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E and eulers formula?

  1. Dec 9, 2007 #1
    can anyone tell me what is the meaning of number e,i mean how it is discovered ?why derivative and antiderivative of this function same?i know it is very practical property of it but where did we get this number? and another thing eulers formula, which is
    i also cant understand this equation. which comes from e^(i*pi)=sinx+i*cosx. where did we get this formula also?
    Last edited: Dec 9, 2007
  2. jcsd
  3. Dec 9, 2007 #2


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    If I'm not mistaken, in the 17th century, banks became interested in the problem of how to compute continual (compound?) interest, and that pretty much drove the number "e" forth, since it is the limit of the expression [tex](1+\frac{1}{n})^{n}[/tex] as n grows beyond bound.

    [tex]why derivative and antiderivative of this function same?[/tex]
    I assume you are talking about the exponential, with "e" as its base.
    Proofs of this is somewhat tricky.

    It is not an equation, but an equality. It is a staggering result.

    The complex exponential was originally shown to look like that by Euler, who used a power series representation of the exponential, and then saw how the trigonometric functions cropped up.
  4. Dec 10, 2007 #3
    Okay, in order to understand a little more about the number e, one must analyze first
    the limit presented above,[tex] \lim_{n\rightarrow \infty}(1+\frac{1}{n})^n[/tex]. In particular,
    one must prove that such limit exists, that means, that when you take n really big,
    the number doesn't go to infinity. To do this, you may use the binomial theorem: in this
    case, we obtain: [tex](1+\frac{1}{n})^n = \sum_{k=0}^{n}\binom{n}{k}\frac{1}{n^k}[/tex].
    But the binomial factor we can express it in a more convenient way and write it like this:
    [tex]\sum_{k=0}^{n}\frac{1}{k!}n(n-1)...(n-k+1)\frac{1}{n^k}[/tex]. Now, the number of those factors n(n-1)...(n-k+1) is k, so we can take the n^k and "share" it between all of them,

    we obtain: [tex]\sum_{k=0}^{n}\frac{1}{k!}1(1-\frac{1}{n})...(1-\frac{1-k}{n})[/tex].

    Let's call this sum S(n). Now, in orden to prove that the limit exists, whe have to prove that S(n+1)>S(n), this is, that the sequence of the S(n) is a increasing one, and then we may prove that it is smaller than a given number, so then, the limit must exist.

    Following the same steps as before, we find that S(n+1) is equal to:
    Now dividing each one of the factors in one of the n's from [tex](n+1)^k[/tex] we get:

    Now if we compare S(n) and S(n+1), due to the fact that [tex]1-\frac{x}{n}[/tex] is smaller than [tex]1-\frac{x}{n+1}[/tex] (when x varies from 0 to k-1), we have found the proof that [tex](1+\frac{1}{n})^{n}\leq(1+\frac{1}{n+1})^{n+1}[/tex]

    Now that we have proved that the sequence is monotone increasing, we must prove that there is a "top" or a "roof" (a limit) for the sum. We must see that beacuse all of the factors [tex](1-\frac{1}{n})...(1-\frac{k-1}{n})[/tex] are smaller than 1, we note that if we replace all of them by 1, we would obtain something that is bigger than S(n), this is:

    So the question is really if [tex]\sum_{k=0}^{n}\frac{1}{k!}[/tex] is convergent or not. If it is convergent, then S(n) also is because we have proved that it is smaller.
    Now we must consider the fact that [tex]k!\geq2^k[/tex] for k=0,1... so that means that: [tex]\frac{1}{k!}\leq\frac{1}{2^{k-1}}[/tex] and of course that if we take the sum of the smallers, it will be smaller than the sum of the largests:

    Finally, we see that this last sum, [tex]\sum_{k=0}^{\infty}\frac{1}{2^{k-1}}=2+\sum_{k=1}^{\infty}\frac{1}{2^{k-1}}[/tex]
    and the LAST sum, [tex]\sum_{k=1}^{\infty}\frac{1}{2^{k-1}}=1+1/2+1/4+...=2[/tex] so that finally we obtain:

    And obviously [tex] \lim_{n\rightarrow\infty}(1+\frac{1}{n})^n\leq3[/tex]

    So, the number e is in fact finite and smaller than 3. There is no "analytical" way of actually finding the exact value of e, so you just ask a computer to calculate it.

