Luận án Các định lý thuận và đảo của xấp xỉ Padé

CHUaNG IV: PHU LUC. . -> #THIS IS PROGRAMUSED TO CALCULATE A KIND OF RATIONAL APPROXIMATION - CLASSICAL PADE APPROXIMATION OF A H(O) FUNCTION (MORE DETAILS AND MORE EXPLAINS) PadeApproximant := proc(f,m,n,value) option operator; local taylorArray,i,coefficients,seqCoeffsTaylor,seqMatrix1, seqMatrix2,j,k,t,distanceMN, padeMatrix, detPadeMatrix,rightOfSystem,Q,P, sQ, setRootQ1,setRootQ, sP, setRootP, a, b, c ,r, temp,fQ,t1,t2 ; print('THIS IS A PROGRAMUSED TO CALCULATE CLASSICAL PADE APPROXIMANT OF AFUNCTION') ; printf (' \n\n') ; printf('\tThe function f(z) is:',f); print(f) ; #EXPAND TAYLOR OF F(Z) taylorArray := taylor(f, z =0, n+m+1); printf('\tThe Taylor expand of f(z) is: '); print (taylorArray) ; for i from 0 to n+m do coefficients[i] := coeff(taylorArray,z,i) ; od; #SEQUENCEOF COEFFICIENTS seqCoeffsTaylor:= seq(coefficients[i],i=O..m+n); printf('\tSo that the cofficients(from 0 to %d) of Taylor expand of f(z) is: ',m+n) ; print (seqCoeffsTaylor) ; temp := 0; for i from 1 to n+m+1do temp:=temp + seqCoeffsTaylor[i]*zA(i-1); od; distanceMN:= m-n; if (distanceMN -1 > 0) then for j from (1 - distanceMN) coefficients[j] := 0; by 1 to -1 do cd; fi; #CALCULATE COEFFICIENTS TO ESTABLISH MATRIX for t from 0 by 1 to m-1 do seqMatrix1[t] := seq(coefficients[k],k = I-distanceMN+t. .n+t); Page 1 od; seqMatrix2 := [seq(seqMatrix1[t],t = 0..m-1)]; printf('\tThis is sequence of numbers to form matrix for system of equations: '); print (seqMatrix2) ; nops(seqMatrix2) ; #MATRIX OF COFFICIENTS padeMatrix := linalg[matrix] (m,m,seqMatrix2); printf('\tSo that the matrix formed is: '); print (padeMatrix) ; with (linalg) ; detPadeMatrix := det(padeMatrix) ; #RIGHT SIDE OF SYSTEM OF EQUATIONS rightOfSystem := vector([seq(-1*coefficients[1],1=n+1. .n+m)]); print('The right of system of linear equation is: '); print (rightOfSystem) ; sQ := linsolve(padeMatrix, rightOfSystem); setRootQ1[m+1] := 1; for i from 1 to m do setRootQ1[i] := sQ[i]; od; for i from 1 to m+1 do setRootQ[i] := setRootQ1[m+2-i]; od; #COEFFICIENTS OF Q(Z) POLYNOMIAL setRootQ :=[seq(setRootQ[i],i=1. .m+1)]; printf('\tSet of coeffs of Q: '); print (setRootQ) ; Q :=0; #ESTABLISH Q(Z) for i from 1 to m+1 do Q := Q + setRootQ[i]*zA(i-1); od; printf('\tPolynomial print(Q) ; #CALCULATE f(Z) .Q(Z) fQ:=expand(temp*Q) ; Q(z):'); for i from 0 to n do sP[i] := coeff(fQ, z,i); od; #ONLYGET N+1 FIRST COEFFICIENrS OF sP setRootP := [seq(sP[i],i=O..n)]; printf('\tSet of coeffs of p:'); page 2 print (setRootP) ; printf('\tPolynomial P(z) :'); P:=o; #ESTABLISH P (Z) for i from 1 to n+l do P := P + setRootP[i] * zA(i-l); od; print(P) ; printf('\tPade approximant of f with order (%d,%d) is: ',m,n); print(P/Q) ; tl:=evalf(normal(subs(z=value,f») ; t2:=evalf(subs(z=value,P/Q» ; printf('The value of f(z) at %.8f is: %.8f\n', value,tl); printf('The value of (p/Q) (z) at %.8f is %.8f\n', value,t2); printf('The error of approximants: %.8f', abs(tl-t2»; end; PadeApproximant:=proc(f,m,n, value) localtaylorArray,i, coefficients,seqCoeffsTaylor,seqMatrix1,seqMatrix2,j,k, t,distanceMN, padeMatrix,detPadeMatrix,rightOfSystem,Q, P, sQ, setRootQ1,setRootQ,sF, setRootP,a, b, c,r, temp,fQ,t1,t2; optionoperator; print('THIS IS A PROGRAM USED TO CALCULATE CLASSICAL PADE APPROXIMA\ NT OF AFUNCTION'); printf('Inln'); printf('ItThefunctionf(z) is:" f); print(f); taylorArray:=taylor(f,z =0,n +m+ 