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Why is the number e special?

  1. Jul 24, 2008 #1
    Why is 2.71828182846...etc special? Is there any reason why e should equal 2.71828182846 and not some other number? I read that e is "transcendental" (what the hell does that mean- some kind of hippie spiritualism thing?). There must be some reason for why e=2.71828182846.

    i know that the function e^x=y is special in that y=y'. (which after reasoning through it i realize that there must be such a number, because for 2^x=y, y>y' for all x values, and for 3^x=y, y<y' for all x values, so there should be a number between 2 and 3 such that y=y')

    But how is e=(1+1/∞)^∞? (i dont know how to use that latex thing for limit notation, but you get the idea. actually i kind of prefer this definition over the definition that uses limit notation)
    Some interesting things i discovered while playing with my calculator and want to know why:
    (1+1/∞)^-∞=e^-1 (which makes sense- no questions here.)
    (1-1/∞)^∞=e^-1 (that's really, really weird. why is this?)
    (1-1/∞)^-∞=e (apparently the negatives cancel out somehow, but why? subtraction and inverse exponentation are two completely different things!)

    Some other questions about e:

    Why is e = ∑(1/n!) where n goes from zero to infinity?

    And why does e^x= ∑(x^n/n!) where n goes from zero to infinity?

    Why does the natural logarithm function work when finding the derivative of a^x? I always think of the logarithm as just a way to solve for x in the situation a^x=y

    Are there any other weird things about e that i should know?

    Also, a kind of unrelated thing, why does 0!=1?
  2. jcsd
  3. Jul 24, 2008 #2


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    One can show that the derivative of the function f(x)= ax, for a any positive number, is of the form Caax where Ca is constant (with respect to x) but depends on a. One can also show that, for example, if a= 2, then C2< 1 while if a= 3, then C3> 1. "e" is defined as the number such that Ce is exactly equal to 1. For that reason, the derivative of ex is ex.

    Yes, you can use a logarithm to solve ax= y precisely because "loga(x) is defined as the inverse function to ax so that loga(ax)= x. And, because of the property loga(xy)= y loga(x), you can use any base logarithm: If ax= y, then, taking the natural logarithm of both sides, ln(ax)= x ln(a)= ln(y) so x= ln(y)/ln(x). And, because loga(a^x)= x (from the definition above), loge(ex)= ln(ex)= x. From that
    [tex]a^x= e^{ln(a^x)}= e^{x ln(a)}[/tex]
    Now, the chain rule shows that
    [tex]\frac{da^x}{dx}= e^{x ln(a)}\frac{d x ln(a)}{dx}= a^x ln(a)[/itex]

    The fact that [itex]e^x= \sum x^n/n![/itex] follows from the fact that repeatedly differentiating ex always gives ex which is equal to 1 at x= 0, together with the formula for the Taylor's series.

    And, of course, taking x= 1 in [itex]e^x= \sum x^n/n![/itex] immediately gives [itex]e= \sum 1/n![/itex].

    These are simply wrong. Those should all be limits, not numerical values.
    Last edited: Jul 25, 2008
  4. Jul 24, 2008 #3
    A number is transcendental if it is not algebraic. If you're not sure about the meaning of either of those terms then a quick google will clarify them for you. It is nothing to do with 'hippy' anything - maths ain't like that.
  5. Jul 24, 2008 #4


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    You listed several reasons that answer that question, eg.
    e^x is a solution of y = y'
    lim[n\rightarrow\infty] \ (1+1/n)^n = e

    Um, not really, other than it has to equal something.

    The area under the curve y = 1/x is given by


    If you know that 3! is 6, then:
    Divide 3! by 3 to get 2! = 6/3 = 2, then
    Divide 2! by 2 to get 1! = 2/2 = 1.
    Repeating this algorithm:
    Divide 1! by 1 to get 0! = 1/1 = 1.

    So 0! = 1 is consistent with properties of factorials for positive integers.

    And moreover:
    Try dividing 0! by 0 to get (-1)! is undefined, as are factorials for all negative integers.
  6. Jul 24, 2008 #5
    Because it is useful and consistent to so declare it. Unless you say what you think the definition of n! is for n>0, we're guessing a little as to what you'd accept as justification.

    A reasonable one is that n! for n>0 counts the number of strings made from n distinct symbols, with order important.

    1!=1, since a is the unique string on one symbol, and 3!=6 since abc, acb, bac, bca, cab, cba are the 6 orderings.

    0! should then count the strings on 0 symbols. And since there is a unique empty string, 0!=1 makes sense here.

    It also fits into other arguments, and a nice recurrence relation.
  7. Jul 24, 2008 #6
    What? 1!/1 is not 0!

    To the poster, 0! is defined to be 1 because of the following:

    n! = n x (n-1)!

    dividing both sides by n,

    [tex]\frac{n!}{n}[/tex] = (n-1)!

    Choosing n=1,

    [tex]\frac{1!}{1}[/tex] = 1 = (1-1)! = 0!

    As for e? Who says its special. It just happens to be a constant with some nice properties. Just like pi is a constant with nice properties (one being that pi times diameter = circumference). As to where the value comes from? Well, you will have to read up on that.
  8. Jul 24, 2008 #7


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    Well, which is it then? 1!/1 either is or isn't 0!
  9. Jul 24, 2008 #8
    Since mathematicians use the incluse or, it is both!

    (no not really. i just didn't find an algorithm as convincing as a general equality, even though in essence they were the same thing)
    Last edited: Jul 25, 2008
  10. Jul 25, 2008 #9


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    Don't be ridiculous. Please.
  11. Jul 25, 2008 #10
    Search up Euler's number on wikipedia

  12. Jul 25, 2008 #11
    1!/1 = 1/1 = 0! also

    0! is said to be equal to 1 as we can define n! as the number of ways in which n objects can be arranged (say, in a row), and since here you have 0 objects, there's only one way, and that is empty.

    Because, for example in a permutation combination question, if you have used all the entities, you can multiply it with 0! any number of times (for zero objects remaining) , and the answer won't be affected.
  13. Jul 26, 2008 #12
    Ok then, in limit language:

    limit of y for y=(1+1/x)^x as x approaches infinity=e
    limit of y for y=(1-1/x)^x as x approaches infinity=e^-1

    limit of y for y=(1-1/x)^-x as x approaches infinity=e

    Why? how does a negative 1/x cause a multiplicative inverse of e, and a negative exponent x with a negative 1/x equals e?
    Last edited by a moderator: Jul 27, 2008
  14. Jul 26, 2008 #13
    You've posted many requests on here for people to help you in your understanding of lots of concepts. One of the simplest things you can learn is to be precise, and to think things through properly and clearly: how are we supposed to know what is your ignorance and your completely nonstandard and unacceptable abuse of notation? A little gratitude wouldn't go amiss, but I doubt anyone expects that anymore.
    Last edited by a moderator: Jul 27, 2008
  15. Jul 26, 2008 #14
    The main reason for 0!=1 is that it fits many different situations, like the ones that everyone posted above. But actually, the factorial is a function that is defined as:

    [tex]n! = \prod_{k=1}^{n}k, \forall n \in \mathbb{N}[/tex]

    (in an informal language: n! = 1 x 2 x ... x n)

    Since the domain of the function is [tex]\mathbb{N}[/tex], you don't have 0! defined here. That's why you need to define it. Since it's very useful for it to be 1,

    \prod_{k=1}^{n}k & \text{if } n \in \mathbb{N} \\
    1 & \text{if } n=0[/tex]

    It's all about being useful.
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