# Proof by using Taylor's Remainder Term: Truncation Error $=\frac{(b-a)h^2}{12} \max{|f''(z)|}$ for Trapezoidal rule of integration

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 $$=\frac{(b-a)h^2}{12} \max{|f''(z)|}$$,

• Where is your $\int$? You are estimating the truncation error of an integral yet no integral appears here. – xbh Dec 15 '18 at 5:32
• Using the Taylor's approximation is messy. An easier way is to apply integration by parts to estimate the local error $E_n = \frac{h}{2}[f(x_n) + f(x_{n+1})] - \int_{x_n}^{x_{n+1}} f(x) \, dx$ – RRL Dec 15 '18 at 5:38

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

$$|GE| \leqslant \frac{Mmh^3}{12}= \frac{(b-a)h^2}{12}M$$

• @Suresh: You're welcome. You can start by looking at this and I'll see if I can find a better reference or derive it for you. – RRL Dec 15 '18 at 6:32
• I do however think this is the "best" approach and is worth learning. – RRL Dec 15 '18 at 6:33
• That is correct. Now you can write the first term as $-cf(x)|_{x_n}^{x_{n+1}} = -c\int_{x_n}^{x_{n+1}} f'(x) \, dx$ and you will get $E_n=\int_{x_n}^{x_{n+1}} (x-c)f'(x) \, dx$. – RRL Dec 15 '18 at 13:14
• +1: Not a criticism of the above, but I would like to see an error analysis that makes the error 'obvious'. The integration by parts twice certainly yields the desired result, but personally I find the (textbook) derivation a bit opaque in terms of intuition. – copper.hat Dec 15 '18 at 20:23
• @copper.hat: Thank you. I agree. I had a more transparent way in mind showing $\mathcal{O}(h^3)$ local error but could get the coefficient $1/12$. Not sure if that is important to OP. – RRL Dec 15 '18 at 23:31