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carlodelmundo
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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 f4(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]f4(x) = [tex]\sum[/tex][tex]\frac{(n-1)(n-2)(n-3)x^{n-4}}{2}[/tex]
Therefore:
f4(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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