On the inverse of two variables power series(Received 27.5.1993

M. Pt a k , A. Ru t k o w s k a, A. So w a

Krakow

On the inverse of two variables power series

A bstract. We show a new recurrent formula for coefficients of an inverse of a power series of two variables with partially vanishing coefficients.

In the following paper we will consider an inverse function to a power series of two variables. Then the inverse function is also a power series of two variables (see [PRS]). The purpose of the paper is a new formula for coefficients of the inverse series when certain coefficients of the starting series are equal zero. This result can be straight applied, for example, in geodesy where the Gauss-Kriiger mapping, a function from the surface of the earth’s ellipsoid to a subset of R 2, is considered. It is usually given by a power series of two variables, the longitude and the latitude. One of the problems in geodesy is to find the inverse mapping to the Gauss-Kriiger mapping for different ellipsoids (eg. Bessel’s, Hayford’s and Krasowski’s ellipsoids). The coefficients of the Gauss-Kriiger mapping for all cases fulfill assumptions of our new formulas. The result, we will obtain, is much more useful in applications comparing with the result from [PRS] since it reduces a quantity of components (see last paragraphs). Especially we can take an advantage of the new formulas calculating, using computers, more coefficients of the inverse series in the same computing time.

Choose the norm in R 2 which gives rectangles as balls, i.e. 11(x,y)\\ =

max{|a;|, A|y|}, where A > 0 is fixed. Let us recall that a power series

Yl^>,q=oapqfPlq (apq £ R 2) converges on an open set G to a function :

G — R 2 if q=n avqfPl<1 ~~ i i j',1) if

00

Th e o r e m 1. Suppose that a power series J2^q=o avqfP^q (apq £ R 2)

converges in a neighbourhood U of 0 in R 2 to a function if : U f'(U) C

«io R “. Assume also that ago = 0 and clet

«oi

7

*■ def

is invertible on the ball K{0,r) C U and the inverse ip : V = ’if (Ii (0,r)) — A’(0,r) is given as a power series Y^Tt=o ast:L‘syt (ctst £ R 2) converging on V . Moreover, a0o = 0, = ( 1 ) =

Y

—

,...,

E

x

Y

)x(i)

where \ sets the number of a coordinate exStti (i.e. {aSiti )x(j) is the x(i) coordinate of aSt t, ) •

Now our main result can be formulated.

Th e o r e m 2. Suppose that, a power series ( X ^ g=o Apqf plq,

^^°g=o Bpqfplq) converges in a neighbourhood U of 0 in R 2 to a function

Aio Rio

if : U —rif(U) C R “. Let (/loo, Aoo) = 0 and let

also that Tqi Ri01

/ 0. Assume

* *)

Apq = 0, when q is odd and Bpq = 0, when q is even.

Then exist, r such that if is invertible on A'(0, r) C U and the inverse <p> :

V d= if( A (0, r)) — K (0, r) is given as a power series YlTt=o QstA'syt (a st G

R 2) converging on V. Moreover ip satisfies the following conditions: (

2

3

(ast)i = 0, when t is odd,

(4) (ci'Si h = _A

10

£

£

£

1 B S t ) 2 Ty i>(y\ { a sk 2rk+ x ( k ) - l ) x { k ) ^ j A p q ' j , (5)

£

£

P r o o f of tlieorem 2: The existence of <p given as a. power series fol-lows from Theorem 1. Condition (2) is proved in Theorem 1 too. Condition (3) will be proved by induction on m = s -f- /. The case m = 1 is straight-forward from (2). Assuming (3) for m < ??., we will show it for m = n + 1. Applying (*) to Theorem 1 one can easily see that

Assume that t is odd. Suppose (ast)i ^ 0. Thus there is a function

X : Ik {1,2} and decompositions t,\ + . .. + tk = T •?! + . .. + Sk = s such that (aSltl)x{i) ' • • •' (asktk)x(k) i1 0. We will show that S{ + < n for any i = 1 , 2 , . . . , k. Assume for a while that there is j such that s.j + tj > n. Then s + t = X )L i s* + = s.i + tj + E i = i (s* + U )• But S{ + /,; > 1 for any

^ 3

i = 1, 2 , . . . , k. Since k > 2 thus .5 -f t > n + 1 ■ (k — 1) > n + 2 — 1 = n + 1 and this leads to contradiction because s + t = n + 1. Hence we can use the inductive assumption for any si -f ti where i = 1, 2 , . . . , k. If there is isuch that x(i) = 1 and tt is odd then by inductive assumption (o t, )x(q = 0, thus (aSlt J x(i) • • {<xSktk)x(k) = 0- For \(/) = 2 and tt even we have the same conclusion. Then we can assume that if

x(‘i)

