Discussion about math, puzzles, games and fun. Useful symbols: ÷ × ½ √ ∞ ≠ ≤ ≥ ≈ ⇒ ± ∈ Δ θ ∴ ∑ ∫ π -¹ ² ³ °

You are not logged in.

- Topics: Active | Unanswered

Pages: **1**

**Harold****Guest**

What is the solution of this problem?and how is the problem solved-

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,274

Hi Harold;

This one has been around for a long time. The standard answer starts with raising both sides to the power of

after that it is a maxima-minima problem.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

**Online**

**Harold****Guest**

You mean e^e is always bigger than pi^pi?but why?

**bobbym****Administrator**- From: Bumpkinland
- Registered: 2009-04-12
- Posts: 84,274

Hi;

e^e is not greater than π^π. That is not what I said. You did something wrong with the first step.

From here it is an ugly calculus problem.

**In mathematics, you don't understand things. You just get used to them.I have the result, but I do not yet know how to get it.All physicists, and a good many quite respectable mathematicians are contemptuous about proof.**

**Online**

**bob bundy****Moderator**- Registered: 2010-06-20
- Posts: 6,294

Looks to me that y = x^(1/x) has a single maximum at x = e.

See graph and derivative graph.

Bob

You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei

Offline

**scientia****Member**- Registered: 2009-11-13
- Posts: 222

Let ; then when . So is decreasing for ; as , , i.e. .

Offline

Pages: **1**