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What is e?

  1. Dec 16, 2004 #1

    J7

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    Hi, can anyone explain "e" to me? It's used all the time in calc and I don't understand what it represents or it's value. Thanks
     
  2. jcsd
  3. Dec 16, 2004 #2
    e is a number, approximately equal to 2.71828183. It's irrational and transcendental. The function f: R -> R, f(x) = e^x satisfies f'(x) = f(x).
     
  4. Dec 16, 2004 #3

    dextercioby

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    Along with [tex]\pi [/tex],[tex] 1 [/tex] and [tex] i [/tex],it is the most important number in mathematics,hence in science.
    It is a real number,mathematicians call it irrational and transcendent.It has an infinite number of decimals,the first fifteen (hopefully i haven't forgotten them :tongue2: ) are:
    [tex] 2.718281828459045... [/tex] and it is defined as the limit of the sequence:
    [tex] e=:lim_{n\rightarrow +\infty} (1+\frac{1}{n})^{n} [/tex]

    It has an interesting history and some mathematicians call it "Euler's number",hence the letter "e".

    Daniel.
     
  5. Dec 16, 2004 #4

    matt grime

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    It is also

    [tex]\sum_{r=0}^{\infty}\frac{1}{r!}[/tex]

    (can we ignore the "infinite number of decimals" thing - that's neither here nor there, as well as incorrect English 1/3 has also not a finitely long decimal expansion in base 10, and? And there were no rational transcendental numbers. A number is transcendental if it is not a root of any polynomial with integer (whole number) coefficients.The complementary notion is 'algebraic'; all integers are algebraic, sqrt(2) is algebraic, pi isn't.)

    I, e, t arises as the (unique) solution to f'=f, which tells you that it is important in "the real world" since we model that with differential equations, and importantly, once we've defined e, we can solve f'=kf, as well as a whole load of other differential equations without needing to define anything else.
     
    Last edited: Dec 16, 2004
  6. Dec 16, 2004 #5

    HallsofIvy

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    It's just this number, y'know? Any function of the form f(x)= ax has the property that its derivative (rate of change) is proportional to ax itself.
    e (which, as Muzza said, is "approximately equal to 2.71828183.") has the nice property that the constant of proportionality is 1- that is, the rate of change of the function ex is precisely ex.
     
  7. Dec 16, 2004 #6

    Great explanation! I know some calculus, mostly all elementary. And I was thinking about what you said, it makes sense. I'm trying to figure out partial derivatives.

    Could you explain some real-life explainations of [tex]e[/tex]? Like an example, then solve it? That'd be real helpful to me, if you could think of one.

    Before you posted this, I thought e was actually just related to the logaritam.

    Since its proportion is 1:1, would this mean that..

    [tex]\int \frac{dy}{dx} e^2 dx = [tex]1/3^3e + C[/tex]
     
    Last edited: Dec 16, 2004
  8. Dec 17, 2004 #7
    What about this? I remember this from a class.

    [tex]e = (1 + \frac{1}{x})^x[/tex]
     
  9. Dec 17, 2004 #8

    Integral

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    [tex] 1 = \int_1^e \frac {dx} x [/tex]

    I believe that it this relationship which gives the number physical significance.
     
  10. Dec 17, 2004 #9

    shmoe

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    You probably mean:

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

    Or more generally:

    [tex]e^a=\lim_{x\rightarrow\infty}(1 + \frac{a}{x})^x[/tex]
     
  11. Dec 17, 2004 #10

    matt grime

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    nb.your backslashes and forward slashes are wonky

    It is related to the logarithm, and the examples ought to tell you you've underestimated the importance of log.

