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I seem to be having trouble with part of this question (Reference: Apostol Volume 1, Section 7.8, Question 8). The full question states:

(a) If $0 \leq x\leq \frac{1}{2}$, show that $\sin x = x - x^3/3! + r(x)$, where $|r(x)| \leq (\frac{1}{2})^5/5!$

(b) Use the estimate in part (a) to find an approximate value for the integral $\int_0^{\sqrt{2}/2} \sin(x^2)\, dx$. Make sure you give an estimate for the error.

I have successfully solved part (a) and the approximate value for the integral in part (b), but it's the error estimate in part (b) that's got me stumped. My working for part (b) has been as follows:

Since $\sin x = x - x^3/3! + r(x)$, we can say $\sin x^2 = x^2 - x^6/3! + r(x^2)$. We already know that $$ r(x^2) = E_4(x^2) \leq \frac{(x^2)^5}{5!} $$

so it then follows:

$$ \int_0^{\sqrt{2}/2} \sin(x^2)\, dx = \left(\int_0^{\sqrt{2}/2} x^2 - \frac{x^6}{3!}\, dx\right) + E_4(x^2) $$ $$ = \left[\frac{x^3}{3} - \frac{x^7}{42}\right]_0^{\sqrt{2}/2}+ E_4(x^2) $$ $$ = \sqrt{2}(\frac{55}{672}) + E_4(x^2) $$ Since $$ E_4(x^2) \leq \frac{(x^2)^5}{5!} $$ then (AFAIK) the error estimate would be $((\sqrt{2}/2)^2)^5/120$ which is $\approx 2.6\times 10^{-4}$. This doesn't match the answer in Apostol, which gives the error estimate as $\frac{\sqrt{2}}{7680} < 2\times 10^{-4}$.

I'm clearly missing the point somewhere so any pointers in the right direction would be appreciated.

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  • $\begingroup$ $\[\int_{0}^{{\scriptstyle{}^{\sqrt{2}}\!\!\diagup\!\!{}_{2}\;}}{\left| r(x) \right|}dx\le \left[ \frac{{{\left( \tfrac{1}{2} \right)}^{5}}}{5!}x \right]_{0}^{{\scriptstyle{}^{\sqrt{2}}\!\!\diagup\!\!{}_{2}\;}}=\frac{\sqrt{2}}{7680}\]$ $\endgroup$ – Paul Sep 30 '14 at 8:47
  • $\begingroup$ Sorry about the scruffy TeX - you get the gist I'm sure! $\endgroup$ – Paul Sep 30 '14 at 8:48
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Write $\sin x = x - x^3/3! + E(x)$, where $E(x)$ is the error and you have shown that $\lvert E(x) \rvert \leq 1/2^5 5!$ for $0 \leq x \leq \frac{1}{2}$. Equivalently, $\sin x^2 = x^2 - x^6/3! + E(x^2)$, where $\lvert E(x^2) \rvert \leq 1/2^5 5!$ for $0 \leq x \leq 1/\sqrt 2$.

Then $$\int_0^{1/\sqrt{2}} \sin x^2 dx = \int_0^{1/\sqrt 2} [x^2 - x^6/3!] dx + \int_0^{1/\sqrt 2} E(x^2) dx.$$

You know how to evaluate the first integral. The error comes from the second interval. A good start is to simply try to bound the error, which leads us to try to understand

$$ \left \lvert \int_0^{1/\sqrt 2} E(x^2) dx \right \rvert \leq \int_0^{1/\sqrt 2} \lvert E(x^2) \rvert dx \leq \int_0^{1/\sqrt 2} \frac{1}{2^5 5!} dx = \frac{1}{2^{5.5} 5!} = \frac{\sqrt 2}{2^6 5!}.$$

So we can conclude that the error is bounded by $\sqrt 2 / 2^6 5! = \sqrt 2 / 7880$. $\diamondsuit$

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