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Revision difference : algebraic function |
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| \PMlinkescapeword{algebraic}% |
\PMlinkescapeword{algebraic}% |
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| A function of one variable is said to be \emph{algebraic} if it satisfies a polynomial equation whose coefficients are polynomials in the same variable. |
A function of one variable is said to be \emph{algebraic} if it satisfies a polynomial equation whose coefficients are polynomials in the same variable. |
| Namely, the function $f(x)$ is algebraic if $y=f(x)$ is a solution of an equation of the form |
Namely, the function $f(x)$ is algebraic if $y=f(x)$ is a solution of an equation of the form |
| \[ |
\[ |
| p_n(x) y^n + \cdots + p_1(x) y + p_0(x) = 0, |
p_n(x) y^n + \cdots + p_1(x) y + p_0(x) = 0, |
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\] |
| where the $p_0(x), p_1(x), \ldots, p_n(x)$ are polynomials in $x$. A function that satisfies no such equation is said to be \emph{transcendental}. |
where the $p_0(x), p_1(x), \ldots, p_n(x)$ are polynomials in $x$. A function that satisfies no such equation is said to be \emph{transcendental}. |
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| The graph of an algebraic function is an algebraic curve, which is, loosely speaking, the solution set of a polynomial equation in two variables. |
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| \subsection*{Examples} |
\subsection*{Examples} |
| Any rational function $f(x) = P(x)/Q(x)$ is algebraic, since $y=f(x)$ is a solution to $Q(x)y - P(x) = 0$. |
Any rational function $f(x) = P(x)/Q(x)$ is algebraic, since $y=f(x)$ is a solution to $Q(x)y - P(x) = 0$. |
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| The function $f(x)=\sqrt{x}$ is algebraic, since $y=f(x)$ is a |
The function $f(x)=\sqrt{x}$ is algebraic, since $y=f(x)$ is a |
| solution to $y^2 - x = 0$. The same is true for any power function |
solution to $y^2 - x = 0$. The same is true for any power function |
| $x^{n/m}$, with $n$ and $m$ integers, it satisfies the equation $y^m-x^n=0$. |
$x^{n/m}$, with $n$ and $m$ integers, it satisfies the equation $y^m-x^n=0$. |
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| It is known that the functions $e^x$ and $\ln x$ are transcendental. Many special functions, such as Bessel functions, elliptic integrals, and others are known to be transcendental. |
It is known that the functions $e^x$ and $\ln x$ are transcendental. Many special functions, such as Bessel functions, elliptic integrals, and others are known to be transcendental. |
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| \textbf{Remark}. There is also a version of an algebraic function defined on algebraic systems. Given an algebraic system $A$, an \emph{$n$-ary algebraic function} on $A$ is an $n$-ary operator $f(x_1,\ldots,x_n)$ on $A$ such that there is an \PMlinkname{$(n+m)$-ary polynomial}{PolynomialsInAlgebraicSystems} $p(x_1,\ldots,x_n,x_{n+1},\ldots, x_{n+m})$ on $A$ for some non-negative integer $m$, and elements $a_1,\ldots, a_m\in A$ such that $$f(x_1,\ldots,x_n) = p(x_1,\ldots,x_n,a_1,\ldots, a_m).$$ |
\textbf{Remark}. There is also a version of an algebraic function defined on algebraic systems. Given an algebraic system $A$, an \emph{$n$-ary algebraic function} on $A$ is an $n$-ary operator $f(x_1,\ldots,x_n)$ on $A$ such that there is an \PMlinkname{$(n+m)$-ary polynomial}{PolynomialsInAlgebraicSystems} $p(x_1,\ldots,x_n,x_{n+1},\ldots, x_{n+m})$ on $A$ for some non-negative integer $m$, and elements $a_1,\ldots, a_m\in A$ such that $$f(x_1,\ldots,x_n) = p(x_1,\ldots,x_n,a_1,\ldots, a_m).$$ |
| For example, in a ring $R$, a function $f$ on $R$ given by $f(x)=a_nx^n+\cdots + a_1x+a_0$ where $a_i\in R$ is a unary algebraic function on $R$, as $f(x)=p(x,a_0,\ldots,a_n)$, where $p$ is an $(n+2)$-ary polynomial on $R$ given by $p(x,x_0,\ldots,x_n)=x_nx^n+\cdots + x_1x+x_0$. |
For example, in a ring $R$, a function $f$ on $R$ given by $f(x)=a_nx^n+\cdots + a_1x+a_0$ where $a_i\in R$ is a unary algebraic function on $R$, as $f(x)=p(x,a_0,\ldots,a_n)$, where $p$ is an $(n+2)$-ary polynomial on $R$ given by $p(x,x_0,\ldots,x_n)=x_nx^n+\cdots + x_1x+x_0$. |
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