# Thread: Equating Coefficients of Polynomials Proof

1. ## Equating Coefficients of Polynomials Proof

Hi,

I intuitively understand Equating Coefficients of Polynomials but I was hoping to find a proof of the method. Does anyone know where I might find one?

Cheers,

Scott.

2. I think I have figured it out.

Given say three arbitrary functions of x, $f_1(x)$, $f_2(x)$ and $f_3(x)$, the constants $a,b,c,i,j,k$ and the following equation:

$af_1(x)+bf_2(x)+cf_3(x)=if_1(x)+jf_2(x)+kf_3(x)$

Can be written:

$(a-i)f_1(x)+(b-j)f_2(x)+(c-k)f_3(x)=0$

If the functions are linearly independent then the only solution to the equation is when the coefficients of the functions are all zero. This leads to:

$a=i$
$b=j$
$c=k$

Which shows the coefficients of each side of the original equation are equated. The idea can be extended to any number of linearly independent functions.

The "x" terms of a polynomial are linearly independent so the coefficients can be equated also.

3. Of course, to complete the proof, one would have to show that the different powers of x, 1, x, $x^2$, etc. are "independent". That means showing that $a_1+ a_2x+ a_3x^2+ \cdot\cdot\cdot+ a_{n+1}x^n= 0$ only if $a_1= a_2= a_3= \cdot\cdot\cdot= a_{n+1}= 0$.

One way to do that is to take n different values for x to get n equations for the coefficients. If we take x= 0 we get $a_1= 0$ immediately. If we take x= 0 we get [tex]a_1+ a+2+ a_3+ \cdot\cdot\cdot+ a_{n+1}= 0[/itex], etc.

A simpler, but more sophisticated, way would be to take derivatives: since $a_1+ a_2x+ a_3x^2+ \cdot\cdot\cdot+ a_{n+1}x^n= 0$ for all x, it is a constant so its derivative, $a_2+ 2a_3x+ \cdot\cdot\cdot+ na_{n+1}$ is also 0 for all x. Taking x= 0, we now get $a_2= 0$. Of course, since $a_2+ 2a_3x+ \cdot\cdot\cdot+ na_{n+1}= 0$ for all x, it is a constant and its derivative is 0. Continuing through the nth derivative gives us 0 for all n coefficients.

4. The way I see the linear independence of the polynomial terms is to take any two of them, say $ax^1$ and $bx^0$. If they were linearly dependent then there would be some a and b not zero and:

$ax^1 + bx^0=0$
$ax^1=-bx^0$
$x=-b/a$

But x is not a constant which shows a contradiction with the original assumption that the terms are linearly dependent. The only solution is when a and b are zero.

This can be applied to any two terms of a polynomial of any degree which implies that the terms of any polynomial are linearly independent.

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