User:Egm6341.s11.team4.BM/HW4

=Using Simple Simpson's Rule to Integrate a 3rd Order Polynomial Exactly=

Problem Given on Lecture Slide 18-2

Given
Given an arbitrary 3rd degree polynomial
 * $$ \displaystyle f(x)=P_{3}(x)=c_{0}+c_{1}x^{1}+c_{2}x^{2}+c_{3}x^{3} $$

which can be expressed in sigma notation as
 * $$\displaystyle \sum_{i=0}^{3} c_{i}x^{i} $$

examine the interval
 * $$ \displaystyle [a,b]=[-2,1] $$

With the following coefficients
 * $$\displaystyle c_{0}=1, c_{1}=3, c_{2}=-9, c_{3}=12 $$

Objective
1. Evaluate the integral using simple Simpson's Rule

2. Evaluate the integral exactly

3. Compare the integral calculations to show that simple Simpson's rule exactly integrated the 3rd degree polynomial.

Solution
First, we will evaluate the integral exactly. Since this is a simple polynomial, the integral evaluation can be performed analytically with ease.


 * $$ I(f)=\displaystyle \int_{a}^{b} 12x^{3}-9x^{2}+3x+1 dx $$
 * $$ I(f)=[3x^{4}-3x^{3}+\frac{3}{2}x^{2}+x]_{a}^{b} $$
 * $$ I(f)=[3(1)^{4}-3(1)^{3}+\frac{3}{2}(1)^{2}+(1)]-[3(-2)^{4}-3(-2)^{3}+\frac{3}{2}(-2)^{2}+(-2)] $$
 * $$ I(f)=[3-3+\frac{3}{2}+1]-[48+24+\frac{3}{2}4-2] $$
 * $$ I(f)=\displaystyle [2.5]-[76] $$
 * $$ I(f)= \displaystyle -73.5 $$

So, if simple Simpson's rule exactly integrates a 3rd degree polynomial as expected we expect that it will produce a value of -73.5 through quadrature.

Simple Simpson's is expressed below


 * $$ I_{n}(f)=\frac{b-a}{6}[f(a)+4f(\frac{a+b}{2})+f(b)] $$

Substituting in for a and b, we get
 * $$ I_{n}(f)=\frac{1+2}{6}[f(-2)+4f(\frac{-2+1}{2})+f(1)] $$

Evaluating our function at the given points, we have
 * $$ I_{n}(f)=\frac{1}{2}[-137+4(-4.25)+7] $$
 * $$ I_{n}(f)=\frac{1}{2}[-137-17+7] $$
 * $$ I_{n}(f)=\frac{-147}{2} $$

Now we have a final answer through quadrature, comparing it to our exact integral, we see
 * $$ I_{n}(f)=I(f)=\frac{-147}{2}=-73.5 $$

Thus simple Simpson's rule, though it approximates the function as a simple 2nd order parabola, can exactly integrate a polynomial of degree 3. Though the quadrature rule does not exactly approximate the function it exactly integrates the function through cancellation of areas. This principle is shown in the graph below.



Simpson's rule underestimates the area under the curve initially and then overestimates the curve. These errors in the approximation exactly cancel each other out for a 3rd order polynomial and the simple Simpson's rule exactly integrates the function.

Given

 * $$ e^{(1)}(t)= [F(-t)+F(t)]-\frac{1}{3}[F(-t)+4F(0)+F(t)]-\frac{t}{3}[-F^{(1)}(-t)+F^{(1)}(t)] $$

Objective
Show that
 * $$ e^{(3)}(t)=-\frac{t}{3}[-F^{(3)}(-t)+F^{(3)}(t)] $$

Solution
Starting with our given equation for $$ e^{(1)} $$ which was given in class on lecture slide 20-2 and is proven in homework problem 4.2.


 * $$ e^{(1)}(t)= [F(-t)+F(t)]-\frac{1}{3}[F(-t)+4F(0)+F(t)]-\frac{t}{3}[-F^{(1)}(-t)+F^{(1)}(t)] $$

We take the derivative with respect to t and obtain.


 * $$ e^{(2)}(t)=[-F^{(1)}(-t)+F^{(1)}(t)]   -     \frac{1}{3}[-F^{(1)}(-t)+F^{(1)}(t)]    -    \frac{1}{3}[-F^{(1)}(-t)+F^{(1)}(t)] -\frac{t}{3}[F^{(2)}(-t)+F^{(2)}(t)]] $$

This function can be simplified to


 * $$ e^{(2)}(t)=\frac{1}{3}[-F^{(1)}(-t)+F^{(1)}(t)] - \frac{t}{3}[F^{(2)}(-t)+F^{(2)}(t)] $$

Now, if we take the derivative again, we obtain


 * $$ e^{(3)}(t)=\frac{1}{3}[F^{(2)}(-t)+F^{(2)}(t)] -\frac{1}{3}[F^{(2)}(-t)+F^{(2)}(t)] -\frac{t}{3}[-F^{(3)}(-t)+F^{(3)}(t)] $$

Simplifying again, we obtain


 * {| style="width:100%" border="0"

$$ \displaystyle
 * style="width:92%; padding:10px; border:2px solid #8888aa" |
 * style="width:92%; padding:10px; border:2px solid #8888aa" |

e^{(3)}(t)= -\frac{t}{3}[-F^{(3)}(-t)+F^{(3)}(t)]

$$
 * 
 * }

Which is the desired equation.