Method of Tschirnhaus

The method of Tschirnhaus , imagined and developed by Ehrenfried Walther von Tschirnhaus, is a general method of resolution of the polynomial equations.

This method tries to bring back the equation which one wants to solve with other equations of less low degrees. This method fails in an unquestionable way for the equations of degrees equal to or higher than five which have a Groupe of nonresolvable Welshman.

Principle of the method

Let us consider an equation of degree N:

$\ qquad a_n x^n + a_ {N - 1} x^ {N - 1} + \ cdots + a_1 X + a_0 = 0$

The principle of the method consists in making a change of variable while posing:

$\ qquad there = b_ {n-1} x^ {n-1} +b_ {N2} x^ {N2} + \ cdots+b_1 X + b_0$

A transformation of this type names transformation of Tschirnhaus .

By eliminating X between this relation and the equation to be solved, one obtains an equation of degree N and unknown factor there whose coefficients depend on $b_ {n-1}, b_ {N2}, b_ {n-3}, \ ldots, b_1, b_0 \,$ . One then will try to determine $b_ {n-1}, b_ {N2}, b_ {n-3}, \ ldots, b_1, b_0 \,$ in order to obtain an equation in there form:

$\ qquad y^n - C = 0$

For that, in the equation in there, one poses equal to 0, all the coefficients of the students' rag processions of degree 1 with n-1. One thus obtains a system of n-1 unknown equations to N $b_ {n-1}, b_ {N2}, b_ {n-3}, \ ldots, b_1, b_0 \,$ . These values, once obtained, are deferred in the relation:

$\ qquad there = b_ {n-1} x^ {n-1} +b_ {N2} x^ {N2} + \ cdots+b_1 X + b_0$

Where will successively take there for value one of N roots of C.

We were thus reduced to the resolution of N equations in X of n-1 degree. We can thus renew the operation until obtaining equations of sufficiently low degree to be able to solve them.

Application to the resolution of the cubic equations

We will expose the method on the following example:

$x^3+x-2=0 ~$

Let us pose:

$there = ax^2+bx+c \ qquad (*) ~$

The two preceding equations are put in the form:

$\ left \ {\ begin {matrix} x^3=2-x \ \ y-c-bx=ax^2 \ end {matrix} \ right.$

We must eliminate X between these two equations. For that, we replace the first equation by the product member with member of these two equations. after simplification, we obtain:

$\ left \ {\ begin {matrix} bx^2=ax+xy-cx-2a \ \ y-c-bx=ax^2 \ end {matrix} \ right.$

This way of proceeding makes it possible to decrease the degree of the one of the equations compared to X . We thus will reiterate the process until X disappeared from the one of the equations. In addition, as we make products member with member, we are likely to introduce parasitic solutions. It will be thus necessary to us to the end of the resolution to check that all the found solutions check the equation well to be solved.

After a new product member with member, we obtain:

$\ left \ {\ begin {matrix} a^2x+axy-acx+b^2x=2a^2+by-bc \ \ y-c-bx=ax^2 \ end {matrix} \ right.$

After a new product member with member by replacing this time the second equation, we obtain:

$\ left \ {\ begin {matrix} a^2x+axy-acx+b^2x=2a^2+by-bc \ \ a^2y+ay^2-2acy+b^2y-a^2c+ac^2-b^2c=2a^3x+2abxy-2abcx+a^2x+b^3x \ end {matrix} \ right.$

A last product member with member gives us after reduction of the similar terms and simplification by a²x:

$y^3+ (2a-3c) y^2+ (a^2-6ab-4ac+3c^2+b^2) y=4a^3-6abc+2a^2b+2b^3+a^2c-2ac^2+c^3+b^2c \ qquad (**) ~$

We must now determine has, B, C so that:

$\ left \ {\ begin {matrix} 2a-3c=0 \ \ a^2-6ab-4ac+3c^2+b^2=0 \ end {matrix} \ right.$

By drawing C from the first equation and while deferring in the second equation, we obtain:

$\ left \ {\ begin {matrix} 2a-3c=0 \ \ a^2+18ab-3b^2=0 \ end {matrix} \ right.$

We see whereas a/b report/ratio is root of the equation:

$X^2+18X-3=0 ~$

One of the roots of this equation being:

$X=2 \ sqrt {21} - 9= \ frac {6 \ sqrt {21} - 27} {3} ~$

One can deduce some for has, B, C the choice of the following values:

$\ left \ {\ begin {matrix} a=6 \ sqrt {21} - 27 \ \ b=3 \ \ c=4 \ sqrt {21} - 18 \ end {matrix} \ right.$

By deferring these values on the one hand in (*), one obtains:

$there = (6 \ sqrt {21} - 27) x^2+3x+4 \ sqrt {21} - 18 \ qquad (***) ~$

And in addition in (**), one obtains:

