Maclaurin Series: Find Value of f(4)(0)

[carlodelmundo]

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?

 

[Focus]

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]

I think you are making this much too complicated!

The n^th 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.

 

[carlodelmundo]

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!