Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

In an attempt to actually grok sine, I came across the $y''= -y$ definition.

This is incredibly cool, but it leads me to a whole new series of questions. Sine seems pretty prevalent everywhere in life (springs, sound, circles...) and I have to wonder, what's so special about the second derivative in this scenario?

In other words, why does nature / math seem to care more about the scenario where $y'' = -y$ instead of, say, $y' = -y$ or $y''' = y$?

Why is acceleration equal to negative the magnitude such a recurring theme in math and nature, while velocity equal to negative the magnitude ($y'=-y$) or jerk equal to negative the magnitude ($y'''=-y$) are seemingly unimportant?

In other words, what makes sine so special?

Note that this question also sort of applies to $e$, which satisfies $y'' = y$.

(Edit: Yes, I understand that $e$ and $\sin$ are closely related. I'm not looking for a relationship between $e$ and $\sin$.

Rather, I'm wondering why these functions in particular, which both arise from a relationship between a function and its own second derivative, are so prevalent. For example, do functions satisfying $y'''=-y$ also recur frequently, and I just haven't noticed them? Or is the second derivative in some way 'important'?)

share|cite|improve this question
There is a reasonable case that $y'=-y$ is just as important, or more important. – André Nicolas Feb 28 '12 at 18:33
Note that more fundamentally $y=e^{ix}$ satisfies $y''=-y$. – Alex Becker Feb 28 '12 at 18:34
Among the solutions of $y'''=-y$ is $y = e^{-t/2}(\cos(\sqrt{3}t/2) + i\sin(\sqrt{3}t/2) )$. Notice that this is in a sense built up from solutions of first- and second-degree differential equations. – Michael Hardy Feb 28 '12 at 19:10
A solution to $y'=ky$ is $e^{kx}$, hardly unimportant. – guy Feb 28 '12 at 19:28
I recently had a question on things like $y'''=-y$. See here. Very nice answer included. – draks ... Feb 28 '12 at 19:35
up vote 20 down vote accepted

This answer is perhaps more suited to physics.SE, but anyways.

I can see why you're wondering about the "specialness" of y'' = -y (or more generally -ky). Let me try to give an explanation of why it shows up all over the place.

Imagine a ball sitting at the bottom of a round well (looking like a U), its equation of motion is $y = 0$, its height is constant and not changing (let's say it is 0). Imagine you disturb this ball very slightly, you raise it to a very small height $\epsilon$ along the wall of the well. There's a force that will act on it to return it to 0. This force is a function of $\epsilon$, $F(\epsilon)$.

We can expand $F(\epsilon) = a_0 + a_1 \epsilon + a_2 \epsilon^2 + ...$ Noting that F(0) = 0, we get $a_0 = 0$. Since $\epsilon$ is small, $\epsilon^2$ and all higher powers are just very small.. So we will ignore them. So we get $F(\epsilon) \approx - k \epsilon$ (the negative sign is there because the function pulls the ball down). We can conclude that $F(y) = -k y$, around the equilibrium.

We know that the force is proportional to accleration (at low speeds). Therefore, $m y'' = -k y$ and thus $y'' = -c y$. This means that any small motion around a stable equilibrium is approximately a sinusoid. Now a circular motion is just two oscillations (one on each axis).

This is all just a consequence of the fact that our universe seems to favor equations of the second degree, since the force is proportional to the acceleration (not to the velocity and not to the change of acceleration). Why is this the case? One can trace that to the action (Hamiltonian) of the physical system or even to the conservation laws (which follow from the invariants, for example, that the physics won't change if we shift everything up or down, or if we try the experiment at a later time). Still these are just features of our universe.

share|cite|improve this answer
Thanks! I suppose I probably should take this question over to physics, I'm also curious as to why physics seems to care so much about the second derivative (in everything from F=ma to the position/momentum interaction). – Nate Feb 29 '12 at 13:11

From a physical perspective, this comes from the prevalence of simple harmonic oscillators. A SHO is a system in which the force is linear in $\mathbf{x}$ and directed opposite to the displacement from equilibrium. If you put that into Newton's second law, you get

$$\mathbf{F}_\text{net} = -k\mathbf{x} = m\ddot{\mathbf{x}}$$

With a suitable change of variables, this is just your differential equation $y''=-y$.

