# Minimize function

ark13
Posers and Puzzles 31 Oct '05 23:27
1. ark13
Enola Straight
31 Oct '05 23:27
f(x)=x^x
What positive value of x minimizes the function? Just an interesting thing I discovered...
2. XanthosNZ
Cancerous Bus Crash
31 Oct '05 23:38
0.36788 (5 sf)
3. ark13
Enola Straight
31 Oct '05 23:43
Originally posted by XanthosNZ
0.36788 (5 sf)
Right, but what is that exactly equal to?
4. 01 Nov '05 03:131 edit
Originally posted by ark13
f(x)=x^x
What positive value of x minimizes the function? Just an interesting thing I discovered...
x = 1/e is the solution.

used log differentiation to get f'(x) = (1 + ln(x)) x^x. Set this to 0, and x^x cannot be 0, so (1+ln(x)) must be 0, so x = 1/e. I am not going to test for concavity, because it is pretty clear the x^x is concave up.

Is that a satisfactory answer for you, ark?
5. ark13
Enola Straight
01 Nov '05 21:56
Originally posted by rheymans
x = 1/e is the solution.

used log differentiation to get f'(x) = (1 + ln(x)) x^x. Set this to 0, and x^x cannot be 0, so (1+ln(x)) must be 0, so x = 1/e. I am not going to test for concavity, because it is pretty clear the x^x is concave up.

Is that a satisfactory answer for you, ark?
Correct! Great job. ðŸ™‚

What about the positive value of x (in terms of e) which minimizes f(x)=x^(x^x) ? This one I don't know the answer to yet.
6. PBE6
Bananarama
01 Nov '05 22:16
Originally posted by ark13
Correct! Great job. ðŸ™‚

What about the positive value of x (in terms of e) which minimizes f(x)=x^(x^x) ? This one I don't know the answer to yet.
I just checked it out. That function increases monotonically, and the limit of the function as you approach 0 is 0, so there is no actual minimum. You can say 0 is the greatest lower bound for this function, however.
7. XanthosNZ
Cancerous Bus Crash
01 Nov '05 22:45
Originally posted by PBE6
I just checked it out. That function increases monotonically, and the limit of the function as you approach 0 is 0, so there is no actual minimum. You can say 0 is the greatest lower bound for this function, however.
f(0) doesn't exist. Neither does f(x) for any negative non-integer value (and x=-0.2).
The limit as x approaches zero from the positive side (there is no limit approaching from the negative side) is 1.

The lower bound of the function is in fact -1.379 (at x=-0.2). The lower bound in the positive domain is 1/e.
8. PBE6
Bananarama
01 Nov '05 23:13
Originally posted by XanthosNZ
f(0) doesn't exist. Neither does f(x) for any negative non-integer value (and x=-0.2).
The limit as x approaches zero from the positive side (there is no limit approaching from the negative side) is 1.

The lower bound of the function is in fact -1.379 (at x=-0.2). The lower bound in the positive domain is 1/e.
You're checking out the wrong function, XanthosNZ. ark13 is asking about x^(x^x), not (x^x)^x. x^(x^x) tends to 0 as x approaches 0 from the positive side, whereas (x^x)^x tends to 1 as you stated. ark13 is also asking about the positive value(s) that minimize said function. Although x=-0.2 does give a value less than 0 for both functions, it does not minimize either function, as any odd negative integer for x will give you a more negative value. But you are right when you say f(0) does not exist (for either function).
9. ark13
Enola Straight
01 Nov '05 23:461 edit
Whoops, sorry I meant f(x)=(x^x)^x ðŸ˜³ Sorry about that. And yes, I did want the positive value of x.

My calculations gave the positive x min at approx. 0.60653176 and the value at .83198595

I was just curious if anyone could use logarithms to find the x value in terms of e as was done on the previous problem.
10. PBE6
Bananarama
02 Nov '05 00:50
Originally posted by ark13
Whoops, sorry I meant f(x)=(x^x)^x ðŸ˜³ Sorry about that. And yes, I did want the positive value of x.

My calculations gave the positive x min at approx. 0.60653176 and the value at .83198595

I was just curious if anyone could use logarithms to find the x value in terms of e as was done on the previous problem.
Hmm, it looks like I was checking out the wrong function. Curses!!
11. ark13
Enola Straight
02 Nov '05 02:07
Originally posted by PBE6
Hmm, it looks like I was checking out the wrong function. Curses!!
Yep, my bad. It was what I said, but not what I meant to say. And although quoting Dr. Seuss will only confuse people further, as it's contrary to what I just said, I will anyway. "I meant what I said, and I said what I meant."
12. 02 Nov '05 02:10
Originally posted by ark13
Whoops, sorry I meant f(x)=(x^x)^x ðŸ˜³ Sorry about that. And yes, I did want the positive value of x.

My calculations gave the positive x min at approx. 0.60653176 and the value at .83198595

I was just curious if anyone could use logarithms to find the x value in terms of e as was done on the previous problem.
So, your function is x^(x^2) then. I only get one min at 1/(sqrt(e)).
13. ark13
Enola Straight
02 Nov '05 02:15
Originally posted by rheymans
So, your function is x^(x^2) then. I only get one min at 1/(sqrt(e)).
Nice, you did it again! It's 1/ sqrt (e)

The result is easy enough to get, the trick is to find it in terms of e.
14. 02 Nov '05 02:21
Originally posted by ark13
Nice, you did it again! It's 1/ sqrt (e)

The result is easy enough to get, the trick is to find it in terms of e.
ðŸ˜€
15. ark13
Enola Straight
02 Nov '05 02:34
Originally posted by rheymans
ðŸ˜€
From these results, we can assume that the positive minimizing value of x in x^x^x^x is 1/cubrt (e), and etc. Almost makes you wonder why names of operations stop at exponents. Why isn't there an operation that is the number of times a number is put to it's own power? Well, I guess they had to stop somewhere.