Problem expanding algebraic function via Newton polygon

In summary, Jack suggests that you substitute the z\to z+z_0 expansion for the original function around the ramification point r_i\neq 0. He also says that there is a modified method which can overcome some of the problems with the standard Newton polygon.
  • #1
jackmell
1,807
54
Hi,

I've run into a problem with expanding algebraic functions via Newton polygons. Consider the function:

[tex]f(z,w)=a_0(z)+a_1(z)w+a_2(z^2)w^2+\cdots+a_{10}(z)w^{10}=0[/tex]

and say the degree of each [itex]a_i(z)[/itex] is ten.

Now suppose I wish to expand the function around some ramification point of the function, say [itex]r_i\neq 0[/itex]

Now, in general, I won't be able to compute exactly that ramification point so that the standard means of expanding around this point, that is, by letting [itex]z\to z+r_i[/itex] and expanding around zero, would seem to fail because I won't be able to exactly compute the new expansion center. That is, the new center of expansion will be slightly off from the ramification point. This point would then be just a regular point and so the algorithm would just compute regular expansions (cycle-1 sheets) when the branching at this point may in fact be ramified.

I just do not understand how it is possible, using this method, to compute ramified power series centered away from zero of not only this function, but any sufficiently high degree function in which the ramification points cannot be computed exactly because of this problem.

Am I missing something with this?

Thanks,
Jack
 
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  • #2
you seem to be asking how to proceed when you need to know the solution of some polynomial equation of high degree, when such an equation has no obvious solution formula. as far as i know there is no remedy for this. that's why all methods taught in calculus say for integration by partial fractions, e.g. that begin by saying: first find all the roots of this polynomial, are bogus.
 
  • #3
mathwonk said:
as far as i know there is no remedy for this.

I find this disappointing as all the references I have about Newton polygon suggest making that substitution ([itex]z\to z+z_0[/itex]) to expand the function about the point [itex]z_0[/itex]. Then if there is no solution to this problem, such a substituion in a degree 4 or higher function likely will encounter round-off error and thus fail to pick up the ramifications about [itex]z_0[/itex] if [itex]z_0[/itex] is a ramification point -- basically that's saying the method is pretty much useless for expanding high degree functions about ramification points except the origin.

I just cannot believe that's not mentioned anywhere. Not even Walker says anything about this.But thanks for replying. :)

Edit:

Dang it. I re-read some of my reference and this point is in fact emphasized in several of them. For example in:
http://www.lifl.fr/~poteaux/fichiers/monodromy_snc07.pdf
. This fact is not surprising since any close approximation, [itex]\overline{c}_k[/itex] of a critical point [itex]c_k[/itex] is a regular point and expansions above [itex]\overline{c}_k[/itex] have a very small convergence radius
and in fact would lead to incorrect ramifications for [itex]c_k[/itex]. Basically, the calculations have to be done exactly in the standard Newton polygon but there appears to be modified methods which can overcome some of these problems.

Guess I didn't understand before. Sorry guys.
 
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  • #4
the substitution method is perfectly correct. this problem has nothing to do with that method, nor with Newton polygons. the problem is with solving polynomials.
 
  • #5
mathwonk said:
the substitution method is perfectly correct. this problem has nothing to do with that method, nor with Newton polygons. the problem is with solving polynomials.

Ok, I understand that now.
 
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  • #6
this is why we have existence theorems in mathematics. e.g. every polynomial f of odd degree has a root, by the IVT, since f(x) goes to infinity and aLSO TO MINUS INFINITY as |x|-->infinity.

but just try finding the roots of a complicated polynomial of degree 9. all you can do is give an iterative procedure for writing down any finite number of decimals places of a root.

of course it also depends on what you mean by "writing down" a root. if your polynomial has integer coefficients and is irreducible then you could try to work in the field Q[X]/(f), where in fact the root is {X}!

but an element of this field is represented by a vector in nine dimensional Q -space. i.e. an element of this field is represented by a polynomial of degree 8 with rational coefficients.
 

Related to Problem expanding algebraic function via Newton polygon

What is the Newton polygon method for expanding algebraic functions?

The Newton polygon method is a graphical technique used to expand algebraic functions. It involves drawing a polygon on a coordinate grid and using the slopes of the polygon’s sides to determine the terms in the expanded function.

How does the Newton polygon method work?

The Newton polygon method works by breaking down a function into its individual terms and organizing them in a specific way based on the slopes of the polygon. This allows for a systematic and efficient way to expand algebraic functions.

What are the benefits of using the Newton polygon method?

The Newton polygon method offers several benefits, including the ability to quickly and easily expand algebraic functions without having to use complex algebraic techniques. It also provides a visual representation of the function, making it easier to understand and analyze.

What types of functions can be expanded using the Newton polygon method?

The Newton polygon method can be used to expand any algebraic function, including polynomials, rational functions, and exponential functions. It is particularly useful for functions with multiple variables and high degrees.

Are there any limitations to using the Newton polygon method?

While the Newton polygon method is a powerful tool for expanding algebraic functions, it does have some limitations. It can only be used for functions with real coefficients, and it may not work for all types of functions, such as trigonometric functions or functions with complex roots.

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