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Are polynomial functions with fractional exponents transcendental?
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I'm having trouble categorizing fractional--order systems, that means functions like
$$ f(x) = K cdot (1 + sqrtx)$$
or more generally, e.g.
$$ g(x) = dfraca_0 + a_1 x^1/2 + a_2 x^1 + a_3 x^3/2b_0 + b_1 x^1/2 + b_2 x^1 + b_3 x^3/2$$
where the exponents are rational numbers $gamma in mathbbQ$. All coefficients and $K$ are real numbers. $x$ is, in my case, complex.
I doubt they can still be refered as rational functions, can they? But on the other hand, are these functions transcendental functions?
The definition on Wikipedia is not entirely clear to me on that matter:
A transcendental function is an analytic function that does not satisfy a polynomial equation, in contrast to an algebraic function.1[2] In other words, a transcendental function "transcends" algebra in that it cannot be expressed in terms of a finite sequence of the algebraic operations of addition, multiplication, and root extraction.
Would the definition change, if we include real exponents $gamma in mathbbR$?
And what happens in case of the implicit form:
$$ h(x) = K cdot sqrt1 + x$$
I'm a little confused about all these definitions and I hope you can shed some light on my problem.
polynomials definition rational-functions transcendental-functions
asked Jul 16 at 8:51
thewaywewalk
183211
add a comment |Â
up vote
0
down vote
favorite
I'm having trouble categorizing fractional--order systems, that means functions like
$$ f(x) = K cdot (1 + sqrtx)$$
or more generally, e.g.
$$ g(x) = dfraca_0 + a_1 x^1/2 + a_2 x^1 + a_3 x^3/2b_0 + b_1 x^1/2 + b_2 x^1 + b_3 x^3/2$$
where the exponents are rational numbers $gamma in mathbbQ$. All coefficients and $K$ are real numbers. $x$ is, in my case, complex.
I doubt they can still be refered as rational functions, can they? But on the other hand, are these functions transcendental functions?
The definition on Wikipedia is not entirely clear to me on that matter:
A transcendental function is an analytic function that does not satisfy a polynomial equation, in contrast to an algebraic function.1[2] In other words, a transcendental function "transcends" algebra in that it cannot be expressed in terms of a finite sequence of the algebraic operations of addition, multiplication, and root extraction.
Would the definition change, if we include real exponents $gamma in mathbbR$?
And what happens in case of the implicit form:
$$ h(x) = K cdot sqrt1 + x$$
I'm a little confused about all these definitions and I hope you can shed some light on my problem.
polynomials definition rational-functions transcendental-functions
asked Jul 16 at 8:51
thewaywewalk
183211
add a comment |Â
up vote
0
down vote
favorite
up vote
0
down vote
favorite
I'm having trouble categorizing fractional--order systems, that means functions like
$$ f(x) = K cdot (1 + sqrtx)$$
or more generally, e.g.
$$ g(x) = dfraca_0 + a_1 x^1/2 + a_2 x^1 + a_3 x^3/2b_0 + b_1 x^1/2 + b_2 x^1 + b_3 x^3/2$$
where the exponents are rational numbers $gamma in mathbbQ$. All coefficients and $K$ are real numbers. $x$ is, in my case, complex.
I doubt they can still be refered as rational functions, can they? But on the other hand, are these functions transcendental functions?
The definition on Wikipedia is not entirely clear to me on that matter:
A transcendental function is an analytic function that does not satisfy a polynomial equation, in contrast to an algebraic function.1[2] In other words, a transcendental function "transcends" algebra in that it cannot be expressed in terms of a finite sequence of the algebraic operations of addition, multiplication, and root extraction.
Would the definition change, if we include real exponents $gamma in mathbbR$?
And what happens in case of the implicit form:
$$ h(x) = K cdot sqrt1 + x$$
I'm a little confused about all these definitions and I hope you can shed some light on my problem.
polynomials definition rational-functions transcendental-functions
asked Jul 16 at 8:51
thewaywewalk
183211
I'm having trouble categorizing fractional--order systems, that means functions like
$$ f(x) = K cdot (1 + sqrtx)$$
or more generally, e.g.
$$ g(x) = dfraca_0 + a_1 x^1/2 + a_2 x^1 + a_3 x^3/2b_0 + b_1 x^1/2 + b_2 x^1 + b_3 x^3/2$$
where the exponents are rational numbers $gamma in mathbbQ$. All coefficients and $K$ are real numbers. $x$ is, in my case, complex.
