Science:Math Exam Resources/Courses/MATH105/April 2012/Question 01 (c)

MATH105 April 2012

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q2 (a)  •  Q2 (b)  •  Q2 (c)  •  Q3 (a)  •  Q3 (b)  •  Q4 (a)  •  Q4 (b)  •  Q5 (a)  •  Q5 (b)  •  Q6 (a)  •  Q6 (b)  •  Q7 (a)  •  Q7 (b)  •  Q8 (a)  •  Q8 (b)  •  Q8 (c)  •  Q8 (d)  •  Q8 (e)  •  Q8 (f)  •  Q8 (g)  •  Q8 (h)  •  Q8 (i)  •  

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Question 01 (c)
Express the Taylor series of the function $\displaystyle f(x)=\ln(1+2x)$ about x = 0 in summation notation.

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
First, use the fact from part (b) that ${\frac {1}{1-r}}=1+r+r^{2}+r^{3}+\dots$ provided |r| < 1, to get the power series of $\ln(1-r)$ per integration. Then make a proper substitution for r.

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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. For |r|<1 we use that ${\frac {1}{1-r}}=1+r+r^{2}+r^{3}+\dots$ and integrate both sides to obtain ${\begin{aligned}-\ln |1-r|&=\int {\frac {1}{1-r}}dr\\&=\int (1+r+r^{2}+r^{3}+\dots )dr\\&=r+r^{2}/2+r^{3}/3+r^{4}/4+...+C\end{aligned}}$ Since |r| < 1 we have |1-r| = 1-r, therefore $\displaystyle -\ln(1-r)=r+r^{2}/2+r^{3}/3+r^{4}/4+\dots +C$ To solve for the integration constant C we plug in r = 0: ${\begin{aligned}-\ln(1-r)&=-\ln 1=0\\&=0+0^{2}/2+0^{3}/3+0^{4}/4+...+C=C\end{aligned}}$ so C = 0. So we obtain the intermediate result that ${\begin{aligned}\ln(1-r)=-r-r^{2}/2-r^{3}/3-r^{4}/4+\dots =-\sum _{k=1}^{\infty }r^{k}/k.\end{aligned}}$ To arrive at the summation notation, we noted that every term has the same sign (so there are no terms like $(-1)^{k}$ floating around), every integer power of $r$ is included starting from $r^{1}$, and each $r^{k}$ is divided by k. As a last step, the Taylor series for $\ln(1+2x)$ can be found by replacing $r$ by $-2x$ in the series above. We find: $\ln(1+2x)=-\sum _{k=1}^{\infty }{\frac {(-2x)^{k}}{k}}$.

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