Joseph Louis Lagrange 1736 1813 Born in Turin

Joseph Louis Lagrange (1736 -1813) • • Born in Turin at that time the capitol of Sardinia-Piemont as Giuseppe Lodovico Lagrangia 1774 Started correspondence with Euler and in 1775 sent him his results on the tautochrone containing his method of maxima and minima. 1755 was appointed professor of mathematics at the Royal Artillery School in Turin. 1756 he sent Euler results that he had obtained on applying the calculus of variations to mechanics. 1766 after having refused arrangements made by d’Alembert once and once before that Euler he accepts a post at the Berlin Academy of Science and became the successor of Euler as Director of Mathematics. He won prizes of the Académie des Sciences of Paris. He shared the 1772 prize on the three body problem with Euler, won the prize for 1774, another one on the motion of the moon, and he won the 1780 prize on perturbations of the orbits of comets by the planets. In 1770 he also presented his important work Réflexions sur la résolution algébrique des équations which made a fundamental investigation of why equations of degrees up to 4 could be solved by radicals 1787 he left Berlin to become a member of the Académie des Sciences in Paris, where he remained for the rest of his career, escaping the turmoil of the French Revolution and being decorated by Napoleon.

Reflections on the Algebraic Solution of Equations 1. Lagrange re-derives the solution of del Ferro-Cardano-Tartaglia. Terminology: • • The proposed equation is the original cubic equation The reduced equation is the equation of order six which is derived from the original equation. 2. He shows that vice versa with two assumptions one is lead to their method. Assumptions: 1. The roots of the reduced equation have linear expressions in terms of the roots of the proposed equation. 2. The reduced equation involves only multiples of third powers. 3. He gives a general approach for solving higher order equations by finding reduced equations such that 1. The roots of the proposed equation are rational functions of the roots of the reduced equation. (Linear with coefficients being roots of unity in the cases of n=3, 4) 2. The reduced equations are solvable. A reduced equation with these properties is called a Lagrange resolvent.

Reflections on the Algebraic Solution of Equations 1. 2. 3. 4. 5. 6. The cubic Start with the proposed equation x 3+nx+p=0 Set x=y+z, so y 3+z 3+p+(y+z)(3 yz+n)=0 Say (1) y 3+z 3+p=0 and (2) 3 yz+n=0 From (2): (3) z=-n/(3 y) Inserting (3) into (1) one obtains the reduced equation: (4) y 6+py 3 -n 3/27=0 From (4): 7. Taking the 3 rd root: 8. For any solution y: x=y+z=y-n/(3 y) Also get other 3 rd roots! Let 1, a, b be the 3 rd roots of unity and set q= p 2/4+n 3/27, then the six solutions for y are 9.

Expressing the roots of the reduced equation by the roots of the proposed equation 1. Start with the proposed equation x 3+mx 2+nx+p=0 and Call the roots of the equation a, b, c. (x-a)(x-b)(x-c)=x 3+mx 2+nx+p 2. Get rid of the quadratic term by substituting x=x’-m/3 to obtain: x’ 3+n’x+p’=0 n’=n-m 2/3, p’=p-mn/3+2 m 3/27 3. Get the reduced equation y 6+p’y 3 -n’ 3/27=0 Set: which gives three values for y: y=r, y=ar, y=br and three values for x’: x’=r-n’/(3 r) x’=ar-n’/(3 ar) x’=br-n’/(3 br) 4. Using x=x’-m/3 and setting s= n’/(3 r) one gets a= -m/3+r-s b= -am/3+ar-s/a c= -bm/3+br-s/b 5. Solving for r: r=(a+ab+bc)/3 6. With a 2=b get six solutions for y: y=(a+ab+a 2 c)/3 y=(a+ac+a 2 b)/3 y=(b+aa+a 2 c)/3 y=(b+ac+a 2 a)/3 y=(c+ab+a 2 a)/3 y=(c+aa+a 2 b)/3 Notice that these solutions are obtained from the first two by multiplying with a and a 2.

Finding and solving the reduced equation 1. Say one has the proposed equation x 3+mx 2+nx+p=0 with roots: a, b, c. 2. Suppose that a root of the reduced equation are given by Aa+Bb+Cc. 3. If the reduced equation only depends on the coefficients of the proposed equation, then since m =-(a+b+c) n=(ab+bc+ac) p=-(abc) then all permutations of the a, b, c will also give solutions. So there will be six roots and the reduced equation will have order 6. 4. If the reduced equation only has terms of order a multiple of 3, the if r is a root, so is ar and a 2 r. This yields C=a. A, B=a 2 A, a 3 A=A 5. Setting A=1, r=a+ab+a 2 c and s=a+ac+a 2 b we obtain the six solutions as: r, a 2 r and s, a 2 s 6. With these roots from (y-r)(y-a 2 r)= y 3 -r 3 and (y-s)(y-a 2 s)= y 3 -r 3 we obtain the reduced equation y 6 -(r 3+s 3)y 3+r 3 s 3=0 7. Use the equations 3 to get the coefficients of the reduced equation in terms of the roots of the proposed equation. 8. Since the resulting expressions are symmetric in the roots, one can express the coefficients of the reduced equation in terms of the coefficients of the proposed equation. 9. Find the equation for the roots of the proposed equation in terms of the roots of the reduced equation.