I doubt they can still be refered as rational functions, can they? But on the other hand, are these functions transcendental functions?
The definition on Wikipedia is not entirely clear to me on that matter:
A transcendental function is an analytic function that does not satisfy a polynomial equation, in contrast to an algebraic function.1[2] In other words, a transcendental function "transcends" algebra in that it cannot be expressed in terms of a finite sequence of the algebraic operations of addition, multiplication, and root extraction.
Would the definition change, if we include real exponents $gamma in mathbbR$?
And what happens in case of the implicit form:
$$ h(x) = K cdot sqrt1 + x$$
I'm a little confused about all these definitions and I hope you can shed some light on my problem.
polynomials definition rational-functions transcendental-functions
asked Jul 16 at 8:51
thewaywewalk
183211
edited Jul 16 at 9:13
asked Jul 16 at 8:51
thewaywewalk
183211
asked Jul 16 at 8:51
thewaywewalk
183211
asked Jul 16 at 8:51
thewaywewalk
183211
183211
add a comment |Â
add a comment |Â
2 Answers
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oldest
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1
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accepted
$$f(x)=1+sqrt x$$ is algebraic because it satisfies the polynomial equation
$$(f(x)-1)^2-x=0.$$
Same goes with your function $g(x)$, though the polynomial will be of a much higher degree.
Transcendence begins with irrational exponents, as they would require a polynomial of infinite degree.
answered Jul 16 at 9:17
Yves Daoust
111k665204
add a comment |Â
up vote
2
down vote
The confusion arises because it is quite unclear in which ring the coefficients of the polynomial are supposed to be. The notion of algebraicity can only be defined properly if it is clear what is meant by a polynomial equation. Visiting the French article on the same subject, we see that the coefficients are supposed to be polynomials over the ground field, which in this case are the complex numbers.
Based on that definition, I would say that the first function is indeed algebraic, because it satisfies a quadratic equation, as is easily seen. (Given that $K$ denotes a constant.)
The second function should also be algebraic, because one may use polynomials to counteract growing exponents in the denominator. (I don't see the equation immediately.)
answered Jul 16 at 9:08
AlgebraicsAnonymous
69111
In general, transcendental means precisely "not algebraic". Once one settles for a polynomial ring, the definitions become clear.
– AlgebraicsAnonymous
Jul 16 at 9:10
1
I gave some more information about the other variables.
– thewaywewalk
Jul 16 at 9:11
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
$$f(x)=1+sqrt x$$ is algebraic because it satisfies the polynomial equation
$$(f(x)-1)^2-x=0.$$
Same goes with your function $g(x)$, though the polynomial will be of a much higher degree.
Transcendence begins with irrational exponents, as they would require a polynomial of infinite degree.
answered Jul 16 at 9:17
Yves Daoust
111k665204
add a comment |Â
up vote
1
down vote
accepted
$$f(x)=1+sqrt x$$ is algebraic because it satisfies the polynomial equation
$$(f(x)-1)^2-x=0.$$
Same goes with your function $g(x)$, though the polynomial will be of a much higher degree.
Transcendence begins with irrational exponents, as they would require a polynomial of infinite degree.
answered Jul 16 at 9:17
Yves Daoust
111k665204
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
$$f(x)=1+sqrt x$$ is algebraic because it satisfies the polynomial equation
$$(f(x)-1)^2-x=0.$$
Same goes with your function $g(x)$, though the polynomial will be of a much higher degree.
Transcendence begins with irrational exponents, as they would require a polynomial of infinite degree.
answered Jul 16 at 9:17
Yves Daoust
111k665204
$$f(x)=1+sqrt x$$ is algebraic because it satisfies the polynomial equation
$$(f(x)-1)^2-x=0.$$
Same goes with your function $g(x)$, though the polynomial will be of a much higher degree.
Transcendence begins with irrational exponents, as they would require a polynomial of infinite degree.
answered Jul 16 at 9:17
Yves Daoust
111k665204
answered Jul 16 at 9:17
Yves Daoust
111k665204
answered Jul 16 at 9:17
Yves Daoust
111k665204
answered Jul 16 at 9:17
Yves Daoust
111k665204
111k665204
add a comment |Â
add a comment |Â
up vote
2
down vote
The confusion arises because it is quite unclear in which ring the coefficients of the polynomial are supposed to be. The notion of algebraicity can only be defined properly if it is clear what is meant by a polynomial equation. Visiting the French article on the same subject, we see that the coefficients are supposed to be polynomials over the ground field, which in this case are the complex numbers.
