How Do You Prove Equivalence of Two Polynomials?

[jeremy22511]

Can somebody prove the equivalence statement of two real polynomials in one variable x for me? My Math teacher just told us to remember it as a definition and so I didn't get any proof for it; I attempted to prove it myself and ended up confusing myself with a lot of symbols.

So, can somebody help me with this?

Thanks
Jeremy

 

[micromass]

So, what exactly does this "equivalence statement" state?

 

[HallsofIvy]

The only thing I can think of is that if two polynomials are equal for all x, then they have the same degree and corresponding coeffcients are equal.

If $a_0+ a_1x+ a_2x^2+$$+ \cdot\cdot\cdot\+ a_nx^n=$$b_0+ b_1x+ b_2x^2+ \cdot\cdot\cdot+ b_nx^n$ for all x, then we must have
$(a_0- b_0)+ (a_1- b_1)x$$+ (a_2- b_2)x^2+$$\cdot\cdot\cdot+ (a_n- b_n)x^n= 0$ so it is sufficient to show that if
$a_0+ a_1x+ a_2x^2+ \cdot\cdot\cdot+ a_nx^n= 0$ for all x then $a_0= a_1= a_2= \cdot\cdot\cdot= a_n= 0$. That is, prove that the functions, $1$, $x$, $x^2$, ..., $x^n$ are "independent".

One way to do that is to take n different values for x, say x= 0, 1, 2, ..., n, to get n equations to solve and show that those equations are independent: x= 0 gives $a_0= 0$ so that's easy, x= 1 gives $a_0+ a_1+ a_2+ \cdot\cdot\cdot+ a_n= 0$, x= 2 gives $a_0+ 2a_1+ 4a_2+ \cdot\cdot\cdot 2^n a_n= 0$, etc.

More sophisticated but simpler is to note that if $a_0+ a_1x+ a_2x^2+ \cdot\cdot\cdot+ a_nx^n= 0$ for all x, then it is a constant so its derivative, $a_1+ 2a_2x+ \cdot\cdot\cdot+ na_nx^{n-1}$ i also equal to 0 for all x and so its derivative if 0 for all x, etc. Setting x= 0 in the formula for the polynomial and all of its derivatives gives $a_0= 0$, $a_1= 0$, $2a_2= 0$, ..., $n! a_n= 0$ which again say that all coefficients are 0.

 

[jeremy22511]

Thanks. That really helped.

Jeremy