Franklin’s Blog

More Functional Equations

2017 Aug 15

NOTE: I will be using superscripts in three different ways in this post, so be careful not to get confused. The first way is as an exponent. The expression
\[f(x)^n\]
reads “f of x, raised to the nth power.” The second way is to represent iteration. The expression
\[f^n(x)\]
reads “the nth iterate of f of x.” The last way is to represent multiple derivatives. The expression
\[f^{(n)}(x)\]
reads “the nth derivative of f of x.”

It’s time for some more functional equations!

In this previous post, I looked at a few interesting functional equations, and in retrospect, it doesn’t seem like they were very difficult. The answers were mostly polynomial functions that could be solved by differentiating a few times. This time, however, I have some very hard functional equations to try.

Find non-constant functions (if they exist) with the following properties:
\[\alpha(x)^2+\alpha(2x)^2=1\]
\[\beta(x)+2\beta(2x)=3\beta(3x)\]
\[\gamma(x)+\gamma(2x)=e^x\]
\[\delta(5x)^5\delta(10x)^{10}=x+1\]
\[\epsilon(x)+\epsilon\bigg(\frac{x\cos\frac{\pi}{50}-\sin\frac{\pi}{50}}{x\sin\frac{\pi}{50}+\cos\frac{\pi}{50}}\bigg)=\ln(x)\]
\[\zeta(x+1)=4\zeta(x)-4\zeta(x)^2\]

Let’s start with the first problem. It is rather easy, but it is a good problem to start with.
\[\alpha(x)^2+\alpha(2x)^2=1\]
This can be simplified with a technique that is analogous to substitution in normal equations. To simplify the equation, we can define
\[\psi_1(\log_2 x)=\alpha(x)\]
so that we have
\[\psi_1(\log_2 x)^2+\psi_1(1+\log_2 x)^2=1\]
and making the substitution \(\log_2 x\to x\), we have
\[\psi(x)_1^2+\psi(x+1)_1^2=1\]
At this point, you should be able to recognize a similarity between this equation and another functional equation involving the trigonometric functions:
\[\sin^2 x+\cos^2 x=1\]
or
\[\sin^2 x+\cos^2 \bigg(x-\frac{\pi}{2}\bigg)=1\]
Noticing this yields the obvious solution
\[\psi_1(x)=\sin \frac{\pi x}{2}\]
or
\[\alpha(x)=\sin \frac{\pi \log_2 x}{2}\]

Now it’s time for the hard problems!

I used differentiation in my previous post about functional equations, but only to solve functional equations with polynomial solutions. I would like to show how to use differentiation to solve other types of functional equations. For example, the second problem:
\[\beta(x)+2\beta(2x)=3\beta(3x)\]
This one stumped me for a while, and I was rather annoyed to find out that there was no non-constant solution. Watch what happens when we differentiate:
\[\beta'(x)+4\beta'(2x)=9\beta'(3x)\]
And if we differentiate again:
\[\beta''(x)+8\beta''(2x)=27\beta''(3x)\]
and so on:
\[\beta^{(n)}(x)+2^{n+1}\beta^{(n)}(2x)=3^{n+1}\beta^{(n)}(3x)\]
Which means that the nth derivative of \(\beta\) at \(x=0\) is
\[\beta^{(n)}(0)+2^{n+1}\beta^{(n)}(0)=3^{n+1}\beta^{(n)}(0)\]
\[(3^{n+1}-2^{n+1}-1)\beta^{(n)}(0)=0\]
and so, for all positive integer \(n\),
\[\beta^{(n)}(0)=0\]
And, of course, the only function all of whose derivatives are equal to zero is the constant function, meaning that there exists no non-constant \(\beta\) satisfying the functional equation.

On to the third one:
\[\gamma(x)+\gamma(2x)=e^x\]
The solution to this one is rather complicated. You could try the method that we used on the first problem, but you would end up with
\[\psi_1(x)+\psi_1(x+1)=e^{\log_2 x}\]
Yuck.

