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

MATH103 April 2009

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[hide] 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.

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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!

[show] 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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[show] Solution

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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