All entries for Thursday 04 October 2007
October 04, 2007
Mathematical Methods For Physicists – Lecture 1 – 02.10.07
Functions of More Than One Variable
e.g. Given region of xy plane 0≤y≤1, 0≤x≤1 - call this M (see diagram 1)
We say a ral function of real variables x,y is defined in M if a rule is given according to which exactly one real number is assigned to each point (x, y) of M is called the domain of definition of the function.
We write z = f(x, y)
Ex 1 z = x2y defined for whole xy plane – domain is xy plane.
Ex 2 z = √1-x2-y2 domain is all points (x, y) such that 1 – x2 – y2 ≥ 0
i.e. inside and on x2 + y2 = 1
Ex 3 z = √(1-x12-x22-…-xN2) domain is x12 + x22 + … + xN2 ≥ 1
Closed since inside boundary.
Viewing A Function
Think of z as a height about the xy plane (see diagram 2)
z is a function of two functions, each of which is a function of xy
z = h(f(x,y)g(x,y))
We say we have a composite function of U = f(x,y) and V = g(x,y).
Such that z = h(u, v)
u, v defined over xy plane.
We require to define z over a set N such that U, V Ç N
Ç in this context is used to denote 'contains', since the blog doesn't support the correct symbol.
Ex 1 z = (1+x2+y2)sin xy → composite function of U = 1+x2+y2
V = sin xy
Write z = UV
M à whole of xy plane
N could be xy such that 0 ≤ xy ≤π gives > 0
or π ≤ xy ≤ 2π gives < 0
z = √(1-x2sinx) composite of u = x2, v = sin x
z = √(1-UV)
only has domain for UV ≥ 1 i.e. only on one side of hyperbola UV = 1 and on it
(See diagram 3)
The Distance Between Two Points
In Cartesian coordinates the dsiance between (x1, y1) and (x2, y2) is given by
d = √((x1-x2)2 + (y1-y2)2)
In n dimension for 2 points (x1, x2, …, xN) and (y1, y2, …, yN)
d = √((x1-y1)2 + (x2-y2)2 + … + (xN-yN)2)
The δ Neighbourhood Of A Point P
The set of all point whose distance from P is <δ
e.g. in xy plane δ neighbourhood of a point (x0, y0) is all points within circle given by
(x-x0)2 + (y-y0)2 + δ2
The limits, A, of a function z = f(x, y) at P(x0, y0) UNINTELLIGIBLE NONESENSE!
We say z has a linit A at P to every UNINTELLIGIBLE NONESENSE!
Then exists a δ > 0.
Such that for all points in the d neighbourhood UNINTELLIGIBLE NONESENSE!
f has a limit, A, at P if f is sufficiently close to A. For all points sufficiently close to A UNINTELLIGIBLE NONESENSE!
f is continuous at (x0, y0) if
1) it is defined at this point
2) If, corresponding to an area ε > 0 there exists a δ > 0.
We have |f(x, y) - f(x0, y0)| < ε
A) Can say that f(x, y) is defined at (x0, y0) then it is continuous at that point if, and only if,
lim f(x, y) = f(x0, y0)
(x, y) à (x0, y0)
We say it is continuous in M.
Limits is f(x, y) à A and g(x, y) à B
k f(x,y) = kA
lim f(x,y) ± g(x,y) = A±B
(x, y) à (x0, y0) f(x,y)g(x,y) = AB
f(x,y)/g(x,y) = A/B g ¹ 0
Theorems Without Proof
1) A continuous function of a continuous function is continuous
2) If f(x,y) is continuous in a region R and if (x1, y1) and (x2, y2) are any two points of R then f(x, y) then in R every value between f(x, y) and f(x0, y0).
3) A function hat is continuous in a SOMETHING closed region R’, takes on a greatest value at least at one point (x0, y0) Ç R’
i.e. f(x0, y0) ≥ f(x, y) for all points (x, y) Ç R’ at least value at limit of the point (x, y) Ç R.