Rational function

In mathematics, a rational function is a function that can be represented as the quotient of two polynomial functions. It therefore has the form

{\displaystyle f(x)={\frac {a_{m}x^{m}+a_{m-1}x^{m-1}+\dotsb +a_{1}x+a_{0}}{b_{n}x^{n}+b_{n-1}x^{n-1}+\dotsb +b_{1}x+b_{0}}}={\frac {P_{m}(x)}{Q_{n}(x)}}}

with natural numbers mand n. The numbers {\displaystyle a_{m},\dotsc ,a_{0},b_{n},\dotsc ,b_{0}}can be any real numbers (or complex numbers); the only restriction is that {\displaystyle Q_{n}\neq 0}must be. The highest coefficients a_mand b_{n}shall not be zero.

More abstractly, for the coefficients can {\displaystyle a_{m},\dotsc ,a_{0},b_{n},\dotsc ,b_{0}}admit elements of any body. The rational functions with complex coefficients belong to the meromorphic functions.

More generally, one can consider rational functions in several variables as well as rational functions on algebraic varieties over arbitrary solids.

Zoom

red: Graph of fractional function f(x)={\tfrac {2(x+2)(x+1)(x-1)^{2}}{(x+1)(2x-1)}}
blue: Pole line through the pole at
x=0{,}5
green: asymptote function
g(x)=x^{2}+x/2-11/4, continuously rectifiable definition gap at x=-1

Division

  • If the denominator polynomial Q_ndegree n = 0, i.e. constant, then one speaks of a completely rational function or of a polynomial function.
  • If the function term can be n>0represented exclusively with a denominator polynomial of degree , then it is a fractional rational function.
    • If n>0< n m < nthen it is a true fractional function.
    • If n>0m\geq nthen it is a non-genuinely fractional function. It can be split into an integer function and a true fractional function via polynomial division (see below).

Examples of rational functions with different numerator degrees mand denominator degrees n:

Example

alternative spelling

m =

n =

Function type

f\colon x\mapsto {\frac {3x^{3}-4x+5}{2}}

f\colon x\mapsto {\frac {3}{2}}x^{3}-2x+{\frac {5}{2}}

3

0

fully rational

f\colon x\mapsto {\frac {2x-1}{x^{2}+1}}

1

2

Genuinely broken-rational

f\colon x\mapsto {\frac {(x-1)^{2}\cdot (x+2)}{x\cdot (2-3x^{2})}}

f\colon x\mapsto {\frac {x^{3}-3x+2}{2x-3x^{3}}}

3

3

fake broken rational

f\colon x\mapsto x+1+{\frac {1}{x-1}}

f\colon x\mapsto {\frac {x^{2}}{x-1}}

2

1

fake broken rational

Curve discussion

Using the function term of the rational function f={p \over q}\colon x\mapsto {\frac {p(x)}{q(x)}}following statements can be made about the function graph (curve discussion).

Definition range, zeros and poles

The fractional function is not defined at the zeros of the denominator function q

The zeros of a fractional rational function are pdetermined by those zeros of the numerator function which belong to the definition range of the entire function.

A special case arises when a real number a\in \mathbb{R} is simultaneously zero of the numerator polynomial and of the denominator polynomial. Then the numerator and denominator polynomials are divisible by the corresponding linear factor x-a(possibly even several times), i.e. the function term can be truncated with this factor (possibly several times).

  • If x-ain the denominator occurs ntimes more often than in the numerator (with natural number n, n>0is a pole (n {\displaystyle nthe multiplicity of the pole);
  • otherwise the rational function at the position has aa definition gap that can be continuously eliminated, and one can continue the function continuously

Examples:

