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What does e look like?

  1. Jul 13, 2004 #1
    I hope all of this makes sense. e, the eularian constant, has always bothered me. It comes up all the time but I never have a feel for what it is doing there.

    The way I look at pi is that it occurs to relate a straight and (as far as we can measure) "round" distance. The definition of pi that makes perfect sense to me is that it is the ratio of an infinity-gon's diagonal to its perimeter. If you know how many points you are dealing with you can use the exact "pi" to the proper accuracy. Pi for a square is 2*sqrt2 because that's the perimeter when the diagonal is a unit (don't get me started on sqrt2). A 5000-gon's perimieter is probably something like 3.1416. I can see how pi starts and evolves. When I see pi in a function I know its purpose. Or maybe I'm just nuts and/or wrong. But I can't see any of this with e and would like to be able to. What grows e-ishly? Why is it that variables are exponents of e all the time? Is ln a better way of looking at e-ing? Do any of those questions even make any sense?

    I hope all that wasn't too kooky and would really appreciate some help. Thanks.
     
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  3. Jul 13, 2004 #2

    matt grime

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    strictly speaking "pi for a circle" isn't a term people use. but you're getting at approximations for pi, fine, but there's no reason why pi for a 5000 gon is rational when it isn't even rational for a square.

    anyway, pi is only pi for a circle, so lets leave that alone.

    e is 1+1+1/2+1/3!+1/4!+....

    so if you want rational appoximations you may stop summing at any point.

    i don't understand what on earth grows e ishly might mean.

    e occurs all the time because it appears naturally in all kinds of situations, in particular the solution of differential equations, and fourier series.

    it's also related to trigonometry
     
    Last edited: Jul 13, 2004
  4. Jul 13, 2004 #3

    HallsofIvy

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    If you mean to think of e in a similar way to thinking of pi by increasing the number of sides of a polygon, you might try compound interest.


    A principal of P, at interest r, compounded n times a year is worth, at the end of the year, P(1+ r/n)n. As n goes to infinity, in the same way as the number of sides of the polygon increases, this goes to Per. In particular, investing $1 at 100% (r= 1), compounded n times a year, for one year gives a value of (1+1/n)n and that approaches e as n increases
     
  5. Jul 13, 2004 #4

    loseyourname

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    The base e for the function [itex]f(x) = a^x[/itex] was first used in calculus because it greatly simplifies the formula for differentiating an exponential function. If [itex]f(x) = a^x[/itex] is differentiable at 0 (which is the y-intercept of the graph of [itex]f(x) = a^x[/itex]), then [itex]f'(x) = f'(0)a^x[/itex]. This formula obviously becomes extremely simple if [itex]f'(0) = 1[/itex]. From the definition of the derivative, if [itex]f(x) = a^x[/itex], then

    [tex]f'(x) = \lim_{h \rightarrow 0}{\frac{a^{x + h} - a^x}{h}[/tex]

    which reduces to

    [tex]a^x\lim_{h \rightarrow 0}{\frac{a^h - 1}{h}[/tex]

    so that

    [tex]f'(0) = \lim_{h \rightarrow 0}{\frac{a^h - 1}{h}[/tex]

    Now we simply have to set

    [tex]\lim_{h \rightarrow 0}{\frac{a^h - 1}{h} = 1[/tex]

    so that e is now defined as the number such that

    [tex]\lim_{h \rightarrow 0}{\frac{e^h - 1}{h} = 1[/tex]

    If you want a graphical representation, simply look to the graph of the function [itex]f(x) = a^x[/itex]. You should be able to see now that e is the base of this function such that the slope of the function at the y-intercept is 1.
     
    Last edited: Jul 13, 2004
  6. Jul 13, 2004 #5

    Integral

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    It is my belief, though I do not fully understand why it is significant, that the reason nature loves e, is because of its relationship to the inverse function:

    [tex] \int _1 ^e \frac {dx} {x} =1 [/tex]

    The area under the inverse curve is unity between 1 and e.
     
  7. Jul 13, 2004 #6
    I don't have any calculus textbooks but I have one precalc book and I read about compound interest and I don't really get it. Say you are quoted 12% interest compounded monthly. You deposit 1000 dollars. I would expect that at the end of the first month I receive 120 dollars in interest and it is reinvested. At the end of the next month the total would be 1254.50. P(1+r)^n. But with the compound interest formula I'm really getting 1% per month over the 12 months. I guess that's why they call it ANNUAL percentage rate. It seems to me like the bankers are saying, "Customers, this is how nature does it's growth and so we're gonna do it this way too. And hey banker buddy, we can quote much higher rates than we're going to really pay because these morons don't know the difference between addition and mulitplication. I convinced that last guy his rate was 12% but it was really 1!! Even if we have to show them the calculations by hand they'll see the fake rate at the top of the fraction and think that's the one they're getting." Maybe I'm just stuck on a wrong defininition of compounding, but I don't think nature is playing games as often as e occurs.

