# 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.

G c |C; f : G -> |C; b:-|C
Suppose that function f has a pole at point b.
Can it also have a removable singularity at b?
NO.
By virtue of our supposition:
(1) a doesn't belong to G
(2) there exists B = {z:-|C : 0 < |z-a| < r} c G such that f is diffable on B
(3) lim(z->b) |f(z)| = oo
G c |C; f : G -> |C; b:-|C
What does it mean that function f has an essential isolated singularity at point b?
(1) f has an isolated singularity at b
(2) the singularity is not removable
(3) the singularity is not a pole
Consider the complex function f(z)=exp(1/z).
What kind of isolated singularity does it have at 0?
essential
(1) lim(n->oo) |f(1/n)| = oo
(2) lim(n->oo) |f(i/n)| = 1 < oo
(1) not removable
(2) not a pole
Let m be a positive integer.
Consider the complex function f(z)=z^m * sin(1/z).
What kind of isolated singularity does it have at 0?
essential
(1) lim(n->oo) |f(i/n)| = oo
(2) lim(n->oo) |f(1/n)| = 0
(1) not removable
(2) not a pole
Consider the complex function f(z) = sin(z)/z.
What kind of isolated singularity does it have at 0?
removable
====================
You can prove your answer by referring to the definition only.
This is also an opportunity to test the unproven theorem. (MRW 2001.02.27)
THEOREM: G c |C; f : G -> |C; b:-|C
thesis: If function f has an isolated singularity at poin
Consider the complex function f(z) = ( cos(z) - 1 ) / z.
What kind of isolated singularity does it have at 0?
removable
====================
You can prove your answer by referring to the definition only.
This is an opportunity to test the unproven theorem. (MRW 2001.02.04)
THEOREM: G c |C; f : G -> |C; b:-|C
thesis: If function f has an isolated singularity at poin
lim(x->0,y->0) (x-sin(y))*y^2 / (y^2 + sin(x-y)*sin(x-y) ) = ???
= 0
hint: y^2 <= y^2 + sin(x-y)*sin(x-y)
lim(x->0) x*sin(x) / ( x^2 + sin(x)*sin(x) ) = ???
= 1/2
hint: use the squeeze theorem
Consider the complex function f(z) = (z+1)/z.
Does it have a removable isolated singularity at 0 ?
Justify your answer in two different ways.
NO. It actually has a pole. Notice that lim(z->0) |f(z)| = oo.
---------------------------------
(1) lim(z->0) z*f(z) = 1
hence not (2) lim(z->0) (z-0)*f(z) = 0.
And (2) is a necessary condition for a removable isolated singularity.
(page 103, Conway, "Functions of One Complex Variable", cha
w:-|C
2*sin^2(w) = ???
2*sin^2(w) = 1 - cos(2*w)