Mixing
Is there a function $f:[0;1]rightarrow Bbb R ^n$ of class $C^ infty$ with $f^{(i)}(0)=10^i.f^{(i)}(1) ,...
$begingroup$
I have the function $g:Bbb R_{>0} rightarrow [0.1;1)$ given by
$$g(x)=10^{-1-lfloor log_{10} (x) rfloor}.x$$
Which kind of "erase" the coma in the decimal expresion of $x$. For example $$g(123,539)=0,123539 qquad; qquad g(0,00012)=0,12$$
And I'm trying to find a function $f:[0.1;1]rightarrow Bbb R ^n$ of class $C^ infty$ such that $f circ g$ is a function of class $C^ infty$ (because $g$ it's not a pretty function and I want to make it pretty in order to be able to work with it). I've managed to prove $f circ g$ is $C^ infty$ if and only if $f$ satisfy the following property
$$f^{(i)}(0.1)=10^i.f^{(i)}(1) , forall , i in Bbb N_0$$
I've also manage to prove this is equivalent to finding a function $h:[0;1]rightarrow Bbb R ^n$ of class $C^ infty$ with the same property
$$h^{(i)}(0)=10^i.h^{(i)}(1) , forall , i in Bbb N_0$$
Because we can take $f(x)=h((10x-1)/9)$ and it will satisfy the property we were looking for. In order to avoid trivial solutions such as $h$ beeing constant, I will also ask for the following property
$$xneq y qquad h(x)=h(y) Rightarrow x=0 ; , ; y=1$$
So $h$ is injective in $(0;1)$. I haven't been able to find such a function and I don't know how to search for one so I'm kind of lost.
analysis derivatives
asked Feb 3 at 4:32
Marcos Martínez WagnerMarcos Martínez Wagner
1548
$endgroup$
add a comment |
$begingroup$
I have the function $g:Bbb R_{>0} rightarrow [0.1;1)$ given by
$$g(x)=10^{-1-lfloor log_{10} (x) rfloor}.x$$
Which kind of "erase" the coma in the decimal expresion of $x$. For example $$g(123,539)=0,123539 qquad; qquad g(0,00012)=0,12$$
And I'm trying to find a function $f:[0.1;1]rightarrow Bbb R ^n$ of class $C^ infty$ such that $f circ g$ is a function of class $C^ infty$ (because $g$ it's not a pretty function and I want to make it pretty in order to be able to work with it). I've managed to prove $f circ g$ is $C^ infty$ if and only if $f$ satisfy the following property
$$f^{(i)}(0.1)=10^i.f^{(i)}(1) , forall , i in Bbb N_0$$
I've also manage to prove this is equivalent to finding a function $h:[0;1]rightarrow Bbb R ^n$ of class $C^ infty$ with the same property
$$h^{(i)}(0)=10^i.h^{(i)}(1) , forall , i in Bbb N_0$$
Because we can take $f(x)=h((10x-1)/9)$ and it will satisfy the property we were looking for. In order to avoid trivial solutions such as $h$ beeing constant, I will also ask for the following property
$$xneq y qquad h(x)=h(y) Rightarrow x=0 ; , ; y=1$$
So $h$ is injective in $(0;1)$. I haven't been able to find such a function and I don't know how to search for one so I'm kind of lost.
analysis derivatives
asked Feb 3 at 4:32
Marcos Martínez WagnerMarcos Martínez Wagner
1548
$endgroup$
add a comment |
$begingroup$
I have the function $g:Bbb R_{>0} rightarrow [0.1;1)$ given by
$$g(x)=10^{-1-lfloor log_{10} (x) rfloor}.x$$
Which kind of "erase" the coma in the decimal expresion of $x$. For example $$g(123,539)=0,123539 qquad; qquad g(0,00012)=0,12$$
And I'm trying to find a function $f:[0.1;1]rightarrow Bbb R ^n$ of class $C^ infty$ such that $f circ g$ is a function of class $C^ infty$ (because $g$ it's not a pretty function and I want to make it pretty in order to be able to work with it). I've managed to prove $f circ g$ is $C^ infty$ if and only if $f$ satisfy the following property
$$f^{(i)}(0.1)=10^i.f^{(i)}(1) , forall , i in Bbb N_0$$
I've also manage to prove this is equivalent to finding a function $h:[0;1]rightarrow Bbb R ^n$ of class $C^ infty$ with the same property
$$h^{(i)}(0)=10^i.h^{(i)}(1) , forall , i in Bbb N_0$$
Because we can take $f(x)=h((10x-1)/9)$ and it will satisfy the property we were looking for. In order to avoid trivial solutions such as $h$ beeing constant, I will also ask for the following property
$$xneq y qquad h(x)=h(y) Rightarrow x=0 ; , ; y=1$$
So $h$ is injective in $(0;1)$. I haven't been able to find such a function and I don't know how to search for one so I'm kind of lost.
