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Definition Of Natural Logarithm

  1. Jul 18, 2012 #1
    Hello,

    The definition is [itex]ln(x) = \int_1^x\frac{1}{t}dt[/itex]

    I have read several sources regarding this, but what I can't seem to find is why it was defined this way. What is the justification for defining it this way, and how was ln (x) found to be the same as the that particular integral?
     
  2. jcsd
  3. Jul 18, 2012 #2
    All of the usual properties of the natural log can be verified if you use this definition: ln(ab)=ln(a)+ln(b), ln(a^b)=b ln(a), ln(1)=0, etc. And then we can define e as the number such that ln(e)=1, and use that definition to prove that e is equal to the limit of (1 + 1/n)^n as n goes to infinity.
     
  4. Jul 18, 2012 #3

    jbunniii

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    Another way to look at it, assuming that the exponential function exp has already been defined, it is that ln is the inverse function of exp. One key characteristic of exp is that if

    y = exp(x)

    then

    dy/dx = exp(x) = y

    Hence the inverse function

    x = ln(y)

    must satisfy

    dx/dy = 1/y

    Writing t instead of y gives us

    dx/dt = 1/t

    Now integrate both sides:

    [tex]\ln(a) = x(a) = \int_c^a \frac{1}{t} dt[/tex]

    where c is some constant. To find out its value, recognize that we must have ln(1) = 0 [because exp(0) = 1], so c must satisfy

    [tex]\int_c^1 \frac{1}{t} dt = 0[/tex]

    which forces c = 1.
     
  5. Jul 18, 2012 #4

    Fredrik

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    If you go with jbunniii's approach, you will want to know more about why the exponential function is defined the way it is. My posts here can help you with that.
     
  6. Jul 18, 2012 #5

    jbunniii

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    Nice writeup! I added a few comments at the end of that thread in case you're interested: one is just to note a typo, another is to clarify that a smooth function isn't necessarily analytic, and finally I wondered aloud if there could be a nondifferentiable function satisfying f(x+y) = f(x)f(y) for all x,y.
     
  7. Jul 18, 2012 #6

    Fredrik

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    Cool. Thanks for checking the details, and for letting me know that there are non-analytic smooth functions. I must have learned that at some point, and then quickly forgotten all about it.
     
  8. Jul 18, 2012 #7
    Non-analytic smooth functions are important in distribution theory, because typically the test function space is the set of bump functions, i.e. smooth functions of compact support (except for tempered distributions whose test functions just need to have derivatives vanish sufficiently quickly).
     
  9. Jul 19, 2012 #8

    Bacle2

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    I think the standard example is:

    e-1/x2 , x≠0

    0, if x=0 .
     
    Last edited: Jul 19, 2012
  10. Jul 19, 2012 #9

    jbunniii

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    Yes, this function has the striking feature that

    [tex]f^{(n)}(0) = 0[/tex]

    for all [itex]n[/itex], so its Taylor series is identically zero.
     
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