Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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:
  2. jcsd
  3. Jun 6, 2006 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    What do you mean by "evaluate"?

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


    User Avatar
    Science Advisor

    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    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


    User Avatar
    Science Advisor
    Homework Helper


    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


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

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


    User Avatar
    Homework Helper

    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


    User Avatar
    Homework Helper


    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

    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
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook