# How to find the full Taylor expansion of the following:

I need to find the full Taylor expansion of $$f(x)=\frac{1+x}{1-2x-x^2}$$

Any help would be appreciated. I'd prefer hints/advice before a full answer is given. I have tried to do partial fractions\reductions. I separated the two in hopes of finding a known geometric sum but I could not.

Edit: I guess you could say that I did not have the.... insight to take the path with the partial decomposition mentioned. I have done some work (I had to go to the gym that is why it took a while)

$$\frac{1+x}{1-2x-x^2}=\frac{1}{2(\sqrt{2}-x-1)}-\frac{1}{2(\sqrt{2}+x+1)}$$ I am going to work with this to go further.

I got to this:

$$\frac{1}{2}\left(\sum_{n=0}^\infty\frac{x^n}{(\sqrt{2}-1)^{n+1}}+\sum_{n=0}^\infty\frac{x^n}{(-\sqrt{2}-1)^{n+1}}\right)$$ But I think this is wrong for some reason.

Edit: Figured it out.

\begin{align*} \implies\frac{1+x}{1-2x-x^2}&=\frac{1}{2(\sqrt{2}-x-1)}-\frac{1}{2(\sqrt{2}+x+1)} \\[2mm] &=\frac{1}{2}\left(\frac{1}{a-x}-\frac{1}{b+x}\right) \mbox{where a=\sqrt{2}-1 and b=\sqrt{2}+1}. \\[2mm] &=\frac{1}{2}\left(\frac{1}{a} \frac{1}{1-\frac{x}{a}}-\frac{1}{b} \frac{1}{1-\frac{x}{-b}}\right) \\[2mm] &=\frac{1}{2}\left(\frac{1}{a}\sum_{n=0}^\infty \frac{1}{a^n}x^n-\frac{1}{b}\sum_{n=0}^\infty\frac{1}{(-b)^n}x^n\right) \\[2mm] &=\frac{1}{2}\left(\frac{1}{\sqrt{2}-1}\sum_{n=0}^\infty \frac{1}{(\sqrt{2}-1)^n}x^n-\frac{1}{\sqrt{2}+1}\sum_{n=0}^\infty\frac{1}{(-\sqrt{2}-1)^n}x^n\right) \\[2mm] &=\frac{1}{2}\left(\sum_{n=0}^\infty\frac{x^n}{(\sqrt{2}-1)^{n+1}}+\sum_{n=0}^\infty\frac{x^n}{(-\sqrt{2}-1)^{n+1}}\right) \\ &=1+3x+7x^2+17x^3+\ldots \end{align*}

• Your partial fraction decomposition is...? – Martín-Blas Pérez Pinilla May 1 '14 at 21:05
• Well I expanded to $$-\frac{x}{x^2+2x-1}-\frac{1}{x^2+2x-1}$$ and also $$\frac{-x-1}{x^2+2x-1}$$ – H5159 May 1 '14 at 21:08
• A decomposition is... see the answer of gt6989b. – Martín-Blas Pérez Pinilla May 1 '14 at 21:09
• @Frumpy partial fractions should reduce denominator to degree 1 – gt6989b May 1 '14 at 21:09
• It is easier in powers of $x+1$. For about $x=0$, partial fractions should work, we get $\frac{1/2}{\sqrt{2}-1-x}+\frac{1/2}{\sqrt{2}+1+x}$, and two parts are not too bad. – André Nicolas May 1 '14 at 21:12

Hint: $1 - 2x - x^2 = 2 - (1 + 2x + x^2) = 2 - (x + 1)^2 = (\sqrt{2} - (x + 1))(\sqrt{2} + (x + 1))$

• How was I supposed to know that this was the method to decompose it? I would have never guessed to go this path.. – H5159 May 2 '14 at 0:01

Hint For the function $$g(x) = \frac{1}{1-x} \text{ over } x \in (-1,1),$$ the Taylor series is the ordinary geometric series $$\frac{1}{1-x} = 1 + x + x^2 + \ldots = \sum_{k=0}^\infty x^k.$$

If the partial fraction decomposition is given by $$\frac{A}{x-B} + \frac{C}{x-D}$$ how can you get the decomposition you are seeking?

• @user88595 fixed... – gt6989b May 1 '14 at 21:19
• I don't know how I missed your answer until I was finished with mine. (+1) – robjohn May 1 '14 at 23:33
• I changed my answer in case my previous comment is confusing :-) – robjohn May 2 '14 at 0:29

Here is an alternate approach. Suppose that $$\frac{1+x}{1-2x-x^2}=\sum_{k=0}^\infty a_kx^k$$ Multiply by $1-2x-x^2$ to get \begin{align} 1+x &=\underbrace{a_0}_{1}+\underbrace{(a_1-2a_0)}_1x+\sum_{k=2}^\infty\underbrace{(a_k-2a_{k-1}-a_{k-2})}_{0}x^k \end{align} Equating coefficients of $x^k$ gives $$a_0=1,a_1=3,\text{ and }a_k=2a_{k-1}+a_{k-2}\text{ for }k\ge2$$ Therefore, $$\frac{1+x}{1-2x-x^2}=1+3x+7x^2+17x^3+41x^4+99x^5+\dots$$

If we wish a formula for $a_n$ we can solve the recurrence $$a_k=2a_{k-1}+a_{k-2}$$ using the roots of $x^2-2x-1=0$, which are $1\pm\sqrt2$, and $a_0=1,a_1=3$ to get $$a_k=\frac{(1+\sqrt2)^{k+1}+(1-\sqrt2)^{k+1}}{2}$$ which is the integer closest to $\dfrac{(1+\sqrt2)^{k+1}}{2}$.