1
$\begingroup$

My textbook gives me the error term for the Composite Trapezoidal Rule as this:

$-\frac{b-a}{12}h^2f''(\mu)$, where $\mu \in(a,b)$ and $f \in C^2 [a,b]$

I am using MatLab to produce approximations with the Composite Trapezoidal Rule for $\int_0^{0.99} \frac{1}{\sqrt{1-x^2}}{\rm d}x$ with the intervals $h = 0.01, 0.005, 0.0025, 0.00125, 0.000625$

Below is my table of the approximations by my code and the absolute error for each interval:

 ....h............S(h)...........abs. err...
 0.010000    1.432052842622   0.002795989152
 0.005000    1.429980957924   0.000724104453
 0.002500    1.429439827337   0.000182973867
 0.001250    1.429302728001   0.000045874530
 0.000625    1.429268330467   0.000011476997

Evaluating the error with the error formula, however, gives me a very different number than what my code is spitting out. For example, evaluating the error term for $h = 0.01, a = 0, b = 0.99$, I end up with $0.437161725$. Should my approximation of the error be that off? Am I not using the error term properly?

$\endgroup$
  • $\begingroup$ How did you compute the error? If you go into the details, the actual error is more $\frac{h^2}{12}\int_a^b|f''(s)|ds$ than $\frac{h^2}{12}(b-a)\max_{s\in[a,b]}|f(s)|$ which makes a difference when there are large changes in magnitude as in this example. $\endgroup$ – LutzL May 18 '18 at 5:30
  • $\begingroup$ I computed the error using the latter method. $\endgroup$ – user-2147482428 May 18 '18 at 5:39
  • $\begingroup$ How do you do the trapezoidal summation when the interval length is not a multiple of $h$? Do you cut the last interval short? $\endgroup$ – LutzL May 18 '18 at 6:18
1
$\begingroup$

You should be careful with this expression:

$$ {\rm err} = -\frac{b-a}{12}h^2f''(\mu) \tag{1} $$

The meaning is: there is a point $\mu \in (a,b)$ such that the error is given by this expression. To show this is true I calculate $S(h)$ for various values of $h$ and the absolute error $\epsilon$. I then find the value of $\mu$ guaranteed by Eq. (1), that is, the value of $\mu$ such that ${\rm err} = \epsilon$

enter image description here

$\endgroup$
  • $\begingroup$ Thank you. I guess I did misunderstand the purpose of the error term after all. $\endgroup$ – user-2147482428 May 18 '18 at 5:52
1
$\begingroup$

Note that the error is in the derivation the sum over the error of all the sub-intervals. For each of these intervals you get $$ \int_{x_k}^{x_{k+1}}f(s)\,ds-(x_{k+1}-x_k)\frac{f(x_{k+1})+f(x_k)}2=\frac{(x_{k+1}-x_k)^3}{12}f''(μ_k) $$ With constant length of the sub-intervals, $x_{k+1}-x_k=h$, the error is thus a Riemann sum for $$ \frac{h^2}{12}\int_a^bf''(s)\,ds=\frac{h^2}{12}(f'(b)-f'(a)) $$ For $f(x)=(1-x^2)^{-1/2}$ we get $f'(x)=x(1-x^2)^{-3/2}$ so that with $[a,b]=[0, 0.99]$ the constant in the error formula is $C=(f'(b)-f'(a))/12=29.38829..$. The experimental table extended to include estimates of this second order error constant is \begin{array}{r|lllllccc} n&h&S(h)&E(h)=S(h)-I&E(h)/h^2&E(h)/h^2-C\\\hline 100 &0.0099 &1.4319994706 & 0.00274261713177 &27.983033688 &-1.40525747436 \\ 200 &0.00495 &1.42996674039 & 0.000709886919313 &28.9720199699 & -0.416271192477 \\ 400 &0.002475 &1.42943619958 & 0.000179346106281 &29.2780093918 & -0.110281770601 \\ 800 &0.0012375 &1.42930181596 & 4.49624937289e{-}05 &29.3602652653 & -0.0280258971486 \\ 1600 &0.00061875 &1.42926810213 & 1.1248659324e{-}05 &29.3812548408 & -0.00703632162522 \\ 3200 &0.000309375 &1.42925966614 & 2.81266974667e{-}06 &29.386530156 & -0.00176100644524 \\ \end{array} which shows that indeed the error behaves like $E(h)=29.38829\cdot h^2+O(h^4)$.

$\endgroup$

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.