This post is in ref. to my earlier post's row #10.
- $\{n^\frac 1n: n \in \mathbb{N} \}$:
Domain of values is in the set of naturals.
I am unable to find minimum, maximum (in range), and list a few values below:
$$\begin{array}{c|c|} & \text{$n\in \mathbb{N}$}& \text{$n^{\frac1n}$}\\ \hline
a & 1& 1\\ \hline
b & 2& \sqrt{2}\\ \hline
c & 3& 3^{\frac13}\\ \hline
d & 4& 4^{\frac14}\\ \hline
\end{array}$$
The max. /min. value in range of function is unknown by me, hence it needs finding derivative (both first & second).
The eqn. would be :
$y = n^\frac 1n: n \in \mathbb{N}$, with steps to solve being in-complete:
$ \implies \ln y = \frac 1n \ln n \implies \frac 1y y' = \frac 1{n^2}(1-\ln n ) \implies y' = n^\frac 1n\frac 1{n^2}(1-\ln n )$
Need to double differentiate the above, but don't know how to pursue further.
Edit:
Based on the responses (comments, answers) have modified my attempt, that is still incomplete. Request vetting the contents also.:
As the function is exponential, so continuous one; but consider restricted domain of natural numbers, as given:
$y = n^\frac 1n: n \in \mathbb{N}$
As $\log$ is a monotonic function, so $\log y$ will be too.
$ \implies \ln y = \frac 1n \ln n \implies \frac 1y y' = \frac 1{n^2}(1-\ln n ) \implies y' = n^\frac 1n\frac 1{n^2}(1-\ln n )$
In $3$ product terms of $y' = n^\frac 1n\frac 1{n^2}(1-\ln n )$, only last term $(1-\ln n)$ can reduce to $0$ for finite values, i.e. at $x=e$, as $\ln e = 1$.
First approach is to confirm that at $e$ if there is a maxima / minima, & need find by 2nd derivative.
Second approach (as shown in the selected answer) is to take value of fn. at integers surrounding $e$ at $x=2,3$, i.e. $3^{\frac13}, 2^{\frac12}$; it shows max. value at $x=e$.
Coming back to the first approach:
if $y'$ max at $x=e$, then $y''$ is negative there, & vice versa.
$ \implies \ln y = \frac 1n \ln n \implies \frac 1y y' = \frac 1{n^2}(1-\ln n ) \implies y' = n^\frac 1n\frac 1{n^2}(1-\ln n )$
Need to double differentiate the above.
$y' = n^\frac 1n\frac 1{n^2}(1-\ln n) \implies \ln y' = \frac 1n \ln n\frac 1{n^2}(1-\ln n)$
Differentiating w.r.t. $n$ again:
$y'' = \frac{d}{dn}(y'.\frac 1n \ln n\frac 1{n^2}(1-\ln n))\implies \frac{d}{dn}(n^\frac 1n\frac 1{n^2}(1-\ln n).\frac 1n \ln n\frac 1{n^2}(1-\ln n))$
Need help in completing finding the second derivative.
Best Answer
We have $1^{\frac11}=1$ and for any $n>1$, $n^\frac1n > 1$, the minimun and infimum is $1$.
Let $y = x^{\frac1x}$, $$\ln y = \frac{\ln x}{x}$$
$$\frac{d\ln y}{dx}=\frac{d}{dx}\left(\frac{\ln x}{x}\right)= \frac{1-\ln x}{x^2}$$
The value of $y$ increases when $\ln y$ increases. That is when $\frac{d\ln y}{dx}>0$, which is equivalent to $1-\ln x > 0$which is just $\ln x < 1$, taking exponential both sides give us $x < e$.
Hence $y$ increases up to $e$ and then decreases.
Hence for any $x_1, x_2 \in (0,e)$ $x_1 < x_2$ implies that $x_1^{\frac1{x_1}}< x_2^{\frac1{x_2}}$.
For any $x_1, x_2 \in (e, \infty)$ $x_1 < x_2$ implies that $x_1^{\frac1{x_1}}> x_2^{\frac1{x_2}}$.
The only two possible value that could have attained the maximum values are $2$ and $3$.
Since $3^\frac13 > 2^\frac12$, the maximum and supremum is $3^\frac13$.
Remark:
I am working with $\{ n^\frac1n: n \in \mathbb{N} \}$ which is a subset of the real number.
I am not working with $\{ x^\frac1x: x \in \mathbb{R}, x>0 \}$.