Todd Trimble
Characterization of sine


Functional equation characterization

The nLab page on pi characterizes the sine function as the unique continuous function f:f: \mathbb{R} \to \mathbb{R} such that

In this section we will check that this really is a definite description (i.e., that only one function satisfies this). It’s clear that f(x)=sin(x)f(x) = \sin(x) indeed satisfies these conditions.

By far the more significant condition is the second one, which we’ll call the functional equation. The first condition could be considered a tameness condition; its utility here is that it implies f(0)=0f(0) = 0 and f(0)=1f'(0) = 1, which effectively eliminates pathological solutions.

The functional equation implies that ff is uniquely determined from its germ at x=0x = 0 (i.e., that the restriction of ff to any neighborhood UU of 00 determines ff globally: this is because \mathbb{R} is the union of neighborhoods of the form 3 nU3^n U).

Also we may deduce that |f(x)|1{|f(x)|} \leq 1 for all xx. Certainly 1f(x)1-1 \leq f(x) \leq 1 is true for all xx in some small neighborhood UU of 00. One then proves that 1f(x)1-1 \leq f(x) \leq 1 for all x3 nUx \in 3^n U by induction on nn and with the help of the functional equation, using the fact that the function uu defined by u(x)3x4x 3u(x) \coloneqq 3 x - 4 x^3 maps [1,1][-1, 1] to itself.

The plan of attack will be to reconstruct from ff (which plays the role of the sine) a second function gg that plays the role of the cosine. (You can’t just define g=1f 2g = \sqrt{1 - f^2} since we know the sign has to change at appropriate spots.) The trick is first to get the germ of gg at zero, and invoke a companion functional equation which can be used to propagate out to a function gg defined on all of \mathbb{R}. Then, one considers the corresponding complex exponential g+ifg + i f, with its own companion functional equation which is used to deliver the coup de grace.


Let a>0a \gt 0 be chosen small enough so that |f(x)|<1/2{|f(x)|} \lt 1/2 whenever x[a,a]x \in [-a, a]. Define gg on [a,a][-a, a] by g(x)=1f(x) 2g(x) = \sqrt{1 - f(x)^2}. Then

g(3x)=4g(x) 33g(x)g(3 x) = 4g(x)^3 - 3g(x)

for all x[a/3,a/3]x \in [-a/3, a/3].


For x[a/3,a/3]x \in [-a/3, a/3] we have

4g(x) 33g(x) = g(x)(g(x) 23(1g(x) 2)) = g(x)(1f(x) 23f(x) 2) = g(x)(14f(x) 2)\array{ 4g(x)^3 - 3g(x) & = & g(x)(g(x)^2 - 3(1 - g(x)^2)) \\ & = & g(x)(1 - f(x)^2 - 3f(x)^2) \\ & = & g(x)(1 - 4f(x)^2) }

which is positive by choice of aa. Hence both g(3x)g(3 x) and 4g(x) 33g(x)4g(x)^3 - 3g(x) are positive for x[a/3,a/3]x \in [-a/3, a/3], so it suffices to show their squares are equal. In other words, that

1f(3x) 2=(4g(x) 33g(x)) 2=g(x) 2(14f(x) 2) 2.1 - f(3 x)^2 = (4g(x)^3 - 3g(x))^2 = g(x)^2(1 - 4f(x)^2)^2.

This boils down to checking

1(3f(x)4f(x) 3) 2=(1f(x) 2)(14f(x) 2) 21 - (3f(x) - 4f(x)^3)^2 = (1 - f(x)^2)(1 - 4f(x)^2)^2

but this follows from the polynomial identity 1(3t4t 3) 2=(1t 2)(14t 2) 21 - (3t - 4t^3)^2 = (1 - t^2)(1 - 4t^2)^2, which is easily checked.

Now extend gg to a function \mathbb{R} \to \mathbb{R} by recursion: put g 0=g:[a,a]g_0 = g: [-a, a] \to \mathbb{R} and define g n:[3 na,3 na]g_n: [-3^n \cdot a, 3^n \cdot a] \to \mathbb{R} by g n(x)4g n1(x/3) 33g n1(x/3)g_n(x) \coloneqq 4g_{n-1}(x/3)^3 - 3g_{n-1}(x/3).


