Xth root of X question

Discussion in 'Math' started by Mike M., Oct 15, 2007.

  1. Mike M.

    Thread Starter Active Member

    Oct 9, 2007
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    Why is x^(1/x) maximum when x=the number "e"? The graph starts at (0,0) rises to (e, about 1.4) and then becomes asymptotic to 1. I am not very strong in the math department past geometry and algebra 1 so, if someone has an answer, please try to dumb it down a little for me.
     
  2. Dave

    Retired Moderator

    Nov 17, 2003
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    I have just plotted this in Matlab and indeed you are correct.

    I haven't been through the rigourous proof of this, but for y = x^(1/x), the reason it peaks where x = e is because at this point dy/dx is equal to 0, i.e. the gradient is constant and d^2y/dx^2 (the second derivative) at x = e is negative indicating a maximum point. I may try and test the maths at a later point by this is the crux of the mathematical analysis of the function.

    Dave
     
  3. thingmaker3

    Retired Moderator

    May 16, 2005
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    Idle curiosity prompts me to ask if "about 1.4" means "square root of 2?"
     
  4. Mike M.

    Thread Starter Active Member

    Oct 9, 2007
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    1.44466786101........ and I assume it goes on forever since it is related to "e". Just do e^(1/e).
     
  5. Dave

    Retired Moderator

    Nov 17, 2003
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    I calculated it to be 1.444667860997891, but I except there is a small error in my calculations as I am working on a sample density of 0.0001. Root-2 is ~1.4142 so it is most probably a coincidence that they are similar.

    Dave
     
  6. Mike M.

    Thread Starter Active Member

    Oct 9, 2007
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    That is a little over a 2.15% discrepency.......seems like a lot to be related. If you raise it to the 16th power (1.444667860997891^(2^(2^2))), you get almost exactly 360, like degrees in a circle.......maybe just a coincidence though because 360 is invented by humans. You can divide 360 by every number from 1-10 evenly, with the exception of 7. Really, a circle should have 2520 degrees so that it can be evenly divided by EVERY number between 1-10.
     
  7. Eduard Munteanu

    Active Member

    Sep 1, 2007
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    The point where the first derivative is zero and changes sign is a local maximum. So one needs to show e is the global maximum.
     
  8. Dave

    Retired Moderator

    Nov 17, 2003
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    I concur with the assessment that the discrepancy is to great to be correlated. I must say, like thingmaker3, when I saw you write that it rises to about 1.4 I did think about root-2 as the max value.

    I'm not sure about the significance of the 360. Interesting point.

    Indeed so, I truncated my calculations from 0-15 so that I could see the local maximum around x = e. The global maximum can be assumed if one can accurately deduce that as x \rightarrow \infty then y (I'll assume we're calling it y) tends to zero asymptotically - simple enough for the mathematicians amongst us!

    Remember the second derivative must be negative for it to be a maximum. If its positive its a minimum, and if its zero it is a point of inflection. Therefore, it is a little more than a change of sign.

    Dave
     
  9. randomEE

    New Member

    Oct 16, 2007
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  10. Mike M.

    Thread Starter Active Member

    Oct 9, 2007
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    If I were a mathematician it probably would help because I do see a pattern similarity but............

    Damn that hurt my brain!!!!!:D
     
  11. recca02

    Senior Member

    Apr 2, 2007
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    i have tried this (many may have already solved it)
    one reason i didnt solve it was it can't be solved on calculator and i have rusted my once sharp mathematics skills + the pain to type it without the availability of logs and sq etc as fonts.
    Y=x^(1/x)
    we first take natural log
    logy = (1/x)logx---------(A)
    note log means ln here

    d/dx of above eqn gives.
    1/y.dy/dx = (1/sq.x)(1-logx)----(1)

    equate to zero we have e as answer(or minima/maxima).
    0= (1/sq.x)(1-logx)
    gives possible roots as x= +-infinity, and logx=1 meaning x=e.

    i think infinity will lead us to minimum in this case so lets discuss only x=e.

    again d/dx of (1)
    d2(logy)/dx2= -3/cu.(x) + 2logx/cu.(x)
    substituting we get,

    d2(logy)/dx2 at x=e = -3/cu.e + 2/cu.e= -1/cu.e
    which has to be -ve (-.049 something)

    hence we get a maxima at 'e' for the above equation.

    and since log y will only have a max at max y since log is an cont increasing function hence this point of maxima is pt of maximum for y.
    substituting value of x=e in eqn (A) we get y(max)=1.44466

    cu means raised to 3 and sq means raised to 2.

    i think i have already embarassed myself;
    if possible please point out corrections.
     
  12. Mike M.

    Thread Starter Active Member

    Oct 9, 2007
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    I am confused. I didn't follow the logs..........or really anything else.
     
  13. Dave

    Retired Moderator

    Nov 17, 2003
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    Where is the 1 from here recca?

    Dave
     
  14. recca02

    Senior Member

    Apr 2, 2007
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  15. Mike M.

    Thread Starter Active Member

    Oct 9, 2007
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    I apologize recca. I am better than average with the meanings and applications of integrals, minimum/maximum, and logarithms but I just have been out of the game so long by using my damn calculator as a crutch that I need a good kick in the a** 10-12 week crash course just to get back to where I left off from High School before I went into the Navy. Math is a language as well as an art and I think I was born with a deficiency in both, with the exception of english. I thought this question would have a simple enough answer to easily understand when I asked it but I can tell now that I was in over my head. It isn't that your explaination isn't good, it's just that I have gone math dumb over the years and now HAVE to resort to a calculator for complex things instead of it being OPTIONAL. I really need a crash course to get me back on track from the retard mode I was forced into in the Navy........only sleeping 2 hours a day for 6 years didn't work out so well with the math section in my brain.
     
  16. recca02

    Senior Member

    Apr 2, 2007
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    math is just a way of bringing figures, equations and graphs into paper/writing.
    and i personally believe calculus can be learned just by understanding what they mean formulaes can be derived later and much much more can be learned by trying to apply these concepts in real life situations.there is nothing more to it ,it is just a tool by no means does its knowledge make a human superior or inferior-this is for those who get such an impression.
    another method is to somehow make a plot (rough one shud do). and find the peak of the plot.
    point of maxima and minima on a graph are nothing but where the rate (here is where derivative comes handy-it gives nothing but rate with respect to the 'dx' part) of change in function (can be increasing rate or decreasing) becomes zero. thus meaning there is no further change in that direction for that curve/graph (increasing or decreasing) thus defining it as either a point of maxima of minima.

    edit: I have tried to make the above solution simpler to understand.
     
  17. Dave

    Retired Moderator

    Nov 17, 2003
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    Ok, I understand the derivatives however was confuse by the spurious number 1. I'll have a work through the rest of the derivation.

    Dave
     
  18. recca02

    Senior Member

    Apr 2, 2007
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    i know that 1 did seem a bit spurious but now that i think of it, it seems sort of ironical that out of all the calculus one was confused with a number from which he started learning maths.:D ROTFL
     
  19. Dave

    Retired Moderator

    Nov 17, 2003
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    Lol! The smallest cog will always break the largest gearbox! It is much clearer now, and from a brief browse through I cannot see at the moment any obvious errors. When I get a minute later I'll jot it down on paper and try it myself.

    Dave
     
  20. recca02

    Senior Member

    Apr 2, 2007
    1,211
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    look what i found?
    links:
    http://www.ebtx.com/ntx/ntx33a1.htm
    http://answers.yahoo.com/question/index?qid=1005120802746

    there is some relation betn e and pi those eqn tell me not to bother (i've got another examination coming up) so i'm off.
     
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