$y^3=73920 \ sqrt {21} - 338688 ~$

From where one deduces the three possible values from there:

$y_1=10 \ sqrt {21} - 42 ~$

$y_2= (10 \ sqrt {21} - 42) e^ {2i \ pi/3} = (10 \ sqrt {21} - 42) (- \ frac {1} {2} +i \ frac {\ sqrt {3}} {2}) = 21-5 \ sqrt {21} + (15 \ sqrt {7} - 21 \ sqrt {3}) I ~$

$y_3= (10 \ sqrt {21} - 42) e^ {- 2i \ pi/3} = (10 \ sqrt {21} - 42) (- \ frac {1} {2} - I \ frac {\ sqrt {3}} {2}) = 21-5 \ sqrt {21} - (15 \ sqrt {7} - 21 \ sqrt {3}) i~$

It is enough for us to defer these three values of there in (***) successively obtaining the three quadratic equations following:

$(6 \ sqrt {21} - 27) x^2+3x+4 \ sqrt {21} - 18 = 10 \ sqrt {21} - 42 ~$

$(6 \ sqrt {21} - 27) x^2+3x+4 \ sqrt {21} - 18 = 21-5 \ sqrt {21} + (15 \ sqrt {7} - 21 \ sqrt {3}) I ~$

$(6 \ sqrt {21} - 27) x^2+3x+4 \ sqrt {21} - 18 = 21-5 \ sqrt {21} - (15 \ sqrt {7} - 21 \ sqrt {3}) I ~$

Who are simplified in the form:

$(2 \ sqrt {21} - 9) x^2+x+8-2 \ sqrt {21} = 0 ~$

$(2 \ sqrt {21} - 9) x^2+x+3 \ sqrt {21} - 13 (5 \ sqrt {7} - 7 \ sqrt {3}) I = 0 ~$

$(2 \ sqrt {21} - 9) x^2+x+3 \ sqrt {21} - 13+ (5 \ sqrt {7} - 7 \ sqrt {3}) I = 0 ~$

It does not remain us any more which has to solve these three equations to deduce the possible values from them from X. The three discriminants of these quadratic equations are respectively:

$\ triangle_1 = 625-136 \ sqrt {21} = (17-4 \ sqrt {21}) ^2 ~$

$\ triangle_2 = 212 \ sqrt {21} - 971 - 4 (87 \ sqrt {7} - 133 \ sqrt {3}) I = (10-2 \ sqrt {21} +14i \ sqrt {3} - 9i \ sqrt {7}) ^2 ~$

$\ triangle_3 = 212 \ sqrt {21} - 971 + 4 (87 \ sqrt {7} - 133 \ sqrt {3}) I = (10-2 \ sqrt {21} - 14i \ sqrt {3} +9i \ sqrt {7}) ^2 ~$

One from of respectively deduced the six possible values for X:

$\ frac {- 12-2 \ sqrt {21}} {3}, 1, \ frac {- 1+i \ sqrt {7}} {2}, \ frac {- 4 \ sqrt {21} - 15-3i \ sqrt {7}} {6}, \ frac {- 1-i \ sqrt {7}} {2}, \ frac {- 4 \ sqrt {21} - 15+3i \ sqrt {7}} {6} ~$

As we made products member with member at the beginning, we are likely to have introduced parasitic roots. We must thus check that the values obtained for X check the equation well to be solved. We note that only three of the six values obtained are well solution of the equation. These values are:

$x_1 = 1 ~$

$x_2 = \ frac {- 1+i \ sqrt {7}} {2} ~$

$x_3 = \ frac {- 1-i \ sqrt {7}} {2} ~$

Special method for the equations of the fourth degree

Let us consider the general equation of the fourth degree following: $\ qquad a_4 x^4 + a_3 x^3 + a_2 x^2 + a_1 X + a_0 = 0$

While dividing by $a_4 \,$ and while posing

$\ qquad X = Z - \ frac {a_3} {4a_4}$

one brings back oneself to an equation of the form:

$\ qquad z^4 + C z^2 + D z+ E = 0$

Let us consider the following transformation of Tschirnhaus:

$y = z^2 + pz + \ frac {C} {2} ~$

By eliminating Z between the two preceding relations, we obtain the equation of the fourth degree in there following:

$y^4 + (cp^2- \ frac {c^2} {2} +3dp+2e) y^2 + (dp^3+4ep^2-c^2p^2-2cdp-d^2) there + ep^4- \ frac {cdp^3} {2} + \ frac {c^3p^2} {4} - cep^2+ \ frac {c^2dp} {4} - dep+ \ frac {c^4} {16} - \ frac {c^2e} {2} + \ frac {cd^2} {2} +e^2 = 0 ~$

We see whereas we can obtain on this level an equation biquadratic of the fourth degree if p checks the relation:

$dp^3+4ep^2-c^2p^2-2cdp-d^2 = 0~$

I.e. if p is solution of the cubic equation:

$dx^3+ (4th-c^2) x^2-2cdx-d^2 = 0~$

We were thus reduced to the solution of a cubic equation.