1); printf('ItTheTaylorexpandoff(z) is: '); print(taylorArray); for i from 0 to n +mdo coefficients[i] :=coeff(taylorArray,z, i) od; seqCoeffsTaylor:=seq(coefficients[i], i =0 .. n +m); printf('ItSothatthecofficients(from0to%d)ofTaylorexpandoff(z) is: " n +m); print(seqCoeffsTaylor); temp:=0; for i ton +m+1dotemp:=temp+seqCoeffsTaylor[i]*zJ\(i- 1)od; distanceMN:=m- n; if 0<distanceMN- 1thenforj from1- distanceMNto-1docoefficients[j]:=0odfi; fortfrom0tom- 1do page 3 seqMatrixl[t] :=seq(coefficients[k],k=1- distanceMN+t .. n+t) od; seqMatrix2:=[seq(seqMatrixl[t],t=O.. m-l)]; printf('ItThisis sequenceof numberstoform matrixfor systemof equations:'); print(seqMatrix2); nops(seqMatrix2); padeMatrix:=linalg[matrix](m,m,seqMatrix2); printf('ItSothatthematrixformedis: '); print(padeMatrix); withelinalg); detPadeMatrix:=det(padeMatrix); rightOjSystem:=vector([seq(-coefficients[I], 1=n+1.. n+m)]); print('Therightof systemof linearequationis: '); print(rightOjSystem); sQ :=linsolve(padeMatrix,rightOjSystem); setRootQl[m+1] :=1; for ito m dosetRootQl[i] :=sQ[i] od; for i tom+1dosetRootQ[i]:=setRootQl[m+2- i] od; setRootQ:=[seq(setRootQ[i],i =1.. m +1)]; printf('ItSetofcoeffsofQ: '); print(setRootQ); Q '=0'. , fori tom+1doQ :=Q+setRootQ[i]*zl\(i- 1)od; printf('ItPolynomialQ(z):'); print(Q) ; fQ :=expand(temp*Q); for i from0ton dosP[i] :=coeff(jQ,z, i) od; setRootP:=[seq(sP[i],i =0.. n)]; printf('ItSetofcoeffsofP: '); print(setRootP); printf('ItPolynomialP(z):'); P :=0; fori ton+1doP :=P +setRootP[i]*zl\(i- 1)od; print(P) ; printf('ItPadeapproximantoffwithorder(%d,%d)is: " m,n); print(P/ Q); t1:=evalf(normal(subs(z=value,!))); t2:=evalf(subs(z=value,P / Q)); printf('Thevalueoff(z)at%.8fis:%.8f\n" value,t1); Page 4 printf('Thevalueof (P/Q)(z)at %.8f is %.8f\n" value,t2); print£\:'Theerrorofapproximants:%.8f, abs(t1- t2)) -end j> #USEMY FUNCTION TO DO SOME TEST ON SOME BASIC l EXPONENTIAL, SIN, POWER, RATIONAL FUNCTIONS. >#THE FISRT FUNCTION IS EXPONENTIAL FUNCTION FUNCTION SUCH AS: PadeApproximant(exp(z) ,3,2,1); print('--------------------------------------------------------- - ') ; PadeApproximant(exp(z) ,3,3,1); THISIS A PROGRAM USED TO CALCULATE CLASSICAL PADE APPROXIMANT OF AA UNCTION The function f(z) is: z e The Taylor expand of f(z) is: 1 2 1 3 1 4 1 5 6 1 + z +- z +- z + - z + - z + O(z )2 6 24 120 So that the cofficients(from 0 to 5) of Taylor expand of f(z) 1 1 1 1 11---- , , 2' 6' 24' 120 This is sequence of numbers to form matrix for system of equations: is: [ 1 11111 ] 11-1----- , , 2' '2'6'2' 6'24 So that the matrix formed is: 1 1 1 2 1 6 1 24 1 1 2 1 6 1 2 Warning, new definition for norm Warning, new definition for trace Therightof systemof linearequationis: [ -1 -1 -1 :6' 24' 120 Set of coeffs of Q: [ -3 3 -1 ]1, 5' 20' 60 . Polynomial Q (z) : 3 3 2 1 3l--z+-z --z 5 20 60 Page 5 I Set of coeffs of P: [ 2 1 ] 1 -- , 5'20 Polynomial P (z) : 2 1 21+-z+-z 5 20 Pade approximant of f with order (3,2) is: The value of The value of The error of 2 1 2 1 +-z+-z 5 20 3 3 2 1 3l--z+-z --z 5 20 60 f(z) at 1.00000000 is: 2.71828182 (P/Q) (z) at 1.00000000is 2.71875000 approximants: .00046817 ---------------------------------------------------------- THISIS A PROGRAM USED TO CALCULATE CLASSICAL PADE APPROXIMANT OF AF\ UNCTION The function f(z) is: ze The Taylor expand of f(z) is: 1 2 1 3 1 4 1 5 1 6 7 1 + z + - z + - z + - z + - z + - z + D( z ) 2 6 24 120 720 So that the cofficients(from 0 to 6) of