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Posers and Puzzles

Posers and Puzzles

  1. Standard member 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. Standard member XanthosNZ
    Cancerous Bus Crash
    31 Oct '05 23:38
    0.36788 (5 sf)
  3. Standard member 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 03:13 / 1 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. Standard member 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.

    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. Standard member PBE6
    Bananarama
    01 Nov '05 22:16
    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.
    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. Standard member 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. Standard member 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. Standard member ark13
    Enola Straight
    01 Nov '05 23:46 / 1 edit
    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.
  10. Standard member PBE6
    Bananarama
    02 Nov '05 00:50
    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.
    Hmm, it looks like I was checking out the wrong function. Curses!!
  11. Standard member 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 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. Standard member 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.
  15. Standard member ark13
    Enola Straight
    02 Nov '05 02:34
    Originally posted by rheymans
    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.