The Taylor expansion of
$\ln(1+x)$ is defined as $\ln(1+x)=x-\frac{x^2}{2}+\frac{x^3}{3}-\cdots$. Expand $\ln\!\left(\frac{1-x}{1+x^2}\right)$ up to and including the 4th power of $x$.
-
What properties of
$\log$can you use break down$\ln\!\left(\frac{1-x}{1+x^2}\right)$? -
How would you relate each term in the above breakdown to the given identity?
-
Although this question involves Taylor expansion, we do not need to know how to formally expand a function into a Taylor series to answer it. Let
$f(x)=\ln(1+x)$. Notice that$\ln\!\left(\frac{1-x}{1+x^2}\right) = \ln(1-x)-\ln(1+x^2)=$$f(-x)-f(x^2).$Since the Taylor expansion of$f(x)$is given, substitute$-x$and$x^2$to obtain the terms up to$x^4$:$$ \begin{aligned} \ln\!\left(\frac{1-x}{1+x^2}\right) &=(-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-\ldots) \\ &\qquad-(x^2-\frac{x^4}{2}+\ldots)\\ &=-x-\frac{3x^2}{2}-\frac{x^3}{3}+\frac{x^4}{4}+\ldots \end{aligned} $$