UPDATE: Dump of initial files
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Nicolás Hatcher
329
base/src/functions/engineering/transcendental/bessel_jn_yn.rs
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329
base/src/functions/engineering/transcendental/bessel_jn_yn.rs
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// https://github.com/JuliaLang/openlibm/blob/master/src/e_jn.c
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/*
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* ====================================================
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* Copyright (C) 1993 by Sun Microsystems, Inc. All rights reserved.
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*
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* Developed at SunSoft, a Sun Microsystems, Inc. business.
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* Permission to use, copy, modify, and distribute this
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* software is freely granted, provided that this notice
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* is preserved.
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* ====================================================
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*/
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/*
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* __ieee754_jn(n, x), __ieee754_yn(n, x)
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* floating point Bessel's function of the 1st and 2nd kind
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* of order n
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*
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* Special cases:
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* y0(0)=y1(0)=yn(n,0) = -inf with division by 0 signal;
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* y0(-ve)=y1(-ve)=yn(n,-ve) are NaN with invalid signal.
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* Note 2. About jn(n,x), yn(n,x)
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* For n=0, j0(x) is called,
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* for n=1, j1(x) is called,
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* for n<x, forward recursion us used starting
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* from values of j0(x) and j1(x).
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* for n>x, a continued fraction approximation to
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* j(n,x)/j(n-1,x) is evaluated and then backward
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* recursion is used starting from a supposed value
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* for j(n,x). The resulting value of j(0,x) is
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* compared with the actual value to correct the
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* supposed value of j(n,x).
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*
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* yn(n,x) is similar in all respects, except
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* that forward recursion is used for all
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* values of n>1.
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*
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*/
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use super::{
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bessel_j0_y0::{j0, y0},
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bessel_j1_y1::{j1, y1},
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bessel_util::{split_words, FRAC_2_SQRT_PI},
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};
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// Special cases are:
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//
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// $ J_n(n, ±\Infinity) = 0$
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// $ J_n(n, NaN} = NaN $
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// $ J_n(n, 0) = 0 $
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pub(crate) fn jn(n: i32, x: f64) -> f64 {
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let (lx, mut hx) = split_words(x);
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let ix = 0x7fffffff & hx;
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// if J(n,NaN) is NaN
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if x.is_nan() {
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return x;
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}
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// if (ix | (/*(u_int32_t)*/(lx | -lx)) >> 31) > 0x7ff00000 {
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// return x + x;
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// }
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let (n, x) = if n < 0 {
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// hx ^= 0x80000000;
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hx = -hx;
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(-n, -x)
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} else {
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(n, x)
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};
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if n == 0 {
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return j0(x);
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}
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if n == 1 {
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return j1(x);
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}
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let sign = (n & 1) & (hx >> 31); /* even n -- 0, odd n -- sign(x) */
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// let sign = if x < 0.0 { -1 } else { 1 };
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let x = x.abs();
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let b = if (ix | lx) == 0 || ix >= 0x7ff00000 {
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// if x is 0 or inf
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0.0
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} else if n as f64 <= x {
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/* Safe to use J(n+1,x)=2n/x *J(n,x)-J(n-1,x) */
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if ix >= 0x52D00000 {
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/* x > 2**302 */
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/* (x >> n**2)
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* Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
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* Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
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* Let s=x.sin(), c=x.cos(),
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* xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
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*
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* n sin(xn)*sqt2 cos(xn)*sqt2
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* ----------------------------------
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* 0 s-c c+s
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* 1 -s-c -c+s
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* 2 -s+c -c-s
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* 3 s+c c-s
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*/
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let temp = match n & 3 {
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0 => x.cos() + x.sin(),
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1 => -x.cos() + x.sin(),
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2 => -x.cos() - x.sin(),
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3 => x.cos() - x.sin(),
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_ => {
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// Impossible: FIXME!
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// panic!("")
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0.0
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}
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};
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FRAC_2_SQRT_PI * temp / x.sqrt()
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} else {
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let mut a = j0(x);
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let mut b = j1(x);
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for i in 1..n {
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let temp = b;
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b = b * (((i + i) as f64) / x) - a; /* avoid underflow */
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a = temp;
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}
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b
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}
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} else {
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// x < 2^(-29)
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if ix < 0x3e100000 {
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// x is tiny, return the first Taylor expansion of J(n,x)
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// J(n,x) = 1/n!*(x/2)^n - ...
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if n > 33 {
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// underflow
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0.0
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} else {
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let temp = x * 0.5;
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let mut b = temp;
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let mut a = 1;
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for i in 2..=n {
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a *= i; /* a = n! */
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b *= temp; /* b = (x/2)^n */
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}
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b / (a as f64)
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}
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} else {
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/* use backward recurrence */
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/* x x^2 x^2
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* J(n,x)/J(n-1,x) = ---- ------ ------ .....
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* 2n - 2(n+1) - 2(n+2)
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*
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* 1 1 1
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* (for large x) = ---- ------ ------ .....
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* 2n 2(n+1) 2(n+2)
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* -- - ------ - ------ -
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* x x x
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*
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* Let w = 2n/x and h=2/x, then the above quotient
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* is equal to the continued fraction:
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* 1
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* = -----------------------
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* 1
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* w - -----------------
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* 1
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* w+h - ---------
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* w+2h - ...
