# 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.

Integral(0,pi,cos(2x),dx) = ???
Integral(0,pi,cos(2x),dx) = 0
hint: [ sin(2x)/2 ]' = cos(2x)
Integral(0,pi,sin(x)sin(x),dx) = ???
Integral(0,pi,sin(x)sin(x),dx) = pi/2
There are at least two ways:
(1) integrate by parts: page 87 in OLDTIMER
(2) use: 2sin(x)sin(x) = 1-cos(2x), and then Integral(0,pi,cos(2x),dx)=0
Integral(-1,1,sqrt(1-x^2),dx) = ???
Integral(-1,1,sqrt(1-x^2),dx) = pi/2
hint: x=cos(t), later use: Integral(0,pi,sin(x)sin(x),dx) = pi/2
Notice the significance of this integral.
It calculates half the area of the unit circle, which is pi.
Integral(0,pi,cos(x)cos(x),dx) = ???
Integral(0,pi,cos(x)cos(x),dx) = pi/2
Use:
(1) cos(x)cos(x) = 1 - sin(x)sin(x)
(2) Integral(0,pi,sin(x)sin(x),dx) = pi/2
Or you can integrate by parts: page 87 in OLDTIMER
Let F be an algerba in X.
Let f : X -> |R\{oo,-oo} be F-measurable.
What can we conclude about function [f] ?
([x] denotes the integer part of real number x)
g:|R->|R, g(x)=[x] is an increasing function,
hence gof:X->|R is measurable.
page 91,90 in 2nd measure
Let (X,d) be a metric space.
Let A be an uncountable subset of X satisfying:
(1) \/(a>0) /\(x,y:-A) x!=y => d(x,y)>=a
What can we conclude?
X does not contain a countable dense subset.
Every dense subset of X is uncountable.
Every countable subset of X is not dense.
page 193 in OLDTIMER
(1) f : |R x |R -> |R
(2) /\x,y:-|R f(y,x) = f(x,y)
(3) a,b :- |R
(4) /\x,y:-|R f(x,y) <= f(a,b)
Can we conclude that a=b ?
NO.
Let f(1,0)=f(0,1)=7 and otherwise: f(x,y)=0.
Let a=1, b=0.
(1) f : |R x |R -> |R
(2) /\x,y:-|R f(y,x) = f(x,y)
(3) a,b :- |R
(4) /\x,y:-|R (x,y)!=(a,b) ==> f(x,y) < f(a,b)
Can we conclude that a=b ?
YES.
Transform (4) into (5) /\x,y:-|R f(x,y) >= f(a,b) ==> (x,y)=(a,b).
Then by (2), f(b,a)=f(a,b), and f(b,a)>=f(a,b).
By (5), (b,a)=(a,b). Hence a=b.
Draw the set { (x,y):-|R^2 : x^2 + y^2 <= x and x^2 + y^2 <= y }.
It is the intersection of two closed disks.
(1) center (0 , 1/2) radius 1/2
(2) center (1/2 , 0) radius 1/2
Let (x,y):-|R^2 satisfy x^2 + y^2 = y and x>=0.
How can we express the distance of point between (x,y) and (0,0)?
There are two ways:
1) sqrt(y)
2) sin(a), where a = arctan(y/x) = sin( arg(x+iy) )
page 91 in OLDTIMER