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

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MATH103 April 2012

  •  Q1 (a)  •  Q1 (b)  •  Q1 (c)  •  Q1 (d)  •  Q1 (e)  •  Q1 (f)  •  Q1 (g)  •  Q2 (a)  •  Q2 (b)  •  Q2 (c)  •  Q2 (d)  •  Q2 (e)  •  Q2 (f)  •  Q3 (a)  •  Q3 (b)  •  Q4 (a)  •  Q4 (b)  •  Q4 (c)  •  Q4 (d)  •  Q4 (e)  •  Q4 (f)  •  Q5 (a)  •  Q5 (b)  •  Q5 (c)  •  

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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.

Checking a solution serves two purposes: helping you if, after having used the hint, you still are stuck on the problem; or if you have solved the problem and would like to check your work. If you are stuck on a problem: Read the solution slowly and as soon as you feel you could finish the problem on your own, hide it and work on the problem. Come back later to the solution if you are stuck or if you want to check your work. If you want to check your work: Don't only focus on the answer, problems are mostly marked for the work you do, make sure you understand all the steps that were required to complete the problem and see if you made mistakes or forgot some aspects. Your goal is to check that your mental process was correct, not only the result.

Solution 1
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

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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