Præface: Okay, having somehow survived without proper Unicode input (Ctrl-Shift-U) for five(?) months, I still feel like killing myself every time I have to type in Greek stuff or mathematical operators, or whatever. (Also, “everytime” is wrong, and “every time” is definitely spelt as two words, yet I constantly see some people using it. I’m not really interested in derailing every conversation by being a grammar nazi, but gosh.)
Let’s start with the factorial-over-double-odd-factorial series for π/2. You know, the one that Euler guy came up with centuries ago.
Now let’s work modulo 2. Notice that the terms from 
So there.
Now, it seems that we can even use this series to calculate the 2-adic representation of π/2 (and hence π or any other rational multiple thereof), because not only does it converge under the usual metric, it also converges under the 2-adic metric, as the number of times 2 divides into the numerator increases by at least one every two terms, while the denominator stays odd forever.
Just as an aside, there’s one pretty cool algorithm for calculating the inverse of a nonzero number modulo a prime, which is kinda related to the Euclidean algorithm if you stare at it enough. See pages 131-132 of The Book of Numbers. To wit, in code:
def modinv(a,p):
'''Calculate the multiplicative inverse of a modulo p.'''
inv = 1
a %= p
while a != 1:
q = (p+a-1)//a # ceil(p/a)
a = a*q-p
inv = inv*q%p
return inv
Because the modulus is prime (and by the loop condition, 
Anyway.
Working out the first sixteen terms of the series, we get π/2=…10111110001011102. What an utterly fascinating result. we get π/2=…0000000000000002. (I fixed something wrong with my calculation.) Does it actually tend to all zeroes?
Not every series for π converges with the 2-adic metric though (obviously). The Gregory-Leibniz series reduces to Grandi’s series modulo 2, which can hardly be called convergent. On the other hand, there (probably?) are other series with only odd denominators that converge under the 2-adic metric, and converge to 
Stealth edit: fixed the value of 
Edit #2: tried using the Euler-transformed series for arctan(1/2)+arctan(1/3), which also seemed to give a few thousand binary digits of 0s. Now, why exactly would it converge to zero, if it actually does?
samples around the one we want to interpolate as an 
exactly, with zero first, second, …, 
th derivatives at 0 frequency. (Of course, the number of zero derivatives is exactly the same as the degree of polynomial interpolated.)![f_n[k]](http://s0.wp.com/latex.php?latex=f_n%5Bk%5D&bg=ffffff&fg=444444&s=0 "f_n[k]")
as 
, you get ![\lim_{n\to\infty}f_n[k]=(-1)^{k+1}](http://s0.wp.com/latex.php?latex=%5Clim_%7Bn%5Cto%5Cinfty%7Df_n%5Bk%5D%3D%28-1%29%5E%7Bk%2B1%7D&bg=ffffff&fg=444444&s=0 "\lim_{n\to\infty}f_n[k]=(-1)^{k+1}")
for nonzero 
. Which is yet another appearance of Grandi’s series, but I digress.![k:g_n[k]](http://s0.wp.com/latex.php?latex=k%3Ag_n%5Bk%5D&bg=ffffff&fg=444444&s=0 "k:g_n[k]")
such that their sum is 1, and the sum of squares is minimised. Do we know how to solve that? Of course we do. You just give every coefficient equal value.
) from all the other samples. What do you do?
samples at 
, we can form ![f_n[k]=(-1)^{k+1}\frac{n!^2}{(n-k)!(n+k)!}](http://s0.wp.com/latex.php?latex=f_n%5Bk%5D%3D%28-1%29%5E%7Bk%2B1%7D%5Cfrac%7Bn%21%5E2%7D%7B%28n-k%29%21%28n%2Bk%29%21%7D&bg=ffffff&fg=444444&s=0 "f_n[k]=(-1)^{k+1}\frac{n!^2}{(n-k)!(n+k)!}")
for nonzero ![f_n[0]=0](http://s0.wp.com/latex.php?latex=f_n%5B0%5D%3D0&bg=ffffff&fg=444444&s=0 "f_n[0]=0")
), where we’re approximating our sample ![X[0]](http://s0.wp.com/latex.php?latex=X%5B0%5D&bg=ffffff&fg=444444&s=0 "X[0]")
with ![\sum_{k\in\mathbb Z}X[k]f_n[-k]](http://s0.wp.com/latex.php?latex=%5Csum_%7Bk%5Cin%5Cmathbb+Z%7DX%5Bk%5Df_n%5B-k%5D&bg=ffffff&fg=444444&s=0 "\sum_{k\in\mathbb Z}X[k]f_n[-k]")
, which is really just a convolution. Note that we can rewrite the kernel, again for nonzero ![f_n[k]=(-1)^{k+1}\binom{2n}{n+k}/\binom{2n}n](http://s0.wp.com/latex.php?latex=f_n%5Bk%5D%3D%28-1%29%5E%7Bk%2B1%7D%5Cbinom%7B2n%7D%7Bn%2Bk%7D%2F%5Cbinom%7B2n%7Dn&bg=ffffff&fg=444444&s=0 "f_n[k]=(-1)^{k+1}\binom{2n}{n+k}/\binom{2n}n")
, where we can factor out the 
denominator if we want, but the main highlight is that such a kernel would be exactly a row of Pascal’s triangle with alternating sign and the middle term set to zero (then normalised to unit sum).
iirc, which grows at 
. Which is unbounded, yeah. So it’s not just bad, it gets worse for large 
equalling 
, 
and 
, but not the general solution), I posted a new one about linear recurrences. I know, old topic. I covered that before somewhere here or on zznq. Maybe. I can’t remember and don’t feel like checking.
. You can actually work this out with brute force and notice that it has a period of six terms. This one got answered pretty fast, so I followed it up with something a bit less trivial. Define our recurrence by 
. For what values of 
and 
is this recurrence periodic, regardless of initial values?
, and we have the condition that the two zeros must satisfy 
for some integer 
, that forces the two zeros to be reciprocals of each other, and letting one of them by 
, we get 
and 
.
. In this case, we can either have a doubled zero at 
and one zero at 
. The latter gives the characteristic polynomial 
, or equivalently, the recurrence 
. Yup, periodic with period 2, no doubt. What about the former? Turns out that having doubled zeros makes them non-solutions; those give characteristic polynomials 
, or equivalently the recurrence 
. In the 
case, this is equivalent to linear extrapolation which obviously isn’t periodic, and in the 
case, this is equivalent to linear extrapolation with signs flipping every term, which also isn’t periodic.
and 
. All the other solutions have 
.
, so the sum has 
error. Let’s improve it by one order. Observe that the sequence 
isn’t quite centred on the same value as the (one-valued) sequence 
. We can shift the positive terms by one-eighth of a term to compensate.
.
? We should expect it to be higher. And it is, by exactly one order. 
, so the sum has 
error.
