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Math Help - Prove

  1. #1
    Super Member dhiab's Avatar
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    Prove

    n is a naturel number
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  2. #2
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    So the problem is to show that 2^{n+1}- 2^n+ 1= 0 has at least on solution between 2 and \frac{2n}{n+1}.

    I'm sorry but that makes no sense. If n is "unknown" what does it mean to say that a solution is between 2 and a function of n?
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  3. #3
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    ?

    The question is lacking a bit in definition, (is n an integer and is it greater than 1 ?), but it does make sense.

    (BTW, I'm still learning LaTex, can anyone tell me how to line up these n's with the rest of the text ?)

    Assuming that n is an integer greater than 1, it is saying, by way of example, that if

    n=2, the equation x^3-2x^2+1=0 has a root between x=2 and x=4/3. It does, 1.618 (3d).

    n=3, the equation x^4-2x^3+1=0 has a root between x=2 and x=6/4. It does, 1.839 (3d).

    To show that the result is correct in general, define

    f(x)=x^{n+1}-2x^n+1.

    Then, f(2)=2^{n+1}-2.2^n+1=1>0, and showing that f(\frac{2n}{n+1})<0 will be sufficient.

    f(\frac{2n}{n+1})=\frac{(2n)^{n+1}}{(n+1)^{n+1}}-2\frac{(2n)^n}{(n+1)^n}+1=1-\frac{2^{n+1}n^n}{(n+1)^{n+1}}=1-\frac{2^{n+1}/(n+1)}{(1+1/n)^{n}}.

    Without going into the analysis,the second term is always greater than 1, ( n\geq2)
    (Check it out numerically for n=2 and n=3 and then use the fact that the numerator is an increasing function, while the denominator tends to e from below).
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