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How can e be 2.71828183?

  1. May 14, 2009 #1
    I mean, e is [tex](1+(1/n))^n[/tex] as n approaches infinity.
    Now, if you put [tex]\infty[/tex] instead of n, you get [tex](1+(1/\infty))^\infty=(1+0)^\infty=1[/tex]
    So how can e be 2.71828183... when you get 1?
    Note that I am 12, so if it is obvious, don't attack me
  2. jcsd
  3. May 14, 2009 #2


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    Welcome to PF - don't worry we only attack people who talk about perpetual motion machines.

    You have to be a little carefull putting infinity in equations, the normal rules of arithmatic don't really apply - you can only put approaching infinity.
  4. May 14, 2009 #3
    The thing to note is the word approaches infinity. this does not mean that if you put infinity into the equation you will get the right value (like you have tried).

    It means that e is equal to the theoretical maximum possible value that you could get by putting a value into the equation (called the limit). If you just try putting larger and larger values in then you will see that it gets closer and closer to 2.71828183~.

    (1+1/1,000)^1000 = 2.71692
    (1+1/10,000)^10,000 = 2.71815

    Also, if you are very keen on putting infinity in, then you could say:

    1/infinity = a non-zero value but still infinitly small.

    As 1+1/infinity is raised to higher and higher powers, it is still an infinitly small number away from 1. So if you raise it to the infinite power, the distance will become finite.

    This is not a method of calculating the value of e, but is just a helpful way of looking at it.

    beware though, this is not a rigorous method at all, just a way of looking at it. So if your reading this and you are a real mathematician, don't attack me either. :).
    Last edited: May 14, 2009
  5. May 14, 2009 #4
    Thanks for the quick reply.
  6. May 14, 2009 #5

    matt grime

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    As has been pointed out you can't just 'put in infinity' to a limit. This is nothing special about infinity, just limits. To evaluate the limit as x goes to 0 you can't just put in zero and hope.

    But, there is a second point to make - although you shouldn't just put in infinity (or zero or whatever), doesn't mean you can't. Sometimes it does work out. But what you can't do is substitute in separately. You assert that 1/infinity is 0, then raise 1 to the power infinity, and assume that makes sense (which it does). But look at the expression:


    the denominator and the power are linked - you took the limits separately. It is correct to say

    lim_{m} lim_{n} (1+1/n)^m =1

    if we let n tend to infinity then let m tend to infinity. But that is very different from the original limit you wrote down.
  7. May 14, 2009 #6
    Yes, just to agree with this, it's about limits. For example the equation:

    y = 1/x when x is 0 is undefined

    however as x tends to zero from the positive side it tends to +infinity (try plugging in a small x if you don't believe me!)
    from the negative x side it tends to -infinity (again plug in a small negative x)

    This is quite normal.
  8. May 14, 2009 #7
    This is one the problem I was facing with understanding e. Like the OP, I was wondering why the limit didn't equal 1^inf. I was so used to plugging in infinity for n in limits such as those for the test for divergence of series and improper integrals that I didn't realized that the limit of a quantity to a power is not the same as the limit of what is inside the quantity to a power. I've seen the limit outside of a compound function being "brought into" the inner function: lim sin(pi/x) = sin(lim (pi/x)), but I guess this doesn't work in this situation. By linked, I assume you mean the same variable. This post was very helpful.
  9. May 14, 2009 #8
    I like most of what you point out...however rather than saying "plug in infinity", I prefer to say that in mathematics infinity is just about how an expression behaves as some quantity gets arbitrarily large.

    The empirical illustrations nicely get the point accross but of course to actually evaluate the limit we need to use L'Hospital's rule. This type of indeterminate form is routinely treated in most calculus texts.

    Hey one more thing....my hats off to you Pumpkineater, when I was twelve the most math I did was to compute baseball stats....Keep on asking those questions!
  10. May 15, 2009 #9
    e is not [tex](1+(1/n))^n[/tex], it is just the way to calculate the "numerical value" of e.
    e is something else, you can define e as the base of natural logarithm.
    e can also be defined as the unique real number such that area of region bounded by hyperbola
    [tex]y = 1/x[/tex] and [tex] 1 \leq x \leq e [/tex] is 1.
    I give you a link, it explains e in very simple way.
    http://betterexplained.com/articles/an-intuitive-guide-to-exponential-functions-e/" [Broken]
    Last edited by a moderator: May 4, 2017
  11. May 15, 2009 #10


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    Also, e can be defined as [itex]\lim_{n \rightarrow +\infty} (1 + 1/n)^n[/itex]. :tongue:
    Last edited by a moderator: May 4, 2017
  12. May 15, 2009 #11


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    Do you remember the word "continuous"?
  13. May 15, 2009 #12

    matt grime

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    If the power is fixed this is probably going to work, but in the (1+1/n)^n the power isn't fixed, is it?

    In the (1+1/n)^n case they (the 'ns') are the same, but that is not the main point - you have taken limits selectively, which you can't do: you can't let n go to infinity here, then there, then somewhere else.

    As Hurkyl says, sin is continuous, i.e. you can take the limit inside by definition.
  14. May 15, 2009 #13
    Maybe the best way to get the numerical value is to use the binomial expansion of the expression. It is also a great way to pass the time on a rainy day.

    I only attack people who believe in ddwfttw, so you are safe, I hope!
  15. May 15, 2009 #14
    Oh my god. DDWFTTW. I had no idea people believed in things like this.

    I just lost faith in humanity.
  16. May 15, 2009 #15
    The YouTube videos ARE quite convincing.
  17. May 15, 2009 #16
    So was the movie Robocop, but you don't see me checking my sock drawer for cyborgs.
  18. May 15, 2009 #17
    Is there a particular reason why the integral of 1/x is equal to lnx? I know it is defined as such, but it seems like a rather arbitrary definition. I know that the power of x in this case makes finding the anti-derivative in the traditional sense impossible, but why lnx? Does it have to do with the base e of the natural logarithm, or with Taylor series?
  19. May 15, 2009 #18
    y = ln(x)

    x = e^y

    dx/dy = e^y ...and dy/dx = 1/(dx/dy)

    dy/dx = 1/(e^y) = 1/x ...because x = e^y as stated earlyer
  20. May 15, 2009 #19
    Thank you, Georgepowell, Mattgrime and Hurkyl for your help. Sry for hijacking this thread.
  21. May 16, 2009 #20

    matt grime

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    As well as the other answer you got, you can also try the following:

    define F(x) as

    [tex]\int_1^x \frac{1}{t}dt[/tex]

    Now, exercise, show that


    which shows you that F behaves 'just like a logarithm'.
  22. May 16, 2009 #21
    Yeah! Sorry my bad.:frown:
    Thanks for correction. :smile:
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