How Can We Prove That x^x Reaches a Minimum at x = e^-1?

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Discussion Overview

The discussion revolves around the mathematical exploration of the function x^x and its minima, specifically questioning how to prove that x^x reaches a minimum at x = e^-1. Participants also consider related expressions like (x^x)^x and their minima, suggesting a potential pattern in these minima.

Discussion Character

  • Exploratory
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • Some participants assert that x^x has a minimum at x = e^-1 and seek a proof for this claim.
  • Others propose that (x^x)^x has a minimum at x = e^-(1/2) and that ((x^x)^x)^x has a minimum at x = e^-(1/3), suggesting a pattern.
  • One participant expresses uncertainty about differentiating x^x and finding turning points, relying on calculators instead.
  • Another participant notes that while (x^x) does not have a minimum for the lowest possible x, if y = (x^x), then (y^y) does have a minimum for the lowest possible y, raising questions about the relationship between these functions.
  • There is mention of a method for differentiating x^x available online, with a claim that the minimum occurs at x = 1/e, and a reference to a transcendental equation x ln x = -1 that cannot be solved numerically.
  • Some participants express a desire for clarification on properties of local extrema and how to derive the minima mathematically.

Areas of Agreement / Disagreement

Participants generally agree that x^x reaches a minimum at x = e^-1, but there is no consensus on the methods for proving this or on the behavior of related expressions. Multiple competing views and uncertainties remain regarding the calculus involved and the nature of the minima.

Contextual Notes

Participants express limitations in their calculus skills, which affects their ability to engage with the mathematical proofs and reasoning fully. There are also unresolved questions regarding the properties of local extrema and the relationships between different expressions involving x.

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I've noticed that x^x is a minimum for x = e^-1

I put it as a high school problem because I presume it's one of those simple differential proofs/identities, but I can't really see how to get to e^-1. Too long since I did any calculus. Can someone please show me how to arrive at that?

How about (x^x)^x is a minimum for x = e^-(1/2)?

and ((x^x)^x)^x is a minimum for x = e^-(1/3)

I presume the pattern goes on.
 
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cmb said:
I've noticed that x^x is a minimum for x = e^-1

I put it as a high school problem because I presume it's one of those simple differential proofs/identities, but I can't really see how to get to e^-1. Too long since I did any calculus. Can someone please show me how to arrive at that?

How about (x^x)^x is a minimum for x = e^-(1/2)?

and ((x^x)^x)^x is a minimum for x = e^-(1/3)

I presume the pattern goes on.
How did you find those minima? And what do you remember about the properties of local extrema?
 
If you're asking me to differentiate and find the turning point, sure, I just don't know how to do that.

I did this with a calculator.

In the meantime, it is curious that;-

(x^x) is NOT a minimum for the lowest possible x
BUT
if y=(x^x) then; (y^y) IS a minimum for the lowest possible y

(i.e. (x^x)^(x^x) IS a minimum for the lowest possible (x^x) but NOT the lowest possible x)

Can calculus show that too? I'm just no good at calculus.
 
cmb said:
If you're asking me to differentiate and find the turning point, sure, I just don't know how to do that.

I did this with a calculator.

In the meantime, it is curious that;-

(x^x) is NOT a minimum for the lowest possible x
BUT
if y=(x^x) then; (y^y) IS a minimum for the lowest possible y

(i.e. (x^x)^(x^x) IS a minimum for the lowest possible (x^x) but NOT the lowest possible x)

Can calculus show that too? I'm just no good at calculus.

You can find the method for differentiating ##x^x## on line. There are several videos showing how it's done.

The minimum is when ##x = 1/e##.

If ##y = x^x## then the minimum of ##y^y## is when ##y = 1/e##, hence ##x^x = 1/e##.

This can be simplified to ##x \ln x = -1##, which is a transcendental equation and can ot be solved numerically.
 
cmb said:
I've noticed that x^x is a minimum for x = e^-1

I put it as a high school problem because I presume it's one of those simple differential proofs/identities, but I can't really see how to get to e^-1. Too long since I did any calculus. Can someone please show me how to arrive at that?

How about (x^x)^x is a minimum for x = e^-(1/2)?

and ((x^x)^x)^x is a minimum for x = e^-(1/3)

I presume the pattern goes on.

Yes, this is not too hard to prove using the same technique as minimizing ##x^x##.
 

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