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, $\displaystyle f_1(x)$, $\displaystyle f_2(x)$ and $\displaystyle f_3(x)$, the constants $\displaystyle a,b,c,i,j,k$ and the following equation:

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

Can be written:

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

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:

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

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 $\displaystyle a_1+ a_2x+ a_3x^2+ \cdot\cdot\cdot+ a_{n+1}x^n= 0$ only if $\displaystyle a_1= a_2= a_3= \cdot\cdot\cdot= a_{n+1}= 0$.

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 $\displaystyle a_1= 0$ 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 $\displaystyle a_1+ a_2x+ a_3x^2+ \cdot\cdot\cdot+ a_{n+1}x^n= 0$ for all x, it is a constant so its derivative, $\displaystyle a_2+ 2a_3x+ \cdot\cdot\cdot+ na_{n+1}$ is also 0 for all x. Taking x= 0, we now get $\displaystyle a_2= 0$. Of course, since $\displaystyle a_2+ 2a_3x+ \cdot\cdot\cdot+ na_{n+1}= 0$ 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 $\displaystyle ax^1$ and $\displaystyle bx^0$. If they were linearly dependent then there would be some a and b not zero and:

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

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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