  • The function f\colon x\mapsto {\frac {x-1}{(2x-4)^{2}}}has the domain of definition {\mathbb {D}}={\mathbb {R}}\setminus \{2\}, since the denominator function q\colon x\mapsto (2x-4)^{2}has the zero x=2and the zero x=1, since that is the only zero of the counter function p\colon x\mapsto x-1(and x=1belongs to\mathbb {D} ). x=2is a (double) pole.
  • The function f\colon x\mapsto {\frac {x^{2}-x}{x^{2}-1}}has the domain of definition {\mathbb {D}}_{f}={\mathbb {R}}\setminus \{\pm 1\}. But here x=1is a zero of the numerator and the denominator function. To truncate the corresponding linear factor (x-1), first factorise the numerator and denominator (by factoring out or using the binomial formula). This leads to f\colon x\mapsto {\frac {x\cdot (x-1)}{(x+1)\cdot (x-1)}}respectively. after shortening to f\colon x\mapsto {\frac {x}{x+1}}. Thus it follows: x=-1is a (simple) pole, x=1on the other hand is a continuously recoverable definition gap of fand fhas the zero x=0(note: x=1is not a zero of fsince this value does not belong to \mathbb {D} !). For the continuous continuation of fget: {\tilde {f}}(x)={\frac {x}{x+1}}and {\mathbb {D}}_{{{\tilde {f}}}}={\mathbb {R}}\setminus \{-1\}.

Asymptotic behaviour

For the behaviour of xtowards infinity, the degrees mor nthe numerator or denominator polynomial are decisive:

For x\to\inftygoes f(x)

  • (case 1) against {\displaystyle \operatorname {sgn} \left({\tfrac {a_{m}}{b_{n}}}\right)\cdot \infty }, if m>nwhere \sgnrepresents the sign function.
  • (case 2) against {\displaystyle {\tfrac {a_{m}}{b_{n}}}}if m=n(the asymptote is parallel to the xaxis),
  • (case 3) towards {\displaystyle 0}(the xaxis is horizontal asymptote), if m<n,

For x\to -\inftysame limit value results in cases 2 and 3 as for x\to \infty . In case 1, one has to consider the numerator and denominator degrees even more precisely:

  • If {\displaystyle m-n}is even, the same limit value results as for x\to \infty .
  • If is {\displaystyle m-n}odd, the sign of the limit changes compared to x\to \infty

Examples:

  • For the fractional function f\colon x\mapsto {\frac {2x-1}{x^{2}+1}}the numerator degree m=1and the denominator degree n=2the limit value for {\displaystyle x \to \pm \infty}is therefore {\displaystyle 0}.
  • The fractional rational function f\colon x\mapsto {\frac {x^{3}-3x+2}{2x-3x^{3}}}has the numerator degree m=3and also the denominator degree n=3; since here a_{3}=1and b_{3}=-3, the equation of the horizontal asymptote gives: y=-{\frac {1}{3}}.
  • The fractional rational function f\colon x\mapsto {\frac {x^{2}}{x-1}}has the numerator degree m=2and the denominator degree n=1; with the coefficients a_{2}=1and b_{1}=1thus obtain: f(x)\to \operatorname{sgn} \left({\tfrac {1}{1}}\right)\cdot \infty =+\infty for x\to \infty . Since here {\displaystyle m-n=1}is odd, it follows for the limit for x\to -\infty the reversed sign, i.e. f(x)\to -\infty . This function can also be written as f\colon x\mapsto x+1+{\frac {1}{x-1}}that is, the (oblique) asymptote has the equation y=x+1(and this also easily results in the limit behaviour just described).

Investigation with polynomial division

In the above case 1 ( m>ndecompose the function term by means of polynomial division into a sum of a polynomial and a true fractional term; the polynomial then describes a so-called asymptote curve. The behaviour of the function values for {\displaystyle x \to \pm \infty}described above can also be obtained more simply by examining only the behaviour of this asymptote curve. In the special case m=n+1an oblique asymptote results.

As above, mstands for the degree of the numerator polynomial p(x)and ndegree of the denominator polynomial q(x). Again, all cases are considered (not only

By means of polynomial division of p(x)by q(x)one first obtains a representation

{\displaystyle p(x)=g(x)\cdot q(x)+r(x)}

with polynomials g(x)and r(x)where the degree of r(x)is q(x)genuinely greater than that of From this follows the useful equation

{\displaystyle f(x)={p(x) \over q(x)}=g(x)+{r(x) \over q(x)}}.

The asymptotic behaviour of f(x)is now the same as the asymptotic behaviour of the integral function ("asymptote function") g(x). The quotient {\displaystyle r(x) \over q(x)}plays no role.

If you have taken the trouble of polynomial division and set up the useful equation described above, you will find it easier to distinguish between cases. It applies:

Case 1: m < nx-axis is asymptote: g(x) = 0

Case 2: m=n→ horizontal asymptote: {\displaystyle g(x)={\frac {a_{m}}{b_{n}}}}

Case 3: {\displaystyle m=n+1}→ oblique asymptote: {\displaystyle g(x)=bx+c}with {\displaystyle b={\frac {a_{m}}{b_{n}}}}and {\displaystyle c={\frac {a_{m-1}}{b_{n}}}-{\frac {a_{m}b_{n-1}}{b_{n}^{2}}}}

Case 4: {\displaystyle m>n+1}g(x)is a polynomial of degree m-n; the leading coefficient of this polynomial is equal to {\displaystyle {\frac {a_{m}}{b_{n}}}}.

Symmetry

A polynomial function (integer function) is even/odd if all exponents are even/odd. If numerator polynomial pand denominator polynomial qof one of these two types, the rational function fis also even or odd:

  • If pand are qboth even or both odd, then is feven (i.e. the graph is symmetrical to the y-axis).
  • If is peven and qis odd, then fis odd (i.e. the graph is point-symmetric with respect to the origin); the same applies if podd and qeven.

In all other cases, i.e. when numerator or denominator function or both are neither even nor odd, symmetry properties of are fmore difficult to decide. (See also discussion of curves and symmetry in geometry).

Examples:

  • The graph of the function fwith f(x)={\frac {2x^{3}-3x}{x^{2}+1}}is symmetric about the origin, since podd and qeven, so the function as a whole is odd.
  • The graph of the function f\colon x\mapsto {\frac {x^{5}-x^{3}}{x^{3}+x}}is symmetrical to the y-axis, because pand qboth odd, so the function as a whole is even. This can also be seen differently: If you exclude x from the numerator and denominator, you can shorten the function term to f(x)={\frac {x^{4}-x^{2}}{x^{2}+1}}; now pand are qeven, so the function as a whole is even again.
  • The graph of the function with the term f(x)={\frac {x}{x-1}}not show any symmetry at first ( pis odd, but qis neither even nor odd); however, it can be shown that the graph is symmetrical to the point P(1|1); namely, it holds:

f(1+x)-1={\frac {1+x}{(1+x)-1}}-1={\frac {1+x}{x}}-{\frac {x}{x}}={\frac {1}{x}}and

1-f(1-x)=1-{\frac {1-x}{(1-x)-1}}={\frac {x}{x}}+{\frac {1-x}{x}}={\frac {1}{x}},

so in total: f(1+x)-1=1-f(1-x)which means straight symmetry to the point P(1|1). Alternatively, one can also show that the graph of fconsists of the graph of the function g\colon x\mapsto {\frac {1}{x}}(which is symmetrical to the origin) by shifting it by 1 in the xdirection and by 1 in the ydirection.

Derivation

To derive fractional rational functions, one must generally use the quotient rule; in addition, the chain rule can often be useful, for example, if the denominator function is a power of a binomial. Before deriving, it is often advisable to first rewrite the function term with the help of a polynomial division and to shorten the remaining true fractional term.

Examples:

  • For the function f\colon x\mapsto {\frac {2x-1}{(x^{2}+1)^{2}}}it makes sense to apply the chain rule in addition to the quotient rule instead of first applying the first binomial formula in the denominator. With the chain rule, the derivative of the denominator function q(in the quotient rule usually vdenoted by ):

q'(x)=2(x^{2}+1)\cdot 2x=4x(x^{2}+1),

and thus altogether for the derivative function of f:

f'(x)={\frac {2\cdot (x^{2}+1)^{2}-(2x-1)\cdot 4x(x^{2}+1)}{(x^{2}+1)^{4}}}.

