Mixing
Showing no algebraic solution exists for a given equation
$begingroup$
Let $f(x)=g(x)$ be an equation (1) where at least one of $f$ and $g$ are transcendental functions. Let $h(x)=f(x)-g(x)$. If it can be shown that $h^{-1}(0)$ is non-algebraic, that implies that there is no algebraic solution to (1). How exactly does one go about doing this?
The answer here shows that there is no general method, but presumably there's a way to do this for individual equations - say, $xe^x=k, kinmathbb{R}$ (2). Obviously, the solution for, say, $k=3e^3$ is the algebraic value $x=3$, but this isn't an algebraic solution - I get it from sight. So in this sense, I'm using the term 'algebriac solution' to describe arriving explicitly at the form of $x=text{constant}$ via some combination of arithmetic operations, as opposed to merely stumbling upon an algebraic number that satisfies an equation. Do we get a computer to run through every possible combination of operations on (2) to get that no solution exists, or is there a more sophisticated method available?
If such a method can't be generalised to all transcendental functions, can it be generalised to equations between specific families of functions? Like, $k^x=P_n(x)$ (where $k in mathbb{R}^{+}backslash{1}$ and $P_n$ is a polynomial of degree $n>0$), doesn't have an algebraic solution. For example, it seems as though $2^x=3x^2-1$ is only solvable numerically (again, in spite of the solution $x=1$). How do we know this?
There's a very real possibility that I've abused mathematical vocabulary throughout. Apologies for length and/or if this has already been answered. If the answer is that it involves maths beyond what I've likely already been exposed to, could you point me in the right direction at least?
transcendental-equations
edited Apr 13 '17 at 12:19
Community♦
1
asked Dec 18 '14 at 12:29
cluelessclueless
263
$endgroup$
add a comment |
$begingroup$
Let $f(x)=g(x)$ be an equation (1) where at least one of $f$ and $g$ are transcendental functions. Let $h(x)=f(x)-g(x)$. If it can be shown that $h^{-1}(0)$ is non-algebraic, that implies that there is no algebraic solution to (1). How exactly does one go about doing this?
The answer here shows that there is no general method, but presumably there's a way to do this for individual equations - say, $xe^x=k, kinmathbb{R}$ (2). Obviously, the solution for, say, $k=3e^3$ is the algebraic value $x=3$, but this isn't an algebraic solution - I get it from sight. So in this sense, I'm using the term 'algebriac solution' to describe arriving explicitly at the form of $x=text{constant}$ via some combination of arithmetic operations, as opposed to merely stumbling upon an algebraic number that satisfies an equation. Do we get a computer to run through every possible combination of operations on (2) to get that no solution exists, or is there a more sophisticated method available?
If such a method can't be generalised to all transcendental functions, can it be generalised to equations between specific families of functions? Like, $k^x=P_n(x)$ (where $k in mathbb{R}^{+}backslash{1}$ and $P_n$ is a polynomial of degree $n>0$), doesn't have an algebraic solution. For example, it seems as though $2^x=3x^2-1$ is only solvable numerically (again, in spite of the solution $x=1$). How do we know this?
There's a very real possibility that I've abused mathematical vocabulary throughout. Apologies for length and/or if this has already been answered. If the answer is that it involves maths beyond what I've likely already been exposed to, could you point me in the right direction at least?
transcendental-equations
edited Apr 13 '17 at 12:19
Community♦
1
asked Dec 18 '14 at 12:29
cluelessclueless
263
$endgroup$
1
$begingroup$
Galois, of course, proved that there are algebraic numbers (solutions of some quintics, in particular) that can't be found by "usual algebraic operations" including taking roots. I suspect this is enough to answer any possible refinement of your question with "no"...
$endgroup$
– HTFB
Dec 18 '14 at 12:48
$begingroup$
Yeah, reading it back I wasn't especially clear on what I was asking. So take $xe^x$, the inverse Lambert-W function. How do I know it's not the case that we just haven't tried hard enough to find a combination of operations to perform on it that allow us to get a nice, explicit $x=text{constant}$?
$endgroup$
– clueless
Dec 18 '14 at 13:18
add a comment |
$begingroup$
Let $f(x)=g(x)$ be an equation (1) where at least one of $f$ and $g$ are transcendental functions. Let $h(x)=f(x)-g(x)$. If it can be shown that $h^{-1}(0)$ is non-algebraic, that implies that there is no algebraic solution to (1). How exactly does one go about doing this?
