# What's so special about sine? (Concerning $y'' = -y$)

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'?)

• 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

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.

• 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.

• Darn, you beat me too it :) The simple harmonic oscillator is the perfect motivation for the $y''=-y$ definition – William 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 !

Theorem
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

• 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 $\gamma^{\,\prime}(t)$.

$\hspace{4cm}$

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.

• +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...

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.

• 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: home.scarlet.be/math/ellipse.htm – 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.

$\exp x$ 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.

Second derivative can be represented as following. Let write $$\mathrm f'(x)=\lim_{\Delta x\to 0}{\frac{\mathrm f(x)-\mathrm f(x-\Delta x)}{\Delta x}}$$ for current infinitesimal interval and $$\mathrm f'(x-\Delta x)=\lim_{\Delta x\to 0}{\frac{\mathrm f(x-\Delta x)-\mathrm f(x-2\cdot\Delta x)}{\Delta x}}$$ for previous, so \begin{align*} \mathrm f''(x)=\lim_{\Delta x\to 0}{\frac{\mathrm f'(x)-\mathrm f'(x-\Delta x)}{\Delta x}}=\lim_{\Delta x\to 0}{\frac{\frac{\mathrm f(x)-\mathrm f(x-\Delta x)}{\Delta x}-\frac{\mathrm f(x-\Delta x)-\mathrm f(x-2\cdot\Delta x)}{\Delta x}}{\Delta x}}&=\\=\lim_{\Delta x\to 0}{\frac{\mathrm f(x)-2\cdot \mathrm f(x-\Delta x)+\mathrm f(x-2\cdot\Delta x)}{\Delta x^2}} \end{align*} This formula can be checked using CAS for any function. Let use some trick from digital signal processing. Let $y_{n}=\mathrm f(x),\,y_{n-1}=\mathrm f(x-\Delta x),\,y_{n-2}=\mathrm f(x-2\cdot \Delta x)$. Using $\mathrm f''(x)=-\mathrm f(x)$ and $T_d=\Delta x$ (period of discretization) we can write: $$-y_n=\frac{1}{T_d^2}\cdot(y_n-2\cdot y_{n-1}+y_{n-2})$$ This implicit equation can be solved as $$y_n=\frac{2}{1+T_d^2}\cdot y_{n-1}-\frac{1}{1+T_d^2}\cdot{y_{n-2}}{}$$ Using non-zero initial conditions (IC) $y_{0}=1,\,y_{1}=0$ we can plot impulse response for given $T_d=0.1$ (e.g.): (Transients) we can see exponential attenuation. This recurrence equation represents 2-nd order IIR filter without using input signal (impulse response is formed with non-zero initial conditions only). Using relationship between coefficients of discrete filter $$\mathrm {Fd}(z)=\frac{1}{1+a_1\cdot z^{-1}+a_2\cdot z^{-2}}=\frac{z^2}{z^2+z\cdot a_1+a_2}$$ and its recurrence equation $y_n=-a_1\cdot y_{n-1}-a_2\cdot y_{n-2}$ we can find (according to $a_1=-\frac{2}{1+T_d^2}$ and $a_2=\frac{1}{1+T_d^2}$) poles $z_{1,2}=\frac{1}{1+T_d^2}\pm I\cdot \frac{T_d}{1+T_d^2}$ (always at internal area of unit circle for any $T_d$ - stable system) and complex frequency response of this transfer function using $z=\mathrm e^{I\cdot 2\cdot \pi \cdot f\cdot T_d}$ substitution where $f$ is test frequency. For given transfer function $\mathrm {Fd}(z)$ and coefficients we can see non-dissipative process with discrete sine only at $T_d=0$ value where $y_n=2\cdot y_{n-1}- y_{n-2}$ (straight line response for any IC) and transfer function becomes $\frac{z^2}{z^2-2\cdot z+1}=\frac{z^2}{(z-1)^2}$ with 2-nd order pole so this method is not allow to produce exact sine samples. To "correct" transfer function to generate exact discrete