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## Homework Statement

The Maclaurin series for a function f is given by [tex]\sum\frac{x^n}{2n}[/tex]. 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 + [tex]\frac{f''(0)x^2}{2!}[/tex] + [tex]\frac{f^n (0)x^n}{n!}[/tex]

## The Attempt at a Solution

I tried differentiating the series starting from f(x) to f

^{4}(x) Here is my work:

f(x) = [tex]\sum[/tex][tex]\frac{x^n}{2n}[/tex]

f'(x) = [tex]\sum[/tex][tex]\frac{x^{n-1}}{2}[/tex]

f''(x) = [tex]\sum[/tex][tex]\frac{(n-1)x^{n-2}}{2}[/tex]

f'''(x) = [tex]\sum[/tex][tex]\frac{(n-1)(n-2)x^{n-3}}{2}[/tex]

f

^{4}(x) = [tex]\sum[/tex][tex]\frac{(n-1)(n-2)(n-3)x^{n-4}}{2}[/tex]

Therefore:

f

^{4}(0) = [tex]\sum[/tex][tex]\frac{(n-1)(n-2)(n-3)(0)}{2}[/tex] = 0.

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

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