Is $f(f^{-1}(x))=x quad land quad f^{-1}(f(x))=x$ true for all inverse trigonometric functions? [closed]
$begingroup$
Based on:
$f(f^{-1}(x))=x quad quadlandquad quad f^{-1}(f(x))=x$
Are all the following definitions true?
$arcsin(sin x)=x$
$sin(arcsin x) = x$
$arccos (cos x) = x$
$cos (arccos x) = x$
$arctan(tan x) = x$
$tan(arctan x) = x$
$text{arccot} (cot x) = x$
$cot (text{arccot} x)=x$
That's make everything easier if they all were true. Thanks for clarification.
analysis trigonometry inverse-function
asked Dec 5 '18 at 3:23
wenoweno
1349
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closed as too broad by Abcd, Cesareo, Shailesh, Lee David Chung Lin, Adrian Keister Jan 10 at 14:51
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
$begingroup$
Based on:
$f(f^{-1}(x))=x quad quadlandquad quad f^{-1}(f(x))=x$
Are all the following definitions true?
$arcsin(sin x)=x$
$sin(arcsin x) = x$
$arccos (cos x) = x$
$cos (arccos x) = x$
$arctan(tan x) = x$
$tan(arctan x) = x$
$text{arccot} (cot x) = x$
$cot (text{arccot} x)=x$
That's make everything easier if they all were true. Thanks for clarification.
analysis trigonometry inverse-function
asked Dec 5 '18 at 3:23
wenoweno
1349
$endgroup$
closed as too broad by Abcd, Cesareo, Shailesh, Lee David Chung Lin, Adrian Keister Jan 10 at 14:51
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
3
$begingroup$
A serious fault of the high-school treatment of the direct and inverse trigonometric functions (and in many elementary calculus treatments as well) is that they do not give due emphasis to the importance of specifying the domain of definition of each function. In particular, it makes no sense to ask about the “truth” of such statements as those you’ve made without specifying the domain of definition of each function. This is particularly important for the inverse trig functions. As you’ve posed it, your question has no answer.
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– Lubin
Dec 5 '18 at 3:50
$begingroup$
These statements are true on the restricted domains of the periodic functions.
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– symplectomorphic
Dec 5 '18 at 3:54
1
$begingroup$
That’s true as far as it goes, @symplectomorphic. But the periodic functions may be restricted in any number of ways… This is yet more obvious when we try to define an inverse of the complex periodic function exp. Once you’ve restricted it to a set small enough for exp to be one-to-one, you’ve got your logarithm. But there are too too many possible such restrictions…
$endgroup$
– Lubin
Dec 5 '18 at 4:02
$begingroup$
Oh, agreed entirely @Lubin — I upvoted your comment. Mine was addressed to the OP. By “the” restricted domains, I meant to point OP to the standard ones for these functions on the reals.
$endgroup$
– symplectomorphic
Dec 5 '18 at 4:52
add a comment |
$begingroup$
Based on:
$f(f^{-1}(x))=x quad quadlandquad quad f^{-1}(f(x))=x$
Are all the following definitions true?
$arcsin(sin x)=x$
$sin(arcsin x) = x$
$arccos (cos x) = x$
$cos (arccos x) = x$
$arctan(tan x) = x$
$tan(arctan x) = x$
$text{arccot} (cot x) = x$
$cot (text{arccot} x)=x$
That's make everything easier if they all were true. Thanks for clarification.
analysis trigonometry inverse-function
asked Dec 5 '18 at 3:23
wenoweno
1349
$endgroup$
Based on:
$f(f^{-1}(x))=x quad quadlandquad quad f^{-1}(f(x))=x$
Are all the following definitions true?
$arcsin(sin x)=x$
$sin(arcsin x) = x$
$arccos (cos x) = x$
$cos (arccos x) = x$
$arctan(tan x) = x$
$tan(arctan x) = x$
$text{arccot} (cot x) = x$
$cot (text{arccot} x)=x$
That's make everything easier if they all were true. Thanks for clarification.
analysis trigonometry inverse-function
analysis trigonometry inverse-function
asked Dec 5 '18 at 3:23
wenoweno
1349
asked Dec 5 '18 at 3:23
wenoweno
1349
asked Dec 5 '18 at 3:23
wenoweno
1349
asked Dec 5 '18 at 3:23
wenoweno
1349
asked Dec 5 '18 at 3:23
wenoweno
1349
1349
closed as too broad by Abcd, Cesareo, Shailesh, Lee David Chung Lin, Adrian Keister Jan 10 at 14:51
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
closed as too broad by Abcd, Cesareo, Shailesh, Lee David Chung Lin, Adrian Keister Jan 10 at 14:51
Please edit the question to limit it to a specific problem with enough detail to identify an adequate answer. Avoid asking multiple distinct questions at once. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.