Now you can factor out (x^{2}+1)in the numerator and shorten it:

f'(x)={\frac {2\cdot (x^{2}+1)-(2x-1)\cdot 4x}{(x^{2}+1)^{3}}}.

Simplifying the counter finally leads to

f'(x)={\frac {-6x^{2}+4x+2}{(x^{2}+1)^{3}}}.

  • The function term f(x)={\frac {x^{4}+x^{3}-7x^{2}-12x-4}{3x^{3}+12x^{2}+12x}}is first brought to the form with the help of a polynomial division.

f(x)={\frac {1}{3}}x-1+{\frac {x^{2}-4}{3x^{3}+12x^{2}+12x}},

from which you can also read off the equation of the oblique asymptote:

y={\frac {1}{3}}x-1.

Factorising numerator and denominator then leads to

f(x)={\frac {1}{3}}x-1+{\frac {(x+2)(x-2)}{3x(x+2)^{2}}},

one can therefore (x+2)truncate a factor Finally one has:

f(x)={\frac {1}{3}}x-1+{\frac {x-2}{3x^{2}+6x}};

in this form it is now much easier to derive the function than in the one originally given.

With the help of the quotient rule we get:

f'(x)={\frac {1}{3}}+{\frac {1\cdot (3x^{2}+6x)-(x-2)\cdot (6x+6)}{(3x^{2}+6x)^{2}}}={\frac {1}{3}}+{\frac {-3x^{2}+12x+12}{(3x^{2}+6x)^{2}}}={\frac {1}{3}}+{\frac {-x^{2}+4x+4}{3x^{2}(x+2)^{2}}}.

If one sets the first derivative equal to zero in order to search for the extreme points, it is advisable to combine the two fractions again beforehand:

f'(x)={\frac {x^{2}(x+2)^{2}-x^{2}+4x+4}{3x^{2}(x+2)^{2}}}={\frac {x^{4}+4x^{3}+3x^{2}+4x+4}{3x^{2}(x+2)^{2}}}.

Stem function

In contrast to the integral functions, it is often relatively difficult to find a root function for fractional functions. Depending on the form of the fractional function, the following rules can be applied (usually the function term must first be brought into a suitable form by transformations and/or substitutions):

\int {\frac {1}{mx+a}}dx={\frac {1}{m}}\cdot \ln(mx+a)+Cfor m,a\in {\mathbb {R}},m\neq 0

\int {\frac {1}{(mx+a)^{n}}}dx={\frac {1}{m}}\cdot {\frac {-1}{n-1}}\cdot {\frac {1}{(mx+a)^{{n-1}}}}+Cfor m,a\in {\mathbb {R}},m\neq 0,n\in {\mathbb {N}}\setminus \{0;1\}

\int {\frac {1}{x^{2}+1}}dx=\arctan(x)+Cor {\displaystyle =-\operatorname {arccot}(x)+C}

\int {\frac {1}{x^{2}-1}}dx=\operatorname {artanh}(x)+C={\frac {1}{2}}\ln \left({\frac {1+x}{1-x}}\right)for |x| < 1

\int {\frac {1}{x^{2}-1}}dx=\operatorname {arcoth}(x)+C={\frac {1}{2}}\ln \left({\frac {x+1}{x-1}}\right)for |x| > 1

\int {\frac {u'(x)}{u(x)}}dx=\ln |u(x)|+Cfor u(x)\neq 0

Often the partial fraction decomposition can also be helpful for determining a parent function. Examples:

  • Let us look for a primitive function of f(x)={\frac {5x-1}{3x+2}}. By means of a polynomial division, this can first be rewritten as:

f(x)={\frac {5}{3}}-{\frac {13}{9x+6}}.

Applying the first rule then yields as possible root function:

F(x)={\frac {5}{3}}x-{\frac {13}{9}}\ln(9x+6).

  • Let us find a primitive function for f(x)={\frac {x^{2}+1}{x^{2}-1}}where xshould lie between -0.5 and 0.5. Again, the function term can first be rewritten using polynomial division:

f(x)=1+{\frac {2}{x^{2}-1}}.

