The discussion revolves around finding the 13th Taylor coefficient of the function f(x) = e^(7x) at x = 3. Participants are exploring the Taylor series expansion and the relevant equations for exponential functions.
The discussion is ongoing, with participants questioning the original poster's approach and suggesting a deeper examination of the Taylor series terms and derivatives. There is no explicit consensus yet, but guidance is being offered regarding the evaluation of derivatives and the identification of terms.
Participants are navigating the complexities of Taylor series and the specific requirements for determining coefficients, including the evaluation of derivatives at a particular point. The original poster's confusion about their calculations indicates potential misunderstandings about the series expansion process.
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
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##?