{t\, . . . ,tk} are odd. But since q is even we have t\ + . . . + //. = / is even. This is contradiction with the assumption. The proof of the second condition in (3) can be made similarly taking q odd and t even. To show (4) and (5) one can see first that t — q > 0 because q elements are odd and p elements are even in set {A , • • •, tk} and hence t, > q ■ 1 + p ■ 0 = q. By Theorem 1, (6) and (7) it is enough to prove now that

( a sk 2 rk +x(k)-l)x(k)

We know that (aSiti )i ^ 0 for t% even and {aSiti h / 0 for / odd. Hence we can assume that ti — x(*) + 1 is even for i - 1,2 ,.. .,/r. Since \—1 (1) = p

and W 1

(2)

x(*)

-1 • p — 2 • (k - p) = t — q. Let r ■ d=f tj — x(i) + 1 for any i = 1 , 2 , . . . , k.

Then 5 3 ) x ( i ) ' • • •' ( a sktk ) x ( k) + —t Si -\~ti ^ 1 ~ 5 3 ( CL'sar'^-f-x(l) —1 )x (l) ‘ ' { a skr'k+ x ( k ) - l )x(>') r'i + ---+r'k = t - q r\ even • s . + H + x D - 1 ^ 1 /

( a V 2n + x ( l ) - l )x (l) ' • • • ' ( a sk2rk+ x ( k ) - \ ) x ( k ) r l + ...+ r2= 1-^ lL

s t+ 2 r i > 2 - x ( i )

and the proof has been finished.

All the algorithms needed for generating indexes in the sums are the classical one, which can be taken, for example, from [R.ND]. We would only like to make a remark on the algorithm for generating Vi, m,. . ., rp in the last sum of (4) and (5) in Theorem 2. Since t—fr- is integer, the algorithm we need can be modified easily from the algorithm for all decompositions of positive integer as a sum of non-negative integer. However it would be difficult to find the formula for the numbers of components in the formulas of Theorem 1 and 2, one can easily notice that the number of components in the formulas in Theorem 2 is less then in Theorem 1.

Observe firstly that half of components are equal 0 (see (3)), secondly that the sums (4) and (5) run only over half of q's. Except the above the main influence for reduction of the numbers of computations has the index running over all compositions of in (4) and (5) instead of all compositions of t in (1) in the most inner summation. To estimate the difference we fix k

(for example k even), take very large t (/ even and much bigger than k) and compare the number of /..'-compositions of t and |. By [R.ND] the number of

^-elements compositions of n is so we can calculate that

j def ( l - - i ) _ 2 k ~ 1 + k ~ !)••(* + k — 2 ) ! ! " ( H * " 1) " ( / - ! ) ! ! ( / + 2 k — 2 ) ! ! ’

where (2m)!! = 2 • 4 • . . . • 2m and (2m - 1)!! — 1 • 3 • . . . • (2m, - T). Using approximate formulas (2m)!! « 2TO( ™ )rn\/2wm and (2m —1)!! & 2m+5( 2L)TO we obtain 2 1V ( i ) l 2 1±f = 1 (i± | f= i)i±f = 1 v/(( + 2fc - 2 )tt = 2 k . /, t s t + k k t + k — 2 l. rf t , /+A’ — 2 k 2 /--- :---, <= ((1 + t )‘ ) ~ ' T • t s - ( ( 1 + ) " T — — + k - 2 t ' - ^ K 1 + i i f i ) ^ )

Since our t is very large one can see that

t 2 k 2 2 k — 2 \/{ t -(- 2 k — 2 ) 7T L « 2k- 1 t “\-k fc t ^ k — 2 k — 2 : 2 ' t • c 2 t t ~h2fe — 2 2k — 2

we have computed the number of the most interior products that should be calculated to find all ast for .s - f / < n. Using Theorem 1 and Theorem 2 for n = 5 the number is equal to 4327 and 375, respectively. For n — 10 it is equal to 20 977 816 and 183 646, respectively.

References

[PRS] M. Pt.ak, A. R u tk ow sk a , J. Szczurek, A note on inverses of pouter series, Journal of Computational and Applied Mathematics 39 (1992), 95-101.

[RND] E. M. R e in g o ld , J. Nievergelt, N. Deo, Combinatorial Algorithms. Theory

and Practice, Prantice-Hall, Inc., Englewood Cliffs, New Jersey, 1977.

KATEDRA ZASTOSOWAN M AT EM AT Y K1 AKADEMIA ROLNICZA