Taylor expand of f(z) is: 1 1 1 1 1 11----- , '2'6'24' 120' 720 This is sequence of numbers to form matrix for system [ 11111111 ]1, 2' 6' 2' 6' 24' 6' 24' 120 So that the matrix formed is: of equations: Therightof systemof linearequationis: [ -1 -1 -1 ]24' 120'720 Set of coeffs of Q: [ -1 1 -1 ]1, 2' 10' 120 Polynomial Q(z): Page 6 1 1 1 - - 2 6 1 1 1- - - 2 6 24 1 1 1- - - 6 24 120 1 1 2 1 31--z+-z --z 2 10 120 Set of coeffs of P: [ 1 1 1 ]1, 2' 10' 120 I Polynomial P (z) : 1 1 2 1 31+-z+-z +-z 2 10 120 Fade approximant of f with order (3,3) is: 1 1 2 1 31+-z+-z +-z 2 10 120 1 1 2 1 31--z+-z --z 2 10 120 The value of f(z) at 1.00000000 is: 2.71828182 The value of (P/Q) (z) at 1.00000000 is 2.71830985 The error of approximants: .00002803 > #THE SECOND FUNCTION IS SIN FUNCTION PadeApproximant(sin(z) ,1,5,evalf(Pi/4»; PadeApproximant(sin(z) ,3,3,evalf(Pi/4»; THISIS A PROGRAM USED TO CALCULATE CLASSICAL PADE APPROXIMANT OF AF\ UNCTION The function f(z) is: sin(z) The Taylor expand of f(z) is: 1 3 1 5 7 Z - - z +- z + O( z ) 6 120 So that the cofficients(from 0 to 6) of Taylor expand of f(z) is: -1 1 0,1,0,-,0, -,0 6 120 II This is sequence of numbers to form matrix for system of equations: [1~0] So that the matrix formed is: [ 1~0 ] The right of system of linear equation is: [0] Set of coeffs of Q: [ 1, 0] . Polynomial Q(z): 1 Set of coeffs of P: page 7 [-1 1 ] 0,1,0,-,0,- 6 120 Polynomial P(z): 1 3 1 5 Z--Z +-Z 6 120 Fade approximant of f with order (1,5) is: The value of The value of The error of 1 3 1 5Z--Z +-z 6 120 f (z) at .78539816 is: .70710678 (p/Q) (z) at .78539816 is .70714304 approximants: .00003626 THISIS A PROGRAM USED TO CALCULATE CLASSICAL PADE APPROXIMANT OF AF\ UNCTION The function f(z) is: sineZ) The Taylor expand of f(z) is: 1 3 1 5 7 Z - - Z +- Z + O( Z )6 120 So that the cofficients(from 0 to 6) of Taylor expand of f(z) lS: -1 1 0,1,0,-,0, -,0 6 120 This is sequence of numbers to form matrix for system of equations: [ -1 -1 -1 1 ] 10-0-0-0- "6"6"6"120 So that the matrix formed is: Therightofsystemof linearequationis: [ -1 ] 0 -0 , 120' Set of coeffs of Q: [ 1,O'2~'O] Polynomial Q (z) : 1 2 1 +-Z' 20 Set of coeffs of P: [ 0, 1, 0, ~~] page 8 -1 1 0 - 6 -1 0 - 0 6 -1 1- 0 - 6 120 Polynomial P(z): 7 3z--z 60 Fade approximant of f with order (3,3) is: 7 3z--z 60 1 21+-z 20 The value of f(z) at .78539816 is: .70710678 The value of (p/Q) (z) at .78539816 is .70706853 The error of approximants: .00003824 > #THE THIRD FUNCTION IS POWER FUNCTION PadeApproximant(sqrt(l+z) ,1,1,1); print(' jjjj------------------------------- --') ; PadeApproximant(sqrt(l+z) ,3,3,1); THIS IS A PROGRAM USED TO CALCULATE CLASSICAL PADE APPROXIMANT OF AA UNCTION The function f(z) is: ,JI;; The Taylor expand of f(z) is: 1 1 2 1 3 5 4 5 1 + - z - - z + - z - - z + D( z )2 8 16 128 So that the cofficients(from 0 to 2) of Taylor expand of f(z) is: 1 -1 1 -- , 2' 8 This is sequence of numbers to form matrix for system of equations: [;] So that the matrix formed is: [ 1 2 ] Set of coeffs of Q: The right of system of linear equation is: [t [ 1, ~] Polynomial Q (z) : I Set of coeffs of P: '1 1 +-z 4 page 9 Polynomial P(z): [ 1,~] 3 1+-z 4 Pade approximant of f with order (1,1) is: 3 1 +-z 4 1 1+-z 4 f(z) at 1.00000000 is: 1.41421356 (p/Q) (z) at 1.00000000 is 1.40000000 approximants: .01421356 The value of The value of The error of ////--------------------------------- THISISA