A little help understanding math.... - Printable Version +- HP Forums (https://www.hpmuseum.org/forum) +-- Forum: HP Calculators (and very old HP Computers) (/forum-3.html) +--- Forum: General Forum (/forum-4.html) +--- Thread: A little help understanding math.... (/thread-131.html) A little help understanding math.... - Thomas Klemm - 12-16-2013 11:56 PM There was a recent thread in the old forum where Cyrille asked a question concerning the implementation of the complex inverse trigonometric functions. I must admit that it took me a while to understand what it was about. I will only deal with the complex inverse of cosine. Let's assume $$w=u+iv$$. Then \begin{align} z&=&x+iy=cos(w)=cos(u+iv) \\ &=&cos(u)cos(iv)-sin(u)sin(iv) \\ &=&cos(u)cosh(v)-i sin(u)sinh(v) \\ \end{align} But we would like to find $$w=arccos(z)$$. Better don't ask me who came up with this solution: \begin{align} |z+1|^2 &=& (x+1)^2+y^2 \\ &=& x^2+2x+1+y^2 \\ &=& cos^2(u)cosh^2(v)+2cos(u)cosh(v)+1+sin^2(u)sinh^2(v) \\ &=& cos^2(u)(1+sinh^2(v))+2cos(u)cosh(v)+1+sin^2(u)sinh^2(v) \\ &=& cos^2(u)+cos^2(u)sinh^2(v)+2cos(u)cosh(v)+1+sin^2(u)sinh^2(v) \\ &=& cos^2(u)+2cos(u)cosh(v)+1+(cos^2(u)+sin^2(u))sinh^2(v) \\ &=& cos^2(u)+2cos(u)cosh(v)+1+sinh^2(v) \\ &=& cos^2(u)+2cos(u)cosh(v)+cosh^2(v) \\ &=& (cos(u)+cosh(v))^2 \\ |z+1| &=& cosh(v)+cos(u) \end{align} And similarly we can conclude: $|z-1| = cosh(v)-cos(u)$ Add the two equations to get: $|z+1|+|z-1| = 2cosh(v) \\ cosh(v)=\frac{|z+1|+|z-1|}{2}$ From $$x=cos(u)cosh(v)$$ we conclude: $2x=2cos(u)cosh(v) \\ cos(u)=\frac{2x}{|z+1|+|z-1|}$ This leads to the following function: Code: from math import sqrt, acos, acosh def arccos(z):     x, y = z.real, z.imag     y2 = y**2     U = sqrt((x+1)**2 + y2)     V = sqrt((x-1)**2 + y2)     W = (U+V)/2     u = acos(x/W)     v = acosh(W)     return u - 1j*v Let's compare this to the builtin function: Code: >>> import cmath >>> z=3+4j >>> arccos(z) (0.9368124611557199-2.305509031243477j) >>> cmath.acos(z) (0.9368124611557198-2.305509031243477j) >>> z/=10 >>> arccos(z) (1.2901667645030908-0.40511233717803047j) >>> cmath.acos(z) (1.2901667645030908-0.4051123371780309j) >>> z/=10 >>> arccos(z) (1.5408158285382982-0.04000730997058477j) >>> cmath.acos(z) (1.5408158285382982-0.04000730997058378j) It appears we loose some accuracy when calculating $$v=acosh(W)$$ for values of W close to 1. Why is that? (substitute $$t=e^v$$) \begin{align} W &=& cosh(v) \\ &=& \frac{e^v+e^{-v}}{2} \\ W &=& \frac{t+\frac{1}{t}}{2} \\ 2Wt &=& t^2+1 \\ t^2-2Wt+1 &=& 0 \\ t &=& W\pm\sqrt{W^2-1} \\ \end{align} Now we can see what's happening: when W is close to 1 we experience cancellation and thus loss of accuracy. We'd like to calculate W-1 without loosing digits. Can we do better? Let's see what the HP-48 suggests: Code: * Q := y^2/(sqrt((|x|+1)^2 + y^2) + (|x|+1)) * R := sqrt((|x|-1)^2 + y^2) +||x|-1| * S := y^2/R if R<>0, 0 otherwise * M := Q+R if |x|>=1, Q+S otherwise * P := Q+S if |x|>=1, Q+R otherwise * B := 2*x/(M+2) * C := sqrt((P/(M+2))*(2-(P/(M+2)))) * sg(y,x) := sgn(y) if y<>0, -sgn(x) otherwise * IM := sg(y,x)*lnp1((M/2) + sqrt((M/2)*((M/2)+2))) (sign replacement) * *        { arccos(B)            |B| <= (1/sqrt(2)) * RE1 := { arcsin(C)            |B| > (1/sqrt(2)) and B >= 0 *        { pi - arcsin(C)       |B| > (1/sqrt(2)) and B < 0 * * RE2 := { arcsin(B)            |B| <= (1/sqrt(2)) *        { sgn(B)*arccos(C)     |B| > (1/sqrt(2))   (sign replacement) * I've taken the liberty to add some new variables: Code: def sgn(x):     return cmp(x, 0.0) def sg(y,x):     return sgn(y) if y != 0 else -sgn(x) def inverse(z):     x, y = z.real, z.imag     _x_ = abs(x)     y2 = y**2     U = sqrt((_x_+1)**2 + y2)     T = U + (_x_+1)     Q = y2/T  # = U - (|x|+1)     V = sqrt((_x_-1)**2 + y2)     R = V + abs(_x_-1)     S = y2/R if R != 0 else 0  # = V - ||x|-1|     M = Q+R if _x_ >= 1 else Q+S  # M = U + V - 2     P = Q+S if _x_ >= 1 else Q+R  # P = U + V - 2|x|     B = 2*x/(M+2)     C = sqrt((P/(M+2))*(2-(P/(M+2))))     IM = sg(y,x)*log1p((M/2) + sqrt((M/2)*((M/2)+2))) # sign replacement     _B_ = abs(B)     sqrt_2 = 1/sqrt(2)     RE1 = acos(B) if _B_ <= sqrt_2 else asin(C) if B >= 0 else pi - asin(C)     RE2 = asin(B) if _B_ <= sqrt_2 else sgn(B)*acos(C) # sign replacement     return IM, RE1, RE2 Please note that M/2 = W-1 since M = U+V-2. That's exactly the value we were looking for. Instead of ln we use lnp1 (or log1p in Python) for values close to 1: $v=ln(t)=lnp1(t-1)=lnp1(W-1\pm\sqrt{(W-1)(W+1)}$ Substitute W-1 by M/2 and we end up with the formula for IM. Well nearly, we still have to deal with the sign of the expression. What about $$\frac{1}{\sqrt{2}}$$? That was Cyrille's question. For small values of B close to 0 I assume both acos(B) and asin(B) are rather accurate. However with values close to 1 we experience again cancellation and thus loss of accuracy. In this case we'd prefer to use $$\sqrt{1-B^2}$$ instead. That's exactly what is calculated with C but without cancellation. I'm lazy and leave that as an exercise. So where exactly should we switch? Why not just in the middle between 0 and $$\frac{\pi}{2}$$? And: $$cos(\frac{\pi}{4})=sin(\frac{\pi}{4})=\frac{1}{\sqrt{2}}$$. But in this area we don't experience cancellation at all thus .7 will do as well. For those interested in the SYSRPL code I've added the stack-diagrams: Code: ::                     ; z   C%>%%                ; x y   2DUP                 ; x y x y   DUP                  ; x y x y y   %%*SWAP              ; x y y^2 x   %%ABS                ; x y y^2 |x|   DUP                  ; x y y^2 |x| |x|   %%1+                 ; x y y^2 |x| |x|+1   DUPDUP               ; x y y^2 |x| |x|+1 |x|+1 |x|+1   %%*                  ; x y y^2 |x| |x|+1 (|x|+1)^2   4PICK                ; x y y^2 |x| |x|+1 (|x|+1)^2 y^2   %%+                  ; x y y^2 |x| |x|+1 (|x|+1)^2+y^2   %%SQRT               ; x y y^2 |x| |x|+1 √((|x|+1)^2+y^2)   %%+                  ; x y y^2 |x| √((|x|+1)^2+y^2)+(|x|+1)   3PICKSWAP            ; x y y^2 |x| y^2 √((|x|+1)^2+y^2)+(|x|+1)   %%/                  ; x y y^2 |x| Q   OVER                 ; x y y^2 |x| Q |x|   %%1                  ; x y y^2 |x| Q |x| 1   %%-                  ; x y y^2 |x| Q |x|-1   %%ABS                ; x y y^2 |x| Q ||x|-1|   DUPDUP               ; x y y^2 |x| Q ||x|-1| ||x|-1| ||x|-1|   %%*                  ; x y y^2 |x| Q ||x|-1| ||x|-1|^2   5PICK                ; x y y^2 |x| Q ||x|-1| ||x|-1|^2 y^2   %%+                  ; x y y^2 |x| Q ||x|-1| ||x|-1|^2+y^2   %%SQRT               ; x y y^2 |x| Q ||x|-1| √(||x|-1|^2+y^2)   %%+                  ; x y y^2 |x| Q R   4ROLLOVER            ; x y |x| Q R y^2 R   DUP                  ; x y |x| Q R y^2 R R   %%0<>                ; x y |x| Q R y^2 R   ITE                  ; x y |x| Q R y^2 R   %%/                  ; x y |x| Q R y^2/R   ::                   ; x y |x| Q R y^2 R     2DROP              ; x y |x| Q R     %%0                ; x y |x| Q R 0   ;                    ; x y |x| Q R S   UNROTOVER            ; x y |x| S Q R Q   %%+                  ; x y |x| S Q R+Q   UNROT                ; x y |x| R+Q S Q   %%+                  ; x y |x| R+Q S+Q   ROT                  ; x y R+Q S+Q |x|   %%1                  ; x y R+Q S+Q |x| 1   %%<                  ; x y R+Q S+Q   ?SKIPSWAP            ; x y P M   DUP                  ; x y P M M   %%2                  ; x y P M M 2   %%+                  ; x y P M M+2   ROTOVER              ; x y M M+2 P M+2   %%/                  ; x y M M+2 P/(M+2)   %%2                  ; x y M M+2 P/(M+2) 2   OVER                 ; x y M M+2 P/(M+2) 2 P/(M+2)   %%-                  ; x y M M+2 P/(M+2) 2-P/(M+2)   %%*                  ; x y M M+2 P/(M+2)*(2-P/(M+2))   %%SQRT               ; x y M M+2 C   5PICK                ; x y M M+2 C x   DUP                  ; x y M M+2 C x x   %%+                  ; x y M M+2 C 2x   ROT                  ; x y M C 2x M+2   %%/                  ; x y M C B   ROT                  ; x y C B M   %%2                  ; x y C B M 2   %%/                  ; x y C B M/2   DUP                  ; x y C B M/2 M/2   %%2                  ; x y C B M/2 M/2 2   %%+                  ; x y C B M/2 M/2+2   OVER                 ; x y C B M/2 M/2+2 M/2   %%*                  ; x y C B M/2 (M/2+2)*M/2   %%SQRT               ; x y C B M/2 √((M/2+2)*M/2)   %%+                  ; x y C B M/2+√((M/2+2)*M/2)   %%lnp1               ; x y C B lnp1(M/2+√((M/2+2)*M/2))   5ROLL                ; y C B lnp1(M/2+√((M/2+2)*M/2)) x   5ROLL                ; C B lnp1(M/2+√((M/2+2)*M/2)) x y   DUP                  ; C B lnp1(M/2+√((M/2+2)*M/2)) x y y   %%0=                 ; C B lnp1(M/2+√((M/2+2)*M/2)) x y   ITE                  ; C B lnp1(M/2+√((M/2+2)*M/2)) x y   ::                   ; C B lnp1(M/2+√((M/2+2)*M/2)) x y     DROP               ; C B lnp1(M/2+√((M/2+2)*M/2)) x     %%0<               ; C B lnp1(M/2+√((M/2+2)*M/2))   ;                    ; C B lnp1(M/2+√((M/2+2)*M/2))   ::                   ; C B lnp1(M/2+√((M/2+2)*M/2)) x y     SWAPDROP           ; C B lnp1(M/2+√((M/2+2)*M/2)) y     %%0>=              ; C B lnp1(M/2+√((M/2+2)*M/2))   ;                    ; C B lnp1(M/2+√((M/2+2)*M/2))   ?SKIP                ; C B lnp1(M/2+√((M/2+2)*M/2))   %%CHS                ; C B sg(y,x)*lnp1(M/2+√((M/2+2)*M/2))   UNROTDUP             ; IM C B B   %%ABS                ; IM C B |B|   %%.7                 ; IM C B |B| .7   %%<=                 ; IM C B   case                 ; IM C B   ::                   ; IM C B     SWAPDROPDUP        ; IM B B     %%ACOSRAD          ; IM B RE1     SWAP               ; IM RE1 B     %%ASINRAD          ; IM RE1 RE2   ;                    ; IM RE1 RE2   SWAPDUP              ; IM B C C   %%ASINRAD            ; IM B C arcsin(C)   SWAP                 ; IM B arcsin(C) C   %%ACOSRAD            ; IM B arcsin(C) arccos(C)   %%ABS                ; IM B arcsin(C) |arccos(C)|   ROT                  ; IM arcsin(C) |arccos(C)| B   %%0>=                ; IM arcsin(C) |arccos(C)| ?SEMI                  ; IM RE1 RE2   %%CHS                ; IM arcsin(C) -|arccos(C)|   %%PI                 ; IM arcsin(C) RE2 pi   ROT                  ; IM RE2 pi arcsin(C)   %%-                  ; IM RE2 pi-arcsin(C)   SWAP                 ; IM RE1 RE2 ;                      ; IM RE1 RE2 It's much easier to understand this program thanks to the comments. I guess I'd never have figured that out without them. Thus my appeal to Cyrille and Tim: please make sure that this information doesn't get lost. Send the source-code of HP-48 (and whatever you have) to a computer museum if that's legally possible. There's so much we can learn from that. Kind regards Thomas PS: There are these two lines somewhere close to the end: Code: %%ACOSRAD            ; IM B arcsin(C) arccos(C) %%ABS                ; IM B arcsin(C) |arccos(C)| However the use of the absolute value is missing in the comment: Code: * RE2 := { arcsin(B)            |B| <= (1/sqrt(2)) *        { sgn(B)*arccos(C)     |B| > (1/sqrt(2))   (sign replacement) Since %%ACOSRAD should never return negative values I don't understand why this was added. IMHO the command %%ABS could be removed. Any ideas why this is needed? RE: A little help understanding math.... - Paul Dale - 12-17-2013 10:19 AM Thanks for the analysis. It is like we suspected. This looks a lot like one of Kahan's little gems. - Pauli RE: A little help understanding math.... - Namir - 12-17-2013 01:17 PM Thomas, Your Python code tells me that you use it and most likely the numpy module. How can you install the numpy module? I appreciate your help (including links to sites that make the process very clear to the