3
$begingroup$
A serious fault of the high-school treatment of the direct and inverse trigonometric functions (and in many elementary calculus treatments as well) is that they do not give due emphasis to the importance of specifying the domain of definition of each function. In particular, it makes no sense to ask about the “truth” of such statements as those you’ve made without specifying the domain of definition of each function. This is particularly important for the inverse trig functions. As you’ve posed it, your question has no answer.
$endgroup$
– Lubin
Dec 5 '18 at 3:50
$begingroup$
These statements are true on the restricted domains of the periodic functions.
$endgroup$
– symplectomorphic
Dec 5 '18 at 3:54
1
$begingroup$
That’s true as far as it goes, @symplectomorphic. But the periodic functions may be restricted in any number of ways… This is yet more obvious when we try to define an inverse of the complex periodic function exp. Once you’ve restricted it to a set small enough for exp to be one-to-one, you’ve got your logarithm. But there are too too many possible such restrictions…
$endgroup$
– Lubin
Dec 5 '18 at 4:02
$begingroup$
Oh, agreed entirely @Lubin — I upvoted your comment. Mine was addressed to the OP. By “the” restricted domains, I meant to point OP to the standard ones for these functions on the reals.
$endgroup$
– symplectomorphic
Dec 5 '18 at 4:52
add a comment |
3
$begingroup$
A serious fault of the high-school treatment of the direct and inverse trigonometric functions (and in many elementary calculus treatments as well) is that they do not give due emphasis to the importance of specifying the domain of definition of each function. In particular, it makes no sense to ask about the “truth” of such statements as those you’ve made without specifying the domain of definition of each function. This is particularly important for the inverse trig functions. As you’ve posed it, your question has no answer.
$endgroup$
– Lubin
Dec 5 '18 at 3:50
$begingroup$
These statements are true on the restricted domains of the periodic functions.
$endgroup$
– symplectomorphic
Dec 5 '18 at 3:54
1
$begingroup$
That’s true as far as it goes, @symplectomorphic. But the periodic functions may be restricted in any number of ways… This is yet more obvious when we try to define an inverse of the complex periodic function exp. Once you’ve restricted it to a set small enough for exp to be one-to-one, you’ve got your logarithm. But there are too too many possible such restrictions…
$endgroup$
– Lubin
Dec 5 '18 at 4:02
$begingroup$
Oh, agreed entirely @Lubin — I upvoted your comment. Mine was addressed to the OP. By “the” restricted domains, I meant to point OP to the standard ones for these functions on the reals.
$endgroup$
– symplectomorphic
Dec 5 '18 at 4:52
3
3
$begingroup$
A serious fault of the high-school treatment of the direct and inverse trigonometric functions (and in many elementary calculus treatments as well) is that they do not give due emphasis to the importance of specifying the domain of definition of each function. In particular, it makes no sense to ask about the “truth” of such statements as those you’ve made without specifying the domain of definition of each function. This is particularly important for the inverse trig functions. As you’ve posed it, your question has no answer.
$endgroup$
– Lubin
Dec 5 '18 at 3:50
$begingroup$
A serious fault of the high-school treatment of the direct and inverse trigonometric functions (and in many elementary calculus treatments as well) is that they do not give due emphasis to the importance of specifying the domain of definition of each function. In particular, it makes no sense to ask about the “truth” of such statements as those you’ve made without specifying the domain of definition of each function. This is particularly important for the inverse trig functions. As you’ve posed it, your question has no answer.
$endgroup$
– Lubin
Dec 5 '18 at 3:50
$begingroup$
These statements are true on the restricted domains of the periodic functions.
$endgroup$
– symplectomorphic
Dec 5 '18 at 3:54
$begingroup$
These statements are true on the restricted domains of the periodic functions.