PROGRAMUSEDTOCALCULATECLASSICALPADEAPPROXIMANTOFAF\ UNCTION The function f(z) is: [1;; The Taylor expand of f(z) is: 1 1 2 1 3 5 4 7 5 21 6 33 7 429 8 9 1 +-z--z +-z --z +-z --z +-z --z +O(z) 2 8 16 128 256 1024 2048 32768 So that the cofficients(from 0 to 6) of Taylor expandof f(z) is: 1 -1 1 -5 7 -21 1------ , 2' 8' 16'128'256'1024 This is sequence of numbers to form matrix for system of equations: [ 1 -1 1 -1 1 -5 1 -5 7 ]2' 8' 16' 8' 16'128'16'128'256 So that the matrix formed is: 1 2 -1 8 1 16 -1 8 1 16 -5 128 1 16 -5 128 7 256 Therightof systemof linearequationis: [ 5 -7 21 ]128'256'1024 Set of coeffs of Q: Polynomial Q(z): [ 5} 1 ]1,4'8'64 page10 5 3 2 1 31+-z+-z +-z 4 8 64 Set of coeffs of P: [ 7 7 7 ]1, 4' 8' 64 Polynomial P(z): 7 7 2 7 31+-z+-z +-z 4 8 64 Fade approximant of f with order (3,3) is: 7 7 2 7 31+-z+-z +-z 4 8 64 5 3 2 1 31+-z+-z +-z 4 8 64 The value of f(z) at 1.00000000 is: 1.41421356 The value of (p/Q) (z) at 1. 00000000 is 1. 41420118 The error of approximants: .00001237 > #THE FOURTH FUNCTION IS RATIONAL FUNCTION PadeApproximant((z+1)/(zA2+1) ,1,1,0.5); print(' ////------------------------------- --') ; PadeApproximant((z+1)/(zA2+1) ,2,1,0.5); THISIS A PROGRAM USED TO CALCULA TE CLASSICAL PADE APPROXIMANT OF AF\ UNCTION The function f(z) is: l+z i+ 1 The Taylor expand of f(z) is: 2 3 1 + z - z + O(Z ) So that the cofficients(from 0 to 2) of Taylor expand of f(z) is: 1, 1, -1 This is sequence of numbers to form matrix for system of equations: [ 1] So that the matrix formed is: [ 1] The right of system of linear equation is: I I Set of coeffs of Q: II [ 1] [ 1, 1] Polynomial Q (z) : 1+z I Set of coeffs of P: [ 1, 2] Polynomial P (z) : 1+2z Page 11 1+2z 1+z z z ////--------------------------------- THISIS A PROGRAM USED TO CALCULATE CLASSICAL PADE APPROXIMANT OF API UNCTION T z l+z l+ 1 2 3 4 1+z - z - z +O(z ) 3 1,1,-1,-1 [1,1,1,-1] [~ -~] Theright(ifsystemof linearequationis: [ 1,1] [1, 0, 1] () l+ 1 [ 1, 1] () l+z Pa ea 1'0imantoff ithorer IS: l+z l+ 1 ( ) 0000000 120000000 ( )() 0000000 120000000 0 00000000 #THIS IS PROGRAMUSE CLASSICAL PADE APPROXIMATION TO DO APPROXIMATE CALCULATION FOR SOME SPECIAL QUANTITIES (e & Pi, ONLY GET RESULT NOT EXPLAIN OBVIOU$LY!) PadeApproximantTB := proc(f,m,n) option operator; local taylorArray,i,coefficients,seqCoeffsTaylor,seqMatrix1, seqMatrix2,j,k,t,distanceMN, padeMatrix, detPadeMatrix,rightOfSystem,Q,P, sQ, setRootQ1,setRootQ, sP, Page12 setRootP, a, b, c ,r, temp,fQ, last; taylorArray := taylor(f, z =0, n+m+1); for i from 0 to n+m do coefficients[i] := coeff(taylorArray,z,i); od; seqCoeffsTaylor:= seq(coefficients[i],i=O..m+n); temp := 0; for i from 1 to n+m+1do temp:=temp + seqCoeffsTaylor[i]*zA(i-1); od; distanceMN:= m-n; if (distanceMN -1 > 0) then for j from (1 - distanceMN) coefficients[j] := 0; by 1 to -1 do od; fi; for t from 0 by 1 to m-1 do seqMatrix1[t] := seq(coefficients[k],k = 1-distanceMN+t. .n+t); od; seqMatrix2 := [seq(seqMatrix1[t],t = O..m-1)]; nops(seqMatrix2) ; padeMatrix := linalg[matrix] (m,m,seqMatrix2); with (linalg) ; detPadeMatrix := det(padeMatrix) ; rightOfSystem := vector([seq(-1*coefficients[1],1=n+1..n+m)]); sQ := linsolve(padeMatrix, rightOfSystem); setRootQ1[m+1] := 1; for i from 1 to m do setRootQ1[i] := sQ[i]; od; for i from 1 to m+1 do setRootQ[i] := setRootQ1[m+2-i]; od; setRootQ :=[seq(setRootQ[i],i=1..m+1)]; Q :=0; for i from 1 to m+1do Q := Q + setRootQ[i]*zA(i-1); od; fQ:=expand(temp*Q) ; for