# Applied BBP-formula for the n-th digit of pi

Hello everyone,

I have trouble understanding how to apply the BBP formula, i.e. actually compute the n-th digit of pi.

$\pi$=$\sum$$\frac{1}{16^{k}}$($\frac{4}{8k+1}$-$\frac{2}{8k+4}$-$\frac{1}{8k+5}$-$\frac{1}{8k+6}$)

where the sum uses k from 0 to ∞.

I've read a few explanations how to adapt it, but have always failed to understand the following issues:
1) what is should be the expected returned value, i.e. what should I extract from it to get my n-th digit (I expect something fractional like $\frac{p}{q}$)?
2) what should I do with the infinite part of the sum? It seems illogical to me to approximate it somehow, since I expect the exact value of the desired digit
3) the above formula is supposed to calculate the hexadecimal formula of $\pi$. What does that mean for the calculated digit? Is in that case the fractional part of $\pi$ regarded as a hexadecimal number?

Basically, I'm looking for an explanation how to apply the above formula and would be very thankful if someone of you could provide this explanation or a suitable link.

I'm not a professional mathematician, I'm an engineer, so I apologize in advance for any "foolish" questions/statements or if I've misplaced the question on the wrong forum.

Thanks for your understanding.

Regards
scienalc

## Answers and Replies

arildno
Homework Helper
Gold Member
Dearly Missed
The Wikipedia article goes into quite some detail on how to extract a particular digit. Essentially, it requires a clever use of the modulo function.

Yes, the wiki article was one of the sources I consulted. However, it did not explain the issues I listed.

I did get the idea behind the use of the modulo function, though

arildno
Homework Helper
Gold Member
Dearly Missed
"2) what should I do with the infinite part of the sum? It seems illogical to me to approximate it somehow, since I expect the exact value of the desired digit"
--
1. As for the infinite part of the sums, they simply won't contribute to the value of the higher digits.

If you add three 0.0007s together, do those 7s add anything to the first three zeros?

No, they don't!

2. Thus, by careful budget balancing, you may IGNORE that infinite part as irrelevant for the digit value.

1 person
Thanks a lot, that helped me.

Could you clarify issue 3 a bit (since it effectively summarizes number 1 as well)?

arildno
Homework Helper
Gold Member
Dearly Missed
"3) the above formula is supposed to calculate the hexadecimal formula of π. What does that mean for the calculated digit? Is in that case the fractional part of π regarded as a hexadecimal number?"

Here, you need to understand the difference between a NUMBER, and the ways we may REPRESENT a number.

In binary REPRESENTATION, we have a number 1000. Now, AS A NUMBER, this is identical with its DECIMAL representation 8.

All properties that hold for 1000 holds for 8 as well, and vice versa.
----

Another way to understand the difference between a number and its representations is to have a meter stick as you "number".

Now, there are many different ways to chop up that stick, leading to widely different amounts of bits of stick and lengths of such bits.

But, a bag containing all the bits gained from one such method of chopping up the meter stick is a different REPRESENTATION of the number, but the same number as the unchopped stick.

Hm... that part is perfectly clear. As you state, a 1000 binary and a 8 decimal are the same number, but a different representation. So, if we take the first few decimals of pi 3.1415926535... and I wanted to extract the fourth digit, in this case 5 (decimal), would I be correct to assume using the above formula would yield F (hexadecimal) as the hexadecimal representation of pi is 3.243F6A8885...?

When the formula is applied, the result is a fraction (sometimes even negative, if I'm not mistaken),
e.g. for the above case of the fourth digit, my result was 1.AE18C8D4F02D7 (1.68006568144499 decimal) using 200 members of the infinite progression. How do I extract my digit from this value?

PS: I really appreciate your patience, thanks a lot.