The CSAT and Practice[+] are designed by the Climb Foundation to help candidates. We are advocates for more opportunity to shine and less opportunity to fail, and we strive to level the playing field. 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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