Mixing
Existence of a function whose derivative of inverse equals the inverse of the derivative
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I've been thinking about the calculation of inverse functions through Taylor series expansions.
My hypothesis was that if we had: $$ f(x) =sum_{n=0}^infty frac{(x-x_0)}{n!}f^{n}(x_0),$$ then $$ f^{-1}(x) =sum_{n=0}^infty frac{(x-x_0)}{n!}({f^{-1}})^{(n)}(x_0).$$
Here it'd be possible to calculate the derivatives $ ({f^{-1}})^{(n)}(x_0)$ through the inverse function derivation theorem but for this to work we should have the derivative of the inverse of$ f$ equal the inverse of the derivative of $ f$, i.e. $$ (f^{-1})' = (f')^{-1} quad (1).$$ In general this does not hold; take for example $ f(x) = x^2 $.
To formulate my question, assume we have a differentiable bijective map $ f:Arightarrow B$ with bijective derivative. Its inverse $ f^{-1}$ is also a differentiable bijection. Assume the derivative of the inverse is also bijective.
To try and find a function for which$ (1) $ holds, I was able to deduce (from$ (1) $) that if such a function exists, we must have $$ f(x) = (f' circ f')(x) quad (2).$$
Clearly this does not hold for any polynomial nor (I'm pretty sure) for any other elementary functions. So, does there exist a function for which this condition holds?
Deriving (2):
$ (f')^{-1}(x) = (f^{-1})'(x) = frac{1}{(f' circ f^{-1})(x)} iff (f' circ f^{-1})(x) = frac{1}{(f')^{-1}(x)}$. Then mapping by $ f$ from the right gives $ f'(x)=frac{1}{(f')^{-1}(f(x))}$ and mapping by $ frac{1}{f'}$ from the left $ frac{1}{f(x)} = (frac{1}{f'} circ f')(x) iff f(x)= frac{1}{(frac{1}{f'} circ f')(x)}=frac{1}{frac{1}{(f' circ f')(x)}}=(f' circ f')(x)$
derivatives power-series inverse-function
edited Jan 20 at 13:32
Wesley Strik
2,113423
asked Jun 17 '16 at 8:28
CleggsteinCleggstein
316
$endgroup$
add a comment |
$begingroup$
I've been thinking about the calculation of inverse functions through Taylor series expansions.
My hypothesis was that if we had: $$ f(x) =sum_{n=0}^infty frac{(x-x_0)}{n!}f^{n}(x_0),$$ then $$ f^{-1}(x) =sum_{n=0}^infty frac{(x-x_0)}{n!}({f^{-1}})^{(n)}(x_0).$$
Here it'd be possible to calculate the derivatives $ ({f^{-1}})^{(n)}(x_0)$ through the inverse function derivation theorem but for this to work we should have the derivative of the inverse of$ f$ equal the inverse of the derivative of $ f$, i.e. $$ (f^{-1})' = (f')^{-1} quad (1).$$ In general this does not hold; take for example $ f(x) = x^2 $.
To formulate my question, assume we have a differentiable bijective map $ f:Arightarrow B$ with bijective derivative. Its inverse $ f^{-1}$ is also a differentiable bijection. Assume the derivative of the inverse is also bijective.
To try and find a function for which$ (1) $ holds, I was able to deduce (from$ (1) $) that if such a function exists, we must have $$ f(x) = (f' circ f')(x) quad (2).$$
Clearly this does not hold for any polynomial nor (I'm pretty sure) for any other elementary functions. So, does there exist a function for which this condition holds?
Deriving (2):
$ (f')^{-1}(x) = (f^{-1})'(x) = frac{1}{(f' circ f^{-1})(x)} iff (f' circ f^{-1})(x) = frac{1}{(f')^{-1}(x)}$. Then mapping by $ f$ from the right gives $ f'(x)=frac{1}{(f')^{-1}(f(x))}$ and mapping by $ frac{1}{f'}$ from the left $ frac{1}{f(x)} = (frac{1}{f'} circ f')(x) iff f(x)= frac{1}{(frac{1}{f'} circ f')(x)}=frac{1}{frac{1}{(f' circ f')(x)}}=(f' circ f')(x)$
derivatives power-series inverse-function
edited Jan 20 at 13:32
Wesley Strik
2,113423
asked Jun 17 '16 at 8:28
CleggsteinCleggstein
316
$endgroup$
2
$begingroup$
How did you derive the fact $f(x)=(f' circ f')(x)?$
$endgroup$
– JasonM
Jun 17 '16 at 8:43
$begingroup$
@JasonM I edited the post now.
