This is a problem from a textbook:

By taking the 4th degree Maclaurin polynomial for $e^{-x^2}$ find an approximation to $\int^1_0 e^{-x^2} \text{dx}$. Place bounds on the error in this approximation.

The first part is done by substituting $-x^2$ in series $$ e^{x} = 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!} + \cdots $$ yielding $$e^{-x^2} = 1 - x^2 + \frac{x^4}{2!} + \cdots $$ Integrating gives $$ \int^1_0 e^{-x^2} \approx x - \frac{x^3}{3} + \frac{x^5}{10} \bigg|^1_0 = \frac{23}{30}$$ Which is correct according the answer. However, I do not know how they have computed the bound,

error < $2.38 \times 10$

May someone explain? Thank you so much!

  • $\begingroup$ I think you want this. Also, the answer should be $\frac{23}{30}$.. $\endgroup$ – Mattos Apr 11 '15 at 9:40
  • $\begingroup$ @CWL: Probably your error bound is missing an exponent $-2$? :) I didn't edit, however, because the bound in John's answer is a bit larger than $2.38 \times 10^{-2}$, yet is likely what your book intended. Did you round off when posting, by chance? $\endgroup$ – Andrew D. Hwang Apr 11 '15 at 12:34

The reason is that your sequence

$$1 - x^2 + \frac{x^4}{2!}- \frac{x^6}{3!} + .... = \sum_{k=0}^\infty \frac{(-1)^k x^{2k}}{k!}$$

is alternating. In this case, if you use the first $n$-th term to approximate the value, then the error will be bounded by the $n+1$- term. (I hope this is proved in your textbook). In your case, the next term will be (forgetting the negative sign)

$$\frac{x^7}{7\cdot 3!}\bigg|_0^1 = \frac{1}{42} \sim 0.0238095....$$

  • $\begingroup$ Thank you so much! I wonder how we should bound the sum if it was not an alternating sequence. If the bound for the original summation is smaller than $|R_n(x)|$, the error term, what would be the bound of it's integral? $\endgroup$ – CL. Apr 11 '15 at 9:56
  • 1
    $\begingroup$ @CWL: If it wasn't an alternating series, you must use the definition of the remainder term (also known as the error term), $R_n(x)$ (use the Lagrange form; it is much easier than the integral form.) All you need to do is to find the values that maximizes the error and hence bounds it (invoke the extreme value theorem to find the maximum/minimum values on the interval of definition of the series.) $\endgroup$ – Meshal Apr 11 '15 at 10:42
  • 2
    $\begingroup$ (+1), but note that "alternating" here means "satisfies the conditions of the alternating series test", i.e., the terms alternate in sign and decrease to $0$ in absolute value. If the original upper bound of integration had been larger than $1$, a bit more work would have been required to show the terms decrease in size. :) $\endgroup$ – Andrew D. Hwang Apr 11 '15 at 12:29

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.