# Does Euler number $e$ have a role in kinematics?

Euler number $$e$$ is often explained with the example of compound continuous interest.

I was wondering if it could also be illustrated with an example about the displacement of a body (although not an oscillating one, with which I have less difficulty in finding e, but extending endlessly in a straight line).

My attempts so far to find an answer:

• One of the ways to calculate Euler’s number is with the series of the inverse factorials:

$$e = \sum\limits_{n = 0}^\infty {(\frac{1}{{n!}})} = \frac{1}{{0!}} + \frac{1}{{1!}} + \frac{1}{{2!}} + \frac{1}{{3!}} + \frac{1}{{4!}} + \frac{1}{{5!}} + ... = 1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{{24}} + \frac{1}{{120}} + ... % MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaamyzaiabg2 % da9maaqahabaGaaiikamaalaaabaGaaGymaaqaaiaad6gacaGGHaaa % aiaacMcaaSqaaiaad6gacqGH9aqpcaaIWaaabaGaeyOhIukaniabgg % HiLdGccqGH9aqpdaWcaaqaaiaaigdaaeaacaaIWaGaaiyiaaaacqGH % RaWkdaWcaaqaaiaaigdaaeaacaaIXaGaaiyiaaaacqGHRaWkdaWcaa % qaaiaaigdaaeaacaaIYaGaaiyiaaaacqGHRaWkdaWcaaqaaiaaigda % aeaacaaIZaGaaiyiaaaacqGHRaWkdaWcaaqaaiaaigdaaeaacaaI0a % GaaiyiaaaacqGHRaWkdaWcaaqaaiaaigdaaeaacaaI1aGaaiyiaaaa % cqGHRaWkcaGGUaGaaiOlaiaac6cacqGH9aqpcaaIXaGaey4kaSIaaG % ymaiabgUcaRmaalaaabaGaaGymaaqaaiaaikdaaaGaey4kaSYaaSaa % aeaacaaIXaaabaGaaGOnaaaacqGHRaWkdaWcaaqaaiaaigdaaeaaca % aIYaGaaGinaaaacqGHRaWkdaWcaaqaaiaaigdaaeaacaaIXaGaaGOm % aiaaicdaaaGaey4kaSIaaiOlaiaac6cacaGGUaaaaa!69B7!$$

• You get a similar series if you consider the displacement, after 1 second, of an object starting at position = 1 m, with v = 1 m/s and subjected to acceleration = 1 m/s^2 and to successive derivatives of position with respect to time (jerk, snap, crackle, pop... and so on ad infinitum), all of value = 1:

$$\Delta r = {r_o} + {v_o}\Delta t + a\frac{{\Delta {t^2}}}{2} + j\frac{{\Delta {t^3}}}{6} + c\frac{{\Delta {t^4}}}{{24}} + p\frac{{\Delta {t^5}}}{{120}}... = 1 + 1 + \frac{1}{2} + \frac{1}{6} + \frac{1}{{24}} + \frac{1}{{120}} + ... % MathType!MTEF!2!1!+- % feaagKart1ev2aqatCvAUfeBSjuyZL2yd9gzLbvyNv2CaerbuLwBLn % hiov2DGi1BTfMBaeXatLxBI9gBaerbd9wDYLwzYbItLDharqqtubsr % 4rNCHbGeaGqiVu0Je9sqqrpepC0xbbL8F4rqqrFfpeea0xe9Lq-Jc9 % vqaqpepm0xbba9pwe9Q8fs0-yqaqpepae9pg0FirpepeKkFr0xfr-x % fr-xb9adbaqaaeGaciGaaiaabeqaamaabaabaaGcbaGaeyiLdqKaam % OCaiabg2da9iaadkhadaWgaaWcbaGaam4BaaqabaGccqGHRaWkcaWG % 2bWaaSbaaSqaaiaad+gaaeqaaOGaeyiLdqKaamiDaiabgUcaRiaadg % gadaWcaaqaaiabgs5aejaadshadaahaaWcbeqaaiaaikdaaaaakeaa % caaIYaaaaiabgUcaRiaadQgadaWcaaqaaiabgs5aejaadshadaahaa % WcbeqaaiaaiodaaaaakeaacaaI2aaaaiabgUcaRiaadogadaWcaaqa % aiabgs5aejaadshadaahaaWcbeqaaiaaisdaaaaakeaacaaIYaGaaG % inaaaacqGHRaWkcaWGWbWaaSaaaeaacqGHuoarcaWG0bWaaWbaaSqa % beaacaaI1aaaaaGcbaGaaGymaiaaikdacaaIWaaaaiaac6cacaGGUa % GaaiOlaiabg2da9iaaigdacqGHRaWkcaaIXaGaey4kaSYaaSaaaeaa % caaIXaaabaGaaGOmaaaacqGHRaWkdaWcaaqaaiaaigdaaeaacaaI2a % aaaiabgUcaRmaalaaabaGaaGymaaqaaiaaikdacaaI0aaaaiabgUca % RmaalaaabaGaaGymaaqaaiaaigdacaaIYaGaaGimaaaacqGHRaWkca % GGUaGaaiOlaiaac6caaaa!6F1D!$$

But this situation looks very unreal.

