# Where x^x is a minimum

• B
• cmb

#### cmb

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.

Last edited:

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.

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.

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##.