Approximation and interpolation sum of exponential functions

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1 Approximation and interpolation sum of exponential functions József Dombi Department of Informatics University of Szeged, Hungary 1 Learning pliant operator Recall Dombi operator[] o(x) = o(x 1,,x n ) = 1 ( ( ) α ) 1/α (1) 1 + (1 xi ) x i x i [0, 1] If α > 0 then o(x 1,,x n ) is a conjunctive operator, and if α < 0 then o(x 1,, x n ) is a disjunctive operator and α 0 From (1) ( ) α o(x) 1 = ( ) α 1 xi () o(x) i=i x i Let us suppose, that α is known (In most cases is α = ±1) In the pliant concept we use the Dombi operator with sigmoid function σ i (t) = e λ i(t d i ) (3) so σ i (t) = x i Substituting (3) into ()

2 ( ) α o(x) 1 y = = e λ iα(t d i ) = a i e α it o(x) where a i = e αλ id i and α i = αλ i Interpolation sum of exponential functions Our task is now to approximate or interpolate the following function: ϕ(t) = a 1 e α 1t + a e α t + + a n e αnt (4) Let us given (t 1, y 1 ), (t, y ),, (t n, y n ) (and let us suppose that the abscissas are equidistant i e t i+1 t i = h, i = 1,,, n 1) First we deal with the interpolation case Because in (1) we have n variables (a 1, a,,a n, α 1, α,,α n ) we have to give n equations: y 1 = a 1 e α 1t 1 + a e α t a n e αnt 1 y = a 1 e α 1t + a e α t + + a n e αnt y n = a 1 e α 1t n + a e α t n + + a n e αnt n (5) Let us introduce new variables: p k = a k e α kt k, z k = e αkh, k = 1 n (6) Using this notation one term is in (4): a k e α kt j = a k e α k(t 1 +(j 1)h = a k e α kt 1 (e αkh ) j 1 = p k z j 1, (j = 1,,n), (k = 1,,n) Now (5) has the following form and we get the equations: y 1 = p 1 + p + p p n y = p 1 z 1 + p z + p 3 z p n z n y 3 = p 1 z 1 + p z + p 3z p nz n y 4 = p 1 z p z 3 + p 3z p nz 3 n (7) y n = p 1 z n p z n 1 + p 3 z n p n z n 1 n where p 1, p, p 3,,p n and z 1, z, z 3,, z n are n unknown quantities

3 3 Ramanujan solution Ramanujan find an interesting solution of (7)[1] In the following we make a detailed construction of the proof It is easy to verify that p 1 1 θz 1 = p 1 + p 1 z 1 θ + p 1 z 1 θ + p 1 θz = p + p z θ + p z θ + p n = p n + p n z n θ + p n zn 1 θz θ + n (8) The sum of (8) is: φ(θ) = 1 θz i = + θ z i + θ z i + Using (8) we get: φ(θ) = 1 θz i = y 1 + θy + θ y 3 + θ n y n θ n 1 y n + (9) The left hand side of (9) is a rational expression, and when it is simplified, then its numerator is an expression of the (n 1)-th degree in θ, and its denominator is an expression of the n-th degree in θ φ(θ) = 1 θz i = p 1 (1 θzi ) 1 θz 1 + p (1 θzi ) 1 θz + + p (1 θzi ) n 1 θz n (1 θzi ) = D(θ) N(θ) = = Y 1 + Y θ + Y 3 θ + + Y n θ n Z 1 θ + Z θ + Z 3 θ Z n θ n = y 1 + y θ + y 3 θ + + y n θ n 1, (10) and so (1+Z 1 θ+z θ + +Z n θ n )(y 1 +y θ+y 3 θ + +y n θ n 1 ) = Y 1 +Y θ+y 3 θ + +Y n θ n 1 Equating the coefficiens of the powers of θ, we have Y 1 = y 1 Y = y + y 1 Z 1 Y 3 = y 3 + y Z 1 + y 1 Z (11) Y n = y n + y n 1 Z 1 + y n Z + + y 1 Z n 1

4 and 0 = y n+1 + y n Z y 1 Z n 0 = y n+ + y n+1 Z y Z n 0 = y n+3 + y n+ Z y 3 Z n 0 = y n + y n 1 Z y n Z n (1) From (1) Z 1, Z,,Z n can easily be found, and since Y 1, Y,,Y n in (11) depend upon these values they can also be found Now, our task is splitting φ(θ) = 1 θz i = Y 1 + Y θ + Y 3 θ + + Y n θ n Z 1 θ + Z θ + Z 3 θ Z n θ n = (where Y 1,, Y n, Z 1,,Z n are known) into partial fractions, ie and comparing with (1), we see that which is the desired solution φ(θ) = q 1 1 r 1 θ + q 1 r θ + q 3 1 r 3 θ + + q n 1 r n θ, p 1 = q 1, z 1 = r 1 ; p = q, z = r ; p 3 = q 3, z 3 = r 3 ; p n = q n, z n = r n The question, how we can get this partial fraction form The denominator of (10) is: D(θ) = 1 + Z 1 θ + Z θ + + Z n θ n = n (1 r i θ) Let us substitude: θ = 1 ξ Z 1 ξ + Z 1 ξ + + Z 1 n ξ = (1 r 1 n 1 ξ )(1 r 1 ξ ) (1 r 1 n ξ )) = = 1 ξ n n (ξ r i )

