10
$\begingroup$

Assume that $e^x$ has not been defined, so please do not refer to $e$ in an answer.

Given the D.E $y' =ry$, we substitute a power series and arrive at the solution:

$$y = \sum_{k =0}^\infty \frac{(rx)^k}{k!} $$

I am trying to gain a more precise physical intuition for the mechanism of this type of growth.

As such, I am trying to intuitively understand the meaning of each term in the expression for $y$.

Question

How can one physically interpret each term in the infinite series in the context of the simple growth process above?

$\endgroup$
5
  • 5
    $\begingroup$ Interesting question. I'm looking forward to answers. You might get better ones for the differential equation $y' = ry$ for a fixed rate constant $r$. Then the powers of $r$ appear in the power series and may have meaning. $\endgroup$ Jun 17, 2018 at 0:04
  • 2
    $\begingroup$ @EthanBolker In fact this constant needs to be there if one wants a physical explanation, in oder for the units to add up. (dimensional analysis) $\endgroup$
    – Hyperplane
    Jun 17, 2018 at 0:55
  • $\begingroup$ Consider a weight in a force-field. Let the acceleration our force-field exerts on our weight be linear to the distance the weight has travelled, i.e. let our force-field be fully described the function $$a(s) = s$$ (Per traveled meter, the acceleration exerted grows by one) $$ $$ If we now observe the traveled distance of our weight dependent on the passed time, it is precisely $$s(t) = e^t$$ So in theory, if you choose to approximate the traveled distance dependent on passed time, you'll arrive at the given power series. $\endgroup$
    – Sudix
    Jun 17, 2018 at 2:23
  • $\begingroup$ While we are at it: How can one physically interpret each digit in the infinite decimal expansion of $e\,$? $\endgroup$ Jun 17, 2018 at 19:00
  • $\begingroup$ @ChristianBlatter, what's the difference between your question and: How can one physically interpret each digit in the infinite hexadecimal expansion of $e$? $\endgroup$
    – jaslibra
    Jun 17, 2018 at 19:42

3 Answers 3

3
$\begingroup$

Suppose each step $k$ gives you $x^k$ solutions. However, of those $x^k$ solutions, not all are unique, because some are permutations of others, or in other words re-ordering of others. So, if we count solutions that are permutations of one another as same solutions, then there are $x^k/k!$ solutions at each step $k$. This explains dividing by $k!$. https://en.wikipedia.org/wiki/Permutation#k-permutations_of_n

Why would each step produce $x^k$ solutions? Assume your set $S$ has $x$ elements. If you're asking how many $k$-tuplets there are on $S$, or in how many ways you can arrange $x$s in packets of $k$ of them, then the answer is $x^k$. https://en.wikipedia.org/wiki/Permutation#Permutations_with_repetition

So, if you pack $x$ elements in packets of $k$, but don't mind the way elements are ordered in packets, then you can make this many packets of all the different lengths altogether:

$\sum\limits_{k=1}^{\infty} \frac{x^k}{k!}$

The fact that $x$ is not an integer in general is just an echo of the weird possibility that this result makes sense for any $x\in\mathbb{R}$ and even $x\in\mathbb{C}$.

The weirder possibility is that $x$ represents the number of infinitely many elements. One can form packets of any sizes then. This would explain why $k$ may range over $\mathbb{N}$. And still weirder, the smallness of $x$ may be explained as re-normalization: you squeeze entire $\mathbb{N}$ line so that it becomes $\mathbb{R}$ or, say, interval $\left[0,1\right]$. This is possible in non-standard analysis where infinitesimals live, numbers that are smaller than any positive real number. So $x$ is just re-normalized number of elements of the set $S$. And $S$ has infinitely many discrete elements, countably many if you will. https://en.wikipedia.org/wiki/Non-standard_analysis

Suppose you have $x$ of some weird rabbits in heaven, who are all hermaphrodites: can act as male and female, but cannot breed with themselves, only with other rabbits. Suppose every rabbit breeds with every other rabbit. How many geneticly diverse offspring will be created? Well, you pack them in pairs, and don't mind if the pair is (male,female) or (female, male), obviously. In how many ways can you pack them this way? Well, $x^2/2!$.

