# Practice Paper 1 Question 7

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 \(x\).

## Related topics

## Hints

- Hint 1What properties of \(\log\) can you use break down \(\ln\!\left(\frac{1-x}{1+x^2}\right)\)?
- Hint 2How would you relate each term in the above breakdown to the given identity?

## Solution

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 \(x^4\): \[ \begin{align} \ln\!\left(\frac{1-x}{1+x^2}\right) &=(-x-\frac{x^2}{2}-\frac{x^3}{3}-\frac{x^4}{4}-\ldots) \\ &\qquad-(x^2-\frac{x^4}{2}+\ldots)\\ &=-x-\frac{3x^2}{2}-\frac{x^3}{3}+\frac{x^4}{4}+\ldots \end{align} \]

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