# Degrees of polynomials?

Let $A\left(x\right)$ represent a polynomial with a degree of $n-1$.

Split $A\left(x\right)$ into odd and even powers. For example:

$A\left(x\right) = 3 + 4x+6x^2+2x^3+x^4+10x^5$

$= \left(3+6x^2 + x^4\right)+x\left(4+ 2x^2 + 10x^4\right)$

More generally:

$A\left(x\right) = A_e\left(x^2\right) + \left(x\right)A_o\left(x^2\right)$

where $A_e\left(∙\right)$ are the even-numbered coefficients and $A_o\left(∙\right)$are the odd-numbered coefficients.

Are the degrees of $A_e\left(∙\right)$ and $A_o\left(∙\right)$ necessarily $≤\frac{n}{2} -1$? If so why?

This isn't homework, just a book I'm reading.

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Try induction on degree of A? –  iloveinna May 13 '12 at 2:32
@Warwicker I have a limited mathematical background in using proofs- I don't know how to perform induction. –  user26649 May 13 '12 at 2:34
Except that it seems that this is kind of false... If $A(x)=x^2$ then $A_e(x)=x$. And $\deg A=2$ and $\deg A_e=1$. –  Yongyi Chen May 13 '12 at 2:35
@YongyiChen Sorry, my fault. I forgot to mention $n > 1$ –  user26649 May 13 '12 at 2:38
If $A(x)=x^2+x+1$ then $A_e(x)=x+1$, and $\deg A=2$ and $\deg A_e=1$. But I know what you're trying to say, so I will post an answer soon. Actually the error lies in the phrase "$n$ monomials." I'm not sure why you say that instead of just saying that $A(x)$ has degree $n-1$. –  Yongyi Chen May 13 '12 at 2:40

## 1 Answer

Take $A_e(x)$ for example; that polynomial is constructed by taking the monomials of $A(x)$ with even degree and halving the exponents. The highest possible even exponent in $A(x)$ is $n-1$ itself in the case that $n-1$ is even, so the highest exponent in $A_e(x)$ is $\frac{n-1}2$. If $n-1$ is odd, then the highest possible exponent in $A_e(x)$ is only $\frac{n-2}2=\frac n2-1$.

Similarly, $A_o(x)$ is constructed by taking the monomials of $A(x)$ with odd degree, subtracting 1 from each exponent, and then halving them. The highest possible odd exponent in $A(x)$ is $n-1$ if $n-1$ is odd, so the highest exponent in $A_o(x)$ is $\frac{n-2}2=\frac n2-1$. If $n-1$ is even, then the highest possible exponent in $A_o(x)$ is $\frac{n-3}2$.

So a minor edit to your statement: The degrees of $A_e(x)$ and $A_o(x)$ are at most $\frac{n-1}2$. Now it's true.

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Thanks, that helps! –  user26649 May 13 '12 at 3:01