Re: Rational approximation of exponential function
From: Alex.Lupas (alexandru.lupas_at_ulbsibiu.ro)
Date: 06/15/04
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Date: 14 Jun 2004 23:58:18 -0700
kariwala@ualberta.ca (Vinay Kariwala) wrote in message news:<200406112211.i5BMBDa10382@proapp.mathforum.org>...
> Dear all,
> This problem would be trivial for most of you. I am looking for an
> approximation of the exponential function, e^(-x) with the following properties:
> 1. The nth order approximation f(x,n) should be rational, i.e. it
> should be possible to write f(x,n) as ratio of polynomials in x. f
> (x,n) = g(x,n)/h(x,n), where g(x,n) has less number of zeros than h(x,n).
>
> 2. As n approaches infinity, f(x,n) should converge back to the
> exponential function.
> Any comments or suggestions will be very helpful.
> Thanks Vinay
Try in following way : integrating by parts one finds
(Ch.Hermite,K. Petr, Obreschkoff,....) : for f in C^{2n+1}[a,b] , x in [a,b],
there exists a point c in [a,b] such that
==========================================================================
(1) f(x)= f(a) +
+ SUM_{k=1 to k=n}C(n,k)(2n-k)!(x-a)^k( f^{(k)}(a)-(-1)^kf^{(k)}(x) )/(2n)! +
+ r_n(f;x)
================================================================
where C(n,k):= n!/(k!(n-k)!) and
r_n(f;x)=((-1)^n/(2n) !) INT_{t=a to t=x}(t-a)^n(x-t)^n f^{(2n+1)}(t) dt =
=(-1)^n\frac{(x-a)^{2n+1}f^{(2n+1)}(c)/(C(2n,n) (2n+1) ! ) .
Remark: Above formula (1) may be obtained by integrating, on [a,x],
the Hermite interpolation polynomial H_{2n-1}(g;t) on the knots
a,a,a,...,a , x,x,x, ...,x , that is integrating the
\ n-times/ \ n -times/
equality
g(t)=H_{2n-1}(g;t) +R(g;t) , t in [a,x]
where g=f' , R(g;t) being the remainder term.
Further consider in (1) f(t)= e^{t} and a=0 , x in R .
Denote ======================
c_k(n)=C(n,k)(2n-k)!/n! , k in { 0,1,...,n } ,
=======================
=======================================
(2) E_n(x) = SUM_{k=0 to k=n}c_k(n)x^k .
=======================================
According to (1) you find the equality
================================
(3) E_n(-x)*e^x = E_n(x)+ e_n(x)
================================
where
(4) e_n(x)=((-1)^n x^{2n+1})/n !)INTEGRAL_{t=0 to t=1}t^n(1-t)^n e^{xt} dt .
Moreover there exists c=c(n,x) in the interval ( min{(0, x)},max{(0,x)})
such that
(4.1) e_n(x)= (-1)^n x^{2n+1} n !)e^c )/(2n+1)! .
Polynomials E_n(x) are connected with :
- Legendre Polynomials , Laguerre polynomials
- Bessel Polynomials
- Confluent hypergeometric functions
- Func'tiile lui Whittaker functions.
The first polynomials E_n(x) are :
E_0(x)= 1 , E_1(x)=x+2 , E_2(x)=x^2 +6x+12
E_3(x)=x^2+60x +120
E_5(x)=x^5+ 30x^4+420x^3+3360x^2+15120x+30240
E_6(x)=x^6+42x^5+840x^4+10080x^3+75600x^2+332640x+665280 .
Observe that E_n(0)=(2n)!/n! .
========================================
E_{n+1}(x)=2(2n+1)E_n(x)+x^2*E_{n-1}(x)
========================================
By dividing in (3) one finds
======================================
e^x = E_n(x)/E_n(-x) + remainder_n(x)
====================================
where remainder_n(x):= e_n(x)/E_n(-x) .
To do this we must be sure that E_n(-x)=/=0 . Therefore arise the problem
to investigate the roots of E_n(x).
It may be shown (Kakeya Theorem) that E_n(x) has at most one real root.
A more precis result is : if E_n(z_k)=0 , k in {1,2,...,n}, then
n+1 =< |z_k|=< n(n+1) , (n >= 1)
Therefore we have the approximative formula
=================================
e^x =approx=E_n(x)/E_n(-x) for |x| < 2 .
========================
Consider n=6 , Q = ln{2} = 0.69314718055994530942... and use formula
e^y = 2^N e^{x(y)} , |y|<300 , |x(y)| < Q ,
with N=[x/Q] , x(y)= Q*{x/Q}. Here [.] denotes integral part and
{.}=fractionary part.
I try to send you a .pdf file with detailed proofs. Regards/Alex
=============
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