$endgroup$
– symplectomorphic
Dec 5 '18 at 3:54
1
1
$begingroup$
That’s true as far as it goes, @symplectomorphic. But the periodic functions may be restricted in any number of ways… This is yet more obvious when we try to define an inverse of the complex periodic function exp. Once you’ve restricted it to a set small enough for exp to be one-to-one, you’ve got your logarithm. But there are too too many possible such restrictions…
$endgroup$
– Lubin
Dec 5 '18 at 4:02
$begingroup$
That’s true as far as it goes, @symplectomorphic. But the periodic functions may be restricted in any number of ways… This is yet more obvious when we try to define an inverse of the complex periodic function exp. Once you’ve restricted it to a set small enough for exp to be one-to-one, you’ve got your logarithm. But there are too too many possible such restrictions…
$endgroup$
– Lubin
Dec 5 '18 at 4:02
$begingroup$
Oh, agreed entirely @Lubin — I upvoted your comment. Mine was addressed to the OP. By “the” restricted domains, I meant to point OP to the standard ones for these functions on the reals.
$endgroup$
– symplectomorphic
Dec 5 '18 at 4:52
$begingroup$
Oh, agreed entirely @Lubin — I upvoted your comment. Mine was addressed to the OP. By “the” restricted domains, I meant to point OP to the standard ones for these functions on the reals.
$endgroup$
– symplectomorphic
Dec 5 '18 at 4:52
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
No.
To elaborate a bit on the previous answer, inverse trig functions have a limited range, so they’ll give the angle within their allowed quadrants. For example:
$$arcsinbigg(sin frac{7pi}{6}bigg) = -frac{pi}{6}$$
This happens since the range of $arcsin x$ is $big[-frac{pi}{2}, frac{pi}{2}big]$.
Also, $tan x$ and $cot x$ are not defined for all real numbers, unlike their inverse counterparts. You might want to learn the domain and range of the $6$ main trig functions then switch their domains and ranges for their inverses. (Knowing these can help greatly.)
answered Dec 5 '18 at 3:43
KM101KM101
5,9281523
$endgroup$
$begingroup$
Thanks. So the conclusion is $f^{-1}(f(x)) = a$ where $a$ is $x$ after the shift ($x$ has to be in $[0; pi]$ range for $arccos$,$[frac{- pi}{2}; frac{pi}{2}] $ for $arcsin$)? What about $arctan$ and $text{arccot}$?
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– weno
Dec 5 '18 at 4:06
$begingroup$
Yes, you could say something like that! The same applies to those.
$endgroup$
– KM101
Dec 5 '18 at 4:11
$begingroup$
But unlikely to $arcsin(sin x)$ and $arccos(cos x)$, I don't have to limit $arctan(tan x) $ and $text{arccot}(cot x) $ domain since they're defined for $x = mathbb{R}$? Hence $arctan(tan x) = x$ where x can be any $mathbb{R} $ and same about $text{arccot}$?
$endgroup$
– weno
Dec 5 '18 at 4:16
$begingroup$
@weno Note that $tan x$ is defined for any $x$ not of the form $pi/2+npi.$ So it is defined e.g. for $x=100pi.$ But $arctan x$ always is in the open interval $(-pi/2,pi/2)$ So it couldn't work for $x=100pi$ that $arctan(tan x)=x.$
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– coffeemath
Dec 5 '18 at 6:39
add a comment |
$begingroup$
The ones like $sin(arcsin(x)=x$ [with the inverse function "inside"] are true, not the others. Try experimenting with large angles for $x.$
answered Dec 5 '18 at 3:30
coffeemathcoffeemath
2,7361415
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add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
No.
To elaborate a bit on the previous answer, inverse trig functions have a limited range, so they’ll give the angle within their allowed quadrants. For example:
$$arcsinbigg(sin frac{7pi}{6}bigg) = -frac{pi}{6}$$
This happens since the range of $arcsin x$ is $big[-frac{pi}{2}, frac{pi}{2}big]$.
