MacLaurin Series for $f(x)=\frac{1}{1+x^2}$

Question

I would need some help with a MacLaurin Series. The function for which I should find a MacLaurin Series looks like $$f(x) = \frac{1}{1+x^2}$$ and I also know the answer how the MacLaurin Series should look like $$f(x)=1−x^2+x^4−x^6+x^8+O(x^9)$$ but I do not not the steps in between.

Until know I have the following $$f'(x) = -\frac{2x}{(1+x^2)^2}$$ $$f''(x) = \frac{4x^2}{(1+x^2)^3}$$ $$f''(x) = -\frac{8x^3}{(1+x^2)^4}$$

So the series should look something like this $$f(x) = \frac{1}{1+x^2} -\frac{2x^2}{(1+x^2)^2} + \frac{4x^3}{(1+x^2)^3} -\frac{8x^4}{(1+x^2)^4} + ...$$

or I am completely wrong? I hope someone can help me!

Answer 1

Remember the easiest expansion

$$\color{green}{\frac{1}{1-x}=1+x+x^2+x^3+...+x^n(1+\epsilon(x))}$$

then

as when $x\to 0$, $\; -x^2 \to 0$, we can replace $x\;$ by $\; -x^2$ to get

$$\frac{1}{1+x^2}=1-x^2+x^4-x^6+...(-1)^nx^{2n}(1+\epsilon(x)).$$

Answer 2

Plug in $u=-x^2$ into the series for $\frac{1}{1-u} = \sum_{i=0}^\infty u^i$.

Answer 3

We have $$f(x)=\frac{1}{1+x^2}$$ The particular series you provided $$T_f(x)=1−x^2+x^4−x^6+x^8+O(x^9)$$ is a Taylor series centered at $x_0=0$. Thus you have to plug that value into the derivatives, as the taylor series is defined by $$T_f(x)=\sum_{i=0}^{\infty}\frac{d^i}{dx^i}f(x_0)\frac{(x-x_0)^i}{i!}$$

Also, the derivatives you computed are wrong.

$$\frac{d^2}{dx^2}=\frac{6x^2-2}{(x^2+1)^3}$$

Evaluated at $x_0=0$ we get $$\frac{d^2}{dx^2}f|_{x=0}=-2$$

As you see this matches the second coefficient in your series. Looking at the formula for $T_f$ we see that the second term in the series should be $$\frac{d^2}{dx^2}f(x_0)\frac{(x-x_0)^i}{2!}\vert_{x_0=0}=-x^2$$

We proceed identically for higher order terms.

The same can be achieved by evaluating the geometric series $$\frac{1}{1-t}=\sum_{i=0}^{\infty}t^i$$ at $t=-x^2$ Thus $$\frac{1}{1+x^2}=\sum_{i=0}^{\infty}(-1)^ix^{2i}$$