What does e look like?


by shoryuken
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nickdanger
#19
Aug24-04, 08:15 PM
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Quote Quote by Integral
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.
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.
mathwonk
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Aug24-04, 09:43 PM
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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.
CRGreathouse
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Sep1-04, 01:28 PM
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Quote Quote by shoryuken
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.
The formula you're looking for here is [tex]p_n=\frac{n}{2}\sin\frac{2\pi}{n}[/tex]. [tex]p_4=2[/tex] (not [tex]2\sqrt{2}[/tex]), since the area of a square is [tex]A=2r^2=2\left(\frac{s\sqrt{2}}{2}\right)^2=2\left(\frac{s}{\sqrt{2}}\ri ght)^2=s^2[/tex].

[tex]p_{5000}\approx3.1415918[/tex], by the way....
mathwonk
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Sep18-04, 11:07 PM
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for all linear and other transformations, fixed points are a natural means of understanding them. the most important linear transformation in mathematics is the derivative operator taking f to Df. for this operator, the only fixed points are scalar multiples of e^x.


gee i just noticed a gave a much more robust answer a month ago. i guess the summer is really over.

this time though i tried to distinguish e^x from a^x.

to expand on the claims made above about generating other functions.

assume the basic operations on functions are algebraic ones, and also integration, and inversion.

then start with just one function x. then reciprocation gives us 1/x, then integration gives us ln(x), and inversion gives us e^x. then we get e^(rln(x)) = x^r, and e^(xln(a)) = a^x. then inversion gives loga(x), and by complex algebra we get (1/2i)(e^(ix) - e^(-ix)) = sin(x), and
(1/2)(e^(ix) + e^(-ix)) = cos(x). As stated above, thus we can generate all the power, trig, log, and exponential functions, starting just from x, and the natural log function (base e) is the first really interesting step after 1/x.

Integral also hinted at this above, more elegantly.


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