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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 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 Did your results concur? 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? • Answer: • 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? • Answer: • 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