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    Find the maximum value of n^(1/n) and prove that it is indeed the maximum

    I think basically youve got to find a differentiate it in some form.

    Ive got it to xlny=lyx and y^x=x

    I dont know if this helps, any ideas anyone
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    (Original post by r_perry)
    Find the maximum value of n^(1/n) and prove that it is indeed the maximum

    I think basically youve got to find a differentiate it in some form.

    Ive got it to xlny=lyx and y^x=x

    I dont know if this helps, any ideas anyone
    n=e^ln n
    =>n^(1/n) = (e^ln n)^1/n =e^((ln n)/n)
    derivative of e^f(n) = f'(n)e^f(n)
    derivative of (ln n)/n = [1 - (ln n)(-n^-2)]/n^2 = [1+(ln n)/n^2]/n^2
    =>derivative of n^(1/n) = [[1+(ln n)/n^2]/n^2]e^((ln n)/n)
    then set that equal to 0 and then show that its a maximum by differentiating again or by drawing a graph
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    *has no idea what he's doing in a thread entitled "for the good mathematician" *

    Good luck mate!!
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    (Original post by r_perry)
    Find the maximum value of n^(1/n) and prove that it is indeed the maximum

    I think basically youve got to find a differentiate it in some form.

    Ive got it to xlny=lyx and y^x=x

    I dont know if this helps, any ideas anyone
    Maximum value occurs when n = e :cool:

    y = n^(1/n) = e^[(1/n)(ln n)]
    y' = [(1-ln n)/n^2]e^[(1/n)(ln n)]
    y' = 0 => (1-ln n)/n^2 = 0 => (1-ln n) = 0 => ln n = 1 => n = e

    To prove its a maximum, differentiate y twice and set n = e and notice it will be < 0:

    y'' = n^(1/n)*(-ln(n)/(n^2) + 1/(n^2))^2 + n^(1/n)*(2*ln(n)/(n^3) - 3/(n^3))

    y''(e) = [-e^(1/e)]/(e^3) = -0.7193 hence is a maximum QED

    Euclid.
 
 
 
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