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

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In summary: The minimum of ##(x^x)^x## occurs when ##x = 1/e^{1/2} = e^{-1/2}##, and for ##((x^x)^x)^x## it is at ##x = 1/e^{1/3} = e^{-1/3}##. The pattern seems to continue, with the minimum occurring at ##x = 1/e^{1/n} = e^{-1/n}## where n is the number of nested exponents.
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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?
 
  • #3
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.
 
  • #4
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.
 
  • #5
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##.
 

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

1. What does it mean for x^x to be a minimum?

When we say that x^x is a minimum, it means that the function x^x has the lowest possible value at that point. This is also known as the local minimum, as it is only the lowest value within a small range of values.

2. How do you find where x^x is a minimum?

To find where x^x is a minimum, we need to take the derivative of the function and set it equal to 0. This will give us the critical points, which are potential minimums. We then need to test these points to see which one is the actual minimum.

3. Can x^x have more than one minimum?

Yes, x^x can have more than one minimum. This is because the function is continuous and can have multiple points where the derivative is equal to 0. However, there can only be one global minimum, which is the lowest value of x^x for all possible values of x.

4. What is the difference between a local minimum and a global minimum?

A local minimum is the lowest value of a function within a small range of values, while a global minimum is the lowest value of a function for all possible values. A local minimum can occur multiple times within a function, but there can only be one global minimum.

5. Can x^x have a minimum if x is negative?

Yes, x^x can have a minimum even if x is negative. This is because the negative sign will be raised to the power of x, which can result in a positive value. However, it is important to note that the minimum may not be a real number, as x^x can result in complex numbers when x is negative.

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