- #1

jamesb1

- 22

- 0

## Homework Statement

I want to express the following expression in its Taylor expansion about x = 0:

$$

F(x) = \frac{x^{15}}{(1-x)(1-x^2)(1-x^3)(1-x^4)(1-x^5)}

$$

## The Attempt at a Solution

First I tried to rewrite the function in partial fractions (its been quite a while since I've last covered these).

So far I have:

\begin{align*}

F(x) = &\frac{A}{1-x} &&+ \\

&\frac{Bx + C}{1-x^2} &&+ \\

&\frac{Dx^2 + Ex + F}{1-x^3} &&+ \\

&\frac{Gx^3 + Hx^2 + Ix + J}{1-x^4} &&+ \\

&\frac{Kx^4 + Lx^3 + Mx^2 + Nx + O}{1-x^5}

\end{align*}

Then:

\begin{align*}

x^{15} = &A(1-x^2)(1-x^3)(1-x^4)(1-x^5) &&+ \\

&(Bx + C)(1-x)(1-x^3)(1-x^4)(1-x^5) &&+ \\

&(Dx^2 + Ex + F)(1-x)(1-x^2)(1-x^4)(1-x^5) &&+ \\

&(Gx^3 + Hx^2 + Ix + J)(1-x)(1-x^2)(1-x^3)(1-x^5) &&+ \\

&(Kx^4 + Lx^3 + Mx^2 + Nx + O)(1-x)(1-x^2)(1-x^3)(1-x^4)

\end{align*}

As you can see, this does not seem to make sense. For example, in the equation above I'm getting $1=0$ when $x=1$.

What am I doing wrong?