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$$e^{-x^2}$$

I've had a hard time understanding power series since as long as I can remember. To my understanding, the question is asking me to write out the terms in the formula for Taylor series, then find out how to write that in sigma notation, then evaluate the Riemann sum at the given point above. Is that the correct way to look at this?

My attempt:

I thought it seemed similar to $e^{-x}$ so I just did

$\displaystyle\sum_{n = 0}^{\infty}\frac{(-x^2)^n}{n!}$ and then plugged in $n=39$ to get

$$ \frac{(-x^2)^{39}}{39!}$$

I don't feel like this is right but I don't really understand how to go about this in the first place...

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The Taylor series for $f(x) = e^{-x^2}$, centered at $x = 0$, is indeed $\displaystyle\sum_{n = 0}^{\infty}\dfrac{(-x^2)^n}{n!}$.

However, to evaluate $f^{(39)}(0)$, you need to look at the coefficient of the $x^{39}$ term (and then multiply that coefficient by $39!$). The term $\dfrac{(-x^2)^{39}}{39!} = -\dfrac{x^{78}}{39!}$ is not the correct term to look at to determine $f^{(39)}(0)$, as that term has $x$ raised to the $78$th power, not the $39$th power.

Hint: Are there any terms of $\displaystyle\sum_{n = 0}^{\infty}\dfrac{(-x^2)^n}{n!}$ which have $x$ raised to the $39$th power? If not, what is the coefficient of $x^{39}$?

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  • $\begingroup$ I would guess that there isn't because you can't multiply $x^2$ by any power to get 39. So the coefficient would be zero? $\endgroup$ – whatwhatwhat Jan 26 '16 at 0:48
  • $\begingroup$ ^Yes, that is correct. $\endgroup$ – JimmyK4542 Jan 26 '16 at 0:50
  • $\begingroup$ Ok so I'll face palm myself later. But in regards to the first part - if I didn't have an easy $e^x$ function to compare to, how would I find the expansion/sum? Do people really just plug it into the Taylor series equation, then find the term that looks to be repeating? Taking the derivatives to do that seems like it could be potentially fatal.... $\endgroup$ – whatwhatwhat Jan 26 '16 at 0:54

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