- 01 Nov '05 03:13 / 1 edit

x = 1/e is the solution.*Originally posted by ark13***f(x)=x^x**

What positive value of x minimizes the function? Just an interesting thing I discovered...

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? - 01 Nov '05 21:56

Correct! Great job.*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?

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. - 01 Nov '05 22:16

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.*Originally posted by ark13***Correct! Great job.**

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. - 01 Nov '05 22:45

f(0) doesn't exist. Neither does f(x) for any negative non-integer value (and x=-0.2).*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.**

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. - 01 Nov '05 23:13

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).*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. - 01 Nov '05 23:46 / 1 editWhoops, sorry I meant f(x)=(x^x)^x 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. - 02 Nov '05 00:50

Hmm, it looks like I was checking out the wrong function. Curses!!*Originally posted by ark13***Whoops, sorry I meant f(x)=(x^x)^x 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. - 02 Nov '05 02:07

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."*Originally posted by PBE6***Hmm, it looks like I was checking out the wrong function. Curses!!** - 02 Nov '05 02:10

So, your function is x^(x^2) then. I only get one min at 1/(sqrt(e)).*Originally posted by ark13***Whoops, sorry I meant f(x)=(x^x)^x 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. - 02 Nov '05 02:34

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.*Originally posted by rheymans*