四次函数 - 王朝网络宽屏版

形如y=ax^4+bx^3+cx^2+dx+e(a≠0,b,c,d,e为常数)的函数叫做四次函数。

四次函数的图像

a*x^4+b*x^3+c*x^2+b*x+a=0的求解方法，对于一般的四次方程a*x^4+b*x^3+c*x^2+d*x+e=0,先求解三次方程8y^3-4cy^2+(2bd-8e)y+e(4c-b^2)-d^2=0,得到的y的任一实根分别代入下面两个方程：

x^2+(b+sqrt(8y+b^2-4c))x/2+(y+(by-d)/sqrt(8y+b^2-4c))=0

及x^2+(b-sqrt(8y+b^2-4c))x/2+(y-(by-d)/sqrt(8y+b^2-4c))=0

就可得到原方程的四个根。

在数学中, 四次方程是令一个四次函数等于零的结果.四次方程的一个例子如下

它的通式是

<math>a_0x^4+a_1x^3+a_2x^2+a_3x+a_4=0,qquadmboxa_0

e0.</math>

代数基本定理告诉我们, 一个四次方程总有四个解 (根). 它们可能是复数而且可能有等根.

[编辑]解决四次方程

自然,人们为了找到这些根做了许多努力. 就像其它多项式, 有时可能对一个四次方程分解出因式;但更多的时候这样的工作是极困难的,尤其是当根是无理数或复数时.因此找到一个通式解法或运算法则 (就像二次方程那样, 能解所有的一元二次方程)是很有用的. After much effort, such a formula was indeed found for quartics — but since then it has been proven (by Evariste Galois) that such an approach dead-ends with quartics; they are the highest-degree polynomial equations whose roots can be expressed in a formula using a finite number of arithmetic operators and n-th roots. From quintics on up, one requires more powerful methods if a general algebraic solution is sought, as explained under quintic equations.

Given the complexity of the quartic formulae (see below), they are not often used. If only the real rational roots are needed, they can be found (as is true for polynomials of any degree) via trial and error, using Ruffini's rule (so long as all the polynomial coefficients are rational). In the modern age of computers, furthermore, good numerical approximations for the roots are rapidly obtainable via Newton's method. But if the quartic must be solved entirely and precisely, the procedures are outlined below.

特殊情况

名义上的四次方程

如果a4 = 0，那么其中一个根为x = 0，其它根可以通过消去四次项，并解产生的三次方程,

双二次方程

四次方程式中若 a3 和 a1 均为 0 者有下列型态：

因此它是一个双二次方程式。解双二次方程式非常容易，只要设 <math>z=x^2</math> ，我们的方程式便成为：

这是一个简单的二次方程式，其根可用二次方程式的求根公式来解：

<math>z={{-a_2pmsqrt{a_2^2-4a_0a_4}} over },!</math>

当我们求得 z 的值以後，便可以从中得到 x 的值：

若任何一个 z 的值为负数或复数，那麼一些 x 的值便是复数。

一般情况,沿着费拉里的路线

To begin, the quartic must first be converted to a depressed quartic.

转变成减少次数的四次方程

Let

<math> A x^4 + B x^3 + C x^2 + D x + E = 0 qquadqquad(1')</math>

be the general quartic equation which it is desired to solve. Divide both sides by A,

The first step should be to eliminate the x3 term. To do this, change variables from x to u, such that

<math> x = u - {B over 4 A} </math>.

