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Intutive explanation of Euler's constant (e)

  1. Nov 6, 2011 #1
    Hi,

    I am facinated by the http://en.wikipedia.org/wiki/E_(mathematical_constant)" [Broken] and how it shows up in natural processes.

    However, I have never been able to comprehend its defination.
    e = lim n→ ∞ (1 + 1 / n) ^n

    I think like this ( and I know this is not correct): If n becomes very large, the term 1/n should become infinitesimaly small and therefore 0. And therefore the answer should be 1.

    Is there an example or applet that may help me intutively understand this equation?

    Thanks.
     
    Last edited by a moderator: May 5, 2017
  2. jcsd
  3. Nov 6, 2011 #2

    HallsofIvy

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    Have you tried simply doing the calculation, for large n, and see what you get?

    Take n= 1000. What is [itex](1+ 1/1000)^{1000}= (1.001)^{1000}[/itex]?

    Try n= 1000000. What is [itex](1+ 1/1000000)^{1000000}= (1.000001)^{1000000}[/itex]?
     
  4. Nov 6, 2011 #3
    [tex]
    1^{\infty}
    [/tex]
    is an indeterminate form.
     
  5. Nov 6, 2011 #4
    Why are you assuming that you can can just look at the the expression inside the parenthesis first? That is not valid because you are supposed to simultaneously keep raising it to higher and higher powers. The guy inside the parentheses is getting closer to 1, but it's not clear that that is pulling it down as fast as raising it to the nth power is pulling it up.

    One way to understand what's going on more clearly is to use Euler's method.

    This is a way to solve differential equations numerically.

    In this case, what you want is a function that is its own derivative, so that means you are looking for a solution of the differential equation

    [itex]\frac{dx}{dt} = x [/itex]

    subject to the initial condition

    x(0) = 1

    Geometrically, this means that you want a function whose slope is equal to its height. How can we find such a function? One approach is to imagine a "slope field". Set up coordinates on a plane so that x is the vertical direction and t is the horizontal. Now, for each point, we want a line whose slope is the x coordinate (remember, slope = height). We start out at the point (0,1). Then we go up the slope at that point for some time t/n. Then, when that time has passed, we end up at some other point. Now, we go up the slope at this new point. This procedure gives you the formula

    [itex](1+t/n)^n[/itex]

    As n goes to infinity, we should get a function that is closer and closer to what we want--its slope will be equal to its height. That's [itex]e^t[/itex].

    It's clear that this thing is increasing, not staying equal to 1.

    To get e, set t = 1 and you get your formula.

    Draw a picture and work it out.

    This construction is a special case of Euler's method.

    http://en.wikipedia.org/wiki/Euler_method

    Another way to visualize it is in terms of a vector field, rather than a slope field. In this case, you have a vector field on the real number line. The length of the vectors is proportional to the distance from 0. You start at 1 at time 0. So [itex]e^t[/itex] is obtained by flowing along this vector field. And you could do the same thing we did with the slope field to get an approximation.

    Yet another way to think of it is in terms of interest rates compounded at smaller and smaller time intervals, which gives the same formula, and also makes it clear that the thing is increasing, so that e has to be greater than 1.
     
  6. Nov 7, 2011 #5
  7. Nov 7, 2011 #6

    mathman

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    One easy way to see what happens is to use the binomial theorem.

    (1 + 1/n)n = 1 + 1 + {(n-1)/n}/2! + {(n-1)(n-2)/n2}/3! + ....

    which converges to the power series for e as n becomes infinite.
     
  8. Nov 11, 2011 #7
    My favorite definition of e is the one obtained from the prime number theorem,

    [tex]e = \lim_{n\to\infty} n^{\frac{\pi(n)}{n}}[/tex]

    ,where π(x) is the prime counting function. So the exponent is the density of prime numbers among natural numbers.

    It is painfully slow to converge because even for n = a billion trillion you still have a notable 2% error. But is elegant as it links the most important constant of calculus with the primes.

    This also gives one more proof of the infinity of primes because if the primes where finite their density would be zero and e would be 1. On the other hand, their density is so miniscule that the exponent becomes so small as to compress infinite n to just 2.71828... .
     
  9. Nov 11, 2011 #8
    [tex]e = \sum_{n=0}^\infty \frac{1}{n!} = 1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{24}...[/tex]

    This is pretty much just a Taylor expansion of [itex]e^x[/itex] where [itex]x = 1[/itex].
     
  10. Nov 12, 2011 #9

    Deveno

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    i think the definition in the original post is bernoulli's definition (so how come it's not called bernoulli's constant? life is so unfair....), and if i'm not mistaken (and i rather could be) he arrived at that definition from studying compound interest (other expressions involving e occur quite naturally in games of chance).

    here are some sample calculations run on a hand-held calculator:

    (1 + 1)1 = 2
    (1 + 1/10)10 = 2.5937424601
    (1 + 1/100)100 = 2.7048138294215260932671947108075
    (1 + 1/10,000)10,000 = 2.7181459268252248640376646749131
    (1 + 1/1,000,000)1,000,000 = 2.7182804693193768838197997084544
     
  11. Nov 12, 2011 #10

    lavinia

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    Intuitively the n'th power is exponential growth but ocurring only at n points in time. As n gets large the number of points in time increases and the amount of time between them deceases. in the limit becomes continuous exponential growth.
     
    Last edited by a moderator: May 5, 2017
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