The equation g n+1(x)=g n(x)g_{n+1}(x) = g_n(x) holds for x[3 n1a,3 n1a]x \in [-3^{n-1} a, 3^{n-1}a].


The case for n=0n=0 is just the preceding lemma. The rest is by induction: let vv be the function v(x)=4x 34xv(x) = 4x^3 - 4x, and use the fact that for n>0n \gt 0 and x[3 n1a,3 n1a]x \in [-3^{n-1}a, 3^{n-1}a] we have

g n+1(x)=(vg n)(x/3)=(vg n1)(x/3)g n(x)g_{n+1}(x) = (v \circ g_n)(x/3) = (v \circ g_{n-1})(x/3) \coloneqq g_n(x)

where the second equation follows from the inductive hypothesis.

By this lemma, the g ng_n patch together to form a continuous global function g:g: \mathbb{R} \to \mathbb{R} that satisfies the functional equation g(3x)=4g(x) 3g(x)g(3x) = 4g(x)^3 - g(x).


The continuous function h=g+if:h = g + i f: \mathbb{R} \to \mathbb{C} satisfies h(0)=1h(0) = 1, h(3x)=h(x) 3h(3 x) = h(x)^3, and |h(x)|=1{|h(x)|} = 1 for all xx.


By our earlier choice of aa, we have g(x) 2+f(x) 2=1g(x)^2 + f(x)^2 = 1 for x[a,a]x \in [-a, a]. One shows by induction that the equations hold for x[3 na,3 na]x \in [-3^n a, 3^n a], using the functional equations and g(x) 2+f(x) 2=1g(x)^2 + f(x)^2 = 1 for x[3 n1a,3 n1a]x \in [-3^{n-1}a, 3^{n-1}a]:

(g(x)+if(x)) 3 = g(x) 33g(x)f(x) 2+i(3g(x) 2f(x)f(x) 3) = g(x) 33g(x)(1g(x) 2)+i(3(1f(x) 2)f(x)f(x) 3) = 4g(x) 33g(x)+i(3f(x)4f(x) 3) = g(3x)+if(3x)\array{ (g(x) + i f(x))^3 & = & g(x)^3 - 3g(x) f(x)^2 + i(3g(x)^2 f(x) - f(x)^3) \\ & = & g(x)^3 - 3g(x)(1 - g(x)^2) + i(3(1 - f(x)^2)f(x) - f(x)^3) \\ & = & 4g(x)^3 - 3g(x) + i(3f(x) - 4f(x)^3) \\ & = & g(3 x) + i f(3 x) }

so that h(3x)=h(x) 3h(3 x) = h(x)^3, and in particular h(3x)h(3 x) is on the unit circle provided that h(x)h(x) is, so that the induction goes through.

We also have of course h(0)=ih'(0) = i. Intuitively, this means that for any tt we have

h(t)=h(t3 n) 3 n(1+it3 n) 3 nh(t) = h(\frac{t}{3^n})^{3^n} \sim (1 + \frac{i t}{3^n})^{3^n}

where the right side tends to exp(it)\exp(i t) as nn \to \infty. In other words, h(t)=exp(it)h(t) = \exp(i t), whence f(t)=sin(t)f(t) = \sin(t) by matching imaginary parts. The goal now is to make this intuitive argument rigorous.


For any given tt \in \mathbb{R} and η>0\eta \gt 0, there exists NN such that

|h(t)(1+it3 n) 3 n|<η{|h(t) - (1 + \frac{i t}{3^n})^{3^n}|} \lt \eta

whenever nNn \geq N.

Since (as is well-known) lim m(1+itm) m=exp(it)lim_{m \to \infty} (1 + \frac{i t}{m})^m = \exp(i t), it follows at once from this proposition that h(t)=exp(it)h(t) = \exp(i t).


