Science:Math Exam Resources/Courses/MATH103/April 2012/Question 01 (f)

MATH103 April 2012

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Question 01 (f)
Answer the following multiple choice question. Check your answer very carefully. Your answer will be marked right or wrong (work will not be considered for this problem). In the Taylor series of $\displaystyle x\ln(1-x)$ (about the point x = 0), the coefficient of the x3 term is A. 1/3 B. -1/3 C. 1/2 D. -1/2

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
The Taylor series of $\displaystyle x\ln(1-x)$ is the Taylor series of $\displaystyle \ln(1-x)$ multiplied by x.

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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. Recall that $\displaystyle \ln(1-x)=-x-{\frac {x^{2}}{2}}-{\frac {x^{3}}{3}}-\cdots ,$ therefore $\displaystyle x\ln(1-x)=-x^{2}-{\frac {x^{3}}{2}}-{\frac {x^{4}}{3}}-\cdots$ The coefficient of x3 term is -1/2 and so the correct answer is D.

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. If you don't remember the Taylor series about x = 0 of $\displaystyle \ln(1-x)$ you can simply re-compute it. Recall that the Taylor series of a function ƒ about the point x = a is $f(a)+f'(a)(x-a)+{\frac {f''(a)}{2!}}(x-a)^{2}+{\frac {f^{(3)}(a)}{3!}}(x-a)^{3}+\ldots$ For the function $\displaystyle f(x)=\ln(1-x)$ we have $\displaystyle f'(a)=-{\frac {1}{1-a}}\quad f''(a)=-{\frac {1}{(1-a)^{2}}}\quad f^{(3)}(a)=-{\frac {2}{(1-a)^{3}}}$ and so at a = 0 we obtain $\displaystyle f'(0)=-1\quad f''(0)=-1\quad f^{(3)}(x)=-2$ And so the Taylor series of $\displaystyle \ln(1-x)$ around x = 0 is $\displaystyle -x-{\frac {1}{2}}x^{2}-{\frac {1}{3}}x^{3}+\ldots$ And hence the Taylor series of $\displaystyle x\ln(1-x)$ around x = 0 is $\displaystyle -x^{2}-{\frac {1}{2}}x^{3}-{\frac {1}{3}}x^{4}+\ldots$ And we can observe that the x3 coefficient is -1/2 which is the answer D.

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