Inverse of sum of two functions in terms of individual inverse functions
When we can express the inverse of sum of two functions for example $f=f_1+f_2$ in terms of inverse of two functions $(f_1^{-1},f_2^{-1})$?
functions inverse
asked Jul 15 '15 at 16:13
user51780user51780
937
add a comment |
When we can express the inverse of sum of two functions for example $f=f_1+f_2$ in terms of inverse of two functions $(f_1^{-1},f_2^{-1})$?
functions inverse
asked Jul 15 '15 at 16:13
user51780user51780
937
Technically, always. Just let $g = f^{-1}$ and observe that $g = f^{-1} + 0(f_1^{-1} + f_2^{-2})$. You should be more clear in the question formulation. What do you mean 'in terms of'?
– user217285
Jul 15 '15 at 16:20
I doubt that you will get a satisfactory answer, since in many many cases $f$ is not invertible, even though $f_1,f_2$ are (think $f_1=f_2 + c$).
– PhoemueX
Jul 15 '15 at 16:21
@PhoemueX Perhaps you want $f_1=c-f_2$.
– Michael Burr
Jul 15 '15 at 16:23
@MichaelBlurr: Oh, yes of course. Thanks.
– PhoemueX
Jul 15 '15 at 16:26
add a comment |
When we can express the inverse of sum of two functions for example $f=f_1+f_2$ in terms of inverse of two functions $(f_1^{-1},f_2^{-1})$?
functions inverse
asked Jul 15 '15 at 16:13
user51780user51780
937
When we can express the inverse of sum of two functions for example $f=f_1+f_2$ in terms of inverse of two functions $(f_1^{-1},f_2^{-1})$?
functions inverse
functions inverse
asked Jul 15 '15 at 16:13
user51780user51780
937
asked Jul 15 '15 at 16:13
user51780user51780
937
asked Jul 15 '15 at 16:13
user51780user51780
937
asked Jul 15 '15 at 16:13
user51780user51780
937
asked Jul 15 '15 at 16:13
user51780user51780
937
937
Technically, always. Just let $g = f^{-1}$ and observe that $g = f^{-1} + 0(f_1^{-1} + f_2^{-2})$. You should be more clear in the question formulation. What do you mean 'in terms of'?
– user217285
Jul 15 '15 at 16:20
I doubt that you will get a satisfactory answer, since in many many cases $f$ is not invertible, even though $f_1,f_2$ are (think $f_1=f_2 + c$).
– PhoemueX
Jul 15 '15 at 16:21
@PhoemueX Perhaps you want $f_1=c-f_2$.
– Michael Burr
Jul 15 '15 at 16:23
@MichaelBlurr: Oh, yes of course. Thanks.
– PhoemueX
Jul 15 '15 at 16:26
add a comment |
Technically, always. Just let $g = f^{-1}$ and observe that $g = f^{-1} + 0(f_1^{-1} + f_2^{-2})$. You should be more clear in the question formulation. What do you mean 'in terms of'?
– user217285
Jul 15 '15 at 16:20
I doubt that you will get a satisfactory answer, since in many many cases $f$ is not invertible, even though $f_1,f_2$ are (think $f_1=f_2 + c$).
– PhoemueX
Jul 15 '15 at 16:21
@PhoemueX Perhaps you want $f_1=c-f_2$.
– Michael Burr
Jul 15 '15 at 16:23
@MichaelBlurr: Oh, yes of course. Thanks.
– PhoemueX
Jul 15 '15 at 16:26
Technically, always. Just let $g = f^{-1}$ and observe that $g = f^{-1} + 0(f_1^{-1} + f_2^{-2})$. You should be more clear in the question formulation. What do you mean 'in terms of'?
– user217285
Jul 15 '15 at 16:20
Technically, always. Just let $g = f^{-1}$ and observe that $g = f^{-1} + 0(f_1^{-1} + f_2^{-2})$. You should be more clear in the question formulation. What do you mean 'in terms of'?
– user217285
Jul 15 '15 at 16:20
I doubt that you will get a satisfactory answer, since in many many cases $f$ is not invertible, even though $f_1,f_2$ are (think $f_1=f_2 + c$).
– PhoemueX
Jul 15 '15 at 16:21
I doubt that you will get a satisfactory answer, since in many many cases $f$ is not invertible, even though $f_1,f_2$ are (think $f_1=f_2 + c$).
– PhoemueX
Jul 15 '15 at 16:21
@PhoemueX Perhaps you want $f_1=c-f_2$.
– Michael Burr
Jul 15 '15 at 16:23
@PhoemueX Perhaps you want $f_1=c-f_2$.
– Michael Burr
Jul 15 '15 at 16:23
@MichaelBlurr: Oh, yes of course. Thanks.
– PhoemueX
Jul 15 '15 at 16:26
@MichaelBlurr: Oh, yes of course. Thanks.
– PhoemueX
Jul 15 '15 at 16:26
add a comment |
2 Answers
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There is no easy way to do so.
(There is an obvious hard way: invert $f_1^{-1}$ and $f_2^{-1}$ to get $f_1$ and $f_2$, sum these to get $f$ and then invert.)
It can well happen that $f_1$ and $f_2$ are invertible but $f$ is not.
Or that neither of $f_1$ and $f_2$ is invertible but $f$ is.
And you can find an invertible function $f_1$ and a non-invertible function $f_2$ so that $f$ is invertible — and another pair of such functions $f_1$ and $f_2$ so that $f$ is not invertible.
Counterexamples shouldn't be too hard to find once you know they are there.
