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Taking the Nth root of some real number $N$ (ie: $R(N) = N^{1/N}$), generally $R(X) > R(Y)$ when $X < Y$. This obviously isn't the case though when $ X< Y < 3$. Put another way, starting at $N = 0.1$ and increasing $N$ by $0.1$ if we plot each R(N) we see that the result increases and peaks just before it reaches $3$ and then steadily decreases. Can anyone shed some light on this behavior?

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3 Answers 3

Write $x^{1/x} = \exp (\frac1x \log x)$. Since the exponential function is monotonic (increasing), it is sufficient to look at the behaviour of $f(x) = \frac1x \log x$. Its derivative is

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

and that is positive for $\log x < 1 \iff x < e$, and negative for $x > e$, hence it has a (unique) maximum at $x = e$, and therefore $x^{1/x}$ also has a (unique) maximum at $x = e$.

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Yes, that makes perfect sense. Thank you! –  SeaBass Jul 14 '13 at 20:41

The peak is at $e$. $$x^{1/x}=\exp(\frac1x\ln x)$$ is maximal when $\frac1x\ln x$ is maximal. The derivative of this is $-x^{-2}\ln x + x^{-2}$ and becomes zero when $\ln x=1$.

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Got it, thanks. –  SeaBass Jul 14 '13 at 20:44

Let $y=x^{1/x}$ for $x>0$. Then $\ln y=\frac{\ln x}x$. Now differentiate with respect to $x$:

$$\frac{y'}y=\frac{1-\ln x}{x^2}\;,$$

so

$$y'=\frac{x^{1/x}}{x^2}(1-\ln x)\;.$$

This is $0$ precisely when $\ln x=1$, i.e., when $x=e\approx2.718$. Moreover, $\frac{x^{1/x}}{x^2}>0$, and $\ln x$ is an increasing function of $x$, so $y'>0$ for $0<x<e$ and $y'<0$ for $e<x$, and therefore $y=x^{1/x}$ attains its maximum at $x=e$.

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Thanks, another great explanation! –  SeaBass Jul 14 '13 at 20:41
    
@SeaBass: You're welcome. –  Brian M. Scott Jul 14 '13 at 20:45

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