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

MATH103 April 2015

  •  Q1 (a) (i)  •  Q1 (a) (ii)  •  Q1 (a) (iii)  •  Q1 (b) (i)  •  Q1 (b) (ii)  •  Q1 (b) (iii)  •  Q1 (c) (i)  •  Q1 (c) (ii)  •  Q1 (c) (iii)  •  Q1 (d) (i)  •  Q1 (d) (ii)  •  Q1 (e) (i)  •  Q1 (e) (ii)  •  Q1 (e) (iii)  •  Q2 (a)  •  Q2 (b)  •  Q2 (c)  •  Q2 (d)  •  Q2 (e)  •  Q3 (a)  •  Q3 (b)  •  Q3 (c)  •  Q4  •  Q5 (a)  •  Q5 (b)  •  Q5 (c)  •  Q6 (a)  •  Q6 (b)  •  Q6 (c)  •  Q7 (a)  •  Q7 (b)  •  Q7 (c)  •  Q7 (d)  •  Q8  •  Q9  •  Q10 (a)  •  Q10 (b)  •  

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Question 01 (e) (ii)
Match the given Taylor series and functions below $\sin x,\quad \cos x,\quad {\frac {1}{(1-x)^{2}}},\quad \ln(1+x),\quad e^{-x}$ (i) $\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{(2n-1)!}}x^{2n-1}=x-{\frac {x^{3}}{6}}+{\frac {x^{5}}{120}}\cdots$

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
Recall the Taylor Series for $\sin x.$

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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 know that the Taylor series of $\sin x$ is given by $\sin x=x-{\frac {x^{3}}{3!}}+{\frac {x^{5}}{5!}}+...+{\frac {(-1)^{n-1}x^{2n-1}}{(2n-1)!}}+...=\sum _{n=1}^{\infty }{\frac {(-1)^{n-1}}{(2n-1)!}}x^{2n-1}.$ If you do not memorize the $\sin x$ series, you can check by eliminating the other options. We know that $\sin 0=0$, the only series with 0 value at 0 is the second option. Among the given functions $\ln(x+1)$ is also 0 at $x=0$, however, the value of its second derivative at $x=0$ is non-zero, so its Taylor series must have a second degree term, so the only option remains is $\sin x$ .