Let us take an example to study in a more precise way the method.

That is to say to solve the equation:

$x^4 + 4x^3 + 3x^2 - 8x - 10 = 0 ~$

Let us pose:

$X = Z - 1 \ qquad (*) ~$

While replacing in the equation, one obtains:

$z^4 - 3z^2 - 6z - 2 = 0 ~$

Let us consider the transformation of Tschirnhaus:

$y = z^2 + pz - \ frac {3} {2} ~$

By eliminating Z by products successive member with member (see preceding paragraph) between the two preceding relations, we obtain:

$y^4 - (3p^2+18p+ \ frac {17} {2}) y^2 - (6p^3+17p^2+36p+36) there - (2p^4+9p^3+ \ frac {51} {4} p^2+ \ frac {51} {2} p+ \ frac {575} {16}) =0 ~$

If it is wanted that this equation is an equation biquadratic of the fourth degree, we see that we must choose p among the roots of the equation:

$6x^3+17x^2+36x+36=0 ~$

This equation admits for obvious root:

$x=- \ frac {3} {2} ~$

We will thus choose:

$p=- \ frac {3} {2} ~$

The transformation of Tschirnhaus considered is thus:

$y = z^2 - \ frac {3} {2} Z - \ frac {3} {2} ~$

And by elimination of Z with the equation:

$z^4 - 3z^2 - 6z - 2 = 0 ~$

One obtains:

$8y^4+94y^2-49=0 ~$

While posing:

$X = y^2 ~$

One is reduced to the quadratic equation:

$8X^2+94X-49=0 ~$

Who has as roots:

$X_1 = \ frac {1} {2} ~$

$X_2 = - \ frac {49} {4} ~$

From where one deduces the four values from there following:

$y_1 = \ frac {1} {\ sqrt {2}} ~$

$y_2 = - \ frac {1} {\ sqrt {2}} ~$

$y_3 = \ frac {7i} {2} ~$

$y_4 = - \ frac {7i} {2} ~$

These four values of deferred in the transformation of Tschirnhaus considered gives us four quadratic equations there:

$z^2 - \ frac {3} {2} Z - \ frac {3} {2} = \ frac {1} {\ sqrt {2}} ~$

$z^2 - \ frac {3} {2} Z - \ frac {3} {2} = \ frac {7i} {2} ~$

$z^2 - \ frac {3} {2} Z - \ frac {3} {2} = - \ frac {7i} {2} ~$

Who are simplified respectively in the form:

$2z^2 - 3z - 3 - \ sqrt {2} = 0 ~$

$2z^2 - 3z - 3 + \ sqrt {2} = 0 ~$

$2z^2 - 3z - 3 - 7i = 0 ~$

$2z^2 - 3z - 3 + 7i = 0 ~$

These four equations have respectively as discriminate:

$\ triangle_1 = 33 + 8 \ sqrt {2} = (1+4 \ sqrt {2}) ^2 ~$

$\ triangle_2 = 33 - 8 \ sqrt {2} = (1-4 \ sqrt {2}) ^2 ~$

$\ triangle_3 = 33 + 56i = (7+4i) ^2 ~$

$\ triangle_4 = 33 - 56i = (7-4i) ^2 ~$

Each of the four quadratic equations providing two roots, one from of deduced eight possible values for Z:

$1+ \ sqrt {2}, \ frac {1} {2} - \ sqrt {2}, 1 \ sqrt {2}, \ frac {1} {2} + \ sqrt {2}, \ frac {5} {2} +i, - 1-i, \ frac {5} {2} - I, - 1+i ~$

Only four values:

$z_1 = 1+ \ sqrt {2} ~$

$z_2 = 1 \ sqrt {2} ~$

$z_3 = -1-i ~$

$z_4 = -1+i ~$

Check the equation:

$z^4 - 3z^2 - 6z - 2 = 0 ~$

The other values are parasitic roots appeared at the time of the products member to member carried out to eliminate Z higher.

While carrying the four valid values of Z in (*), one obtains:

$x_1 = \ sqrt {2} ~$

$x_2 = - \ sqrt {2} ~$

$x_3 = -2-i ~$

$x_4 = -2+i ~$

Who are the four roots of the equation which one had been given to solve.

Notice historical

This method is the first general method of resolution of the equations to be published. Its publication goes up with 1683.

Method of Tschirnhaus

Principle of the method

Application to the resolution of the cubic equations

Special method for the equations of the fourth degree

Notice historical

Other methods of solution of equations