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I'm working with functions themselves, and I have learned that functional powers mean composition so:

$f^3 = f \circ f \circ f$

But I'm looking for something that means $fff$. So $(fff)(x) = (f(x))^3$ Is there a de facto standard notation I can use when mixing the two? Preferably without having to write out the $(x)$ part?

It seems we have a good deal ambiguity here. Right at the question title: The power of a function may mean two things. Sometimes even the circle is omitted to make it look like multiplication.

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    $\begingroup$ $f^k$ is commonly used for both, the $k$-fold product ($\sin^k x$) and the $k$-fold iteration of $f$. Just say what you denote that way. $\endgroup$ – Daniel Fischer Jun 8 '15 at 22:25
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    $\begingroup$ @DanielFischer I don't think that the example $\sin^n x$ is a good argument that $f^n(x)$ is sometimes interpreted as $(f(x))^n$ instead of the usual compositional power. "Named" functions, and trig functions especially, often get special treatment in notation (note the lack of a parenthesis around $x$ as well). I'd love to see an example of usage of $f^n(x)=(f(x))^n$ where they are using an actual $f$ (or other variable such as $g,h$) instead of a "named" function like $\sin$ or $\log$. $\endgroup$ – Mario Carneiro Jun 8 '15 at 23:09
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    $\begingroup$ @MarioCarneiro I do that, and I don't think it's uncommon in my area (machine learning) in contexts where function iteration would be very unusual. For example, in the "powers" section of this CV answer, or in the paper I'm currently editing.... (Though in those cases it happens to be with two-argument functions.) $\endgroup$ – Dougal Jun 9 '15 at 1:06
  • $\begingroup$ Possible duplicate of math.stackexchange.com/questions/795334/…. $\endgroup$ – lhf Jun 9 '15 at 3:08
  • $\begingroup$ Theoretically, the notation $f^n$ should be used, and, also theoretically, $f^n(x)$ is not the same as $\big[f(x)\big]^n$. But, in practice, this is thrown out the window. Personally, I would introduce the notation $f^{[n]}$, similar to $f^{(\pm n)}$, which means repeated differentiation or integration with regard to the argument. $\endgroup$ – Lucian Jun 9 '15 at 3:10
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I've already seen the following, which I adopted: $$\begin{align} f^n(x) &= (f(x))^n \\ f^{(n)}(x) &= \frac{{\rm d}^nf}{{\rm d}x^n} \\ f^{\circ n}(x) &= (f \circ f \circ \cdots \circ f)(x)\end{align}$$ However $f^{\circ n}$ does not seem to be standard, so you should always warn the reader when using it.


The power of a function may mean two things. Sometimes even the circle is omitted to make it look like multiplication.

That's because the collection of all bijections from a set to itself form a group with composition of functions - and the operation in a group is usually seen as multiplication.

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    $\begingroup$ I would warn the reader about the first notation more than the last - if the reader is familiar with $f^{\circ n}$, then it is unambiguous, whereas $f^n$ is still ambiguous (and I think is more commonly composition) - though, really, you ought to (gracefully) warn the reader about all of them. $\endgroup$ – Milo Brandt Jun 8 '15 at 22:33
  • $\begingroup$ Yes, agreed. ${}{}$ $\endgroup$ – Ivo Terek Jun 8 '15 at 22:33
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    $\begingroup$ I have come across this and also find it to be vastly superior. I hope it sticks as a useful notation. $\endgroup$ – Alfred Yerger Jun 9 '15 at 2:27
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    $\begingroup$ I like $f^{\circ n}$, and it makes me want to use $f^{\times n}$ or $f^{\cdot n}$ for multiplication, which would avoid the ambiguity of $f^n$, but I've never seen it used. $\endgroup$ – JiK Jun 9 '15 at 11:15
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The most common notation I've seen for $n$-fold composition is $$f(f(\ldots f(x)\ldots ))=f^{n}(x)$$ However this is generally always accompanied by a remark explaining that this is what the notation means. I would recommend you include such a remark.

I'm pretty sure there isn't a standard notation for raking a function to the power $n $, but again if you define some notation in the text then you're unlikely to be criticised.