The answer here shows that there is no general method, but presumably there's a way to do this for individual equations - say, $xe^x=k, kinmathbb{R}$ (2). Obviously, the solution for, say, $k=3e^3$ is the algebraic value $x=3$, but this isn't an algebraic solution - I get it from sight. So in this sense, I'm using the term 'algebriac solution' to describe arriving explicitly at the form of $x=text{constant}$ via some combination of arithmetic operations, as opposed to merely stumbling upon an algebraic number that satisfies an equation. Do we get a computer to run through every possible combination of operations on (2) to get that no solution exists, or is there a more sophisticated method available?
If such a method can't be generalised to all transcendental functions, can it be generalised to equations between specific families of functions? Like, $k^x=P_n(x)$ (where $k in mathbb{R}^{+}backslash{1}$ and $P_n$ is a polynomial of degree $n>0$), doesn't have an algebraic solution. For example, it seems as though $2^x=3x^2-1$ is only solvable numerically (again, in spite of the solution $x=1$). How do we know this?
There's a very real possibility that I've abused mathematical vocabulary throughout. Apologies for length and/or if this has already been answered. If the answer is that it involves maths beyond what I've likely already been exposed to, could you point me in the right direction at least?
transcendental-equations
edited Apr 13 '17 at 12:19
Community♦
1
asked Dec 18 '14 at 12:29
cluelessclueless
263
$endgroup$
Let $f(x)=g(x)$ be an equation (1) where at least one of $f$ and $g$ are transcendental functions. Let $h(x)=f(x)-g(x)$. If it can be shown that $h^{-1}(0)$ is non-algebraic, that implies that there is no algebraic solution to (1). How exactly does one go about doing this?
The answer here shows that there is no general method, but presumably there's a way to do this for individual equations - say, $xe^x=k, kinmathbb{R}$ (2). Obviously, the solution for, say, $k=3e^3$ is the algebraic value $x=3$, but this isn't an algebraic solution - I get it from sight. So in this sense, I'm using the term 'algebriac solution' to describe arriving explicitly at the form of $x=text{constant}$ via some combination of arithmetic operations, as opposed to merely stumbling upon an algebraic number that satisfies an equation. Do we get a computer to run through every possible combination of operations on (2) to get that no solution exists, or is there a more sophisticated method available?
If such a method can't be generalised to all transcendental functions, can it be generalised to equations between specific families of functions? Like, $k^x=P_n(x)$ (where $k in mathbb{R}^{+}backslash{1}$ and $P_n$ is a polynomial of degree $n>0$), doesn't have an algebraic solution. For example, it seems as though $2^x=3x^2-1$ is only solvable numerically (again, in spite of the solution $x=1$). How do we know this?
There's a very real possibility that I've abused mathematical vocabulary throughout. Apologies for length and/or if this has already been answered. If the answer is that it involves maths beyond what I've likely already been exposed to, could you point me in the right direction at least?
transcendental-equations
transcendental-equations
edited Apr 13 '17 at 12:19
Community♦
1
asked Dec 18 '14 at 12:29
cluelessclueless
263
edited Apr 13 '17 at 12:19
Community♦
1
asked Dec 18 '14 at 12:29
cluelessclueless
263
edited Apr 13 '17 at 12:19
Community♦
1
edited Apr 13 '17 at 12:19
Community♦
1
edited Apr 13 '17 at 12:19
Community♦
1
1
asked Dec 18 '14 at 12:29
cluelessclueless
263
asked Dec 18 '14 at 12:29
cluelessclueless
263
asked Dec 18 '14 at 12:29
cluelessclueless
263
263
1
$begingroup$
Galois, of course, proved that there are algebraic numbers (solutions of some quintics, in particular) that can't be found by "usual algebraic operations" including taking roots. I suspect this is enough to answer any possible refinement of your question with "no"...
$endgroup$
– HTFB
Dec 18 '14 at 12:48
$begingroup$
Yeah, reading it back I wasn't especially clear on what I was asking. So take $xe^x$, the inverse Lambert-W function. How do I know it's not the case that we just haven't tried hard enough to find a combination of operations to perform on it that allow us to get a nice, explicit $x=text{constant}$?
$endgroup$
– clueless
Dec 18 '14 at 13:18
add a comment |
1
$begingroup$
Galois, of course, proved that there are algebraic numbers (solutions of some quintics, in particular) that can't be found by "usual algebraic operations" including taking roots. I suspect this is enough to answer any possible refinement of your question with "no"...
$endgroup$
– HTFB
Dec 18 '14 at 12:48
$begingroup$
Yeah, reading it back I wasn't especially clear on what I was asking. So take $xe^x$, the inverse Lambert-W function. How do I know it's not the case that we just haven't tried hard enough to find a combination of operations to perform on it that allow us to get a nice, explicit $x=text{constant}$?
