Use the Trapezoidal Rule in order to approximate $\displaystyle \int_0^1 \frac{\sin(x)}{x}~ dx$, with error bounded by $10^{-4}$.

I've tried to bound $|f''(x)|$ at $[0,1$]...I want to prove it's monotonic at $[0,1]$ but I don't know how I can prove it easily...



The Trapezoidal Rule is given by:

$$\int_a^b f(x) ~ dx = \dfrac{b-a}{2n}(f(x_0) + 2f(x_1) + \ldots + 2f(x_{n-1}) + f(x_n))$$

The error term is given by:

$$|e_n| \le \dfrac{max_{a,b} |f''(x)|}{12 n^2} (b-a)^3$$

We have:

$$f(x) = \dfrac{\sin(x)}{x}, x \in (0,1)$$

We find the second derivative:

$$f''(x) = \frac{2 \sin (x)}{x^3}-\frac{2 \cos (x)}{x^2}-\frac{\sin (x)}{x}$$

A plot of $f(x), f'(x), f''(x)$ shows:

enter image description here

We need to find the maximum of $f''(x), x \in (0,1)$, which yields $|-0.34375|$ at $x = 0$.

To find the number of iterations, we find $n$ from the error bound, thus:

$$|e_n| \le \dfrac{max_{a,b} |~f''(x)|~}{12 n^2} (b-a)^3 = \dfrac{|-0.34375|}{12 n^2} (1-0)^3\le 10^{-4} \implies n \ge 16.9251 $$

So, we choose $n = 17$.

Doing $17-$steps of the Trapezoidal Rule yields:

$$\int_0^1 \dfrac{\sin(x)}{x} ~dx \approx 0.9459962252$$

Using WA, we get the value as:

$$ \int_0^1 \dfrac{\sin(x)}{x} ~dx \approx 0.9460830704$$

So, our error estimate produces:

$$ \Delta =0.9460830704 - 0.9459962252 = 0.000086845$$

This satisfies our requirement of less than $10^{-4}$ error.

Aside: Sometimes using the second derivative as an error bound can have issues, see Deriving the Trapezoidal Rule Error. Sometimes it is better to just take a much more pessimistic value for the max when using second derivative estimates, for example, $1$ in this problem and just avoid the second derivative altogether.

  • $\begingroup$ Very nicely done, Amzoti! $\endgroup$
    – amWhy
    Mar 30 '14 at 12:23

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.