# Trapezodial Rule Error Proof (taylor)

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.

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.

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