$endgroup$
– clueless
Dec 18 '14 at 13:18
1
1
$begingroup$
Galois, of course, proved that there are algebraic numbers (solutions of some quintics, in particular) that can't be found by "usual algebraic operations" including taking roots. I suspect this is enough to answer any possible refinement of your question with "no"...
$endgroup$
– HTFB
Dec 18 '14 at 12:48
$begingroup$
Galois, of course, proved that there are algebraic numbers (solutions of some quintics, in particular) that can't be found by "usual algebraic operations" including taking roots. I suspect this is enough to answer any possible refinement of your question with "no"...
$endgroup$
– HTFB
Dec 18 '14 at 12:48
$begingroup$
Yeah, reading it back I wasn't especially clear on what I was asking. So take $xe^x$, the inverse Lambert-W function. How do I know it's not the case that we just haven't tried hard enough to find a combination of operations to perform on it that allow us to get a nice, explicit $x=text{constant}$?
$endgroup$
– clueless
Dec 18 '14 at 13:18
$begingroup$
Yeah, reading it back I wasn't especially clear on what I was asking. So take $xe^x$, the inverse Lambert-W function. How do I know it's not the case that we just haven't tried hard enough to find a combination of operations to perform on it that allow us to get a nice, explicit $x=text{constant}$?
$endgroup$
– clueless
Dec 18 '14 at 13:18
add a comment |
1 Answer
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You are looking for general methods for solving a given equation $h(x)=0$ by applying $h^{-1}$, where $h$ is a function of only one unknown and $h(x)$ is a closed-form expression in dependence of $x$ (Wikipedia: Closed-form expression).
a) Let us allow the elementary functions as closed form:
The elementary functions are according to Liouville and Ritt those functions of one variable which are obtained in a finite number of steps by performing algebraic operations and taking exponentials and logarithms (Wikipedia: Elementary function).
The incomprehensibly unfortunately hardly noticed theorem of Joseph Fels Ritt in Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90 and of Robert H. Risch in Risch, R. H.: Algebraic Properties of the Elementary Functions of Analysis. Amer. J.
Math. 101 (1979) (4) 743-759 answer which kinds of Elementary functions can have an inverse which is an Elementary function. You can also take the method of
Rosenlicht, M.: On the explicit solvability of certain transcendental equations. Publications mathématiques de l'IHÉS 36 (1969) 15-22.
b) You could allow algebraic expressions of known Standard functions as closed form:
If $h$ can be decomposed into a composition of algebraic functions and other known Standard functions than $exp$ and $ln$, an analog theorem to the theorem of Ritt of [Ritt 1925] could be applied. I hope to prove such a generalization of Ritt's theorem for this class of functions.
answered Jul 19 '17 at 20:23
IV_IV_
1,138522
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add a comment |
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$begingroup$
You are looking for general methods for solving a given equation $h(x)=0$ by applying $h^{-1}$, where $h$ is a function of only one unknown and $h(x)$ is a closed-form expression in dependence of $x$ (Wikipedia: Closed-form expression).
a) Let us allow the elementary functions as closed form:
The elementary functions are according to Liouville and Ritt those functions of one variable which are obtained in a finite number of steps by performing algebraic operations and taking exponentials and logarithms (Wikipedia: Elementary function).
The incomprehensibly unfortunately hardly noticed theorem of Joseph Fels Ritt in Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90 and of Robert H. Risch in Risch, R. H.: Algebraic Properties of the Elementary Functions of Analysis. Amer. J.
Math. 101 (1979) (4) 743-759 answer which kinds of Elementary functions can have an inverse which is an Elementary function. You can also take the method of
Rosenlicht, M.: On the explicit solvability of certain transcendental equations. Publications mathématiques de l'IHÉS 36 (1969) 15-22.
b) You could allow algebraic expressions of known Standard functions as closed form:
If $h$ can be decomposed into a composition of algebraic functions and other known Standard functions than $exp$ and $ln$, an analog theorem to the theorem of Ritt of [Ritt 1925] could be applied. I hope to prove such a generalization of Ritt's theorem for this class of functions.
answered Jul 19 '17 at 20:23
IV_IV_
1,138522
$endgroup$
add a comment |
$begingroup$
You are looking for general methods for solving a given equation $h(x)=0$ by applying $h^{-1}$, where $h$ is a function of only one unknown and $h(x)$ is a closed-form expression in dependence of $x$ (Wikipedia: Closed-form expression).
a) Let us allow the elementary functions as closed form:
The elementary functions are according to Liouville and Ritt those functions of one variable which are obtained in a finite number of steps by performing algebraic operations and taking exponentials and logarithms (Wikipedia: Elementary function).
