polynomials: sum of coefficients?

Question

Given a polynomial, p, let A be the sum of the coefficients of the even powers and let B be the sum of the coefficients of the odd powers. Prove that A^2 + B^2 = p(1)p(-1). Ideas?

the correct equation is A^2 - B^2 = p(1)p(-1)! Sorry!!!

Answer 1

Let the polynomial 'p' be of the form:

p(x) = a0 + a1*x + a2*(x^2) + a3*(x^3) + a4*(x^4) + ......

Now, we find that:

A = a0 + a2 + a4 + a6 + ...

B = a1 + a3 + a5 + a7 + ...

A^2 + B^2 = ( a0 + a2 + a4 + a6 + ... )^2 + ( a1 + a3 + a5 + a7 + ... )^2


Now, calculate p(1) and p(-1):

p(1) = a0 + a1*1 + a2*(1^2) + a3*(1^3) + a4*(1^4) + ......
p(1) = a0 + a1 + a2 + a3 + a4 + a5 + a6 + a7 + ......
p(1) = ( a0 + a2 + a4 + a6 + ... ) + ( a1 + a3 + a5 + a7 + ... )
p(1) = A + B

p(-1) = a0 + a1*(-1) + a2*(-1^2) + a3*(-1^3) + a4*(-1^4) + ......
p(-1) = a0 - a1 + a2 - a3 + a4 - a5 + a6 - a7 + ......
p(-1) = ( a0 + a2 + a4 + a6 + ... ) - ( a1 + a3 + a5 + a7 + ... )
p(-1) = A - B

We can see that:

p(1)*p(-1) = (A + B) * (A - B) = A^2 - B^2

A^2 - B^2 = p(1)*p(-1)

I think your question has a small mistake, it should be A^2 - B^2 instead of A^2 + B^2

Answer 2

For the general case, start by writing p = a(1) + b(1)x +a(2)x^2 +b(2)x^3 + a(3)x^4 +b(3)x^5 . . . (It would be better to use subscripts instead of the numbers in brackets but I don't know how to do that on a keyboard.) Then write down what this would be with x replaced by 1, that is p(1) and then replaced by -1 i.e. p(-1). You should be able to take it from there.

Answer 3

Well, it would work if we were trying to prove that A^2-B^2=p(1)p(-1).

Assuming a polynomial with, say, 3 terms,

LHS=(a+c)^2-b^2

=((a+c)-b)((a+c)+b)

=(a-b+c)(a+b+c)

RHS=(a-b+c)(a+b+c)

Your equation, however, does not work.