Now, what if weird rabbits can breed in trios too, each possible trio producing just 1 offspring? You count that in too, and get $x^3/3!$ of genetic varieties, if you assume that a baby rabbit can have 2 fathers. I told you these rabbits are weird...

And so on. When you count all the possible varieties in, you get $e^x$. Number of rabbits $x$ has to be infinite for this to make sense, but if when passing from heaven to Earth numbers become squeezed, so that infinite numbers become finite, this becomes ordinary usual $e^x$.

So, if the number of rabbits $x$ grows, and it surely does after all this, then the number of genetic varieties among rabbits, grows as $e^x$ as weird rabbits keep reproducing in heaven.

To demonstrate the way one may squeeze an infinite natural number of rabbits $x$ to become a real finite number $r$, consider a unit interval $\left[ 0,1 \right] $ of real numbers on the real number line, and assume this:

$\bf{Assumption}$: The number of elements $c$ of the unit interval $\left[ 0,1 \right] $ is constant.

This seems a fairly reasonable assmuption. Why shouldn't the number of reals in the unit interval be constant? Do real numbers multiply?

What's the minimal distance between numbers of the unit interval? Well, the length of the unit interval is $1$, the number of elements is $c$, which is obviously an infinitely large number, and hence the minimal distance of equidistributed, equadistant elements is the length divided by the number of elelments, $1/c=0$. Hence, there is no minimal distance. This is so because $\mathbb{R}$ is dense: there's a real number between any two real numbers.

OK, now stretch the unit interval by factor $c$! The new, stretched interval is now of length $c$. Number of elements is constant, so still $c$. What is the minimal length between elements of this new interval? Length divided by the number of elements is $c/c=1$. The distance between two consecutive elements is now $1$. So, by stretching the unit interval by $c$, we have created $\mathbb{N}$.

So, any infinitely large natural number $x$ is just some real number $r$ from the unit interval, stretched by $c$, really...

Now, newly created $\mathbb{N}$ has a unit interval as well, so if we stretch this new unit interval by, say, $c$, then...

So, you see, $e^x$ is a function from heaven...

This is a question suitable for physics forums too.

Maybe you may find this answer suitable? Click a like or something if you do, please, I need 50 reputations in order to be able to comment, and I have serious questions... Thanks! :D

$\endgroup$
10
  • 1
    $\begingroup$ I really, really like this answer. It's really got me starting to think. I'd also like to wait for some other ideas as well, before declaring this my official favourite. $\endgroup$
    – jaslibra
    Jun 17, 2018 at 20:55
  • $\begingroup$ @jaslibra Yes, it is tempting to think of all the bosons and fermions being distributed with distributions that involve $e^x$ as a sea of infinitely many particles, that is then renormalized to become finite. If this can be done, this would solve the renormalization problem of physics. Planck factor: en.wikipedia.org/wiki/Planck%27s_law and en.wikipedia.org/wiki/Fermi%E2%80%93Dirac_statistics and en.wikipedia.org/wiki/Boltzmann_distribution and en.wikipedia.org/wiki/Bose%E2%80%93Einstein_statistics . Problems: en.wikipedia.org/wiki/Renormalization . $\endgroup$
    – anonymous
    Jun 17, 2018 at 21:12
  • $\begingroup$ Do you have any insight relating to this somewhat related question? math.stackexchange.com/q/2766802/338817 $\endgroup$
    – jaslibra
    Jun 18, 2018 at 3:37
  • $\begingroup$ Consider $e^x$. Now replace $x$ by $x^2$. Now weird rabbit are even weirder: they come in squares! Do notice that $e^x$ makes sense for negative $x$ as well. So does $e^{-x^2}$. Now rabbits come in negative squares... Think of it like this: $e^{-x}$ is the same as $e^x$, but with negative number of rabbits... And they're dying, because $e^{-x}$ is decreasing... These are rabbits from hell then I guess... So $e^{-x^2}$ describes square rabbits from hell? Basically, replace $x\to x^2$ in $e^{-x}$ and you're there. $\endgroup$
    – anonymous
    Jun 18, 2018 at 4:00
  • 1
    $\begingroup$ @anonymous But that gives no motivation for CLT, which makes it unsuitable for probability. Also note that really this explanation above is a reskin of the traditional one: "how far would you go if your position were fixed, how far would you go if your velocity was fixed, how far would you go if your acceleration were fixed, ..., how far would you go if your nth derivative were fixed, ..." $\endgroup$
    – Ian
    Jun 18, 2018 at 16:24
1
$\begingroup$

