For a part of my half-year school project I am trying to make a program to calculate pi to large precision. Now python (my preferred language) doesn't have that high precision, just like most other languages. What programming language would be good for my usage?

You can use the decimal module in Python for computing decimal numbers to high precision.

- Jelte2357
- Advanced Newbie (Posts: 56)

- 25 Oct 2022 02:56:24 pm
- Last edited by Jelte2357 on 25 Oct 2022 03:01:17 pm; edited 1 time in total

that is correct, but then my program that uses sqrt() or factorial or basically anything the program goes wrong after 15 decimals, the same amount i get when i dont use the module.

**Jelte2357 wrote:**

that is correct, but then my program that uses sqrt() goes wrong after 15 decimals, the same amount i get when i dont use the module.

Did you set the precision correctly? Also, how are you utilizing square roots to calculate pi?

Maker of things. Co-founder of TI-Toolkit.

I am using multiple different ways to calculate pi, including the earliest one (archimedes) and that one uses square roots. is used getcontext().prec=50 for my program, and the output is 3.14159265358979 [42262209880937007255852222442626952]. the text between [] is incorrect.

it sounds like you are using math.sqrt instead of Decimal.sqrt.

Example:

outputs

the second one using Decimal.sqrt() being the expected output.

Example:

**Code:**```
import decimal,math
```

decimal.getcontext().prec = 53

print(decimal.Decimal(math.sqrt(2)))

print(decimal.Decimal(2).sqrt())

**Code:**```
1.4142135623730951454746218587388284504413604736328125
```

1.4142135623730950488016887242096980785696718753769481

Negativity: the best way to deal with my square imaginary friends. - LogicalJoe [2018]

Co-founder of TI-Toolkit.

thank you logicaljoe, how would this work with powers and factorials? (sorry if I am asking easy/dumb questions I'm quite new to programming)

**Jelte2357 wrote:**

thank you logicaljoe, how would this work with powers and factorials? (sorry if I am asking easy/dumb questions I'm quite new to programming)

Decimal objects support all the usual arithmetic operations, provided that the arguments are integers or other Decimal objects so that precision isn't lost (i.e. no floats). Powers can thus be computed using the ** operator.

Factorials of integers can be calculated using the function in the math module; since the factorial of an integer is also an integer, you can perform operations on them and Decimals with no issue. Taking factorials of Decimals, though, is much harder, as you'd need to implement the Gamma function (the continuous extension of the factorial) with arbitrary precision (and, surprisingly, the decimal module doesn't do this for you).

Maker of things. Co-founder of TI-Toolkit.

**kg583 wrote:**

Decimal objects support all the usual arithmetic operations, provided that the arguments are integers or other Decimal objects so that precision isn't lost (i.e. no floats). Powers can thus be computed using the ** operator.

There is no decimal.pow or decimal.power or decimal.**, and even if i do Decimal(Decimal([pi with 100 digits]**2)).sqrt() the answer result will only ever have 15 correct decimal places. How do I fix this?

**Jelte2357 wrote:**

There is no decimal.pow or decimal.power or decimal.**, and even if i do Decimal(Decimal([pi with 100 digits]**2)).sqrt() the answer result will only ever have 15 correct decimal places. How do I fix this?

Decimal objects have their operators as bound methods, so if you wanted to use pow instead of **, it'd be [object].pow([power]) or decimal.Decimal.pow([object], [power]). The ** operator hooks into the pow implementation for Decimals, though, so it works just as well (and decimal.** isn't even valid as a method name).

The reason you are not getting the result you want is because you are squaring before the object is cast to a Decimal, so you're bound to lose precision. Also, all the operators return Decimal objects, so there's no need to cast more than once.

Maker of things. Co-founder of TI-Toolkit.

there is a Decimal.__pow__ that does work but it doesnt work how i want it to.

the X is where the output goes wrong.

