lecture 10 (Starting Numerical Integration, Trapezoid Rule, and Simpson's Rule)

Analytic vs Numerical Method

In calculus, you do something like:
$f^{(n)}(x) \leftarrow f(x) \rightarrow \int_a^bf(x)dx$
In Numerical, we start with the Lagrange Interpolant, and do some other stuff:
$????? \leftarrow\textrm{Lagrange Interpolant: }f(x) = P(x) + error \rightarrow ????$
Actually, it’s pretty simple, we essentially just transform the function into a polynomial using the Lagrange Interpolant, and then we can differentiate that polynomial. It’s really easy to differentiate polynomials, so that’s a pretty solid method.
So for our differentiation, we simply have:
$$

f’(x)=P’(x) + error’ \leftarrow \textrm{Lagrange Interpolant: }f(x) = P(x) + error \rightarrow ???? $Where P is our Lagrange Polynomial. Integration, should be obvious, is just given by:$ f’(x)=P’(x) + error’ \leftarrow \textrm{Lagrange Interpolant: }f(x) = P(x) + error \rightarrow \int_a^bf(x)dx=\int_a^bP(x)dx+\int_a^b\textrm{error }dx $$

Numerical Integration

Today we’ll be looking at the (n+1) point formula again, but for integration. This is called the Newton’s-Cotes formula, which converges as a rate of \(O(h^{2n+1})\).

Suppose our goal is to compute
$\int_a^bf(x)dx$

Graphically, our goal is to calculate the area under the curve between a and b. If we look at this for n = 1, i.e. using only two points to approximate, you can see we get a trapezoid! This is known as the Trapezoid Rule.
The area of this trapezoid is simply \(\frac{f(a)+f(b)}{2}(b-a)\), so it’s pretty easy to calculate. Wowza. You can already do numerical integration. But it’s not very accurate. How inaccurate? Let’s find out!
Back to the Lagrange Interpolant:
$f(x) = f(x_0)L_{1,0} + f(x_1)L_{1,1}(x) + \frac{f''(\xi(x))}{2}(x-x_0)(x-x_1)$
Now we just have to integrate it from \(x_0\) to \(x_1\). To make that more obvious, let’s expand the Lagrange Interpolants
$f(x) = f(x_0)\frac{x-x_1}{x_0-x_1} + f(x_1)\frac{x-x_0}{x_1-x_0} + \frac{f''(\xi(x))}{2}(x-x_0)(x-x_1)$
Finally let’s replace \(x_1 - x_0\) with h
$f(x) = f(x_0)\frac{x-x_1}{-h} + f(x_1)\frac{x-x_0}{h} + \frac{f''(\xi(x))}{2}(x-x_0)(x-x_1)$
\(f(x_n)\) and \(h\) are both constants, so we can pull them out of the integral we’re about to do:
$\frac{-f(x_0)}{h}\int_{x_0}^{x_1}(x-x_1)dx+\frac{f(x_1)}{h}\int_{x_0}^{x_1}(x-x_0)dx + \frac{1}{2!}\int_{x_0}^{x_1}f''(\xi(x))(x-x_0)(x-x_1)dx$
If we solve that integral:
$\int_{x_0}^{x_1}f(x)dx = \frac{-f(x_0)}{h}\left[\frac{x^2}{2}-x_1x\right]_{x_0}^{x_1} + \frac{f(x_1)}{h}\left[\frac{x^2}{2}-x_0x\right]_{x_0}^{x_1} + \textrm{error}$
If we plug in the x’s, and then simplify, we get something like:
$\int_{x_0}^{x_1}f(x)dx = \frac{h}{2}f(x_0)+\frac{h}{2}f(x_1) + \textrm{error}$
If you’re having trouble doing that simplification, don’t forget that \(x_1-x_0 = h\). That might help.
Now we just have that pesky error term remaining.
$\frac{1}{2!}\int_{x_0}^{x_1}f''(\xi(x))(x-x_0)(x-x_1)dx$
The \(f’’\) term is definitely the most ugly part of that integral, so let’s just bring it out. We can do this because \(\xi\) is bound by the \([x_0,x_1]\) interval, so we can use some Mean Value Theorem magic to just bring it out: We’ll end up with:
$\frac{-h^3}{12}f''(c) \textrm{ where } c \in [x_0,x_1]$
as our error. Remember that was all for n = 1, aka using two points, aka the Trapezoid rule. If we want to get a little bit fancier, we can use Simpson’s rule, which is using n = 2, aka 3 points. Now, instead of having a straight line to cross the gap between \(a,f(a)\) and \(b,f(b)\), we’ll make a quadratic polynomial using the three points. You can solve for the error in Simpson’s Rule in the same way as we did for the Trapezoid Rule, but the integration is a lot uglier, because there’s so many points. It’s not really particularly difficult, and is a great exercise to do yourself, but, it is rather long. Anyway, here’s the final result of those mathematics. Note the error here converges at order 5 when using three points (n = 2), as we mentioned at the very beginning, these formulas converage at a rate of \(O(h^{2n+1})\).
$\int_{x_0}^{x_1}f(x)dx = \frac{h}{3}(f(x_0)+4f(x_1)+f(x_2))-\frac{h^5}{90}f^4(c)$

Example

$\int_0^2x^2dx=?$
using

1. Exact
2. Trapezoid
3. Simpson’s Exact is pretty simple, just integrate to get
   $\int_0^2x^2dx=\left[\frac{x^3}{3}\right]_0^2 = \frac{8}{3}$
   For Trapezoid, we just plug in to the trapezoid formula from above:
   $\int_0^2x^2dx \approx \frac{2}{2}(f(0)+f(2)) = 4$
   For Simpson’s Rule, we plug in again:
   $\int_0^2x^2dx \approx \frac{1}{3}(0 + 4 + 4) = \frac{8}{3}$
   The take-away? Simpson’s Rule is exact for a polynomial of degree three. Also, it’s generally more accurate than the trapezoid rule, without being terribly more difficult. These are all called Newton’s-Cotes methods. In Newton’s-Cotes, you can have open or closed.
   • Closed Newton’s-Cotes: Take endpoints into account
   • Open Newton’s-Cotes: No endpoints (aka midpoint)