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    $\begingroup$ The OP mentions this and asks for something that means $[f(x)]^3$. How does this answer the question? $\endgroup$ – JiK Jun 9 '15 at 9:19
  • $\begingroup$ Oh sorry, I misread! $\endgroup$ – preferred_anon Jun 9 '15 at 10:16
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FWIW, $f\,f\,f$ is also not quite unabiguous: it could be read as $f(f(f))$, or $f\:(f,f)$ – though both are really a bit strange and only make sense for polymorphic / higher-order functions.

Do use $f^n$ if you need this a lot, but also make some quick clarification about it. Or use something completely different – how about $\prod^n f$? That should be pretty unambiguous (at least unless you also have a symbol $\Pi$ around...).

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What you seem to want is the composition of a function with cubing.

Although a little verbose, I think this is correct:

$$ (x \mapsto x^3) \circ f $$

I don't think $^3 \circ f$ or $^3f$ is standard notation, but you may find it useful if you are repeating it a lot.

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This is a negative consequence of the pigeon-hole principal. There are more mathematical concepts to express than there are (uncomplicated) symbols. There has to be some trade off between clarity and simplicity.

In 'Surely You're Joking Mr. Feynman', Richard Feynman talks about studying trigonometry.

"While I was doing all this trigonometry, I didn't like the symbols for sine, cosine, tangent, and so on. To me, "sin f" looked like s times i times n times f! So I invented another symbol, like a square root sign, that was a sigma with a long arm sticking out of it, and I put the f underneath. For the tangent it was a tau with the top of the tau extended, and for the cosine I made a kind of gamma, but it looked a little bit like the square root sign. Now the inverse sine was the same sigma, but left -to-right reflected so that it started with the horizontal line with the value underneath, and then the sigma. That was the inverse sine, NOT sink f--that was crazy! They had that in books! To me, sin_i meant i/sine, the reciprocal. So my symbols were better."

$\cdots$

I thought my symbols were just as good, if not better, than the regular symbols--it doesn't make any difference what symbols you use--but I discovered later that it does make a difference. Once when I was explaining something to another kid in high school, without thinking I started to make these symbols, and he said, "What the hell are those?" I realized then that if I'm going to talk to anybody else, I'll have to use the standard symbols, so I eventually gave up my own symbols.

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As far as I am aware notation $f^n$ would only be taken to mean repeated (point-wise) multiplication of a function with itself without further mention if your setting is a ring of functions, for instance in $\mathcal C^\infty(\Bbb R)$. This is because multiplication is already defined to mean point-wise multiplication in such rings, and this just uses the usual relation between multiplication and exponentiation. (Convention also favours this interpretation when $f$ is a trigonometric function and $n\neq-1$, but that is really just a bad habit.)

In any other setting the more usual interpretation of $f^n$ would be $n$-fold self-composition. Because of that I think you should probably be more explicit to avoid the ambiguity, so just write $$ x\mapsto f(x)^n $$ which is a couple of symbols more than $f^n$, but self-explanatory and unambiguous.

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  • $\begingroup$ Your suggestion can still appear ambiguous: there are examples of omitting parentheses for arguments: $\sin x$. Considering this, RHS of your definition could be interpreted as $f((x)^n)=f(x^n)$ or as $(f(x))^n$. $\endgroup$ – Ruslan Jun 9 '15 at 13:59
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    $\begingroup$ @Ruslan: When function application is written in the usual style by writing the name of the function followed by argument(s) delimited by parentheses, then the parentheses umm.. delimit the argument. This is mathematical notation, not Haskell or LISP. It would be somewhat weird to assume that (1) the author considers parentheses around the argument to a function optional, so omits them, but (2) for no other reason than the fundamental right to write redundant parentheses writes $(x)$ for $x$, and (3) in taking the square of that as argument of $f$ puts the exponent outside the parentheses. $\endgroup$ – Marc van Leeuwen Jun 11 '15 at 14:28
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The notation $f^n(x)$ is ambigious. Sometimes it means $f^n(x)=\underbrace{f(f(f(\ldots)))}_{n}$, while other times, it means $f^n(x)=(f(x))^n$.

For example, $\sin^2 x=(\sin x)^2$

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  • $\begingroup$ That's acknowledged in the OP. $\endgroup$ – OJFord Jun 9 '15 at 13:24
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The way i read it, raising a number gives a power of the measure, so

$\sin^2 x$ corresponds to x sin square, and one is supposed to read down the tower of powers, not up them.

Multiple application, like $\sin \sin x$ = x, sin, sin, is not that common, and is best served by sequencing the terms from inner right to outer left.

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