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Question about formula for natural log

  1. Oct 26, 2011 #1
    Any information on the following formula for natural logarithm (I looked in wikipedia and Mathworld but didn't see it). It came from another equation I was working on a bit ago, and I was curious about it as I didn't recall seeing it before (which doesn't mean I haven't), although it reminded me of some equations for e.

    [tex] \ln{x} =\lim_{n\to\infty} \left[ \left (1- \frac{1}{x^{\frac{1}{n}}} \right) \times n \right ] [/tex]
    For better visibility (bottom of the fraction is the nth root of x):
    [tex]\ln{x} =\lim_{n\to\infty} \left[ \left (1- \frac{1}{ \sqrt[n]{x}} \right) \times n \right ] [/tex]

    Or yet another form:

    [tex] \ln{x} =\lim_{n\to\infty} \left[ \left (1- x^{- \frac{1}{n}} \right) \times n \right ] [/tex]
    And I might have answered my own question with this last one... sheesh... anyways, still would like to read about it.
    [tex] \ln{x} =\lim_{n\to\infty} \left[n- n\times x^{- \frac{1}{n}} \right ] [/tex]

    Makes the derivative readily apparent, ehh? :D
     
    Last edited: Oct 26, 2011
  2. jcsd
  3. Oct 26, 2011 #2

    lurflurf

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    It follows from
    [tex] e^x =\lim_{n\to\infty} \left (1+\frac{x}{n}\right)^n [/tex]
     
  4. Oct 26, 2011 #3
    Thanks, I realized that post-post, right after I reformulated it a last time and powered down the computer. Ended up writing it out on paper and deriving this particular formula for [itex]e^x[/itex]:

    [tex] x = \lim_{n\to\infty} \left(1- \frac{\ln{x}}{n} \right )^{-n} [/tex]
    which is basically the following reformulated~~~
    [tex] e^x = \lim_{n\to\infty} \left(1- \frac{x}{n} \right )^{-n} [/tex]
     
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