The inverse of many variable Taylor serie
Janne Pesonen
Hi,
Suppose we have two sets of variables x1, x2,..,xN and y1, y2,..,yN,
where now y:s are functions of x:s, that is
y1 = f1(x1,x2,..,xN)
y2 = f2(x1,x2,..,xN)
.....
Furthermore, these functions can be inverted, so the x.s can be written
(at least in theory if not in practice) as functions of the y:s
x1 = g1(y1,y2,..,yN)
x2 = g2(y1,y2,..,yN)
Here is my question:
If we have expressed the y:s as Taylor series of of the x:s, ie.
y1 = L[1;i]*xi + (1/2!)*L[1;ij]*xi*xj + ...
y2 = L[2;i]*xi + (1/2!)*L[2;ij]*xi*xj + ...
...
(Einstein summation convention is used. Now L[1;i] is the value of
partial derivative dy1/dxi at point of expansion, L[1;ij] is the second
partial derivative d^2y1/dxidxj at the point of expansion etc.)
Can we use these coefficients to express x:s as Taylor series of y:s,
x1 = £[1;i]*yi + (1/2!)*£[1;ij]*yi*yj + ...
x2 = £[2;i]*yi + (1/2!)*£[2;ij]*yi*yj + ...
...
(where £[1;i] is the value of partial derivative dx1/dyi at point of
expansion, £[1;ij] is the second partial derivative d^2x1/dyidyj at the
point of expansion etc).
ie. can we somehow calculate the £ coefficients from the L coefficients?
Any help is appreciated. Thanks in advance.
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Janne Pesonen
http://fkmarilyn.pc.helsinki.fi/Janne/index.html
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Paul Abbott
Janne Pesonen wrote:
Here is a polynomial example in 2D (computed using Mathematica). For the
function
In[1]:= f[{x_, y_}] = {2x y^2 + x, 3x y^2 + y};
consider compuing the Taylor series at (0,0) of the inverse of f, i.e.,
finding x and y as series in u and v. The Jacobian determinant is
In[2]:= Det[Outer[D, f[{x, y}], {x, y}]]
Out[2]=
2
1 + 6 x y + 2 y
which is nonzero at (0,0),
In[3]:= % /. {x -> 0, y -> 0}
Out[3]= 1
so the inverse exists. After rearranging the mapping to
In[4]:= g[{x_, y_}] = {u, v} + {x, y} - f[{x, y}]
Out[4]=
2 2
{u - 2 x y , v - 3 x y }
iteration (or successive substitution) yields the Taylor series for the
inverse mapping about (0,0):
In[5]:= Transpose[Expand[NestList[g, {x, y}, 3] /.
{x -> 0, y -> 0}]]
Out[5]=
2 2 2 3 4
{{0, u, u - 2 u v , u - 2 u v + 12 u v + 4 u v -
3 4 2 5 3 6
18 u v - 24 u v + 36 u v },
2 2 2 3 4
{0, v, v - 3 u v , v - 3 u v + 18 u v + 6 u v -
3 4 2 5 3 6
27 u v - 36 u v + 54 u v }}
A complete algorithm for such polynomial inversion is given in J
McDonald, Journal of Pure and Applied Algebra 104, 213, 1995.
Cheers,
Paul
____________________________________________________________________
Paul Abbott Phone: +61-8-9380-2734
Department of Physics Fax: +61-8-9380-1014
The University of Western Australia
Nedlands WA 6907 mailto:pa...@physics.uwa.edu.au
AUSTRALIA http://www.pd.uwa.edu.au/~paul
God IS a weakly left-handed dice player
____________________________________________________________________
David Ziskind
Janne Pesonen <Janne....@Helsinki.FI> in article
<3553097F...@Helsinki.fi addressed the subject of expanding the
inverse of an n-dimensional map in terms of the partials of the base
map F.
This is generally possible if one is content to leave the expansion
in terms of products of (x-Fa)_i, where a is some n-point selectable
at will. (However, choosing a=(Inverse(F))0 leads to a circular
exercise.) The calculation is summarized below.
