Derivative of a Taylor Series f(x) is unknown

  • Thread starter Brilliant
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  • #1
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Homework Statement



If [tex]\sum_{n=0}^{\infty} a_{n}x^n[/tex] is a Taylor series that converges to f(x) for all real x, then f'(1) = ?


Homework Equations


A Taylor series:
[tex]\sum_{n=0}^{\infty} \frac {f^{(n)}(c)}{n!}(x-c)^n[/tex]
and the dirv of a Taylor series:
[tex]f'(x)=\sum_{n=0}^{\infty} na_{n}(x-c)^{n-1}[/tex]

The Attempt at a Solution


I started by taking the derivative of the basic Taylor serie:
[tex]f'(x)=\sum_{n=0}^{\infty} na_{n}(x-c)^{n-1}[/tex]
Then simply plugging in 1. I guess I was just thinking this is might be centered at 0, so I put in 0 for c, although the problem doesn't specify.
I end up with the expression:
[tex]f'(1)=\sum_{n=0}^{\infty}n\frac {f^{(n)}(0)}{n!}[/tex]

At this point I'm not sure what to do because I don't know the function f(x). It seems like the answer should be numerical, but the question is quite vague. Maybe this is blatantly simple and I'm just failing to see it, but any tips would be appreciated.
 

Answers and Replies

  • #2
Dick
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I don't think the answer is numerical. Just differentiate the series and plug in x=1.
 
  • #3
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Interesting. It is an odd question.

So you think
[tex]\sum_{n=0}^{\infty} \frac {f^{(n)}(0)}{(n-1)!}[/tex]
is all it's looking for?

Also, it seems like that should maybe be: f(n-1)(0). If you applied the chain rule when you differentiate, but then you would need to multiply by the derivative of the inside function? that I don't know?
EDIT:Actually maybe I'm just confused. :P
EDITEDIT: Yes, because the whole an term is constant, so no chain rule or anything funny?

Anyway, thanks.
 
  • #4
Dick
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I think you are over complicating this. If f(x)=a0+a1*x+a2*x^2+a3*x^3, then f'(1)=a1+2*a2+3*a3, right? It's the same problem but infinite.
 
  • #5
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Yes, you're right. I do this all the time. Thank you very much! Sometimes I just need to be slapped.
 

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