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26th Derivative of a Function

by Frillth
Tags: 26th, derivative, function
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Frillth
#1
Dec16-06, 11:17 AM
P: 80
The problem statement, all variables and given/known data

Given that f(x) = sin(x) for x =/= 0 and f(x) = 1^x for x=0, find the 26th derivative of f at 0. Hint: can you find a power series for f(x)?

The attempt at a solution

I have no idea how to solve this problem. Since 1^x is always 1, the first derivative at 0 is 0, so ALL derivatives must be 0, right? I'm confused as to how a power series even comes into play in this problem.
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StatusX
#2
Dec16-06, 11:38 AM
HW Helper
P: 2,567
That's a very strange defintion. Note that f(0) is just a number, so all they had to say was f(x)=1 for x=0, the 1^x bit is superfluous. But moreover, the function is not continuous at x=0, so doesn't have any derivatives, let alone 26. Which leads me to ask, are you sure you copied the question correctly?
Frillth
#3
Dec16-06, 11:45 AM
P: 80
I just noticed that somebody erased a line in my book! It should have been sin(x)/x for x=/=0 and 1 for x=0. That makes a lot more sense.

arildno
#4
Dec16-06, 12:24 PM
Sci Advisor
HW Helper
PF Gold
P: 12,016
26th Derivative of a Function

I advise you to use the hint given!


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