Happy New Year to all of you! Let us start the year with something for your inner maths nerd ðŸ™‚

For those of you who don’t yet know Rosetta Code: it is a really cool site where you can find lots of interesting code examples in all kinds of different languages for many different tasks. Of course R is also present big time (at the time of writing 426 code examples!): Rosetta Code for R.

The name of the site is inspired by the famous Rosetta Stone of Ancient Egypt which is inscribed with three different versions of the same text: in Ancient Egyptian hieroglyphs, Demotic script, and Ancient Greek script which proved invaluable in deciphering Egyptian hieroglyphs and thereby opening the window into ancient Egyptian history.

Now, a few days ago I again added an example (for the other tasks I solved I will write more posts in the future, so stay tuned!). The task is to verify the correctness of Machin-like formulae using exact arithmetic.

A little bit of mathematical background is in order, so Wikipedia to the rescue:

Machin-like formulae are a popular technique for computing to a large number of digits. They are generalizations of John Machin]s formula from 1706:

which he used to compute to 100 decimal places.

Machin-like formulae have the form

where and are positive integers such that , is a signed non-zero integer, and is a positive integer.

The exact task is to verify that the following Machin-like formulae are correct by calculating the value of tan (right hand side) for each equation using exact arithmetic and showing they equal one:

The same should be done for the last and most complicated case…

… but it should be confirmed that the following, slightly changed, formula is incorrect by showing tan (right hand side) is **not** one:

This is what I contributed to Rosetta Code:

library(Rmpfr) prec <- 1000 # precision in bits `%:%` <- function(e1, e2) '/'(mpfr(e1, prec), mpfr(e2, prec)) # operator %:% for high precision division # function for checking identity of tan of expression and 1, making use of high precision division operator %:% tanident_1 <- function(x) identical(round(tan(eval(parse(text = gsub("/", "%:%", deparse(substitute(x)))))), (prec/10)), mpfr(1, prec)) tanident_1( 1*atan(1/2) + 1*atan(1/3) ) ## [1] TRUE tanident_1( 2*atan(1/3) + 1*atan(1/7)) ## [1] TRUE tanident_1( 4*atan(1/5) + -1*atan(1/239)) ## [1] TRUE tanident_1( 5*atan(1/7) + 2*atan(3/79)) ## [1] TRUE tanident_1( 5*atan(29/278) + 7*atan(3/79)) ## [1] TRUE tanident_1( 1*atan(1/2) + 1*atan(1/5) + 1*atan(1/8) ) ## [1] TRUE tanident_1( 4*atan(1/5) + -1*atan(1/70) + 1*atan(1/99) ) ## [1] TRUE tanident_1( 5*atan(1/7) + 4*atan(1/53) + 2*atan(1/4443)) ## [1] TRUE tanident_1( 6*atan(1/8) + 2*atan(1/57) + 1*atan(1/239)) ## [1] TRUE tanident_1( 8*atan(1/10) + -1*atan(1/239) + -4*atan(1/515)) ## [1] TRUE tanident_1(12*atan(1/18) + 8*atan(1/57) + -5*atan(1/239)) ## [1] TRUE tanident_1(16*atan(1/21) + 3*atan(1/239) + 4*atan(3/1042)) ## [1] TRUE tanident_1(22*atan(1/28) + 2*atan(1/443) + -5*atan(1/1393) + -10*atan(1/11018)) ## [1] TRUE tanident_1(22*atan(1/38) + 17*atan(7/601) + 10*atan(7/8149)) ## [1] TRUE tanident_1(44*atan(1/57) + 7*atan(1/239) + -12*atan(1/682) + 24*atan(1/12943)) ## [1] TRUE tanident_1(88*atan(1/172) + 51*atan(1/239) + 32*atan(1/682) + 44*atan(1/5357) + 68*atan(1/12943)) ## [1] TRUE tanident_1(88*atan(1/172) + 51*atan(1/239) + 32*atan(1/682) + 44*atan(1/5357) + 68*atan(1/12944)) ## [1] FALSE

As you can see all statements are `TRUE`

except for the last one!

In the code, I make use of the `Rmpfr`

package (from Martin Maechler of ETH ZÃ¼rich, Switzerland) which is based on the excellent GMP (GNU Multiple Precision) library. I define a new infix operator `%:%`

for high-precision division and after that convert all standard divisions in the formulae to high-precision divisions and calculate the tan. Before I check if the result is identical to one I round it to 100 decimal places which is more than enough given the precision of , so about 300 decimal places, in the example.

Please let me know in the comments what you think of this approach and whether you see room for improvement for the code – Thank you!