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*
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* To determine how many terms needed, let
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* Q(0) = w, Q(1) = w(w+h) - 1,
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* Q(k) = (w+k*h)*Q(k-1) - Q(k-2),
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* When Q(k) > 1e4 good for single
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* When Q(k) > 1e9 good for double
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* When Q(k) > 1e17 good for quadruple
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*/
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let w = ((n + n) as f64) / x;
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let h = 2.0 / x;
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let mut q0 = w;
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let mut z = w + h;
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let mut q1 = w * z - 1.0;
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let mut k = 1;
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while q1 < 1.0e9 {
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k += 1;
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z += h;
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let tmp = z * q1 - q0;
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q0 = q1;
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q1 = tmp;
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}
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let m = n + n;
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let mut t = 0.0;
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for i in (m..2 * (n + k)).step_by(2).rev() {
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t = 1.0 / ((i as f64) / x - t);
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}
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// for (t=0, i = 2*(n+k); i>=m; i -= 2) t = 1/(i/x-t);
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let mut a = t;
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let mut b = 1.0;
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/* estimate log((2/x)^n*n!) = n*log(2/x)+n*ln(n)
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* Hence, if n*(log(2n/x)) > ...
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* single 8.8722839355e+01
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* double 7.09782712893383973096e+02
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* long double 1.1356523406294143949491931077970765006170e+04
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* then recurrent value may overflow and the result is
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* likely underflow to 0
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*/
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let mut tmp = n as f64;
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let v = 2.0 / x;
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tmp = tmp * f64::ln(f64::abs(v * tmp));
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if tmp < 7.097_827_128_933_84e2 {
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// for(i=n-1, di=(i+i); i>0; i--){
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let mut di = 2.0 * ((n - 1) as f64);
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for _ in (1..=n - 1).rev() {
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let temp = b;
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b *= di;
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b = b / x - a;
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a = temp;
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di -= 2.0;
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}
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} else {
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// for(i=n-1, di=(i+i); i>0; i--) {
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let mut di = 2.0 * ((n - 1) as f64);
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for _ in (1..=n - 1).rev() {
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let temp = b;
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b *= di;
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b = b / x - a;
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a = temp;
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di -= 2.0;
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/* scale b to avoid spurious overflow */
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if b > 1e100 {
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a /= b;
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t /= b;
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b = 1.0;
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}
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}
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}
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let z = j0(x);
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let w = j1(x);
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if z.abs() >= w.abs() {
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t * z / b
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} else {
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t * w / a
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}
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}
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};
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if sign == 1 {
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-b
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} else {
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b
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}
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}
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// Yn returns the order-n Bessel function of the second kind.
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//
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// Special cases are:
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//
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// Y(n, +Inf) = 0
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// Y(n ≥ 0, 0) = -Inf
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// Y(n < 0, 0) = +Inf if n is odd, -Inf if n is even
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// Y(n, x < 0) = NaN
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// Y(n, NaN) = NaN
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pub(crate) fn yn(n: i32, x: f64) -> f64 {
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let (lx, hx) = split_words(x);
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let ix = 0x7fffffff & hx;
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// if Y(n, NaN) is NaN
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if x.is_nan() {
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return x;
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}
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// if (ix | (/*(u_int32_t)*/(lx | -lx)) >> 31) > 0x7ff00000 {
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// return x + x;
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// }
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if (ix | lx) == 0 {
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return f64::NEG_INFINITY;
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}
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if hx < 0 {
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return f64::NAN;
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}
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let (n, sign) = if n < 0 {
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(-n, 1 - ((n & 1) << 1))
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} else {
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(n, 1)
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};
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if n == 0 {
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return y0(x);
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}
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if n == 1 {
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return (sign as f64) * y1(x);
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}
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if ix == 0x7ff00000 {
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return 0.0;
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}
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let b = if ix >= 0x52D00000 {
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// x > 2^302
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/* (x >> n**2)
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* Jn(x) = cos(x-(2n+1)*pi/4)*sqrt(2/x*pi)
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* Yn(x) = sin(x-(2n+1)*pi/4)*sqrt(2/x*pi)
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* Let s=x.sin(), c=x.cos(),
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* xn=x-(2n+1)*pi/4, sqt2 = sqrt(2),then
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*
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* n sin(xn)*sqt2 cos(xn)*sqt2
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* ----------------------------------
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* 0 s-c c+s
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* 1 -s-c -c+s
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* 2 -s+c -c-s
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* 3 s+c c-s
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*/
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let temp = match n & 3 {
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0 => x.sin() - x.cos(),
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1 => -x.sin() - x.cos(),
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2 => -x.sin() + x.cos(),
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3 => x.sin() + x.cos(),
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_ => {
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// unreachable
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0.0
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}
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};
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FRAC_2_SQRT_PI * temp / x.sqrt()
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} else {
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let mut a = y0(x);
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let mut b = y1(x);
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for i in 1..n {
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if b.is_infinite() {
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break;
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}
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// if high_word(b) != 0xfff00000 {
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// break;
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// }
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(a, b) = (b, ((2.0 * i as f64) / x) * b - a);
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}
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b
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};
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if sign > 0 {
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b
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} else {
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-b
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}
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}
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