 # 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 $$x$$.

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## Hints

• Hint 1
What properties of $$\log$$ can you use break down $$\ln\!\left(\frac{1-x}{1+x^2}\right)$$?
• Hint 2
How 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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