# Maclaurin Series

## Homework Statement

The Maclaurin series for a function f is given by $$\sum\frac{x^n}{2n}$$. What is the value of f(4)(0), the fourth derivative of at x = 0?

a.) 1
b.) 2
c.) 3
d.) 4
e.) 5

## Homework Equations

The Maclaurin Series is the infinite series centered at x = 0 with the following formula:

f(x) = f(0) + f'(0)x + $$\frac{f''(0)x^2}{2!}$$ + $$\frac{f^n (0)x^n}{n!}$$

## The Attempt at a Solution

I tried differentiating the series starting from f(x) to f4(x) Here is my work:

f(x) = $$\sum$$$$\frac{x^n}{2n}$$

f'(x) = $$\sum$$$$\frac{x^{n-1}}{2}$$

f''(x) = $$\sum$$$$\frac{(n-1)x^{n-2}}{2}$$

f'''(x) = $$\sum$$$$\frac{(n-1)(n-2)x^{n-3}}{2}$$

f4(x) = $$\sum$$$$\frac{(n-1)(n-2)(n-3)x^{n-4}}{2}$$

Therefore:

f4(0) = $$\sum$$$$\frac{(n-1)(n-2)(n-3)(0)}{2}$$ = 0.

0 is not in the answer choice. I would like hints please, not answers... what am I doing wrong?

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## Answers and Replies

Depends on the limit of the sum. Say I gave you $x^5+x^4+x^3+x^2+x+1=\sum_{n=0}^5 x^n$ and asked you what the derivative at zero was, you wouldn't say its zero would you?

HallsofIvy
Homework Helper
I think you are making this much too complicated!

The nth term of a MacLaurin series is, as you say,
$$\frac{f^{(n)}(0)}{n!}x^n$$.

If a power series is $\sum a_nx^n$, then
$$\frac{f^{(n)}(0)}{n!}x^n= a_n x^n$$
so
$$f^{(n)}(0)= n!a_n$$

In your example, $a_n= 1/2n$.

Thank you. I see what I did wrong now then. I didn't know we can be able to equate the nth term of the MacLaurin with the series.

Thanks again!