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From: Gottfried Helms on 2 Jan 2010 09:13
Am 22.12.2009 00:45 schrieb mike3:
> Hi.
>
> I saw this:
>
> http://en.wikipedia.org/wiki/Finite_difference#Calculus_of_finite_differences
>
> which shows that the "forward difference operator" can be expressed as
> a "Taylor series of operators" in the differentiation operator:
>
> Delta_h = hD + hD^2/2! + hD^3/3! + ... = "exp(hD) - 1"
>
> How does one come up with this (it mentions "applying Taylor's theorem
> to h", but I'm not sure how exactly that'd be done here for
> operators)? Also, is there a similar expression for the summation
> operator Delta^-1_h in terms of powers of the integration operator? If
> so, what is it?

Hi Mike -

a bit late - but since I'm not familiar with the
common use of the "operator"-term in function-analysis
I step aside if I see it mentioned, especially, if it
is involved in operations like you mention.

But anyway - I tried to translate this to my operator-concept
of matrix-operators on formal powerseries. There we have
the (upper triangular) Pascal-matrix, which performs on a vandermonde
vector V(x) = rowvector(1,x,x^2,x^3,x^4,x^5,...)

V(x) * P = V(x+1)

which apparently mimics that Delta_h (h=1) of the functions-analysis.
(note, that I used the transposed V and P here compared to my usual notation
in tetration-forum etc for notational ease)

Moreover with a matrix-operator F implementing a function f(x) on
a vandermondevector

V(x) * F = V(f(x))

we have also

V(x) * P * F = V(f(x+1))

and can read this as

PF = P * F

as application of the "differenceoperator" at a "functionoperator"
giving a new object.
-------------------------

Now the matrix(operator) P can be written as a matrix-exponential

P = exp(L) = I + L + L^2/2! + L^3/3! + ...

where L is the first upper subdiagonal containing
updiag(1)(1,2,3,4,..)

Also we have

V(x)* L* F = row(0,f(x)', (f(x)^2)', (f(x)^3)', ...
\\ f()' meaning the derivative
\\ = diff V(x) / dx

So again this seems to match the hD-notation above.

The inverse operation, the (formal) integral on powerseries,
can be expressed by the transposed of L, doubly factorially
rescaled:

M = fac^2 * L~ * FAC^2

where FAC=diag(0!,1!,2!,3!,...) and fac = FAC^-1 and
M = lowdiag(1)(1,1/2,1/3,1/4,...)

Then
V(x) * M F = integral V(f(x)) dx

However, M cannot be fully understood as "the inverse" of L because
M*L = diag(0, 1,1,1,1,..) \\ having leading zero
and
L*M has a trailing zero if the matrices are finite.

Gottfried

==========================================================================


Codeexample Pari/GP, using functions from PariTTY

dim=6
\\ ---------------
V(x) = vector(dim,r,x^(r-1)) \\ rowvector
FAC = matdiagonal(vector(dim,r,(r-1)!))
fac = FAC^-1
P = matpascal(dim-1)~
fS2F = fac*makemat_stirling2(dim)*FAC \\ implements operator for exp(x)-1
MEpx(M) = \\ implementation of matrixexponential of M
MLog(M) = \\ implementation of matrixlogarithm of M
\\ ----------------

V(x)
% [1, x, x^2, x^3, x^4, x^5]


V(x) * P
% [1,
x + 1,
x^2 + 2*x + 1,
x^3 + 3*x^2 + 3*x + 1,
x^4 + 4*x^3 + 6*x^2 + 4*x + 1,
x^5 + 5*x^4 + 10*x^3 + 10*x^2 + 5*x + 1 ]
\\ = [ 1, x+1, (x+1)^2 , (x+1)^3, ...]
\\ = V(x+1)

%pri L=MLog(P)
% L:
0 1 0 0 0 0
. 0 2 0 0 0
. . 0 3 0 0
. . . 0 4 0
. . . . 0 5
. . . . . 0

V(x) * fS2F
% [1,
1/120*x^5 + 1/24*x^4 + 1/6*x^3 + 1/2*x^2 + x,
1/4*x^5 + 7/12*x^4 + x^3 + x^2,
5/4*x^5 + 3/2*x^4 + x^3,
2*x^5 + x^4,
x^5]
\\ for lim dim->inf
\\ = row( 1 , exp(x)-1 , (exp(x)-1)^2, ,...)


V(x)*L*fS2F
% [0,
1/24*x^4 + 1/6*x^3 + 1/2*x^2 + x + 1,
5/4*x^4 + 7/3*x^3 + 3*x^2 + 2*x,
25/4*x^4 + 6*x^3 + 3*x^2,
10*x^4 + 4*x^3,
5*x^4]
\\ for lim dim->inf
\\ = row( 1 , (exp(x)-1)' , ((exp(x)-1)^2)', ,...)
\\ = V( exp(-1)) '

%pri M = fac^2 * L~ * FAC^2
0 . . . . .
1 0 . . . .
0 1/2 0 . . .
0 0 1/3 0 . .
0 0 0 1/4 0 .
0 0 0 0 1/5 0

V(x)*M*fS2F
% [x,
1/120*x^5 + 1/24*x^4 + 1/6*x^3 + 1/2*x^2,
7/60*x^5 + 1/4*x^4 + 1/3*x^3,
3/10*x^5 + 1/4*x^4,
1/5*x^5,
0]
\\ for lim dim->inf
\\ = [ x, int(exp(x)-1)dx, int( (exp(x)-1)^2)dx, ... ]
\\ = int(V(exp(x)-1))dx

%pri M*L
0 . . . . .
. 1 . . . .
. . 1 . . .
. . . 1 . .
. . . . 1 .
. . . . . 1

%pri L*M
1 . . . . .
. 1 . . . .
. . 1 . . .
. . . 1 . .
. . . . 1 .
. . . . . 0

====================================================


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