Applying the fourth rule then yields as possible root function:

F(x)=x+2\cdot \operatorname {artanh}(x).

  • Let us find a root function of f(x)={\frac {x+2}{x^{2}+4x+5}}. This can also be written as

f(x)={\frac {1}{2}}{\frac {2x+4}{x^{2}+4x+5}}={\frac {1}{2}}{\frac {u'(x)}{u(x)}}with u(x)=x^{2}+4x+5.

Applying the last rule then yields as possible root function:

F(x)={\frac {1}{2}}\ln(x^{2}+4x+5).

  • A primitive function to f(x)={\frac {1}{x^{2}+2x+2}}can be y=x+1determined using the substitution after transforming the denominator using quadratic completion:

{\begin{aligned}\int {\frac {1}{x^{2}+2x+2}}dx&=\int {\frac {1}{(x+1)^{2}+1}}dx=\int {\frac {1}{y^{2}+1}}dy\\&=\arctan(y)+C=\arctan(x+1)+C\end{aligned}}

  • A primitive function to f(x)={\frac {1}{x^{2}-x-6}}can be obtained by partial fraction decomposition after first factorising the denominator:

{\begin{aligned}\int {\frac {1}{x^{2}-x-6}}dx&=\int {\frac {1}{(x-3)(x+2)}}dx=\int {\frac {1}{5}}\left({\frac {1}{x-3}}-{\frac {1}{x+2}}\right)dx\\&={\frac {1}{5}}\left(\ln(x-3)-\ln(x+2)\right)+C={\frac {1}{5}}\ln \left({\frac {x-3}{x+2}}\right)+C\end{aligned}}

Rational functions in several variables

A rational function in variables x_{1},\ldots ,x_{n}is a function of the form {\displaystyle f(x_{1},\ldots ,x_{n})={\frac {P(x_{1},\ldots ,x_{n})}{Q(x_{1},\ldots ,x_{n})}}}where Pand x_{1},\ldots ,x_{n}are Qpolynomials in the indefinites and {\displaystyle Q\not =0}.

Examples

  • {\displaystyle f(x_{1},\ldots ,x_{n})={\frac {x_{1}^{2}+\ldots +x_{n}^{2}}{1-x_{1}\ldots x_{n}}}}
  • {\displaystyle f(x,y)={\frac {xy}{x+y}}}
  • {\displaystyle f(m,M)={\frac {m}{2M+m}}g}

Continuity

The domain of fconsists of those points (x_{1},\ldots ,x_{n})which are either not a zero of Qor whose multiplicity as a zero of Pat least as large as the multiplicity as a zero of Q. Rational functions are continuous in all points of their domain.

Applications

Rational functions have many applications in science and technology:

