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Does there exist a concept of fractional composition for functions? Continuous or differentiable functions?

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    $\begingroup$ Related: Fractional Composite of Functions $\endgroup$ – Surb Jun 7 '16 at 20:54
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    $\begingroup$ I mean if $f(x)=Dx$ where $D=\operatorname{diag}(d_1,\ldots,d_n)$ is some diagonal matrix with nonnegative entries, then $f^{[r]}(x):=\operatorname{diag}(d_1^{r},\ldots,d_n^{r})$. Such trick can be generalized relatively easily to positive semidefinite matrices. (I know more or less nothing about category theory) $\endgroup$ – Surb Jun 7 '16 at 20:58
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    $\begingroup$ Well, if you look for something like $f^{[r]}$ for rational (or real) $r$ and $f$ a morphism of a category, you preferably want $f$ to be an endomorphism, and thus suddenly we are talking about $r$th power in a monoid. $\endgroup$ – Berci Jun 9 '16 at 22:53
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    $\begingroup$ Wha'ts a fractional composition in your mind ? Standard composition states $(g\circ f)(x):=g(f(x))$. But I can't even imagine what "integer composition" would mean: for example, $(g\circ^2 f)(x):= ?$". $\endgroup$ – Yves Daoust Jun 13 '16 at 12:14
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    $\begingroup$ @JacobWakem May I edit your question to push it in the same direction as proposed in my previous comment? $\endgroup$ – idm Jun 13 '16 at 12:47
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Yes, fractional function iteration is defined.

Like it works for natural iteration

$$f^{(n)}(x):=f(f(\cdots f(x)\cdots)),$$

and (negative) integer iteration

$$f^{-n}(x):=(f^{-1})^{(n)}(x)$$

you can generalize to rationals with the functional equation

$$g(x):=f^{(n/m)}(x)\iff g^{(m)}(x)=f^{(n)}(x).$$

For example, the iterative square root is defined as

$$g(x):=f^{1/2}(x)\iff g(g(x))=f(x).$$

In the case of the identity function, you get the so-called Babbage equation,

$$g(g(x))=x.$$

A particular solution is $g(x)=-x$, and as you can check, more solutions are found as

$$g(x)=h^{-1}(-h(x))$$ where $h$ is an arbitrary invertible function.

For instance, with $h(x)=\ln(x+1)$,

$$g(x)=\exp(-\ln(x+1))-1=-\frac x{x+1}.$$

So this square root isn't unique.


If $f(x)=x^r$ is a power law, the $n^{th}$ iterate is $((x^r)^{r\cdots})^r=x^{r^n}$ so that a particular solution of

$$g^{(m)}(x)=(x^r)^{(n)}$$ is another power law

$$g(x)=x^s$$ with $s^m=r^n$, or $s=r^{n/m}$.

$$(x^r)^{(q)}=x^{r^q}.$$


As another example, the iteration of a linear law gives

$$(ax+b)^{(n)}=a(\cdots a(ax+b)+b\cdots)+b=a^nx+\frac{a^n-1}{a-1}b,$$ so that a fractional iterate can also be linear

$$(ax+b)^{(n/m)}=g(x)=Ax+B$$ with

$$A^m=a^n,\frac{A^m-1}{A-1}B=\frac{a^n-1}{a-1}b$$

or

$$A=a^{n/m},B=b\frac{a^n-1}{a-1}\frac{A-1}{A^m-1}=\frac{a^{n/m}-1}{a-1}.$$

$$(ax+b)^{(q)}=a^qx+\frac{a^q-1}{a-1}b.$$

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    $\begingroup$ It's a bit abusive to say that "fractional function iteration is defined". What is defined is what it means for a function $g$ to be a fractional iterate of a function $f$. Otherwise it's like saying "the complex square root is defined". It makes sense to say that $i$ is a square root of $-1$, it doesn't make sense to talk about "the square root of $-1$". Similarly, when you write $g(x) := f^{(n/m)}(x)$ as if it were a definition of $g$, that's abusive IMO. $\endgroup$ – Najib Idrissi Jun 13 '16 at 13:37
  • $\begingroup$ Your iteration is not uni uely defined! I guess if we had a metric space or a particular continuous process that the function performs to get from x to f(x) it might be uni uely defined. $\endgroup$ – Jacob Wakem Jun 13 '16 at 13:38
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    $\begingroup$ @JacobWakem: it is not "my iteration", and this definition looks quite natural. It ends-up in multiple solutions. Remember that $z^n=w$ doesn't define a unique $n^{th}$ root either. $\endgroup$ – Yves Daoust Jun 13 '16 at 13:41

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