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Book defined e as the number

  1. Jun 22, 2004 #1
    I just started calc 2. And my book defined e as the number such that: [tex] \lim_{h \rightarrow 0} \frac{e^h-1}{h}=1 [/tex] I’m having trouble picturing what e is. Is there another definition of e? Is a actual physical ratio, (like pi or the golden ratio) or is it just some random number?
  2. jcsd
  3. Jun 22, 2004 #2


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    For one, the exponential function [itex]f(x) = e^x[/itex] is the same as all of its derivatives (and anti-derivatives if you ignore the constant). There are other definitions for e, perhaps look it up at wikipedia and mathworld.
  4. Jun 22, 2004 #3
    e = lim as h->infinity of (1+1/h)^h
  5. Jun 22, 2004 #4


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    Another popular version is:
    [tex]e=\sum_{i=0}^{\infty} \frac{1}{i!}[/tex]
  6. Jun 22, 2004 #5
    Thanks nate that was exactly what I was looking for, I can picture that one
  7. Jun 22, 2004 #6


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    But that's (the series expansion) only saying how big e is (2.718281...). It doesn't tell you anything more about why e is useful or interesting.
    Last edited: Jun 22, 2004
  8. Jun 23, 2004 #7
    Other than the limits, and the mclaurin series, all i can think of at the moment is euler's (identity?)

    e^(i*Pi) - 1 = 0

    For a calc 2 perspective, you'll probably use all three.

    Here's the easiest way to think of e.

    d e^x / dx = e^x

    The function is its own derivative (...and integral)
    Last edited: Jun 23, 2004
  9. Jun 23, 2004 #8


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    If I'm not mistaken, I believe that the function Exp(x) defined by:
    is necessary in order to rigourously define the exponentiation process in general (for example, to introduce the concept of an irrational number raised to an irrational exponent (I think this is the rational way to do it..)).

    Besides, the sequence of finite series approximations to Exp(1) converges quite fast.
    So, Exp(1) is perhaps a form of e worth mentioning.
    Last edited: Jun 23, 2004
  10. Jun 23, 2004 #9


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    The magic of e lies in this expression

    [tex] \int ^e _1 \frac {dx} x =1 [/tex]

    so the area under the inverse curve between 1 and e is exactly 1 square unit.
  11. Jun 23, 2004 #10


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    The point of [itex] \lim_{h \rightarrow 0} \frac{e^h-1}{h}=1 [/itex] is that it is easy to show that the derivative of ax is
    [itex] (\lim_{h \rightarrow 0} \frac{a^h-1}{h})a^x [/itex].

    Since [itex] \lim_{h \rightarrow 0} \frac{e^h-1}{h}=1 [/itex],
    ex has the nice property that its derivative is just ex again.

    Think of it this way: the derivative of ax is C ax where C is a constant (i.e. does not depend on x) the does depend on C. e is defined as the number for which that C is 1.
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