Science:Math Exam Resources/Courses/MATH103/April 2015/Question 02 (b)

MATH103 April 2015

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Question 02 (b)
Consider the function and its series $f(x)={\frac {1}{1-x^{3}}}=a_{0}+a_{1}x+a_{2}x^{2}+\cdots$. Find $a_{99}$ and $f^{(100)}(0)$. (i.e. the 100th derivative of $f(x)$ evaluated at $x=0$).

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
What is Maclaurin series for ${\frac {1}{1-z}}$?

Hint 2
In the Maclaurin series $\sum _{n=0}^{\infty }a_{n}x^{n}$ for $f(x)$, what's the relation between $a_{n}$ and $f^{(n)}$?

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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. Note that the Maclaurin series for ${\frac {1}{1-z}}$ is ${\frac {1}{1-z}}=\sum _{n=0}^{\infty }z^{n}=1+z+z^{2}+z^{3}+...$ and the equality holds when $|z|<1$. Plugging $x^{3}$ into $z$, we obtain the series expression of $f(x)$ as ${\frac {1}{1-x^{3}}}=\sum _{n=0}^{\infty }x^{3n}=1+x^{3}+x^{6}+\cdots .$ Since $a^{99}$ is the coefficient of $x^{99}$ in the series and $99=3\cdot 33$, we have $a_{99}=1$ when $|x|<1$. On the other hand, $a_{100}$ has the following relation with $f^{(100)}(0)$: $a_{100}={\frac {f^{(100)}(0)}{100!}}$. Since $a_{100}=0$ (because 100 is not a multiple of 3), we have $f^{(100)}(0)=0$. Answer: $\color {blue}a_{99}=1,f^{(100)}(0)=0.$