As it turns out, the answer may not even have a closed-form representation, and the answe has to be expressed as a Taylor Series. In case you are not familiar with the idea, the Taylor Series of a function \(f\) is
\[\sum_{n=0}^\infty \frac{(x-a)^n}{n!}f^{(n)}(a)\]
and Taylor’s Theorem states that
\[f(x)=\sum_{n=0}^\infty \frac{(x-a)^n}{n!}f^{(n)}(a)\]
This can be very helpful with some functional equations. If we differentiate both sides of our functional equation \(n\) times, we end up with
\[\gamma^{(n)}(x)+2^n\gamma^{(n)}(2x)=e^x\]
and so
\[\gamma^{(n)}(0)+2^n\gamma^{(n)}(0)=1\]
\[\gamma^{(n)}(0)=\frac{1}{2^n+1}\]
and so
\[\gamma(x)=\sum_{n=0}^\infty \frac{x^n}{n!(2^n+1)}\]

The next functional equation is
\[\delta(5x)^5\delta(10x)^{10}=x+1\]
and it uses the same strategy, but required a bit of manipulation before iterated differentiation can be applies. First we must take the natural logarithm of both sides:
\[\ln \big(\delta(5x)^5\delta(10x)^{10}\big)=\ln(x+1)\]
\[5\ln \delta(5x)+ 10\ln\delta(10x)=\ln(x+1)\]
Now we must make a substitution. Define \(\psi_1\) as \(\ln\delta(x)=\psi(x)\) so that our functional equation is transformed into
\[5\psi_1(5x)+ 10\psi(10x)=\ln(x+1)\]
Now we can differentiate both sides \(n\) times to get
\[5^{n+1}\psi_1^{(n)}(5x)+ 10^{n+1}\psi_1^{(n)}(10x)=\frac{(-1)^{n-1} (n-1)!}{x+1}\]
for \(n\ge 1\). Then we have
\[5^{n+1}\psi_1^{(n)}(0)+ 10^{n+1}\psi_1^{(n)}(0)=(-1)^{n-1} (n-1)!\]
\[\psi_1^{(n)}(0)=\frac{(-1)^{n-1} (n-1)!}{10^{n+1}+5^{n+1}}\]
for \(n\ge 1\). Then, since we have all but the first term of the Taylor Series (which is zero anyways since \(\ln (1)=0\), so we don’t need to worry about it) we can express \(\psi_1\) as
\[\psi_1(x)=\sum_{n=1}^\infty \frac{x^n}{n!}\frac{(-1)^{n-1} (n-1)!}{10^{n+1}+5^{n+1}}\]
\[\psi_1(x)=\sum_{n=1}^\infty \frac{x^n(-1)^{n-1}}{n(10^{n+1}+5^{n+1})}\]
and so, when we reverse our substitution, we have
\[\ln\delta(x)=\sum_{n=1}^\infty \frac{x^n(-1)^{n-1}}{n(10^{n+1}+5^{n+1})}\]
\[\delta(x)=\exp\Bigg(\sum_{n=1}^\infty \frac{x^n(-1)^{n-1}}{n(10^{n+1}+5^{n+1})}\Bigg)\]

Of course, this solution isn’t perfect - since we used the natural logarithm and obtained \(\ln(x+1)\) on the right side of one of our equations, we basically invalidated our answer for any values of \(x\) for which \(\ln(x+1)\) is nonreal; namely, \(x\le 0\).

Now let’s try and tackle those last two functional equations.