Then

<math> left( u - {B over 4 A}

ight)^4 + {B over A} left( u - {B over 4 A}

ight)^3 + {C over A} left( u - {B over 4 A}

ight)^2 + {D over A} left( u - {B over 4 A}

ight) + {E over A} = 0. </math>

Expanding the powers of the binomials produces

<math> left( u^4 - {B over A} u^3 + {6 u^2 B^2 over 16 A^2} - {4 u B^3 over 64 A^3} + {B^4 over 256 A^4}

ight)

+ {B over A} left( u^3 - {3 u^2 B over 4 A} + {3 u B^2 over 16 A^2} - {B^3 over 64 A^3}

ight) + {C over A} left( u^2 - {u B over 2 A} + {B^2 over 16 A^2}

ight) + {D over A} left( u - {B over 4 A}

ight) + {E over A}. </math> Collecting the same powers of u yields

<math> u^4 + left( {-3 B^2 over 8 A^2} + {C over A}

ight) u^2 + left( {B^3 over 8 A^3} - {B C over 2 A^2} + {D over A}

ight) u + left( {-3 B^4 over 256 A^4} + {C B^2 over 16 A^3} - {B D over 4 A^2} + {E over A}

ight) = 0. </math>

Now rename the coefficients of u. Let

<math> alpha = {-3 B^2 over 8 A^2} + {C over A}, </math>

<math> gamma = {-3 B^4 over 256 A^4} + {C B^2 over 16 A^3} - {B D over 4 A^2} + {E over A}. </math>

The resulting equation is

<math> u^4 + alpha u^2 + eta u + gamma = 0 qquad qquad (1) </math>

which is a depressed quartic equation.

If <math>eta=0</math> then we have a Biquadratic equation, which (as explained above) is easily solved; using reverse substitution we can find our values for <math>x</math>.

费拉里的解法

Otherwise, the depressed quartic can be solved by means of a method discovered by Ferrari. Once the depressed quartic has been obtained, the next step is to add the valid identity

<math> (u^2 + alpha)^2 - u^4 - 2 alpha u^2 = alpha^2 </math>

to equation (1), yielding

<math> (u^2 + alpha)^2 + eta u + gamma = alpha u^2 + alpha^2. qquad qquad (2) </math>

The effect has been to fold up the u4 term into a perfect square: (u2 + α)2. The second term, α u2 did not disappear, but its sign has changed and it has been moved to the right side.

The next step is to insert a variable y into the perfect square on the left side of equation (2), and a corresponding 2y into the coefficient of u2 in the right side. To accomplish these insertions, the following valid formulas will be added to equation (2),

<math>

egin (u^2+alpha+y)^2-(u^2+alpha)^2 & = & 2y(u^2+alpha)+ y^2 \ & = & 2yu^2+2yalpha+y^2, end </math>

and

<math> 0 = (alpha + 2 y) u^2 - 2 y u^2 - alpha u^2 </math>

These two formulas, added together, produce

<math> (u^2 + alpha + y)^2 - (u^2 + alpha)^2 = (alpha + 2 y) u^2 - alpha u^2 + 2 y alpha + y^2 qquad qquad (y-hbox) </math>

which added to equation (2) produces

<math> (u^2 + alpha + y)^2 + eta u + gamma = (alpha + 2 y) u^2 + (2 y alpha + y^2 + alpha^2). </math>

This is equivalent to

<math> (u^2 + alpha + y)^2 = (alpha + 2 y) u^2 - eta u + (y^2 + 2 y alpha + alpha^2 - gamma). qquad qquad (3) </math>

The objective now is to choose a value for y such that the right side of equation (3) becomes a perfect square. This can be done by letting the discriminant of the quadratic function become zero. To explain this, first expand a perfect square so that it equals a quadratic function:

The quadratic function on the right side has three coefficients. It can be verified that squaring the second coefficient and then subtracting four times the product of the first and third coefficients yields zero:

Therefore to make the right side of equation (3) into a perfect square, the following equation must be solved:

<math> (-eta)^2 - 4 (2 y + alpha) (y^2 + 2 y alpha + alpha^2 - gamma) = 0. </math>

Multiply the binomial with the polynomial,

<math> eta^2 - 4 (2 y^3 + 5 alpha y^2 + (4 alpha^2 - 2 gamma) y + (alpha^3 - alpha gamma)) = 0 </math>

Divide both sides by −4, and move the −β2/4 to the right,

<math> 2 y^3

+ 5 alpha y^2 + ( 4 alpha^2 - 2 gamma ) y + left( alpha^3 - alpha gamma - {eta^2 over 4}

ight) = 0 qquad qquad </math> This is a cubic equation for y. Divide both sides by 2,