i from 0 to n do sp [i ] : = coeff (fQ, z, i) ; od; setRootP := [seq(sP[i],i=O..n)]; P:=O; Page 13 for i from 1 to n+1 do p := P + setRootP[i] * zA(i-1}; od; p/Q; end; PadeApproximantTB:=proc(f,m,n) localtaylorArray,i coefficients,eqCoeffsTaylor,seqMatrixl,seqMatrix2,j,k,t,distanceMN, padeMatrix,detPadeMatrix,rightOfSystem,Q,P, sQ,setRootQl,setRootQ,sF,setRootP,a,b, c,r, temp,jQ,last; optionoperator; taylorArray:=taylor(f,z=0,n+m+1); for i from0ton +mdocoefficients[i]:=coeff(taylorArrcry,z, i) od; seqCoeffsTaylor:=seq(coefficients[i],i =0" n+m); temp:=0; for i ton +m+1dotemp:=temp+seqCoeffsTcrylor[i ]*zt\(i-I) od; distanceMN:=m- n; if 0 <distanceMN- 1thenforj from1- distanceMNto-1docoefficientsU]:=0odfi; fortfrom0tom- 1do seqMatrixl[t]:=seq(coefficients[k],= I-distanceMN+t.. n+t) od; seqMatrix2:=[seq(seqMatrixl[t],t=0..m- 1)]; nops(seqMatrix2); padeMatrix:=linalg[matrix](m,m,seqMatrix2); withelinalg); detPadeMatrix:=det(padeMatrix); rightOfSystem:=vector([seq(-coefficients[I],1=n+1..n+m)]); sQ :=linsolve(padeMatrix,ightOfSystem); setRootQl[m+1] :=1; for i tomdosetRootQl[i]:=sQ[i] od; fori tom+1dosetRootQ[i]:=setRootQl[m+2- i] od; setRootQ:=[seq(setRootQ[i],i =1.. m+1)]; Q :=0; fori tom+1doQ :=Q+setRootQ[i]*zt\(i- 1)od; fQ :=expand(temp*Q); for i from0ton dosP[i] :=coeff(fQ,z,i) od; setRootP:=[seq(sP[i], i =0.. n)]; P :=0; for i ton +1doP :=P +setRootP[i]*zt\(i- 1)od; P/Q page14 c end > printf('TEST FOR USING PADE APPROXIMANT TO CALCULATE NUMBER E (%.10f)\n' ,evalf(exp(l»); printf('Order\t\t\t\tApproximant\t\t\t\t\t\t\t for i from 1 to 6 do printf(' (%d,%d)\t\t\t\t%.10f\t\t\t\t\t%.10f',i,i,evalf(subs(z=1, PadeApproximantTB(exp(z) ,i,i,l») ,evalf(subs(z=l,PadeApproximant TB (exp ( z) , i , i , 1) ) - exp ( 1) ) ) ; printf ( '\n' ) ; print(PadeApproximantTB(exp(z),i,i,l» ; oct; TEST FOR USING PADE APPROXIMANTTO CALCULATENUMBERE (2.7182818280) Order Approximant Error (1,1) 3.0000000000 .2817181720 1 1 2 1 31+-z+-z +-z 2 10 120 1 1 2 1 3l--z+-z --z 2 10 120 -.0000001100 1321314 1+-z+-z +-z +-z 2 28 84 1680 1321314 l--z+-z --z +-z 2 28 84 1680 .0000000010 1 1 2 1 3 1 4 1 51+-z+-z +-z +-z +-z 2 9 72 1008 30240 1 1 2 1 3 1 4 1 5l--z+-z --z +-z --z 2 9 72 1008 30240 2.7182818280 0.0000000000 152131415 16 1+-z+-z +-z +-z + :--z + z 2 44 66 792 15840 665280 152131415 16 l--z+-z --z +-z --z + z 2 44 66 792 15840 665280 I> printf('TEST FOR USING PADE APPROXIMANT TO CALCULATE Page 15 (2,2) 2.7142857140 (3,3) 2.7183098590 (4,4) 2.7182817180 (5,5) 2.7182818290 (6,6) Error\n') ; 1 1+-z 2 1 l--z 2 -.0039961140 1 1 21+-z+-z 2 12 1 1 2l--z+-z 2 12 .0000280310 NUMBER Pi (%.10f)\n',evalf(arctan(1)*4» ; printf('Order\t\t\t\tApproximant\t\t\t\t\t\t\t for i from 1 to 5 do printf(' (%d,%d)\t\t\t\t%.10f\t\t\t\t\t%.10f',i,i,evalf(subs(z=1, PadeApproximantTB(arctan(z) ,i,i»)*4,evalf(subs(z=1,PadeApproxim antTB(arctan(z) ,i,i» -arctan(1» *4); printf ( '\n' ) ; print(PadeApproximantTB(arctan(z) ,i,i»; oct; TEST FOR USING PADE APPROXIMANTTO CALCULATENUMBERPi (3.1415926540) Order Approximant Error (1,1) 4.0000000000 .8584073460 Error\n') ; (2,2) 3.0000000000 z -.1415926540 z 1 21+-z 3 (3,3) 3.1666666670 .0250740128 (4,4) 3.1372549020 4 3z+-z 15 3 21+-z 5 -.0043377520 11 3z+-z 21 (5,5) 3.1423423420 6 2 3 41+-z +-z 7 35 .0007496884 7 3 64 5z+-z +-z 9 945 10 2 5 41+-z +-z 9 21 '> #Integral FUNCTION Integral:=proc(f,omega) option operator; local setRootOmega, i, total Integral, temp; readlib(residue) ; setRootOmega1 := {solve(omega)}; for i from 1 to nops(setRootOmega1) do setRootOmega[i] :=evalf(setRootOmega1[i]); oct; totalIntegral:= 0; temp := f/omega; for i from 1 to nops(setRootOmega1) do totalIntegral:=totalIntegral + 2*Pi*I*residue(temp, page 16 z=setRootOmega[i]) ; od; normal (evalf(totalIntegral» ; #totalIntegral; end; Warning, 'setRootOmegal' is implicitly declared local Integral:=proc(f,co) localsetRootOmega,i, totalIntegral,temp,setRootOmegal; optionoperator; readlib(residue); setRootOmegal:={solve(co)}; for i tonops(setRootOmegal)dosetRootOmega[i] :=evalf(setRootOmegal[i] ) od; totalIntegral:=0; temp:=i/ co; for i tonops(setRootOmegal)do totalIntegral:=totalIntegral+2*I*rc*residue(t mp,z=setRootOmega[i]) od; normal(evalf(totalIntegral)) Lend r> #PadeApproximant FUNCTION PadeApproximantK := proc(f,m,n,omega,value) option operator; local seqIntegral,i,t, seqMatrix1, seqMatrix2, padeMatrix, Q,k, setRootQ1, setRootQ, sQ, rightOfSystem, setRootP, setRootP1, P, t1, t2 ; print('THIS IS A PROGRAMUSED TO CALCULATE CLASSICAL PADE APPROXIMANT OF AFUNCTION'); printf('\n\n') ; printf('\tThe function f(z) is:',f); print(f) ; seqIntegral := [seq(evalf(Integral(f*zAi, omega», i=O..2*m-1)]; print (seqIntegral) ; for t from 1 to m do seqMatrix1[t] := seq(seqIntegral[k+t] ,k = 1. .m); od; seqMatrix2 := [seq(seqMatrix1[t],t = 1..m)]; printf('\tThis is sequence of numbers to form of equations: '); print (seqMatrix2) ; matrix for system Page17 nops(seqMatrix2) ; padeMatrix := linalg[matrix] (m,m,seqMatrix2); printf('\tSo that the matrix formed is: '); print (padeMatrix) ; with (linalg) ; rightOfSystem := vector([seq(-1*seqlntegral[1],1=1..m)]); printf('The right of system of linear equation is: '); print (rightOfSystem) ; sQ := linsolve(padeMatrix, rightOfSystem); for i from 1 to nops(sQ) do print(whattype(sQ[i]» ; cd; print (sQ) ; setRootQ1[0] :=1; for i from 1 to m do setRootQ1[i] := sQ[i]; cd; setRootQ :=[seq(setRootQ1[i],i=0..m)]; printf('\tSet of coeffs of Q: '); print (setRootQ) ; Q :=0; for i from 1 to m+1 do Q := Q + setRootQ[i]*zA(i-1); cd; printf('\tPolynomial Q(z) :'); print(Q) ; setRootP1 := [seq(Integral(Q*f*zAi,omega) ,i=m..m+n)]; print(setRootP1); setRootP:=[seq(setRootP1[n+2-i]/(2*Pi*I) ,i=1. .n+1)]; printf('\tSet of coeffs of P: '); print (setRootP) ; P :=0; for i from 1 to n+1 do P := P + setRootP[i]*zA(i-1); cd; printf('\tPolynomial P(z) :'); print(P) ; printf('\tPade approximant of print(normal(evalf(P/Q») ; f with order (%d,%d) is: ',m,n); t1:=evalf(subs(z=value,f» ; t2:=evalf(subs(z=value,P/Q» ; printf('The value of f(z) at %f is: %f \n',value,t1); printf('The value of P(z)/Q(z) at %f is: %a \n',value,t2); printf('The error of approximants: %.8f', abs(t1-t2»; end; Page 18 PadeApproximantK:=proc(f,m,n,0),value) localseqIntegral,i, t,seqMatrix1,seqMatrix2,padeMatrix,Q,k,setRootQ1,setRootQ,sQ, rightOfSystem,setRootP,setRootP1,P, t1,t2; optionoperator; print('THISIS A PROGRAMUSEDTOCALCULATECLASSICBLPADEAPPROXIMA\ NT OFAFUNCTION'); printf('Inln'); printf('liThefunctionf(z) is:',f); print(f); seqIntegral:=[seq(evalf(Integral(f*zl\i,0)), i =0" 2*m- 1)]; print(seqIntegral); for t to m doseqMatrix1[t] :=seq(seqIntegral[k +t], k = 1 .. m) od; seqMatrix2:=[seq(seqMatrix1[t],= 1.. m)]; printtr:'ItThisissequenceofnumberstoformmatrixfor systemofequations:'); print(seqMatrix2); nops(seqMatrix2); padeMatrix:=linalg[matrix](m,m,seqMatrix2); printf('ItSothatthematrixformedis: '); print(padeMatrix); withelinalg); rightOfSystem:=vector([seq(-seqIntegral[I], 1=1" m)]); printf('Therightofsystemoflinearequationis: '); print(rightOfSystem); sQ :=linsolve(padeMatrix,rightOfSystem); for i tonops(sQ) doprint(whattype(sQ[i] )) od; print(sQ); setRootQ1[0]:=1; for i tomdosetRootQ1[i]:=sQ[i] od; setRootQ:=[seq(setRootQ1[i],=0.. m)]; printf('ItSetofcoeffsofQ: '); print(setRootQ); Q '=0', , for i tom+1doQ :=Q +setRootQ[i]*zl\(i- 1)od; printtr:'ItPolynomialQ(z):'); print(Q); setRootP1:=[seq(Integral(Q*f*zl\i,0),i =m'" n+m)]; print(setRootP1); setRootP:=[seq(- 1/ 2*I*setRootPl[n +2- i] / n, i =1..n+1)]; printf('ItSetofcoeffsofP: '); page19 print(setRootP); P :=0; fori ton+1doP :=P +setRootP[i]*zJ\(i- 1)od; printf('ItPolynomialP(z):'); print(P); print£{'ItPadeapproximantoffwithorder(%d,%d)is: " m,n); print(normal(evalf(P / Q) )); t1:=evalf(subs(z=value,f)); t2:=evalf(subs(z=value,P / Q)); printf('Thevalueoff(z)at%fis: %fIn" value,t1); printf('ThevalueofP(z)/Q(z)at %fis: %a In" value,t2); i printf('Theerrorofapproximants:%.8j', abs(tl - t2)) lend r> PadeApproximantK(exp(z+l) ,2,1,zA4-l,O); #IN CASE THE OMEGA FUNCTION: zAn. WE HAVE THE CLASSICAL PADE APPROXIMANT ION PadeApproximantK(exp(z) ,1,2,zA4,1); THISISA PROGRAMUSEDTOCALCULATECLASSICALPADEAPPROXIMANTOFAF\ UNCTION The function f(z) is: e(z+l) [2.8499672801,8.563460414I, 17.22184443I, 17.79153661I] II This is sequence of numbers to form matrix for system of equations: [8.563460414I, 17.22184443I, 17.22184443I, 17.79153661I] SO that the matrix formed is: [ 8.563460414I 17.22184443I ]17.22184443I 17.79153661I II The right of system of linear equation is: [-2.849967280I, -8.563460414I] float [-.6709426697,.1681366781] Set of coeffs of Q: [1, -.6709426697,.1681366781] Polynomial Q(z): 1- .6709426697z+.1681366781 i [5.7639273881,17.319203731] Set of coeffs of P: [ 8.659~OI865.2.881:63694] Polynomial P(z): Page 20 8.659601865 z + 2.881963694- n n Fade approximant of f with order (2,1) is: The value of The value of The error of 2.756436883+.9173575352z 1. - .6709426697Z +.1681366781i f(z) at 0 is: 2.718281 P(z)/Q(z) at 0 is: 2.756436883 approximants: .03815505 THISISA PROGRAMUSEDTOCALCULATECLASSICALPADEAPPROXIMANTOFAA UNCTION The function f(z) is: ze [1.0471975511,3.1415926541] This is sequence of numbers to form matrix for system of equations: [3.1415926541] So that the matrix formed is: [3.1415926541] The right of system of linear equation is: [ -1.0471975511] float [ -.3333333332] Set of coeffs of Q: [1, -.3333333332] Polynomial Q(z): 1- .3333333332 Z Set of coeffs of [1.0471975521,4.1887902071,6.2831853081] P: [ 3.141592654,2.094395104,.5235987760 ]n n n I PolynomialP(z): . 3.141592654 z i I +2.094395104- +.5235987760-n n n Fade approximant of f with order (1,2) is: .9999999999+ .6666666670Z + .1666666668i -1. +.3333333332Z The value of f(z) at 1 is: 2.718281 The value of P(z)/Q(z) at 1 is: 2.750000000 The error of approximants: .03171817 ] > PadeApproximantK(sqrt(z+2) ,1,2,ZA4-1,-1); PadeApproximantK(sqrt(z+2) ,2,1,ZA4-1,-1); PadeApproximantK(sqrt(z+2) ,1,2,ZA4-1,O); I PadeApproximantK(sqrt(z+2) ,2,1,ZA4-1,O); I I THISISA PROGRAMUSEDTOCALCULATECLASSICALPADEAPPROXIMANTOFAA page 21 UNCTION The function f(z) is: 0 [.0705747922 I, -.280611096 I] This is sequence of numbers to form matrix for system of equations: [ -.280611096 I] So that the matrix formed is: [-.2806110961] The right of system of linear