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.

Let W c P(X) and let W contain the empty set.

Let y : W -> [0,oo] and y(O)=0.

In this broad setting, define an outer measure on X.

Prove that it is indeed an outer measure.

Let y : W -> [0,oo] and y(O)=0.

In this broad setting, define an outer measure on X.

Prove that it is indeed an outer measure.

y*(A) = inf { +(k=1 to oo) y(E[k]) : E[k]:-W and A c \\//(k:-
|N)_E[k] }

inf O = oo

page 95 in 2nd measure

inf O = oo

page 95 in 2nd measure

Let f : X -> Y.

Let J be an ideal in Y.

Let K = {f-1(A) : A :- J}.

Can we conclude that K is also an ideal?

Let J be an ideal in Y.

Let K = {f-1(A) : A :- J}.

Can we conclude that K is also an ideal?

NO

f : {1,2} -> {3}

J = { O, {3} }

K = { O, {1,2} }

J is an ideal but K is not

f : {1,2} -> {3}

J = { O, {3} }

K = { O, {1,2} }

J is an ideal but K is not

(1) f : X -> Y

(2) D,E c Y

(3) f-1(D) c f-1(E)

Can we conclude that D c E ?

(2) D,E c Y

(3) f-1(D) c f-1(E)

Can we conclude that D c E ?

NO

f : {1,2} -> {3,4}, f(1)=f(2)=3

D={3,4}, E={3}

f-1(D) = f-1(E) = {1,2}

----------------------------

f : {1} -> {2,3}, f(1)=3

D={2}, E={3}

f : {1,2} -> {3,4}, f(1)=f(2)=3

D={3,4}, E={3}

f-1(D) = f-1(E) = {1,2}

----------------------------

f : {1} -> {2,3}, f(1)=3

D={2}, E={3}

f : X -> Y; g : Y -> Z; E c Z

f-1(g-1(E)) = { x:-X : ????? }

f-1(g-1(E)) = { x:-X : ????? }

f-1(g-1(E)) = { x:-X : g(f(x)):-E }

f : X -> Y; g : Y -> Z; E c Z

{ x:-X : g(f(x)):-E } = ???

{ x:-X : g(f(x)):-E } = ???

{ x:-X : g(f(x)):-E } = f-1(g-1(E))

Let f : |C\{0} -> |C be f(z)=1/z.

Derive the formula for the nth derivative of f.

Derive the formula for the nth derivative of f.

f|n| (z) = (-1)^n * n! / z^(n+1)

Give the power series expansion of exp(z)*sin(z) about 0 and
find its radius of convergence.

exp(z)*sin(z) = +(n=0 to n=oo) a[n]*z^n

a[n] = ( (1+i)^n - (1-i)^n ) / ( 2i * n! )

radius = oo

hint: sin(z) = ( exp (iz) - exp(-iz) ) / 2i

a[n] = ( (1+i)^n - (1-i)^n ) / ( 2i * n! )

radius = oo

hint: sin(z) = ( exp (iz) - exp(-iz) ) / 2i

Let G be an open subset of |C.

Let f : G -> |C be diffable.

Let z0 :- G and let m:-|N, m>=1.

What does it mean that z0 is a zero of function f of multiplicity m ?

Let f : G -> |C be diffable.

Let z0 :- G and let m:-|N, m>=1.

What does it mean that z0 is a zero of function f of multiplicity m ?

(1) f(z0) = 0

there exists g:G->|C diffable such that

(2) /\z:-G f(z) = (z-z0)^m * g(z)

(3) g(z0) != 0

page 87 in OLDTIMER

there exists g:G->|C diffable such that

(2) /\z:-G f(z) = (z-z0)^m * g(z)

(3) g(z0) != 0

page 87 in OLDTIMER

Let G be an open subset of |C. Let f : G -> |C be diffable.

Let z0 :- G and let m:-|N, m>=1.

There exists g:G->|C diffable such that

(1) f(z0) = 0

(2) /\z:-G f(z) = (z-z0)^m * g(z)

(3) g(z0) != 0

Let z0 :- G and let m:-|N, m>=1.

There exists g:G->|C diffable such that

(1) f(z0) = 0

(2) /\z:-G f(z) = (z-z0)^m * g(z)

(3) g(z0) != 0

z0 is a zero of function f of multiplicity m

page 87 in OLDTIMER

multiplicity [m^l t(i) 'pli s(i) ti(:)]

page 87 in OLDTIMER

multiplicity [m^l t(i) 'pli s(i) ti(:)]

f(z) = z^2 * ( exp(z^2) - 1 )

Find the multiplicity of 0.

Find the multiplicity of 0.

4

page 87 in OLDTIMER

page 87 in OLDTIMER