It turns out that nearly every bounded system in physics can be modeled as a simple harmonic oscillator, at least as a leading approximation. This is related to the tendency of physical systems to seek out a stable equilibrium in some potential function $V(\mathbf{x})$. At a stable equilibrium position, the first derivative of $V(\mathbf{x})$ vanishes, and because only potential differences are physically meaningful, the actual value at equilibrium can be set to zero, so the leading nontrivial term in a Taylor series expansion of the potential is the quadratic term, $V(\mathbf{x}) \approx V''(\mathbf{x}_0)(\mathbf{x}-\mathbf{x}_0)^2$. And because $\mathbf{F}=-\nabla V$, this potential gives you the behavior of a SHO for small displacements.

share|cite|improve this answer
Darn, you beat me too it :) The simple harmonic oscillator is the perfect motivation for the $y''=-y$ definition – you Feb 28 '12 at 19:20
"Nearly every"? Which ones can't? Also, given the taylor series approximation for sine, can't anything approximated by a harmonic oscillator be modeled by even simpler functions? – Nate Mar 3 '12 at 1:42
(1) There are such things as quartic oscillators where the leading term in the potential is $x^4$; these would not be considered harmonic. Or in classical mechanics, you can have damped oscillators which are subject to significant drag forces or friction; those are not considered simple. (2) I'm afraid I don't see what you're getting at with the Taylor series approximation for the sine function. How do you propose to approximate a harmonic oscillator by "simpler" functions, and perhaps more importantly, why? – David Z Mar 3 '12 at 2:00

I guess this is not quite the expected answer, but it's definitely something very special about sine.

"What makes sine so special?" It is simply the wisest and the most moderate real function that could possibly exist !

Let $f : \Bbb R \to \Bbb R$ a function of class $\mathcal C^\infty$. Assume that :

  1. $f'(0) = 1$ ;
  2. for all non-negative integer $n$, the function $|f^{(n)}|$, the modulus of the $n$th derivative of $f$, is bounded by 1.

Then $f$ is the function sine. #

There are slightly stronger forms of this theorem, but I don't remember them... Everything is in the RMS, issue 116-3. Table of contents

share|cite|improve this answer
Such a beauty ! – overcoder Feb 28 '12 at 23:10

In the following animation, the red arrow inside the circle traces out $\gamma(t)=(\cos(t),\sin(t))$ and red arrow outside the circle shows the tangent to the circle at $\gamma(t)$.

The blue arrow inside the circle shows $\gamma^{\,\prime}(t)$, which is parallel to the tangent at $\gamma(t)$. The blue arrow outside the circle shows the tangent to the circle at $\gamma^{\,\prime}(t)$.

The green arrow shows $\gamma^{\,\prime\prime}(t)$, which is parallel to the tangent at $\gamma^{\,\prime}(t)$.

$\hspace{4cm}$harmonic motion

Note that the green arrow is pointed in precisely the opposite direction as the red arrow inside the circle. Thus, we have $\gamma^{\,\prime\prime}(t)=-\gamma(t)$. Therefore, $\cos^{\,\prime\prime}(t)=-\cos(t)$ and $\sin^{\,\prime\prime}(t)=-\sin(t)$.

Periodic functions can be broken down into circular components. This is the main idea behind Fourier Analysis. If one of the circular components has a larger amplitude than the others, the whole function appears essentially circular. In two dimensions, if the amplitude in one of the directions is larger, the motion will appear one-dimensional, but it is still harmonic motion, where $x''=-kx$.

