# 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.

Prove that if x <= c <= y, then c = ax+by, for some a,b>=0, a+b=1.
a = (y-c)/(y-x)
b = (c-x)/(y-x)
Show a function which is defined on the real line and continuous at only one point.
0 for rationals,
x for irrationals
continuous only at 0
Show a function which is defined on the real line and differentiable only at one point.
0 for rationals,
x*x for irrationals
differentiable only at 0, because continuous only at 0
Sketch the proof of Rolle's theorem.
First prove that the function is either constant or has a local extremum.
Later use the fact that if a differentiable function has a local extremum at a point,
then its derivative is equal to zero at that point.
page 154 in Golden Gate
State and prove Cauchy's Mean-value Theorem.
You can refer to Rolle's theorem.
page 155 in Golden Gate
Consider a real-valued differentiable function defined on a bounded closed interval. Does it have to be a Lipschitz function?
NO. g:[0,1]->|R, g(0) = 0, g(x) = x*x*sin(exp(1/x))
This function is differentiable, g'(0) = 0, but g' is not bounded. Hence it is not Lipschitz.
Put x[n]=1/log(n*2*pi). Show that lim(n->oo) g'(x[n]) = -oo.
Thus the derivative is not bounded.
Let f be a real-valued function defined on an interval on the real line. Suppose that it is monotonic and that its image is an interval. Does the function have to be continuous?
YES.
Use the fact that a bounded monotonic function has a limit.
Let f be a function from a metric space, say X, to a complete metric space. Let A be a nonclosed subset of X. Let f be a uniformly continuous function on A. Can f be extended to the closure of A so that the extended function is continuous?
YES. And what's more, the extension is uniformly continuous.
Completeness is essential. In the proof, we use the Cauchy criterion for the existence of the limit of a function at a point.
Consider a function from one metric space into another. Suppose it is continuous and uniformly continuous on a dense set. Does this function have to be uniformly continuous?
YES. The proof can be performed routinely.
(1) f : R -> R
(2) f has n derivatives
(3) the n-th derivative is constantly zero