analysis derivatives
asked Feb 3 at 4:32
Marcos Martínez WagnerMarcos Martínez Wagner
1548
$endgroup$
I have the function $g:Bbb R_{>0} rightarrow [0.1;1)$ given by
$$g(x)=10^{-1-lfloor log_{10} (x) rfloor}.x$$
Which kind of "erase" the coma in the decimal expresion of $x$. For example $$g(123,539)=0,123539 qquad; qquad g(0,00012)=0,12$$
And I'm trying to find a function $f:[0.1;1]rightarrow Bbb R ^n$ of class $C^ infty$ such that $f circ g$ is a function of class $C^ infty$ (because $g$ it's not a pretty function and I want to make it pretty in order to be able to work with it). I've managed to prove $f circ g$ is $C^ infty$ if and only if $f$ satisfy the following property
$$f^{(i)}(0.1)=10^i.f^{(i)}(1) , forall , i in Bbb N_0$$
I've also manage to prove this is equivalent to finding a function $h:[0;1]rightarrow Bbb R ^n$ of class $C^ infty$ with the same property
$$h^{(i)}(0)=10^i.h^{(i)}(1) , forall , i in Bbb N_0$$
Because we can take $f(x)=h((10x-1)/9)$ and it will satisfy the property we were looking for. In order to avoid trivial solutions such as $h$ beeing constant, I will also ask for the following property
$$xneq y qquad h(x)=h(y) Rightarrow x=0 ; , ; y=1$$
So $h$ is injective in $(0;1)$. I haven't been able to find such a function and I don't know how to search for one so I'm kind of lost.
analysis derivatives
analysis derivatives
asked Feb 3 at 4:32
Marcos Martínez WagnerMarcos Martínez Wagner
1548
asked Feb 3 at 4:32
Marcos Martínez WagnerMarcos Martínez Wagner
1548
edited Feb 3 at 15:30
Marcos Martínez Wagner
asked Feb 3 at 4:32
Marcos Martínez WagnerMarcos Martínez Wagner
1548
asked Feb 3 at 4:32
Marcos Martínez WagnerMarcos Martínez Wagner
1548
asked Feb 3 at 4:32
Marcos Martínez WagnerMarcos Martínez Wagner
1548
1548
add a comment |
add a comment |
1 Answer
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$begingroup$
To avoid confusion, the $g$ function from the problem will be called $G_{10}$
$$G_{10}(x)=10^{-1-lfloor log_{10}(x) rfloor}.x quad forall x in Bbb R_{>0}$$
First, let's get some motivation for this problem. Let $f:Bbb R rightarrow Bbb R_{>0}$ be a function, although $G_{10} circ f$ give us a lot of information about the exponential growth of the function (we can interpret it as a way of comparing $f$ with the function $g(x)=10^x$, i won't go into details to make the answer shorter), it has the inconvenience of not (usually) being continuous even when $f$ is continuous. We can solve this by "sticking" the edges of $[0.1;1)$ together vía, for example, the exponential function $g:[0.1;1] rightarrow S^1$ given by
$$g(x)=e^{2pi i.(10x-1)/9} tag 1\$$
Note that $g$ is inyective over $(0.1;1)$ so the composition $h=g circ G_{10}$ has the same "information" as $G_{10}$ (meaning: knowing the value of $G_{10}(x)$ it's equivalent to knowing the value of $h(x)$). Also $g(0.1)=g(1)$ and $g$ is continuous so we conclude $h$ is continuous and, therefore, $h circ f$ si continuous for every continuous $f$. So, $h$ has the same information as the function $G_{10}$ but has the property of being continuous and, therefore, is an improvement ($h$ is better and more manageable than $G_{10}$). We can get this idea further by replacing "continuous" with "$C^1$". If $f:Bbb R rightarrow Bbb R_{>0}$ is $C^1$ we would like $G_{10} circ f$ to be $C^1$ so we ask if there is a function $g:[0.1;1] rightarrow S^1$ inyective over $(0.1;1)$ (so we don't lose information) with $g circ G_{10} circ f in C^1$ for every $f in C^1$. It's easy to prove the last condition is equivalent to asking $g circ G_{10}in C^1$. You can check the function $(1)$ does not have this property. If we want even more we can replace "$C^1$" with "$C^k$" with $k in Bbb N$ and if we are really ambitious we can replace "$C^1$" with "$C^infty$"
So, we are looking for a function $f:[0.1;1] rightarrow S^1$ inyective over $(0.1;1)$ with $f circ G_{10} in C^infty$. To make it easier to find, we will suppose $f in C^infty$. Note $f circ G_{10}$ must be continuous and that happens iff $f(0.1)=f(1)$ and, to make the most natural choice, we will ask $f(0.1)=f(1)=1$. We will show there is a function that satisfys all of this properties, an interesting question would be if the solution is unique. Well, it couldn't be unique since if $f$ si a solution, then $g(x)=f(x)^{-1}$ is also a solution. Then we could ask if there are more than two solutions (I think there are only two solutions but I haven't been able to prove it).
