Notices
Results 1 to 4 of 4

Thread: (1+x)^1/x near zero

  1. #1 (1+x)^1/x near zero 
    Forum Freshman
    Join Date
    Jun 2009
    Posts
    11
    I noticed some unexpected behavior in the real-valued f(x)=(1+x)^1/x, as a function of real numbers, when plotting it on wolfram alpha. I inputed:

    plot (1+x)^1/x from x=-0.0000001 to x=0.0000001

    and saw that it unexpectedly seemed to oscillate near zero. I took a closer look with:

    plot (1+x)^1/x from x=-0.00000000001 to x=0.00000000001

    and saw that it definitely seems to oscillate near zero. My original rough graph on paper using a hand calculator suggested the curve was smooth near zero, and even windows calculator's 32 decimal places were unable to reveal the oscillation when I manually calculated many different values near zero.

    I don't think f(x) is smooth near zero, though, since

    d/dx[f(x)]=f(x)d/dx[1/xln(1+x)]

    so all of f(x)'s derivatives have a discontinuity at zero. Since Leonhard Euler showed limf(x)=e as x approaches zero, and this limit is the same as you approach zero from the left or from the right, then the discontinuity of f(x) at zero is removable and the function

    f(x)=(1+x)^1/x, x not = 0
    f(0)=0 x=0

    is continous. But the discontinuities in the derivatives don't appear to be removable. Each derivative will have a term containing the factor 1/x (not the 1/x in the exponent of f(x)) which cannot be removed. E.g.

    d/dx[f(x)]=f(x)[(1/x)(1/(1+x)+(-1/x^2)(ln(x+1))]

    the 1/x in the 1st term is not removable.

    I'm not saying it doesn't oscillate near zero, I'm just saying that this is unexpected. I need confirming opinions and confirming logic to explain it so I can believe it.
    So the questions are:
    1. Does f(x) oscillate near zero?
    2. Why does f(x) oscillate near zero?
    3. Why does this behavior not appear until |x|<~10^-7?
    4. Is it a glitch in Mathimatica? [doubt it]
    5. If it doesn't oscillate [which it probably does] then how is the behavior of the derivatives explained near zero?
    6. If it oscillates, then why is it so hard to replicate this oscillation manually on a calculator?

    Note#1: This behavior concerns the real-valued f(x). I don't think Re[f(z)] or Im[f(z)], z complex, are of interest here.
    Note#2: This behavior concerns (1+x)^1/x. I dont think either (1+1/x)^x or lim(1+1/x)^x as x goes to infinty are of interest here.

    Thx in advance and Kudos to the person with the explanation!


    Reply With Quote  
     

  2.  

    Related Discussions:

       

    • #2  
      Forum Professor
      Join Date
      Jul 2008
      Location
      New York State
      Posts
      1,027
      The oscillations of f(x) are an artifact of the plotting program.

      Your expression for the derivative is incorrect. It should read:
      d/dx[f(x)]=f(x)d/dx[(ln(1+x))/x]

      To first order the function is e(1 - x/2).


      Reply With Quote  
       

    • #3  
      Forum Bachelors Degree
      Join Date
      Apr 2007
      Location
      Veles,Macedonia
      Posts
      473
      y=(1+x)^(1/x)

      You need to find y as x --> 0.

      ln(y)=(1/x)*ln(1+x)



      we got indeterminate form so we can apply the L'Hopitals rule:



      So ln(y) --> 1 as x --> 0. That means as x-->0
      or y --> e as x-->0
      Reply With Quote  
       

    • #4  
      Forum Freshman
      Join Date
      Jun 2009
      Posts
      11
      Thx all for the replies. I should have known the problem was mathematica's [wolfram alpha] behavior when doing certain calculations.
      Reply With Quote  
       

    Bookmarks
    Bookmarks
    Posting Permissions
    • You may not post new threads
    • You may not post replies
    • You may not post attachments
    • You may not edit your posts
    •