Question

Use this list of Basic Taylor Series to find the Taylor Series for f(x) = 2x 1+x2 based at 0. Give your answer using summation notation and give the largest open interval on which the series converges. (If you need to enter ∞ , use the ∞ button in CalcPad or type "infinity" in all lower-case.) The Taylor series for f(x)= 2x 1+x2 is: ∞ k=0 The Taylor series converges to f(x) for |x|< . Use the Taylor series you just found for f(x)= 2x 1+x2 to find the Taylor series for ln(1+x2) based at 0. Give your answer using summation notation and give the largest open interval on which the series converges. The Taylor series for ln(1+x2) is: ∞ k=0 The Taylor series converges to ln(1+x2) for |x|< .

Answer 1

Recall that for $|x|$, we have

$\displaystyle\frac1{1-x}=\sum_{n=0}^\infty x^n$

Then for $|-x^2|$, or $|x|$, we have

$\displaystyle\frac1{1+x^2}=\frac1{1-(-x^2)}=\sum_{n=0}^\infty(-x^2)^n=\sum_{n=0}^\infty(-1)^nx^{2n}$

Multiply this series by $2x$ to get the Taylor series for $f(x)$:

$f(x)=\dfrac{2x}{1+x^2}=\displaystyle2\sum_{n=0}^\infty(-1)^nx^{2n+1}$

Notice that

$\dfrac{\mathrm d(\ln(1+x^2))}{\mathrm dx}=\dfrac{2x}{1+x^2}$

so to find the Taylor series for $g(x)=\ln(1+x^2)$, we integrate the Taylor series for $f(x)$:

$g(x)=\displaystyle\int f(x)\,\mathrm dx=C+2\sum_{n=0}^\infty\frac{(-1)^nx^{2n+2}}{2n+2}$

Since $g(0)=\ln(1+0^2)=\ln1=0$, it follows that $C=0$ and

$g(x)=\displaystyle\sum_{n=0}^\infty\frac{(-1)^nx^{2n+2}}{n+1}$

which converges for $|x|$ as well.