Also, $tan x$ and $cot x$ are not defined for all real numbers, unlike their inverse counterparts. You might want to learn the domain and range of the $6$ main trig functions then switch their domains and ranges for their inverses. (Knowing these can help greatly.)
answered Dec 5 '18 at 3:43
KM101KM101
5,9281523
$endgroup$
$begingroup$
Thanks. So the conclusion is $f^{-1}(f(x)) = a$ where $a$ is $x$ after the shift ($x$ has to be in $[0; pi]$ range for $arccos$,$[frac{- pi}{2}; frac{pi}{2}] $ for $arcsin$)? What about $arctan$ and $text{arccot}$?
$endgroup$
– weno
Dec 5 '18 at 4:06
$begingroup$
Yes, you could say something like that! The same applies to those.
$endgroup$
– KM101
Dec 5 '18 at 4:11
$begingroup$
But unlikely to $arcsin(sin x)$ and $arccos(cos x)$, I don't have to limit $arctan(tan x) $ and $text{arccot}(cot x) $ domain since they're defined for $x = mathbb{R}$? Hence $arctan(tan x) = x$ where x can be any $mathbb{R} $ and same about $text{arccot}$?
$endgroup$
– weno
Dec 5 '18 at 4:16
$begingroup$
@weno Note that $tan x$ is defined for any $x$ not of the form $pi/2+npi.$ So it is defined e.g. for $x=100pi.$ But $arctan x$ always is in the open interval $(-pi/2,pi/2)$ So it couldn't work for $x=100pi$ that $arctan(tan x)=x.$
$endgroup$
– coffeemath
Dec 5 '18 at 6:39
add a comment |
$begingroup$
No.
To elaborate a bit on the previous answer, inverse trig functions have a limited range, so they’ll give the angle within their allowed quadrants. For example:
$$arcsinbigg(sin frac{7pi}{6}bigg) = -frac{pi}{6}$$
This happens since the range of $arcsin x$ is $big[-frac{pi}{2}, frac{pi}{2}big]$.
Also, $tan x$ and $cot x$ are not defined for all real numbers, unlike their inverse counterparts. You might want to learn the domain and range of the $6$ main trig functions then switch their domains and ranges for their inverses. (Knowing these can help greatly.)
answered Dec 5 '18 at 3:43
KM101KM101
5,9281523
$endgroup$
$begingroup$
Thanks. So the conclusion is $f^{-1}(f(x)) = a$ where $a$ is $x$ after the shift ($x$ has to be in $[0; pi]$ range for $arccos$,$[frac{- pi}{2}; frac{pi}{2}] $ for $arcsin$)? What about $arctan$ and $text{arccot}$?
$endgroup$
– weno
Dec 5 '18 at 4:06
$begingroup$
Yes, you could say something like that! The same applies to those.
$endgroup$
– KM101
Dec 5 '18 at 4:11
$begingroup$
But unlikely to $arcsin(sin x)$ and $arccos(cos x)$, I don't have to limit $arctan(tan x) $ and $text{arccot}(cot x) $ domain since they're defined for $x = mathbb{R}$? Hence $arctan(tan x) = x$ where x can be any $mathbb{R} $ and same about $text{arccot}$?
$endgroup$
– weno
Dec 5 '18 at 4:16
$begingroup$
@weno Note that $tan x$ is defined for any $x$ not of the form $pi/2+npi.$ So it is defined e.g. for $x=100pi.$ But $arctan x$ always is in the open interval $(-pi/2,pi/2)$ So it couldn't work for $x=100pi$ that $arctan(tan x)=x.$
$endgroup$
– coffeemath
Dec 5 '18 at 6:39
add a comment |
$begingroup$
No.
To elaborate a bit on the previous answer, inverse trig functions have a limited range, so they’ll give the angle within their allowed quadrants. For example:
$$arcsinbigg(sin frac{7pi}{6}bigg) = -frac{pi}{6}$$
This happens since the range of $arcsin x$ is $big[-frac{pi}{2}, frac{pi}{2}big]$.
Also, $tan x$ and $cot x$ are not defined for all real numbers, unlike their inverse counterparts. You might want to learn the domain and range of the $6$ main trig functions then switch their domains and ranges for their inverses. (Knowing these can help greatly.)
answered Dec 5 '18 at 3:43
KM101KM101
5,9281523
$endgroup$
No.
To elaborate a bit on the previous answer, inverse trig functions have a limited range, so they’ll give the angle within their allowed quadrants. For example:
$$arcsinbigg(sin frac{7pi}{6}bigg) = -frac{pi}{6}$$
This happens since the range of $arcsin x$ is $big[-frac{pi}{2}, frac{pi}{2}big]$.
