Science:Math Exam Resources/Courses/MATH105/April 2013/Question 01 (k)

MATH105 April 2013

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Question 01 (k)
Short-Answer Questions. Put your answer in the box provided but show your work also. Each question is worth 3 marks, but not all questions are of equal difficulty. Find the Maclaurin series for $\displaystyle f(x)={\frac {1}{2x-1}}$ .

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
Recall the geometric series summation given by ${\begin{aligned}{\frac {1}{1-x}}=\sum _{n=0}^{\infty }x^{n}\end{aligned}}$ valid when $\displaystyle |x|<1$

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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 proceed by using the geometric series summation given by ${\begin{aligned}{\frac {1}{1-x}}=\sum _{n=0}^{\infty }x^{n}\end{aligned}}$ valid for $|x|<1$. To mimic this, we proceed via ${\begin{aligned}{\frac {1}{2x-1}}&=-{\frac {1}{1-2x}}\\&=-\sum _{n=0}^{\infty }(2x)^{n}\end{aligned}}$ valid when $\displaystyle |2x|<1$, that is, when $\displaystyle |x|<1/2$.

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