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.

a,b :- |C; a!=1, b!=1
(a-b) / (1-a)(1-b) = ???
(a-b) / (1-a)(1-b) = 1/(1-a) - 1/(1-b)
a,b :- |C; a!=1, b!=1
1/(1-a) - 1/(1-b) = ???
(1/(1-a) - 1/(1-b)) = (a-b) / (1-a)(1-b)
For z:-|C, |z|<1, let f(z) = 1 / (z+1)(z-2).
Give the power series expansion of function f about 0.
Find its radius of convergence.
f(z) = +(n=0 to oo) a[n] * z^n
a[n] = -1/(6 * 2^n) - ((-1)^n / 3)
R=1
hint: (1/(z-2) - 1/(z+1))/3
x :- |R
Im( i / (1 + ix) ) = ???
Im( i / (1 + ix) ) = 1 / (1 + x^2)
For z:-|C, |z|<1, let f(z) = 1 / (1+z^3)^2.
Give the power series expansion of function f about point 0.
Find the radius of convergence.
f(z) = +(n=0 to n=oo) (-1)^n * (n+1) * z^(3n)
R=1
hint: for |w|<1, 1 / (1+w) = +(n= .... ; then differentiate the series.
For z:-|C, |z|<1, let f(z) = +(n=0 to n=oo) (-1)^n * (n+1) * z^n.
Find a direct formula for f(z).
f(z) = 1 / (1+z)^2
hint: f(z) = +(n=1 to n=oo) (-1)^(n+1) * n * z^(n-1)
g(z) = +(n=0 to n=oo) (-1)^(n+1) * z^n
g'(z) = f(z)
g(z) = -1/(1+z)
Let w be a complex number.
1 + cos(2w) = 2* ???
1 + cos(2w) = 2*cos^2(w)
hint: cos(2w) = cos^2(w) - sin^2(w)
Let f : |R -> |R be defined f(x) = x * sqrt(1+x^2).
Find a differentiable function F : |R -> |R such that
/\x:-|R F'(x) = f(x).
F(x) = sqrt(1+x^2) * (1+x^2) * 1/3
hint: Integral( 0, A, f(x), dx ) = ???
u(x) = sqrt(1+x^2)
/\a:-|R Integral( 0, a, sqrt(a^2 - x^2), dx ) = ???
/\a:-|R Integral( 0, a, sqrt(a^2 - x^2), dx ) = a*|a|*pi/4
Notice that this integral calculates one fourth of the area of a circle of radius a.
a:-|C
lim(z->0) sin(az)/z = ???
/\a:-|C lim(z->0) sin(az)/z = a