0
$\begingroup$

I search for a proof of the (local) error of trapezodial rule using taylor series. I can only find proofs for the error of the rectangle rule and for trapezodial it's always just "similar" whatever this means... I tried to start like this:

$I_{Ti} - err = I_i \quad I_i = \int\limits_{x_i}^{x_{i+1}} f(x)dx$

$I_{Ti}=\frac{f(x_i)+f(x_{i+1})}{2}(x_{i+1}-x_i)$

taylor series of f(x) at $x_{i+1/2} (= \frac{x_i + x_{i+1}}{2})$

$f(x)=f(x_{i+1/2})+f'(x_{i+1/2})(x-x_{i+1/2})+f''(x_{i+1/2})(x-x_{i+1/2})^2 + O((x-x_{i+1/2})^3)$

$\int\limits_{x_i}^{x_{i+1}}f(x) dx = f(x_{i+1/2})(x_{i+1}-x_i)+\frac{1}{24}f''(x_{i+1/2})(x_{i+1}-x_i)^3+O((x_{i+1}-x_i)^5)$

and I am already stuck here.

$\endgroup$
0
$\begingroup$

For setup, you have a function $f$ which is twice continuously differentiable on $[a,b]$, and you want to estimate the error in estimating $I(f)=\int_a^b f(x) dx$ by $T(f)=\frac{f(b)+f(a)}{2} (b-a)$.

It is convenient to think about this as exactly integrating the function $L(x)=f(a)+\frac{f(b)-f(a)}{b-a}(x-a)$ over $[a,b]$. This is just the linear function which passes through $(a,f(a))$ and $(b,f(b))$.

So the idea is to show that $f(x)-L(x)$ is small. One thing to notice is that it vanishes at $a$. Thus it is productive to Taylor expand about $a$. You have

$$f(x)-L(x)=(f'(a)-L'(a))(x-a)+\frac{f''(\xi_x)(x-a)^2}{2}.$$

Here $\xi_x$ is some element of $(a,b)$ that we don't know. So if $|f''|$ is bounded by $M$, we get the error estimate

$$|I(f)-T(f)| \leq |f'(a)-L'(a)| \int_a^b (x-a) dx + M \int_a^b (x-a)^2 dx.$$

The two integrals can be done exactly. It then remains to estimate $|f'(a)-L'(a)|$. This can again be done by Taylor expansion:

$$L'(a)=\frac{f(b)-f(a)}{b-a}=\frac{f(a)+f'(a)(b-a)+\frac{f''(\xi_b)(b-a)^2}{2}-f(a)}{b-a}=f'(a)+\frac{f''(\xi_b)(b-a)}{2}.$$

So you're left with

$$|I(f)-T(f)| \leq \frac{M(b-a)}{2} \int_a^b (x-a) dx + M \int_a^b (x-a)^2 dx.$$

You recover a more familiar estimate by actually computing these two integrals. You can get a slightly sharper estimate if you also expand about $b$ and balance the two (for example, by using the estimate at $a$ on the left half of the interval and the estimate at $b$ on the right half of the interval).

My estimate here is actually quite suboptimal. The optimal estimate can be proven by using the interesting fact:

$$T(f)-I(f)=\int_a^b (x-c) f'(x) dx$$

where $c=\frac{a+b}{2}$. Knowing that this is true, it can be checked with integration by parts; deriving it is probably more difficult.

$\endgroup$
  • $\begingroup$ got it on my own just now (more or less). thanks for the complete and "clean" answer. $\endgroup$ – southpole Aug 13 '15 at 21:27

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.