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Math Help - Problem 43

  1. #1
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    Problem 43

    1)Let a_1,a_2,...,a_n be real (or complex) distinct numbers where n\geq 2. Define P_k = \prod_{j\not = k} (a_k - a_j). Can it be that P_1=P_2=...=P_n?
    (For example, let a_1=1,a_2=2,a_3=3 then P_1 = (1-2)(1-3)=2, P_2 = (2-1)(2-3)=-1, and P_3=(3-1)(3-2)=2 but all three are not the same in this case).

    The next problem is for the younger kids so please give them a chance.

    2)Let f(x) and g(x) be functions which you can differenciate. Note that [f(x)g(x)]' = f'(x)g(x)+f(x)g'(x). Can you find a formula for [f(x)g(x)]^{(n)} where by ^{(n)} means to preform differenciation n times repeatedly.
    Last edited by ThePerfectHacker; December 6th 2007 at 09:37 AM.
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  2. #2
    Senior Member DivideBy0's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    2)Let f(x) and g(x) be functions which you can differenciate. Note that [f(x)g(x)]' = f'(x)g(x)+f(x)g'(x). Can you find a formula for [f(x)g(x)]^{(n)} where by ^{(n)} means to preform differenciation n times repeatedly.
    After a little experimenting:

    [f(x)g(x)]'=f'(x)g(x) + f(x)g'(x)

    [f(x)g(x)]''=f''(x)g(x)+2f'(x)g'(x)+f(x)g''(x)

    [f(x)g(x)]''' = f'''(x)g(x)+3f''(x)g'(x)+3f'(x)g''(x)+f(x)g'''(x)

    This reminds me of the binomial theorem:

    (x+y)^n = \sum_{k=0}^n {n \choose k}x^{n-k} y^{k}

    So perhaps

    [f(x)g(x)]^{(n)} = \sum_{k=0}^n {n\choose k} f^{n-k}(x) g^{k}(x)

    Where h^n(x) denotes the nth derivative of h(x)

    Nice problem

    P.S Is there any way to hide text?
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by DivideBy0 View Post
    P.S Is there any way to hide text?
    Color the text white.

    -Dan
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    Quote Originally Posted by DivideBy0 View Post
    Nice problem
    Good job. This result is called "Leibniz's Rule".
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    Forum Admin topsquark's Avatar
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    Quote Originally Posted by ThePerfectHacker View Post
    1)Let a_1,a_2,...,a_n be real (or complex) distinct numbers where n\geq 2. Define P_k = \prod_{j\not = k} (a_k - a_j). Can it be that P_1=P_2=...=P_n?
    (For example, let a_1=1,a_2=2,a_3=3 then P_1 = (1-2)(1-3)=2, P_2 = (2-1)(2-3)=-1, and P_3=(3-1)(3-2)=2 but all three are not the same in this case).
    Since I know the way your mind works, there is something wrong with what I am about to say...

    Consider the case n = 3. Then the question is can
    a_1a_2 = a_1a_3 = a_2a_3
    where all three are distinct?

    Obviously not since a_1a_2 = a_1a_3 \implies a_2 = a_3.

    Now, there's got to be something screwy here because I can generalize this argument to larger n and come up with similar results. But I can't believe it would be this easy...

    -Dan
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  6. #6
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    I came up with problem #1 accidently when I playing around with polynomials. Here is my original solution, however it seems to me that this problem is easy even if approached directly.

    Proof:
    Let a_1,...,a_n be real (or complex numbers) and define f(x) = (x-a_1)...(x-a_n). The key step is to note that f'(x_k) = P_k by using the general product rule for derivatives. Now, \deg f(x) = n\geq 2 thus, \deg f'(x) = n-1\geq 1, this means that f'(x) cannot attain the same values at a_1,a_2,...,a_n (meaning f(a_1)=f(a_2)=...=f(a_n)) because otherwise the situation is that a degree n-1 polynomial attains the same value n times for n distinct numbers. Which is impossible. Thus, P_1,P_2,...,P_n cannot all be the same.
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