# 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.

lim(x->0+) log(x^x) = ???
lim(x->0+) log(x^x) = 0
hint: Use L'Hospital's Rule for ln(x)/(1/x)
lim(x->0+) x^x = ???
lim(x->0+) x^x = 1
hint: lim(x->0+) log(x^x) = 0, and then exp is continuous at 0
lim((x,y)->(0,0)) x*y*log(x^2 + y^2) = ???
lim((x,y)->(0,0)) x*y*log(x^2 + y^2) = 0
hint: x=r*cos(a), y=r*sin(a)
+(n=1 to n=oo) 1/(n^2 + n) = ???
+(n=1 to n=oo) 1/(n^2 + n) = 1
hint: 1/(n^2 + n) = 1/n - 1/(n+1)
inf { M:-|R : /\x:-|R |sin(x)cos(x)|<M } = ???
inf { M:-|R : /\x:-|R |sin(x)cos(x)|<M } = 1/2
A c |R; f,g : A -> |R
Suppose that functions f and g are increasing.
Can we conclude that function f*g is increasing?
NO
A = {1,2}
f(1) = g(1) = -7
f(2) = g(2) = 0
A c |R; f,g : A -> |R
Suppose that functions f and g are increasing.
Suppose that /\x:-A f(x) >= 0.
Can we conclude that function f*g is increasing?
NO
A = {1,2}
f(1)=10, f(2)=20
g(1)=g(2)= (-1)
f(1) = 1; f(2) = 2; g(1) = -7; g(2) = -5
1<2, f(1)<f(2), g(1)<g(2), but f(1)g(1) > f(2)g(2)
A c |R; f,g : A -> |R
Suppose that functions f and g are increasing.
Suppose that /\x:-A f(x) >= 0 and g(x) >= 0.
Can we conclude that function f*g is increasing?
YES
A c |R; f,g : A -> |R
Suppose that functions f and g are increasing.
Suppose that /\x:-A f(x) >= 0 and g(x) >= 0.
What can we conclude?
We can conclude that function f*g is increasing.
Let A = { (x,y) : 0<=x, 0<=y, x^2+y^2 <= 1 }.
Let f : A -> |R, be f(0,0)=0 and f(x,y) = x*y*log(x^2+y^2).
Investigate max_f(A) and min_f(A).
First, check that f is continuous on compact A.
Hence both max and min exist.
max_f(A) = 0
min_f(A) = -1/(2e) = f( 1/sqrt(2e) , 1/sqrt(2e) )
hint: x=r*cos(a), y=r*sin(a)
g(a,r) = r^2*cos(a)*sin(a)*log(r^2)