I have dedicated my life to mathematics, and I spend my time in the study of such. My main contributions are to the theory of numbers, analysis, and analytic and celestial mechanics. Here is some of my work:
"Lagrange developed the calculus of variations which was later expanded by Weierstrass. Lagrange also established the theory of differential equations, and provided many new solutions and theorems in number theory, including Wilson's theorem. Lagrange's classic Théorie des fonctions analytiques laid some of the foundations of group theory, anticipating Galois. Lagrange also invented the method of solving differential equations known as variation of parameters." (quote from Scienceworld.wolfram.com )
I'm not really interested in meeting people. I just thought that I’d fill the great gap that was the result of my absence, lol. Hilarious, I know.
Lagrange's Group Theorem:
Also known as Lagrange's lemma. The most general form of Lagrange's theorem states that for a group G, a subgroup H of G , and a subgroup K of H,
(G: K) = (G: H)(H: K) , where the products are taken as cardinalities (thus the theorem holds even for infinite groups) and (G: H) denotes the subgroup index for the subgroup H of G. A frequently stated corollary (which follows from taking K = {e}, where e is the identity element) is that the order of G is equal to the product of the order of H and the subgroup index of H.
The corollary is easily proven in the case of G being a finite group, in which case the left cosets of H form a partition of G, thus giving the order of G as the number of blocks in the partition (which is (G: H)) multiplied by the number of elements in each partition (which is just the order of H).
For a finite group G, this corollary gives that the order of H must divide the order of G. Then, because the order of an element x of G is the order of the cyclic subgroup generated by x, we must have that the order of any element of G divides the order of G.
The converse of Lagrange's theorem is not, in general, true (Gallian 1993, 1994).
Source: Bray, Nicolas. "Lagrange's Group Theorem." From
MathWorld--A Wolfram Web Resource, created by Eric W. Weisstein. http://mathworld.wolfram.com
Lagrange Multiplier:
Lagrange multipliers, also called Lagrangian multipliers, can be used to find the extremum of a multivariate function f (x 1 , x 2 , x 3 ,..., x n ) subject to the constraint
g (x 1 , x 2 , x 3 ,..., x n ) = 0, where f and g are functions with continuous first partial derivatives on the open set containing the curve
g (x 1 , x 2 , x 3 ,..., x n ) = 0, and at any point on the curve (where is the gradient).
For an extremum of f to exist on g, the gradient of f must line up with the gradient of g. In the illustration above, f is shown in red, g in blue, and the intersection of and is indicated in light blue. The gradient is a horizontal vector (i.e., it has no z-component) that shows the direction that the function increases; for it is perpendicular to the curve, which is a straight line in this case. There is a red arrow representing the direction of the gradient of both f and g at the optimal point. If the two gradients are in the same direction, then one is a multiple (- l ) of the other, so
The two vectors are equal, so all of their components are as well, giving
for all k = 1,2,3, ...,n, where the constant l is called the Lagrange multiplier.
The extremum is then found by solving the n + 1 equations in n + 1 unknowns, which is done without inverting g, which is why Lagrange multipliers can be so useful.
For multiple constraints, g 1 = 0 , g 2 = 0 , ...
Source: Gluss, David and Weisstein, Eric W. "Lagrange Multiplier." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/LagrangeMultiplier.html
Variation of Parameters:
For a second order ODE of the form
1) y''+ P(x)y'+Q(x)y = f(x)
Let y 1 (x) and y 2 (x) be the solutions to the homogeneous differential equation
2) y''+P(x)y'+Q(x)y = 0
Then let y* be the solution of the nonhomogeneous ODE such that
3) y* = u 1 y 1 (x)+ u 2 y 2 (x)
4) y'* = u' 1 y 1 (x)+u 1 y' 1 (x)+ u' 2 y 2 (x)+ u 2 y' 2 (x)
5) y''* = u'' 1 y 1 (x)+ 2u' 1 y' 1 (x)+
u 1 y'' 1 (x)+
u'' 2 y 2 (x)+ 2u' 2 y' 2 (x)+
u 2 y'' 2 (x)
Imposing the restriction
6) u' 1 y 1 + u' 2 y 2 = 0
Plugging y* and its derivatives into the nonhomogeneous differential equation yields
7) u'' 1 y 1 (x)+ 2u' 1 y' 1 (x)+
u 1 y'' 1 (x)+
u'' 2 y 2 (x)+ 2u' 2 y' 2 (x)+
u 2 y'' 2 (x) +
P(x)<[i>u' 1 y 1 (x)+u 1 y' 1 (x)+ u' 2 y 2 (x)+ u 2 y' 2 (x)] + Q(x)<[i>u 1 y 1 (x)+ u 2 y 2 (x)] = f(x)
Simplifying and using equation six gives
8) u' 1 y' 1 + u' 2 y' 2 = f(x)
Which leaves two equations and two unknowns
9) u' 1 y' 1 + u' 2 y' 2 = f(x)
10) u' 1 y 1 + u' 2 y 2 = 0,
Source: Weisstein, Eric W. "Variation of Parameters." From MathWorld--A Wolfram Web Resource. http://mathworld.wolfram.com/VariationofParameters.html
The real number p 2 is irrational:
Proof:
Assume that p 2 where a and b are integers and b does not equal zero. Define
f(x)= x n (1-x) n /n!
and
G(x)= b n { p 2n f(x)- p 2n-2 f''(x)+...+(-1) n p 0 f (2n) (x)}
Where the superscript indicates derivatives. Now one claims that any derivative of f takes on integer values at both 0 and 1.
If both factors x n and (1-x) n are differentiated fewer than n the values of the corresponding terms is 0 whenever x = 0 or 1. If one factor is differentiated n or more times then the denominator n! is canceled out. Hence G(0) and G(1) are integers. Now
d / dx [G'(x)sin( p x)- p G(x)cos( p x)]={G''(x)+ p 2 G(x)}sin( p x)
= b n p 2n+2 f(x)sin( p x)
since f(x) is a polynomial in x of degree 2n, so that f (2n+2) (x) = 0 . And this expression is equal to
p 2 a n sin( p x) f(x)
Therefore
the integral of the above function is equal to G(0)+G(1)
Which is an integer. The integral is not zero. But
tends to zero as n tends to infinity. This is a contradiction.
Source: Stewart, Ian "Galois Theory." 2nd ed. New York: Chapman & Hall/CRC.
This is not an exact copy; I am responsible for errors.
MORE TO COME
Math books.
Math folk.