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What is maximum value among $1, 2^{1/2}, 3^{1/3}, 4^{1/4},....$ ?

My approach: let $f(x)=x^{1/x}$ then I found out the derivative of $f$. Since $f(x)$ is maximum where $f'(x)=0$ and $f''(x)<0$

But it's not working. Is there any other way?

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  • $\begingroup$ Taking log should make the calculus easier. $\endgroup$
    – Sandeep Silwal
    Apr 18, 2014 at 3:18
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    $\begingroup$ math.stackexchange.com/questions/116112/… $\endgroup$
    – lab bhattacharjee
    Apr 18, 2014 at 3:19
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    $\begingroup$ The above link shows that the maximum of $f(x)$ occurs at $x=e$, so you need to figure out which of $2^{1/2}$ and $3^{1/3}$ is larger. $\endgroup$
    – eeeeeeeeee
    Apr 18, 2014 at 3:22
  • $\begingroup$ I tried taking log. Answer is coming as x=10. where as the answer should ideally be 3^(1/3). $\endgroup$
    – Rudstar
    Apr 18, 2014 at 3:24
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    $\begingroup$ Perhaps you should show us your calculations, so we can see if you made a mistake. $\endgroup$
    – Greg Martin
    Apr 18, 2014 at 3:33

2 Answers 2

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Your approach is good but, may be, you have had problem with the derivatives. Even if it is easier using logarithms, let us use $$f(x)=x^{\frac{1}{x}}$$ Standard derivation leads, after factoring, to $$f'(x)=-x^{\frac{1}{x}-2} (\log (x)-1)$$ and $$f''(x)=-x^{\frac{1}{x}-4} (-3 x+\log (x) (2 x+\log (x)-2)+1)$$ The first derivative cancels if $\log (x)=1$, that is to say for $x=e$. For this value of $x$, the value of the second derivative is $-e^{\frac{1}{e}-3}$ which is negative. So $x=e$ is a maximum and $e$ is close to $3$.

This is the results for you discrete list of numbers since you can easily show that $2^{\frac{1}{2}}$ and $4^{\frac{1}{4}}$ are both equal to $\sqrt 2$

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  • $\begingroup$ "$e$ is close to $3$" is not a rigorous justification that $3^{1/3}$ is the maximum. $\endgroup$
    – Greg Martin
    Apr 18, 2014 at 7:45
  • $\begingroup$ Yes, it is since $2^{\frac{1}{2}}=4^{\frac{1}{4}}=\sqrt 2 \lt 3^{\frac{1}{3}}$ $\endgroup$
    – Claude Leibovici
    Apr 18, 2014 at 7:58
  • $\begingroup$ What he's saying is that the REASON why you choose $n=3$ being bigger than $n=2$ is not simply because $e$ is close to $3$. This is not rigorous. $\endgroup$
    – Zhuli
    Apr 18, 2014 at 8:12
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We have $f(x)=x^{1/x}$, so $\log(f(x))=\frac{\log(x)}{x}$. The derivative of this is, by the quotient rule:

$f'(x)=\frac{1-\log(x)}{x^2}=0 \iff \log(x)=1 \iff x=e.$

Therefore, the maximum of $f(x)$ is $e^{1/e}$. The closest number in your list to $e^{1/e}$ is $3^{1/3}$.

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  • $\begingroup$ The closest number is not necessary the greater. $\endgroup$
    – Anixx
    Apr 18, 2014 at 3:57
  • $\begingroup$ $e^{1/e}$ is not on the list. Since this is the maximum, the number on the list with the greatest absolute value from $e^{1/e}$ is therefore the maximum of the sequence above. $\endgroup$
    – Fred
    Apr 18, 2014 at 4:01
  • $\begingroup$ what do you mean by "greatest absolute value from"? $\endgroup$
    – Anixx
    Apr 18, 2014 at 4:06
  • $\begingroup$ "The closest number in your list to $e^{1/e}$ is $3^{1/3}$" needs justification. $\endgroup$
    – eeeeeeeeee
    Apr 18, 2014 at 4:10
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    $\begingroup$ Since the only maximum occurs at $e$, that means it's strictly decreasing from this point in either direction. That means the maximum can only be at either $n=2$ or $n=3$. Just evaluate both and find the bigger one. $\endgroup$
    – Zhuli
    Apr 18, 2014 at 4:22

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