Dear Martin,

Thank you very much for your analysis.

I add only a small comment:
- the purpose of the modified formula was to apply l'Hospital;
- there are other ways to transform the formula; although applying l'Hospital once is probably more robust than simple transformations (but the computations are also more tedious and error prone);

After more careful thinking, I believe that it is a limitation due to floating points:
x = 1E-4
1/(-x^2/2 - x^4/24) + 2/x^2
1/6

y = 1 - x^2/2 - x^4/24;
1/(cos(x) - 1) + 2/x^2
1/(y - 1) + 2/x^2
# -1.215494
# correct: 1/6

We need the 3rd term for the correct computation of cos(x) in this problem: but this is x^4 / 24, which for 1E-4 requires precision at least up to 1E-16 / 24, or ~ 1E-18. I did not thought initially about that. The trigonometric functions skip one term, and are therefore much uglier than the log. The problem really stems from the representation of 1 - x^2/2 as shown below:
x = 1E-4
print(1 - x^2/2, digits=20)
print(0.999999995, digits=20) # fails
# 0.99999999500000003039

Maybe some functions of type cos1p and cos1n would be handy for such computations (to replace the manual series expansion):
cos1p(x) = 1 + cos(x)
cos1n(x) = 1 - cos(x)
Though, I do not have yet the big picture.


Sincerely,


Leonard


On 8/17/2023 1:57 PM, Martin Maechler wrote:
Leonard Mada
     on Wed, 16 Aug 2023 20:50:52 +0300 writes:
     > Dear Iris,
     > Dear Martin,

     > Thank you very much for your replies. I add a few comments.

     > 1.) Correct formula
     > The formula in the Subject Title was correct. A small glitch swept into
     > the last formula:
     > - 1/(cos(x) - 1) - 2/x^2
     > or
     > 1/(1 - cos(x)) - 2/x^2 # as in the subject title;

     > 2.) log1p
     > Actually, the log-part behaves much better. And when it fails, it fails
     > completely (which is easy to spot!).

     > x = 1E-6
     > log(x) -log(1 - cos(x))/2
     > # 0.3465291

     > x = 1E-8
     > log(x) -log(1 - cos(x))/2
     > # Inf
     > log(x) - log1p(- cos(x))/2
     > # Inf => fails as well!
     > # although using only log1p(cos(x)) seems to do the trick;
     > log1p(cos(x)); log(2)/2;

     > 3.) 1/(1 - cos(x)) - 2/x^2
     > It is possible to convert the formula to one which is numerically more
     > stable. It is also possible to compute it manually, but it involves much
     > more work and is also error prone:

     > (x^2 - 2 + 2*cos(x)) / (x^2 * (1 - cos(x)))
     > And applying L'Hospital:
     > (2*x - 2*sin(x)) / (2*x * (1 - cos(x)) + x^2*sin(x))
     > # and a 2nd & 3rd & 4th time
     > 1/6

     > The big problem was that I did not expect it to fail for x = 1E-4. I
     > thought it is more robust and works maybe until 1E-5.
     > x = 1E-5
     > 2/x^2 - 2E+10
     > # -3.814697e-06

     > This is the reason why I believe that there is room for improvement.

     > Sincerely,
     > Leonard

Thank you, Leonard.
Yes, I agree that it is amazing how much your formula suffers from
(a generalization of) "cancellation" --- leading you to think
there was a problem with cos() or log() or .. in R.
But really R uses the system builtin libmath library, and the
problem is really the inherent instability of your formula.

Indeed your first approximation was not really much more stable:

## 3.) 1/(1 - cos(x)) - 2/x^2
## It is possible to convert the formula to one which is numerically more
## stable. It is also possible to compute it manually, but it involves much
## more work and is also error prone:
## (x^2 - 2 + 2*cos(x)) / (x^2 * (1 - cos(x)))
## MM: but actually, that approximation does not seem better (close to the 
breakdown region):
f1 <- \(x) 1/(1 - cos(x)) - 2/x^2
f2 <- \(x) (x^2 - 2 + 2*cos(x)) / (x^2 * (1 - cos(x)))
curve(f1, 1e-8, 1e-1, log="xy" n=2^10)
curve(f2, add = TRUE, col=2,   n=2^10)
## Zoom in:
curve(f1, 1e-4, 1e-1, log="xy",n=2^9)
curve(f2, add = TRUE, col=2,   n=2^9)
## Zoom in much more in y-direction:
yl <- 1/6 + c(-5, 20)/100000
curve(f1, 1e-4, 1e-1, log="x", ylim=yl, n=2^9)
abline(h = 1/6, lty=3, col="gray")
curve(f2, add = TRUE, n=2^9, col=adjustcolor(2, 1/2))

Now, you can use the Rmpfr package (interface to the GNU MPFR
multiple-precision C library) to find out more :

if(!requireNamespace("Rmpfr")) install.packages("Rmpfr")
M <- function(x, precBits=128) Rmpfr::mpfr(x, precBits)

(xM <- M(1e-8))# yes, only ~ 16 dig accurate
## 1.000000000000000020922560830128472675327e-8
M(10, 128)^-8 # would of course be more accurate,
## but we want the calculation for the double precision number 1e-8

## Now you can draw "the truth" into the above plots:
curve(f1, 1e-4, 1e-1, log="xy",n=2^9)
curve(f2, add = TRUE, col=2,   n=2^9)
## correct:
curve(f1(M(x, 256)), add = TRUE, col=4, lwd=2, n=2^9)
abline(h = 1/6, lty=3, col="gray")

But, indeed we take note  how much it is the formula instability:
Also MPFR needs a lot of extra bits precision before it gets to
the correct numbers:

xM <- c(M(1e-8,  80), M(1e-8,  96), M(1e-8, 112),
         M(1e-8, 128), M(1e-8, 180), M(1e-8, 256))
## to and round back to 70 bits for display:
R <- \(x) Rmpfr::roundMpfr(x, 70)
R(f1(xM))
R(f2(xM))
## [1]                         0                          0  
0.15407439555097885670915
## [4] 0.16666746653133802175779  0.16666666666666666749979  
0.16666666666666666750001

## 1. f1() is even worse than f2() {here at x=1e-8}
## 2. Indeed, even 96 bits precision is *not* sufficient at all, ...
##    which is amazing to me as well !!

Best regards,
Martin

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