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Ok, so this has been bothering me for a while.Integral of (1/x)dx

  1. Sep 24, 2011 #1
    Ok, so this has been bothering me for a while.

    Integral of (1/x)dx = log of x to the base e.

    My question is, why e? Why can't it be instead to a base 3.14? Or any other number.

    I am guessing there is some reason. 2.71 was not chosen at random.

    I have searched a lot but can not seem to find it.

    Any ideas?
  2. jcsd
  3. Sep 24, 2011 #2
    Re: e=2.71

    An interesting history of the number can be found on wikipedia. Its something that is very intricately ingrained in nature, not just something that was made up for convenience sake. The first person to actually come across it was Bernoulli when he tried to evaluate an interesting limit. It was then shown my Euler to have relevance also on the complex plane; and is also became more and more recognised as a natural rate function, meaning that the rates of many natural occurances such as bacterial growth or radioactive decay, can be modelled on this number.
  4. Sep 24, 2011 #3
    Re: e=2.71

    Yes, I saw that history but I still do not know WHY "Integral of (1/x)dx = log of x to the base e" and not some other number.
  5. Sep 24, 2011 #4

    D H

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    Re: e=2.71

    It's not 2.71. It is 2.718281828459045235360287471352662497757247093... e is an irrational number.

    Let's look at this function and its inverse:

    g(x) = \int_1^x \frac{dt}{t} \\ \\
    \int_1^{f(x)} \frac {dt}{t} = x

    I have not attached names other than the generic f(x) and g(x) to these functions. I'm not going to derive the following, but these functions are intimately coupled with the logarithm to some base and exponentiation functions:

    \log_a(x) = \frac{g(x)}{g(a)} \\ \\
    a^x = f(x\cdot g(a))

    The number a for which g(a)=1 is obviously special; it simplifies the above considerably. Call this number e. Note that e=f(g(e))=f(1). With this e defined as such,

    g(x) &= \log_e(x) \\ \\
    f(x) &= e^x

    This function f(x) is arguably the most important function in all of mathematics.
  6. Sep 24, 2011 #5
    Using the inverse function derivative theorem, you can show that [tex]\ln' = \frac{1}{\exp' \circ \ln} = \frac{1}{\exp \circ \ln} = \frac{1}{\text{Id}}[/tex]
    Thus [tex]\ln'(x) = \frac{1}{x}[/tex]
    And by integrating
    [tex]\int_1^x \frac{dt}{t} = \int_1^x \ln'(t)dt = \ln(x)[/tex]
  7. Sep 24, 2011 #6


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    Re: e=2.71

    So, as DH explained, you're basically asking, 'If I take the function 1/x and integrate it from 1 out to some number, how far do I have to go such that the area under the curve is 1?' If you draw out the curve and play a little, or use a graphing calculator, you'll quickly see where the number e comes from.
  8. Sep 24, 2011 #7
    Re: e=2.71

    In fact, e can be defined this way. As other people have pointed out, e turns up in a number of really unexpected areas. For example, have you heard of the Hat Check Problem? Basically, imagine some sort of Ball in which every man leaves his hat with the hat check upon entering (OK, imagine this is the '60s when people wore hats). What is the probability that no one gets his hat back? That is, everyone gets someone else's hat? It turns out that as the number of people grow, this probability goes to e^-1. Pretty cool, isn't it?
  9. Sep 24, 2011 #8
    Re: e=2.71

    I have to leave for class soon but thank you so much for your replies. Basically, I wanted confirmation that there is some significance with 2.71...... and there is, I am gonna read all the posts when i get back.
  10. Sep 25, 2011 #9
    Re: e=2.71

    Okay, so I went through all. I somewhat understood what DH said (I am gonna have to look at it all more closely), what sachav said went over my head abut I think phyzguy got what I was looking for.
    Last edited: Sep 25, 2011
  11. Sep 27, 2011 #10


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    Re: e=2.71

    From the integral given it is easy to see that a function thus defined has the property
    f(x y)=f(x)+f(y)
    This together with f'(1)=1 tells us
    in particular (limit h->0)
    f'(1)=lim f(1+h)/h
    since a f(x)=f(x^a)
    =lim f((1+h)^(1/h))
    since f is continuous at 1
    =f(lim (1+h)^(1/h))
    since lim (1+h)^(1/h)=e
  12. Sep 28, 2011 #11


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    2017 Award

    Re: e=2.71

    The exponential is the inverse function of the natural log. e is the value of this function at 1. Y
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