# Polynomial coefficient equivalence

• Aug 18th 2011, 08:39 PM
mathnerd15
Polynomial coefficient equivalence
what is the fundamental algebra result that allows you to equate coefficients for 2 polynomials if they have equal values for all values of x?
and how can you replace this expression-
F(x)=A(x-alpha1)(x-alpha2)....)x-alphar)

with F(x)=A(x-alpha1)^m1(x-alpha2)^m2....(x-alphar)^mr, with A=an
mk are integers >=1 and are multiplicities of the roots mk being the multiplicity of alphak
thanks very much!
• Aug 19th 2011, 04:40 AM
Drexel28
Re: Polynomial coefficient equivalence
Quote:

Originally Posted by mathnerd15
what is the fundamental algebra result that allows you to equate coefficients for 2 polynomials if they have equal values for all values of x?

If I am interpreting you correctly, you are wondering how you can say that if two polynomials over $\mathbb{C}$ agree for all values of $x$ then they must be equal. In other words, if two polynomials induce the same function they are the same polynomial. This is true more generally over any infinite integral domain $R$. First prove it's true for an infinite field $k$ by noting first that this problem is evidently equivalent to showing that a non-zero polynomial may not be simultaneously zero, which is clear since if $p(x)$ is such with $\deg(p(x))>0$ then we know that $0\leqslant \text{number of zeroes of p over }k\leqslant \deg(p(x))$, and so in particular $p$ cannot vanish for every value of $k$ (since $k$ is infinite). For the more general case merely show that if a non-zero polynomial over an integral domain $R$ induced the zero function then it would induce the zero function over[ $\text{Frac}(R)$, and apply the previous result.

Quote:

and how can you replace this expression-
F(x)=A(x-alpha1)(x-alpha2)....)x-alphar)

with F(x)=A(x-alpha1)^m1(x-alpha2)^m2....(x-alphar)^mr, with A=an
mk are integers >=1 and are multiplicities of the roots mk being the multiplicity of alphak
thanks very much!
What does this mean?
• Aug 19th 2011, 06:48 AM
HallsofIvy
Re: Polynomial coefficient equivalence
If polynomial $p(x)= a_nx^n+ a_{n-1}x^{n-1}+ \cdot\cdot\cdot+ a_1x+ a_0$ and polynomial $q(x)= b_nx^n+ b_{n-1}x^{n-1}+ \cdot\cdot\cdot+ b_1x+ b_0$ are equal, for all x, then
$a_nx^n+ a_{n-1}x^{n-1}+ \cdot\cdot\cdot+ a_1x+ a_0= b_nx^n+ b_{n-1}x^{n-1}+ \cdot\cdot\cdot+ b_1x+ b_0$
for all x. In particular, taking x= 0, [itex]a_0= b_0[/tex]. Since those are equal, we can cancel them leaving
$a_nx^n+ a_{n-1}x^{n-1}+ \cdot\cdot\cdot+ a_1x= b_nx^n+ b_{n-1}x^{n-1}+ \cdot\cdot\cdot+ b_1x$
which is the same as
$x(a_nx^{n-1}+ a_{n-1}x^{n-2}+ \cdot\cdot\cdot+ a_1)= x(b_nx^{n-1}+ b_{n-1}x^{n-2}+ \cdot\cdot\cdot+ b_1)$

For x not 0, we can divide by x and have
$a_nx^{n-1}+ a_{n-1}x^{n-2}+ \cdot\cdot\cdot+ a_1= b_nx^{n-1}+ b_{n-1}x^{n-2}+ \cdot\cdot\cdot+ b_1$
and now, because polynomials are continuous, taking the limit as x goes to 0, $a_1= b_1$.

Just repeat that to get $a_k= b_k$ for all k up to n.

Notice that I had to say "x not 0" at one step so that I could not just let x= 0 in the next. But because polynomials are continuous I can say instead "take the limit as x goes to 0".

Quote:

and how can you replace this expression-
F(x)=A(x-alpha1)(x-alpha2)....)x-alphar)

with F(x)=A(x-alpha1)^m1(x-alpha2)^m2....(x-alphar)^mr, with A=an
mk are integers >=1 and are multiplicities of the roots mk being the multiplicity of alphak
thanks very much!
You can't! Those are very different functions. Now, if you did NOT use "alpha1" to "alphar" in both, you could if some of the "alpha"s are the same, combine them. That is, if alpha1= alpha2, you could write (x- alpha1)(x- alpha2) as (x- alpha1)^2. But then you would not have the "(x- alpha2)" term.
• Sep 10th 2011, 06:39 PM
mathnerd15
Re: Polynomial coefficient equivalence
thanks so much for the replies! sorry I have been having serious health problems
this is from p. 4 of the Riley/Hobson math. methods:
the condition f(Ak)=0 for k=1,2,...r could be met if
F(x)=A(x-alpha1)(x-alpha2)....(x-alphar) were replaced by
F(x)=A(x-alpha1)^m1(x-alpha2)^m2....(x-alphar)^mr
• Sep 10th 2011, 10:39 PM
mathnerd15
Re: Polynomial coefficient equivalence
I know that for cubic equations you can solve by the partial differentiation method, but I'm sure you already know this...