Can you find them?
answered Jul 15 '15 at 16:22
Joonas IlmavirtaJoonas Ilmavirta
20.6k94282
add a comment |
The Cantor staircase function can be seen as a sum of functions with an explicit inverse in some places. The philosophy is obtaining some basis decomposition for your functions and then finding an inverse.
answered Nov 28 '18 at 1:37
Thomas KojarThomas Kojar
1155
add a comment |
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2 Answers
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2 Answers
2
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There is no easy way to do so.
(There is an obvious hard way: invert $f_1^{-1}$ and $f_2^{-1}$ to get $f_1$ and $f_2$, sum these to get $f$ and then invert.)
It can well happen that $f_1$ and $f_2$ are invertible but $f$ is not.
Or that neither of $f_1$ and $f_2$ is invertible but $f$ is.
And you can find an invertible function $f_1$ and a non-invertible function $f_2$ so that $f$ is invertible — and another pair of such functions $f_1$ and $f_2$ so that $f$ is not invertible.
Counterexamples shouldn't be too hard to find once you know they are there.
Can you find them?
answered Jul 15 '15 at 16:22
Joonas IlmavirtaJoonas Ilmavirta
20.6k94282
add a comment |
There is no easy way to do so.
(There is an obvious hard way: invert $f_1^{-1}$ and $f_2^{-1}$ to get $f_1$ and $f_2$, sum these to get $f$ and then invert.)
It can well happen that $f_1$ and $f_2$ are invertible but $f$ is not.
Or that neither of $f_1$ and $f_2$ is invertible but $f$ is.
And you can find an invertible function $f_1$ and a non-invertible function $f_2$ so that $f$ is invertible — and another pair of such functions $f_1$ and $f_2$ so that $f$ is not invertible.
Counterexamples shouldn't be too hard to find once you know they are there.
Can you find them?
answered Jul 15 '15 at 16:22
Joonas IlmavirtaJoonas Ilmavirta
20.6k94282
add a comment |
There is no easy way to do so.
(There is an obvious hard way: invert $f_1^{-1}$ and $f_2^{-1}$ to get $f_1$ and $f_2$, sum these to get $f$ and then invert.)
It can well happen that $f_1$ and $f_2$ are invertible but $f$ is not.
Or that neither of $f_1$ and $f_2$ is invertible but $f$ is.
And you can find an invertible function $f_1$ and a non-invertible function $f_2$ so that $f$ is invertible — and another pair of such functions $f_1$ and $f_2$ so that $f$ is not invertible.
Counterexamples shouldn't be too hard to find once you know they are there.
Can you find them?
answered Jul 15 '15 at 16:22
Joonas IlmavirtaJoonas Ilmavirta
20.6k94282
There is no easy way to do so.
(There is an obvious hard way: invert $f_1^{-1}$ and $f_2^{-1}$ to get $f_1$ and $f_2$, sum these to get $f$ and then invert.)
It can well happen that $f_1$ and $f_2$ are invertible but $f$ is not.
Or that neither of $f_1$ and $f_2$ is invertible but $f$ is.
And you can find an invertible function $f_1$ and a non-invertible function $f_2$ so that $f$ is invertible — and another pair of such functions $f_1$ and $f_2$ so that $f$ is not invertible.
Counterexamples shouldn't be too hard to find once you know they are there.
Can you find them?
answered Jul 15 '15 at 16:22
Joonas IlmavirtaJoonas Ilmavirta
20.6k94282
answered Jul 15 '15 at 16:22
Joonas IlmavirtaJoonas Ilmavirta
20.6k94282
answered Jul 15 '15 at 16:22
Joonas IlmavirtaJoonas Ilmavirta
20.6k94282
answered Jul 15 '15 at 16:22
Joonas IlmavirtaJoonas Ilmavirta
20.6k94282
20.6k94282
add a comment |
add a comment |
The Cantor staircase function can be seen as a sum of functions with an explicit inverse in some places. The philosophy is obtaining some basis decomposition for your functions and then finding an inverse.
answered Nov 28 '18 at 1:37
Thomas KojarThomas Kojar
1155
add a comment |
The Cantor staircase function can be seen as a sum of functions with an explicit inverse in some places. The philosophy is obtaining some basis decomposition for your functions and then finding an inverse.
answered Nov 28 '18 at 1:37
Thomas KojarThomas Kojar
1155
add a comment |
The Cantor staircase function can be seen as a sum of functions with an explicit inverse in some places. The philosophy is obtaining some basis decomposition for your functions and then finding an inverse.
answered Nov 28 '18 at 1:37
Thomas KojarThomas Kojar
1155
The Cantor staircase function can be seen as a sum of functions with an explicit inverse in some places. The philosophy is obtaining some basis decomposition for your functions and then finding an inverse.
answered Nov 28 '18 at 1:37
Thomas KojarThomas Kojar
1155
answered Nov 28 '18 at 1:37
Thomas KojarThomas Kojar
1155
answered Nov 28 '18 at 1:37
Thomas KojarThomas Kojar
1155
answered Nov 28 '18 at 1:37
Thomas KojarThomas Kojar
1155
1155
add a comment |
add a comment |
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Technically, always. Just let $g = f^{-1}$ and observe that $g = f^{-1} + 0(f_1^{-1} + f_2^{-2})$. You should be more clear in the question formulation. What do you mean 'in terms of'?
– user217285
Jul 15 '15 at 16:20
I doubt that you will get a satisfactory answer, since in many many cases $f$ is not invertible, even though $f_1,f_2$ are (think $f_1=f_2 + c$).
– PhoemueX
Jul 15 '15 at 16:21
@PhoemueX Perhaps you want $f_1=c-f_2$.
– Michael Burr
Jul 15 '15 at 16:23
@MichaelBlurr: Oh, yes of course. Thanks.
– PhoemueX
Jul 15 '15 at 16:26