1. Polynomial proof

Hello MHF!
This problem is again from I.M Gelfand's Algebra book.

Problem 157. Prove that for any polynomial of degree 2 all second differences are equal.

My "proof":
Let $\displaystyle P(x)=x^2$

$\displaystyle P(x+1)-P(x)=x^2+2x+1-x^2=2x+1=q(x)$

$\displaystyle q(x+1)-q(x)=2(x+1)+1-(2x+1)=2x+2+1-2x-1=2$.

This proof isn't correct right? since $\displaystyle x^2$ isn't general case, if not, should I use induction to prove it?

2. Re: Polynomial proof

Just generalize your proof with $\displaystyle P(x)=ax^2+bx+c$.

3. Re: Polynomial proof

Originally Posted by MarkFL2
Just generalize your proof with $\displaystyle P(x)=ax^2+bx+c$.
Oh of course! Thanks for the answer, I don't know what happened to me haha.

4. Re: Polynomial proof

Prove that for any polynomial of degree 2 all second differences are equal.
Is this what you mean?

$\displaystyle f(x) \:=\:ax^2 + bx + c$

$\displaystyle \begin{array}{c|c|c|c|c|c|c|c|} x & n && n+1 && n+2 && n+3 \\ \hline \text{Function} & an^2+bn + c && a(n\!+\!1)^2+b(n\!+\!1) + c && a(n\!+\!2)^2 + b(n\!+\!2)+c && a(n\!+\!3)^2 + b(n\!+\!3) + c \\ \hline \text{1st diff.} && a(2n\!+\!1)+b && a(2n\!+\!3) + b && a(2n\!+\!5) + b & \\ \hline \text{2nd diff.} &&& 2a &&2a && \\ \hline\end{array}$

5. Re: Polynomial proof

Originally Posted by Soroban

Is this what you mean?

$\displaystyle f(x) \:=\:ax^2 + bx + c$

$\displaystyle \begin{array}{c|c|c|c|c|c|c|c|} x & n && n+1 && n+2 && n+3 \\ \hline \text{Function} & an^2+bn + c && a(n\!+\!1)^2+b(n\!+\!1) + c && a(n\!+\!2)^2 + b(n\!+\!2)+c && a(n\!+\!3)^2 + b(n\!+\!3) + c \\ \hline \text{1st diff.} && a(2n\!+\!1)+b && a(2n\!+\!3) + b && a(2n\!+\!5) + b & \\ \hline \text{2nd diff.} &&& 2a &&2a && \\ \hline\end{array}$
Hey Soroban, yes, that's it, and I got 2a as an answer for 2nd diff.

6. Re: Polynomial proof

And here's solution if anyone is interested:

Let $\displaystyle \left P(x)=ax^2+bx+c$

$\displaystyle P(x+1)-P(x)=q(x)=a(x+1)^2+b(x+1)+c-(ax^2+bx+c)=$
$\displaystyle =a(x^2+2x+1)+bx+b+c-ax^2-bx-c=$
$\displaystyle =ax^2+2ax+a+bx+b+c-ax^2-bx-c=2ax+a+b$

$\displaystyle q(x+1)-q(x)=2a(x+1)+a+b-(2ax+a+b)=$
$\displaystyle =2ax+2a+a+b-2ax-a-b=2a.$

So, every 2nd difference polynomial is constant and is equal to 2 times first coefficient.

7. Re: Polynomial proof

Although this problem has been thoroughly solved, here is the outline of a different approach you might find interesting.

(1) Show that for a first degree polynomial all first differences are equal. (You may already have shown this, or it might be a theorem in your book.)

(2) Show that the difference operator applied to a second degree polynomial yields a first degree polynomial.

(1) and (2) combined show that the second differences of a second degree polynomial are equal.

8. Re: Polynomial proof

Originally Posted by awkward
Although this problem has been thoroughly solved, here is the outline of a different approach you might find interesting.

(1) Show that for a first degree polynomial all first differences are equal. (You may already have shown this, or it might be a theorem in your book.)

(2) Show that the difference operator applied to a second degree polynomial yields a first degree polynomial.

(1) and (2) combined show that the second differences of a second degree polynomial are equal.
$\displaystyle (1) P(x+1)-P(x)=ax+a+b-ax-b=a$

$\displaystyle (2) P(x+1)-P(x)=q(x)=a(x+1)^2+b(x+1)+c-(ax^2+bx+c)=$
$\displaystyle =a(x^2+2x+1)+bx+b+c-ax^2-bx-c=$
$\displaystyle =ax^2+2ax+a+bx+b+c-ax^2-bx-c=2ax+a+b$

You're right, this is another method. Thanks for the idea.