Understanding Trigonometric Functions for Negative Integers and Rational Numbers

[thereddevils]

I know that $$\sin^2 b= (\sin b)^2$$ and in general $$\sin^n b=(\sin b)^n$$ if n is a positive integer .

What if n is a negative integer , would it be

$$\sin^{-1}b=(\sin b)^{-1}=\frac{1}{\sin b}$$

I don't think this is right , because properties of indices only works for numbers and NOT function , but why it works for case 1 above ?

What if it's $$sin^{-3}b$$

How bout if n is rational ?

 

[D H]

[itex]x^{-n} = 1/x^n[/tex] for all integers n. It doesn't really matter what x is. This x might just be some variable x, or it might be $\sin b$. Things get a bit trickier when n is not necessarily an integer. That $x^{-n}=1/x^n$ still works so long as n is real and x is a positive real. Things get a bit more complex with negative real x and complex number x and/or n.However, and this is a big however, there is a bit of ambiguity regarding $\sin^{-1} b$. This might mean $1/\sin b$ or it might mean $\arcsin b$. To avoid this ambiguity, people write $1/\sin b$ or $(\sin b)^{-1}$ but never $\sin^{-1} b$ when they want to express $1/\sin b$.

 

[thereddevils]

[itex]x^{-n} = 1/x^n[/tex] for all integers n. It doesn't really matter what x is. This x might just be some variable x, or it might be $\sin b$. Things get a bit trickier when n is not necessarily an integer. That $x^{-n}=1/x^n$ still works so long as x and n are real. Complex numbers are a bit more complex.


However, and this is a big however, there is a bit of ambiguity regarding $\sin^{-1} b$. This might mean $1/\sin b$ or it might mean $\arcsin b$. To avoid this ambiguity, people write $1/\sin b$ or $(\sin b)^{-1}$ but never $\sin^{-1} b$ when they want to express $1/\sin b$.

thanks ! So $$sin^{-1}b$$ can be taken as arcsin b ?

But these do not work for functions right , where $$f^{-1}(x)\neq \frac{1}{f(x)}$$

Also , does it apply to logarithms ?

$$\log^5(x)=(\log x)^5$$ ??

 

[Martin Rattigan]

EDIT: See corrections in later post.

I would say that $f^n(x)=(f(x))^n$ for $n\in \mathbb{N}$ applies to sin, cos, tan, cot, sinh, cosh, tanh and coth only. Though I can't ever remember seeing it I'd understand the same for $n\in \mathbb{Z}-\{1\}$.

$f^{-1}$ is a function selected from the converse, $\breve{f}$, of $f$ (as a relation) doesn't apply to anything outside the same set, and does apply to sin, cos, tan and cot. (I wouldn't like to be dogmatic about this use with the hyperbolic functions - I think you might have to go with the context in these cases.)

Otherwise I think for $n\in \mathbb{Z}$ $f^n(x)$ would generally mean $x$ for $n=0$, $f(f^n(x))$ for $n>0$ and $\breve{f}^{-n}(x)$ when $n<0$ and would imply that $f$ is an invertible function for the last case.

So for example $log^5(x)$ should mean $log(log(log(log(log(x)))))$, but even here I think you would generally see $log(log(x))$ in preference to $log^2(x)$.

 

[D H]

You will see in many papers things like $\ln^2 x$ and $f^4(x)$. There is yet another ambiguity with the latter: Does $f^4(x)$ mean $(f(x))^4$ or $d^4 f(x)/dx^4$? The latter is quite non-standard, but it is out there. More typical is $f^{(iv)}(x)$ to denote the fourth derivative and $f^{(n)}(x)$ to denote the n^th derivative.

Bottom line:

• When you see something like $f^n(x)$ you had better look for a nomenclature or read the text to decipher what the author wrote.
• Never use $f^{-1}(x)$ to denote the multiplicative inverse. That notation is almost always reserved for the inverse function.
• Never use $f^{n}(x)$ to denote the n^th derivative. You are going to confuse your readers mightily.
• Take care and think twice when you use $f^{n}(x)$. Ask yourself whether this usage might be confusing to your readers.

There is a mantra regarding computer programming that also applies to writing a technical paper. The programming mantra is "Always code and comment as if the person who ends up maintaining your code will be a psychopath who knows where you live."

 

[Martin Rattigan]

Unfortunately things are actually even more ambiguous.

With general functions, the same notation is used for the result of applying a function to an element of its domain and applying it to a subset of its domain (and presumably therefore to a subset of the power set of its domain etc.).

So if $\alpha\in S$ and $\alpha\subset S$ and $f$ is a function with domain $S$, then $f(\alpha)$ could mean the value of $f$ for the argument $\alpha$, or $\{f(x):x\in \alpha\}$ where $f(x)$ is here the value of $f$ for the argument $x$.

