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Math Help - alternative generalisation of binominals law

  1. #1
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    alternative generalisation of binominals law

    I have problems with proving some generalisation of binominal law.
    It is commonly known that

    x^2-2x+1\geqslant 0

    But, I have problems with proving that

    x^{k+1}-(k+1)x+k >= 0

    is true for k\geqslant0.
    Can anybody help me?
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  2. #2
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    Hi

    The inequality is true for x \geq 0
    For instance for k=2 : x^3-3x+2 \leq 0 for x=-3

    There are several ways to show the inequality

    You can study f(x) = x^{k+1}-(k+1)x+k
    f is differentiable and f'(x) = (k+1)(x^k-1)
    The minimum of f is obtained for x=1 and f(1)=0 therefore f(x) \geq 0

    Second way : since f(1)=0 and f'(1)=0, f(x) can be factorized by (x-1)
    You can show that f(x) = (x-1) \left(\sum_{j=1}^{k}x^j-k\right)
    and f(x) = (x-1)^2 \left(\sum_{j=0}^{k-1}(k-j)x^j\right)
    Both factors are positive for x positive
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  3. #3
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    Quote Originally Posted by running-gag View Post
    Hi

    The inequality is true for x \geq 0
    For instance for k=2 : x^3-3x+2 \leq 0 for x=-3

    There are several ways to show the inequality

    You can study f(x) = x^{k+1}-(k+1)x+k
    f is differentiable and f'(x) = (k+1)(x^k-1)
    The minimum of f is obtained for x=1 and f(1)=0 therefore f(x) \geq 0

    Second way : since f(1)=0 and f'(1)=0, f(x) can be factorized by (x-1)
    You can show that f(x) = (x-1) \left(\sum_{j=1}^{k}x^j-k\right)
    and f(x) = (x-1)^2 \left(\sum_{j=0}^{k-1}(k-j)x^j\right)
    Both factors are positive for x positive
    Nice.

    A third way: induction over k. The case k = 2 has already been established.

    Suppose now that
    x^k -kx +k-1 \geq 0
    Then
    x^k \geq kx - k + 1
    Multiplying by x, since x \geq 0,
    x^{k+1} \geq kx^2 -kx + x\geq k (2x-1) -kx + x ... by the k=2 case
    x^{k+1} \geq (k+1)x -k
    x^{k+1} -(k+1)x + k \geq 0
    QED
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