However, there should be a real-life example, because you can easily get displacement = e (if I am not mistaken) when the acceleration affects only direction, not modulus. Take for example a satellite that is 1 unit away from a gravitation source and that moves tangentially by inertia at v = 1 of whatever units. If gravity acts perpendicularly to the tangential motion and the acceleration caused by gravity is v^2/r (that is to say, with this numbers, 1), then the displacement after 1 time unit will be circular and of modulus 1 radian = e^i, wouldn’t it? This situation looks to me very much like compound interest, because you start with a capital (radius) and an interest rate (velocity) and then you make the interest compound and continuous, although only in terms of direction (centripetal acceleration).

But what if the satellite were heading towards the earth and what changed were the modulus of its velocity, do we lose then the possibility to involve number e?

• Your going to love the Fourier transform and Fourier analysis & synthesis. ;) – PM 2Ring Dec 3 '18 at 23:02
• @PM 2Ring I had been studying it during latest months, I do love it and, well, that is what inspired the example about the satellite! I am afraid that I was inaccurate in talking about kinematics. Of course kinematics also describes waves and oscillators, but I have no problem in seeing it there; what I meant is that I am looking for it in non-oscillating phenomena. – Sierra Dec 3 '18 at 23:31

Of course, $$e$$ is ubiquitous in kinematics. For example, consider a repulsive force proportional to $$x$$, $$F = kx.$$ Then the acceleration is $$a = \frac{k}{m} x = \omega^2 x, \quad \omega = \sqrt{\frac{k}{m}}.$$ This differential equation has solutions of the form $$e^{\omega t}$$ and $$e^{- \omega t}$$. In particular, suppose that $$x(0) = 1$$ and $$v(0) = \omega$$. In that case the solution is exactly $$x(t) = e^{\omega t}.$$ In general, for any linear force law, the solutions will be exponentials or complex exponentials, so it's honestly hard to avoid using $$e$$.

• I wrongly asked about a role in kinematics in general, while I should have probably referred to kinematics in a straight line. I have less problem for identifying e in oscillatory phenomena, such as waves or harmonic oscillators. But your answer is very helpful anyhow. – Sierra Dec 3 '18 at 23:58
• @Sierra This is motion in a straight line -- everything is along the $x$-axis. – knzhou Dec 4 '18 at 12:06
• You are right of course. I am being very inaccurate… Actually what I meant is: in a non-oscillatory context, that is to say, a body moving without go-and-return trip, for example one moving in a straight line without change of direction. I have edited the question in this sense. – Sierra Dec 4 '18 at 12:26
• @Sierra This is not oscillatory motion, the function $e^{\omega t}$ doesn't oscillate. – knzhou Dec 4 '18 at 12:28
• @AaronStevens You're completely right, not sure where I was going with that statement. – knzhou Dec 10 '18 at 14:24

A simple example is just an object starting at rest falling with a drag force proportional to the velocity of the object, $$F_D=-bv$$. Then the acceleration is given by

$$a=\frac{dv}{dt}=g-\frac bmv$$

Therefore, the velocity over time is given by $$v(t)=\frac{mg}{b}(1-e^{-bt/m})$$

Typically you get $$e$$ popping up when the rate of change of something is proportional to itself.

• Thanks, in the end I accepted the other answer because there e appears alone in the solution and signifying displacement. But can you please clarify this? In the last comment I asked knzhou if he could build a similar example with a car speeding up. I could ask from you a similar clarification: what if the drag force were proportional to displacement, what if a car is progressively stepping on the brake in proportion to x? If x starts at 1 and v starts at 1, after t=1, would x be e? – Sierra Dec 10 '18 at 8:34
• @Sierra I mean getting the actual position or velocity to be $e$ is pretty contrived. In my example you can find a time where $x$ or $v$ is equal $e$ (with appropriate units). Even objects moving at a constant velocity will have a time where the distance traveled is equal to $e$. For example, for a particle moving with a constant velocity $v=e$, then after one second $x=e$. For a particle moving at constant velocity $v=e/2$, then after two seconds $x=e$. There are many possibilities where you can find some point in time where a value is $e$. – Aaron Stevens Dec 10 '18 at 14:07
• @Sierra What is more interesting, and what I thought you were going for, is looking at instances where $e$ is involved in the underlying dynamics of the problem. When there is something changing whose rate of change is proportional to itself. In other words, as just a number there really is nothing special about $e$. It is just a real number like all other real numbers. It becomes important when you start considering "special" functions like $e^x$ whose rate of change is equal to itself. – Aaron Stevens Dec 10 '18 at 14:09
• I am aware that e is just 2.718 + infinite decimals... and you can find a similar number anyhwhere (eg in simple interest), although probably not with exactly with same digits if the underlying situation does not match with what you call its dynamics. I am also aware that its dynamics are a function whose rate of change is itself. In fact, the Newton series that is included in the OP has the characteristic that for each term the derivative is the previous one. This series assumes 1 in x, v and t and gives e but you can tweak e's exponent and adapt to any other situation... – Sierra Dec 10 '18 at 17:05
• My question is that this series, with value e or e tweaked however you may imagine, reflects a situation with infinite derivatives that looked unreal. I then wondered if the case mentioned by knzhou does fit in spite of all, why is it and if another example would be one of those I mention (someone stepping on gas or brakes in proportion to displacement…). Do you get in those case jerk, snap, crackle, pop and so on ad infinitum and of dwindling values? – Sierra Dec 10 '18 at 17:10