Worse, $f^{-1}$ can refer either to the inverse function of $f$ if it exists, or to a function $f^{-1}:\mathfrak{P}(\mathcal{R})\rightarrow \mathfrak{P}(\mathcal{D}_f)$, where $\mathfrak{P}$ denotes the power set, $\mathcal{D}_f$ is the domain of $f$ and $\mathcal{R}$ is at least the range of $f$ (which may itself be ambiguous) s.t. $f^{-1}:\beta\subset \mathcal{R}\mapsto \{\gamma\in\mathcal{D}_f:f(\gamma)\in\beta\}$ (here $f(\gamma)$ is the value of $f$ for the argument $\gamma$).

No doubt the ambiguities inherent in the foregoing could be confabulated to arbitrary heights, so it's quite surprising that it works at all. In practice it causes little confusion.

 

[Martin Rattigan]

$f^{-1}$ is a function selected from the converse, $\breve{f}$, of $f$ (as a relation) doesn't apply to anything outside the same set, and does apply to sin, cos, tan and cot. (I wouldn't like to be dogmatic about this use with the hyperbolic functions - I think you might have to go with the context in these cases.)

Actually I think I managed to confuse myself here. When $f$ is invertible, the usage $f^{-1}$ to mean the converse would be the normal use. In this case it would of course also be "a function selected from the converse" viz. all of it.

Since apart from cosh and sech the hyperbolic functions are essentially invertible anyway I would guess that $sinh^{-1}$ etc. would also refer to the inverse functions.

Not only that, I missed out sec, cosec, sech and csch from the list. For these also $sec^2(x)=(sec(x))^2$ etc.

So all in all a pretty good job.

By the way I've never seen $f^{-1}(x)$ used to mean $1/f(x)$ for anything.

 

[Mentallic]

I know that $$\sin^2 b= (\sin b)^2$$ and in general $$\sin^n b=(\sin b)^n$$ if n is a positive integer .

What if n is a negative integer , would it be

$$\sin^{-1}b=(\sin b)^{-1}=\frac{1}{\sin b}$$

I don't think this is right , because properties of indices only works for numbers and NOT function , but why it works for case 1 above ?

What if it's $$sin^{-3}b$$

How bout if n is rational ?

In the case of trigonometry, for all n>0 we can write the LHS which is understood as equalling the RHS $$sin^nx=(sinx)^n$$
If n<0 we use the fact that $$(sinx)^{-1}=cscx\neq sin^{-1}x$$ to show all n<0. For n>0, $$(sinx)^{-n}=((sinx)^{-1})^{n}=(cscx)^n=csc^nx$$

This avoids the ambiguity of the inverse sin function $$sin^{-1}x$$ being confused with the reciprocal of sin.

 

[thereddevils]

thanks all ! So in conclusion ,

tan^-1x , f^-1 , log^-1(x) are all meant to be inverses most of the time .

I understood now !

 

[Mentallic]

Yep!

But I don't know where you would see $$log^{-1}x$$ since it has another form entirely to express that, e^x.

 

[Martin Rattigan]

Yep!

Apart from:

(a) What Mentallic said about log applies to many other functions, so for named functions the $^{-1}$ notation is probably little used except for trignometric and hyperbolic functions and even here arcsin, arsinh etc. seem to be preferred these days.

(b) When used with ad hoc function names, e.g. $f^{-1}$, it probably most often means something other than the inverse, viz. one of the ambiguous possibilities I mentioned in an earlier post.

So if f is defined by:

$a\mapsto 1$
$b\mapsto 2$
$c\mapsto \{1,2\}$

then $f^{-1}( \{1,2\})$ could mean variously:

(i) $c$ - the image of $\{1,2\}$ under the function inverse to $f$.
(ii) $\{c\}$ - the set of elements that map to the element $\{1,2\}$ under $f$.
(iii) $\{a,b\}$ - the set of elements that map to a member of the set of elements $\{1,2\}$ under $f$.

The meaning (i) is the one that you suggested would be meant most of the time, but had $f$ included $d\mapsto 1$ it would no longer have been invertible and that meaning disappears.

This you just have to live with, but as I said it doesn't cause too much confusion in practice.

 

[thereddevils]

Yep!

Apart from:

(a) What Mentallic said about log applies to many other functions, so for named functions the $^{-1}$ notation is probably little used except for trignometric and hyperbolic functions and even here arcsin, arsinh etc. seem to be preferred these days.

(b) When used with ad hoc function names, e.g. $f^{-1}$, it probably most often means something other than the inverse, viz. one of the ambiguous possibilities I mentioned in an earlier post.

So if f is defined by:

$a\mapsto 1$
$b\mapsto 2$
$c\mapsto \{1,2\}$

then $f^{-1}( \{1,2\})$ could mean variously:

(i) $c$ - the image of $\{1,2\}$ under the function inverse to $f$.
(ii) $\{c\}$ - the set of elements that map to the element $\{1,2\}$ under $f$.
(iii) $\{a,b\}$ - the set of elements that map to a member of the set of elements $\{1,2\}$ under $f$.

The meaning (i) is the one that you suggested would be meant most of the time, but had $f$ included $d\mapsto 1$ it would no longer have been invertible and that meaning disappears.

This you just have to live with, but as I said it doesn't cause too much confusion in practice.

thank !