# [R] Numerical stability of: 1/(1 - cos(x)) - 2/x^2

Bert Gunter bgunter@4567 @end|ng |rom gm@||@com
Sat Aug 19 01:06:19 CEST 2023

```"The ugly thing is that the error only gets worse as x decreases. The
value neither drops to 0, nor does it blow up to infinity; but it gets
worse in a continuous manner."

If I understand you correctly, this is wrong:

> x <- 2^(-20) ## considerably less then 1e-4 !!
> y <- 1 - x^2/2;
> 1/(1 - y) - 2/x^2
[1] 0

It's all about the accuracy of the binary approximation of floating point
numbers (and their arithmetic)

Cheers,
Bert

On Fri, Aug 18, 2023 at 3:25 PM Leonard Mada via R-help <
r-help using r-project.org> wrote:

> I have added some clarifications below.
>
> On 8/18/2023 10:20 PM, Leonard Mada wrote:
> > [...]
> > After more careful thinking, I believe that it is a limitation due to
> > floating points:
> > [...]
> >
> > 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
>
> The floating point representation of 1 - x^2/2 is the real culprit:
> # 0.99999999500000003039
>
> The 3039 at the end is really an error due to the floating point
> representation. However, this error blows up when inverting the value:
> x = 1E-4;
> y = 1 - x^2/2;
> 1/(1 - y) - 2/x^2
> # 1.215494
> # should be 1/(x^2/2) - 2/x^2 = 0
>
>
> The ugly thing is that the error only gets worse as x decreases. The
> value neither drops to 0, nor does it blow up to infinity; but it gets
> worse in a continuous manner. At least the reason has become now clear.
>
>
> >
> > 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:
> >>>>>>>      on Wed, 16 Aug 2023 20:50:52 +0300 writes:
> >>      > Dear Iris,
> >>      > Dear Martin,
> >>
> >>
> >>      > 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")
> >>
> >> 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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