Also, $tan x$ and $cot x$ are not defined for all real numbers, unlike their inverse counterparts. You might want to learn the domain and range of the $6$ main trig functions then switch their domains and ranges for their inverses. (Knowing these can help greatly.)
answered Dec 5 '18 at 3:43
KM101KM101
5,9281523
edited Dec 11 '18 at 15:29
answered Dec 5 '18 at 3:43
KM101KM101
5,9281523
answered Dec 5 '18 at 3:43
KM101KM101
5,9281523
answered Dec 5 '18 at 3:43
KM101KM101
5,9281523
5,9281523
$begingroup$
Thanks. So the conclusion is $f^{-1}(f(x)) = a$ where $a$ is $x$ after the shift ($x$ has to be in $[0; pi]$ range for $arccos$,$[frac{- pi}{2}; frac{pi}{2}] $ for $arcsin$)? What about $arctan$ and $text{arccot}$?
$endgroup$
– weno
Dec 5 '18 at 4:06
$begingroup$
Yes, you could say something like that! The same applies to those.
$endgroup$
– KM101
Dec 5 '18 at 4:11
$begingroup$
But unlikely to $arcsin(sin x)$ and $arccos(cos x)$, I don't have to limit $arctan(tan x) $ and $text{arccot}(cot x) $ domain since they're defined for $x = mathbb{R}$? Hence $arctan(tan x) = x$ where x can be any $mathbb{R} $ and same about $text{arccot}$?
$endgroup$
– weno
Dec 5 '18 at 4:16
$begingroup$
@weno Note that $tan x$ is defined for any $x$ not of the form $pi/2+npi.$ So it is defined e.g. for $x=100pi.$ But $arctan x$ always is in the open interval $(-pi/2,pi/2)$ So it couldn't work for $x=100pi$ that $arctan(tan x)=x.$
$endgroup$
– coffeemath
Dec 5 '18 at 6:39
add a comment |
$begingroup$
Thanks. So the conclusion is $f^{-1}(f(x)) = a$ where $a$ is $x$ after the shift ($x$ has to be in $[0; pi]$ range for $arccos$,$[frac{- pi}{2}; frac{pi}{2}] $ for $arcsin$)? What about $arctan$ and $text{arccot}$?
$endgroup$
– weno
Dec 5 '18 at 4:06
$begingroup$
Yes, you could say something like that! The same applies to those.
$endgroup$
– KM101
Dec 5 '18 at 4:11
$begingroup$
But unlikely to $arcsin(sin x)$ and $arccos(cos x)$, I don't have to limit $arctan(tan x) $ and $text{arccot}(cot x) $ domain since they're defined for $x = mathbb{R}$? Hence $arctan(tan x) = x$ where x can be any $mathbb{R} $ and same about $text{arccot}$?
$endgroup$
– weno
Dec 5 '18 at 4:16
$begingroup$
@weno Note that $tan x$ is defined for any $x$ not of the form $pi/2+npi.$ So it is defined e.g. for $x=100pi.$ But $arctan x$ always is in the open interval $(-pi/2,pi/2)$ So it couldn't work for $x=100pi$ that $arctan(tan x)=x.$
$endgroup$
– coffeemath
Dec 5 '18 at 6:39
$begingroup$
Thanks. So the conclusion is $f^{-1}(f(x)) = a$ where $a$ is $x$ after the shift ($x$ has to be in $[0; pi]$ range for $arccos$,$[frac{- pi}{2}; frac{pi}{2}] $ for $arcsin$)? What about $arctan$ and $text{arccot}$?
$endgroup$
– weno
Dec 5 '18 at 4:06
$begingroup$
Thanks. So the conclusion is $f^{-1}(f(x)) = a$ where $a$ is $x$ after the shift ($x$ has to be in $[0; pi]$ range for $arccos$,$[frac{- pi}{2}; frac{pi}{2}] $ for $arcsin$)? What about $arctan$ and $text{arccot}$?
$endgroup$
– weno
Dec 5 '18 at 4:06
$begingroup$
Yes, you could say something like that! The same applies to those.
$endgroup$
– KM101
Dec 5 '18 at 4:11
$begingroup$
Yes, you could say something like that! The same applies to those.