<math> y^3 + {5 over 2} alpha y^2 + (2 alpha^2 - gamma) y + left( {alpha^3 over 2} - {alpha gamma over 2} - {eta^2 over 8}

ight) = 0. qquad qquad (4) </math>

转化嵌套的三次方程为减少次数的三次方程

Equation (4) is a cubic equation nested within the quartic equation. It must be solved in order to solve the quartic. To solve the cubic, first transform it into a depressed cubic by means of the substitution

<math> y = v - {5 over 6} alpha. </math>

Equation (4) becomes

<math> left( v - {5 over 6} alpha

ight)^3 + {5 over 2} alpha left( v - {5 over 6} alpha

ight)^2 + (2 alpha^2 - gamma) left( v - {5 over 6} alpha

ight) + left( {alpha^3 over 2} - {alpha gamma over 2} - {eta^2 over 8}

ight) = 0. </math>

Expand the powers of the binomials,

<math> left( v^3 - {5 over 2} alpha v^2 + {25 over 12} alpha^2 v - {125 over 216} alpha^3

ight) + {5 over 2} alpha left( v^2 - {5 over 3} alpha v + {25 over 36} alpha^2

ight) + (2 alpha^2 - gamma) v - {5 over 6} alpha (2 alpha^2 - gamma ) + left( {alpha^3 over 2} - {alpha gamma over 2} - {eta^2 over 8}

ight) = 0. </math>

Distribute, collect like powers of v, and cancel out the pair of v2 terms,

<math> v^3 + left( - {alpha^2 over 12} - gamma

ight) v + left( - {alpha^3 over 108} + {alpha gamma over 3} - {eta^2 over 8}

ight) = 0. </math>

This is a depressed cubic equation.

Relabel its coefficients,

<math> P = - {alpha^2 over 12} - gamma, </math>

<math> Q = - {alpha^3 over 108} + {alpha gamma over 3} - {eta^2 over 8}. </math>

The depressed cubic now is

<math> v^3 + P v + Q = 0. qquad qquad (5)</math>

解嵌套的减少次数的三次方程

The solutions (any solution will do, so pick any of the three complex roots) of equation (5) are

let <math>U=sqrt[3]{{Qover 2}pm sqrt{{Q^over 4}+{P^over 27}}}</math>

(taken from Cubic equation)

<math>v = {Pover 3U} - U</math>

therefore the solution of the original nested cubic is

<math>y = - {5 over 6} alpha + {Pover 3U} - U qquad qquad (6)</math>

Remember 1: <math>P=0 Longleftarrow {Qover 2} + sqrt{{Q^over 4}+{P^over 27}}=0</math>

Remember 2: <math>lim_{Po 0}{P over sqrt[3]{{Qover 2} + sqrt{{Q^over 4}+{P^over 27}}}}=0</math>

配成完全平方项

With the value for y given by equation (6), it is now known that the right side of equation (3) is a perfect square of the form

<math>(s^2)u^2+(2st)u+(t^2) = left(left(sqrt{(s^2)}

ight)u + {(2st) over 2sqrt{(s^2)}}

ight)^2</math>

This is correct for both signs of square root, as long as the same sign is taken for both square roots. A ± is redundant, as it would be absorbed by another ± a few equations further down this page.

so that it can be folded:

<math> (alpha + 2 y) u^2 + (- eta) u + (y^2 + 2 y alpha + alpha^2 - gamma ) = left( left(sqrt{(alpha + 2y)}

ight)u + {(-eta) over 2sqrt{(alpha + 2 y)}}

ight)^2</math>.

Note: If β ≠ 0 then α + 2y ≠ 0. If β = 0 then this would be a biquadratic equation, which we solved earlier.

Therefore equation (3) becomes

<math>(u^2 + alpha + y)^2 = left( left(sqrt{alpha + 2 y}

ight)u - {eta over 2sqrt{alpha + 2 y}}

ight)^2 qquadqquad (7)</math>.

