# P/q continuity

1. Nov 15, 2008

### kathrynag

1. The problem statement, all variables and given/known data
Let p and q be a polynomial and x0 be a zero of q of multiplicity m. Prove that p/q can be assigned a value at x0 such that the function thus defined will be continuous there iff x0 is a zero of p of multiplicity greater than or equal to m.

2. Relevant equations

3. The attempt at a solution
I'm not quite sure how to even get started. This question just confuses me in what it's asking.

2. Nov 15, 2008

### Office_Shredder

Staff Emeritus
If f(x) is a polynomial, and has a zero at x0 of degree m, then there exists a polynomial g(x) such that f(x) = g(x)*(x-x0)m

As written, p/q (if q has a zero at x0) is not defined at x0, but is defined continuously in a ball around x0. A standard trick is to extend functions by finding what their limit is as you approach points that aren't in the original domain, and then defining a new function that equals the original function on the original domain, and if you're on one of the non-defined points, you define it to be the limit of the original function.

So try writing p and q in the form I gave at the top of my post, and then use the tip to consider the relative degrees of the roots and figure out if the limit as x approaches x0 exists

3. Nov 15, 2008

### kathrynag

Ok so p: there exists a polynomial q such that p=q*(x-x0)m
Is this right then do the same for q?

4. Nov 16, 2008

### kathrynag

or p=r*(x-x0)m?

5. Nov 16, 2008

### HallsofIvy

Staff Emeritus
Office Shredder said p(x)= g(x)(x- x0)m, NOT "q(x)(x- x0)m". You cannot use "q" to mean two different polynomials.

Don't write "p(x)= q(x)(x- x0)m" or "p= r(x- x0)m). Do what Office Shredder suggested!

6. Nov 19, 2008

### kathrynag

Ok, so I have p=g(x)(x-$$x_{0}$$)$$^{m}$$
q=g(x)(x-$$x_{0}$$)$$^{n}$$
p/q=(x-$$x_{0}$$)$$^{m-n}$$

7. Nov 19, 2008

### kathrynag

I'm just confused on where to go next... Like do I do a delta, epsilon proof?