# Prove two polynomials are equal in R^n

1. Nov 6, 2016

### lep11

1. The problem statement, all variables and given/known data
The task is to prove that $$\lim_{x\rightarrow0}\frac{Q_1(x)-Q_2(x)}{\|x\|^k}=0 \implies Q_1=Q_2,$$ where $Q_1,Q_2$ are polynomials of degree $k$ in $\mathbb{R}^n$.

2. Relevant equations

$$\lim_{x\to 0} \frac{a x^\alpha}{\|x\|^n}=\left\{\begin{array}{c} 0 \textrm{ if } |\alpha|>n \\ a \textrm { if } |\alpha|=n \\ \infty \textrm { if } |\alpha|<n \textrm{ and } a\neq 0 \\ 0 \textrm{ if } a=0 \end{array}\right.$$

$$|\alpha|=k=\alpha_1!\alpha_2!\cdot...\cdot\alpha_n!$$

3. The attempt at a solution Proof by contradiction. Assume that $Q_{1}\neq{Q_2}$ and let's denote $Q_1(x)-Q_2(x)=F(x)+G(x)$ where $F$ is lowest degree ($l$) polynomial and and $G$ contains the rest. Then let's consider the limit $$\lim_{t\rightarrow0}\frac{F(tx)+G(tx)}{\|tx\|^l},$$ where $b\neq{0}$ and $F(b)\neq{0}$.

$$\lim_{t\rightarrow0}\frac{F(tb)+G(tb)}{\|tb\|^l}=\lim_{t\rightarrow0}\frac{G(tb)}{\|tb\|^l}+\lim_{t\rightarrow0}\frac{F(tb)}{\|tb\|^l}=...\neq{0}$$ which is contradiction. Therefore it must hold that $Q_1=Q_2$.

I have problem expanding the limit expression.

Last edited: Nov 6, 2016
2. Nov 6, 2016

### Staff: Mentor

Are you sure you can split up the limit that way?
If yes (which is not trivial), you can just split it into k+1 limits and show that every term of the polynomial has to be zero.

I would look for all polynomials of degree <= k which satisfy the given limit. The difference between two polynomials of degree k has to be such a polynomial.

3. Nov 6, 2016

### Stephen Tashi

What does "in $\mathbb{R}^n$" mean in this context? Are we talking about polynomials in n-variables ?

4. Nov 6, 2016

### lep11

Yes.
$F(tb)=t^{\alpha}F(b)$, but how to use that?
Is $||tb||^l=|t|^l||b||$?

$$\lim_{t\rightarrow0}\frac{F(tb)+G(tb)}{\|tb\|^l}=\lim_{t\rightarrow0}\frac{t^lF(b)+t{^\alpha}G(b)}{|t|^l\|b\|^l}=\lim_{t\rightarrow0}(\frac{t^lF(b)}{|t|^l\|b\|^l}+\frac{t{^\alpha}G(b)}{|t|^l\|b\|^l})=...?$$, where $|\alpha|>l$
Or
$$\lim_{t\rightarrow0}\frac{F(tb)+G(tb)}{\|tb\|^l}=(\lim_{t\rightarrow0}\frac{a(tb)^l}{\|tb\|^l}+\frac{a(tb)^{\alpha}}{\|tb\|^l})=a+\lim_{t\rightarrow0}\frac{a(tb)^{\alpha}}{\|tb\|^l}=a+0\neq{0}$$ for some $a_i$?

Last edited: Nov 6, 2016