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Evaluating expression for e

  1. Jun 6, 2006 #1
    How to you evaluate the expression for e (the limit) ? I don't see how you could do this unless you do it numerically since e is irrational:confused:
     
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  3. Jun 6, 2006 #2

    Hurkyl

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    What do you mean by "evaluate"?

    (And you realize the square root of 2 is irrational, right)
     
  4. Jun 6, 2006 #3

    HallsofIvy

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    What limit are you talking about? There exist an infinite number of functions or sequences that have limit e. Of course, how you evaluate a limit has nothing to do with whether the limit is rational or irrational.
     
  5. Jun 7, 2006 #4
    Ok, I'll just ask this:

    [tex]\lim_{x\rightarrow\infty} (1 + \frac{1}{n})^n[/tex]

    How would I find that limit?
     
  6. Jun 7, 2006 #5

    Hurkyl

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    That limit's easy: it's (1 + 1/n)^n. :smile:

    That limit, once you fix the typo, will be equal to e. How you prove it depends on what you use as your definition of e. (Some people use that limit as the definition of e, so it's a rather trivial proof!)
     
  7. Jun 7, 2006 #6
    The only way I've ever been able to prove it (without really thinking about the topic - I'm sure there are other proofs) is to use logarithms and L'H[tex]\hat{o}[/tex]pital's rule. I'll start it off and you can fill in the rest.

    [tex]y = (1 + \frac{1}{n})^n[/tex]

    [tex]ln(y) = ln(1 + \frac{1}{n})^n[/tex]

    [tex]ln(y) = n \cdot ln(1 + \frac{1}{n})[/tex]

    [tex]\lim_{n\rightarrow \infty} ln(y) = \lim_{n\rightarrow \infty} n \cdot ln(1 + \frac{1}{n})[/tex]

    Now the right side is [tex] \infty \cdot 0[/tex] so you can apply L'H[tex]\hat{o}[/tex]pital's rule.
     
  8. Jun 7, 2006 #7

    Tide

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    BSMS,

    But you're using the value of e to derive the value of e with the natural logarithm - that's somewhat circular.
     
  9. Jun 7, 2006 #8

    Hurkyl

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    If you never use the value of e, you cannot possibly prove that the limit is e. :tongue:
     
  10. Jun 7, 2006 #9

    StatusX

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    You can prove that the limit is some number, and that it's, say, less than 3. Or you can compute it numerically by plugging in larger and larger values of n to get better and better approximations. But, as has been mentioned, you need a definition of e to show it equals e. One thing you could do is show different definitions are equivalent. For example, prove

    [tex]\lim_{n\rightarrow \infty} (1+\frac{1}{n})^n = \sum_{n=0}^{\infty} \frac{1}{n!}[/tex]
     
    Last edited: Jun 8, 2006
  11. Jun 9, 2006 #10
    So we can use,

    [tex]\lim_{h\rightarrow0} \frac{e^h - 1}{h} = 1[/tex]

    as our definition and with it we can show the limit in my above post is equal to e?
     
  12. Jun 10, 2006 #11

    lurflurf

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    yes

    we can make any number of definitions for e and prove they are equivelent. For each definition to be valid we should prove that there exist at least one number satisfying the definition (existence), and that there do not exist more than one number satisfying the definition (uniqueness).

    Your definition is my prefered one, but a possible problem is the function a^x must be defined (again any number of definitions are possible).

    I personally patch this up this way.

    Theorem:there exist a function f:R->R such that for all real numbers x,y
    a) f(x)*f(y)=f(x+y)
    b) lim x(real)->0 [f(x)-1]/x=1
    Theorem: if f and g are two functions as above f=g

    Definition: if f is a function as above
    e:=f(1)

    Several potential definitions of e are
    e=lim n(natural)->infinity (1+1/n)^n
    e=sum n(natural) 1/n!
    1=lim x(real)->0 (e^h-1)/h (having defined a^x)
    if f'(x)=f(x) f(0)=1 e=f(1)
    log(e)=1 (having defined log(x))

    There are other likely more interesting possibilities
     
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