    I hope it will help you a bit okay, and sory about the english, i don't know it very well yet
    Last edited: Dec 10, 2007
  5. Dec 10, 2007 #4
    aaaa sory about the symbols... just starting with latex...
  6. Dec 10, 2007 #5
    You can edit your posts for 24h after the posting. Long posts with latex usually have typos, but it's not a problem because you can fix it quickly once you see how the post looks like.
  7. Dec 10, 2007 #6
    if you look at the taylor expansions of sinx, cosx, and e^x you get eulers equation when you replace x in the exponential with i*theta.
  8. Dec 10, 2007 #7
    I put some similar calculations into this thread.


    Check in particular my post #8. There's one way to "find e".

    Some words of criticism:

    Consider a following question. Let [itex]f:[0,1]\to\mathbb{R}[/itex] be [itex]f(x)=1[/itex] for all [itex]x\in [0,1][/itex]. Now, prove what the extension [itex]\tilde{f}:[0,2]\to\mathbb{R}[/itex] is.

    Okey, that's a dumb question. There's no way of proving what the extension should be.

    But then suppose we know what the exponential map [itex]\mathbb{R}\to\mathbb{R}[/itex], [itex]x\mapsto e^x[/itex] is. How do you prove that the extension of this mapping to the complex plane, [itex]\mathbb{C}\to\mathbb{C}[/itex], [itex]z\mapsto e^z[/itex], is something?

    That is almost equally dumb question! You cannot prove what [itex]e^z[/itex] is, without some definition what it is supposed to be. Concepts such as analytic continuation or series are quite complicated concepts, so it wouldn't be unjustified to simply define

    e^{x+iy} := e^x (\cos y + i\sin y)

    This definition implies

    e^{z_1} e^{z_2} = e^{z_1+z_2}

    and agrees with the old real exponentiation when y=0, so the exponentiation notation is justified.
    Last edited: Dec 10, 2007
  9. Dec 10, 2007 #8


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    [tex] f(x) = e^x [/tex]

    is the inverse function to

    [tex] g(x) = \log(x) [/tex]

    where [itex] \log(x) [/itex] is the "natural" logarithm and is defined to be this area or integral:

    [tex] \log(x) = \int_{1}^{x} \frac{1}{u} du [/tex]

    note that it is a "1" in the numerator and not some other constant (which is what differentiates the natural logarithm from other bases of logarithm). we can see easily that

    [tex] \log(ab) = \int_{1}^{ab} \frac{1}{u} du = \int_{1}^{a} \frac{1}{u} du + \int_{a}^{ab} \frac{1}{u} du = \int_{1}^{a} \frac{1}{u} du + \int_{1}^{b} \frac{1}{u} du = \log(a) + \log(b) [/tex]

    so it fits the main property of the logarithm and its inverse must be an exponential.

    so e (a base of an exponential function) is that number so that

    [tex] \log(e) = \int_{1}^{e} \frac{1}{u} du = 1 [/tex]

    so the area under the 1/x curve from 1 to e is 1 unit of area.

    a consequence of all of this is that, although it is the case that if

    [tex] f(x) = A^x [/tex]


    [tex] f'(x) = \mathrm{(some constant)} \cdot A^x [/tex]

    that constant of proportionality is equal to 1 only if A = e = 2.71828182845905...
  10. Dec 10, 2007 #9

    Ben Niehoff

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    Mostly right, except that having another constant instead of 1 will not change the base; it will merely add a constant:

    [tex]\int_{C}^{x} \frac{1}{u} du = \int_{1}^{x} \frac{1}{u} du - \int_{1}^{C} \frac{1}{u} du = \ln x - \ln C[/tex]

    Edit: Oh, wait, you said "numerator". I thought you were referring to the other "1". Whoops.
  11. Dec 10, 2007 #10