Based on that definition, I would say that the first function is indeed algebraic, because it satisfies a quadratic equation, as is easily seen. (Given that $K$ denotes a constant.)
The second function should also be algebraic, because one may use polynomials to counteract growing exponents in the denominator. (I don't see the equation immediately.)
answered Jul 16 at 9:08
AlgebraicsAnonymous
69111
In general, transcendental means precisely "not algebraic". Once one settles for a polynomial ring, the definitions become clear.
– AlgebraicsAnonymous
Jul 16 at 9:10
1
I gave some more information about the other variables.
– thewaywewalk
Jul 16 at 9:11
add a comment |Â
up vote
2
down vote
The confusion arises because it is quite unclear in which ring the coefficients of the polynomial are supposed to be. The notion of algebraicity can only be defined properly if it is clear what is meant by a polynomial equation. Visiting the French article on the same subject, we see that the coefficients are supposed to be polynomials over the ground field, which in this case are the complex numbers.
Based on that definition, I would say that the first function is indeed algebraic, because it satisfies a quadratic equation, as is easily seen. (Given that $K$ denotes a constant.)
The second function should also be algebraic, because one may use polynomials to counteract growing exponents in the denominator. (I don't see the equation immediately.)
answered Jul 16 at 9:08
AlgebraicsAnonymous
69111
In general, transcendental means precisely "not algebraic". Once one settles for a polynomial ring, the definitions become clear.
– AlgebraicsAnonymous
Jul 16 at 9:10
1
I gave some more information about the other variables.
– thewaywewalk
Jul 16 at 9:11
add a comment |Â
up vote
2
down vote
up vote
2
down vote
The confusion arises because it is quite unclear in which ring the coefficients of the polynomial are supposed to be. The notion of algebraicity can only be defined properly if it is clear what is meant by a polynomial equation. Visiting the French article on the same subject, we see that the coefficients are supposed to be polynomials over the ground field, which in this case are the complex numbers.
Based on that definition, I would say that the first function is indeed algebraic, because it satisfies a quadratic equation, as is easily seen. (Given that $K$ denotes a constant.)
The second function should also be algebraic, because one may use polynomials to counteract growing exponents in the denominator. (I don't see the equation immediately.)
answered Jul 16 at 9:08
AlgebraicsAnonymous
69111
The confusion arises because it is quite unclear in which ring the coefficients of the polynomial are supposed to be. The notion of algebraicity can only be defined properly if it is clear what is meant by a polynomial equation. Visiting the French article on the same subject, we see that the coefficients are supposed to be polynomials over the ground field, which in this case are the complex numbers.
Based on that definition, I would say that the first function is indeed algebraic, because it satisfies a quadratic equation, as is easily seen. (Given that $K$ denotes a constant.)
The second function should also be algebraic, because one may use polynomials to counteract growing exponents in the denominator. (I don't see the equation immediately.)
answered Jul 16 at 9:08
AlgebraicsAnonymous
69111
answered Jul 16 at 9:08
AlgebraicsAnonymous
69111
answered Jul 16 at 9:08
AlgebraicsAnonymous
69111
answered Jul 16 at 9:08
AlgebraicsAnonymous
69111
69111
In general, transcendental means precisely "not algebraic". Once one settles for a polynomial ring, the definitions become clear.
– AlgebraicsAnonymous
Jul 16 at 9:10
1
I gave some more information about the other variables.
– thewaywewalk
Jul 16 at 9:11
add a comment |Â
In general, transcendental means precisely "not algebraic". Once one settles for a polynomial ring, the definitions become clear.
– AlgebraicsAnonymous
Jul 16 at 9:10
1
I gave some more information about the other variables.
– thewaywewalk
Jul 16 at 9:11
In general, transcendental means precisely "not algebraic". Once one settles for a polynomial ring, the definitions become clear.
– AlgebraicsAnonymous
Jul 16 at 9:10
In general, transcendental means precisely "not algebraic". Once one settles for a polynomial ring, the definitions become clear.
– AlgebraicsAnonymous
Jul 16 at 9:10
1
1
I gave some more information about the other variables.
– thewaywewalk
Jul 16 at 9:11
I gave some more information about the other variables.
– thewaywewalk
Jul 16 at 9:11
add a comment |Â
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