confused folks like Mois!!) Namir RE: A little help understanding math.... - Thomas Klemm - 12-17-2013 03:04 PM One of the reasons I use Python is that it comes with "batteries included". Until now I never had to install additional Python modules. Thus stackoverflow or some Python related forums might be more helpful. And then I'm not familiar with Windows. Cheers Thomas Addendum: To my surprise numpy is already installed. You just have to import it: Code:  >>> import numpy as np Or: Code:  >>> import numpy.polynomial.polynomial as P >>> P.polyroots((1,-1,-1)) array([-1.61803399,  0.61803399]) I've just skimmed through the Python Scientific lecture notes but it appears to give a nice introduction. RE: A little help understanding math.... - Albert Chan - 08-15-2021 03:48 AM (12-16-2013 11:56 PM)Thomas Klemm Wrote:  Let's assume $$w=u+iv$$. Then \begin{align} z&=&x+iy=cos(w)=cos(u+iv) \\ &=&cos(u)cos(iv)-sin(u)sin(iv) \\ &=&cos(u)cosh(v)-i sin(u)sinh(v) \\ \end{align}... Code:     u = acos(x/W)     v = acosh(W)     return u - 1j*v There is a flaw with the code, arc function only give principle branch. We have to match sign of imaginery part. From above code, 0 ≤ u ≤ pi, v ≥ 0, sign(-sin(u) * sinh(v)) = negative If y is also negative, no correction needed, else we flip sign. This is the patch: < return u - 1j*v > return complex(u, v if y<0 else -v) RE: A little help understanding math.... - Albert Chan - 08-15-2021 12:25 PM Another proof of acos(z) algorithm, and showed why we need |z+1|, |z-1| XCas> z := cos(u+i*v) XCas> x, y := re(z), im(z) cos(u)*cosh(v), -sin(u)*sinh(v) |z|² = x²+y² = cos(u)²*cosh(v)² + sin(u)²*sinh(v)² = cos(u)²*cosh(v)² + (1-cos(u)²)*(cosh(v)²-1) = cosh(v)² + cos(u)² - 1 |z±1|² = (x±1)² + y² = (x²+y²+1) ± 2x = (cosh(v)² + cos(u)²) ± 2*cosh(v)*cos(u) = (cosh(v) ± cos(u))² cosh(v) ≥ 1, cos(u) ≤ 1, we take square root of both side: |z±1| = cosh(v) ± cos(u) RE: A little help understanding math.... - Albert Chan - 08-26-2021 02:31 PM (12-16-2013 11:56 PM)Thomas Klemm Wrote:  From $$x=cos(u)cosh(v)$$ we conclude: $2x=2cos(u)cosh(v) \\ cos(u)=\frac{2x}{|z+1|+|z-1|}$ There is an issue of recovering u from acos() Accuracy may be bad. Worse, it might hit outside valid range (-1 ≤ cos(u) ≤ 1) Similarly, recovering v from acosh() might hit by same issue, since cosh(v) ≥ 1 >>> z = 1.5430806348152439+0j # acos(z) = u + i*v >>> a, b = abs(1+z), abs(1-z) >>> V = (a+b) / 2 >>> U = z.real / V >>> U, V ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ # U, V = cos(u), cosh(v) (1.0000000000000002, 1.5430806348152437) --- Kahan's algorithm avoided these problems, using atan/asinh for complex acos From complex.c, cacos(z), https://opensource.apple.com/source/Libm/Libm-47.1/complex.c.auto.html Code: real(cacos(z)) = 2.0*atan(real(csqrt(1.0-z)/real(csqrt(1.0+z)))) imag(cacos(z)) = arcsinh(imag(csqrt(1.0-z)*csqrt(cconj(1.0+z)))) z = x+i*y = cos(u+i*v) = cos(u)*cosh(v) - i*sin(u)*sinh(v) ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ..... (1) Let U = cos(u), V = cosh(v), we have x = U*V Let t = tan(u/2) Since u = acos(U) = 0 to pi, non-negative, t = |t| cos(u) = U = (1-t^2)/(1+t^2) ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ → t = |t| = √((1-U)/(1+U)) Again, from the same source, for csqrt(z) Code: sqrt(x + i*y) = sqrt((|z| + Real(z))/2) + i*sqrt((|z| - Real(z))/2) and sqrt(x - i*y) = sqrt((|z| + Real(z))/2) - i*sqrt((|z| - Real(z))/2) Previously, we showed |1±z| = cosh(v) ± cos(u) = V ± U Assumed we have sign-zero, let s = sign(imag(z)) = ± 1 √(1+z) = √(((V+U)+(1+U*V))/2) + i*s*√(((V+U)−(1+U*V))/2) = √((1+U)*(V+1)/2) + i*s*√((1−U)*(V−1)/2) √(1−z) = √(((V−U)+(1−U*V))/2) − i*s*√(((V−U)−(1−U*V))/2) = √((1−U)*(V+1)/2) − i*s*√((1+U)*(V−1)/2) real(√(1-z)) / real(√(1+z)) = √((1-U)/(1+U)) = t = tan(u/2) → u = atan(real(√(1-z))/real(√(1+z))) * 2 real(√(1+z)) * imag(√(1-z)) = -s * (1+U)/2 * √(V*V-1) ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ..... (2) -imag(√(1+z)) * real(√(1-z)) = -s * (1-U)/2 * √(V*V-1) ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ..... (3) (2) and (3) have the same sign, sum is free from subtraction cancellation. (2)+(3) ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ → RHS = -s * √(V*V-1) = -s * |sinh(v)| = sinh(-s*|v|) From (1), sign(v) = sign(-y) = -s: → v = asinh(real(√(1+z))*imag(√(1-z)) - imag(√(1+z)) * real(√(1-z))) Code: Complex Cacos(Complex z) {   Complex a = csqrt(1+z), b = csqrt(1-z);   double x2 = atan (Real(b)/Real(a)) * 2;   double y2 = asinh(Real(a)*Imag(b) - Imag(a)*Real(b));   return CMPLX(x2, copysign(y2, -Imag(z))); } RE: A little help understanding math.... - Albert Chan - 08-26-2021 06:16 PM We can reuse previous post acos(z) = u+i*v result for asin(z) Naive implementation is with identity asin(z) = pi/2 - acos(z) = (pi/2-u) - i*v But, we could avoid cancellation errors of real part. a = √(1+z) = √((1+U)*(V+1)/2) + i*s*√((1−U)*(V−1)/2) b = √(1−z) = √((1−U)*(V+1)/2) − i*s*√((1+U)*(V−1)/2) +Real(a) * Real(b) = (V+1)/2 * √(1-U*U) ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ..... (4) -Imag(a) * Imag(b) = (V-1)/2 * √(1-U*U) ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ..... (5) With V=cosh(v) ≥ 1, both (4) and (5) are non-negative. Summing them is safe from cancellation error. (4)+(5) ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ → S = V * √(1-U*U) x/S = (U*V) / (V*√(1-U*U)) = cos(u) / |sin(u)| = sign(x) * |tan(pi/2 - u)| sign(x) should match sign(pi/2-u), we have pi/2-u = atan(x/S) Except for sign flip, imaginery part of asin(z) and acos(z) are the same. So, I just copy/paste from acos() code, for the imaginery part. Code: Complex Casin(Complex z) {   Complex a = csqrt(1+z), b = csqrt(1-z);   double x2 = atan(Real(z) / (Real(a)*Real(b) - Imag(a)*Imag(b)));   double y2 = asinh(Real(a)*Imag(b) - Imag(a)*Real(b));   return