When does a function have an addition theorem?

Motivating examples

The addition theorem for cosine says that

\cos(x + y) = \cos x \cos y - \sin x \sin y

and the addition theorem for hyperbolic cosine is analogous, though with a sign change.

\cosh(x + y) = \cosh x \cosh y + \sinh x \sinh y

An addition theorem is a theorem that relates a function’s value at x + y to its values at x and at y. The squaring function satisfies a very simple addition theorem

(x + y)^2 = x^2 + 2xy + y^2

and the Jacobi function sn satisfies a more complicated addition theorem.

\text{sn}(x + y) = \frac{ \text{cn}(x)\, \text{cn}(y) - \text{sn}(x) \,\text{sn}(y) \,\text{dn}(x) \,\text{dn}(y) }{ 1 - m \, \text{sn}^2(x) \,\text{sn}^2(y) }

Defining an algebraic addition theorem

Which functions have addition theorems? Before we can answer this question we need to be more precise about what an addition theorem is. We’ve said that an addition theorem for φ relates φ(x + y) to φ(x) and φ(y). But what exactly do we mean by “relate”? What counts as a relation?

Also, the examples above don’t exactly satisfy this definition. The addition law for cosines, for example, relates cos(x + y) to the values of cos(x) and cos(y) but also to sin(x) and sin(y). Somehow that feels OK because sine and cosine are related. But here again we’re talking about things being related without saying exactly what we mean.

Weierstrass (1815–1897) made the idea of an addition theorem precise and classified functions having addition theorems. A function satisfies an algebraic addition theorem if there is a polynomial F in three variables such that

F(\varphi(x + y), \varphi(x), \varphi(y)) = 0

For example, if φ(x) = x² then

\varphi(x+y)^2 - \left(\varphi(x)^2 + 2\varphi(x)\varphi(y) + \varphi(y)^2 \right) = 0

and so we could take F to be

F(a, b, c) = a^2 - b^2 - c^2 - 2bc

Similarly, if φ(x) = cos x then

\left(\varphi(x+ y) - \varphi(x) \varphi(y)\right)^2 - (1 - \varphi(x))^2 (1 - \varphi(y))^2 = 0

and so we could take F to be

Classifying functions with algebraic addition theorems

Now for Weierstrass’ theorem. A meromorphic function φ(z) has an algebraic addition theorem if and only if it is an elliptic function of z, a rational function of z, or a rational function of exp(λz).

A meromorphic function is one that is analytic everywhere except at isolated singularities. To put it another way, we assume φ has a convergent power series everywhere in the complex plane except at isolated points.

The examples above illustrate all three cases of Weierstrass’ theorem. The function sn(z) is elliptic, the function z² is rational, and the functions cos(z) and cosh(z) are rational functions of exp(iz).

Other kinds of addition theorems

Algebraic addition theorems are not the only kind of addition theorems. For example, Bessel functions satisfy a different kind of addition theorem:

J_n(x + y) = \sum_{k=-\infty}^\infty J_{n-k}(x) J_k(y)

This theorem relates the value of a Bessel function at x + y to the values of other Bessel functions at x and at y, but it is not an algebraic addition theorem because the right hand side is an infinite sum and because the Bessel functions are not algebraically related to each other.

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2 thoughts on “When does a function have an addition theorem?

  1. I have a related conjecture: If a complex function f(z) is a solution to a [Homogeneous Linear Differential Equation Of Constant Coefficients], then f(z) has what you call an addition theorem. I have proved it for infinitely many specific cases, but in the general case, my best guess is that the proof has something to do with separation of variables.

    My specific conjecture:

  2. For any addition theorem, you f(a+b)=[some function of a and b].
    If you replace b=c-a where c is a constant, then you get an invariant related to the function, f(c)=[some function of a].
    Invariants are interesting, and a way to think of it is that we can allow a to vary in any way, but the function of a never varies.

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