5 Multipling both side by ξ n D (ξ) = ξ n + Z 1 ξ n Z n = n (ξ r i ) From D(θ) we build D (ξ) polinom, and the roots of D are the r i The nominator of (10) is: N(θ) = Y i θ i 1 = n j=1 (1 θr j) 1 θr i = p 1 (1 r θ)(1 r 3 θ) (1 r n θ)+ +p (1 r 1 θ)(1 r 3 θ) (1 r n θ)+ +p 3 (1 r 1 θ)(1 r θ) (1 r n θ)+ +p n (1 r θ)(1 r 3 θ) (1 r n 1 θ) = Y 1 + Y θ + Y 3 θ + + Y n θ n 1 From this we get the following linear equation systems: Y 1 Y = = i j, r j Y 3 = r j r k (13) i j,i k,j k Y n = i j r j From (14) we get can be determined by n equation of (7) if is given a i = e α it i

6 4 Ramanujan s example As an example we may solve the equations: x + y + z + v + u =, px + qy + rz + su + tv = 3, p x + q y + r z + s u + t v = 16, p 3 x + q 3 y + r 3 z + s 3 u + t 3 v = 31, p 4 x + q 4 y + r 4 z + s 4 u + t 4 v = 103, p 6 x + q 5 y + r 5 z + s 5 u + t 5 v = 35, p 6 x + q 6 y + r 6 z + s 6 u + t 6 v = 674, p 7 x + q 7 y + r 7 z + s 7 u + t 7 v = 1669, p 8 x + q 8 y + r 8 z + s 8 u + t 8 v = 456, p 9 x + q 9 y + r 9 z + s 9 u + t 9 v = 11595, where x, y, z, u, v, p, q, r, s, t are the unknowns Proceeding as before, we have x 1 θp + y 1 θq + z 1 θr + u 1 θs + v 1 θt = + 3θ + 16θ + 31θ θ θ θ θ θ θ 9 + By the method of indeterminate coefficients, this can be shown to be equal to + θ + 3θ + θ 3 + θ 4 1 θ 5θ + θ 3 + 3θ 4 θ 5 Splitting this into partial fractions, we get the values of the unknowns, as follows: x = 3 5, p = 1, y = z = u = , s = v = 8 5 5, t =, q = 3 + 5,, r = 3 5, 5 1, 5 + 1

7 5 New solution We start with (7): y 1 = p 1 + p + p p n y = p 1 z 1 + p z + p 3 z p n z n y 3 = p 1 z 1 + p z + p 3z p nz n y 4 = p 1 z p z 3 + p 3 z p n z 3 n (14) y n = p 1 z n p z n 1 + p 3 z n p n z n 1 n Let us build the polinom z n + s 1 z n 1 + s z n + + s n = 0 (15) The roots of (15) are z 1,,z n The following wellknown identities are valid s 1 = z i s = i j z i z j s n = ( 1) n z i (16) Based on (14) we show that: y 1 s n + y s n y n s 1 + y n+1 = 0 y s n + y 3 s n y n+1 s 1 + y n+ = 0 y n s n + y n+1 s n y n 1 s 1 + y n = 0 (17) The k th row is in (17): Using (14): s n pi z k 1 i Let us find the coefficient of p j : y k s n + y k+1 s n 1 + y k+ s n + + y n+k 1 s 1 + y n+k = 0 + s n 1 pi z k i + + s 1 pi z n+k i p j (s n z k 1 j + s n 1 z k j + + z n+k 1 j ) = 0 + z n+k 1 i = 0

8 because (15) is valid From (17) if y 1,,y n are given we can get s i If s i is given, then the roots of (15) are the solution for z i α i = lnz i h (18) If z i are given, then from (14) we get 6 Approximation sum of exponential functions Now we have more than n points: (t 1, y 1 ), (t, y ),, (t m, y m ) and ϕ(t) = a 1 e α1t + a e αt + + a m e αmt Suppose that the abscissas are equidistant ie t i+1 t i = h, i = 1,,, n 1, and m > n We build the polinom (15) and using the same construction: y 1 s n + y s n y n s 1 + y n+1 = 0 y s n + y 3 s n y n+1 s 1 + y n+ = 0 y n s n + y n+1 s n y m 1 s 1 + y m = 0 (19) we get it instead of (17) The A matrix of (19) and the constant vector b y 1 y y n y y 3 y n+1 A = y m n y m n+1 y m 1 y 1 b = y n

9 The best approximation of (19) is based on the least square method (A T A)s + A T b = 0 (0) The solution of (0) gives the s values: If s is given we can determine (15), and the roots of (15) are the z i values, and based on (16) α i can be determined Now we have to find the a i -values We use also the least square method: (ϕ(t i ) y i ) = m ( a i e α it j y j ) (1) j=1 The matrix is: e α 1t 1 e α t 1 e αnt 1 e α 1t e α t e αnt F = e α 1t m e α t m e αntm And the b = (y 1,,y n ) The best approximation of (1) is: (F T F)a + F T b = 0 where a is unknown

10 Figure 1: φ(x) = (009e x 1303e x 4481e 15x + e 17x e 14x )10 3 References [1] S Ramanujan Note on a set of simultaneous equations Journal of the Indian Mathematical Society, 94-94, 191 [] J Dombi A general class of fuzzy operators, the De Morgan class of fuzzy operators and fuzziness measures induced by fuzzy operators Fuzzy Sets and Systems Vol 8, 198, pp

Construction of Functions by Fuzzy Operators

Acta Polytechnica Hungarica Vol. 4, No. 4, 007 Construction of Functions by Fuzzy Operators József Dombi Árpád tér, H-670 Szeged, Hungary, e-mail: dombi@inf.u-szeged.hu József Dániel Dombi Árpád tér, H-670