Taylor series of $\log\left(1+2x^2\right)$

Question

I want to find the Taylor-MacLaurin series of the following function until we find a term that is different from zero. This is the function:

$$f(x)=\log\left(1+2x^2\right)+4(\cos{x}-1)$$

Finding the derivative of $\left(\cos x-1\right)$ is easy enough. I found:

$$\cos(x)-1=-\frac{x^2}{2}+\frac{1}{4!}x^4+o(x^4)$$

Finding the derivatives of $\log(1+2x^2)$ up to the $4$th order isn't as straightforward. Can I do something like this:

$$\log\left(1+2x^2\right)\sim2x^2$$

And then find the derivatives from there?

• $0$th order: $f(0)=2(0)^2=0$
• $1$st order: $f'(x)=4x\implies f'(0)=0$
• $2$nd order: $f''(x)=4 \implies f''(0)=4$?

Any hints on how to easily find the Tayler series of $\log\left(1+2x^2\right)$? I have also tried chancing the variables but it leads to unexpected results:

$$\log\left({1+2x^2}\right)=\log{t}$$

• $0$th order: $f(t)=\log{t}\implies f(0)=\log{0}$ isn't defined.

Answer 1

Since$$\log(1+x)=x-\frac{x^2}2+\frac{x^3}3-\frac{x^4}4+\cdots,$$you have\begin{align}\log(1+2x^2)&=2x^2-\frac{(2x^2)^2}2+\frac{(2x^2)^3}3-\frac{(2x^2)^4}4+\cdots\\&=2x^2-\frac{4x^4}2+\frac{8x^6}3-\frac{16x^4}4+\cdots\end{align}

Answer 2

Let $g(x)=\ln 1+x$ therefore $$g'(x)=\dfrac{1}{1+x}\\g''(x)=-\dfrac{1}{(1+x)^2}\\g'''(x)=\dfrac{2}{(1+x)^3}\\.\\.\\.\\g^{(n)}(x)=(-1)^{n+1}\dfrac{(n-1)!}{(1+x)^n}$$which means that$$g^{(n)}(0)=(-1)^{n+1}(n-1)!$$and we have $$\ln(1+x)=\sum_{n=1}^{\infty}\dfrac{(-1)^{n+1}}{n}x^n$$therefore $$\ln(1+2x^2)=\sum_{n=1}^{\infty}\dfrac{(-1)^{n+1}}{n}2^nx^{2n}=\sum_{n=1}^{\infty}-\dfrac{(-2)^{n+1}}{n}x^{2n}=$$