Before we try the next one, I would like to revisit a problem from my previous post about functional equations. Here it is:
\[h(x)-2h\bigg(\frac{2}{x}\bigg)=\frac{1}{2x}\]
The method that I used to solve it was to substitute \(x\to \frac{2}{x}\) to get
\[h\bigg(\frac{2}{x}\bigg)-2h\bigg(\frac{2}{\frac{2}{x}}\bigg)=\frac{1}{2\frac{2}{x}}\]
or
\[h\bigg(\frac{2}{x}\bigg)-2h(x)=\frac{1}{4}x\]
I then took the original equation and this new equation:
\[h(x)-2h\bigg(\frac{2}{x}\bigg)=\frac{1}{2x}\]
\[h\bigg(\frac{2}{x}\bigg)-2h(x)=\frac{1}{4}x\]
and treated them like a system of equations to be solved. If the second equation is multiplied by two, we obtain
\[2\bigg(\frac{2}{x}\bigg)-4h(x)=\frac{1}{2}x\]
and when we add this to the original equation, we get
\[h(x)-2h\bigg(\frac{2}{x}\bigg)+2\bigg(\frac{2}{x}\bigg)-4h(x)=\frac{1}{2x}+\frac{1}{2}x\]
\[h(x)-4h(x)=\frac{1}{2x}+\frac{1}{2}x\]
\[-3h(x)=\frac{1}{2x}+\frac{1}{2}x\]
\[h(x)=-\frac{1}{6x}-\frac{1}{6}x\]
\[h(x)=-\frac{x^2+1}{6x}\]

Notice now that if we have any functional equation in the form
\[f(x)+a(f\circ \theta)(x)=g(x)\]
Where we are trying to solve for \(f\), \(g\) is given, \(a\) is a constant, and \(\theta\) has the property
\[\theta^n(x)=x\]
In fact, a formula can be obtained for \(f\) in this equation. Using the same method as we did with my example, through the substitution \(x\to\theta(x)\) repeated \(n-1\) times, we can obtain the system of equations
\[f(x)+a(f\circ \theta)(x)=g(x)\]
\[(f\circ\theta)(x)+a(f\circ \theta^2)(x)=(g\circ\theta)(x)\]
\[(f\circ\theta^2)(x)+a(f\circ \theta^3)(x)=(g\circ\theta^2)(x)\]
\[...\]
\[(f\circ\theta^{n-2})(x)+a(f\circ \theta^{n-1})(x)=(g\circ\theta^{n-2})(x)\]
\[(f\circ\theta^{n-1})(x)+af(x)=(g\circ\theta^{n-1})(x)\]
Now we multiply the second equation by \(-a\), the third by \(a^2\), the fourth by \(-a^3\), and so on, until we multiply the last equation by \((-a)^{n-1}\), which leaves us with
\[f(x)+a(f\circ \theta)(x)=g(x)\]
\[-a(f\circ\theta)(x)-a^2(f\circ \theta^2)(x)=-a(g\circ\theta)(x)\]
\[a^2(f\circ\theta^2)(x)+a^3(f\circ \theta^3)(x)=a^2(g\circ\theta^2)(x)\]
\[...\]
\[(-a)^{n-2}(f\circ\theta^{n-2})(x)-(-a)^{n-2}(f\circ \theta^{n-1})(x)=(-a)^{n-2}(g\circ\theta^{n-2})(x)\]
\[(-a)^{n-1}(f\circ\theta^{n-1})(x)-(-a)^{n-1}f(x)=(-a)^{n-1}(g\circ\theta^{n-1})(x)\]
And now watch what happens when we add this system together. When we add the first two, we have
\[f(x)+a(f\circ \theta)(x)-a(f\circ\theta)(x)-a^2(f\circ \theta^2)(x)=g(x)-a(g\circ\theta)(x)\]
\[f(x)-a^2(f\circ \theta^2)(x)=g(x)-a(g\circ\theta)(x)\]
and when we add this to the third one:
\[f(x)-a^2(f\circ \theta^2)(x)+a^2(f\circ\theta^2)(x)+a^3(f\circ \theta^3)(x)=g(x)-a(g\circ\theta)(x)+a^2(g\circ\theta^2)(x)\]
\[f(x)+a^3(f\circ \theta^3)(x)=g(x)-a(g\circ\theta)(x)+a^2(g\circ\theta^2)(x)\]
and if you keep adding them up until the very last one, you will eventually end up with
\[f(x)-(-a)^{n-1}f(x)=\sum_{k=0}^{n-1} (-a)^{k}(g\circ\theta^k)(x)\]
and
\[f(x)=\frac{1}{1-(-a)^{n-1}}\sum_{k=0}^{n-1} (-a)^{k}(g\circ\theta^k)(x)\]
Which is our formula.

