# Is there a canonical extension of Euler's totient function to the reals?

Assumption: if a set $$X\times Y$$, exists such that $$\forall\textbf{u},\textbf{v}\in X\times Y. \left(u_1=v_1\implies u_2=v_2\right)$$, then a function $$f$$ exists such that $$f:X\to Y$$.

The set of points $$\Phi_\mathbb{N}=\left\{(n,\phi(n))\mid n\in\mathbb{N}\right\}$$ describing the graph/image of the totient function is a subset of the reals.

By definition, $$\forall \textbf{x},\textbf{y}\in\mathbb{R}^2$$, where $$x_1< y_1.\exists \textbf{z}\in\mathbb{R}^2:x_1. Thus, there is a continuous set of points between any two $$\textbf{a},\textbf{b}\in\Phi_\mathbb{N}$$. It follows, that there is a a set $$\Phi_\mathbb{R}\supset\Phi_\mathbb{N}$$ which is the graph of a continuous function $$\phi:\mathbb{R}\to\mathbb{R}$$, ($$n\in\mathbb{N}\implies \phi(n)\in\mathbb{N}$$).

Thus, it is at least possible to extend the totient function to the reals without breaking the original definition. Is there a canonical way to do this? If not, what is the most appropriate way to extend the totient function?

Note: An extension to the complex numbers is perfectly acceptable too, I just assumed it would be best to "keep it real".

Note: tagged analysis because that's where I'm going with this.

Edit: This might help... probably not, though

For coprime $$a$$ and $$n$$ $$a^{\phi(n)}\equiv1\mod{n}\qquad\text{(Euler's Theorem)}$$ For $$x,y\in\mathbb{R}$$ $$x\ \text{mod}\ y^{*}=\frac{|x|}{2\pi}\sum_{k=1}^\infty\frac{1}{k}\Im\left[\exp\left(\frac{2\pi kix}{y}\right)\right]$$ So $$a^{\phi(n)}\ \text{mod}\ n=\frac{|a^{\phi(n)}|}{2\pi}\sum_{k=1}^\infty\frac{1}{k}\Im\left[\exp\left(\frac{2\pi kia^{\phi(n)}}{n}\right)\right]=1\implies$$ $$|a^{\phi(n)}|=2\pi\left(\sum_{k=1}^\infty\frac{1}{k}\Im\left[\exp\left(\frac{2\pi kia^{\phi(n)}}{n}\right)\right]\right)^{-1}=$$ $$-4\pi i\left(\ln\left(1-\exp\left(\frac{2\pi ia^{\phi(n)}}{n}\right)\right)-\ln\left(\exp\left(-\frac{2\pi ia^{\phi(n)}}{n}\right)\left(-1+\exp\left(\frac{2\pi ia^{\phi(n)}}{n}\right)\right)\right)\right)^{-1}$$

Assuming $$a^{\phi(n)}$$ is positive, let $$a^\phi(n)=u$$ (because I'm running out of space)

$$u=-4\pi i\left(\ln\left(1-\exp\left(\frac{2\pi iu}{n}\right)\right)-\ln\left(\exp\left(-\frac{2\pi iu}{n}\right)\left(-1+\exp\left(\frac{2\pi iu}{n}\right)\right)\right)\right)^{-1}$$

Solve for for $$u$$ (not sure how), then substitute $$a^{\phi(n)}$$ back in and take the base-$$a$$ logarithm to get $$\phi(n)$$

Assuming all the algebra is correct, this might work...

$$^*$$ If I understand correctly, we can translate $$x\equiv z\mod{y}$$ to $$x\ \text{mod}\ y=z$$, using $$\text{mod}$$ as a binary operation.

• In the question Euler's $\phi$-function can be replaced by any arithmetic function. We just want to extend $f\colon \Bbb{N}\rightarrow \Bbb{R}$ to the real numbers. – Dietrich Burde Dec 18 '18 at 16:24
• @DietrichBurde Could you please clarify. – R. Burton Dec 18 '18 at 16:36
• There are any number of ways we could extend $\phi$, but the question is whether any specific criteria make one option canonical. For example, is there a generalisation of coprimality for which we want $\phi(xy)=\phi(x)\phi(y)$? – J.G. Dec 18 '18 at 19:18
• @J.G. Ideally, as many properties of the totient function should be conserved as possible. $\phi(xy)=\frac{z\phi(x)\phi(x)}{\phi(z)}$ might only work for integer $x,y,$, and $z$. There might be some formulae which can be applied to rational numbers without alteration. – R. Burton Dec 18 '18 at 21:17

Let's first try to define $$\phi$$ on $$\Bbb Q$$; presumably, any canonical definition on $$\Bbb R$$ would have a restriction to $$\Bbb Q$$ that would meet with our focus-on-$$\Bbb Q$$ approval. It will be convenient to define $$f(n):=n^{-1}\phi(n)$$.
For $$n\in\Bbb N$$ we have $$f(n)=\prod_{p\in\Bbb P\land p|n}(1-p^{-1})$$, and it is natural to define $$\phi(0)=0$$ (since there are no integers from $$1$$ to $$0$$ exclusive to be coprime to $$0$$) and, if we want to extend to negative integers, presumably $$f(-n)=f(n)$$ since the same primes divide $$-n$$. If for coprime $$m,\,n\in\Bbb Z$$ with $$n>0$$ we demand $$f(m/n)=f(m)f(n)$$ [sic], so as to include $$1-1/p$$ factors for every prime appearing in $$m/n$$'s prime factorisation, we have a canonical extension to $$\Bbb Q$$. In particular, this definition implies $$\phi(p^k)=p^k(1-p^{-1})$$ for $$p\in\Bbb P,\,k\in\Bbb Z\backslash\{0\}$$; unsurprisingly, as $$p^k\to 0$$ we also have $$\phi\to 0$$.
However, I don't think we can extend this to $$\Bbb R$$. The only obvious definition is $$\phi(x):=\lim_{n\to\infty}\phi(x_n)$$ for rationals $$x_n$$ satisfying $$\lim_{n\to\infty}x_n=x$$. But I would be very surprised if, for all irrational $$x$$, this definition is independent of the sequence chosen. In fact, $$\phi$$ as defined above isn't even continuous at $$0$$. For example, take your favourite distinct primes $$p,\,q$$, then define $$x_n:=p^n/q^{a_n}$$, with $$a_n$$ just large enough that $$x_n<1/n$$. Then $$\phi(x_n)=(1-1/p)(1-1/q)$$ for all $$n$$.
• Agreed. Given the behavior of the graph on natural number values alone it seems "intuitively" reasonable to expect that the "natural" extension should be a discontinuous dust or otherwise suitably "wild", and thus not extensible to $\mathbb{R}$ through limits. – The_Sympathizer Dec 18 '18 at 23:45