Joseph Lagrange

Who I'd like to meet:

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)[u'..1..y..1..(x)+u..1..y'..1..(x)+ u'..2..y..2..(x)+ u..2..y'..2..(x)] + Q(x)[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
and
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


since f(x) is a polynomial in x of degree 2n, so that f..(2n+2)..(x) = 0. And this expression is equal to
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 ....