1. MATLAB FUNDAMENTALS: INTRO TO LINEAR ALGEBRA NUMERICAL METHODS HP 101 – MATLAB Wednesday, 11/5/2014 www.clarkson.edu/class/honorsmatlab

2. Quote and Video of The Week  “It’s a great party house guys… that’s why we bought it.” Jon Goss  https://www.youtube.com/watch?v=CwhvJ5B4lYg https://www.youtube.com/watch?v=CwhvJ5B4lYg  https://www.youtube.com/watch?v=NBbHRaNNBuY https://www.youtube.com/watch?v=NBbHRaNNBuY

3. Linear (Matrix) Algebra Intro  Math with matrices is a bit different from regular math…  Addition and subtraction are the same.  Multiplication (“Row Column Rule”)  Syntax: A*B(No dot!)  Quick example…  Division and Powers… well, they don’t really exist in the classical sense. MATLAB can compute them, but we won’t worry about what they mean.

4. Dot Product  Dot Product is amount of vector A that lies in plane with vector B  Essentially, the sum of the results when you multiply two vectors together element by element  AB = ABcos( θ) A B A B = 0 B A B = 2 1 2 1 2

5. Dot Product Example  a = [ 1 2 3];  b = [ 4 5 6];  y = a.*b  y = [ 4 10 18]  dot_product = sum(y) = 32  d_p = dot(a,b) = 32  Vectors must be same length – rows or columns

6. Cross Product  A measure of the orthogonality “perpendicular- ness” of two vectors  Opposite of a dot product  A x B = ABsin( θ) A B A x B = 2 B A x B = 0 1 2 1 2 a = [ 1 2 0 ]; b = [ 3 4 0 ]; c = cross(a,b); c = [ 0 0 -2 ] Cross products are used a lot in statics, dynamics, and fluids

7. Systems of Linear Equations  Linear equations are big, nasty, and algebra can be really tricky  Sometimes, you can have tons of equations but actually solving them all by hand would be painful  Examples:  Complex Circuits  Simple Traffic Flow (or other flows)  “Who bought how much candy?”

8.  Suppose: 3x + 2y – z = 10 -x + 3y + 2z = 5 x - y - z = -1  We can say…. A*b = c A = b = c = 10 5 Solving Systems Linear Equations 3 2 -1 -1 3 2 1 -1 -1 xyzxyz Sizes: 3 X 3 3 X 1 3 X 1

9. Example Continued  A = [3 2 -1; -1 3 2; 1 -1 -1];  c = [10; 5; -1];  b = A^(-1) * c; or b=inv(A)*c  Alternatively, can use Gaussian Elimination (preferred) :  b = A\c  You can also view the solution in matrix form:  b = rref[A,c]  Notes:  No dot before the multiplier  Matrix inner dimensions must agree  Gaussian Elimination is more efficient than inv function

10. Matrix Functions  After you have a Dif EQ or a Linear Algebra course, you can use MATLAB to:  Multiply or divide matrices together a*b a/b  Raise a matrix to a power a^2  Find determinants det( )  Create identity Matrices eye( )  Special matrices such as: pascal ( )magic ( )rosser ( )  help(gallery) for list of all special matrices

11. Numerical Methods

12. So What are Numerical Methods?  A tool for efficiently solving complex mathematical problems.  Discrete methods which repeat calculations until the correct answer is obtained.  Very complex… we will only touch the surface. For more, you should consider taking the Numerical Methods course in the math department.  Really, really cool!

13. Discretization  Numerical methods approximate solutions by treating continuous functions as a series of points.  Fortunately, this is how MATLAB works!  Your “step” (often a time-step) is very important. The smaller it is, the more accurate your results will be but the slower your program will run.

14. Iterative Methods  Numerical methods also require repeating calculations until the correct answer is found.  This can be accomplished in a few easy ways…  Loops which use a value or values from a previous run through the loop.  Loops which call a function and use one or more of its outputs as its inputs in the next loop.  Functions which recursively call themselves, giving information to the next level and passing information back down. These can be hard to conceptualize.

15. Why are Numerical Methods Useful?  First and foremost, they are very fast…  You may have noticed that using symbolics is pretty slow in MATLAB. If you have a large number of calculations to do, this can add up quickly.  Using numerical methods to solve equations, or to find say, the roots of an equation is much faster.  Some equations simply can’t be solved explicitly…  eg:

16. Numerical Integration  If we think of an integral as the area beneath a curve, we can approximate it in a few ways…  By taking two discrete points, we can calculate the area of the trapezoid formed by them and the axis.  Then we can add up all the trapezoids to get the total area.  If the distance between our points is small enough, the solution is almost exact!

17. Numerical Integration  That sounds like a bit of a pain… we could do it, but is there an easier way?  Yes! Because the MATLAB developers love you so much, they included the trapz function!  Note that trapz assumes a spacing of 1 between your elements. So we need to multiply the result by the spacing… Lets try some:  x=0:1:3; y=x.^2; 1*trapz(y)ans = 9.5  x=0:0.1:3; y=x.^2; 0.1*trapz(y)ans = 9.05  x=0:0.01:3; y=x.^2; 0.01*trapz(y)ans = 9.005

18. Numerical Differentiation  Differentiation gives the instantaneous slope of a function at a given point. We can come very close to that exact value by looking at a very small piece of the curve and treating it as a straight line…  This straight line (a secant) approaches the tangent line as the distance between the points gets small.

19. Numerical Differentiation  The formula for numerical differentiation is…  Where f(x) is some function and h is the step.  Lets find the derivative of sine at π/3:  y1=sin(pi/3); y2=sin(pi/3+0.01);  y_prime=(y2-y1)/0.01  ans = 0.4957(The exact answer is.5)  Using a step of 0.001 instead gives 0.4996….

20. Newton’s Method  Newton’s Method is useful to find a root or roots of an equation.  It requires knowledge of the function, the derivative of the function (numerical differentiation can come in handy here) and an initial guess of the answer.  It can be a bit picky about the initial guess… the iteration can get stuck “bouncing” or it can converge to a root which is not desired.  Good judgment is required on a case-by-case basis to determine if the method is appropriate and what the initial guess should be.

21. Newton’s Method  Newton’s Method uses the derivative of at an initial guess function to evaluate a new guess for the root of the function until the root is found.  We iterate the above equation until the following is satisfied:  Epsilon is referred to as the error bound. It should be determined by the function you are solving.

22. Example of Newton’s Method  How calculators do division:  We want to compute x=1/d without using a division operator… lets try Newton’s Method!  First, we need to rewrite our equation in a manner that the solution is zero when we find d:  Next, we need the derivative:  Now we can plug these into the Newton’s Method formula:

23. Continuing the Example…  Please go to the course website and download the skeleton code for the example.  We’ll go through it together. After class a detailed and highly commented version will be posted on the website with the skeleton code.