  • Many quantities are inversely proportional to each other, so one of the quantities is a rational function of the other, with the numerator constant and the denominator a (homogeneous) linearfunction. A few examples:
    • velocity vand the time srequired for a fixed distance tare inversely proportional to each other: t(v)={\tfrac {s}{v}}
    • The concentration ca substance is ninversely proportional to the volume Vthe solvent for a fixed amount of substance : C(V)={\tfrac {n}{V}}
    • Acceleration and mass are Finversely proportional to each other for a fixed force : a(m)={\tfrac {F}{m}}.
    • For the capacitance Ca plate capacitor, the following applies as a function of the plate spacing d: C(d)=\epsilon _{0}\epsilon _{r}{\tfrac {A}{d}}with the area Athe plates, the electric field constant \epsilon _{0}and the permittivity \epsilon_r.
  • In many areas of physics, functions of two variables xand ythe following form occur: f(x;y)={\tfrac {xy}{x\pm y}}. If one of the two variables, e.g. yis constant or chosen as a parameter, the result is a rational function (or set of functions) of x. Such functions always occur when the total reciprocal of some quantity results as the sum or difference of the reciprocals of two other functions.
    • Using the lens equation of optics, one can brepresent the focal length fas a function of object width gand image width : f(g;b)={\tfrac {gb}{g+b}}; rearranging to gor byield a very similar function, but with - instead of +.
    • For the total resistance Ra parallel connection of two resistors R_{1}and R_{2}get: R={\tfrac {R_{1}R_{2}}{R_{1}+R_{2}}}; an analogous formula applies to the series connection of two capacitors.
    • In mechanics, if two springs with spring constants D_{1}and D_{2}attached to each other, the following results for the total spring constant Dof the arrangement: D={\tfrac {D_{1}D_{2}}{D_{1}+D_{2}}}
  • For a voltage divider, the total voltage Rdropping across a resistor Ugiven by: U(R)={\tfrac {U_{0}R}{R+R'}}where U_{0}the voltage to be divided and R'the other resistance.
  • For the electrical power PRproduced by a device with resistance connected to a voltage source (voltage U) with internal resistance R_{i}: P(R)={\tfrac {U^{2}R}{(R+R_{i})^{2}}}. The greatest possible power (to be determined with the aid of differential calculus) is therefore obtained when R=R_{i}(power matching).
  • For the inductance Lof a (not too short) coil as a function of its radius rapplies: L(r)={\tfrac {\mu _{0}N^{2}\pi r^{2}}{l+r/1,1}}. Where lthe length of the coil (Lso can also be ltaken as a rational function of ), Nis the number of turns and μ \mu _{0}is the magnetic field constant.
  • The braking force Ban eddy current brake depends on the velocity vas follows: B(v)={\tfrac {av}{b+v^{2}}}with constants aand b.
  • In Atwood's machine, the acceleration depends on the two masses mand Maas follows: a={\tfrac {m}{2M+m}}\cdot g; athus one can Mtake as a rational function of both mand
  • Geometric questions also often lead to rational functions. Example: For a chest consisting of a cuboid (basic side lengths land 2 r, height r) with an attached half-cylinder (height l, radius r), the following applies to the surface area Ofunction of rgiven volume V: O(r)={\tfrac {(\pi +4)r^{3}+2V}{r}}.

Deviant meaning in abstract algebra

Rational functions over any body

Main article: Rational function body

In abstract algebra, the term rational function is used in a more general and somewhat different sense. Namely, one understands by a rational function in nvariables X_1, X_2, \dotsc, X_nover a body Kan element of the quotient body of the polynomial ring K\left[X_{1},X_{2},\dotsc ,X_{n}\right]. This quotient body is called a rational function body.

In general, then, a rational function is not a function of any kind, but a (formal) fraction of two polynomials. The inverse need not hold, but the difference is only noticeable over finitebodies: For example, for any prime pover the finite body \mathbb {F} _{p}(the body of all residue classes of integers modulo p) the fraction {\tfrac {1}{X^{p}-X}}a well-defined rational function in the variable Xbut not a function in the true sense of the term, because you cannot put a single value into this function without the denominator becoming 0. (For if one puts any x\in {\mathbb F}_{p}into this "function", you get {\tfrac {1}{x^{p}-x}}which is undefined because the denominator 0 according tox^{p}-x Fermat's little theorem). Over infinite bodies, however, a rational function is always a function that may have a gap in definition, but this gap in definition is only very small compared to the domain of definition. This idea is formalised by the notion of Zariski topology: The definition gap is a Zariski-closed set, and the closed hull of the definition domain is the whole set.

Rational functions on an algebraic variety

Main article: Rational function body#Function bodies in algebraic geometry

Let Vbe an algebraic variety defined by polynomials f_{1},\dotsc ,f_{m}\in k\left[x_{1},\dotsc ,x_{n}\right], so

V=\{x\in {\mathbb A}^{n}\mid f(x)=0{\text{ für alle }}f\in S\}.

Be

I(V)=\{f\in k[x_{1},\dotsc ,x_{n}]\mid f(x)=0{\text{ für alle }}x\in V\}.

The ring of entire functions is k[x_{1},\dotsc ,x_{n}]/I(V). The body of rational functions is the quotient body of the ring of integer functions.

More generally, there is the notion of rational mappings between (quasi-projective) varieties. Rational functions are the special case of rational mappings from a variety to {\displaystyle \mathbb {A} ^{1}}.


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