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!

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 $\displaystyle \mathbb{C}$ agree for all values of $\displaystyle x$ 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 $\displaystyle R$. First prove it's true for an infinite field $\displaystyle k$ 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 $\displaystyle p(x)$ is such with $\displaystyle \deg(p(x))>0$ then we know that $\displaystyle 0\leqslant \text{number of zeroes of p over }k\leqslant \deg(p(x))$, and so in particular $\displaystyle p$ cannot vanish for every value of $\displaystyle k$ (since $\displaystyle k$ is infinite). For the more general case merely show that if a non-zero polynomial over an integral domain $\displaystyle R$ induced the zero function then it would induce the zero function over[ $\displaystyle \text{Frac}(R)$, 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?

Re: Polynomial coefficient equivalence

If polynomial $\displaystyle p(x)= a_nx^n+ a_{n-1}x^{n-1}+ \cdot\cdot\cdot+ a_1x+ a_0$ and polynomial $\displaystyle q(x)= b_nx^n+ b_{n-1}x^{n-1}+ \cdot\cdot\cdot+ b_1x+ b_0$ are equal, for all x, then

$\displaystyle a_nx^n+ a_{n-1}x^{n-1}+ \cdot\cdot\cdot+ a_1x+ a_0= b_nx^n+ b_{n-1}x^{n-1}+ \cdot\cdot\cdot+ b_1x+ b_0$

for all x. In particular, taking x= 0, [itex]a_0= b_0[/tex]. Since those are equal, we can cancel them leaving

$\displaystyle a_nx^n+ a_{n-1}x^{n-1}+ \cdot\cdot\cdot+ a_1x= b_nx^n+ b_{n-1}x^{n-1}+ \cdot\cdot\cdot+ b_1x$

which is the same as

$\displaystyle x(a_nx^{n-1}+ a_{n-1}x^{n-2}+ \cdot\cdot\cdot+ a_1)= x(b_nx^{n-1}+ b_{n-1}x^{n-2}+ \cdot\cdot\cdot+ b_1)$

For x not 0, we can divide by x and have

$\displaystyle a_nx^{n-1}+ a_{n-1}x^{n-2}+ \cdot\cdot\cdot+ a_1= b_nx^{n-1}+ b_{n-1}x^{n-2}+ \cdot\cdot\cdot+ b_1$

and now, **because polynomials are continuous**, taking the limit as x goes to 0, $\displaystyle a_1= b_1$.

Just repeat that to get $\displaystyle a_k= b_k$ 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.

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

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...