Since h(0)=ih'(0) = i and h(0)=1h(0) = 1, we can for any ϵ>0\epsilon \gt 0 choose NN so large that

|h(t3 n)1it3 n|ϵ|t3 n|{|h(\frac{t}{3^n}) - 1 - i\frac{t}{3^n}|} \leq \epsilon\; {|\frac{t}{3^n}|}

for all nNn \geq N. In other words, for such nn we have h(t3 n)=1+it3 n+ζt3 nh(\frac{t}{3^n}) = 1 + \frac{i t}{3^n} + \zeta\frac{t}{3^n} for some ζ\zeta with |ζ|ϵ{|\zeta|} \leq \epsilon. There we have

h(t)=h(t3 n) 3 n=(1+it3 n+ζt3 n) 3 nh(t) = h(\frac{t}{3^n})^{3^n} = (1 + \frac{i t}{3^n} + \zeta\frac{t}{3^n})^{3^n}

and so we proceed to bound the difference

(1+it3 n+ζt3 n) 3 n(1+it3 n) 3 n.(1 + \frac{i t}{3^n} + \zeta\frac{t}{3^n})^{3^n} - (1 + \frac{i t}{3^n})^{3^n}.

Put m=3 nm = 3^n. We have

(1+itm+ζtm) m= k=0 m(mk)(1+itm) mk(ζtm) k(1 + \frac{i t}{m} + \zeta\frac{t}{m})^m = \sum_{k=0}^m \binom{m}{k} (1 + \frac{i t}{m})^{m-k} (\frac{\zeta t}{m})^k

so that the difference above is bounded above by

k=1 m(mk)(1+|tm|) mk(|ζtm|) k.\sum_{k=1}^m \binom{m}{k} (1 + {|\frac{t}{m}|})^{m-k} ({|\frac{\zeta t}{m}|})^k.

Clearly we have

(1+|tm|) mk(1+|tm|) mexp(|t|).(1 + {|\frac{t}{m}|})^{m-k} \leq (1 + {|\frac{t}{m}|})^m \leq \exp({|t|}).

Hence the difference is bounded above by

exp(|t|) k=1 m(mk)(|ζtm|) k=exp(|t|)[(1+|ζtm|) m1]exp(|t|)[exp(|ζt|)1]exp(|t|)[exp(|ϵt|)1].\exp({|t|}) \cdot \sum_{k=1}^m \binom{m}{k} ({|\frac{\zeta t}{m}|})^k = \exp({|t|}) \cdot [(1 + {|\frac{\zeta t}{m}|})^m - 1] \leq \exp({|t|}) \cdot [\exp({|\zeta t|}) - 1] \leq \exp({|t|}) \cdot [\exp({|\epsilon t|}) - 1].

This can be made as small as desired, by taking ϵ\epsilon small enough. This completes the proof.

Delange’s characterization

A very pretty characterization of the sine is due to Hubert Delange:


Suppose f:f: \mathbb{R} \to \mathbb{R} is a C C^\infty function such that all of its n thn^{th} derivatives satisfy |f (n)(x)|1{|f^{(n)}(x)|} \leq 1 for all xx, and f(0)=1f'(0) = 1. Then f(x)=sin(x)f(x) = \sin(x) for all xx.

A proof may be found in this online text by Jean Saint Raymond, Théorème 14.4.3 (page 156). I will follow that proof here.

First, a few preliminary observations. Put

F(z)= k0f (k)(0)z kk!F(z) = \sum_{k \geq 0} f^{(k)}(0) \frac{z^k}{k!}

which is an entire analytic function. By the Taylor remainder formula for C nC^n functions on \mathbb{R}, we have

|f(x) k=0 n1f (k)(0)x kk!|sup z|f (n)(z)||x| nn!|x| nn!{|f(x) - \sum_{k = 0}^{n-1} f^{(k)}(0) \frac{x^k}{k!}|} \leq \sup_z {|f^{(n)}(z)|}\; \frac{{|x|}^n}{n!} \leq \frac{{|x|}^n}{n!}

so that taking nn \to \infty, we infer f(x)=F(x)f(x) = F(x) for all xx \in \mathbb{R}. Thus the real-valued ff analytically continues to an entire analytic function FF, and we get to invoke the theory of entire analytic functions.