$endgroup$
– KM101
Dec 5 '18 at 4:11
$begingroup$
But unlikely to $arcsin(sin x)$ and $arccos(cos x)$, I don't have to limit $arctan(tan x) $ and $text{arccot}(cot x) $ domain since they're defined for $x = mathbb{R}$? Hence $arctan(tan x) = x$ where x can be any $mathbb{R} $ and same about $text{arccot}$?
$endgroup$
– weno
Dec 5 '18 at 4:16
$begingroup$
But unlikely to $arcsin(sin x)$ and $arccos(cos x)$, I don't have to limit $arctan(tan x) $ and $text{arccot}(cot x) $ domain since they're defined for $x = mathbb{R}$? Hence $arctan(tan x) = x$ where x can be any $mathbb{R} $ and same about $text{arccot}$?
$endgroup$
– weno
Dec 5 '18 at 4:16
$begingroup$
@weno Note that $tan x$ is defined for any $x$ not of the form $pi/2+npi.$ So it is defined e.g. for $x=100pi.$ But $arctan x$ always is in the open interval $(-pi/2,pi/2)$ So it couldn't work for $x=100pi$ that $arctan(tan x)=x.$
$endgroup$
– coffeemath
Dec 5 '18 at 6:39
$begingroup$
@weno Note that $tan x$ is defined for any $x$ not of the form $pi/2+npi.$ So it is defined e.g. for $x=100pi.$ But $arctan x$ always is in the open interval $(-pi/2,pi/2)$ So it couldn't work for $x=100pi$ that $arctan(tan x)=x.$
$endgroup$
– coffeemath
Dec 5 '18 at 6:39
add a comment |
$begingroup$
The ones like $sin(arcsin(x)=x$ [with the inverse function "inside"] are true, not the others. Try experimenting with large angles for $x.$
answered Dec 5 '18 at 3:30
coffeemathcoffeemath
2,7361415
$endgroup$
add a comment |
$begingroup$
The ones like $sin(arcsin(x)=x$ [with the inverse function "inside"] are true, not the others. Try experimenting with large angles for $x.$
answered Dec 5 '18 at 3:30
coffeemathcoffeemath
2,7361415
$endgroup$
add a comment |
$begingroup$
The ones like $sin(arcsin(x)=x$ [with the inverse function "inside"] are true, not the others. Try experimenting with large angles for $x.$
answered Dec 5 '18 at 3:30
coffeemathcoffeemath
2,7361415
$endgroup$
The ones like $sin(arcsin(x)=x$ [with the inverse function "inside"] are true, not the others. Try experimenting with large angles for $x.$
answered Dec 5 '18 at 3:30
coffeemathcoffeemath
2,7361415
answered Dec 5 '18 at 3:30
coffeemathcoffeemath
2,7361415
answered Dec 5 '18 at 3:30
coffeemathcoffeemath
2,7361415
answered Dec 5 '18 at 3:30
coffeemathcoffeemath
2,7361415
2,7361415
add a comment |
add a comment |
3
$begingroup$
A serious fault of the high-school treatment of the direct and inverse trigonometric functions (and in many elementary calculus treatments as well) is that they do not give due emphasis to the importance of specifying the domain of definition of each function. In particular, it makes no sense to ask about the “truth” of such statements as those you’ve made without specifying the domain of definition of each function. This is particularly important for the inverse trig functions. As you’ve posed it, your question has no answer.
$endgroup$
– Lubin
Dec 5 '18 at 3:50
$begingroup$
These statements are true on the restricted domains of the periodic functions.
$endgroup$
– symplectomorphic
Dec 5 '18 at 3:54
1
$begingroup$
That’s true as far as it goes, @symplectomorphic. But the periodic functions may be restricted in any number of ways… This is yet more obvious when we try to define an inverse of the complex periodic function exp. Once you’ve restricted it to a set small enough for exp to be one-to-one, you’ve got your logarithm. But there are too too many possible such restrictions…
$endgroup$
– Lubin
Dec 5 '18 at 4:02
$begingroup$
Oh, agreed entirely @Lubin — I upvoted your comment. Mine was addressed to the OP. By “the” restricted domains, I meant to point OP to the standard ones for these functions on the reals.
$endgroup$
– symplectomorphic
Dec 5 '18 at 4:52