Puzzles: solution to problem 1

In Problem 1, we asked:

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!),


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.)


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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…
    then let x=y-1 so
    multiplying both sides by e gives
    then let y=pi/e to get
    now just raise both sides of the equation to the power e and we get

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