EXPLANATION FILE FOR LIMITED DEVELOPMENTS ========================================= 1. Introduction ------------ This file is made of two parts: - limited development at order 5 of a real function f(x) in one point; - limited developments of f.g, f/g and g o f knowing the limited dev. of functions f(x) and g(x) (3 programs). 2. Limited Development of f(x) at x=x0 ----------------------------------- 2.1 Principle --------- The limited development at order n of a function f(x) at local point, x0 is given by Taylor's formula: f(x0) 2 f'(x0) 3 f"(x0) f(x) = f(x0) + (x-x0)----- + (x-x0) ------ + (x-x0) ------ 1! 2! 3! (n) n f (x0) n+1 + ... + (x-x0) -------- + O((x-x0) ) n! This can be written as: 2 3 n n+1 f(x) = a0 + a1(x-x0) = a2(x-x0) + a3(x-x0) + ... + an(x-x0) + O((x-x0) ) (i) f (x0) with ai = -------- i! In the special case when x0 = 0, the formula becomes: 2 n n+1 f(x) = a0 + a1x +a2x + ... + anx + O(x ) 2.2 Program Ltddev -------------- The program calculates the limited development of f(x) = 1/(1-x) for x=0 at order 5 2 3 4 5 6 f(x) = 1 + x + x + x + x + x + O(x ) The exact values of a1...a5 equal 1. The program Ltddev finds; a1 = 1.0000416250 a2 = 1.0000416249 a3 = 1.0011207937 a4 = 1.0011208001 a5 = 1.0136468275 As we can see, the direct method is not very accurate. 3. Limited development of f.g -------------------------- 3.1 Introduction ------------ The next limited developments will be for x=0. Usual functions have well known limited developments. We can use these known limited developments to find the limited dev. of other functions. The next three programs will calculate, at high orders and with great accu- racy the limited developments of functions f.g, f/g and g o f, knowing the li- mited dev. of f and g. 3.2 Principle --------- Let us begin with the product of two functions, f.g: 2 n n+1 n+1 f(x) = a0 + a1x +a2x + ... + anx + O(x ) = P(x) + O(x ) 2 n n+1 n+1 g(x) = b0 + b1x +b2x + ... + bnx + O(x ) = Q(x= + O(x ) So the limited development of f.g is: 2 n n+1 (f.g) (x) = c0 + c1x +c2x + ... + cnx + O(x ) The ci coefficients can be easily calculated by multiplying polynomials P(x) and Q(x): i ci = Sum a b k=0 k i-k 3.3 Program Ltddev1 --------------- In the given example, f(x) = exp(x)/(1+x). 2 3 n n+1 f(x) = exp(x) = 1 + x + x /2! + x /3! + ... + x /n! + O(x ) 1 2 n n n+1 g(x) = --- = 1 - x + x + ... + (-1) x + O(x ) 1+x Let us take n=5 We input the following coefficients ai and bi: a0 = 1 a1 = 1 a2 = 0.5 a3 = 0.1666667 a4 = 0.0411667 a5 = 0.0083333 b0 = 1 b1 = -1 b2 = 1 b3 = -1 b4 = 1 b5 = -1 The results are: c0 = 1.0000000 c1 = 0.0000000 c2 = 0.5000000 c3 = -0.3333333 c4 = 0.3750000 c5 = -0.3666667 Exact values are: 1, 0, 1/2, 3/8 and -11/30. We can see that these results are much better than those obtained with the direct method. We can still improve these results by taking, for instance, n=15. 4. Limited development of f/g -------------------------- 4.1 Principle --------- 2 n n+1 n+1 f(x) = a0 + a1x +a2x + ... + anx + O(x ) = P(x) + O(x ) 2 n n+1 n+1 g(x) = b0 + b1x +b2x + ... + bnx + O(x ) = Q(x= + O(x ) Here, me must have b0 <> 0, else f/g is not defined for x=0. So we have to divide P(x) by Q(x) at order n with increasing powers. It can be proved that the ci coefficients obey the following recursive formula: i c0 = a0/b0 ci = ai - Sum bk.ci-k k=1 4.2 Program Ltddev2 --------------- Let us take again the example h(x) = exp(x) / (1+x) Let us take n=6 We input the following coefficients ai and bi: a0 = 1 a1 = 1 a2 = 0.5 a3 = 0.1666667 a4 = 0.0411667 a5 = 0.0083333 a6 = 0.0013889 b0 = 1 b1 = 1 b2 = 0 b3 = 0 b4 = 0 b5 = 0 b6 = 0 The results are: c0 = 1.0000000000 c1 = 0.0000000000 c2 = 0.5000000000 c3 = -0.3333333333 c4 = 0.3750000000 c5 = -0.3666666667 c6 = 0.3680555556 The results are very good. 5. Limited Development of g o f ---------------------------- 5.1 Principle --------- 2 n n+1 n+1 g(x) = a0 + a1x +a2x + ... + anx + O(x ) = P(x) + O(x ) 2 n n+1 n+1 f(x) = b0 + b1x +b2x + ... + bnx + O(x ) = Q(x= + O(x ) Here, me must have b0 = 0, so as f(x) --> 0 when x --> 0. Else we cannot define the limited dev. of g o f at x=0. The limited dev. of g o f is: n+1 (gof) (x) = P{Q(x)} + O(x ) or under more explicit form: 2 n (gof) (x) = a0 + a1[b1x+b2x +...+bnx ] 2 n 2 + a2[b1x+b2x +...+bnx ] 2 n 3 + a3[b1x+b2x +...+bnx ] + --------------------- 2 n n N+1 + an[b1x+b2x +...+bnx ] + O(x ) 2 This development has terms of degree 0 to n . We keep only the terms of degree <= n. 2 n i It can be proved that the xm coefficient of [b1x+b2x +...+bnx ] is given by g = Sum b .b ...b i,m alpha1..alphai alpha1 alpha2 alphai the summation being made for all i-uplets (alpha1..alphai), such as: i Sum alpha = m j=1 j By summing up all partial coefficients for each i, we obtain the xm coeffi- cient of the g o f limited development: n cm = Sum g i=1 i,m 5.2 Program Ltddev3 --------------- The amount of calculation is much greater than in previous two programs. For that reason, n is limited to 30. Let us take the example: h(x) = exp(sin(x)) and n=10. The results are: c0 = 1.0000000000 c1 = 1.0000000000 c2 = 0.5000000000 c3 = 0.0000000000 c4 = -0.1250000000 c5 = -0.0666666667 c6 = -0.0041666667 c7 = 0.0111111111 c8 = 0.0053819444 c9 = 0.0004763668 c10 = -0.0008132165 -9 Results are good, accuracy is better than 10 . From [BIBLI 03]. ----------------------------------------------- End of file Ltddev.txt