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Euler's Number

  1. Jul 4, 2014 #1
    I am familiar with the fact that the number e can be defined several ways. One particularly interesting definition is the one based on limits, namely:
    e = limn [itex]\rightarrow[/itex] ∞ (1 + [itex]\frac{1}{n}[/itex])n
    My question is: wouldn't it be equally true to express e as the limit of the expression above as n goes to NEGATIVE infinity?
     
  2. jcsd
  3. Jul 4, 2014 #2
    According to WolframAlpha, that is quite correct. The proof below is for positive infinity (WA won't show the step-by-step solution for the negative version unless I buy a pro subscription), but the steps are equally valid for negative infinity in this case.

    limits-01.png

    (Open the image in a new tab if you see it with a black background.)
     
  4. Jul 5, 2014 #3

    statdad

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    Homework Helper

    "My question is: wouldn't it be equally true to express e as the limit of the expression above as n goes to NEGATIVE infinity?"

    Yes it is. The verification is an exercise in algebra and exponent chasing.

    If [itex] m [/itex] is a negative integer then [itex] m = -n [/itex] for a positive integer [itex] n [/itex]. For [itex] n > 1 [/itex] this gives

    [tex] \begin{align*} \left(1 + \frac 1 m\right)^m & = \left(1 - \frac 1 n \right)^{-n} = \left(\frac{n - 1}n\right)^{-n} \\
    & = \left(\frac n {n-1}\right)^n = \left(\frac n {n-1}\right) \cdot \left(\frac n {n-1}\right)^{n-1} \\
    & = \left(\frac n {n-1} \right) \cdot \left(\frac{n-1}{n-1} + \frac 1 {n-1}\right)^{n-1} \\
    & = \left(\frac n {n-1} \right) \cdot \left( 1 + \frac 1 {n-1}\right)^{n-1} = A_n \cdot B_n \text{ (say)}
    \end{align*}
    [/tex]

    Note that [itex] \lim_{m \to -\infty} \left(1 + \frac 1 m\right)^m [/itex] equals [itex] \lim_{n \to \infty} \left(1 - \frac 1 n \right)^{-n}[/itex]

    Since
    [tex]
    \begin{align*}
    \lim_{n \to \infty} \left( \frac n {n-1}\right) & = 1 \\
    \text{and}\\
    \lim_{n \to \infty} \left(1 + \frac 1 {n-1}\right)^{n-1} & = \lim_{n \to \infty} \left(1 + \frac 1 n%
    \right)^n = e
    \end{align*}
    [/tex]

    putting everything together gives

    [tex]
    \lim_{m \to -\infty} \left(1 + \frac 1 m\right)^m = \lim_{n \to \infty} \left(1 - \frac 1 n\right)^{-n} = e
    [/tex]
     
  5. Jul 5, 2014 #4
    But why should we consider the constraint n>1? Isn't it enough to say n>0?
     
  6. Jul 5, 2014 #5

    statdad

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    "But why should we consider the constraint n>1? Isn't it enough to say n>0?"

    For the limit it doesn't matter. But for the steps in my approach to work I need [itex] n > 1 [/itex] because
    of the introduction of the denominators of [itex] n - 1 [/itex] .
     
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