The order of FF as an entire function is 11 because we have the growth rate

|F(x+iy)|=| n0f (n)(x)(iy) nn!| n0|iy| nn!=exp(|y|).{|F(x + i y)|} = {|\sum_{n \geq 0} f^{(n)}(x) \frac{(i y)^n}{n!}|} \leq \sum_{n \geq 0} \frac{{|i y|}^n}{n!} = \exp({|y|}).

Now let aa be an arbitrary point in the open interval (π2,π2)(-\frac{\pi}{2}, \frac{\pi}{2}); we will prove f(a)=sin(a)f(a) = \sin(a) (whence f(x)=sin(x)f(x) = \sin(x) by uniqueness of analytic continuation).


D x=af(x)cos(x)= k(1) kf(kπ+π2)(kπ+π2a) 2D_{x = a} \frac{f(x)}{\cos(x)} = \sum_{k \in \mathbb{Z}} (-1)^k \frac{f(k\pi + \frac{\pi}{2})}{(k\pi + \frac{\pi}{2} - a)^2}.


Let gg be the meromorphic function defined by

g(z)F(z)(za) 2cos(z)g(z) \coloneqq \frac{F(z)}{(z - a)^2 \cos(z)}

where cos(z)=e iz+e iz2\cos(z) = \frac{e^{i z} + e^{-i z}}{2}. Consider the square contour C nC_n with vertices at the points nπ(±1±i)n\pi(\pm 1 \pm i), n1n \geq 1. We have

  • cos(x+iy)=cos(x)cosh(y)+isin(x)sinh(y)\cos(x + i y) = \cos(x)\cosh(y) + i \sin(x)\sinh(y),

  • |cos(x+iy)| 2=cos 2(x)cosh 2(y)+sin 2(x)sinh 2(y)=cos 2(x)(1+sinh 2(y))+sin 2(x)sinh 2(y)=cos 2(x)+sinh 2(y){|\cos(x + i y)|}^2 = \cos^2(x)\cosh^2(y) + \sin^2(x)\sinh^2(y) = \cos^2(x)(1 + \sinh^2(y)) + \sin^2(x)\sinh^2(y) = \cos^2(x) + \sinh^2(y)

where cos 2(x)+sinh 2(y)=1+sinh 2(y)=cosh 2(y)\cos^2(x) + \sinh^2(y) = 1 + \sinh^2(y) = \cosh^2(y) along the vertical line segments of C nC_n, and cos 2(x)+sinh 2(nπ)sinh 2(nπ)\cos^2(x) + \sinh^2(n\pi) \geq \sinh^2(n\pi) along the horizontal line segments of C nC_n (where y=±nπy = \pm n\pi). Either way we have the easy estimate

|cos(x+iy)|e |y|3{|cos(x + i y)|} \geq \frac{e^{{|y|}}}{3}

over the contour C nC_n. Moreover 1|za| 21(nπ|a|) 2\frac1{{|z-a|}^2} \leq \frac1{(n\pi - {|a|})^2} over the contour, and

|F(z)||cos(z)|e |y|3e |y|3\frac{{|F(z)|}}{{|cos(z)|}} \leq e^{{|y|}} \cdot \frac{3}{e^{{|y|}}} \leq 3

over the contour according to the growth rate of F(z)F(z) observed above, and so

| C ng(z)dz|(sup|g(z)|)length(C n)3(nπ|a|) 2(8nπ){|\int_{C_n} g(z) d z|} \leq (\sup {|g(z)|}) \cdot length(C_n) \leq \frac{3}{(n\pi - {|a|})^2} \cdot (8 n \pi)

which tends to 00 as nn \to \infty. On the other hand, by a simple residue calculation, we calculate

12πi C ng(z)dz=D z=aF(z)cos(z)+ k=n n1(1) k+1F(π2+kπ)(π2+kπa) 2\frac1{2\pi i} \int_{C_n} g(z) d z = D_{z = a} \frac{F(z)}{\cos(z)} + \sum_{k = -n}^{n-1} (-1)^{k+1} \frac{F(\frac{\pi}{2} + k\pi)}{(\frac{\pi}{2} + k\pi - a)^2}

where the first term on the right comes from the double pole of g(z)g(z) at z=az = a and the rest come from the zeroes of cos(z)\cos(z). Thus in the limit we have

0=D x=af(x)cos(x)+ k= (1) k+1f(π2+kπ)(π2+kπa) 20 = D_{x = a} \frac{f(x)}{\cos(x)} + \sum_{k = -\infty}^{\infty} (-1)^{k+1} \frac{f(\frac{\pi}{2} + k\pi)}{(\frac{\pi}{2} + k\pi - a)^2}

which completes the proof.

