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# A unique sum for pi

A friend of mine recently talked to me about the derivative of $$\left| x \right|$$, 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 $$\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ 2 } } } =\frac { { \pi }^{ 2 } }{ 6 }$$ and I decided that the derivative of $$\left| x \right|$$ should be $$sign\left( x \right)$$ (where $$sign\left( x \right) =1$$ for $$x>0$$, $$sign\left( x \right) =-1$$ for $$0>x$$, and $$sign\left( x \right)$$ 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 $$sign\left( x \right)$$ function. I won't go through every detail, but I ended up with

$$1-\sum _{ n=1 }^{ \infty }{ \frac { \left( 2n+1 \right) \left( sign\left( x-n \right) +1 \right) }{ 2{ n }^{ 2 }{ \left( n-1 \right) }^{ 2 } } }$$

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 $$sign\left( x \right)$$ function, obtaining:

$${ \left( x-\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty }{ \frac { \left( 2n+1 \right) \left( \left| x-n \right| +x \right) }{ { n }^{ 2 }{ \left( n+1 \right) }^{ 2 } } } ) \right] }_{ 0 }^{ x }=x-\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty }{ \frac { \left( 2n+1 \right) \left( \left| x-n \right| +x \right) }{ { n }^{ 2 }{ \left( n+1 \right) }^{ 2 } } } -0+\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty }{ \frac { n\left( 2n+1 \right) }{ { n }^{ 2 }{ \left( n+1 \right) }^{ 2 } } }$$

This is incredible! In our original indefinite integral, if $$n>x$$, then the sum can be written as

$$x-\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty }{ \frac { \left( 2n+1 \right) \left( n-x+x \right) }{ { n }^{ 2 }{ \left( n+1 \right) }^{ 2 } } } =x-\frac { 1 }{ 2 } \sum _{ n=1 }^{ \infty }{ \frac { n\left( 2n+1 \right) }{ { n }^{ 2 }{ \left( n+1 \right) }^{ 2 } } }$$

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

$$x-\frac { 1 }{ 2 } \sum _{ n=1 }^{ \left\lfloor x \right\rfloor }{ \frac { \left( 2n+1 \right) \left( 2x-n \right) }{ { n }^{ 2 }{ \left( n+1 \right) }^{ 2 } } }$$

(Note that, instead of $$\infty$$, we have $$\left\lfloor x \right\rfloor$$ 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 $$\frac { { \pi }^{ 2 } }{ 6 }$$ as $$x$$ increases, I calculate $$\pi$$ by the following:

$$\sqrt { 6\left( 2x-\sum _{ n=1 }^{ \left\lfloor x \right\rfloor }{ \frac { \left( 2n+1 \right) \left( 2x-n \right) }{ { n }^{ 2 }{ \left( n-1 \right) }^{ 2 } } } \right) }$$

And somehow, it works. I'm looking for any info or comments on this. It converges to $$\pi$$ very quickly in comparison to the most standard methods, but I know there are some insanely quick methods out there. Different values of $$x- \left\lfloor x \right\rfloor$$ return different results (let's call this difference $$d$$ for further use). From graphing, it appears that $$d\approx 0.75$$ returns the quickest convergence, but I'm still investigating this. I believe $$d=0$$ is safest, but any value $$0\le d<1$$ appears to converge to $$\pi$$, 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:

$$f\left( x \right) =\left| 1-\frac { \sqrt { 6\left( 2\left( \left\lfloor x \right\rfloor +d \right) -\sum _{ n=1 }^{ \left\lfloor x \right\rfloor }{ \frac { \left( 2n+1 \right) \left( 2(\left\lfloor x \right\rfloor +d)-n \right) }{ { n }^{ 2 }{ \left( n+1 \right) }^{ 2 } } } \right) } }{ \pi } \right|$$ where $$d<1$$

$$g\left( x \right) =\left| 1-\frac { \sqrt { 6\sum _{ n=1 }^{ \left\lfloor x \right\rfloor }{ \frac { 1 }{ { n }^{ 2 } } } } }{ \pi } \right|$$

Note that I used $$\left\lfloor x \right\rfloor$$ 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 $$\frac { f\left( x \right) }{ g\left( x \right)}$$

Upon further research, I found that Desmos is not accurate enough for this sum. It continues to converge to $$\pi$$, but Desmos puts it above $$\pi$$ relatively quickly.

Note by Austin Antonacci
1 year, 4 months ago

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