At least to me, this is what makes the circular functions ($\sin$ and $\cos$) so special; in a way, they are the basic periodic functions.

share|cite|improve this answer
+1, If a picture worth a thousand words, a moving picture must be worth a $1000000$ words. – NoChance Feb 29 '12 at 9:02
@Emmad, or maybe 1000*(number of frame in the animation) words – Zeta Two Feb 29 '12 at 9:55

It's good to have the point of vue of the sinus by the definition of the exp. In fact we have :$\sin(x) = \operatorname{Im}(e^{ix})$ and the definition of $e^{ix}$ is very visual by the equation $y'=iy$ : you must take the tangent direction of the circle for all $x$. $\mathrm{sinus}$ is a special function because $\exp$ too...

share|cite|improve this answer

In other words, what makes sine so special?

You may agree that Trigonometric functions are related to circles, spheres and ellipses. If you look at our universe, you will see that we live on earth and the earth's shape, as well as other stars and planets, are likely to be spherical and that we go in circular-like path around the sun, and so on and so forth. So the universe favors this kind of family of shapes. May be it has do with energy preservation or forces equilibrium for some reason, but it is definitely a favorite shape in the universe. Accordingly, functions that describe such shapes are popular in the sciences that study nature.

share|cite|improve this answer
I understand that, but I don't completely buy this. Wherever there is a circle in nature, looking closer we find it not a circle. Orbits are actually ellipses, the earth isn't quite spherical, and so on. – Nate Mar 3 '12 at 1:49
@Nate, thanks for your comment. Ellipse equations use SIN and COS see for example: – NoChance Mar 3 '12 at 16:54

This comes from the fact that in physics, nearly any oscillating system can be approximated to simple harmonic motion (SHM).

Oscillating systems are pretty common all over physics. Now, $y'=-y,y'''=-y$ are not oscillating equations. $y''=-y$ is perfectly oscillating. In physics, we write $\vec{F}=-m\omega^2\vec{x}$ or $\vec{a}=-\omega^2\vec{x}$, but it's the same thing in the end.

Now, one may ask, why not $y''=-y^2$ of something? The great part about SHM is that since on differentiation we get the linear term $dy$, anything of the form $y''=f(y)$ can be approximated to an SHM, like this: (I'm using $a$ for acceleration, $x$ for displacement as the dashes will make it more confusing)

If $$a=-f(x)$$ then $$da=-f'(x)dx$$ For small changes in x,$f'(x)$ is nearly constant. Thus, $$da\propto dx$$ $$ \therefore a \propto x$$ Here, the fact that the RHS has a linear $y$ term and nothing else is obvious from the fact that on differentiation of small changes, you always get something linear in $dy$.

I doubt there's any natural reason for $y'''=-y$ not being there, just that we don't usually have anything to do with jerks in physics. We have forces dependent of other stuff, but usually no "yank" (mass times jerk), except one that is indirectly inferred from equations.

Aside from that, as I said, oscillating systems are pretty common. All of them can be written as $y''=-f(y)$, so they can be all approximated as SHM sine functions as above.

$y'=-y$ is pretty common. In capacitor-resistor or resistor-inductor circuits, we get that equation leading to a $1-e^{-t/k}$ term everywhere. $y'=-y$ comes in most decay situations which are again pretty common. Again, we can approximate any decay to $y'=-y$, but it's not good practice as the approximation isn't nearly as good as the SHM one when applied to everyday scenarios.

So simply: Most decay situations give you $y'=-y$ (an $e^{-x}$ function); most oscillating situations can be approximated to $y''=-y$ (sin or cos). This makes the two differential equations pretty common.

share|cite|improve this answer

$\exp x$ satisfies $y'=y$, $y''=y$, $y'''=y$, ... ad infinitum

$\sin x$ and $\cos x$ satisfy $y''=-y$, $y''''=y$, ...

$-\exp x$ satisfies $y'=-y$, $y''=y$, $y'''=-y$, $y''''=y$, ...

$\sinh x$ and $\cosh x$ satisfy $y''=y$, $y''''=y$, ...

all of these functions are widespread in nature because they satisfy all these differential equations. There is nothing special with 2nd derivative here.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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