It's "easy" to prove by induction that, if $f:[0.1;1] rightarrow S^1$ is $C^k$ with $k in Bbb N$, then $f circ G_{10} in C^k$ if and only if
$$f^{(i)}(0.1)=10^i.f^{(i)}(1) quad forall i in {0,1,ldots,n}$$
So we want this to happen for all $i in Bbb N_0$. Now we can forget about $G_{10}$, we want to find a function $f:[0.1;1]rightarrow S^1$ inyective over $(0.1;1)$ with $f in C^infty, f(0.1)=f(1)=1$ and satisfying
$$f^{(i)}(0.1)=10^i.f^{(i)}(1) quad forall i in Bbb N_0$$
Inspired by the function $(1)$ we will try to find a solution with the structure $f(x)=e^{2 pi i.h(x)}$ with $h:[0.1;1]rightarrow Bbb R$. Observe $f in C^infty Leftrightarrow h in C^infty$ and
$f(0.1)=f(1)=1 Leftrightarrow h(0.1),h(1) in Bbb Z$ in which case it's "easy" to check
$$f^{(i)}(0.1)=10^i.f^{(i)}(1) quad forall i in Bbb N_0 Longleftrightarrow h^{(i)}(0.1)=10^i.h^{(i)}(1) quad forall i in Bbb N$$
Furthermore, you can prove $f$ is inyective over $(0.1;1)$ if and only if $h$ is monotonic and $|f(1)-f(0.1)|=1$. In fact, the sign of $f(1)-f(0.1)$ will determine if $f$ is increasing or decreasing for obvious reasons.
We change the problem a little bit and it seems like we made no progress but in fact, we made a lot of progress because there is a simple solution to this new problem. The function $h:[0.1;1]rightarrow Bbb R$ given by $h(x)=log_{10}(x)$ satisfys $hin C^infty, h(0.1),h(1) in Bbb Z$, it's an increasing function, $|h(1)-h(0.1)|=|0-(-1)|=|1|=1$ and you can check
$$h^{(i)}(x)=frac{1}{ln(10)} . frac{(-1)^{i-1}.(i-1)!}{x^i} quad forall igeq 1$$
And it's easy to check from there $h^{(i)}(0.1)=10^i.h^{(i)}(1) forall i in Bbb N$. We made it! we found a solution for our problem. More explicitly, our solution for our original problem is
$$f(x)=e^{2 pi i.log_{10}(x)}$$
And we can check $f circ G_{10}$ satisfies al the properties we asked for. In fact, it's easy to notice we can extend $f$'s domain to $Bbb R_{>0}$ via the same formula and what might seem crazy is $f circ G_{10}=f$
Interesting Fact: If $g:Bbb R rightarrow Bbb R_{>0}$ is $C^1$, if we know the values of $f circ g:Bbb R rightarrow S^1$ we can recover de values of $g$ just by knowing the amount of digits of $g(x_0)$ for some $x_0 in Bbb R$, in other words, we can recover the funcion $g$ just by looking at the "graph" of $f circ g$ (module the amount of digits of a "start" value). Note $f$ is not inyective over $Bbb R_{>0}$ so this result is not trivial. Moreover for every $h:Bbb R rightarrow S^1$ with $h in C^1$ there's a $g:Bbb R rightarrow Bbb R_{>0}$ with $g in C^1$ and $f circ g = h$. If we get anoter $g' in C^1$ with $f circ g' = h$ then there's a $k in Bbb Z$ with $g'(x)=10^k.g(x) forall x in Bbb R$.
