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Intuitive understanding of e

  1. Sep 10, 2012 #1
    I'm failing to understanding why e shows up in continuous growth and decay. I've read the BetterExplained article, watched the KhanAcademy videos and read about it in Morris Kline's calculus book (which gives a somewhat better explanation than most for intuition).

    I think I get it in terms of compounding interest - say if you have $1 which gains 100% interest over some period t and it's compounded at every instant (so 100/n is the % gain at each instant), then the amount you'll have at the end of the period is e.

    But I can't visualize this for other things, for example, this question:

    The intensity L(x) of light x feet beneath the surface of the ocean satisfies the differential equation dL/dx=kL. The intensity of light cuts to half at 18ft. You cannot work without without artificial light if the intensity falls below 1/10th of the surface value. About how deep can you expect to work without artificial light?

    The computation was easy enough, but I don't understand the meaning of the expression I got L=exp(1/18ln(0.5)x)
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  3. Sep 10, 2012 #2

    Simon Bridge

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    Same reson [itex]\pi[/itex] keeps showing up when you handle circles or anything involving circular symmetry.

    But want a way of visualizing it - eg. pi is the distance you travel if you have a wheel diameter 1 and you roll it one complete revolution along the ground.

    That doesn't look right ... take care with your boundary conditions.

    Your equation is easier to understand in the following form:[tex]L=L_0 e^{-x/\bar{x}}: \bar{x}=\frac{\ln(2)}{18}[/tex] ... that [itex]\bar{x}[/itex] is the depth by which the light would vanish if the fall-off were linear.

    Here's the bit you need:
    When [itex]x=\bar{x}[/itex], the light is 1/e times what it was at the surface.

    (That is 0.36788 - about a third.)
    Last edited: Sep 10, 2012
  4. Sep 10, 2012 #3
    Wouldn't that be 1/2 revolution for pi?
  5. Sep 10, 2012 #4

    Simon Bridge

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    gah: diameter = 1. So edited.
  6. Sep 13, 2012 #5
    Is there a way to visualise the same thing for e for something other than compounding interest? For that, at 100% interest, it doesn't look hard to derive e intuively (like in the KhanAcademy videos) but when it comes to other things, I find it a bit more abstract

    Doesn't it? I'm quite certain the text gave the same answer. I'm also not sure what boundary conditions are.
  7. Sep 13, 2012 #6

    Simon Bridge

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    If it does then it is wrong. The relation given, L=exp(1/18ln(0.5)x), has L(0)=1 ... you really want to take some care making the negative explicit in the exponent too otherwise a casual reader will think you mean that intensity increases exponentially with depth.
    Then you are missing a fundamental idea in calculus - you cannot solve differential equations without them.

    if L(x) is the intensity of light distance x below the surface, then L0=L(0) is the intensity of light at the surface (x=0). That is a boundary condition.

    You start with [tex]\frac{dL}{dx} = kL[/tex]... then you can write[tex]\int\frac{dL}{L}=\int kdx \Rightarrow \ln|L| = kx + c[/tex]... notice the arbitrary constant of integration? This constant is determined by applying the boundary condition ... namely that L(0)=L0 ... which means that [itex]c=\ln|L_0|[/itex].


    You know how when you've rolled a unit diameter wheel all the way round once, then the wheel has traveled exactly pi units along the ground?

    The equivalent for your under-water light example is this: make [itex]\bar{x}[/itex] your unit for depth ...

    The light at the surface is e times brighter than the light at unit depth.

    If you start out under water, going up by one unit makes the light e times brighter.
  8. Sep 13, 2012 #7
    An exponential function only has the same scale as its derivative when it is e^x, just as a sinusoid function only has the same scale as its derivative when it is sin(x).
    So e^x is a kind of the core/invariant/canonical version of exponential growth/decay functions.
  9. Sep 13, 2012 #8

    Simon Bridge

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    I wondered about saying that ... :) it boils down to symmetry - which is why I used the pi analogy instead. Of course we don't have to pick base e for all our exponentials do we? For things that involve half-lives, for instance, wouldn't base 2 be a more natural choice?
  10. Sep 13, 2012 #9
    Might as well use [tex]\tau[/tex]
  11. Sep 14, 2012 #10

    Simon Bridge

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    @Dens: do you mean base ##\tau## is as good as base 2, or do you mean that mean life" is a more natural way to describe time-decay stuff than half-life?
  12. Sep 15, 2012 #11


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    First, as has already been said, all bases are equivalent: [itex]a^x= b^{log_b(a^x)}=b^{x(log_b(a)}[/itex]- only the multiplier is changed. e is convenient because the derivative of [itex]e^x[/itex] is just [itex]e^x[/itex] itself while the derivative of [itex]a^x[/itex] is [itex]ln(a)a^x[/itex].

    It is easy to show that
    [tex]\frac{da^x}{dx}= \lim_{h\to 0}\frac{a^{x+h}- a^x}{h}= a^x\left(\lim_{h\to 0}\frac{a^h- 1}{h}\right)[/tex]
    so that the "intuitive" meaning of e is that it is the number that makes that last limit equal to 1.

    Alternatively, one can define [itex]ln(x)= \int_1^x dt/t[/itex] and define e to be ln(1).
    Last edited by a moderator: Sep 15, 2012
  13. Sep 20, 2012 #12

    Oh...I think that's I know as initial condition...

    But then why e? How would you visualise the 'growth' of the light (this may end up going in circles!)
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