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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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    @YongyiChen Your right. I removed the "$n$ monomials" part.2012-05-13

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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.