This is a shame about M1ā¦
Removing the +2 gave me a 5% more :
g7(n) = 16*(sum(x -> (b = 4float(x)^2; a = b+3; a/(a^2-16b)), 2:4:(n+1)))
So, @giordano, the conclusion is that
function f(rounds)
p = 1.0
@simd for i in 2:(rounds + 2)
x = (-1)^iseven(i)
p += x / (2 * i - 1)
end
p *= 4
return p
end
g7(n) = 16*(sum(x -> (b = 4float(x)^2; a = b+3; a/(a^2-16b)), 2:4:(n+1)))
@btime f(10^8)
@btime g7(10^8)
gives
43.530 ms (0 allocations: 0 bytes)
11.347 ms (0 allocations: 0 bytes)
So you might update your graph Also, iām guessing that the other Julia in your graph might go up a little bit.