# 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 D be an open subset of |R^2.
Let f : D -> |R.
Suppose that the partial derivatives of f exist on D.
Let (a,b) be a point in D.
Suppose that fx(a,b)=fy(a,b)=0.
Does it mean that f has an extremum at (a,b)?
NO.
f(x,y)=x^2-y^2 at (0,0)
Let X be a set, and W c P(X).
Decide and prove which is true:
W* is the _________ collection of sets that generates W- convergence.
(1) largest
(2) smallest
W* is the largest collection of sets that generates W- convergence.
page 51 in gen top
Let X be a set, and W c P(X).
Prove that (W*)* c W*.
Use the fact that W-convergence implies W*-convergence.
page 52 in gen top
X, W c P(X)
What does it mean that W is a semi-ring?
(1) W is non-empty
(2) A,B:-W => AnB:-W
(3) A,B:-W => A\B is a finite disjoint union of sets from W
page 60 in 1st measure
Let W c P(X) be a semi-ring.
Does W have to contain the empty set?
Yes.
W is non-empty, hence A:-W.
Then A\A is a finite union of sets from W.
Hence W has to contain the empty set.
page 60 in 1st measure
X, J c P(X)
J is a semi-ring.
Prove that Js = R(J).
Js denotes the collection of all finite unions of sets from J.
R(J) denotes the ring generated by J.
Show that Js is a ring.
HINTS:
A\B :- Jsds
Jd = J
Jsds = Jdss = Jds = Js
Let W c P(X) satisfy:
(1) /\B,A:-W B u A :- W
(2) /\B,A:-W B n A :- W
Does W have to be a ring?
NO.
W = { {1} }
W = { {4}, {4,5} }
W = { O, {6}, {6,7} }
sup { sqrt(a/b) / (a + 1/b) : a,b>0 } = ???
= 1/2
Let 0 < b < a.
log(a/b) < ???
0 < b < a ==> log(a/b) < (a-b)/b
page 153 in golden gate
Let 0 < b < a.
log(a/b) > ???
0 < b < a ==> log(a/b) > (a-b)/a
page 153 in golden gate