# Finding Taylor polynomial for fractional function

I should find Taylor polynomial of a degree $n \in \mathbb{N}$ for function $f(x) =\frac{x}{9+x^2}$ at the point of 0.

For which $x \in \mathbb{R}$ this polynomial converges to the given function $f$ for $n \rightarrow \infty$?

So there is a general formula for Taylor series: $f(a)+\frac {f'(a)}{1!} (x-a)+ \frac{f''(a)}{2!} (x-a)^2+\frac{f^{(3)}(a)}{3!}(x-a)^3+ \cdots.$ I assume, that in this formula for my given data, a=0 and the only thing I should do is to derive the function?

• Yes, your suggestion will work. Also, the series will only converge to the function until it reaches the pole, i.e. for |x|<3. – sebigu Feb 23 '15 at 12:01
• Do you know the Taylor expansion of other functions, like $\;\log x\;,\;\;\sin x\;$ and etc.? – Timbuc Feb 23 '15 at 12:10

Here are a few pointers:

1. First, find the Taylor polynomial for $\frac{1}{1-x}$. The geometric series should be of some use! And you can find an explicit formula for the error term too.
2. Use this to find the Taylor polynomial for $\frac{1}{1+x^2}$. Just replace the symbol $x$ with the symbol $-x^2$.
3. Use this to find the Taylor polynomial for $\frac{1}{1+(x/3)^2}$. Again, just replace the symbol $x$ with the symbol $x/3$.
4. Now just multiply by $x/9$.

Now just determine for which values of $x$ your error term goes to zero as $n\to\infty$.

• OK. ad1) It looks fine - just simple $1 + x + x^2 + x^3 + \cdots$? ad2) Replacing the symbol will cause $1 - x^2 + x^4 - x^6 + \cdots$ ad3) Another replacement is leading to $1 - \left(\frac{x}{3}\right)^2 + \left(\frac{x}{3}\right)^4 - \left(\frac{x}{3}\right)^6 + \cdots$ ad4) Should I multiply everything like $\frac{x}{9} - \frac{x}{9}\left(\frac{x}{3}\right)^2 + \frac{x}{9}\left(\frac{x}{3}\right)^4 - \frac{x}{9}\left(\frac{x}{3}\right)^6 + \cdots$ – PObdr Feb 23 '15 at 12:08
• Yes. You're on the right track. But the question requires a Taylor polynomial of degree $n$, not the series. And you also want to find the error term. So you have to find the difference between $1+x+...+x^n$ and $\frac{1}{1-x}$. The first term is a finite geometric series, so you can easily find a formula for its sum. Then use this to find the error... – Auslander Feb 23 '15 at 12:13
• Is the only difference between Taylor series and polynomial, that the sereies goes to infinity, but the polynomial has the last member (the $n^{th}$ one)? – PObdr Feb 23 '15 at 13:32
• Yes, that's right. – Auslander Feb 23 '15 at 20:55
• Right. This means you'll have to approach it as above, from what I can tell. – Auslander Feb 24 '15 at 1:30

An idea:

$$f(x)=\frac x{9+x^2}=\frac12\;\frac{\frac29x}{1+\frac{x^2}9}\implies$$

$$\int f(x)\,dx=\frac12\log\left(1+\frac{x^2}9\right)=\frac12\sum_{n=1}^\infty(-1)^{n+1}\frac{x^{2n}}{3^{2n}}\;,\;\;\text{for}\;\;\frac{x^2}9<1\iff -3<x<3\implies$$

$$\frac x{9+x^2}=\left(\int f(x)\,dx\right)'=\frac12\sum_{n=1}^\infty(-1)^{n+1}\frac{2nx^{2n-1}}{9^n}\;,\;\;\text{for}\;\;-3<x<3$$