    THe second bit is wrong. e^2 is just a number so you're asking for

    [tex]e^2 \int \frac{dy}{dx}dx[/tex]

    which is just ye^2
     
    Last edited: Dec 17, 2004
  12. Dec 17, 2004 #11

    HallsofIvy

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    There are many applications in which the "rate of change" of some quantity is proportional to the quantity itself: population growth, radioactivity, etc. That is:
    dy/dx= C y so that y is necessarily an exponential. For example, suppose you have a radioactive element that has a "half-life" of 1000 years. It's easy to see that, if you start with M0 grams, the amount left after T years is M0 (1/2)T/1000[/sub]. The "T/1000" just counts "the number of times you multiply by 1/2".
    A more formal way of deriving that would be to write dM/dt= kM (k is the unknown constant of proportionality) so that (1/M)dM= kdt. Integrating both sides: ln(M)= kt+ C so that M= ekt+C= eCekt= C' ekt (where C'= eC). Setting t= 0 we get M(0)= C' so that the coefficient is in fact M0, the initial amount. Knowing that the half-life is 1000, tells us that M(1000)= M0e1000k= (1/2)M0. We solve for k by taking the logarithm of both sides: 1000k= ln(1/2) so k= ln(1/2)/1000.

    That is: M(t)= M0eln(1/2)t/1000[/sub]. Of course if you are clever, you will recognize that ln(1/2)t/1000 is the same as ln((1/2)t/1000 so that eln(1/2)t/1000= (1/2)t/1000.
     
  13. Dec 17, 2004 #12

    HallsofIvy

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    There are many applications in which the "rate of change" of some quantity is proportional to the quantity itself: population growth, radioactivity, etc. That is:
    dy/dx= C y so that y is necessarily an exponential. For example, suppose you have a radioactive element that has a "half-life" of 1000 years. It's easy to see that, if you start with M0 grams, the amount left after T years is M0 (1/2)T/1000. The "T/1000" just counts "the number of times you multiply by 1/2".
    A more formal way of deriving that would be to write dM/dt= kM (k is the unknown constant of proportionality) so that (1/M)dM= kdt. Integrating both sides: ln(M)= kt+ C so that M= ekt+C= eCekt= C' ekt (where C'= eC). Setting t= 0 we get M(0)= C' so that the coefficient is in fact M0, the initial amount. Knowing that the half-life is 1000, tells us that M(1000)= M0e1000k= (1/2)M0. We solve for k by taking the logarithm of both sides: 1000k= ln(1/2) so k= ln(1/2)/1000.

    That is: M(t)= M0eln(1/2)t/1000[/sub]. Of course if you are clever, you will recognize that ln(1/2)t/1000 is the same as ln((1/2)t/1000 so that eln(1/2)t/1000= (1/2)t/1000.
     
  14. Dec 23, 2004 #13
    isnt there also a relation between pi and e?

    its something like this....

    (pi^5 + pi^5)^(1/6)

    or something of the sort.... please feel free to correct me
     
  15. Dec 23, 2004 #14

    arildno

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    The arguably most famous relation is:
    [tex]e^{i\pi}+1=0[/tex]
     
  16. Dec 23, 2004 #15

    matt grime

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    It is, at least I think so at any rate, an open question as to whether any such algebraic relation exists betweene them
     
  17. Dec 23, 2004 #16

    HallsofIvy

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    Well, there is certainly no question as to whether THAT relation is true:

    (pi^5+ pi^5)^(1/6)= (2pi^5)^(1/6)= approximately 2.9138 which is not particularly close to e!
     
  18. Dec 23, 2004 #17
    The correct one is (pi^4 + pi^5)^(1/6).

    Forgive me for not using latex, I just don't want to go through all that trouble with it now.
     
  19. Dec 23, 2004 #18

    arildno

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    Is this another piece of Ramanurjan magic?
     
  20. Dec 23, 2004 #19

    matt grime

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    well, log of that number, is according to my not very accurate caculator 0.999 something

    so close, but probably no cigar (seriously, it is an open question, and that ain't the solution)
     
  21. Dec 23, 2004 #20

    dextercioby

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    [tex] e\simeq \sqrt[6] {\pi^{4}+\pi^{5}} [/tex]
    ,the fourth decimal is different.

    Matt,it sounds like the Goldbach conjecture... :tongue2: As for that,i still trust mathematicians would solve it,in the next millenium. :wink: But for an algebraic relation beween the most famous transcendental numbers,i frankly don't...Feel free to contradict me. :tongue2: It would mean proving the existance of this relation,a great benefit to mathematics and generally science+humanity+get u a Medal.I would have said Nobel prize,but that dude hated your kind... :tongue2:

    Daniel.
     
    Last edited: Dec 23, 2004
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