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This is my question with context (I posted earlier but needed to change it!)

Calculate: $$\int_0^\frac{1}{2} e^{-x^2} \,dx$$

by using a Taylor approximation of $e^{−x^2}$ up to the term in $x^4$. Then I need to estimate the error, using the correct remainder term, and tell whether the answer will be correct to 2 decimal places.

The first part I did fine: $\int_0^\frac{1}{2} T_4(x) \,dx=\int_0^\frac{1}{2} 1-x^2+\frac{x^4}{2!} \,dx$

Which gives $\frac{443}{960}$. But then I am not sure how to go about estimating the error? I have the formula:

$$R_n(x)=\frac{f^{n+1}(z)}{(n+1)!}(x-a)^{n+1}$$

If anyone can help show me how to do the remainder part, this would be much appreciated!

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The next term in the expansion of the integrand is $-\frac{x^6}{3!}$ Since it is an alternating series, the remainder is $\le \int_0^\frac{1}{2} max\frac{x^6}{3!}dx=\frac{2^{-7}}{3!}=0.00130208333$.

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  • $\begingroup$ How come you don’t consider any terms after $x^6$ because surely the remainder includes those aswell $\endgroup$ – Olly Reynolds Apr 26 at 7:22
  • $\begingroup$ Also how did you integrate that integral? I didn’t not get that? $\endgroup$ – Olly Reynolds Apr 26 at 12:11
  • $\begingroup$ 1. Since he series has alternating and decreasing terms, the term you stop out gives an upper limit to the remainder. 2 $Max \frac{x^6}{3!}=\frac{2^{-6}}{3!}$, since $\frac{1}{2}$ is the maximum for $x$ over the interval. To integrate a constant, multiply by the interval length $\frac{1}{2}$ to get the integral. $\endgroup$ – herb steinberg Apr 26 at 15:55

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