    I guess I need one that's simple. I read about the fourier series at mathworld. It's beyond me. Something like F(x)=ae^x would be a lot more penetrable.

    I see what your'e saying. The rate is one. And I think the added one is the previous amount reinvested with the first being one. I just don't get and can't see the purpose of the n in the denominator. What about the one year part. I know it's just one so you don't have to show it but n is dependent on it. It is n over a trial of length t. Maybe the t part is just the bankers. Either way the n in the denominator still bugs me. If you have any thoughts that's great. If not, I won't feel alone. Thanks again
     
  8. Jul 13, 2004 #7
    Try seeing it this way: r=interest rate (per year). Then if we started with P dollars, P = principal. The interest in one year would be Pr. But we would have to add that to the principal, which we keep, giving a total of P+Pr = P(1+r). Then in two years we would have P(1+r)r for new interest added to P(1+r) = P(1+r)^2, and so on and on.

    Given the rate r, the bank calculation generally proceeds in terms of months, and this gives for a year of unit principal 1x(1+r/12)^12. This way of calculating interest is larger than it would have been if it was only calculated once a year. The ultimate way of calculating this interest is to compound instantaneously, or (1+r/n)^n as n goes to infinity. This gives the total result of Pe^r.

    A side note on this matter is that before computers these calculations took a lot of time and therefore calculating every week or so, would have been too much computation, so that was one reason for the use of the month. (But, sometimes calculations are only done once a year on things like commercial lending, or it could be every six months.) Today, of course, with computers, none of this would mattter, but banks and lenders are consertative and also competative and you have to live with a certain amount of confusion. It is true the matter is sometimes not made completely clear to the prospective customer as a way of getting him into the store or bank.
     
    Last edited: Jul 13, 2004
  9. Jul 13, 2004 #8

    AKG

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    Unity?

    I see the word "unity" used to mean "1" a lot. Is there any reason for this, aside from the obvious? I mean, does "unity" mean anything more than just "1", do we use "unity" only in specific cases, or does it even make sense to say unity x unity + unity = 2(unity)?
     
  10. Jul 13, 2004 #9

    Janitor

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    Another way of looking at it is that the solution to the linear first-order differential equation (D+k)y=0 is y=C exp(-kx), where by "exp" I mean of course "e to the power of..." The second-order linear differential equation of similar form, but with the square of the D operator in place of the D, and the square of k in the place of the k, has solutions that are sums of constants times sin(kx) and cos(kx), where those trigonometric functions can equally well be written as real and imaginary parts of exp(ikx). Then, in the same way that polynomials in algebra can be broken down into products of first-order polynomials and second-order polynomials, any linear differential equation can be broken down into the product of first- and second-order differential equations of the form I listed. The solutions accordingly involve e.

    So to the extent that nature can be nicely modeled in terms of n-th order linear differential equations, Euler's number is predestined to show up in the spotlight.

    (I am going by years-old memory on this, so feel free to correct anything I have bungled.)
     
    Last edited: Jul 14, 2004
  11. Jul 14, 2004 #10

    arildno

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    First, a technical question:

    Isn't Euler's constant equal to: [tex]\lim_{N\to\infty}((\sum_{i=1}^{N}\frac{1}{i})-log(N))[/tex]
    That is, a totally different number than e?

    Secondly, to chime in with Janitor, amongst others:
    "e" occurs "naturally" in biology, for example, since the rate of growth (of a bacterial population) is roughly proportional to the size of the population at that time.
     
  12. Jul 14, 2004 #11
    I am not a mathematician ,but one reason for e is that it's rate of change
    d( e^x)/dx = e^x is the same as itself , this is intimately connected with oscillating systems especially sinusoids y = A.e^jx repesents the sum of sin and cosine terms
    per Euler. This means it's connected to kinetic and potential energy transformations.
     
  13. Jul 14, 2004 #12
    Thank you all very much for all the info. It looks like I have to get a clear idea of what differentiating means. I think there's a lot more I need to know before I can have the understanding I want. I read that ebooks posting and found a calculus text and have started reading it. All your posts have been extremely helpful. Thanks again.

    Oh yeah. I think I read that e was the eularian constant. I might be wrong though too.
     
