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Thread: Prove

  1. #1
    Super Member dhiab's Avatar
    May 2009


    n is a naturel number
    Attached Thumbnails Attached Thumbnails Prove-09.jpg  
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  2. #2
    MHF Contributor

    Apr 2005
    So the problem is to show that $\displaystyle 2^{n+1}- 2^n+ 1= 0$ has at least on solution between 2 and $\displaystyle \frac{2n}{n+1}$.

    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
    Super Member
    Jun 2009

    The question is lacking a bit in definition, (is $\displaystyle n$ 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 $\displaystyle n$ is an integer greater than 1, it is saying, by way of example, that if

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

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

    To show that the result is correct in general, define

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

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

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

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