samples of sine wave with unit amplitude and given frequency we need to substitute $z=\mathrm e^{I\cdot 2\cdot \pi \cdot f\cdot T_d}$ to $\frac{z^2}{z^2+z\cdot a_1+a_2}$ and solve equations relative to $a_1,a_2$ using following conditions. Substituting complex exponent to this transfer function leads to following complex frequency characteristic: $$\frac{1+\cos \left(2\pi f\cdot T_d \right)\cdot a_1+a_2\cdot \cos \left(4\pi f\cdot T_d \right) + I\cdot \left( \sin \left(2 \pi f\cdot T_d \right)\cdot a_1+a_2\cdot \sin \left(4\pi f\cdot T_d \right) \right)}{ 2\cdot \left(a_1\cdot a_2+a_1 \right)\cdot \cos \left(2\pi f\cdot T_d \right)+2\cdot a_2\cdot \cos \left(4\pi f\cdot T_d \right)+a_1^2+a_2^2+1}$$ Now solve denominator of this function relative for example to $a_2$. It means that we have pole at this point which is defined as dependence of $a_2$ from $a_1,f,Td$: $$a_2=-\left(\cos \left(4\pi f\cdot T_d \right)+a_1\cdot\cos \left(2\pi f\cdot T_d \right)\right)+I\cdot\sin\left(2\pi f\cdot T_d \right)\cdot \left(2\cdot\cos \left(2\pi f\cdot T_d \right)+a_1\right)$$ and we have simple condition for $a_2$ to be only real: $2\cdot\cos \left(2\pi f\cdot T_d \right)+a_1\ = 0$ so $a_1=-2\cdot\cos \left(2\pi f\cdot T_d \right)$. Substitute $a_1$ to $a_2$ leads to simple answer $a_1=-2\cdot\cos \left(2\pi f\cdot T_d \right),\,a_2=1$. So recurrence relation $$y_n=-a_1\cdot y_{n-1}-a_2\cdot y_{n-2}=2\cdot\cos \left(2\pi f\cdot T_d \right)\cdot y_{n-1}-y_{n-2}$$ generates discrete sine for given discretization $T_d$. IC can be found as: \begin {align*} y_0=-\sin\left(2\pi f\cdot T_d \right)\\ y_1=\sin\left(4\pi f\cdot T_d \right) \end {align*} using this IC and recurrence relation unit amplitude sine wave is generated for any non-infinitesimal $T_d$ so we calculate exact answer for given ODE at any discrete point. $f$ must be chosen up to value of Nyquist frequency so $f<1/(2\cdot T_d)$. This recurrence relation can be used to generate the cosine with $y_0=1,y_1=\cos\left(2\pi f\cdot T_d \right)$ IC's. To practical issues substitution $f=Fs/T_d$ can be used where $Fs$ is fixed frequency that we need. We can write $$\mathrm f''(x)=\lim_{\Delta x\to 0}{\frac{\mathrm f(x)\cdot\left(1-(2\pi f\Delta x)^2\right)-2\cdot \cos\left(2\pi f\cdot \Delta x \right) \cdot \mathrm f(x-\Delta x)+\mathrm f(x-2\cdot\Delta x)}{\Delta x^2}}$$ and check by CAS for any function. Note that we make difference equations with exact discrete solution only for appropriate ODE. One of the difficulties of this approach that we need to compute coefficients by sine and cosine to generate sine and cosine. It seems to be main property of difference equation is ability to find coefficients of difference equations (analytical or numerically) which are give exact sample points values for analytical solution of linear same order ODE (?).

• "This can be verified with CAS for any function"... How is "any function" made? – user228113 Aug 4 '16 at 13:07
• Of course these formulas for second derivative needs strict proof that they are really define the second derivative but we can check it for some function using direct substitution. For example let f(x)=tan^3(1/x) or something random algebraic expression, we can use CAS to compute f'' and to find limit and obtain zero difference between results. This "proof" seems to be heuristic and word "any" may be mean "some type of" but whatever substitutions were made the difference was not found and for practical purposes, such "proof" may be applied ad hoc. – Timur Zhoraev Aug 4 '16 at 13:58