$endgroup$
– Cleggstein
Jun 17 '16 at 9:26
$begingroup$
Then I have an answer
$endgroup$
– JasonM
Jun 17 '16 at 9:33
$begingroup$
Actually, I'm feeling a bit uneasy about the step where you mapped by $frac{1}{f'}$ from the left. Are we sure $(frac{1}{f'}) circ frac{1}{((f')^{-1})circ f}=frac{1}{f}$?
$endgroup$
– JasonM
Jun 17 '16 at 10:01
$begingroup$
Ah yes, I've mixed multiplication and mapping there. Actual mapping or multiplication don't seem to yield anything useful either. I'll try and work on this a bit more.
$endgroup$
– Cleggstein
Jun 17 '16 at 10:36
add a comment |
$begingroup$
I've been thinking about the calculation of inverse functions through Taylor series expansions.
My hypothesis was that if we had: $$ f(x) =sum_{n=0}^infty frac{(x-x_0)}{n!}f^{n}(x_0),$$ then $$ f^{-1}(x) =sum_{n=0}^infty frac{(x-x_0)}{n!}({f^{-1}})^{(n)}(x_0).$$
Here it'd be possible to calculate the derivatives $ ({f^{-1}})^{(n)}(x_0)$ through the inverse function derivation theorem but for this to work we should have the derivative of the inverse of$ f$ equal the inverse of the derivative of $ f$, i.e. $$ (f^{-1})' = (f')^{-1} quad (1).$$ In general this does not hold; take for example $ f(x) = x^2 $.
To formulate my question, assume we have a differentiable bijective map $ f:Arightarrow B$ with bijective derivative. Its inverse $ f^{-1}$ is also a differentiable bijection. Assume the derivative of the inverse is also bijective.
To try and find a function for which$ (1) $ holds, I was able to deduce (from$ (1) $) that if such a function exists, we must have $$ f(x) = (f' circ f')(x) quad (2).$$
Clearly this does not hold for any polynomial nor (I'm pretty sure) for any other elementary functions. So, does there exist a function for which this condition holds?
Deriving (2):
$ (f')^{-1}(x) = (f^{-1})'(x) = frac{1}{(f' circ f^{-1})(x)} iff (f' circ f^{-1})(x) = frac{1}{(f')^{-1}(x)}$. Then mapping by $ f$ from the right gives $ f'(x)=frac{1}{(f')^{-1}(f(x))}$ and mapping by $ frac{1}{f'}$ from the left $ frac{1}{f(x)} = (frac{1}{f'} circ f')(x) iff f(x)= frac{1}{(frac{1}{f'} circ f')(x)}=frac{1}{frac{1}{(f' circ f')(x)}}=(f' circ f')(x)$
derivatives power-series inverse-function
edited Jan 20 at 13:32
Wesley Strik
2,113423
asked Jun 17 '16 at 8:28
CleggsteinCleggstein
316
$endgroup$
I've been thinking about the calculation of inverse functions through Taylor series expansions.
My hypothesis was that if we had: $$ f(x) =sum_{n=0}^infty frac{(x-x_0)}{n!}f^{n}(x_0),$$ then $$ f^{-1}(x) =sum_{n=0}^infty frac{(x-x_0)}{n!}({f^{-1}})^{(n)}(x_0).$$
Here it'd be possible to calculate the derivatives $ ({f^{-1}})^{(n)}(x_0)$ through the inverse function derivation theorem but for this to work we should have the derivative of the inverse of$ f$ equal the inverse of the derivative of $ f$, i.e. $$ (f^{-1})' = (f')^{-1} quad (1).$$ In general this does not hold; take for example $ f(x) = x^2 $.
To formulate my question, assume we have a differentiable bijective map $ f:Arightarrow B$ with bijective derivative. Its inverse $ f^{-1}$ is also a differentiable bijection. Assume the derivative of the inverse is also bijective.