The incomprehensibly unfortunately hardly noticed theorem of Joseph Fels Ritt in Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90 and of Robert H. Risch in Risch, R. H.: Algebraic Properties of the Elementary Functions of Analysis. Amer. J.
Math. 101 (1979) (4) 743-759 answer which kinds of Elementary functions can have an inverse which is an Elementary function. You can also take the method of
Rosenlicht, M.: On the explicit solvability of certain transcendental equations. Publications mathématiques de l'IHÉS 36 (1969) 15-22.
b) You could allow algebraic expressions of known Standard functions as closed form:
If $h$ can be decomposed into a composition of algebraic functions and other known Standard functions than $exp$ and $ln$, an analog theorem to the theorem of Ritt of [Ritt 1925] could be applied. I hope to prove such a generalization of Ritt's theorem for this class of functions.
answered Jul 19 '17 at 20:23
IV_IV_
1,138522
$endgroup$
add a comment |
$begingroup$
You are looking for general methods for solving a given equation $h(x)=0$ by applying $h^{-1}$, where $h$ is a function of only one unknown and $h(x)$ is a closed-form expression in dependence of $x$ (Wikipedia: Closed-form expression).
a) Let us allow the elementary functions as closed form:
The elementary functions are according to Liouville and Ritt those functions of one variable which are obtained in a finite number of steps by performing algebraic operations and taking exponentials and logarithms (Wikipedia: Elementary function).
The incomprehensibly unfortunately hardly noticed theorem of Joseph Fels Ritt in Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90 and of Robert H. Risch in Risch, R. H.: Algebraic Properties of the Elementary Functions of Analysis. Amer. J.
Math. 101 (1979) (4) 743-759 answer which kinds of Elementary functions can have an inverse which is an Elementary function. You can also take the method of
Rosenlicht, M.: On the explicit solvability of certain transcendental equations. Publications mathématiques de l'IHÉS 36 (1969) 15-22.
b) You could allow algebraic expressions of known Standard functions as closed form:
If $h$ can be decomposed into a composition of algebraic functions and other known Standard functions than $exp$ and $ln$, an analog theorem to the theorem of Ritt of [Ritt 1925] could be applied. I hope to prove such a generalization of Ritt's theorem for this class of functions.
answered Jul 19 '17 at 20:23
IV_IV_
1,138522
$endgroup$
You are looking for general methods for solving a given equation $h(x)=0$ by applying $h^{-1}$, where $h$ is a function of only one unknown and $h(x)$ is a closed-form expression in dependence of $x$ (Wikipedia: Closed-form expression).
a) Let us allow the elementary functions as closed form:
The elementary functions are according to Liouville and Ritt those functions of one variable which are obtained in a finite number of steps by performing algebraic operations and taking exponentials and logarithms (Wikipedia: Elementary function).
The incomprehensibly unfortunately hardly noticed theorem of Joseph Fels Ritt in Ritt, J. F.: Elementary functions and their inverses. Trans. Amer. Math. Soc. 27 (1925) (1) 68-90 and of Robert H. Risch in Risch, R. H.: Algebraic Properties of the Elementary Functions of Analysis. Amer. J.
Math. 101 (1979) (4) 743-759 answer which kinds of Elementary functions can have an inverse which is an Elementary function. You can also take the method of
Rosenlicht, M.: On the explicit solvability of certain transcendental equations. Publications mathématiques de l'IHÉS 36 (1969) 15-22.
b) You could allow algebraic expressions of known Standard functions as closed form:
If $h$ can be decomposed into a composition of algebraic functions and other known Standard functions than $exp$ and $ln$, an analog theorem to the theorem of Ritt of [Ritt 1925] could be applied. I hope to prove such a generalization of Ritt's theorem for this class of functions.
answered Jul 19 '17 at 20:23
IV_IV_
1,138522
edited Jan 1 at 16:02
answered Jul 19 '17 at 20:23
IV_IV_
1,138522
answered Jul 19 '17 at 20:23
IV_IV_
1,138522
answered Jul 19 '17 at 20:23
IV_IV_
1,138522
1,138522
add a comment |
add a comment |
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$begingroup$
Galois, of course, proved that there are algebraic numbers (solutions of some quintics, in particular) that can't be found by "usual algebraic operations" including taking roots. I suspect this is enough to answer any possible refinement of your question with "no"...
$endgroup$
– HTFB
Dec 18 '14 at 12:48
$begingroup$
Yeah, reading it back I wasn't especially clear on what I was asking. So take $xe^x$, the inverse Lambert-W function. How do I know it's not the case that we just haven't tried hard enough to find a combination of operations to perform on it that allow us to get a nice, explicit $x=text{constant}$?
$endgroup$
– clueless
Dec 18 '14 at 13:18