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I am asked to Find the first four terms of the Maclaurin series for f(x)= $$e^{-x^{2}}$$ and use this information to approximate $$\int_{-1}^{1} (e^{-x^{2}})dx$$ Then compare with the “actual” value (also obtained numerically).

I am unsure about the approach I have taken on the last half of the question... I found the first 4 terms of the Maclaurin series to be $$1-x^2$$ (2nd and 3rd term for n=1 and n=3 went to 0). I used that to find the power series representation from comparing it to $e^x$ and subbing in $-x^2$ and got this to be $$\sum_{n=0}^\infty\frac{(-1)^n}{n!}(x^{2n})$$ It is from here which I am unsure of... I calculated the integral $$\int_{-1}^{1}\frac{(-1)^n}{n!}(x^{2n})dx$$ to get $$\sum_{n=0}^\infty\frac{(-1)^n-(-1)^{3n+1}}{n!(1+2n)}$$

I found the first few terms of this series: $2-\frac{2}{3}+\frac{1}{5}-\frac{1}{21}+\frac{1}{108}-\frac{1}{660}$=approximately 1.49

I don't know how I am meant to approximate the value of $$\int_{-1}^{1} (e^{-x^{2}})dx$$ to compare with it's "actual" value? I tried subbing in x=0,1,2,3,4,5,6 into $$e^{{-x}^{2}}$$ and got an approximation of 1.39.

Any help/explanation would be really helpful because I cant think of any other way but I am pretty sure I have done the last part wrong.

User1997
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  • Are you supposed to know about numerical integration ? Or about the error function ? By the way, the exact value is $1.49365$ and your approximation gives $\frac{31049}{20790}=1.49346$ – Claude Leibovici Oct 30 '18 at 09:06
  • See this link https://math.stackexchange.com/questions/2971647/the-error-for-approximation-of-bessel-function/2972976#2972976 can help you. – JV.Stalker Oct 30 '18 at 09:19
  • @ClaudeLeibovici I think we have covered the error function... Could you show me how you know the exact value and where you got those numbers from for the approximation? – User1997 Oct 30 '18 at 09:31
  • @JV.Stalker I did find the error of my approximation to be approximately 0.0002 by taking the absolute value of the first neglected term $$\frac{1}{4680}$$ I still dont get how this helps to compare the the actual value... I dont actually understand how I get the actual value of it because I cant integrate? – User1997 Oct 30 '18 at 09:35
  • @ClaudeLeibovici could you please tell me how you got those numbers for the approximation? – User1997 Oct 31 '18 at 01:14

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