To try and find a function for which$ (1) $ holds, I was able to deduce (from$ (1) $) that if such a function exists, we must have $$ f(x) = (f' circ f')(x) quad (2).$$
Clearly this does not hold for any polynomial nor (I'm pretty sure) for any other elementary functions. So, does there exist a function for which this condition holds?
Deriving (2):
$ (f')^{-1}(x) = (f^{-1})'(x) = frac{1}{(f' circ f^{-1})(x)} iff (f' circ f^{-1})(x) = frac{1}{(f')^{-1}(x)}$. Then mapping by $ f$ from the right gives $ f'(x)=frac{1}{(f')^{-1}(f(x))}$ and mapping by $ frac{1}{f'}$ from the left $ frac{1}{f(x)} = (frac{1}{f'} circ f')(x) iff f(x)= frac{1}{(frac{1}{f'} circ f')(x)}=frac{1}{frac{1}{(f' circ f')(x)}}=(f' circ f')(x)$
derivatives power-series inverse-function
derivatives power-series inverse-function
edited Jan 20 at 13:32
Wesley Strik
2,113423
asked Jun 17 '16 at 8:28
CleggsteinCleggstein
316
edited Jan 20 at 13:32
Wesley Strik
2,113423
asked Jun 17 '16 at 8:28
CleggsteinCleggstein
316
edited Jan 20 at 13:32
Wesley Strik
2,113423
edited Jan 20 at 13:32
Wesley Strik
2,113423
edited Jan 20 at 13:32
Wesley Strik
2,113423
2,113423
asked Jun 17 '16 at 8:28
CleggsteinCleggstein
316
asked Jun 17 '16 at 8:28
CleggsteinCleggstein
316
asked Jun 17 '16 at 8:28
CleggsteinCleggstein
316
316
2
$begingroup$
How did you derive the fact $f(x)=(f' circ f')(x)?$
$endgroup$
– JasonM
Jun 17 '16 at 8:43
$begingroup$
@JasonM I edited the post now.
$endgroup$
– Cleggstein
Jun 17 '16 at 9:26
$begingroup$
Then I have an answer
$endgroup$
– JasonM
Jun 17 '16 at 9:33
$begingroup$
Actually, I'm feeling a bit uneasy about the step where you mapped by $frac{1}{f'}$ from the left. Are we sure $(frac{1}{f'}) circ frac{1}{((f')^{-1})circ f}=frac{1}{f}$?
$endgroup$
– JasonM
Jun 17 '16 at 10:01
$begingroup$
Ah yes, I've mixed multiplication and mapping there. Actual mapping or multiplication don't seem to yield anything useful either. I'll try and work on this a bit more.
$endgroup$
– Cleggstein
Jun 17 '16 at 10:36
add a comment |
2
$begingroup$
How did you derive the fact $f(x)=(f' circ f')(x)?$
$endgroup$
– JasonM
Jun 17 '16 at 8:43
$begingroup$
@JasonM I edited the post now.
$endgroup$
– Cleggstein
Jun 17 '16 at 9:26
$begingroup$
Then I have an answer
$endgroup$
– JasonM
Jun 17 '16 at 9:33
$begingroup$
Actually, I'm feeling a bit uneasy about the step where you mapped by $frac{1}{f'}$ from the left. Are we sure $(frac{1}{f'}) circ frac{1}{((f')^{-1})circ f}=frac{1}{f}$?
$endgroup$
– JasonM
Jun 17 '16 at 10:01
$begingroup$
Ah yes, I've mixed multiplication and mapping there. Actual mapping or multiplication don't seem to yield anything useful either. I'll try and work on this a bit more.
$endgroup$
– Cleggstein
Jun 17 '16 at 10:36
2
2
$begingroup$
How did you derive the fact $f(x)=(f' circ f')(x)?$
$endgroup$
– JasonM
Jun 17 '16 at 8:43
$begingroup$
How did you derive the fact $f(x)=(f' circ f')(x)?$
$endgroup$
– JasonM
Jun 17 '16 at 8:43
$begingroup$
@JasonM I edited the post now.
$endgroup$
– Cleggstein
Jun 17 '16 at 9:26
$begingroup$
@JasonM I edited the post now.
