Accuracy of BigFloat - Calculating sin with high accuracy

Specific Domains Numerics
1

Hello,
I’am trying to calcuate values of the sin-function to high precision. I want to get at least 1 million valid digits.

Using the following code I test the accuracy of BigFloat:

# Set precision to 7 million bits, so I expect about 2 million valid digits
setprecision(7000000)
# Get value of pi with high precision
bigpi=BigFloat(pi)
# Calculate difference of sin(pi/2)-1. Should be zero in exact arithmetic
sin(bigpi/BigFloat(2))-1
# Calualate difference of sin(pi/6)-1/2. Should be zero in exact arithmetic
sin(bigpi/BigFloat(6))-1/BigFloat(2)

For the first difference I get the value of zero. That means the result of sin(pi/2) is valid up to 2 million digits.

For the secons difference I get an value of about 4.76993…e-306923. That means sin(pi/6) is only calculated up to about 300’000 digits and not 2 million digits.

Did I make something wrong in the setup of this test, or are there limitations in the accuray of the sin-function.

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2

I get 0 for the second calculation too.
Did you try restarting Julia? Perhaps you defined different things with different precision or something like that.

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3

If you want a rigorous guarantee of correctness, you can use IntervalArithmetic.jl, e.g.

julia 0.6> using IntervalArithmetic

julia 0.6> @biginterval sin(pi/6) - 0.5
[-4.31809e-78, 8.63617e-78]₂₅₆

julia 0.6> setprecision(10000)
10000

julia 0.6> @biginterval sin(pi/6) - 0.5
[-5.01238e-3011, 5.01238e-3011]₁₀₀₀₀

julia 0.6> setprecision(100000)
100000

julia 0.6> @biginterval sin(pi/6) - 0.5
[-1.001e-30103, 2.002e-30103]₁₀₀₀₀₀

julia 0.6> setprecision(1000000)
1000000

julia 0.6> @biginterval sin(pi/6) - 0.5
[-5.05018e-3010, 1.01004e-30103]₁₀₀₀₀₀₀

Here, @biginterval is a macro that evaluates the following expression in a guaranteed way by converting each literal (e.g. 0.5) and variable to an Interval{BigFloat}.

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4

Oops, I see that there’s an output problem with that last one. You can get the full output with

@format full

julia 0.6> @biginterval sin(pi/6) - 0.5
...

(Output suppressed!)

5

There are times in which sinpi is more accurate than sin. For instance, both the first and second expression are zero using sinpi instead of sin.

setprecision(7_000_000)
sinpi(BigFloat(0.5)) - 1  == BigFloat(0) # returns true
sinpi(1 / BigFloat(6)) - BigFloat(0.5) == BigFloat(0) # returns true

You may need to take a look at the actual implementation to tell when / why that is the case.

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6

Thanks for your replies.
I think this is an issue of the operating system. I used Win10 and there I observe the same problem for all three variants of sin, sind and sinpi.
Using ubuntu (16.04) the generic julia package (0.6.0) for linux I get the exact result.

Is this a known issue that julia returns different results for win and linux?

I’m not so familiar with high precision arithmetic but for IEEE754 I would expect the same results up to rounding error (~1e-16). In my naive point of view I would assume for high precision arithmetic that only that last few digits are wrong, but in the above example 1,6million of about 2 million digits are wrong.

7

It works fine for me on MacOS with Julia 0.6:

julia> setprecision(7000000)
7000000

julia> sin(big(pi)/6) - 0.5 == 0
true

but on Win64 I can confirm that it only gets 300,000 digits instead of 2 million.

There seems to be some problem with the Windows build of the GNU MPFR library (used by Julia for BigFloat). Can you file an issue?

8

Just a guess until it can be debugged in full, but I know MPFR and GMP use the C long type in many APIs for integer sizes, and on Windows that is 32 bits vs most other operating systems where C long is 64 bits on x86_64. So this may be a limitation caused by the C types used in the library.

9

Update: this bug was fixed in MPFR 3.1.5 (GNU MPFR version 3.1.5), and was already patched locally in Julia (https://github.com/JuliaLang/julia/issues/22758)

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