Science:Math Exam Resources/Courses/MATH101/April 2014/Question 01 (i)

MATH101 April 2014

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Question 01 (i)
Short Problem. Show your work. No credit will be given for the answer without the correct accompanying work. Find the coefficient $\displaystyle c_{5}$ of the fifth degree term in the Maclaurin series $\sum _{n=0}^{\infty }c_{n}x^{n}$ for $\displaystyle e^{3x}$.

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 hints below. Read the first one and consider it for a while. Does it give you a new idea on how to approach the problem? If so, try it! If after a while you are still stuck, go for the next hint.

Hint 1
How do the coefficients of the Maclaurin series for a function $f(x)$ relate to that function's derivatives?

Hint 2
What is the fifth derivative of the function?

Hint 3
(Alternative solution) What is the Maclaurin series for $e^{x}$? Using this, find the Maclaurin series for $e^{3x}$.

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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. By the formula for the Maclaurin series, the the fifth coefficient of the Maclaurin series for $f(x)$ is ${\begin{aligned}c_{5}={\frac {f^{(5)}(0)}{5!}}.\end{aligned}}$ Put $f(x)=e^{3x}$ and compute $f^{(5)}(x)$: ${\begin{aligned}f(x)=e^{3x},\quad f'(x)=3e^{3x},\quad f^{(2)}(x)=3^{2}e^{3x},\quad f^{(3)}(x)=3^{3}e^{3x},\quad f^{(4)}(x)=3^{4}e^{3x},\quad f^{(5)}(x)=3^{5}e^{3x}.\quad \end{aligned}}$ Thus, we get $f^{(5)}(0)=3^{5}$ and the answer is ${\begin{aligned}c_{5}={\frac {f^{(5)}(0)}{5!}}={\frac {3^{5}}{5!}}={\frac {243}{120}}.\end{aligned}}$

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. Since the Maclaurin series for $e^{x}$ ${\begin{aligned}e^{x}=\sum _{n=0}^{\infty }{\frac {x^{n}}{n!}},\end{aligned}}$ by plugging $3x$ in the position of $x$, we have the Maclaurin series for $e^{3x}$: ${\begin{aligned}e^{3x}=\sum _{n=0}^{\infty }{\frac {3^{n}x^{n}}{n!}}.\end{aligned}}$ Since the coefficient of the fifth degree term in this Maclaurin series is ${\frac {3^{5}}{5!}}$, the answer is ${\frac {3^{5}}{5!}}={\frac {243}{120}}$.

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