Equation (7) has a pair of folded perfect squares, one on each side of the equation. The two perfect squares balance each other.

If two squares are equal, then the sides of the two squares are also equal, as shown by:

<math>(u^2 + alpha + y) = pmleft( left(sqrt{alpha + 2 y}

ight)u - {eta over 2sqrt{alpha + 2 y}}

ight) qquadqquad (7')</math>.

Collecting like powers of u produces

<math>u^2 + left(mp_s sqrt{alpha + 2 y}

ight)u + left( alpha + y pm_s {eta over 2sqrt{alpha + 2 y}}

ight) = 0 qquadqquad (8)</math>.

Note: The subscript s of <math>pm_s</math> and <math>mp_s</math> is to note that they are dependent.

Equation (8) is a quadratic equation for u. Its solution is

<math>u={pm_ssqrt{alpha + 2 y} pm_t sqrt{(alpha + 2y) - 4(alpha + y pm_s {eta over 2sqrt{alpha + 2 y}})} over 2}.</math>

Simplifying, one gets

<math>u={pm_ssqrt{alpha + 2 y} pm_t sqrt{-left(3alpha + 2y pm_s {2eta over sqrt{alpha + 2 y}}

ight)} over 2}.</math>

This is the solution of the depressed quartic, therefore the solutions of the original quartic equation are

<math>x=-{B over 4A} + {pm_ssqrt{alpha + 2 y} pm_t sqrt{-left(3alpha + 2y pm_s {2eta over sqrt{alpha + 2 y}}

ight)} over 2}. qquadqquad (8')</math>

Remember: The two <math>pm_s</math> come from the same place in equation (7'), and should both have the same sign, while the sign of <math>pm_t</math> is independent.

费拉里方法的概要

Given the quartic equation

its solution can be found by means of the following calculations:

<math> alpha = - {3 B^2 over 8 A^2} + {C over A}, </math>

<math> gamma = {-3 B^4 over 256 A^4} + {C B^2 over 16 A^3} - {B D over 4 A^2} + {E over A}, </math>

if <math>eta=0</math> solve <math>u^4+alpha u^2 + gamma = 0</math> and substitute <math>x=u-{Bover 4A}</math> finding the roots

<math>x=-{Bover 4A}pm_ssqrt{-alphapm_tsqrt{alpha^2-4gamma}over 2},qquadeta=0</math>.

<math> P = - {alpha^2 over 12} - gamma, </math>

<math> Q = - {alpha^3 over 108} + {alpha gamma over 3} - {eta^2 over 8}, </math>

<math> R = {Qover 2} pm sqrt{{Q^over 4}+{P^over 27}}</math>, (either sign of the square root will do)

<math> U = sqrt[3]</math>, (there are 3 complex roots, any one of them will do)

<math> y = - {5 over 6} alpha -U + eginU=0 &o 0\U

e 0 &o {Pover 3U}end, </math>

<math> x = - {B over 4 A} + { pm_s sqrt{ alpha + 2 y} pm_t sqrt{-left(3alpha + 2 y pm_s {2etaoversqrt{alpha +2y}}

ight) }over 2 }.</math>

The two ±s must have the same sign, the ±t is independent. To get all roots, find x for ±s,±t = +,+ and for +,− and for −,+ and for −,−. Double roots will be given twice, triple roots 3 times and quadruple roots would be given 4 times (although then β = 0, which is a special case). The order of the roots depends on which cubic root U one chose. (see note for (8) vis-à-vis (8'))

Quod Erat Faciendum.

There are other methods of solving the quartic equations, perhaps more optimal. Ferrari was the first to discover one of these labyrinthine solutions. The equation which he solved was

which was already in depressed form. It has a pair of solutions which can be found with the set of formulas shown above.

