So in this project, I try to calculate a constant that's supposed to come out to 1.2020569031595942853997381…, according to Wikipedia. Scratch, being Scratch, decided to round this to 1.202057, and I would like a number with a decimal a little bigger than 6 digits.

Apéry's constant:

1 + 1/2^3 + 1/3^3 + 1/4^3

HelpI'mNewToATPleaseDontHurtMe

There are two main things to note:

• Scratch's variable watchers do not show numbers to full precision. To see a number to the full precision that Scratch stores it at, do something like this:
  

  set [watcher v] to (join [ ] (apery's constant)) // where the first argument in the join block is a space.

• Apery's constant takes a huge amount of iterations to calculate to a large precision - I matched the decimal only up to 1.20205690315 with 10 million iterations.
  

TheMonsterOfTheDeep wrote:

There are two main things to note:

• Scratch's variable watchers do not show numbers to full precision. To see a number to the full precision that Scratch stores it at, do something like this:
  

  set [watcher v] to (join [ ] (apery's constant)) // where the first argument in the join block is a space.

• Apery's constant takes a huge amount of iterations to calculate to a large precision - I matched the decimal only up to 1.20205690315 with 10 million iterations.
  

Thanks!! I did 10mil before sharing, and it didn't change - that's when I figured something was probably wrong.

TheMonsterOfTheDeep wrote:

There are two main things to note:

• Scratch's variable watchers do not show numbers to full precision. To see a number to the full precision that Scratch stores it at, do something like this:
  

  set [watcher v] to (join [ ] (apery's constant)) // where the first argument in the join block is a space.

• Apery's constant takes a huge amount of iterations to calculate to a large precision - I matched the decimal only up to 1.20205690315 with 10 million iterations.
  

I don't know about a table of reciprocals of cubes, but for generating a series of cubes there's an efficient algorithm that just uses adds and the value of the previous cube. (and generates a table of squares incidentally as they're also needed in the expansion of (X+1)^3 in terms of X, X^2 and X^3.)

Here's the code in C (sort of), simplified to make it easier to translate into assembly code after getting the algorithm correct in C first…

http://www.gtoal.com/src/squares/cubes.c

With Scratch's interpretive overhead swamping the cost of floating point maths, this probably won't help in a Scratch version but I thought you might find it amusing anyway. As explained in the comment, this was given as a programming exercise for new programmers of the Deuce computer at Kidsgrove in the late 50's. (I wasn't around then of course but I thought I'd try the exercise to see how hard it was…)

G