Joseph Louis Lagrange (1736-1813)

1
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
  dAlembert 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.

2
Reflections on the Algebraic Solution of Equations

• 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.
• He shows that vice versa with two assumptions one
  is lead to their method. Assumptions
• The roots of the reduced equation have linear
  expressions in terms of the roots of the proposed
  equation.
• The reduced equation involves only multiples of
  third powers.
• He gives a general approach for solving higher
  order equations by finding reduced equations such
  that
• 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 n3,4)
• The reduced equations are solvable.
• A reduced equation with these properties is
  called a Lagrange resolvent.

3
Reflections on the Algebraic Solution of Equations

• The cubic
• Start with the proposed equation x3nxp0
• Set xyz, so y3z3p(yz)(3yzn)0
• Say (1) y3z3p0 and (2) 3yzn0
• From (2) (3) z-n/(3y)
• Inserting (3) into (1) one obtains the reduced
  equation (4) y6py3-n3/270
• From (4)

• Taking the 3rd root
• For any solution y xyzy-n/(3y)
• Also get other 3rd roots! Let 1,a,b be the
  3rd roots of unity and set q p2/4n3/27, then
  the six solutions for y are

4
Expressing the roots of the reduced equation by
the roots of the proposed equation

• Start with the proposed equation x3mx2nxp0
  and Call the roots of the equation a,b,c.
• (x-a)(x-b)(x-c)x3mx2nxp
• Get rid of the quadratic term by substituting
  xx-m/3 to obtain x3nxp0
• nn-m2/3,
• pp-mn/32m3/27
• Get the reduced equation y6py3-n3/270
• Set
• which gives three values for y yr, yar,
  ybr

• and three values for x
  xr-n/(3r) xar-n/(3ar)
  xbr-n/(3br)
• Using xx-m/3 and setting s n/(3r) one
  gets a -m/3r-s b -am/3ar-s/a c
  -bm/3br-s/b
• Solving for r r(aabbc)/3
• With a2b get six solutions for y
  y(aaba2c)/3 y(aaca2b)/3 y(baaa2c)/3
  y(baca2a)/3 y(caba2a)/3 y(caaa2b)/3
  Notice that these solutions are
  obtained from the first two by multiplying with a
  and a2.

5
Finding and solving the reduced equation

1. Setting A1, raaba2c and saaca2b
   we obtain the six solutions as
   r,ar,a2r and s,as,a2s
2. With these roots from (y-r)(y-ar)(y-a2r)
   y3-r3 and (y-s)(y-as)(y-a2s) y3-r3 we
   obtain the reduced equation y6-(r3s3)y3r3s30
3. Use the equations 3 to get the coefficients of
   the reduced equation in terms of the roots of the
   proposed equation.
4. 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.
5. Find the equation for the roots of the proposed
   equation in terms of the roots of the reduced
   equation.

1. Say one has the proposed equation x3mx2nxp0
   with roots a,b,c.
2. Suppose that a root of the reduced equation are
   given by AaBbCc.
3. If the reduced equation only depends on the
   coefficients of the proposed equation, then since
   m -(abc) n(abbcac) 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
   a2r. This yields CaA, Ba2A, a3AA