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    That is why I avoided the phrase that Euler "proved" the formula, he SAW how the trigonometric functions would crop up if it were possible to give some rigorous meaning to e^z for complex z.
  12. Dec 10, 2007 #11


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    BTW, Wikipedia has a pretty good page that explains how

    [tex] e^z = e^{\mathrm{Re}\{z\} + i\mathrm{Im}\{z\}} = e^{\mathrm{Re}\{z\}} e^{i\mathrm{Im}\{z\}} = e^{\mathrm{Re}\{z\}} \left( \cos(\mathrm{Im}\{z\}) + i \sin(\mathrm{Im}\{z\}) \right) = e^{\mathrm{Re}\{z\}} \cos(\mathrm{Im}\{z\}) \ + \ i e^{\mathrm{Re}\{z\}} \sin(\mathrm{Im}\{z\}) [/tex]

    i don't think we need to go through that here, do we?
    Last edited: Dec 10, 2007
  13. Dec 11, 2007 #12
    Nowhere does it discuss the definition of the complex exponential function. It just talks about proving certain results. Not top quality explanation IMO.
  14. Dec 11, 2007 #13


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    i've heard this kind of stuff before and i have never gotten what the problem is.

    jostpuur, do you have a pedagogical problem with the concept of adding an imaginary number to a real number? is such an addition defined? can we just treat the imaginary unit [itex] i [/itex] as a multiplicative factor?

    when dealing with the definition of a complex exponential function, can we define it as the result of raising some number (real and positive for the time being) to a complex power? if addition is allowed regarding the real and imaginary numbers that make a complex number, and, if so, we can multiply such complex numbers together and apply the distributive property?

    can we extend the property of exponentials regarding an exponent that is a sum of numbers to this addition of a real number to an imaginary number?

    that's all we need to "assume" and the explanation in the Wiki article is just fine. all three proofs put in there cut the mustard just fine.

    i don't get what could be bothering even the most anally-retentive rigorous mathematician here. whatever we define or assume about have real exponents to real numbers, we apply the same definitions, axioms, and rules to exponents that are a sum of a real and imaginary number. big deel.
  15. Dec 11, 2007 #14
    No, definition of complex numbers isn't the problem, but instead the way how many introductions start proving properties of the complex exponential function without first discussing its definition. This easily gives the appearance, that it would be trivial what the exponential function is.

    A good introduction would, for pedagogical reasons, first make it clear that the complex exponential function does not exists before it is defined. Then it should discuss what kind of properties we would like the complex exponential function to have. The exp(z1)*exp(z2)=exp(z1+z2) is the key property.

    After this I don't care what the definition is. It can be given with the series, or with the real trigonometric functions, or perhaps in some other way that I don't know at the moment. Then it should be checked that the definition gives the desired properties for the exponential function, but the most important part is, that the definition at least exists.

    No. The


    is of course clear when [itex]z\in\mathbb{C}[/itex] and [itex]n\in\mathbb{Z}[/itex], but in general raising one complex number to the power of another goes like this

    z_1^{z_2} := \exp(z_2\textrm{Log}(z_1))

    and we already need the exponential function for this. Unless there exists some other way to define this, that I wasn't aware of.

    (edit) Oh. hmmh.. you were talking about x^z when x is real and positive and z complex. But no.. the meaning of such power isn't obvious either.
    Last edited: Dec 11, 2007
  16. Dec 11, 2007 #15

    Ben Niehoff

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    If you take two complex numbers z1 and z2 on the unit circle and multiply them, you will get a third complex number z3, also on the unit circle. Treating the complex plane as a 2D normed vector space, you can define the angle each number makes with the x-axis. What you will find is that

    angle(z1) + angle(z2) = angle(z1 * z2)

    What you'll notice is that our "angle" function translates products into sums. There is a natural isomorphism between this and the logarithm; the only question that remains is "Which logarithm?"

