# 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, 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
4 years, 2 months ago

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wow this is insane :o good job :D

- 4 years, 2 months ago