# 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 (X,d) be a bounded metric space.
Let H be the set of all closed subsets of X except the empty set.
Define h:HxH->R by h(A,B) = sup{|d(x,A) - d(x,B)| : x:-X}.
(H,h) is the Hausdorff metric space.
Prove that d(a,b) = h({a},{b}).
page 179 in the palace notebook
Let (X,d) be a metric space. f:X->X.
Suppose that f satisfies:
/\x,y:-X [ x!=y ==> d(f(x),f(y)) < d(x,y) ].
Decide if we can conclude that
\/0<K<1 /\x,y:-X d(f(x),f(y)) <= K*d(x,y).
NO.
Consider [2,oo) with the Euclidean metric.
Put f(x) = x + 1/x, for x>=2. Notice that f(x)>=2.
To show that f satisfies the desired condition,
use the fact that f is increasing.
Since f has no fixed point, it cannot be a contraction.
Prove that there exists a continuous function f:[0,1]->R that is neither increasing nor decreasing on any subinterval of [0,1].
Consider the set of all continuous f:[0,1]->R. Equip it with the sup metric. It's a complete metric space. Let I be a subinterval of [0,1]. Let A(I) denote the set of all continuous f:[0,1]->R that are increasing on I. (x<y==>f(x)<=f(y)). A(I) is closed. Let B(I) be the set of all continuous f:[0,1]->R that are decreasing on I. B(I) is closed. Let K = A(I) u B(I). K is closed and Int(K) = 0. Let {I[n]} be a sequenc
Let z be a complex number such that |z|<1. Let m be a natural number.
Express 1 / (1-z)^m as the sum of a power series.
1 / (1-z)^m = +(n=0 to oo) [ z^n * (n+m-1)! / (n!*(m-1)!) ]
page 190 in the palace notebook
Let z be a complex number.
z * conjugate(z) = ?
z * conjugate(z) = |z|*|z|
Let z be a complex number other than zero.
conjugate(z) / |z|^2 = ?
conjugate(z) / |z|^2 = 1/z
Let z be a complex number other than zero.
z / |z|^2 = ?
z / |z|^2 = 1/conjugate(z)
Let h : |C- > |C be defined h(z)=|z|*|z|.
Where is h diffable?
Only at z = 0.
Show a complex function that is diffable only at one point.
f(z)=|z|^2
f(x,y)=x^2+y^2
f is diffable only at zero
Consider a complex function defined on an open subset of |C. Suppose that it is diffable at some point in that set. Does it have to be diffable in a neighborhood of the point?
NO.
There are complex functions that are diffable only at one point.
1) z->|z|*|z| at 0
2) f(x,y)=xy at 0