Math ASCII Notation Demo

Mathematical content on 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 for all real a,b
sqrt(a*a + b*b) <= |a| + |b|
Prove that for all real a,b
1/sqrt(2) * (|a| + |b|) <= sqrt(a*a + b*b)
Investigate the limit
lim xy / (x*x + y*y) as (x,y) approaches (0,0).
doesn't exist
Investigate the limit
lim x*y*y / (x*x + y*y) as (x,y) approaches (0,0).
= 0
hint: |xy/(x^2+y^2)| is bounded
Investigate the limit
lim exp(-x*x-y*y) as (x,y) approaches (oo,oo)
= 0
Investigate the convergence of the sequence {n^p * z^n}n where z is a complex number and p is a real number.
if |z|>1, then diverges
if |z|<1, then converges
if |z|=1, then it depends on p, see page 15 in OLDTIMER for inspiration
State and prove Weierstrass Test for Unifrom Convergence of Series.
(1) /\n:-|N /\x:-A | f[n](x) | <= a[n]
(2) the series a[n] converges
(3) the series f[n](x) converges uniformly on A
(4) the series | f[n](x) | converges uniformly on A
Let R be the radius of convergence of a power series with center at 0.
Prove that the series converges uniformly on
{z: |z| <= r} for all r, 0<r<R.
Use Weierstrass Test.
very important hint: r<q<R.
Find the radius of converges of the series ((-1)^n / n) * z^(n(n+1)).
hint: a[n(n+1)] = (-1)^n / n
Let a[n], b[n] be sequences of complex numbers, indexed from zero.
Suppose that series |a[n]| and series b[n] both converge.
What can we infer from this? (State without proof.)
Let /\n:-|N C[n] = +(k=0 to n) [ a[n-k]*b[k] ] = +(k=0 to n) [ a[k]*b[n-k] ].
Let A = +(n=0 to oo) a[n].
Let B = +(n=0 to oo) b[n].
Then +(n=0 to oo) C[n] = A * B