Question

Care dintre numerele m^n si n^m este mai mare daca m si n sunt nr nat,
n mai mic decat m ?

Answer 1

$\text{Dac\u{a} }n < m, \text{ atunci } n^m \, \square{}?\, m^n |^{\frac{1}{mn}} \Leftrightarrow n^\frac{1}{n} \,\square? \,m^{\frac{1}{m}}$

$\text{Consider\u{a}m }y =x^{\frac{1}{x}}$

$\ln y = \ln x^{\frac{1}{x}} \Rightarrow \ln y = \frac{1}{x}\ln x |' \Rightarrow \frac{1}{y}\cdot y' = -\frac{\ln x}{x^2} + \frac{1}{x^2} \Rightarrow$

$\Rightarrow y' = y \cdot \left(-\frac{\ln x}{x^2} + \frac{1}{x^2} \right) \Rightarrow y' = x^{\frac{1}{x}}\cdot \left(-\frac{\ln x}{x^2} + \frac{1}{x^2} \right) \Rightarrow$

$\Rightarrow y' = \frac{x^{\frac{1}{x}}}{x^2}\cdot (1 - \ln x)$

$y' = 0 \Rightarrow 1 - \ln x = 0 \Rightarrow x = e$

$\Rightarrow y' > 0,\text{ c\^{a}nd } x < e \Rightarrow y \text{ - strict cresc\u{a}toare} \,\, \text{(I)}$

$\Rightarrow y' < 0,\text{ c\^{a}nd } x > e \Rightarrow y \text{ - strict descresc\u{a}toare} \,\, \text{(II)}$

$\text{Astfel:}$

$\boxed{\text{Dac\u{a} }n = 0\text{ si }n < m, \text{ atunci }n^m < m^n}\,\,\text{(1)}$

$\text{Din (I)}\Rightarrow \text{Dac\u{a} } n < m < e, \text{ atunci }n^\frac{1}{n} < m^{\frac{1}{m}} \Leftrightarrow n^m < m^n$

$\Rightarrow \boxed{\text{Dac\u{a} }1\leq n < m \leq 2, \text{ atunci } n^{m} < m^{n}}\,\,\text{(2)}$

$\boxed{\text{Dac\u{a} }n = 2,\,m = 3, \text{ atunci }n^m < m^n} \,\,\text{(3)}$

$\text{Din (II)}\Rightarrow \text{Dac\u{a} }e < n < m, \text{ atunci }n^\frac{1}{n} > m^{\frac{1}{m}} \Leftrightarrow n^m > m^n$

$\Rightarrow \boxed{\text{Dac\u{a} }3 \leq n < m, \text{ atunci } n^{m} > m^{n}}\,\, \text{(4)}$