answered Mar 30 at 6:27
Marcos Martínez WagnerMarcos Martínez Wagner
1548
$endgroup$
add a comment |
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$begingroup$
To avoid confusion, the $g$ function from the problem will be called $G_{10}$
$$G_{10}(x)=10^{-1-lfloor log_{10}(x) rfloor}.x quad forall x in Bbb R_{>0}$$
First, let's get some motivation for this problem. Let $f:Bbb R rightarrow Bbb R_{>0}$ be a function, although $G_{10} circ f$ give us a lot of information about the exponential growth of the function (we can interpret it as a way of comparing $f$ with the function $g(x)=10^x$, i won't go into details to make the answer shorter), it has the inconvenience of not (usually) being continuous even when $f$ is continuous. We can solve this by "sticking" the edges of $[0.1;1)$ together vía, for example, the exponential function $g:[0.1;1] rightarrow S^1$ given by
$$g(x)=e^{2pi i.(10x-1)/9} tag 1\$$
Note that $g$ is inyective over $(0.1;1)$ so the composition $h=g circ G_{10}$ has the same "information" as $G_{10}$ (meaning: knowing the value of $G_{10}(x)$ it's equivalent to knowing the value of $h(x)$). Also $g(0.1)=g(1)$ and $g$ is continuous so we conclude $h$ is continuous and, therefore, $h circ f$ si continuous for every continuous $f$. So, $h$ has the same information as the function $G_{10}$ but has the property of being continuous and, therefore, is an improvement ($h$ is better and more manageable than $G_{10}$). We can get this idea further by replacing "continuous" with "$C^1$". If $f:Bbb R rightarrow Bbb R_{>0}$ is $C^1$ we would like $G_{10} circ f$ to be $C^1$ so we ask if there is a function $g:[0.1;1] rightarrow S^1$ inyective over $(0.1;1)$ (so we don't lose information) with $g circ G_{10} circ f in C^1$ for every $f in C^1$. It's easy to prove the last condition is equivalent to asking $g circ G_{10}in C^1$. You can check the function $(1)$ does not have this property. If we want even more we can replace "$C^1$" with "$C^k$" with $k in Bbb N$ and if we are really ambitious we can replace "$C^1$" with "$C^infty$"
So, we are looking for a function $f:[0.1;1] rightarrow S^1$ inyective over $(0.1;1)$ with $f circ G_{10} in C^infty$. To make it easier to find, we will suppose $f in C^infty$. Note $f circ G_{10}$ must be continuous and that happens iff $f(0.1)=f(1)$ and, to make the most natural choice, we will ask $f(0.1)=f(1)=1$. We will show there is a function that satisfys all of this properties, an interesting question would be if the solution is unique. Well, it couldn't be unique since if $f$ si a solution, then $g(x)=f(x)^{-1}$ is also a solution. Then we could ask if there are more than two solutions (I think there are only two solutions but I haven't been able to prove it).
It's "easy" to prove by induction that, if $f:[0.1;1] rightarrow S^1$ is $C^k$ with $k in Bbb N$, then $f circ G_{10} in C^k$ if and only if
$$f^{(i)}(0.1)=10^i.f^{(i)}(1) quad forall i in {0,1,ldots,n}$$
So we want this to happen for all $i in Bbb N_0$. Now we can forget about $G_{10}$, we want to find a function $f:[0.1;1]rightarrow S^1$ inyective over $(0.1;1)$ with $f in C^infty, f(0.1)=f(1)=1$ and satisfying
$$f^{(i)}(0.1)=10^i.f^{(i)}(1) quad forall i in Bbb N_0$$
Inspired by the function $(1)$ we will try to find a solution with the structure $f(x)=e^{2 pi i.h(x)}$ with $h:[0.1;1]rightarrow Bbb R$. Observe $f in C^infty Leftrightarrow h in C^infty$ and
$f(0.1)=f(1)=1 Leftrightarrow h(0.1),h(1) in Bbb Z$ in which case it's "easy" to check
$$f^{(i)}(0.1)=10^i.f^{(i)}(1) quad forall i in Bbb N_0 Longleftrightarrow h^{(i)}(0.1)=10^i.h^{(i)}(1) quad forall i in Bbb N$$
Furthermore, you can prove $f$ is inyective over $(0.1;1)$ if and only if $h$ is monotonic and $|f(1)-f(0.1)|=1$. In fact, the sign of $f(1)-f(0.1)$ will determine if $f$ is increasing or decreasing for obvious reasons.
We change the problem a little bit and it seems like we made no progress but in fact, we made a lot of progress because there is a simple solution to this new problem. The function $h:[0.1;1]rightarrow Bbb R$ given by $h(x)=log_{10}(x)$ satisfys $hin C^infty, h(0.1),h(1) in Bbb Z$, it's an increasing function, $|h(1)-h(0.1)|=|0-(-1)|=|1|=1$ and you can check
$$h^{(i)}(x)=frac{1}{ln(10)} . frac{(-1)^{i-1}.(i-1)!}{x^i} quad forall igeq 1$$
And it's easy to check from there $h^{(i)}(0.1)=10^i.h^{(i)}(1) forall i in Bbb N$. We made it! we found a solution for our problem. More explicitly, our solution for our original problem is
$$f(x)=e^{2 pi i.log_{10}(x)}$$
And we can check $f circ G_{10}$ satisfies al the properties we asked for. In fact, it's easy to notice we can extend $f$'s domain to $Bbb R_{>0}$ via the same formula and what might seem crazy is $f circ G_{10}=f$
Interesting Fact: If $g:Bbb R rightarrow Bbb R_{>0}$ is $C^1$, if we know the values of $f circ g:Bbb R rightarrow S^1$ we can recover de values of $g$ just by knowing the amount of digits of $g(x_0)$ for some $x_0 in Bbb R$, in other words, we can recover the funcion $g$ just by looking at the "graph" of $f circ g$ (module the amount of digits of a "start" value). Note $f$ is not inyective over $Bbb R_{>0}$ so this result is not trivial. Moreover for every $h:Bbb R rightarrow S^1$ with $h in C^1$ there's a $g:Bbb R rightarrow Bbb R_{>0}$ with $g in C^1$ and $f circ g = h$. If we get anoter $g' in C^1$ with $f circ g' = h$ then there's a $k in Bbb Z$ with $g'(x)=10^k.g(x) forall x in Bbb R$.