We just need one more thing before we try the next functional equation. I’m not going to go over it again in this post, but if you haven’t already, you should go back to this post and read the very end of it.

You should be able to notice now when you look at the next functional equation
\[\epsilon(x)+\epsilon\bigg(\frac{x\cos\frac{\pi}{50}-\sin\frac{\pi}{50}}{x\sin\frac{\pi}{50}+\cos\frac{\pi}{50}}\bigg)=\ln(x)\]
That this fits the form of our formula perfectly, with \(a=-1\), \(g(x)=\ln(x)\), and
\[\theta(x)=\frac{x\cos\frac{\pi}{50}-\sin\frac{\pi}{50}}{x\sin\frac{\pi}{50}+\cos\frac{\pi}{50}}\]
and after careful observation using the information found in my previous post, we have \(n=100\). Additionally, we can determine that
\[\theta^k(x)=\frac{x\cos\frac{\pi k}{50}-\sin\frac{\pi k}{50}}{x\sin\frac{\pi k}{50}+\cos\frac{\pi k}{50}}\]
and so, by plugging into the formula, we have
\[f(x)=\frac{1}{1-(-1)^{99}}\sum_{k=0}^{99} (-1)^{k}\ln\bigg( \frac{x\cos\frac{\pi k}{50}-\sin\frac{\pi k}{50}}{x\sin\frac{\pi k}{50}+\cos\frac{\pi k}{50}} \bigg)\]
\[f(x)=\frac{1}{2}\sum_{k=0}^{99} (-1)^{k}\ln\bigg( \frac{x\cos\frac{\pi k}{50}-\sin\frac{\pi k}{50}}{x\sin\frac{\pi k}{50}+\cos\frac{\pi k}{50}} \bigg)\]
There’s no need to evaluate this sum - this answer is good enough.

The last functional equation is similar to the first in that it is best solved using substitution and intuition, but it is much more difficult.
\[\zeta(x+1)=4\zeta(x)-4\zeta(x)^2\]
This can be rearranged to get
\[\zeta(x+1)=4\zeta(x)(1-\zeta(x))\]
Let us now make the substitution \(\psi_1(2^x)=\zeta(x)\). Then
\[\psi_1(2^{x+1})=4\psi_1(2^x)(1-\psi_1(2^x))\]
and by taking \(2^x\to x\),
\[\psi_1(2x)=4\psi_1(x)(1-\psi_1(x))\]

At this point, you may be wondering how I decided to make those substitutions. This is where the intuition part of the problem comes in.

Upon seeing the problem, it was immediately obvious to me that the answer would not end up being a non-constant polynomial. This can be easily determined by noticing that if it was a polynomial with degree \(d\), then the left side of the functional equation would also have degree \(d\), whereas the right side would have degree \(2d\).

Once I found this out, I began to think that the solution would involve trigonometric functions, since there are so many identities relating the trigonometric functions to their squares. And when I factored the right side as
\[4\zeta(x)(1-\zeta(x))\]
I was reminded of the pythagorean identities almost immediately.

If you were unable to figure this one out before, stop reading now and let what I have revealed so far be a hint to you.

Because of my suspicions about the pythagorean identities, I made another substitution by letting \(\sin^2 \psi_2(x)=\psi_1(x)\). Then I got
\[\sin^2\psi_2(2x)=4\sin^2(\psi_2(x))(1-\sin^2(\psi_2(x))\]
\[\sin^2\psi_2(2x)=4\sin^2(\psi_2(x))\cos^2(\psi_2(x))\]
\[(\sin\psi_2(2x))^2=(2\sin(\psi_2(x))\cos(\psi_2(x)))^2\]
At this point I realized that I could just let \(\psi_2(x)=x\), because my equation had turned into the double-angle identity. When I substituted back in to find \(\zeta\), I ended up with
\[\zeta(x)=\sin^2 (2^x)\]
Which I thought was a rather satisfying solution.

That’s all that I am going to do with functional equations for now.