n x^{2n}\\ a_{15} = 2 (-\frac 12)^{15} + 6 (-1)^7\\ -2^{-14} - 6$

$f^{(15)}(0) = (-2^{-14} - 6)15!$

Answer 2

$$ 2 \sum_{n=0}^{15} \left(-\frac x 2 \right)^n + 6x \sum_{n=0}^{15} (-x^2)^n $$

The $15$th-degree term of this sum is $\dfrac{f^{(15)}(0)}{15!} x^{15}.$

It is also $2\left( -\dfrac x 2 \right)^{15} + 6x(-x^2)^7.$

From that one can find $f^{(15)}(0).$

Answer 3

It is more precise to write $$\begin{align*} f(x) &= \sum_{n=0}^\infty 2\left(-\frac{x}{2}\right)^n + 6x(-x^2)^n \\ &= \sum_{n=0}^\infty 2 (-1/2)^n x^n + \sum_{m=0}^\infty 6(-1)^m x^{2m+1} \\ &= \sum_{k=0}^\infty \frac{f^{(k)}(0)}{k!} x^k. \end{align*}$$ Thus, $f^{(15)}(0)$ is $15!$ times the coefficient of $x^{15}$ in the series expansion. Note this corresponds to the choices $n = 15$ and $m = 7$; i.e., $$f^{(15)}(0) = 15! \left( 2 (-1/2)^{15} + 6 (-1)^{7} \right).$$ This is because the coefficient of $x^{15}$ is found at different values of the indices of the series expansion of each term.

Answer 4

We have \begin{align*} f(x) = \frac{4}{x+2} + \frac{3}{x+i} + \frac{3}{x-i} \end{align*} and hence \begin{align*} f^{15}(x) = -(15!)(4(x+2)^{-16} + 3(x+i)^{-16} + 3(x-i)^{-16} \end{align*} and \begin{align*} f^{15}(0) = -(2^{-14}+6)\cdot 15! \end{align*}