Calculating 13th Taylor Coefficient of f(x)=e^(7x) at x=3

cathy
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Homework Statement


Determine its 13^{th} Taylor coefficient of the Taylor Series generated by f at x = 3.
f(x)=e^(7x)


Homework Equations


I used the fact that the series for e^x was ∑x^n/n!



The Attempt at a Solution


Using that above, and replacing x with 7x, shouldn't my answer be 3x*7^13/13!
But that is incorrect? Please advise where I am going wrong? Thank you
 
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cathy said:

Homework Statement


Determine its 13^{th} Taylor coefficient of the Taylor Series generated by f at x = 3.
f(x)=e^(7x)


Homework Equations


I used the fact that the series for e^x was ∑x^n/n!



The Attempt at a Solution


Using that above, and replacing x with 7x, shouldn't my answer be 3x*7^13/13!
But that is incorrect? Please advise where I am going wrong? Thank you

Here's a few questions for you. If$$
f(x) = \sum_{n=0}^\infty \frac{ f^{(n)}(3)}{n!}(x-3)^n$$what value of ##n## gives the 13th term? Once you have that, what is that many derivatives of ##e^{7x}## and what do you get when you evaluate that at ##x=3##?
 
LCKurtz said:
Here's a few questions for you. If$$
f(x) = \sum_{n=0}^\infty \frac{ f^{(n)}(3)}{n!}(x-3)^n$$what value of ##n## gives the 13th term? Once you have that, what is that many derivatives of ##e^{7x}## and what do you get when you evaluate that at ##x=3##?

Isn't it the 13th value of n that gives the 13th term?
 
No. Give it some thought. Write out some terms.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...
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