Science:Math Exam Resources/Courses/MATH103/April 2015/Question 01 (e) (iii)

MATH103 April 2015

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Question 01 (e) (iii)
Match the given Taylor series and functions below $\sin x,\quad \cos x,\quad {\frac {1}{(1-x)^{2}}},\quad \ln(1+x),\quad e^{-x}$ (iii) $\sum _{n=0}^{\infty }nx^{n-1}=1+2x+3x^{2}\cdots$

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
Use the fact that ${\frac {1}{1-x}}=1+x+x^{2}+x^{3}+...+x^{n}+...=\sum _{n=0}^{\infty }x^{n}.$

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Solution 1

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 first note that ${\frac {1}{1-x}}=1+x+x^{2}+x^{3}+...+x^{n}+...=\sum _{n=0}^{\infty }x^{n}.$ Taking derivative fro both sides, we get that $({\frac {1}{1-x}})'={\frac {1}{(1-x)^{2}}}=(1+x+x^{2}+x^{3}+...+x^{n}+...)'=0+1+2x+3x^{2}+...+nx^{n-1}+...=\sum _{n=0}^{\infty }nx^{n-1}.$ So, we have that ${\frac {1}{(1-x)^{2}}}=1+2x+3x^{2}+...+nx^{n-1}+...=\sum _{n=0}^{\infty }nx^{n-1}.$

Solution 2

Found a typo? Is this solution unclear? Let us know here. Please rate my easiness! It's quick and helps everyone guide their studies. Again, we can find the answer by eliminating the options. The remaining options are $\cos x,\ln(1+x)$ and ${\frac {1}{(1-x)^{2}}}$. At $x=0$, $\ln(1+x)$ is equal to 0, however the given series at $x=0$ is equal to 1. $\cos x$ although is equal to 1 at $x=0$, its Taylor series does not have any odd degree term because its 1st, 3rd, 5th, ... derivative vanish at $x=0$, so the only option is ${\frac {1}{(1-x)^{2}}}$.