equation is: [ -.0705747922 I] float [ .2515039256] Set of coeffs of Q: [1, .2515039256] Polynomial Q (z) : 1 + .2515039256 z [.280049163 I, 4.458461306 I, 8.881351683 I] Set of coeffs of P: [ 4.440675842, 2.229230653, .1400245815 ]n n n Polynomial P(z): 4.440675842 z i + 2.229230653 - + .1400245815 - n n n Pade approximant of f with order (1,2) is: 1.413511021 + .7095861552 z +.04457120859 i 1. + .2515039256 z The value of f(z) at -1 is: 1 The value of P(z)/Q(z) at -1 is: 1.000000001 The error of approximants: .00000000 THISISA PROGRAMUSEDTOCALCULATECLASSICALPADEAPPROXIMANTOFAA UNCTION The function f(z) is: 0 [.0705747922I, -.2806110961,2.229230648I, 8.863601844I] This is sequence of numbers to form matrix for system of equations: [-.2806110961,2.2292306481,2.2292306481,8.863601844I] SO that the matrix formed is: [ -.280611096I 2.229230648I ]2.229230648I 8.863601844I The right of system of linear equation is: [ -.0705747922I, .280611096I] float Page 22 j;~m2 [.07057479221,-.2806110961,2.2292306481,8.863601844I] This is sequence of numbers to form matrix for system of equations: [-.280611096I, 2.229230648I, 2.229230648I, 8.863601844I] So tho. the matr orIn.ed is: [ -.280611096I 2.229230648I J2.229230648I 8.863601844I i The r'i of system of linear equation is: [-.0705747922I, .280611096I] float [.1677812797,-.01053883924] Set 0 coeffs of Q: [1, .1677812797,-.01053883924] Pol Q (z) : 1+.1677812797z - .01053883924i [3.715633335I, 8.878400291I] Set of coeffs of P: [ 4.439~OO146, L857~16668] Polynomial P(z): 4.439200146 z +1.857816668- TC TC Pade approximant of f with order (2,1) is: 1.413041293+.5913614120z -1. - .1677812797z+ .01053883924i The value of f(z) at -1 is: 1 'I'he value of P(z)!Q(z) at -1 is: .9999999999 The error of approximants: .00000000 THIS IS A PROGRAl'vfUSED TO CALCULATE CLASSICAL PADE APPROXLvfANTOF AFU\ NCTION The function f(z) is: ~z+2 [.0705747922I, -.280611096I] This is sequence of numbers to form matrix for system of [-.280611096I] So that the matrix formed is: The r: [-.280611096I] t of system of linear equation is: [-.0705747922I] float [.2515039256] Set of coeffs of Q: [1,.2515039256] Q (z) : Page 23 1+.2515039256z [.2800491631,4.4584613061,8.8813516831] Set of coeffs of [ 4.440675842,.229230653,.1400245815]n n n p (z) : 4.440675842 z i +2.229230653-+ .1400245815- n n n Pade approximant of f with order (1,2) is: 1.413511021+.7095861552z +.04457120859 1.+.2515039256z The value of f(z) at 0 is: 1.414213 The value of P(z) (z) at 0 is: 1.413511021 The error of' armrox!mants: . 0 70254 I TIlLS'IS A PROGRAA1USED TO CALCULATE CLASSICAL PADE APPROXlJ\;fANTOF AFU\ NCTION The Yunction f(z) is: ~z+2 [.07057479221,-.280611096!,2.229230648!, 8.8636018441] This is sequence of numbers to form matrix for system of eQuations: [-.2806110961,2.2292306481,2.229230648J, 8.8636018441] So that the matrix Yorrned is: [ -.28061109612.229230648! J2.22923064818.8636018441 The right of system of linear equaLian is: [-.0705747922J, .280611096!] float [.1677812797,-'(H053883924] Set of caeffs of Q: [1, .1677812797,-.01053883924] Pol"rnamial Q (z) : 1+.1677812797z - .01053883924i [3.715633335J, 8.878400291!] Set of coeffs of P: [ 4.439~OOI46,1.857:16668] Pol 2 (z) : 4.439200146+1.857816668~n Pade approximant of f with order (2,1) is: 1.413041293+.5913614120z -1. - .1677812797z + .01053883924i The value of f(z) at 0 is: 1.414213 i The value of 2(z)/Q(z) at 0 is: 1.413041293 Page 24