In particular, for the case f(x)=sin(x)f(x) = \sin(x), we have


D x=asin(x)cos(x)= k1(π2+kπa) 2D_{x = a} \frac{\sin(x)}{\cos(x)} = \sum_{k \in \mathbb{Z}} \frac1{(\frac{\pi}{2} +k\pi - a)^2}.

In particular, taking a=0a = 0, we have

1= k1(π2+kπ) 21 = \sum_{k \in \mathbb{Z}} \frac1{(\frac{\pi}{2} +k\pi)^2}

and so

0=1f(0)= k1(1) kf(π2+kπ)(π2+kπ) 2.0 = 1 - f'(0) = \sum_{k \in \mathbb{Z}} \frac{1 - (-1)^k f(\frac{\pi}{2} + k\pi)}{(\frac{\pi}{2} + k\pi)^2}.

But since |f(x)|1{|f(x)|} \leq 1 for all xx, we have 1(1) kf(π2+kπ)01 - (-1)^k f(\frac{\pi}{2} + k\pi) \geq 0. Thus the last equation forces

(1) kf(π2+kπ)=1(-1)^k f(\frac{\pi}{2} + k\pi) = 1

for all kk \in \mathbb{Z}. Plugging this back into the lemma, we deduce

D x=af(x)cos(x)= k1(kπ+π2a) 2=D x=asin(x)cos(x).D_{x = a} \frac{f(x)}{\cos(x)} = \sum_{k \in \mathbb{Z}} \frac1{(k\pi + \frac{\pi}{2} - a)^2} = D_{x = a} \frac{\sin(x)}{\cos(x)}.

As this is true for every a(π2,π2)a \in (-\frac{\pi}{2}, \frac{\pi}{2}), we see f(a)sin(a)=λcos(a)f(a) - \sin(a) = \lambda \cos(a) for some constant λ\lambda, for all aa in this range. Thus f(a)=cos(a)λsin(a)f'(a) = \cos(a) - \lambda \sin(a).

Finally, f(π2)=1f(\frac{\pi}{2}) = 1 and also f(π2)=λf'(\frac{\pi}{2}) = -\lambda. But as 11 is a maximum value, this forces λ=0-\lambda = 0, and thus we have shown f(a)=sin(a)f(a) = \sin(a). The proof of Delange’s theorem is now complete.

Roe’s characterization

A slightly different characterization was given by J. Roe (first name is?):


Suppose there is a collection of functions ,f 2,f 1,f 0,f 1,f 2,\ldots, f_{-2}, f_{-1}, f_0, f_1, f_2, \ldots such that f n+1=f nf_{n+1} = f_{n}' and sup|f n|M\sup {|f_n|} \leq M for all nn, for some uniform bound M0M \geq 0. Then f=f 0f = f_0 is of the form f(x)=asin(x+b)f(x) = a\sin(x + b) for some constants a,ba, b.

At first this might appear to be a weakening of Delange’s theorem, in that one considers a sequence {f n} n\{f_n\}_{n \in \mathbb{Z}} that is infinite in both positive and negative directions, rather than the positive direction only. Notice however that Delange assumes that |f|{|f'|} actually attains this bound MM (by shifting the sequence, he could ask merely that one of the |f (n)|{|f^{(n)}|} attains this MM, for some n1n \geq 1). For example, as pointed out by Roe, the function f(x)=sin(x)+sin(x/π)f(x) = \sin(x) + \sin(x/\pi) has all of its derivatives bounded – but observe that no derivative actually hits a uniform upper bound MM.

It seems odd to me that Roe makes no reference whatever to Delange’s earlier work, as their statements are so close in spirit. His method of proof is different to the complex-analytic one given above; Roe uses Fourier analysis and distribution theory instead.


Revised on August 3, 2019 at 09:39:59 by Todd Trimble