$endgroup$
– Cleggstein
Jun 17 '16 at 9:26
$begingroup$
Then I have an answer
$endgroup$
– JasonM
Jun 17 '16 at 9:33
$begingroup$
Then I have an answer
$endgroup$
– JasonM
Jun 17 '16 at 9:33
$begingroup$
Actually, I'm feeling a bit uneasy about the step where you mapped by $frac{1}{f'}$ from the left. Are we sure $(frac{1}{f'}) circ frac{1}{((f')^{-1})circ f}=frac{1}{f}$?
$endgroup$
– JasonM
Jun 17 '16 at 10:01
$begingroup$
Actually, I'm feeling a bit uneasy about the step where you mapped by $frac{1}{f'}$ from the left. Are we sure $(frac{1}{f'}) circ frac{1}{((f')^{-1})circ f}=frac{1}{f}$?
$endgroup$
– JasonM
Jun 17 '16 at 10:01
$begingroup$
Ah yes, I've mixed multiplication and mapping there. Actual mapping or multiplication don't seem to yield anything useful either. I'll try and work on this a bit more.
$endgroup$
– Cleggstein
Jun 17 '16 at 10:36
$begingroup$
Ah yes, I've mixed multiplication and mapping there. Actual mapping or multiplication don't seem to yield anything useful either. I'll try and work on this a bit more.
$endgroup$
– Cleggstein
Jun 17 '16 at 10:36
add a comment |
1 Answer
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Let $f(x)$ be such invertible function s.t. $(f^{-1})'={f'(x)}^{-1}$ which gives a non-linear differential equation $f'^2 +f^2=0$. Clearly, $f=0$ is the only solution to it, which contradicts that $f(x)$ is invertible.
answered Jun 17 '16 at 9:41
MathloverMathlover
3,6181022
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$begingroup$
Let $f(x)$ be such invertible function s.t. $(f^{-1})'={f'(x)}^{-1}$ which gives a non-linear differential equation $f'^2 +f^2=0$. Clearly, $f=0$ is the only solution to it, which contradicts that $f(x)$ is invertible.
answered Jun 17 '16 at 9:41
MathloverMathlover
3,6181022
$endgroup$
add a comment |
$begingroup$
Let $f(x)$ be such invertible function s.t. $(f^{-1})'={f'(x)}^{-1}$ which gives a non-linear differential equation $f'^2 +f^2=0$. Clearly, $f=0$ is the only solution to it, which contradicts that $f(x)$ is invertible.
answered Jun 17 '16 at 9:41
MathloverMathlover
3,6181022
$endgroup$
add a comment |
$begingroup$
Let $f(x)$ be such invertible function s.t. $(f^{-1})'={f'(x)}^{-1}$ which gives a non-linear differential equation $f'^2 +f^2=0$. Clearly, $f=0$ is the only solution to it, which contradicts that $f(x)$ is invertible.
answered Jun 17 '16 at 9:41
MathloverMathlover
3,6181022
$endgroup$
Let $f(x)$ be such invertible function s.t. $(f^{-1})'={f'(x)}^{-1}$ which gives a non-linear differential equation $f'^2 +f^2=0$. Clearly, $f=0$ is the only solution to it, which contradicts that $f(x)$ is invertible.
answered Jun 17 '16 at 9:41
MathloverMathlover
3,6181022
answered Jun 17 '16 at 9:41
MathloverMathlover
3,6181022
answered Jun 17 '16 at 9:41
MathloverMathlover
3,6181022
answered Jun 17 '16 at 9:41
MathloverMathlover
3,6181022
3,6181022
add a comment |
add a comment |
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2
$begingroup$
How did you derive the fact $f(x)=(f' circ f')(x)?$
$endgroup$
– JasonM
Jun 17 '16 at 8:43
$begingroup$
@JasonM I edited the post now.
$endgroup$
– Cleggstein
Jun 17 '16 at 9:26
$begingroup$
Then I have an answer
$endgroup$
– JasonM
Jun 17 '16 at 9:33
$begingroup$
Actually, I'm feeling a bit uneasy about the step where you mapped by $frac{1}{f'}$ from the left. Are we sure $(frac{1}{f'}) circ frac{1}{((f')^{-1})circ f}=frac{1}{f}$?
$endgroup$
– JasonM
Jun 17 '16 at 10:01
$begingroup$
Ah yes, I've mixed multiplication and mapping there. Actual mapping or multiplication don't seem to yield anything useful either. I'll try and work on this a bit more.
$endgroup$
– Cleggstein
Jun 17 '16 at 10:36