So in trigonometry, we have sin, secant (which is one over sin) and arcisn. Why is arcsin sometimes represented with $\sin^{-1}$? $\sin{^2}$ means sin to the second power, but $\sin^{-1}$ explicitly does not mean sin to the negative first power, as that would be the secant, not the arcsin. Why this confusion in notation?
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2In more advanced mathematics, the inverse function is denoted that way. If the function is $f$ then the inverse function is denoted $f^{-1}$. – GEdgar Jan 24 '15 at 18:38
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2In general, if $f(x)$ is a function, we write the inverse function as $f^{-1}(x)$. It's confusing notation at times, since we write $\sin^2(x)$ for $(\sin x)^2$, not $\sin(\sin(x))$. – Thomas Andrews Jan 24 '15 at 18:39
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18Because of this confusion, I recommend writing $\mathrm{arcsin}, x$ in your own writing. But of course you have to recognize $\sin^{-1} x$ when written by others. – GEdgar Jan 24 '15 at 18:40
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1The notation for inverse functions is highly inconsistent in maths. For instance $\log$ is used instead of $\exp^{-1}$, so if you feel the use $\sin^{-1}$ is confusing, you may use $\arcsin$ instead, or maybe write $(\sin(x))^2$ instead of $\sin^2(x)$, but be consistent with your use. Don't use both $\arcsin(x)$ and $\sin^{-1}(x)$. – Frank Vel Jan 24 '15 at 18:47
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That's how you write inverse of a function. – AvZ Jan 24 '15 at 19:09
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1The worst thing is when someone writes $sin^{-2}$, very hard to tell what they mean. My friend linked me a derivation before with that in it. Fair bit of debating over that one – snulty Jan 24 '15 at 19:59
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3Also $\sec x=\frac{1}{\cos x}$ – snulty Jan 24 '15 at 20:03
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Related: https://math.stackexchange.com/questions/30317/arcsin-written-as-sin-1x – Hans Lundmark Aug 30 '18 at 06:30
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It's the secret advanced mathematician question: What does $\sin^{-1}(x)$ mean? – Lee Mosher Dec 15 '23 at 01:18
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3The situation is even worse in that the function denoted $\sin^{-1}$ isn't actually the inverse of $\sin$---that function fails the horizontal line test and so has no inverse. Instead it is the inverse of the restriction of $\sin$ to $\left[-\frac{\pi}{2}, \frac{\pi}{2}\right]$. – Travis Willse Dec 15 '23 at 01:23
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If we agree, that $f^{-1}$ means the inverse of $f$, we may obtain such strange notation. Let us observe, that $x^{-1}=1/x$ is not the inverse of $f(x)=x$, $x^{-3}$ is not the inverse of $x^3$ and so on.
Przemysław Scherwentke
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4The notation wouldn't have that inconsistency problem because in $f^{-1}$ the $-1$ is applied to the name $f$ of the function. In the case $x^{-1}$, the $x$ is not playing the role of the name of a function, but of the value $f(x)$ of the function called $f$ and defined as $f(x)=x$. To be precise $x^{-1}=(f(x))^{-1}$. So, $-1$, when applied as superscript means: The inverse function, when applied to the name of the function, or the reciprocal when applied to a value. – Pp.. Jan 24 '15 at 18:49
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2I prefer to use $\arcsin$, but actually, the notation that causes the confusion is rather $\sin^2(x)$ to mean $(\sin(x))^2$ instead of $\sin(\sin(x))$. But again, I also prefer to use $\sin^2(x)$ with the former meaning. After all, $\sin(\sin(x))$ is a less common expression to stumble upon. – Pp.. Jan 24 '15 at 18:53
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@Pp.. Nevertheless, the notation $f^*$, as some people suggest, seems to be more resistant for students' mistakes. – Przemysław Scherwentke Jan 24 '15 at 18:54
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@PrzemysławScherwentke But $f^*$ is already used for other purposes such as dual mapping… – Divide1918 Dec 15 '23 at 02:08
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@Divide1918 If the students know the dual mapping, they do not have problems with $\operatorname{arc}\sin(x)$, I hope. – Przemysław Scherwentke Dec 16 '23 at 23:05
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@PrzemysławScherwentke : Clearly, we should teach students to use upside-down symbol names for inverse functions, e.g., "u!s" (with a correct upper baseline for the inverted "i", which I am only able to approximate with a "!"). – Eric Towers Apr 14 '24 at 05:19
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There is the multiplicative meaning of the exponent (iterated product), and the functional meaning (iterated function composition).
$x^2$ is the square of $x$ or $x\cdot x$, $x^3$ is the cube, $x^{-1}$ is the inverse of $x$, or $\dfrac1x$, such that $x\,x^{-1}=x^{-1}\,x=1$. Similarly, $x^{1/2}=\sqrt x$, is the number such that $(x^{1/2})^2=x$.
$f^2(x)$ is either $(f(x))^2$ - multiplicative - , or $f(f(x))$ - functional - , depending on context. By contrast, we never use $f^{-1}(x)$ for $\dfrac1{f(x)}$, but for the function such that $f(f^{-1}(x))=f^{-1}(f(x))=x$.
By the way, there is also a functional square root, $f^{1/2}(x)$, such that $f^{1/2}(f^{1/2}(x))=f(x)$, but this is very rarely met.
This exponent convention makes sense because of the similarity of the concepts. And normally, the context avoids ambiguities.
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1It's perhaps worth specifying that $x^{-1}$ is the multiplicative inverse of $x$; and also noting that $(-1)\cdot x$ is the additive inverse of $x$. Mentioning both of these lays a bit more groundwork for associating "$-1$" with inverses of things, so that "of course" the symbol for an inverse function "should" incorporate "$-1$" in some way. (It's just unfortunate that the way that was chosen overloads the reciprocal notation, causing confusion. I support the alternative $f^{\circ(-1)}(x)$, where $f^{\circ n}(x)$ represents iterated composition.) – Blue Aug 21 '24 at 20:50
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@Blue: the additive inverse is usually called the opposite or the negative, and does not adhere to the exponent notation. I'll change algebraic to multiplicative. – Aug 21 '24 at 21:02
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The confusion is because in most cases, It can be inferred, and if not you use $f^{\circ n}(x)$ notation instead (\circ). In most cases, if I mean an exponent, I usually put the number outside the function e.g. $\sin(x)^2$ V.S. $\sin^2(x)$. As my personal notation. It's similar to an individual programmer's variable and function naming scheme. Long story short; Always elaborate on if you're using function repetition or not.
To avoid confusion;
for repeated function usage, use a circle and put the "exponent" before the terms e.g. $f^{\circ-1}(x) $
for raising to the power of an exponent, put the exponent after the terms, e.g. $f^2(x)$
Xander Henderson
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DaPlumer
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