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Hello I was wondering if I could get some help trying to find a sum of the first three terms in series.

The problem statement is:

"Write the first three terms in the series for $y(t) = e^{t^2}$, and us it to approximate y(2)"

I know that I need to use the Taylor series to calculate this, but for some reason when I try to use it I am struggling with the $t^2$ part.

Thanks for the help!

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  • $\begingroup$ Hint: it is easier to start with the series expansion of $e^x$ and substitute $x=t^2$. $\endgroup$ – lulu Jan 18 '18 at 20:47
  • $\begingroup$ You just put in $t^2$ everywhere you would have put in an $x$ in the Taylor series of $e^x$ i.e. $1 + t^2 + \frac {(t^2)^2}{2!} + \cdots$ $\endgroup$ – Doug M Jan 18 '18 at 20:48
  • $\begingroup$ Ok that makes more sense. So then I have 1+4+8 which the answer should be 13 $\endgroup$ – noreturn Jan 18 '18 at 20:53
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$$e^t=1+t+\frac{t^2}{2!}+\frac{t^3}{3!}+...\to \\ e^{t^2}=1+t^2+\frac{t^4}{2!}+\frac{t^6}{3!}+...\approx 1+t^2+\frac{t^4}{2}\to y(2)\approx 13$$

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Let $$x= t^2$$ in the Taylor expansion of $e^x$ to get $$e^{t^2}=1+t^2+\frac{t^4}{2!}+\frac{t^6}{3!}+...$$ The three term approximation is $13$. This approximation is very weak, because we are far away from the center, $t=0$.

The error is $e^4 - 13 = 41.59$ which is not good at all.

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