4 ms·

The thing is that in this particular case it does in fact just copy the pi. You're essentially calculating radians(x)*180/x, which by definition is equal to pi.

The thing is that in this particular case it does in fact just copy the pi. You're essentially calculating radians(x)*180/x, which by definition is equal to pi. The radians function is doing all the heavy lifting, unlike say using the Taylor series or some other approximation approach.

ksaj 2y ago
It is doing this: sin(555555555555555555), where the 5's are an x number of 5s that increasingly converges on pi, and calculates the correct decimal placement. It's not (quite) as trivial as just copying the pi.
- valicord 2y ago
  Nope, that is not what it does. First, sin(x) ~= x for small x. Second, 1000.../555... = 1.8. Neither of these have anything to do with pi. With these in mind, the program can be simplified as sin(radians(repeated)) * (10 ** (repeat + 2)) = radians(repeated) * 100 * 1.8 / repeated = repeated * pi / 180 * 180 / repeated = pi The main and only reason you get pi in the end is that the radians function is defined as radians(x) = pi * x / 180, which requires knowing the value of pi to begin with. So this program is basically an equivalent of multiply_the_argument_by_pi(555) / 555 except with a few layers of completely unnecessary math on top of it to obscure the magic trick.
  - ksaj 2y ago
    If that was the case, any number other than 5 would also work. Try that.
    - valicord 2y ago
      Any number works if you adjust the multiplier to preserve the 1000.../555... = 1.8 property. For example: https://www.online-python.com/Um4qevAzo2 https://www.online-python.com/Um4qevAzo2
      - ksaj 2y ago
        Ah, that's a missing link I hadn't thought of. This was a result of converting degrees to radians. There is another way as well. from math import sin, acos def nines(repeat: int) -> float: repeated = 2 / int('9' * repeat) value = sin(repeated) * acos(repeated) * (10 ** repeat) return value nines(18) I'm sure this can be edited to accept other values in the same way. Took a while to convince me, but you got there.
        valicord 2y ago
        Hint: it works just as well if you replace the acos argument with 0 value = sin(repeated) * acos(0) * (10 ** repeat) (keep in mind that that acos(0) = pi/2)

ksaj 2y ago

It is doing this: sin(555555555555555555), where the 5's are an x number of 5s that increasingly converges on pi, and calculates the correct decimal placement. It's not (quite) as trivial as just copying the pi.

valicord 2y ago
Nope, that is not what it does. First, sin(x) ~= x for small x. Second, 1000.../555... = 1.8. Neither of these have anything to do with pi. With these in mind, the program can be simplified as sin(radians(repeated)) * (10 ** (repeat + 2)) = radians(repeated) * 100 * 1.8 / repeated = repeated * pi / 180 * 180 / repeated = pi The main and only reason you get pi in the end is that the radians function is defined as radians(x) = pi * x / 180, which requires knowing the value of pi to begin with. So this program is basically an equivalent of multiply_the_argument_by_pi(555) / 555 except with a few layers of completely unnecessary math on top of it to obscure the magic trick.
- ksaj 2y ago
  If that was the case, any number other than 5 would also work. Try that.
  - valicord 2y ago
    Any number works if you adjust the multiplier to preserve the 1000.../555... = 1.8 property. For example: https://www.online-python.com/Um4qevAzo2 https://www.online-python.com/Um4qevAzo2
    - ksaj 2y ago
      Ah, that's a missing link I hadn't thought of. This was a result of converting degrees to radians. There is another way as well. from math import sin, acos def nines(repeat: int) -> float: repeated = 2 / int('9' * repeat) value = sin(repeated) * acos(repeated) * (10 ** repeat) return value nines(18) I'm sure this can be edited to accept other values in the same way. Took a while to convince me, but you got there.
      - valicord 2y ago
        Hint: it works just as well if you replace the acos argument with 0 value = sin(repeated) * acos(0) * (10 ** repeat) (keep in mind that that acos(0) = pi/2)