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

Leonard Mada |eo@m@d@ @end|ng |rom @yon|c@eu
Sat Aug 19 00:25:21 CEST 2023

```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

```