Sine Angle Addition Functional Equation 2

In this post, we continue the discussion from this post, where we investigate the sine angle addition functional equation

$\displaystyle f(a + b) = f(a)f\left( \frac{\pi}{2} - b \right) + f\left( \frac{\pi}{2} - a \right)f(b).$

What if we generalized this equation to (for given constant $c$ )

$\displaystyle f(a + b) = f(a)f(c - b) + f(c - a)f(b)?$

Using similar logic as before, we have that $f(a + b) = f\left( 2c - (a + b) \right)$ , so $f(x) = f(2c - x)$ . If we guess that things scale the same way compared to the specific equation, then we may guess that $f$ has period $4c$ .

So how do we transform sine to have period $4c$ ? We just use a coefficient!

$\displaystyle \sin\left( r(x + p) \right) = \sin(rx) \rightarrow rp = 2\pi$

$\displaystyle p = 4c \rightarrow r = \frac{2\pi}{4c} = \frac{\pi}{2c}$

In general, can we show that the solutions of the sine angle additional functional equation and this generalized one are equivalent upon this transformation? In other words, let $f$ be a solution of the specific equation,

$\displaystyle f(a + b) = f(a)f\left( \frac{\pi}{2} - b \right) + f\left( \frac{\pi}{2} - a \right)f(b).$

Now, define $g$ by $g(x) = f\left( \frac{\pi}{2c}x \right)$ . Can we show $g$ satisfies the generalized equation?

$\displaystyle g(a + b) = f\left( \frac{\pi}{2c}(a + b) \right) = f\left( \frac{\pi}{2c}a + \frac{\pi}{2c}b \right)$

$\displaystyle = f\left( \frac{\pi}{2c}a \right)f\left( \frac{\pi}{2} - \frac{\pi}{2c}b \right) + f\left( \frac{\pi}{2} - \frac{\pi}{2c}a \right)f\left( \frac{\pi}{2c}b \right)$

$\displaystyle = g(a)g\left( \frac{\frac{\pi}{2} - \frac{\pi}{2c}b}{\frac{\pi}{2c}} \right) + g\left( \frac{\frac{\pi}{2} - \frac{\pi}{2c}a}{\frac{\pi}{2c}} \right)g(b)$

$\displaystyle = g(a)g(c - b) + g(c - a)g(b).$

Exactly as we desired! Conversely, if $g$ satisfies the generalized equation, then does $f$ defined by $f(x) = g\left( \frac{2c}{\pi}x \right)$ satisfy the specific one?

$\displaystyle f(a + b) = g\left( \frac{2c}{\pi}(a + b) \right) = g\left( \frac{2c}{\pi}a + \frac{2c}{\pi}b \right)$

$\displaystyle = g\left( \frac{2c}{\pi}a \right)g\left( c - \frac{2c}{\pi}b \right) + g\left( c - \frac{2c}{\pi}a \right)g\left( \frac{2c}{\pi}b \right)$

$\displaystyle = f(a)f\left( \frac{c - \frac{2c}{\pi}b}{\frac{2c}{\pi}} \right) + f\left( \frac{c - \frac{2c}{\pi}a}{\frac{2c}{\pi}} \right)f(b)$

$\displaystyle = f(a)f\left( \frac{\pi}{2} - b \right) + f\left( \frac{\pi}{2} - a \right)f(b),$

again exactly as we desired!

Thus, this equation is fully solved based on the sine angle addition one: we just transform those solutions $f$ via $x \rightarrow f\left( \frac{\pi}{2c}x \right)$ . In particular, $f(x) = 0,\frac{1}{2}$ are the only constant solutions of the generalized equation (which is easy to derive on its own too), and if the only non-constant solution of the sine angle addition equation is sine then the only non-constant solution of the generalized equation is $f(x) = \sin\left( \frac{\pi}{2c}x \right)$ .

—

Now, let’s head back to the specific equation.

Can we show $f$ has period $2\pi$ , given $f(x) = f(\pi - x)$ ?

$\displaystyle f(x + 2\pi) = f(x + \pi + \pi) = f\left( \pi - ( - x - \pi) \right) = f( - x - \pi)$

$\displaystyle f(x + 2\pi) = f\left( \pi - (x + 2\pi) \right) = f( - \pi - x)$

For sine, $f$ is actually odd, so this is $- f(\pi + x) = - \left( - f(x) \right) = f(x)$ .

Hmm … so maybe oddness is useful here?

Or let’s try this a different way. Can we show that we can have a function $f$ with $f(x) = f(\pi - x)$ but not period $2\pi$ ?

So what does it mean for a function to have $f(x) = f(\pi - x)$ ? If we let $x = y + \frac{\pi}{2}$ , then this becomes $f\left( \frac{\pi}{2} + y \right) = f\left( \frac{\pi}{2} - y \right)$ ; if we define a function $g$ by $g(x) = f\left( \frac{\pi}{2} + x \right)$ , then this becomes $g( - y) = g(y)$ . So this is just an expression of $g$ being even. All these steps are reversible, so this is exactly what $f(x) = f(\pi - x)$ says: that the function $x \rightarrow f\left( \frac{\pi}{2} + x \right)$ is even. So clearly then, there is no implication of periodicity just from this — we can take any old even function and just translate it.

OK, so how do we get periodicity then? Well, we need to combine something about $x \rightarrow f\left( \frac{\pi}{2} + x \right)$ being even with something about a function being odd; I think this can yield our desired periodicity. In fact, assume we had any function $f$ that was odd and where $f(x) = f(\pi - x)$ . Can we then show periodicity of $2\pi$ ?

We have

$\displaystyle f(x + 2\pi) = f( - x - \pi) = - f(x + \pi) = - f\left( \pi - ( - x) \right) = - f( - x) = - \left( - f(x) \right) = f(x).$

So that’s it! We just need to establish that the function is odd too, then we get periodicity of $2\pi$ .

—

So let’s try to establish that the function is odd.

It remains to try this.

—

Can we use the fact that any even function can yield a solution of $f(x) = f(\pi - x)$ to derive non-sinusoidal solutions to the functional equation?

Let $g$ be even, and define $f$ by $f(x) = g\left( x - \frac{\pi}{2} \right)$ . Must $f$ satisfy our functional equation?

$\displaystyle f(a + b) = g\left( a + b - \frac{\pi}{2} \right) = \ldots$

On second thought, evenness by itself would say nothing about addition of arguments. So we need to do something else here.

—

Let’s try a different approach. If we write $f$ as a power series (so looking at real analytic solutions), can we constrain $f$ to only be sine? Can we then use Hamel bases and whatnot to exclude other (non-constant) solutions? Or even having the real analytic solution uniqueness would be a cool development on its own, even if we’re able to find non-continuous wacky solutions somehow with Hamel bases.

It remains to try this.

Share and like (need WordPress account to like):

Leave a comment Cancel reply