The question was posed in a YouTube video as to which was larger: \(3.14^{\pi}\) or \( \pi^{3.14}\) ? The use of a calculator was not allowed. The solution provided in the video is interesting.$$ \begin{align} &3.14^{\pi} \text{ versus } \pi^{3.14} \\ &\text{raise each to the } 1/3.14 \text{ power} \\ &3.14^{\pi /3.14} \text{ versus } \pi^{3.14/3.14} \\ &3.14^{\pi/3.14} \text{ versus } \pi \\ & \text{raise each to the } 1/ \pi \text{ power} \\ &3.14^{\pi/(3.14 \pi)} \text{ versus } \pi ^{1/\pi} \\ &3.14^{1/3.14} \text{ versus } \pi ^{1/\pi} \end{align} $$Now consider the function \(y=x^{1/x}\) which we will modify as follows:$$ \begin{align} y &=x^{1/x} \\ &= e^{\ln{x^{1/x}}} \\ &=e^{1/x \ln{x}} \end{align}$$Let's find its first derivative in order to determine stationary values:$$ \begin{align} \frac{dy}{dx} &=e^{1/x \ln{x}} \frac{d}{dx} (1/x \ln{x})\\ &= \frac{e^{1/x \ln{x}}}{x^2}(1-\ln{x}) \end{align} $$the numerator and denominator on the left are always positive so it comes down to:$$ \begin{align} 1-\ln{x} = 0 \text{ when } x=e \\ \\ \text{if } x<e, dy/dx > 0 \text{ and if } x>e, dy/dx < 0 \end{align}$$Thus there is a maximum turning point at \(x=e\) and since, after this point, \(y\) decreases as \(x\) increases, we must have:$$3.14^{1/3.14} > \pi^{1/\pi} \text{ since } 3.14 < \pi$$This in turn means that by reversing our original transformation, we have:$$3.14^{\pi} > \pi^{3.14}$$Figure 1 shows what the graph of \(y=x^{1/x}\) look like.
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Figure 1: note the gradual descent as \(x\) gets larger |
The approximate values of our two expressions are as follows:$$ \begin{align} 3.14^{\pi}&\approx 36.4041195357888 \\ \pi^{3.14} &\approx 36.3957438848941 \end{align}$$The solution to this problem also settles the more general issues surrounding when \(a^b\) is larger or smaller than \(b^a\) with \(a,b>1\). Let's consider the problem of:$$a^b \text{ versus } b^a \text{ for the particular case of }2024^{2025} \text{ versus } 2025^{2004}$$We know from earlier that this comparison reduces to:$$ \begin{align} &a^{1/a} \text{ versus } b^{1/b} \text{ or } 2024^{1/2024} \text{ versus } 2025^{1/2005} \\ &\text{and if }a>e, b>e \text{ and } a<b \text{ then:} \\ &\text{because } 2024 < 2025 \text{ we have:}\\ &2024^{2025} > 2025^{2004} \end{align} $$ However, what about:$$ \phi^{1.618} \text{ versus } 1.618^{\phi}$$Here the problem reduces to:$$\phi^{1/\phi} \text{ versus } 1.618^{1/1.618}$$Here \(\phi<e\) and \(1.618<e\) and \(1.618 < \phi\), so:$$ 1.618^{1/1.618} < \phi ^{1/\phi} \text{ and } 1.618^{\phi}<\phi^{1.618}$$Double checking we find that:$$ \begin{align} 1.618^{\phi} \approx 2.17838352536880\\ \phi^{1.618} \approx 2.17842193783803 \end{align}$$If the values of \(a\) and \(b\) fall on either side of \(e\) then the previous methods don't work. I'll leave this problem for another post possibly as I've spent enough time on this.
