Inverse of Newton's method?

In article <m3znfjj291.fsf@pearl.local>,
Scott Hemphill  <hemphill@alumni.caltech.edu> wrote:
>briggs@encompasserve.org writes:

>> If you know the roots of a polynomial, the polynomial is explicitly
>> given by:
 
>> P(x) = k (x - root1) (x - root2) (x - root3) ... (x - rootn)

>An interesting case is polynomials of infinite degree.

There is no such thing.  By definition a polynomial has finite degree.
There are infinite series, and infinite products, but they're not
the same thing.

>  For example,
>sin(pi*x) has roots at all the integers, so:

>  sin(pi*x) = pi * x * (1-x)(1+x)(1-x/2)(1+x/2)(1-x/3)(1+x/3) ...

>(This statement is true, although I haven't proven it.)

This formula is true, but the "so" is not.  How do you know that this 
product isn't, e.g., sin(pi x) exp(x^2)?  In general, for any analytic 
functions f and g, f(x) and f(x) exp(g(x)) have the same roots.
So the roots don't determine the function.

Robert Israel                                israel@math.ubc.ca
Department of Mathematics        http://www.math.ubc.ca/~israel 
University of British Columbia            
Vancouver, BC, Canada V6T 1Z2

israel@math.ubc.ca (Robert Israel) writes:

> In article <m3znfjj291.fsf@pearl.local>,
> Scott Hemphill  <hemphill@alumni.caltech.edu> wrote:
> >briggs@encompasserve.org writes:
> 
> >> If you know the roots of a polynomial, the polynomial is explicitly
> >> given by:
>  
> >> P(x) = k (x - root1) (x - root2) (x - root3) ... (x - rootn)
> 
> >An interesting case is polynomials of infinite degree.
> 
> There is no such thing.  By definition a polynomial has finite degree.
> There are infinite series, and infinite products, but they're not
> the same thing.
> 
> >  For example,
> >sin(pi*x) has roots at all the integers, so:
> 
> >  sin(pi*x) = pi * x * (1-x)(1+x)(1-x/2)(1+x/2)(1-x/3)(1+x/3) ...
> 
> >(This statement is true, although I haven't proven it.)
> 
> This formula is true, but the "so" is not.  How do you know that this 
> product isn't, e.g., sin(pi x) exp(x^2)?  In general, for any analytic 
> functions f and g, f(x) and f(x) exp(g(x)) have the same roots.
> So the roots don't determine the function.

You are correct.  Please forgive my inexact language, leading to false
statements.  I just find it interesting that certain functions can be
so easily characterized by the product of their roots, (e.g. sin, cos,
1/gamma).

Scott
-- 
Scott Hemphill	hemphill@alumni.caltech.edu
"This isn't flying.  This is falling, with style."  -- Buzz Lightyear