1. Notations, etc.
a. F is an n-dim into n-dim map.
b. x,a are points in n-space.
c. I is the identity function in n-space.
d. VF denotes Inverse(F). Ie, (VF)(F) = I.
e. (P_j)F_i denotes the j th partial of the i th component of F.
(P_jk)F_i denotes the k th partial of the j th partial of
the i th component of F.
Where there is sufficient information to indicate the order of the
partial, "P" is written, rather than, say, PP, or P^2.
f. [PF] is a notation for the Jacobian differential matrix, thus:
[PF]_ij =def (P_j)F_i .
Note the reversal of subscripts in both (f.) and (g.).
g. Higher order Jacobian's are defined similarly. Eg;
[P^2F]_ijk =def (P_jk))F_i .
The tensor notation on the left-hand side *does* write the 2 to
indicate second-order because the whole point of the matrix/tensor
notation will be to write subscript-free relations.
h. S{j: H(j)} = the sum of H(j) taken over j = 1 through n.
S{j,k: H(j,k)} = analogous double sum.
i. * may indicate scalar, matrix, or tensor-inner product.
j. The tensor inner product is adapted to generalize the traditional
matrix product. Thus, eg:
If F is a tensor taking 3 indices, and G takes 2 indices;
(F*G) is a tensor taking 3+2-2=3 indices, specifically,
(F*G)_ijk =def S{q: F_ijq * G_qk}.
k. T generalizes the transpose operator, exemplified as follows.
Let F be a tensor taking 3 indices. Then,
((T23)F)_ijk =def F_ikj as well as
((T13)F)_ijk =def F_kji .
l. Composition and evaluation may be indicated by juxtaposition as
clarity and non-ambiguity allows.
m. V may also indicate matrix inverse. Thus, if A is a matrix,
VA denotes the matrix inverse of A; ie, (VA)(A) = mult.-identity.
2. Taylor's Series Expansion of the Scalar Helper Function
a. Define g, a 1-dim into n-dim map by:
(g_i)t = ((VF)_i)[Fa + t*(x-Fa)]
b. (g_i)0 = a_i; (g_i)1 = ((VF)_i)x
c. ((VF)_i)x = a_i + ((VF)_i)x - a_i
= a_i + (g_i)1 - (g_i)0
and by Taylor's expansion,
= a_i + (g_i)'0 +(1/2!)*(g_i)''0 + ...
and by differentiating 2.a with respect to t
repeatedly and setting t=0 and employing 1.g
notation,
d. ((VF)_i)x = a_i + S{q: (([P(VF)]_iq)Fa)*(x-Fa)_q}
+ S{q,r: (([P^2(VF)]_iqr)Fa)*(x-Fa)_q*(x-Fa)_r}/2!
+ ...
2.d gives the inverse of F in terms of the its own partials. It
remains to develop these in terms of the partials of the base
function F (which is where all the work is).
3. Calculation of Partials of Inverse(F)
One starts out with 1.d, (VF)F = I and takes the q th partial.
Employing 1.g notation, obtaining a 1.j type product, varying out
subscripts, and right-multiplying by V[PF], we obtain the elementary
relation [P(VF)]F = V[PF]. This gives the partials in the
first-order sum in 2.d as is desired.
Development of the higher-order partials follows a definite plan, as
illustrated by the second-order calculation:
Take the relation prior obtained as result of taking the q th
partial, and take the r th partial. Employ 1.g notation, and express
everything in terms of tensors. Employ 1.k transpose operation as
required so as to bring repeated subscripts into "inside positions",
and use prior first-order identity. Vary out subscripts, right-
multiply by V[PF], apply 1.k transpose again, right-multiply by V[PF]
again, and obtain:
[P^2(VF)]F = - T23(T23(V[PF]*[(P^2)F])*V[PF])*V[PF], which is the
result wanted for the second-order sum in 2.d.
The higher-order partials are obtained similarly. Basically, the
calculation of [P^m(VF)]F parallels the scalar case, albeit the
necessity of transposing adds an extra complication.
David Ziskind
zis...@ntplx.net