CMPLX(copysign(x2, Real(z)), copysign(y2, Imag(z))); } RE: A little help understanding math.... - Albert Chan - 08-29-2021 12:20 AM Prove: if 0 ≤ θ < pi/2, Re(atan(exp(i*θ))) = pi/4 atan(z) = atanh(z*i)/i = ln((1+i*z)/(1-i*z)) / (2*i) Let z = exp(i*θ) = cos(θ) + i*sin(θ) (1+i*z)/(1-i*z) = ((1-sin(θ)) + i*cos(θ)) / ((1+sin(θ)) - i*cos(θ)) = i * cos(θ) / (1 + sin(θ)) ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ // flip sin/cos = i * sin(pi/2-θ) / (1 + cos(pi/2-θ)) ﻿ ﻿ ﻿ ﻿ ﻿ ﻿ // tan(α/2) = sin(α) / (1 + cos(α)) = i * tan(pi/4-θ/2) For 0 ≤ θ < pi/2, imaginery part is positive. Re(atan(exp(i*θ))) = arg(i*tan(pi/4-θ/2)) / 2 = pi/4 With signed zero, we have atan(±0 + i) = ±pi/4 + Inf*i Code: Complex Catan(Complex z) {   double x = fabs(Real(z));   double y = fabs(Imag(z));   if (x >= 0x1p27 || y >= 0x1p27) { // atan(1/z) ~= 1/z     Complex t = -1/(x+y*I);         // atan(x+y*I) ~= pi/2 + t     x = Real(t) + M_PI_2;           // |re(t^3/3)/re(t)| <= 0x1p-54     y = Imag(t);                    // |im(t^3/3)/im(t)| <= 0x1p-54   } else {     double x2 = x*x, ym = 1-y;     double u = (1+y)*ym - x2;       // 0 implied z on unit circle     x = atan2(2*x + (u==0), u) * 0.5;     y = log1p(4*y / (ym*ym + x2)) * 0.25;   }   return CMPLX(copysign(x, Real(z)), copysign(y, Imag(z))); } Top branch, atan(t = -1/z) ≈ t - t^3/3 + t^5/5 - ... ≈ t, is because of trig identities. S2k+1 = sin((2k+1)θ)/sin(θ) = 2 cos((2k)θ) + 2 cos((2k-2)θ) + 2 cos((2k-4)θ) + ... + 1 C2k+1 = cos((2k+1)θ)/cos(θ) = 2 cos((2k)θ) - 2 cos((2k-2)θ) + 2 cos((2k-4)θ) − ... + (-1)^k --> |S2k+1| ≤ 2k+1 --> |C2k+1| ≤ 2k+1 t = ε * cis(θ) t^n/n = ε^n/n * cis(nθ) |re(t^3/3)/re(t)| = ε^2 * |C3|/3 ≤ ε^2 |im(t^3/3)/im(t)| = ε^2 * |S3|/3 ≤ ε^2 |re(t^5/5)/re(t)| = ε^4 * |C5|/5 ≤ ε^4 |im(t^5/5)/im(t)| = ε^4 * |S5|/5 ≤ ε^4 ... ε^2 below machine epsilon --> t - t^3/3 + t^5/5 - ... ≈ t RE: A little help understanding math.... - Albert Chan - 06-14-2023 04:55 PM (08-29-2021 12:20 AM)Albert Chan Wrote:  With signed zero, we have atan(±0 + i) = ±pi/4 + Inf*i Technically, there is no complex number z, such that tan(z) = i Whatever default we assigned for atan(i) is purely for convenience. Above quoted default assumed (±0 + i) = limit((±ε + i), ε=0) We could also start with ±0 = 0, exactly. atan(z) = ∫(1/(1+t^2), t = 0 .. z) = z - z^3/3 + z^5/5 - z^7/7 + ... atan(i) = i * (1 + 1/3 + 1/5 + 1/7 + ...) = i * Inf --> atanh(1) = atan(i)/i = Inf      (*) IEEE754_2008.pdf (page 45), atan2(±0, +0) = ±0, patched code is also simpler. Code: <    double x2 = x*x, ym = 1-y; <    double u = (1+y)*ym - x2;       // 0 implied z on unit circle <    x = atan2(2*x + (u==0), u) * 0.5; <    y = log1p(4*y / (ym*ym + x2)) * 0.25; >    double x2 = x*x, ym = 1-y; >    x = atan2(2*x, (1+y)*ym - x2) * 0.5; >    y = log1p(4*y / (ym*ym + x2)) * 0.25; (*) atanh(1 ± 0i) = (Inf ± 0i) matched ieee behavior https://en.cppreference.com/w/cpp/numeric/complex/atanh