A friend of mine recently talked to me about the derivative of ∣x∣, and I decided to work with it to see what I could find. What I did find is probably already a well-known or at least well-explored thing, but the method I used to find it is interesting. To me, at least.
I started with the knowledge that ∑n=1∞n21=6π2 and I decided that the derivative of ∣x∣ should be sign(x) (where sign(x)=1 for x>0, sign(x)=−1 for 0>x, and sign(x) is 0 for x=0), with chain rule applied for x. I went about mimicking the terms of the sum I mentioned earlier using my sign(x) function. I won't go through every detail, but I ended up with
1−∑n=1∞2n2(n−1)2(2n+1)(sign(x−n)+1)
This is a function of x whose integral would hypothetically equal the p-series sum of 2. I took a literal integral here using my knowledge of the antiderivative of my sign(x) function, obtaining:
which is the same as the right side of our definite integral! This means that our definite integral can be represented as the indefinite integral bounded where n<x, written as
x−21∑n=1⌊x⌋n2(n+1)2(2n+1)(2x−n)
(Note that, instead of ∞, we have ⌊x⌋ on top)
One thing I'm still trying to understand is that I have to multiply my sum by two to actually get my desired result. I'll figure it out eventually.
Since this sum is approaching 6π2 as x increases, I calculate π by the following:
6(2x−∑n=1⌊x⌋n2(n−1)2(2n+1)(2x−n))
And somehow, it works. I'm looking for any info or comments on this. It converges to π very quickly in comparison to the most standard methods, but I know there are some insanely quick methods out there. Different values of x−⌊x⌋ return different results (let's call this difference d for further use). From graphing, it appears that d≈0.75 returns the quickest convergence, but I'm still investigating this. I believe d=0 is safest, but any value 0≤d<1 appears to converge to π, just at different rates.
At 2000 sums with d=0, this method converges about 1334 times more quickly than the p-series of 2. At 2000 sums with d=0.75, this method converges about 22445300 times more swiftly than the p-series of 2. However, as the number of terms increases, d must decrease or it loses accuracy. There seems to be some relationship between number of terms and optimal d value, which I'm still looking into, but I suggest (in the meantime, at least) the use of d=0 for any practicality. These values were calculated in Desmos (see my test graph here) as the following:
f(x)=∣∣∣∣∣∣1−π6(2(⌊x⌋+d)−∑n=1⌊x⌋n2(n+1)2(2n+1)(2(⌊x⌋+d)−n))∣∣∣∣∣∣ where d<1
g(x)=∣∣∣∣1−π6∑n=1⌊x⌋n21∣∣∣∣
Note that I used ⌊x⌋ as the upper bound of the p series in order to compare the two sums with equal number of terms. So, the number of times quicker is g(x)f(x)
Upon further research, I found that Desmos is not accurate enough for this sum. It continues to converge to π, but Desmos puts it above π relatively quickly.
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Comments
wow this is insane :o good job :D