Newton-Cotes and other mistakes,
Soroosh Yazdani
I appologize once again for being a bit confused today regarding
Newton-Cotes method. Here is the answer to the questions that I
was unable to answer today.
First, let's ignore the composite methods, that is, whatever method
we are going to use, we will only do it in one step.
That means, for example, the trapazoid approximation to int(f(x),a..b)
is given by (f(a)+f(b))*(b-a)/2. Similarly, the Simpsons approximation
to int(f(x),a..b) is given by (f(a)+f(b)+4*f((a+b)/2))*(b-a)/3.
Now, recall that trapazoid approximation formula comes up by
approximation the function f(x) by a line that goes through (a,f(a))
and (b,f(b)). Similarly, the Simpsons approximation is given
by integrating a quadratic polynomial approximating f(x) at
(a,f(a)), ((a+b)/2,f((a+b)/2)), and (b,f(b)).
Realizing this, one can try to generalize this method. That is, we
can interpolate the function f(x) at a=x_0,x_1,...,x_n=b by a
polynomial of degree n, and integrate the interpolating polynomial
instead. Again, as usual, it is easiest to have x_i=x_0+i*h, and
h=(b-a)/n. This technique is called (closed) Newton-Cotes method.
The above can be summarized as, Newton-Cotes is using Lagrange
interpolation to approximate integrals.
Now, the difference between open Newton-Cotes and closed Newton-Cotes
is fairly minor: In the closed Newton-Cotes technique you do include
the endpoints, while in the open Newton-Cotes you don't include the
endpoints. (BTW, the graph in figure 4.6 is incorrect, since it is
referring to the open Newton-Cotes.) The open Newton-Cotes is useful,
since, for example, sometimes we can't evaluate the function at the
endpoints (consider int(1/sqrt(x),x=0..1).
Finally, regarding the even and odd question, one can verify that
any version of Newton-Cotes will give you an error function that is
odd. What I mean, is if we replace a and b, then the error is replaced
by its negative. This is obvious, since the interpolating polynomial stays
the same, and the only thing that changes is the limits of integration
which gives you a negative sign.
(error=int(f(x)-p(x),x=a..b)=-int(f(x)-p(x),x=b..a)
Therefore, if the error is C*(b-a)^alpha, for some constant C dependant
on f, then alpha better be odd. Now, using this fact, one can argue that
Newton-Cotes method will always have even precision, which is why
the errors in theorem 4.2 and 4.3 are divided into two cases depending
on n even or odd.
Soroosh