Now I dont have any idea what is wrong

**Code:**```
from decimal import *
```

from math import *

getcontext().prec = 50

a=Decimal(3.14159265358979323846264338327950288419716939937510)

x=Decimal.__pow__(a,2)

y=Decimal(x).sqrt()

print(y)

3.141592653589793X1159979634685441851615905761718750

3.14159265358979323846264338327950288419716939937510

the X is where the output goes wrong.

Now I dont have any idea what is wrong

You're causing approximation as a binary float when constructing a decimal value from a literal. Writing the literal as a string avoids that:

(You'll also notice idiomatic use of __pow__ via the exponentiation operator in my example.)

**Code:**```
>>> import decimal
```

>>> from decimal import Decimal

>>> decimal.getcontext().prec = 50

>>> a = Decimal(3.14159265358979323846264338327950288419716939937510)

>>> a

Decimal('3.141592653589793115997963468544185161590576171875')

>>> a = Decimal('3.14159265358979323846264338327950288419716939937510')

>>> a # Note that this value is different from the above

Decimal('3.14159265358979323846264338327950288419716939937510')

>>> (a ** 2).sqrt()

Decimal('3.1415926535897932384626433832795028841971693993751')

(You'll also notice idiomatic use of __pow__ via the exponentiation operator in my example.)

Here’s a program I wrote to calculate pi. It displays 7 digits at a time. If only 6 show, then there should be a leading 0. On my iPad, this program (modified for the iPad) shows 200,000 digits in 204 seconds.

On my Ti 84+ CE, it takes 31 seconds to show 100 digits, 181 seconds for 250 digits.

ClrHome

Input "NBR DIGITS ",D

startTmr→Q

iPart(D/7)+1→S

10^7→T

For(Z,1,D

5*(Z-1)+3→L1(Z

End

For(M,1,S

0→C

For(Z,D,1,-1

L1(Z)*T+C→V

27*((Z-1)^2)+27*(Z-1)+6→B

V-iPart(V/B)*B→L1(Z

iPart(V/B)*(2*((Z-1)^2)-(Z-1))→C

End

V-iPart(V/T)*T→L1(1

Disp iPart(V/T

End

Disp "SECONDS "

checkTmr(Q)

On my Ti 84+ CE, it takes 31 seconds to show 100 digits, 181 seconds for 250 digits.

ClrHome

Input "NBR DIGITS ",D

startTmr→Q

iPart(D/7)+1→S

10^7→T

For(Z,1,D

5*(Z-1)+3→L1(Z

End

For(M,1,S

0→C

For(Z,D,1,-1

L1(Z)*T+C→V

27*((Z-1)^2)+27*(Z-1)+6→B

V-iPart(V/B)*B→L1(Z

iPart(V/B)*(2*((Z-1)^2)-(Z-1))→C

End

V-iPart(V/T)*T→L1(1

Disp iPart(V/T

End

Disp "SECONDS "

checkTmr(Q)

Most likely one of my last questions about this topic,

why doesn't this program output more than 15 correct decimals?

for info, this is chykovsky's algorithm.

why doesn't this program output more than 15 correct decimals?

for info, this is chykovsky's algorithm.

**Code:**```
from math import*
```

from decimal import *

getcontext().prec = 500

R=0

a=0

b=0

while R!=500:

b+=Decimal(((((-1)**a)*factorial(6*a)*((545140134*a)+13591409))/((factorial(3*a))*((factorial(a))**3)*((640320)**(3*a+(3/2))))))

c=Decimal(12*b)

d=Decimal(1/c)

a+=1

R+=1

print(d)

You're doing a large computation in terms of types other than Decimal again, so once again you're limited to float precision.

Annotating the important types:

You should take care that all values used in your expressions are actually Decimals to avoid this sort of thing, though only the division is actually problematic here because every other value used is an integer.

Annotating the important types:

**Code:**```
a: int = 0
```

b: int = 0

t0: int = -1 ** a

t1: int = factorial(6 * a)

t2: int = 545140134*a

t3: int = factorial(3 * a)

t4: int = factorial(a) ** 3

# 3/2 is a float, so t5 is also a float

t5: float = 3 * a + (3 / 2)

b += Decimal(

(

(t0*t1*(t2+13591409))

/

(t3*t4*(640320**t5))

)

)

You should take care that all values used in your expressions are actually Decimals to avoid this sort of thing, though only the division is actually problematic here because every other value used is an integer.

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