MA/CS 375

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# MA/CS 375 - PowerPoint PPT Presentation

MA/CS 375. Fall 2002 Lecture 22. Interlude on Norms. We all know that the “size” of 2 is the same as the size of -2, namely: ||2|| = sqrt(2*2) = 2 ||-2|| = sqrt((-2)*(-2)) = 2 The ||..|| notation is known as the norm function. Norm of A Vector.

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## MA/CS 375

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### MA/CS 375

Fall 2002

Lecture 22

MA/CS 375 Fall 2002

Interlude on Norms
• We all know that the “size” of 2 is the same as the size of -2, namely:
• ||2|| = sqrt(2*2) = 2
• ||-2|| = sqrt((-2)*(-2)) = 2
• The ||..|| notation is known as the norm function.

MA/CS 375 Fall 2002

Norm of A Vector
• We can generalize the scalar norm to a vector norm, however now there are more choices for a vector :

MA/CS 375 Fall 2002

Norm of A Vector
• Matlab command:
• norm(x,1)
• norm(x,2)
• norm(x,inf)

MA/CS 375 Fall 2002

Norm of A Matrix
• We can generalize the vector norm to a matrix norm, however now there are more choices for a matrix :
• 1-norm maximum absolute column sum
• infinity-norm, maximum absolute row sum

MA/CS 375 Fall 2002

Norm of A Matrix
• Matlab command:
• norm(A,1)
• norm(A,2)
• norm(A,inf)
• norm(A,’fro’)

MA/CS 375 Fall 2002

Matrix Norm in Action
• Theorem 5: (page 185 van Loan)
• If is the stored version of then where and

i.e. an error of order ||A||1eps occurs when

a real matrix is stored in floating point.

MA/CS 375 Fall 2002

Condition Number
• Recall that when we asked Matlab to invert a matrix which was almost singular: we got this warning

MA/CS 375 Fall 2002

Definition of Condition Number
• rcond = 1/(condition number)
• We are going to use the 1-norm definition of the condition number given as:

MA/CS 375 Fall 2002

What’s With The Condition Number ?
• Theorem 6: (page 235 van Loan)
• If is non-singular and: In addition if then the stored linear system: is nonsingular and also:

MA/CS 375 Fall 2002

In Plain English
• The theorem supposes that we can solve the Ax=b problem without making any mistakes except for the stored approximation of A and b then:
• If the condition number is small enough, then the difference between the exact answer and the calculated answer will be bounded above by: const*condition#*eps
• Matlab value of eps is approximately 1e-16
• So if A has a condition number of 1 then we can expect to solve Ax=b to 16 decimal places at best
• If A has a condition number of 1000 then in the worst case we could make an error of 1e-13
• If A has a condition number of 10^16 then we could make O(1) errors

MA/CS 375 Fall 2002

cond in Matlab
• cond(A,1) will return the 1-norm condition number
• cond(A,2) will return the 2-norm condition number

MA/CS 375 Fall 2002

Team Exercise
• Build
• For delta=1,0.1,0.01,…,1e-16 compute:
• cond(A,1)
• Plot a loglog graph of the
• x=delta, y=condition number
• Figure out what is going on

MA/CS 375 Fall 2002

Team Exercise (theory)

MA/CS 375 Fall 2002

Team Exercise (theory)

MA/CS 375 Fall 2002

Team Exercise (theory)

Pretty big huh 

MA/CS 375 Fall 2002

Team Exercise (theory)

Pretty big huh 

Remark: remember how when we set delta = 2^(-11) and we could actually get the exact inverse. Well in this case the condition number is a little pessimistic as a guide !.

MA/CS 375 Fall 2002

Enough Theory !

MA/CS 375 Fall 2002

Interpolation
• Question:
• someone gives you a function evaluated at 10 points, how does the the function behave between those 10 points?
• guesses

MA/CS 375 Fall 2002

Interpolation
• Question:
• someone gives you a function evaluated at 10 points, how does the the function behave between those 10 points?
• there is no unique answer,but we can make a good guess

MA/CS 375 Fall 2002

Polynomial Interpolation
• What’s the highest order unique polynomial that you can fit through a function evaluated at 1 point?
• That would be a constant function with the same value as the given function value.

MA/CS 375 Fall 2002

Polynomial Interpolation
• What’s the highest order unique polynomial that you can fit through a function evaluated at 2 points?
• That would be a linear function that passes through the two values:

MA/CS 375 Fall 2002

LinearInterpolation

MA/CS 375 Fall 2002

LinearInterpolation

exp(x)

Samples

Linear fit

MA/CS 375 Fall 2002

General Monomial Interpolation
• Given N function values at N distinct points then there is one unique polynomial which passes through these N points and has order (N-1)
• Guess what – we can figure out the coefficients of this polynomial by solving a system of N equations 

MA/CS 375 Fall 2002

2nd Order Polynomial Fit
• we are going to use a 2nd order approximation
• let’s make sure that the approximation agrees with the sample at:
• x1,x2 and x3

MA/CS 375 Fall 2002

2nd Order Polynomial Fit
• We know:
• x1,x2 and x3
• f(x1),f(x2) and f(x3)
• We do not know:
• a1,a2 and a3

MA/CS 375 Fall 2002

2nd Order Polynomial Fit
• We know:
• x1,x2 and x3
• f(x1),f(x2) and f(x3)
• We do not know:
• a1,a2 and a3

re-written as system

MA/CS 375 Fall 2002

Finally A Reason To Solve A System

So we can build the

following then: a = V\f

MA/CS 375 Fall 2002

Expansion Coefficients
• Once we have the coefficients a1,a2,a3 then we are able to evaluate the quadratic interpolating polynomial anywhere.

MA/CS 375 Fall 2002

Class Exercise
• Part 1:
• Build a function called vandermonde.m which accepts a vector of x values and a polynomial order P
• In the function find N=length of x
• Function returns a matrix V which is Nx(P+1) and whose entries are:

V(n,m) = (xn)(m-1)

• Part 2:
• Translate this pseudo-code to Matlab a script:
• for N=1:5:20
• build x = set of N points in [-1,1]
• build f = exp(x)
• build xfine = set of 10N points in [-1,1]
• build Vorig = vandermonde(N-1,x)
• build Vfine = vandermonde(N-1,xfine)
• build Finterp = Vfine*(Vorig\f);
• plot x,f and xfine,Finterp on the same graph
• end

MA/CS 375 Fall 2002

Next Lecture
• If you have a digital camera bring it in
• If not, bring in some jpeg, gif or tif pictures (make sure they are appropriate)
• We will use them to do some multi-dimensional interpolation
• Estimates for accuracy of polynomial interpolation

MA/CS 375 Fall 2002