    I'm not sure if there is a natural way to decide, besides the power series definition of exp(x). I suspect that there is, however. But I don't have time to look into it right now.
  17. Dec 13, 2007 #16


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    yes, that is what i was talking about. not only is x real and positive, it's a particular positive number, e, that has properties mentioned above that make it the base of the "natural" logarithm.

    now this is obvious from the fundamental meaning of an exponential:

    [tex] e^z = e^{\mathrm{Re}(z) + i \mathrm{Im}(z)} = e^{\mathrm{Re}(z)} e^{i \mathrm{Im}(z)}[/tex]

    don't you agree?

    we can deal with multiplying by the real number [itex] e^{\mathrm{Re}(z)} [/itex] later (an operation with meaning that is obvious as long as we can derive a meaning for [itex] e^{i \mathrm{Im}(z)}[/itex], correct?

    so then it is an issue of imparting meaning to

    [tex] e^{i \mathrm{Im}(z)} = e^{i \theta}[/tex]

    where [itex] \theta [/itex] is real since we know, by axiom, that [itex] \mathrm{Im}(z) [/itex] is real.

    at this point, the Wikipedia article does fine. it's plenty rigorous enough. either of the three proofs there completely answer the question of: "if [itex] e^{i \theta}[/itex] equals something, and if [itex]i[/itex] is a constant and has the property that when squared it's -1, what must that something be?".

    what is missing?
    Last edited: Dec 14, 2007
  18. Dec 14, 2007 #17


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    It is not at all "obvious", jotspurr is right.

    How you SHOULD proceed is, for example, to show that the complex function E(x)=cos(x)+isin(x)

    satisfies the basic properties E(x+y)=E(x)E(y), E(0)=1, and so on.
  19. Dec 14, 2007 #18
    If the exponential mappings




    are already defined, then it makes sense to define


    with equation


    I agree on this.

    It is impossible to prove what exp(iy) is, without a definition telling what it is supposed to be. If the proof does not use some definition, then it is not rigorous enough.
  20. Dec 14, 2007 #19


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    okay, so you agree that this


    follows naturally. how about

    \exp(i y_1 + i y_2)=\exp(i y_1)\exp(i y_2).


    now we can define the value of eiy to be the length of time, in milliseconds, of my dog's farts, but will that definition satisfy the existing properities of the exponential (such as the exponential of a sum is the product of exponentials)? if we treat the imaginary unit as a constant that it is (it sure as hell ain't a variable dependent on y), then there are existing rules for differentiating an exponential function that we can apply and then only "definition" for eiy that fits those rules is eiy = cos(y) + i sin(y). it's the only definition that fits. the Wikipedia article spells out 3 different ways that one might come upon that defintion. and then you can just verify that the definition makes sense by applying the fact above (along with the known trig indenties regarding the cosine and sine of a sum of two terms) .

    so what's the big deal? i still don't get what your "missing link" is about.
  21. Dec 14, 2007 #20


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    sure, i agree. but how would you have guessed from no previous hints that

    [tex] e^{i \theta} = \cos(\theta) + i \sin(\theta) [/tex]


    that is where those three proof in the Wikipedia article come in. (edit: the middle proof in the Wikipedia article already assumes you have a hint, a hypothesis, that [itex] e^{i \theta} = \cos(\theta) + i \sin(\theta) [/itex] ) and then proceeds to show that such a hypothesis satisfies the rules for differentiation for a function that is a ratio of the two equated expressions).

    again, if [itex] e^{i \theta} [/itex] equals something (perhaps a real number, we find that [itex] i^i [/itex] is a real number, or maybe it's complex, or maybe it's a 10 dimensional vector, or some element in a Hilbert space, who know?, but it's assumed to be something), what must that "something" be in order for [itex] e^{i \theta} [/itex] to satisfy

    [tex] e^{i (\theta_1 + \theta_2)} = e^{i \theta_1 + i \theta_2} = e^{i \theta_1} e^{i \theta_2} \quad \forall \theta_1,\theta_2 [/tex]


    if [itex]i[/itex] is a constant, (and it axiomatically is not dependent on [itex]\theta[/itex]), then this must be:

    [tex] \frac{d}{d \theta} e^{i \theta} = i e^{i \theta} [/tex]

    and, remember, by axiom, you have [itex] i^2 = -1 [/itex].

    the rest is fiddling.

    what else do you need (as far as definition)?
    Last edited: Dec 14, 2007
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