# Functional Notation.

I have some doubts regarding function notation:

First If I present a function I write:$f(x)$

If I write it's inverse:$f^{-1}(x)$

So why doesn't$f(f(x))=f^2(x)$

Second If $\frac{df(x)}{dx}=f'(x)$ and $\frac{d^2f(x)}{dx^2}=f''(x)$.

So how do you write probably $\frac{d^nf(x)}{dx^n}=f'(x)$.

Does like this:$f'''^{\cdots\text{n times}}(x)$?

Third What difference between $f^n(x),f(x)^n\text{ and }(f(x))^n$

-
The $n$th derivative is often written as $f^{\left(n\right)}\left(x\right)$. $f^2(x)$ is interpreted as $f(x)\times f(x)$ so is not used to denote the function $f\circ f$. – drhab Aug 17 '14 at 14:08
Though I will note that the usage whereby $f^2:=f\circ f$ becomes more common in fields like set theory, where iterates of functions come up a lot and numerical multiplication doesn't make sense. It's good to be aware of this usage. – Malice Vidrine Aug 17 '14 at 14:30
Note that there is an ambiguity. $f^{-1}$ denotes the inverse with respect to the $f\circ g$ operation. However, $f^2$ is ambiguous. It conventionally indication multiplication ($\times$) only when used with $\sin$ and such. – I like Serena Aug 17 '14 at 15:23
Sometimes people write $f^{\circ n}$ for f composed with itself n times. Related: math.stackexchange.com/questions/588875/… – Myself Aug 17 '14 at 19:32

I have some doubts regarding function notation:

First If I present a function I write: $f(x)$

If I write it's inverse: $f^{-1}(x)$

So why doesn't $f(f(x))=f^2(x)$

This is indeed ambiguous.   Unfortunately $f^n$ is sometimes used to refer to the $n$-th iterate of a function and other times is commonly used to refer to the $n$-th exponentiation.   The same notation was adopted by different branches of mathematics to mean different things and both persist in using it so.   Thus when ever you see $f^2$, you should check the context to determine if the author meant the composition $f\circ f$ or the product $f\cdot f$.

More confusingly, the convention that $f^{-1}$ means the iterate inverse of a function has become the standard even when $f^n$ is otherwise used for exponentiation.

Some mathematicians adopt the notation $f^{\circ n}$ to be clear that they mean the $n$ iterate.

Second If $\frac{df(x)}{dx}=f'(x)$ and $\frac{d^2f(x)}{dx^2}=f''(x)$.

So how do you write probably $\frac{d^nf(x)}{dx^n}=f'(x)$.

Does like this:$f'''^{\cdots\text{n times}}(x)$?

We never go passed three primes.   The convention is that $f^{(n)}$ means the $n$-th derivative of a function (with respect to its argument).

Third What difference between $f^n(x),f(x)^n\text{ and }(f(x))^n$

Clarity.   $f^n(x)$ is ambiguous; it may mean $f\circ f^{n-1}(x)$ or commonly $f\cdot f^{n-1}$.   It is possible that $f(x)^n$ could be parsed as $f(x^n)$.   However, $(f(x))^n$ is fairly unambiguous; although it is more chunky.

-
I have never heard of $f(x)^n$ meaning $f(x^n)$... it will almost certainly mean $(f(x))^n$. However, for something like $\sin x^2$, since the parentheses around the argument of $\sin$ may be omitted, it is quite likely to mean $\sin(x^2)$, especially because $\sin^2 x$ is already standard for $(\sin x)^2$. – echinodermata Aug 18 '14 at 3:15
For "$\sin(y+z)^2$", one can't be sure whether the parentheses around the argument of $\sin$ have been omitted ($\sin (x^2) = \sin ((y+z)^2) = \sin (y+z)^2$) or not ($(\sin(x))^2 = (\sin (y+z))^2 = \sin (y+z)^2$). – JiK Aug 18 '14 at 7:21
I've seen some books where after the third prime you start to use roman numbers. – jinawee Aug 18 '14 at 7:54
As an aside: typically, whenever $f^n(x)$ represents $n$ compositions, it is noted explicitly. – apnorton Aug 28 '14 at 3:30

It depends on the context and the conventions. Four your second question, people usually write $f^{(n)}$. For your first question, some people actually use $f^2$ as an abbreviation for $f \circ f$, but it's not that common.

-

When writing $f^2(x)$, it normally refers to $f(f(x)$. In general: $$f^n(x) = \overbrace{f(f(\ldots f}^{n \text{ times}}(x))\ldots)\\f(x)^n=(f(x))^n\text{ is f(x) raised to the nth power}$$

You can certainly write $\dfrac{d^nf(x)}{dx}=f''^\ldots(x)$. In general, one would write $\dfrac{d^nf(x)}{dx}=f^{(n)}(x)$.

Although, be aware that the context will often dictate what $f^n(x)$ is.

-

I hope this helps:

$$f^n(x) = f(f^{n-1}(x)) = \underbrace{f(f(f(\cdots f(x))\cdots)}_{n \text{ times}} = \underbrace{(f^\circ f^\circ f^\circ \cdots f)}_{n \text{ times}}(x)$$

$$\frac{d^n}{dx^n}f(x) = f^{(n)}(x)$$

$$(f(x))^n = \underbrace{f(x)·f(x)\ldots f(x)}_{n \text{ times}}$$

As far as I know, there's no such thing as $f(x)^n$ but it might mean $(f(x))^n$

-

First: The inverse function notation has nothing to do with exponentiation, therefore $f(f(x))$ is not written as $f^2(x)$. Note, that sometimes on calculators you might see $\cos^{-1}{}$ which means $\arccos$, NOT $\frac{1}{\cos}$.

Second: For derivatives of order $n \geq 4$, typical notation is $f^{(n)}(x)$

Third: This I am not entirely sure of, but to avoid any confusion, I'd only use $(f(x))^n$.

-