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Math Help - Polynomial proof

  1. #1
    Junior Member PaulAdrienMauriceDirac's Avatar
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    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 P(x)=x^2

    P(x+1)-P(x)=x^2+2x+1-x^2=2x+1=q(x)

    q(x+1)-q(x)=2(x+1)+1-(2x+1)=2x+2+1-2x-1=2.

    This proof isn't correct right? since x^2 isn't general case, if not, should I use induction to prove it?
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    MHF Contributor MarkFL's Avatar
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    Re: Polynomial proof

    Just generalize your proof with P(x)=ax^2+bx+c.
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    Junior Member PaulAdrienMauriceDirac's Avatar
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    Re: Polynomial proof

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

    Hello, PaulAdrienMauriceDirac!

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


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

    \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}
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    Junior Member PaulAdrienMauriceDirac's Avatar
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    Re: Polynomial proof

    Quote Originally Posted by Soroban View Post
    Hello, PaulAdrienMauriceDirac!

    Is this what you mean?


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

    \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.
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    Junior Member PaulAdrienMauriceDirac's Avatar
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    Re: Polynomial proof

    And here's solution if anyone is interested:

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

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

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

    So, every 2nd difference polynomial is constant and is equal to 2 times first coefficient.
    Last edited by PaulAdrienMauriceDirac; November 4th 2012 at 05:05 AM.
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    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.
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    Junior Member PaulAdrienMauriceDirac's Avatar
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    Re: Polynomial proof

    Quote Originally Posted by awkward View Post
    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.
    (1) P(x+1)-P(x)=ax+a+b-ax-b=a

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