Science:Math Exam Resources/Courses/MATH103/April 2009/Question 02

MATH103 April 2009

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Question 02
Using the fact that ${\frac {1}{1-x}}=\sum _{k=0}^{\infty }x^{k}$ for x near zero, find the Taylor series for the function $f(x)={\frac {1}{(1-x)^{2}}}$ valid for x near zero.

Make sure you understand the problem fully: What is the question asking you to do? Are there specific conditions or constraints that you should take note of? How will you know if your answer is correct from your work only? Can you rephrase the question in your own words in a way that makes sense to you?

If you are stuck, check the hint below. Consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it!

Hint
Note that ${\frac {1}{(1-x)^{2}}}={\frac {1}{1-x}}\cdot {\frac {1}{1-x}}=\left({\frac {1}{1-x}}\right)^{2}$ If you multiply the two corresponding Taylor series, what will be the coefficient in front of the xk term? Start with small values of k and find a pattern.

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Solution

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. We just plug in the functions ${\frac {1}{(1-x)^{2}}}=\left({\frac {1}{1-x}}\right)^{2}=\left(\sum _{k=0}^{\infty }x^{k}\right)^{2}=\sum _{k=0}^{\infty }x^{k}\cdot \sum _{m=0}^{\infty }x^{m}$ Next we start to multiply the two infinite sums term by term, and sort the result by powers of x. Failed to parse (unknown function "\begin{align}"): {\displaystyle \begin{align} \sum_{k=0}^\infty x^k\cdot \sum_{m=0}^\infty x^m &= {\color{red} \left(1+x+x^2+x^3+\dots\right)}{\color{blue} \left(1+x+x^2+x^3+\dots\right)} \\ &= ({\color{red} 1} \cdot {\color{blue} 1}) + ({\color{red} x} \cdot {\color{blue} 1} + {\color{red} 1} \cdot {\color{blue} x}) + ({\color{red} 1} \cdot {\color{blue} x^2} + {\color{red} x} \cdot {\color{blue} x} + {\color{red} x^2} \cdot {\color{blue} 1}) \\ &\quad + ({\color{red} 1} \cdot {\color{blue} x^3} + {\color{red} x} \cdot {\color{blue} x^2} + {\color{red} x^2} \cdot {\color{blue} x} + {\color{red} x^3} \cdot {\color{blue} 1} ) + \dots\\ &= 1+2x + 3x^2 + 4x^3 + \dots \end{align} } Now we can see the pattern: There are always k + 1 combinations that give the term xk. Hence the Taylor series is given by ${\frac {1}{(1-x)^{2}}}={\frac {1}{(1-x)^{2}}}=\sum _{k=0}^{\infty }(k+1)x^{k}$.

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