answered Mar 30 at 6:27
Marcos Martínez WagnerMarcos Martínez Wagner
1548
$endgroup$
add a comment |
$begingroup$
To avoid confusion, the $g$ function from the problem will be called $G_{10}$
$$G_{10}(x)=10^{-1-lfloor log_{10}(x) rfloor}.x quad forall x in Bbb R_{>0}$$
First, let's get some motivation for this problem. Let $f:Bbb R rightarrow Bbb R_{>0}$ be a function, although $G_{10} circ f$ give us a lot of information about the exponential growth of the function (we can interpret it as a way of comparing $f$ with the function $g(x)=10^x$, i won't go into details to make the answer shorter), it has the inconvenience of not (usually) being continuous even when $f$ is continuous. We can solve this by "sticking" the edges of $[0.1;1)$ together vía, for example, the exponential function $g:[0.1;1] rightarrow S^1$ given by
$$g(x)=e^{2pi i.(10x-1)/9} tag 1\$$
Note that $g$ is inyective over $(0.1;1)$ so the composition $h=g circ G_{10}$ has the same "information" as $G_{10}$ (meaning: knowing the value of $G_{10}(x)$ it's equivalent to knowing the value of $h(x)$). Also $g(0.1)=g(1)$ and $g$ is continuous so we conclude $h$ is continuous and, therefore, $h circ f$ si continuous for every continuous $f$. So, $h$ has the same information as the function $G_{10}$ but has the property of being continuous and, therefore, is an improvement ($h$ is better and more manageable than $G_{10}$). We can get this idea further by replacing "continuous" with "$C^1$". If $f:Bbb R rightarrow Bbb R_{>0}$ is $C^1$ we would like $G_{10} circ f$ to be $C^1$ so we ask if there is a function $g:[0.1;1] rightarrow S^1$ inyective over $(0.1;1)$ (so we don't lose information) with $g circ G_{10} circ f in C^1$ for every $f in C^1$. It's easy to prove the last condition is equivalent to asking $g circ G_{10}in C^1$. You can check the function $(1)$ does not have this property. If we want even more we can replace "$C^1$" with "$C^k$" with $k in Bbb N$ and if we are really ambitious we can replace "$C^1$" with "$C^infty$"
So, we are looking for a function $f:[0.1;1] rightarrow S^1$ inyective over $(0.1;1)$ with $f circ G_{10} in C^infty$. To make it easier to find, we will suppose $f in C^infty$. Note $f circ G_{10}$ must be continuous and that happens iff $f(0.1)=f(1)$ and, to make the most natural choice, we will ask $f(0.1)=f(1)=1$. We will show there is a function that satisfys all of this properties, an interesting question would be if the solution is unique. Well, it couldn't be unique since if $f$ si a solution, then $g(x)=f(x)^{-1}$ is also a solution. Then we could ask if there are more than two solutions (I think there are only two solutions but I haven't been able to prove it).
It's "easy" to prove by induction that, if $f:[0.1;1] rightarrow S^1$ is $C^k$ with $k in Bbb N$, then $f circ G_{10} in C^k$ if and only if
$$f^{(i)}(0.1)=10^i.f^{(i)}(1) quad forall i in {0,1,ldots,n}$$
So we want this to happen for all $i in Bbb N_0$. Now we can forget about $G_{10}$, we want to find a function $f:[0.1;1]rightarrow S^1$ inyective over $(0.1;1)$ with $f in C^infty, f(0.1)=f(1)=1$ and satisfying
$$f^{(i)}(0.1)=10^i.f^{(i)}(1) quad forall i in Bbb N_0$$
Inspired by the function $(1)$ we will try to find a solution with the structure $f(x)=e^{2 pi i.h(x)}$ with $h:[0.1;1]rightarrow Bbb R$. Observe $f in C^infty Leftrightarrow h in C^infty$ and
$f(0.1)=f(1)=1 Leftrightarrow h(0.1),h(1) in Bbb Z$ in which case it's "easy" to check
$$f^{(i)}(0.1)=10^i.f^{(i)}(1) quad forall i in Bbb N_0 Longleftrightarrow h^{(i)}(0.1)=10^i.h^{(i)}(1) quad forall i in Bbb N$$
Furthermore, you can prove $f$ is inyective over $(0.1;1)$ if and only if $h$ is monotonic and $|f(1)-f(0.1)|=1$. In fact, the sign of $f(1)-f(0.1)$ will determine if $f$ is increasing or decreasing for obvious reasons.
