Re: ? estimate newton step


In article <1e98c090-2e81-486d-8615-a93ab70d4350@xxxxxxxxxxxxxxxxxxxxxxxxxxx>,
CCC <asecant@xxxxxxxxx> writes:
On Jul 2, 12:01=A0pm, spellu...@xxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxxx
(Peter Spellucci) wrote:
In article <31a630a8-aa04-4c7c-857f-30d834331...@xxxxxxxxxxxxxxxxxxxxxxxx=
com>,
=A0Cheng Cosine <asec...@xxxxxxxxx> writes:



Hi:

Newton method is an old and famous algorithm to solve equations. But
in general we do not have an estimate on the steps we will get the
convergent result. Nevertheless, will we be able to estimate this in
the case that the problem at hand is linear or only 2nd order?

At the 1st though it seems that we can finish iteration in 1 step if
f is s linear function.

Use Newton's method to solve f(x) =3D 0 -> x(k+1) =3D x(k)-df(k)'*f(k)

is this the zero of the tangent???



Since f(x) =3D A*x, then df(x) =3D A'.
wrong

A' : you mean the transpose: no !

f is a vector function with components f(1), f(n)

f(i) = sum a(i,j)*x(j) - b(i)
then
df(i)/dx(j) = a(i,j)


So we have:

=A0x(k+1) =3D x(k)-A'*A*x(k) =3D ( I-A'*A )*x(k)

even more wrong
Don't see how we will go.

=A0Did I do anything wrong?

yes , almost anything which could be done wrong
peter- Hide quoted text -

- Show quoted text -

Well, where did I do wrong and how to fix it then?

The very first question is: can we always get correct result using
Newton method IF the function is linear.

hth
peter

.

Quantcast