# Finding the Taylor Series Expansion using Binomial Series, then obtaining a subsequent Expansion.

Hi all doing some prep work for a course and really struggling to wrap my head around some of the revision questions. Some help would be really appreciated. There are two halfs, part one is as follows:

Consider the function: $$f(x) =\frac{1}{1-x}$$

Find the Taylor series expansion by expanding as a binomial series.

My solution is using $$(1+x)^n = 1 + nx + \frac{n(n-1)}{2!}x^2 + \frac{n(n-1)(n-2)}{3!}x^3 + ...$$

In this case, $$n = -1$$ and $$x = -1$$.

So by substituting I end up with $$1 + x + x^2 + x^3 + ...$$

So the Taylor Series Expansion in this case is:

$$\sum_{n=0}^\infty x^n$$

Ok so far, now the second half. Using your result for f (x) obtain the Taylor series for: $$g(x) =\frac{x^2}{1+x^2}$$

From what I understand reading, I need to convert f(x) into g(x) then apply whatever mathematical steps made to the terms in the series, which I did:

$$f(x) =\frac{1}{1-x} * \frac{1-x}{1} = \frac{1}{1}$$

$$\frac{1}{1} * \frac {1}{x^2+1} = \frac{1}{x^2+1}$$

$$\frac{1}{x^2+1} * x^2 = \frac{x^2}{x^2+1} = g(x)$$

I'll leave out the individual steps, but applying these to the terms of the series, I get:

$$\frac{x^2}{x^2+1} - \frac{x^3}{x^2+1} + \frac{x^3}{x^2+1} + \frac{x^4}{x^2+1} + \frac{x^4}{x^2+1}$$

Which is as far as I'm comfortable going, since this really doesn't feel right at all for a solution. I'd really appreciate if someone could point me in the right direction or let me know if I'm made some catastrophic error. I'm really not sure how to write the final terms in sigma notation either. Feel free to give me a slap on the wrist if I've made any silly errors.

• Replace $x$ with $-x^2$ in $f(x)$ \begin{eqnarray*} \frac{1}{1+x^2}=1-x^2+x^4-x^6+ \cdots \end{eqnarray*} & can you correct the first bit \begin{eqnarray*} \frac{1}{1-x}=1+\color{red}{x}+x^2+x^3 \cdots \end{eqnarray*} – Donald Splutterwit Sep 19 at 19:21
• Oh sorry, edit made. – William Slater Sep 19 at 19:28

We obtain from $$f(x)=\frac{1}{1-x}=\sum_{n=0}^\infty x^n$$

\begin{align*} \color{blue}{x^2f(-x^2)}&=x^2\sum_{n=0}^\infty(-x^2)^n\\ &=x^2\sum_{n=0}^\infty(-1)^nx^{2n}\\ &=\sum_{n=0}^\infty (-1)^nx^{2n+2}\\ &\,\,\color{blue}{=\sum_{n=1}^\infty(-1)^{n-1}x^{2n}} \end{align*}

In the last line we shift the index by $$1$$ in order to have $$x^{2n}$$.

• Not seen it done but are you saying $$f(x)=\frac{1}{1-x}, f(-x) = \frac{1}{1+x}, f(-x^2) = \frac{1}{1+x^2}, x^2 f(-x^2) = \frac{x^2}{1+x^2}$$ – William Slater Sep 19 at 20:39
• @WilliamSlater: The intention was to put you stepwise to the result. Is this fine for you or do you need some other information? – Markus Scheuer Sep 19 at 20:44
• No I think I understand but I keep pressing enter and having to then edit my comment, sorry. Not seen it done this way but are you saying $$f(x)=\frac{1}{1-x}, f(-x) = \frac{1}{1+x}, f(-x^2) = \frac{1}{1+x^2}, x^2 f(-x^2) = \frac{x^2}{1+x^2}$$ And so terms become $$x^2 - x^4 + x^6 ...$$ – William Slater Sep 19 at 20:47
• @WilliamSlater: Correct, that's what you get and finally you might try to write this expression using the sigma notation. – Markus Scheuer Sep 19 at 20:51
• Well I'm really not sure sigma as mentioned above but to my mind this should be $$\sum_{x=0}^\infty (i^2n) * ((-1)^(n-1))$$ Accounting for the power in each term then the alternating negatives. – William Slater Sep 19 at 21:15

$$g(x) =\frac{x^2}{1+x^2}=\frac{x^2+1-1}{1+x^2}=1-\frac{1}{1+x^2}$$ make $$x^2=-t$$ to face $$f(t)$$, simplify and, at the end, replace $$t$$ by $$-x^2$$.