We change the problem a little bit and it seems like we made no progress but in fact, we made a lot of progress because there is a simple solution to this new problem. The function $h:[0.1;1]rightarrow Bbb R$ given by $h(x)=log_{10}(x)$ satisfys $hin C^infty, h(0.1),h(1) in Bbb Z$, it's an increasing function, $|h(1)-h(0.1)|=|0-(-1)|=|1|=1$ and you can check
$$h^{(i)}(x)=frac{1}{ln(10)} . frac{(-1)^{i-1}.(i-1)!}{x^i} quad forall igeq 1$$
And it's easy to check from there $h^{(i)}(0.1)=10^i.h^{(i)}(1) forall i in Bbb N$. We made it! we found a solution for our problem. More explicitly, our solution for our original problem is
$$f(x)=e^{2 pi i.log_{10}(x)}$$
And we can check $f circ G_{10}$ satisfies al the properties we asked for. In fact, it's easy to notice we can extend $f$'s domain to $Bbb R_{>0}$ via the same formula and what might seem crazy is $f circ G_{10}=f$
Interesting Fact: If $g:Bbb R rightarrow Bbb R_{>0}$ is $C^1$, if we know the values of $f circ g:Bbb R rightarrow S^1$ we can recover de values of $g$ just by knowing the amount of digits of $g(x_0)$ for some $x_0 in Bbb R$, in other words, we can recover the funcion $g$ just by looking at the "graph" of $f circ g$ (module the amount of digits of a "start" value). Note $f$ is not inyective over $Bbb R_{>0}$ so this result is not trivial. Moreover for every $h:Bbb R rightarrow S^1$ with $h in C^1$ there's a $g:Bbb R rightarrow Bbb R_{>0}$ with $g in C^1$ and $f circ g = h$. If we get anoter $g' in C^1$ with $f circ g' = h$ then there's a $k in Bbb Z$ with $g'(x)=10^k.g(x) forall x in Bbb R$.
answered Mar 30 at 6:27
Marcos Martínez WagnerMarcos Martínez Wagner
1548
$endgroup$
add a comment |
$begingroup$
To avoid confusion, the $g$ function from the problem will be called $G_{10}$
$$G_{10}(x)=10^{-1-lfloor log_{10}(x) rfloor}.x quad forall x in Bbb R_{>0}$$
First, let's get some motivation for this problem. Let $f:Bbb R rightarrow Bbb R_{>0}$ be a function, although $G_{10} circ f$ give us a lot of information about the exponential growth of the function (we can interpret it as a way of comparing $f$ with the function $g(x)=10^x$, i won't go into details to make the answer shorter), it has the inconvenience of not (usually) being continuous even when $f$ is continuous. We can solve this by "sticking" the edges of $[0.1;1)$ together vía, for example, the exponential function $g:[0.1;1] rightarrow S^1$ given by
$$g(x)=e^{2pi i.(10x-1)/9} tag 1\$$
Note that $g$ is inyective over $(0.1;1)$ so the composition $h=g circ G_{10}$ has the same "information" as $G_{10}$ (meaning: knowing the value of $G_{10}(x)$ it's equivalent to knowing the value of $h(x)$). Also $g(0.1)=g(1)$ and $g$ is continuous so we conclude $h$ is continuous and, therefore, $h circ f$ si continuous for every continuous $f$. So, $h$ has the same information as the function $G_{10}$ but has the property of being continuous and, therefore, is an improvement ($h$ is better and more manageable than $G_{10}$). We can get this idea further by replacing "continuous" with "$C^1$". If $f:Bbb R rightarrow Bbb R_{>0}$ is $C^1$ we would like $G_{10} circ f$ to be $C^1$ so we ask if there is a function $g:[0.1;1] rightarrow S^1$ inyective over $(0.1;1)$ (so we don't lose information) with $g circ G_{10} circ f in C^1$ for every $f in C^1$. It's easy to prove the last condition is equivalent to asking $g circ G_{10}in C^1$. You can check the function $(1)$ does not have this property. If we want even more we can replace "$C^1$" with "$C^k$" with $k in Bbb N$ and if we are really ambitious we can replace "$C^1$" with "$C^infty$"
So, we are looking for a function $f:[0.1;1] rightarrow S^1$ inyective over $(0.1;1)$ with $f circ G_{10} in C^infty$. To make it easier to find, we will suppose $f in C^infty$. Note $f circ G_{10}$ must be continuous and that happens iff $f(0.1)=f(1)$ and, to make the most natural choice, we will ask $f(0.1)=f(1)=1$. We will show there is a function that satisfys all of this properties, an interesting question would be if the solution is unique. Well, it couldn't be unique since if $f$ si a solution, then $g(x)=f(x)^{-1}$ is also a solution. Then we could ask if there are more than two solutions (I think there are only two solutions but I haven't been able to prove it).
