Calculating Taylor Series for e^(x^2) around x=0

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Homework Help Overview

The discussion revolves around finding the Taylor series expansion of the function e^(x^2) around the point x=0. Participants are exploring the terms of the series and their calculations.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • The original poster attempts to identify the first few terms of the Taylor series but expresses confusion regarding the second term, initially miscalculating it as zero. Some participants clarify that the second term involves the second derivative evaluated at zero.

Discussion Status

Participants are actively engaging with the problem, with some providing clarifications and suggestions for further exploration. There is an acknowledgment of a typographical error in the original post, and guidance is being offered regarding the correct approach to finding the series terms.

Contextual Notes

There is a mention of an answer key that suggests the second term is x^2, which has led to confusion. The discussion reflects on the need to correctly apply the Taylor series formula and evaluate derivatives at the specified point.

dantheman57
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Homework Statement


Find the Taylor series of e^(x^2) about x=0


Homework Equations



Taylor Series = f(a) +f'(a)(x-a) + (f''(a)(x-a)^2)/2 ...

The Attempt at a Solution



So, the first term is pretty obvious. It's e^0^2, which is zero.

The second term is what got me. (e^x^2)'=2x*(e^x^2), so at zero that is zero. Multiply by x, still zero. But the answer key says the second term is x^2. I really cannot understand this.

Thanks!
 
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dantheman57 said:
[

So, the first term is pretty obvious. It's e^0^2, which is zero.

My bad. e^0^2 is one. Typo.
 
Keep going. It's the f'' term that is the second term.
 
Thank you so much!
 
You can also take the Taylor series of ex and then just fill in x2.
 

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