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Re: [PrimeNumbers] Primes between consecutive squares

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  • Peter Kosinar
    Hi Kermit, ... Unless I ve made a mistake, the first and second derivative look like this: f (x) = x/log(x) - 1/(2*log^2(x)) f (x) = 1/log(x) -
    Message 1 of 3 , Dec 7, 2010
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      Hi Kermit,

      > f(x) = x**2/ln(x**2) = x**2/(2 ln(x))
      >
      > I calculate the first and second derivatives of this
      > approximation.
      >
      > [ long calculation snipped ]
      >
      > showing that the second derivative is negative for all x.

      Unless I've made a mistake, the first and second derivative
      look like this:

      f'(x) = x/log(x) - 1/(2*log^2(x))
      f''(x) = 1/log(x) - (3/2)*(1/log^2(x)) + 1/log^3(x)

      Unfortunately, they're both positive for x > sqrt(e) :-)

      Peter
    • Kermit Rose
      ... Thank you. I trust your calculations more than my own. Kermit.
      Message 2 of 3 , Dec 7, 2010
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        On 12/7/2010 11:34 PM, Peter Kosinar wrote:
        >
        > Hi Kermit,
        >
        >> f(x) = x**2/ln(x**2) = x**2/(2 ln(x))
        >>
        >> I calculate the first and second derivatives of this
        >> approximation.
        >>
        >> [ long calculation snipped ]
        >>
        >> showing that the second derivative is negative for all x.
        >
        > Unless I've made a mistake, the first and second derivative look like this:
        >
        > f'(x) = x/log(x) - 1/(2*log^2(x))
        > f''(x) = 1/log(x) - (3/2)*(1/log^2(x)) + 1/log^3(x)
        >
        > Unfortunately, they're both positive for x > sqrt(e) :-)


        Thank you.

        I trust your calculations more than my own.

        Kermit.
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