I am trying to derive Truncation Error $=\frac{(b-a)h^2}{12} \max{|f''(z)|}$ for Trapezoidal rule of integration by using Taylor's Remainder term
My Approach:
In trapezoidal rule of integration, we divide the interval of integration $[a,b]$ into $n$ equal length sub-intervals and then approximate the function as first degree polynomial to connect the intermediate points.
Let $f(x)$ be the given function, then the Taylor series of $f(x)$ is given by:$$f(x)=f(\alpha)+ (x-\alpha)f'(\alpha) + \frac{(x-\alpha)^2}{2!}f''(\alpha) + \frac{(x-\alpha)^3}{3!}f''(\alpha) + \dots$$
We know:$$\text{Taylor's Remainder Term}: R_{n+1}(x;c)=\frac{(x-c)^{n+1}}{(n+1)!}f^{n+1}(z)$$
where $z$ is a point between $c$ and $x$
So, here
$(n+1)=2$ , $x \in [a,b]$, $c=(x-h)$and $z\in(x-h,x)$ ,where $h= \frac{b-a}{n}$
$$\implies R_2(x,x-h)=\frac{(x-(x-h))^{2}}{(2)!}f''(z)$$
$$=\frac{(h)^{2}}{2}f''(z) \dots (i)$$
so, the largest possible error $=\max|R_2(x,x-h)|$
$$=\frac{(h)^{2}}{2} \max|f''(z)| \dots(ii)$$
But Truncation Error $=\frac{(b-a)h^2}{12} \max{|f''(z)|}$,
so did i missed anything?? please help anyone…
Best Answer
You need to first obtain a local approximation over a subinterval.
The error in approximating the integral on a subinterval $[x_n, x_{n+1}]$ with $h = x_{n+1} - x_n$ using the trapezoidal formula is
$$E_n = \frac{h}{2}[f(x_n) + f(x_{n+1})] - \int_{x_n}^{x_{n+1}} f(x) \, dx.$$
Integration by parts shows this to be
$$E_n = \int_{x_n}^{x_{n+1}}(x-c)f'(x) \, dx,$$
where $c = (x_{n+1}+x_n)/2$ is the midpoint.
To see this, note that
$$x_{n+1} - c = c - x_n = \frac{x_{n+1} - x_n}{2} = \frac{h}{2},$$
and with $u = (x-c)$ and $dv = f'(x)dx$, integration by parts yields
$$\int_{x_n}^{x_{n+1}}(x-c)f'(x) \, dx = (x-c)f(x)|_{x_n}^{x_{n+1}} - \int_{x_n}^{x_{n+1}} f(x) \, dx = E_n.$$
If the second derivative is bounded with $\max |f''(x)| = M$, then we can demonstrate $O(h^3)$ local accuracy. Another integration by parts yields
$$E_n = \frac1{2} \int_{x_n}^{x_{n+1}} [(h/2)^2 - (x-c)^2]f''(x) \, dx.$$
Using the bound for $f''$ and integrating we obtain the local error bound
$$|E_n| \leqslant \frac{M}{12}h^3$$
Summing over $m$ subintervals where $mh = b-a$ we get the global error bound
$$ |GE| \leqslant \frac{Mmh^3}{12}= \frac{(b-a)h^2}{12}M$$