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.

(1) lim(n->oo) x[n] = 0

(2) \/M>0 /\n:-|N M <= |y[n]|

What can we conclude from this?

(2) \/M>0 /\n:-|N M <= |y[n]|

What can we conclude from this?

lim(n->oo) x[n]/y[n] = 0

Let a > 0.

Integeral( 0, 1, 1/(a+x), dx ) = ???

Integeral( 0, 1, 1/(a+x), dx ) = ???

Integral( 0, 1, 1/(a+x), dx ) = log( (a+1)/a )

Let a > 1.

Integral( 0, 1, 1/(a-x), dx ) = ???

Integral( 0, 1, 1/(a-x), dx ) = ???

Integral( 0, 1, 1/(a-x), dx ) = log( a/(a-1) )

Let a > 1.

Integral( 0, 1, 1/(a^2 - x^2), dx ) = ???

Integral( 0, 1, 1/(a^2 - x^2), dx ) = ???

Integral( 0, 1, 1/(a^2 - x^2), dx ) = 1/(2a) * log( (a+1)/(a-1) )

Use:

Integral( 0, 1, 1/(a+x), dx ) = log( (a+1)/a )

Integral( 0, 1, 1/(a-x), dx ) = log( a/(a-1) )

Use:

Integral( 0, 1, 1/(a+x), dx ) = log( (a+1)/a )

Integral( 0, 1, 1/(a-x), dx ) = log( a/(a-1) )

Integral( 0, pi/2, sin^2(x)*cos(x) / (1+cos^2(x)), dx ) = ???

hint

hint

Integral( 0, pi/2, sin2(x)cos(x) / (1+cos2(x)), dx )

= 1/sqrt(2) * log( (sqrt(2)+1) / (sqrt(2)-1) ) - 1

hint: sinx = u

Integral( 0, 1, 1/(a^2 - x^2), dx ) = 1/(2a) * log( (a+1)/(a-1) )

= 1/sqrt(2) * log( (sqrt(2)+1) / (sqrt(2)-1) ) - 1

hint: sinx = u

Integral( 0, 1, 1/(a^2 - x^2), dx ) = 1/(2a) * log( (a+1)/(a-1) )

(1) a : |N -> |C

(2) 0 < R <= oo

(3) f : {z:-|C : |z| < R} -> |C

(4) f(z) = +(n=0 to n=oo) a(n) * z^(n+1) / (n+1)

What can we conclude from this?

(2) 0 < R <= oo

(3) f : {z:-|C : |z| < R} -> |C

(4) f(z) = +(n=0 to n=oo) a(n) * z^(n+1) / (n+1)

What can we conclude from this?

function f is diffable and

/\|z|<R f'(z) = +(n=0 to n=oo) a(n) * z^n

page 81 in OLDTIMER

/\|z|<R f'(z) = +(n=0 to n=oo) a(n) * z^n

page 81 in OLDTIMER

A = {z:-|C : |z|<1}

Let k be an integer.

f : A -> |C

/\z:-A f(z) = log(|1+z|) + i*Arg(1+z) + k*2*pi*i

Give the power series expansion of function f about 0.

Let k be an integer.

f : A -> |C

/\z:-A f(z) = log(|1+z|) + i*Arg(1+z) + k*2*pi*i

Give the power series expansion of function f about 0.

/\z:-A f(z) = +(n=0 to n=oo) a[n] * z^n

a[0] = k*2*pi*i

a[n] = (-1)^(n+1) / n

page 82 in OLDTIMER

a[0] = k*2*pi*i

a[n] = (-1)^(n+1) / n

page 82 in OLDTIMER

Express log(2) as the sum of a series.

log(2) = +(n=1 to oo) (-1)^(n+1) / n = 1 - 1/2 + 1/3 - 1/4 +
1/5 - 1/6 + ...

page 83 in OLDTIMER

page 83 in OLDTIMER

1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ... = ???

1 - 1/2 + 1/3 - 1/4 + 1/5 - 1/6 + ... = log(2)

page 83 in OLDTIMER

page 83 in OLDTIMER

D = {z:-|C : z:-|R => z > 0}

Let k be an integer.

log : D -> |C, log(z) = log(|z|) + i*Arg(z) + k*2*pi*i

Let a:-|C.

f : D -> |C, f(z) = exp(a*log(z))

What can we conclude about the differentiability of function f?

Let k be an integer.

log : D -> |C, log(z) = log(|z|) + i*Arg(z) + k*2*pi*i

Let a:-|C.

f : D -> |C, f(z) = exp(a*log(z))

What can we conclude about the differentiability of function f?

Notation: if b:-|C, z:-D, let z^b = exp(b*log(z)).

We can conclude that f is diffable and

/\z:-D f'(z) = a * z^(a-1)

In other words: /\b:-|C (z^b)' = b * z^(b-1).

We can conclude that f is diffable and

/\z:-D f'(z) = a * z^(a-1)

In other words: /\b:-|C (z^b)' = b * z^(b-1).