CS205 - Class 7Covered in class: 1, 4, 5, 7, 8 Readings: Heath Chapter 5 1.Nonlinear equations are much more difficult to solve than simple linear equations, thus we will move from systems of equations to scalar equations to get started. a.Before we move from linear to nonlinear equations, let’s first move from systems of equations to scalar equations in the linear case to get warmed up. i.consider ax=b where a and b are merely numbers 1.of course the solution to this is simply x=b/a 2.the only worry here would be if a was small or “zero” in which case the det(A)=det(a) is zero and the matrix [a] is essentially singular or near singular ii.let’s take a different, functional look at this 1.assume that we had a straight line y=ax+b and that we wanted to find the roots where y=0, then we need to solve 0=ax+b or ax=-b 2.This is our linear equation with solution x=-b/m and now we see that a small slope is equivalent to ill-conditioning a.Consider y=(1e-16)x+1e-16 where the root is x=1 but the division x=-(1e-16)/(1e-16) looks like 0/0 and is indeterminate and full of errors. b.b.Now for nonlinear equations we are solving A(x)=b where A is a vector valued function of x. i.Moreover we can write A(x)=0 since the b term can be incorporated into the vector valued function A of x. ii.As a scalar equation, we have a(x)=0 which looks odd, so we switch notation to f(x)=0. iii.Note how this looks like a root finding problem. In general, any g(x)=b can be rewritten as f(x)=0 by moving b to the left hand side. 1.There may be any number of roots from 0,1,2, etc. to infinity. 2.Remember “roots”=”solutions” from here on out.
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