# Math ASCII Notation Demo

Mathematical content on Apronus.com is presented in Math ASCII Notation which can be properly displayed by all Web browsers because it uses only the basic set of characters found on all keyboards and in all fonts.

The purpose of these pages is to demonstrate the power of the Math ASCII Notation. In principle, it can be used to write mathematical content of any complexity. In practice, its limits can be seen when trying to write complicated formulas (containing, for example, variables with many indexes or multiple integrals).

Despite its limitations the Math ASCII Notation has much expressive power, as can be seen from browsing through these pages.

Let F be a s-algebra in X.
Let f : X -> |R \ {-oo,oo} be F-measurable.
Prove that function f*f is F-measurable.
Take any a:-|R. If a<0, then {x: f^2(x) >= a} :- F.
If a>=0, then {x: f^2(x) > a} = {x: |f(x)| > sqrt(a)} =
= {x: f(x) > sqrt(a) or f(x) < -sqrt(a)} =
= {x: f(x) > sqrt(a)} u {x: f(x) < -sqrt(a)} :- F
Let F be a s-algebra in X.
Let f,g : X -> |R \ {-oo,oo} be F-measurable.
Prove that f*g is F-measurable by using a trick.
f*g = 1/4 * [ (f+g)^2 - (f-g)^2 ]
Use the facts that f+g, f-g are F-measurable,
and that if h:X->|R is measurable then so is h^2.
page 138 in 1st measure
Let F be a s-algebra in X.
Let f,g : X -> |R \ {-oo,oo} be F-measurable.
What are the two ways of proving that f*g is F-measurable?
WAY 1:
(1) F : |R^2 -> |R is continuous
(2) h : X -> |R , h(x) = F(f(x),g(x))
thesis: h is F-measurable
Let F(x,y)=x*y (page 138 in 1st measure)
WAY 2:
f*g = 1/4 * [ (f+g)^2 - (f-g)^2 ]
Let W c P(X). Let U c X.
Suppose that /\(a:-U) \/(B:-W) a:-B and B c U.
Can we conclude that U is W-open?
YES.
page 78,79 in gen top
f : [0,oo[ -> |R , f(x) = 1/(1+sqrt(x))
Find a diffable function F : [0,oo[ -> |R
such that /\x>=0 F'(x) = f(x).
F(x) = 2 * ( sqrt(x) - log(1+sqrt(x) )
page 116 in OLDTIMER
(1) 0 < a <= oo
(2) F,f : [0,a) -> |R
(3) /\0<x<a F'(x) = f(x)
(4) F and f are both continuous at 0
(5) F(0) = 0
What can we conclude from this?
F'(0) = f(0)
page 117 in OLDTIMER
Prove that |R is homeomorphic with the bounded open interval (- 1;1).
Do not use trigonometric functions.
f : |R -> (-1;1) , f(x) = x / ( |x|+1 )
f is 1-1, onto, continuously diffable
Let H be a s-algebra in X.
Let f : X -> |R \ {-oo,oo} be H-measurable.
Let F : |R -> |R be a Borel function.
Let g : X -> |R , g(x) = F(f(x)).
What can we conclude?
function g is H-measurable
Take any a:-|R. Let G={x:-|R : F(x)>a}. G is Borel in |R.
Function f is H-measurable, hence f-1(G) :- H.
{x:-X : g(x)>a} = {x:-X : F(f(x))>a} = {x:-X : f(x) :- G} = f- 1(G) :- H
In the context of measure theory, what does it mean that a function X->|R* is simple?
(0) f : X -> |R* is simple
iff
(1) /\x:-X f(x):-|R
(2) f(X) is finite
===================
f(X) = { a[1], a[2], ..., a[n] } c |R
(1) f : X -> |R*.
(2) n:-|N
(3) a[1], ..., a[n] :- |R
(4) A[1], ..., A[n] c X , none of them is empty
(5) X = \\*//(k=1 to n) A[k]
(6) /\x:-X f(x) = +(k=1 to n) a[k]*1(A[k])(x)
f(X) = { a[1] , ... , a[n] } c |R
f is a simple function
page 105 in 2nd measure
page 140 in 1st measure