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Is there a short way to say $f(f(f(f(x))))$?

I know you can use recursion:

$g(x,y)=\begin{cases} f(g(x,y-1)) & \text{if } y > 0, \ \newline x & \text{if } y = 0. \end{cases}$

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    I write $f^4(x)$ but you can write anything you want.2010-10-27

4 Answers 4

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I personally prefer $f^{\circ n} = f \circ f^{\circ n-1} = \dotsb = \kern{-2em}\underbrace{f \circ \dotsb \circ f}_{n-1\text{ function compositions}}$

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    Because if you want $f^{\circ 2}$, you're composing _once_: $f \circ f$. So the $n-1$ is counting compositions, not how many times it says $f$.2010-10-28
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Some will say $f^4(x)$. But it becomes confused with the fourth power or fourth derivative of $f(x)$. I'm not sure what you mean here by "piecewise". To me "piecewise" would be something like a step function:

$f(x)=1$ if $x\gt 0$

$f(x)=0$ if $x\le 0$

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    One think about $f^4$ for the iteration is that it is consistent with $f^{-1}$ for the inverse. This is problematic with $\sin^{-1}$2010-10-28
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You should define it this way:

$ \begin{eqnarray} \text{iterate}_0(f) &:=& id \\ \text{iterate}_{n + 1}(f) &:=& \text{iterate}_{n}(f) \circ f \end{eqnarray} $

Then write $\text{iterate}_4(f)(x)$.

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    id is the identity function, that is, it maps its input to its output.$id$is f(x)=x for R to R functions, and f(x,y)=(x,y) for R^2 to R^2 functions and so on.2010-10-28
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See "function powers" in Wikipedia "Function composition".

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    This type of remark (just a pointer to wikipedia or similar) is best as a comment.2010-10-27