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Does $f^n(x)$ always mean $f(f(f(f(...f(x))))....)$ [n times]?
i.e. $f^3(x)$ always means $f(f(f(x)))$?
Does $f^0(x)$ mean $x$? [where $f\neq id$]

By always, I mean regardless of whether it's for proofs in computer science or for calculus.

Just want to be doubly sure so I don't make any unfounded leaps in my proofs by induction for computer science.

Apologies for this simian question. Many thanks!

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marked as duplicate by amWhy, user61527, m_t_, Asaf Karagila, Davide Giraudo Jun 25 '14 at 16:21

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Were the answers on this similar question not to your liking? – The Chaz 2.0 Mar 1 '12 at 14:07
Also, this – The Chaz 2.0 Mar 1 '12 at 14:12
Thanks for pointing these out. They didn't come up whilst writing the question, but I should have exercised more due diligence. Apologies! – xlm Mar 1 '12 at 14:19
up vote 1 down vote accepted

No. Sometimes $f^n$ refers to multiplication, rather than composition of functions. This is especially true with trigonometric functions: for example, $\sin^2(x)$ always means $\sin(x) \cdot \sin(x)$, never $\sin(\sin(x))$. Outside trigonometry, composition is a more likely meaning, but multiplication is possible.

Do not confuse either of these with $f^{(n)}$, which means the $n$th derivative of $f$.

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Thanks Chris. However does $f^0(x)$ = x? Or does it also have multiple meanings? – xlm Mar 1 '12 at 14:27
Well, I suppose it might mean the constant function $1$. But I wasn't sure about its usage, so I left it out. – Chris Eagle Mar 1 '12 at 14:30

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