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Math Help - Problem in finite differences

  1. #1
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    Problem in finite differences

    Show that the nth divided difference of is 1
    solution:
    We have,
    dash indicating that the term is to be excluded from denominator
    I don't understand this step. Can any math helper,math expert explain me? Especially 1 on R. H. S.
    Last edited by Vinod; February 16th 2014 at 05:18 AM.
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  2. #2
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    Re: Problem in finite differences

    Hey Vinod.

    What is the dummy variable involved for the summation? Is the same variable within the expression directly to the right of sigma?
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    Re: Problem in finite differences

    Quote Originally Posted by chiro View Post
    Hey Vinod.

    What is the dummy variable involved for the summation? Is the same variable within the expression directly to the right of sigma?
    Hey Chiro,
    As per my understanding,L. H. S.-the term next to summation'(sigma') in R. H. S. is equal to 1 as stated in the problem.I think it is the nth divided difference of n arguments()
    Next steps in the solution are as follows:Put x=0 on both sides, we have
    0=1 - f{}

    Hence f{ }= 1

    If I am wrong, tell me.
    Last edited by Vinod; February 17th 2014 at 05:51 AM.
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  4. #4
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    Re: Problem in finite differences

    So basically inside the summation, is n meant to be equal to i and the summation sums over i = 1 to n?

    It's a little hard to make sense of the question when these things aren't clarified.
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    Re: Problem in finite differences

    Hey chiro,

    Your reply suggests me that you are incognizant of divided difference properties. The term next to sigma' is based on the property which states that divided difference is a symmetrical function of all the arguments involved.
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  6. #6
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    Re: Problem in finite differences

    Yes you are right in my unfamiliarity with this topic - which is why I wanted to clarify these things.

    I don't think I will be able to help you if I don't know what the dummy index is and what it refers to.
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  7. #7
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    Re: Problem in finite differences

    Try it when $n=0$.

    It is easily verified that

    $\dfrac{x}{x-x_0} = 1 + \dfrac{x_0}{1}\dfrac{1}{x-x_0}$?

    So, let's check it out when $n=1$:

    $\dfrac{x^2}{(x-x_0)(x-x_1)} = 1+\dfrac{x_0^2}{x_0-x_1}\dfrac{1}{x-x_0} + \dfrac{x_1^2}{x_1-x_0}\dfrac{1}{x-x_1}$

    The common denominator on the RHS is

    $(x_0-x_1)(x-x_0)(x-x_1)$

    So, the question is, is the following equation true?

    $(x_0-x_1)(x-x_0)(x-x_1) + x_0^2(x-x_1) - x_1^2(x-x_0) \stackrel{?}{=} (x_0-x_1)x^2$

    Well, the LHS becomes

    $(x_0-x_1)(x^2-x_0x-x_1x+x_0x_1) + x_0^2x - x_0^2x_1 - x_1^2x + x_1^2x_0$

    $=(x_0-x_1)x^2 - x_0^2x-x_0x_1x+x_0^2x_1+x_0x_1x + x_1^2x - x_0x_1^2 + x_0^2x - x_0^2x_1 - x_1^2x + x_1^2x_0$

    $=(x_0-x_1)x^2$

    Hence, the LHS and the RHS are equal when $n=1$. To prove it is true for all $n$, you can attempt to use induction. You know it is true when $n=0$ (and when $n=1$), so assume it is true when $n=k$ and try to prove it for $n=k+1$.
    Last edited by SlipEternal; February 18th 2014 at 04:17 PM.
    Thanks from Vinod
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  8. #8
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    Re: Problem in finite differences

    Hey SlipEternal,
    Your answer is correct. very good.
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