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I am trying to find the proof/theorem that states:

Given two polynomials in x, if they are equal to eachother, their coefficients must also be equal

For example, in

ax^3 + bx^2 + cx + d = ex^2 + fx + g

a MUST equal 0, and b must equal e, etc...

Does anyone have the actual proof stating this? Thank you!

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    What happens if you start giving values to $x$? Set $x=0$, $x=1$...2011-10-13

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Set $x=0$ and get $0+0+0+d=0+0+g$ so $d=g$. Subtract $d=g$ from both sides, cancel off $x$, and continue...

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    Thank you! Exactly what I needed!2011-10-14
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It's the same thing as saying that if a polynomial is equal to the zero polynomial then its coefficients must all be zero. That can be proved a number of ways. For example, if the constant term is not zero, then the polynomial is not zero at zero, so it isn't the zero polynomial; if the constant term is zero, you can factor a power of the variable out of the polynomial to get one where the constant term is not zero, reducing to a problem already solved.

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    $T$he only reason I didn't choose your answer was because the other answer was more concise. Thank you for taking the time to answer!2011-10-14
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In a general field, the Vandemonde determinant tells all.

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    http://en.wikipedia.org/wiki/Vandermonde_matrix2011-10-18
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The way I do this is to start with

$p(x)=a_nx^n+a_{n-1}x^{n-1}+\cdots+a_1x+a_0\sim b_nx^n+b_{n-1}x^{n-1}+\cdots+b_1x+b_0=q(x).$

If I want these to be equal I want their outputs to be equal and in particular at $x=0$:

$p(0)\overset{!}{=}q(0)\Rightarrow a_0=b_0.$

Also I want all of their derivatives to be equal if they are to be equal. Direct calculation shows that

$\begin{align} p'(x)&=a_1+\mathcal{O}(x) \\ p''(x)&=2a_2+\mathcal{O}(x) \\ \vdots& \\ p^{(k)}(x)&=k!a_k+\mathcal{O}(x), \end{align}$ and similarly $q^{(k)}(x)=k!b_k+\mathcal{O}(x).$

Setting $p^{(k)}(0)\overset{!}{=}q^{(k)}(0)\Rightarrow k!a_k=k!b_k\Rightarrow a_k=b_k,$ for all $k$.