It's "easy" to prove by induction that, if $f:[0.1;1] rightarrow S^1$ is $C^k$ with $k in Bbb N$, then $f circ G_{10} in C^k$ if and only if
$$f^{(i)}(0.1)=10^i.f^{(i)}(1) quad forall i in {0,1,ldots,n}$$
So we want this to happen for all $i in Bbb N_0$. Now we can forget about $G_{10}$, we want to find a function $f:[0.1;1]rightarrow S^1$ inyective over $(0.1;1)$ with $f in C^infty, f(0.1)=f(1)=1$ and satisfying
$$f^{(i)}(0.1)=10^i.f^{(i)}(1) quad forall i in Bbb N_0$$
Inspired by the function $(1)$ we will try to find a solution with the structure $f(x)=e^{2 pi i.h(x)}$ with $h:[0.1;1]rightarrow Bbb R$. Observe $f in C^infty Leftrightarrow h in C^infty$ and
$f(0.1)=f(1)=1 Leftrightarrow h(0.1),h(1) in Bbb Z$ in which case it's "easy" to check
$$f^{(i)}(0.1)=10^i.f^{(i)}(1) quad forall i in Bbb N_0 Longleftrightarrow h^{(i)}(0.1)=10^i.h^{(i)}(1) quad forall i in Bbb N$$
Furthermore, you can prove $f$ is inyective over $(0.1;1)$ if and only if $h$ is monotonic and $|f(1)-f(0.1)|=1$. In fact, the sign of $f(1)-f(0.1)$ will determine if $f$ is increasing or decreasing for obvious reasons.
We change the problem a little bit and it seems like we made no progress but in fact, we made a lot of progress because there is a simple solution to this new problem. The function $h:[0.1;1]rightarrow Bbb R$ given by $h(x)=log_{10}(x)$ satisfys $hin C^infty, h(0.1),h(1) in Bbb Z$, it's an increasing function, $|h(1)-h(0.1)|=|0-(-1)|=|1|=1$ and you can check
$$h^{(i)}(x)=frac{1}{ln(10)} . frac{(-1)^{i-1}.(i-1)!}{x^i} quad forall igeq 1$$
And it's easy to check from there $h^{(i)}(0.1)=10^i.h^{(i)}(1) forall i in Bbb N$. We made it! we found a solution for our problem. More explicitly, our solution for our original problem is
$$f(x)=e^{2 pi i.log_{10}(x)}$$
And we can check $f circ G_{10}$ satisfies al the properties we asked for. In fact, it's easy to notice we can extend $f$'s domain to $Bbb R_{>0}$ via the same formula and what might seem crazy is $f circ G_{10}=f$
Interesting Fact: If $g:Bbb R rightarrow Bbb R_{>0}$ is $C^1$, if we know the values of $f circ g:Bbb R rightarrow S^1$ we can recover de values of $g$ just by knowing the amount of digits of $g(x_0)$ for some $x_0 in Bbb R$, in other words, we can recover the funcion $g$ just by looking at the "graph" of $f circ g$ (module the amount of digits of a "start" value). Note $f$ is not inyective over $Bbb R_{>0}$ so this result is not trivial. Moreover for every $h:Bbb R rightarrow S^1$ with $h in C^1$ there's a $g:Bbb R rightarrow Bbb R_{>0}$ with $g in C^1$ and $f circ g = h$. If we get anoter $g' in C^1$ with $f circ g' = h$ then there's a $k in Bbb Z$ with $g'(x)=10^k.g(x) forall x in Bbb R$.
answered Mar 30 at 6:27
Marcos Martínez WagnerMarcos Martínez Wagner
1548
$endgroup$
To avoid confusion, the $g$ function from the problem will be called $G_{10}$
$$G_{10}(x)=10^{-1-lfloor log_{10}(x) rfloor}.x quad forall x in Bbb R_{>0}$$
First, let's get some motivation for this problem. Let $f:Bbb R rightarrow Bbb R_{>0}$ be a function, although $G_{10} circ f$ give us a lot of information about the exponential growth of the function (we can interpret it as a way of comparing $f$ with the function $g(x)=10^x$, i won't go into details to make the answer shorter), it has the inconvenience of not (usually) being continuous even when $f$ is continuous. We can solve this by "sticking" the edges of $[0.1;1)$ together vía, for example, the exponential function $g:[0.1;1] rightarrow S^1$ given by
$$g(x)=e^{2pi i.(10x-1)/9} tag 1\$$
Note that $g$ is inyective over $(0.1;1)$ so the composition $h=g circ G_{10}$ has the same "information" as $G_{10}$ (meaning: knowing the value of $G_{10}(x)$ it's equivalent to knowing the value of $h(x)$). Also $g(0.1)=g(1)$ and $g$ is continuous so we conclude $h$ is continuous and, therefore, $h circ f$ si continuous for every continuous $f$. So, $h$ has the same information as the function $G_{10}$ but has the property of being continuous and, therefore, is an improvement ($h$ is better and more manageable than $G_{10}$). We can get this idea further by replacing "continuous" with "$C^1$". If $f:Bbb R rightarrow Bbb R_{>0}$ is $C^1$ we would like $G_{10} circ f$ to be $C^1$ so we ask if there is a function $g:[0.1;1] rightarrow S^1$ inyective over $(0.1;1)$ (so we don't lose information) with $g circ G_{10} circ f in C^1$ for every $f in C^1$. It's easy to prove the last condition is equivalent to asking $g circ G_{10}in C^1$. You can check the function $(1)$ does not have this property. If we want even more we can replace "$C^1$" with "$C^k$" with $k in Bbb N$ and if we are really ambitious we can replace "$C^1$" with "$C^infty$"
So, we are looking for a function $f:[0.1;1] rightarrow S^1$ inyective over $(0.1;1)$ with $f circ G_{10} in C^infty$. To make it easier to find, we will suppose $f in C^infty$. Note $f circ G_{10}$ must be continuous and that happens iff $f(0.1)=f(1)$ and, to make the most natural choice, we will ask $f(0.1)=f(1)=1$. We will show there is a function that satisfys all of this properties, an interesting question would be if the solution is unique. Well, it couldn't be unique since if $f$ si a solution, then $g(x)=f(x)^{-1}$ is also a solution. Then we could ask if there are more than two solutions (I think there are only two solutions but I haven't been able to prove it).