Obtaining alternative solutions the hard way

It could happen that one only obtained one solution through the seven formulae above, because one doesn't like trying all four sign patterns to get all four solutions, and the solution one obtained is complex. It may also be the case that one is only looking for a real solution. Let x1 denote the complex solution. If all the original coefficients A, B, C, D and E are real -- which should be the case when one desires only real solutions -- then there is another complex solution x2 which is the complex conjugate of x1. If the other two roots are denoted as x3 and x4 then the quartic equation can be expressed as

but this quartic equation is equivalent to the product of two quadratic equations:

<math> (x - x_1) (x - x_2) = 0 qquad qquad (9) </math>

and

<math> (x - x_3) (x - x_4) = 0. qquad qquad (10) </math>

Since

then

<math>

egin (x-x_1)(x-x_2)&=&x^2-(x_1+x_1^star)x+x_1x_1^starqquadqquadqquadquad \ &=&x^2-2,mathrm(x_1)x+[mathrm(x_1)]^2+[mathrm(x_1)]^2. end </math>

Let

<math> a = - 2 , mathrm(x_1), </math>

<math> b = [ mathrm( x_1) ]^2 + [ mathrm(x_1) ]^2 </math>

so that equation (9) becomes

<math> x^2 + a x + b = 0. qquad qquad (11) </math>

Also let there be (unknown) variables w and v such that equation (10) becomes

<math> x^2 + w x + v = 0. qquad qquad (12) </math>

Multiplying equations (11) and (12) produces

<math> x^4 + (a + w) x^3 + (b + w a + v) x^2 + (w b + v a) x + v b = 0. qquad qquad (13) </math>

Comparing equation (13) to the original quartic equation, it can be seen that

and

Therefore

<math> w = {B over A} - a = {B over A} + 2 mathrm(x_1), </math>

<math> v = {E over A b} = {E over A left(

[ mathrm(x_1) ]^2 + [ mathrm(x_1) ]^2

ight) }. </math> Equation (12) can be solved for x yielding

One of these two solutions should be the desired real solution.

Alternative methods

Reduction to a biquadratic

We may find the roots of

<math>x^4 + cx^2 + dx + e = 0 qquadqquad (1)</math> (see Quartic equation#Converting to a depressed quartic)

by converting it to a biquadratic equation by means of a Tschirnhaus transformation.

If <math>y = x^2 + px + q</math>, we may set <math>p</math> to be a root of

<math>dp^3 + (4e-c^2)p^2 - 2cdp-d^2 = 0</math> (see Cubic equation)

and

<math>q</math> to be <math>cover 2</math>.

This transforms the equation to

which is biquadratic and can be solved using square roots. Solving for <math>x</math> in terms of <math>y</math> entails solving a quadratic equation, also via square roots.

Hence we have a solution in terms of square roots and a root of a cubic polynomial.

This can result in four <math>y</math>'s and consequently in eight <math>x</math>'s; the four roots of (1) can then be determined by trial and error.

加罗华的理论和因式分解

The symmetric group S4 on four elements has the Klein four-group as a normal subgroup. This suggests using a resolvent whose roots may be variously described as a discrete Fourier transform or a Hadamard matrix transform of the roots. Suppose ri for i from 0 to 3 are roots of

If we now set

then since the transformation is an involution we may express the roots in terms of the four si in exactly the same way. Since we know the value s0 = -b, we really only need the values for s1, s2 and s3. These we may find by expanding out the polynomial

which if we make the simplifying assumption that b=0, is equal to

This polynomial is of degree six, but only of degree three in z2, and so the corresponding equation is solvable. By trial we can determine which three roots are the correct ones, and hence find the solutions of the quartic.

We can remove any requirement for trial by using a root of the same resolvent polynomial for factoring; if w is any root of (3), and if <math>F_1 = ^+wx+1/2,^+1/2,c-1/2,{frac {^w}}-1/2,{frac {^}}-{frac {c^}}+2,{frac }</math> <math>F_2 = ^-wx+1/2,^+1/2,c+1/2,{frac {^}}+{frac {c^}}-2,{frac }+1/2,{frac {^w}}</math> then

We therefore can solve the quartic by solving for w and then solving for the roots of the two factors using the quadratic formula.