0. ES 314 Nov 27, 2012 Numerical Analysis
Source: Maurice Herlihy, Brown Univ.

1. topics covered Numerical Analysis Examples - applications
Solving equations
Finding min/max
Integrating
linear systems: eigenvalue, eigenvector computation
Examples - applications

2. Why Numerical Methods are Important
Exact answers are sometimes hard or impossible to achieve in practice
Numerical methods provide an approximation that is generally good enough
Useful in all fields of engineering and physical sciences, but growing in utility in the life sciences and the arts
Movement of planets, stars, and galaxies
Investment portfolio management by hedge funds
Quantitative psychology
Simulations of living cells
Airline ticket pricing, crew scheduling, fuel planning
Figure is a Babylonian clay tablet showing an approximation of 21/2.

3. Equations in One Variable
General form f (x ) = 0
Approximation may be good enough
Iterative approach with the bisection method
Solution where function changes sign and crosses x-axis
Functions can have none, one, several, or an infinite number of solutions.
Can be solved exactly (algebraically) or numerically.

4. Equations in One Variable (cont.)
x = fzero(function, x0)
function: function to be solved x0: initial guess at solution x: the solution
Function can be specified as
a mathematical expression as a string with no predefined variables
function handle
name of an anonymous function
Must be in the form f (x ) = 0
f (x ) = c  f (x ) - c = 0
Initial guess: plot the function first. Sometimes the domain can be estimated.
Some solutions may not have a physical meaning.

5. Solving a Nonlinear Equation
Find the solution of
Rewrite as
Plot it to get estimate
% Initial plot for estimate
fplot('x*exp(-x)-0.2', [0 8])
grid on
pause
% Find the solution between 0 and 1, guess 0.3
x1 = fzero('x*exp(-x)-0.2', 0.3)
% Find the second solution between 2 and 3
% Define an anonymous function F
F x*exp(-x)-0.2
fzero(F, 2.5)
% What if there is no solution?
fzero('x^2+1', 1)
Plot: one solution between 0 and 1, another between 2 and 3
No solution case: NaN not a number
Other fzero options???
Roots: can use to solve functions in the form of polynomials

6. Newton's Method
Solve f (x ) = 0 by repeating the assignment until x is close enough to 0
function y = f(x) y = x^3 + x - 3;
function y = df(x) y = 3*x^2 + 1;
function [x f conv] = newtfun(fh, dfh, x0)
% newtfun solves f(x)=0 using Newton's Method
% Input arguments
% fh handle to f(x) % dfh handle to f'(x) % x0 initial guess
% Output arguments
% x root % f f(x) % conv convergence=1, 0 otherwise
format long e
steps = 0; x = x0; re = 1e-8;
myrel = 1;
while (myrel > re) & (steps < 20)
xold = x;
x = x - fh(x)/dfh(x);
steps = steps + 1;
disp([x fh(x)])
myrel = abs((x-xold)/x);
end
if myrel <= re
conv = 1;
else
conv = 0;
f = fh(x);
[r fr conv] 1)
Use 2])
set breakpoint, note workspace, myrel
clear bkpt
continue
set cond breakpoint step == 3

7. Finding Minima and Maxima
Local minima or maxima occur when the derivative of the function is zero
x = fminbnd(function, x1, x2)
function: function to be solved x1,x2: interval for minimum x: function minimum
Try 0 to 8
Note still local min
Try 0 to 1 and note problem with endpoints
% lect12_1: Minima and maxima
% (single step)
% Plot the function
F (x) x^3 - 12*x^ *x
fplot(F, [0 8])
grid on
% Find the minimum between 5 and 6
[x fval] = fminbnd(F, 5, 6)
% Change interval to 0 to 8
[x fval] = fminbnd(F, 0, 8)
% Change interval to 0 to 1
[x fval] = fminbnd(F, 0, 1)
% Find maxima by multiplying function by -1
% and finding minima
F2 (x) -F(x)
[x fval] = fminbnd(F2, 0, 8)
m-files: lect12_1

8. Example Maximum Viewing Angle
For best view in a theater, sit at distance x such that angle Θ is maximum
Find x with the configuration shown
Applying Law of Cosines

9. Maximum Viewing Angle Angle is between 0 and π/2
Cosine decreases from 1 at Θ = 0, thus maximum angle corresponds to minimum cos(Θ)
Do a quick plot of cos(Θ) as a function of x to estimate the solution range
Find the minimum with fminbnd
Step 1: notice minimum between 10 and 20

10. Numerical Integration
Find the area under the curve of a function
Examples
Area and volume
Velocity from acceleration
Work from force and displacement
Integrand can be a function or set of data points
In calculus, integrad is usually a function. In science and engineering, can be a function or data points
In numerical intergration, area is found by dividing up into small sections and summing them.
Various methods developed to do this.
Matlab has three ways

11. Integration in MATLAB
q = quad(function, a, b) q = quadl(function, a, b)
function: function to be solved a,b: integration limits q: value of integral
Function must be written for element-by-element operations
Function must be well-behaved between a and b (no vertical asymptote)
Difference is in method of integration

12. Integrating with Data Points
% lect12_2: Integration with data points
% Generate the data points
X = 0:pi/100:pi;
Y = sin(X);
plot(X, Y, 'o-’)
% Use trapz to find area under points
Q = trapz(X,Y)
% Integrate with quad
q = 0, pi)
q = trapz(x, y) x,y: vectors of data points q: value of integral
Example

13. Water Flow in a River
To estimate the amount of water that flows in a river during a year, consider the cross section shown.
The height h and speed v are measured on the first of each month.
Day 1 32 60 91
121
152
182
213
244
274
305
335
366
h (m)
2.0
2.1
2.3
2.4
3.0
2.9
2.7
2.6
2.5
2.2
v (m/s) 5
4.7
4.1
3.8
3.7
2.8

14. Water Flow in a River Flow rate (volume per second)
Total amount of water
60x60x24 converts days to seconds