## Puzzles: solution to problem 1

Without using a calculator, work out which is larger, $e^{\pi}$ or $\pi^e$.

First let us think about strictly monotone increasing functions. A function $f(x)$ is strictly monotone increasing if $f(x) > f(y)$ if and only if $x>y$. Now, $e^x$ is strictly monotone increasing, therefore (why?) so is its inverse, $\ln(x)$ for $x>0$. Now let’s go back to the problem.

Suppose we want to prove that $x^y > y^x$, for $x, y > 0$. Since $\ln$ is strictly monotone increasing, $x^y > y^x$ is logically equivalent to $\ln(x^y) > \ln(y^x)$, i.e. $y\ln(x) > x\ln(y)$, which is the same as $(1/x)\ln(x) > (1/y)\ln(y)$, which is the same as $\ln\left(x^{1/x}\right) > \ln\left(y^{1/y}\right)$. But since $\ln$ is strictly monotone increasing, this is the same as $x^{1/x}> y^{1/y}$. So what have we shown? That $x^y < y^x$ if and only if $x^{1/x} > y^{1/y}$. This is good, as now we can use calculus.

Consider $g(x)=x^{1/x}$. Then by the chain rule (check!),

$g'(x)=\dfrac{g(x)}{x^2}(1-\ln(x)).$

So the only critical point of $g(x)$ is at $x=e$. This clearly is a maximum, as $g(e) > 1$ and $g(x) \rightarrow 0$ as $x \rightarrow \infty$ (by l’Hôpital; check!) But then $e^{1/e} > x^{1/x}$ for any $x \neq e$ and so in particular $e^{1/e} > \pi^{1/\pi}$, which as we have shown is equivalent to $e^\pi > \pi^e$. They are pretty close!  $e^\pi \approx 23.14069263$ and $\pi^e \approx 22.45915771$

(Brownie points to Michael Nolan, but no cigar.)

(MG)

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### One Response to Puzzles: solution to problem 1

1. Nigel says:

Nice general proof…but can I not just use the Maclaurin/Taylor series, e^x=1+x+x^2/2+…, so I know that e^x is greater than 1+x…
e^x>1+x
then let x=y-1 so
e^(y-1)>y
multiplying both sides by e gives
e^y>y*e
then let y=pi/e to get
e^(pi/e)>pi
now just raise both sides of the equation to the power e and we get
e^pi>pi^e

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