This may not qualify as an answer, but a key observation is that every term in that series is the antiderivative of the term before it.

To develop this in a "physical" way, consider an initial function that is identically zero, $y(x) = 0$. Let's say you want to make this function grow, and you get the idea to take the antiderivative of $y$ (which is an arbitrary constant $r$) and add that on to $y$. This now means:

$$y(x) = r$$

You now wonder why you can't keep doing this trick; continue taking the antiderivative of $y$ and adding it to the previous version of $y$. Doing it again twice gives you (as you may expect):

$$y(x) = rx + r\frac{x^2}{2}$$

Doing this "infinity times" leads to your definition of the exponential function multiplied by $r$:

$$y(x) = r\sum_{i = 0}^{\infty}\frac{x^n}{n!} = re^{x}$$

One addendum; say you were to get clever and try to remove that little $n!$ factor in the denominator. (It's just stifling growth after all). Doing that leads to what is called in some circles "geometric" growth, which causes $y(x)$ to become infinite for a finite $x=1$!

$$y(x) = r\sum_{i = 0}^{\infty}x^n = \frac{r}{1-x}$$

Hope this helps; I'm happy to update the answer if you feel something's missing or delete it if it's not what you're looking for.

$\endgroup$
0
$\begingroup$

If you want a physical interpretation of the solution to a differential equation, you have to start with a physical interpretation of the equation. What is $y(x)$, and what is the physical meaning of $y'(x) = r\cdot y(x)$? You can't interpret what the answer means if you don't know what the question means in the first place.

So let's try, as an example, the following interpretation: Let $y(t)$ be a position function. Then the equation $y'(t) = r\cdot y(t)$ means, roughly speaking, "The farther it goes, the higher the velocity." A consequence of such a situation would be "the higher the velocity, the higher the acceleration". Not only that, but the more it accelerates, the more jerk there would be, and so forth.

Now we can turn to the power series solution and try to interpret the individual terms. This is discussed (briefly) in an answer at MESE: Physical applications of higher terms of Taylor series. The idea is that each term can be thought of as a correction to the previous terms, introduced to take into account the fact that each successive derivative is non-constant. To be more specific:

  • If the object were not moving at all, its position would just be $y(t) = y(0)$.
  • However, the position is not constant, so we need to take velocity into account. If the velocity were constant, we would do this by writing $y(t) = y(0) + y'(0)t$.
  • However, the velocity is not constant either, so we need to take acceleration into account. If there were constant acceleration, we would do this by writing $y(t) = y(0) + y'(0)t + \frac 12 y''(0)t^2$.
  • However, the acceleration is not constant either, so we need to take the third derivative of position (jerk) into account...

And so on, and so forth. As I wrote in a comment to the linked MESE answer,

...I think of the linear approximation (1st two terms) as "Where the particle would be if it were not accelerating"; the quadratic approx (1st 3 terms) as "Where the particle would be if it had constant acceleration"; etc. Each additional term is a correction that takes into account a change in the previous term.

In the context of motion, we don't have good physical intuition (or even names) for higher-order derivatives; the 4th derivative is sometimes called "jounce" or "snap" but speaking only for myself I couldn't tell you what it feels like to be in a vehicle moving at a constant nonzero jounce. (I imagine it's probably pretty unpleasant.)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.