It's "easy" to prove by induction that, if $f:[0.1;1] rightarrow S^1$ is $C^k$ with $k in Bbb N$, then $f circ G_{10} in C^k$ if and only if
$$f^{(i)}(0.1)=10^i.f^{(i)}(1) quad forall i in {0,1,ldots,n}$$
So we want this to happen for all $i in Bbb N_0$. Now we can forget about $G_{10}$, we want to find a function $f:[0.1;1]rightarrow S^1$ inyective over $(0.1;1)$ with $f in C^infty, f(0.1)=f(1)=1$ and satisfying
$$f^{(i)}(0.1)=10^i.f^{(i)}(1) quad forall i in Bbb N_0$$
Inspired by the function $(1)$ we will try to find a solution with the structure $f(x)=e^{2 pi i.h(x)}$ with $h:[0.1;1]rightarrow Bbb R$. Observe $f in C^infty Leftrightarrow h in C^infty$ and
$f(0.1)=f(1)=1 Leftrightarrow h(0.1),h(1) in Bbb Z$ in which case it's "easy" to check
$$f^{(i)}(0.1)=10^i.f^{(i)}(1) quad forall i in Bbb N_0 Longleftrightarrow h^{(i)}(0.1)=10^i.h^{(i)}(1) quad forall i in Bbb N$$
Furthermore, you can prove $f$ is inyective over $(0.1;1)$ if and only if $h$ is monotonic and $|f(1)-f(0.1)|=1$. In fact, the sign of $f(1)-f(0.1)$ will determine if $f$ is increasing or decreasing for obvious reasons.
We change the problem a little bit and it seems like we made no progress but in fact, we made a lot of progress because there is a simple solution to this new problem. The function $h:[0.1;1]rightarrow Bbb R$ given by $h(x)=log_{10}(x)$ satisfys $hin C^infty, h(0.1),h(1) in Bbb Z$, it's an increasing function, $|h(1)-h(0.1)|=|0-(-1)|=|1|=1$ and you can check
$$h^{(i)}(x)=frac{1}{ln(10)} . frac{(-1)^{i-1}.(i-1)!}{x^i} quad forall igeq 1$$
And it's easy to check from there $h^{(i)}(0.1)=10^i.h^{(i)}(1) forall i in Bbb N$. We made it! we found a solution for our problem. More explicitly, our solution for our original problem is
$$f(x)=e^{2 pi i.log_{10}(x)}$$
And we can check $f circ G_{10}$ satisfies al the properties we asked for. In fact, it's easy to notice we can extend $f$'s domain to $Bbb R_{>0}$ via the same formula and what might seem crazy is $f circ G_{10}=f$
Interesting Fact: If $g:Bbb R rightarrow Bbb R_{>0}$ is $C^1$, if we know the values of $f circ g:Bbb R rightarrow S^1$ we can recover de values of $g$ just by knowing the amount of digits of $g(x_0)$ for some $x_0 in Bbb R$, in other words, we can recover the funcion $g$ just by looking at the "graph" of $f circ g$ (module the amount of digits of a "start" value). Note $f$ is not inyective over $Bbb R_{>0}$ so this result is not trivial. Moreover for every $h:Bbb R rightarrow S^1$ with $h in C^1$ there's a $g:Bbb R rightarrow Bbb R_{>0}$ with $g in C^1$ and $f circ g = h$. If we get anoter $g' in C^1$ with $f circ g' = h$ then there's a $k in Bbb Z$ with $g'(x)=10^k.g(x) forall x in Bbb R$.
answered Mar 30 at 6:27
Marcos Martínez WagnerMarcos Martínez Wagner
1548
edited Mar 31 at 2:53
answered Mar 30 at 6:27
Marcos Martínez WagnerMarcos Martínez Wagner
1548
answered Mar 30 at 6:27
Marcos Martínez WagnerMarcos Martínez Wagner
1548
answered Mar 30 at 6:27
Marcos Martínez WagnerMarcos Martínez Wagner
1548
1548
add a comment |
add a comment |
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