  14. Jul 14, 2004 #13
    Ive never taken differential equations or population biology, but I beleive the reason E comes up so often is arbitratry. Mathematicians use E to make integration and differentiation easyer.
     
  15. Jul 14, 2004 #14

    Janitor

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    Shoryuken, as Arildno points out there is a number called the Eulerian constant (usually denoted by the Greek letter gamma, I think), that is approximately equal to 0.577, and it is completely different from Euler's number e, which is approximately equal to 2.718.
     
  16. Jul 16, 2004 #15
  17. Jul 16, 2004 #16

    arildno

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  18. Jul 16, 2004 #17

    Gokul43201

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    They rounded; you truncated !
     
  19. Jul 16, 2004 #18

    arildno

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    Oh, yes of course, I was so tired I didn't check the next digit..
    Thanks Gokul!
     
  20. Aug 24, 2004 #19
    I thought I would give a stab at a response for the need for exp, and the above formula is the most telling I think. As you know exponentials are fundamental in physics because often the rate of change of a quantity is proportional to the quantity itself and even when we do not know the exact function, we assume this to be true for physical processes and this assumption does indeed hold up. So we use it to approximate any rate of change we do not have an exact equation for within the theory of differential equations.

    So as not to get long winded here...if we take the above formula we could say that the '1' it produces is one of 'something' or 1 [tex]\cdot[\tex][something] and we want to know what that something is. The result is [tex] \Delta x= \ln{x_1} - \ln{x_2}[\tex]. In thermodynamics x is usually a quantity like energy or entropy so the 'something' is change in energy or change in entropy which is what we want to know. So you could think of the range from 1 to exp on the integral as asking the question: what range do I need to integrate my function, 1/x to get a single 'unit' of change for my 'area' function which may be [tex]\Delta S[\tex] or [tex]\Delta E[\tex].

    Now this all doesn't yet seem to come together until you actually do some calculations. However, what will happen is that even though our units on entropy and energy are arbitrary, they will be equal to exponentials that have fundamental constants in the exponents and these fundamental constants will be in the same arbitrary system of units when we chose 'exp' as the base. When we analyze why this is, we realize the calculus said this would be true because of an argument that our exponents and coefficients in calculus will fundamentally be 'integer' counts of [something], where the 'unit' integer count is the beginning of how we count: 1 [tex]\cdot[\tex][something]. In other words calculus 'counts' things for us properly in physics and the exp seems to set the start of how we will count.

    As we know from Newton, fundamental things that are derivative to one another in physics seem to be other fundamental things (e.g. position, velocity, acceleration...). Now to be fair, Newton stacked the deck for us in the beginning by making sure the components from which he built kinematics were all derivative to each other and so 'exp' is fundamental to our classical physics. There is no reason to believe that there is not a more fundamental base when we move on to non-classical physics like quantum mechanics.

    But if you even casually read about the quantum theory development, it is all about determining integer quantum numbers to do exactly what you might guess, count things. So we always need to first set up the 'derivative levels' of the physical quantities (things like momentum, energy) and then determine how will one of each of these 'things' be mathematically related when we consider each of these levels.
     
  21. Aug 24, 2004 #20

    mathwonk

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    I originally liked e for this reason: I skipped trig in high school and when I got to college my professor defined e^z first and then defined sin(x) as something like (1/2i)(e^ix - e^(-ix)). I always loved that, because it meant I had not missed anything.

    Then later the integral of dz/z along a path in the complex plane (with zero deleted) from 1 to z was even more fascinating, as it explained why ln(z) is multivalued. This ocurred in Courant's calculus book, maybe volume 2.

    In differential equations, ce^kx are the eigenfunctions for the derivative operator. They solve ALL linear constant coefficient differential equations, which is sometimes taught as if it were a whole topic in a math course..

    Later still, exponential mappings play a key role in algebraic and lie groups. It seems e just never goes away.

    It also has the most beautiful Taylor series, and even the beginning of the decimal expansion is memorable: 2.718281828......

    Then its growth rate is amazing, and even its graph is elegant.

    In complex analysis I think I recall it turns out to be essentially equivalent to tan(z), i.e. a function which wraps around and around, and with two branch points, only they have shifted location.

    This recalls also the beautiful spiral staircase image of the exponential mapping in action.

    If one understands e^x one can define all exponentials and all power functions in terms of it and all trig functions. This is almost the only function one needs to understand.

    Of course through the magic of inverse functions, the very sophisticated exponential and log functions are captured from the simple beginning of the almost trivial function 1/x.

    I guess I just love this function.
     
    Last edited: Aug 24, 2004
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