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Guest
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Hi,
I asked that in the maxima-newsgroup before, since I thought it
is only a technical problem, but I possibly made also a mathematical
fault.
I wanted to expand the form
x^m
(----- )^(1/m) into a taylor-series
1+x^m
and possibly made also another error in defining the problem.
------------------------------------------------------------------
Here is the answer, which I got, and where the found coefficients
of the gen.func. surprised me, so that also I should attack my
problem from another side?
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Quote: Taylor expands into a series a0 + a1*x + a2*x^2 + ...., where the
exponents of x are constant integers.
It can only expand into series like a0 + a1*x^m + a2*x^(2*m) + ...,
where m remains symbolic, under rather restricted circumstances.
There are two approaches you could take here. One is to expand
1/(q+1)^(1/m) then substitute q=x^m; the other is to use the powerseries
routine.
powerseries can expand some series with closed-form coefficience. In
this case, for example:
powerseries(1/(x^m+1)^(1/m),x,0) =
('sum(x^(i1*m)/beta(-1/m-i1+1,i1+1),i1,0,inf))/(1-1/m)
two-dimensional display:
inf
==== i1 m
\ x
--------------------------
/ 1
==== beta(- - - i1 + 1, i1 + 1)
i1 = 0 m
---------------------------------
1
1 - -
m
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And this was my reply, containing the question, which I want to
ask here:
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Hi ***** ,
yes, this works fine, thanks.
But now I don't know, whether I have a mathematical or
still a technical problem...
If I cancel in the above sum, (taking care not to use m,
which lead to divergences) I get a simple binomial-formula
with the main terms of binomial(1/m , i1)
as
sum(i1=0,inf, binomial(1/m, i1)*(x^m)^i1 )
where the *first* parameter in binomial(1/m,i1) is constant.
What I was expecting was, that the first parameter would be
*varying* and the second would be constant...
Example:
Assume the pascal-matrix, containing the binomial-coefficients,
P=
1
1 1
1 2 1
1 3 3 1
...
PInv = P^-1 =
1
-1 1
1 -2 1
-1 3 -3 1
...
and the powerseries-vector V(x) = column([1,x,x^2,x^3,... ])
Then the result expresses, with m=1,
P * V(x) = V(1+x)
and the first parameters in the binomials, taken from a row of P, are constant,
which is reflecting the maxima-result.
What I wanted was (if m=1)
1/x* V(1/x))~ * PInv = 1/(1+x) * V(1/(1+x))~
and here the coefficients of PInv are read column-wise, thus the
first parameter of the binomials is varying and the second is
constant.
So, to define the generating-function for the first column of
PInv I use (with approprate scaling)
x
TAYLOR( --- ,x,0,4) = x V(x)~ * column([1,-1,1,-1,...])
1+x
for the second column I use
x
TAYLOR( (---)^2,x,0,4 ) = x^2 V(x)~ * column([1,-2,3,-4,...])
1+x
for the second column I use
x
TAYLOR( (---)^3,x,0,6 ) = x^3 V(x)~ * column([1,-3,6,-10,...])
1+x
and so on, where binomials are produced, which are varying in the first
parameter and constant in the second.
The exponent m at x then should provide a variability of the stepwidth,
when PInv(m) is understood as an operator, which transfers 1/x V(1/x) to 1/y V(y),
where y = x+1 or varying stepwidths like y^m = x^m+1
and I thought it would be an appropriate idea to replace x by x^m and revert this
by powering of the whole expression by 1/m.
But how can then the form of the resulting formula change to an
expression, which reflects a P*V(x)-multiplication instead a
V(1/x)~ * PInv - multiplication???
Or is this operation ambiguous, are both ways possible and maxima
just gets the simpler one?
* scratchhead *
Gottfried Helms |
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