A unique sum for pi

A friend of mine recently talked to me about the derivative of x\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 n=11n2=π26\sum _{ n=1 }^{ \infty }{ \frac { 1 }{ { n }^{ 2 } } } =\frac { { \pi }^{ 2 } }{ 6 } and I decided that the derivative of x\left| x \right| should be sign(x)sign\left( x \right) (where sign(x)=1sign\left( x \right) =1 for x>0x>0, sign(x)=1sign\left( x \right) =-1 for 0>x0>x, and sign(x)sign\left( x \right) is 00 for x=0x=0), with chain rule applied for xx. I went about mimicking the terms of the sum I mentioned earlier using my sign(x)sign\left( x \right) function. I won't go through every detail, but I ended up with

1n=1(2n+1)(sign(xn)+1)2n2(n1)21-\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 xx whose integral would hypothetically equal the pp-series sum of 22. I took a literal integral here using my knowledge of the antiderivative of my sign(x)sign\left( x \right) function, obtaining:

(x12n=1(2n+1)(xn+x)n2(n+1)2)]0x=x12n=1(2n+1)(xn+x)n2(n+1)20+12n=1n(2n+1)n2(n+1)2{ \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>xn>x, then the sum can be written as

x12n=1(2n+1)(nx+x)n2(n+1)2=x12n=1n(2n+1)n2(n+1)2x-\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<xn<x, written as

x12n=1x(2n+1)(2xn)n2(n+1)2x-\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 x\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 π26\frac { { \pi }^{ 2 } }{ 6 } as xx increases, I calculate π\pi by the following:

6(2xn=1x(2n+1)(2xn)n2(n1)2)\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 π\pi very quickly in comparison to the most standard methods, but I know there are some insanely quick methods out there. Different values of xxx- \left\lfloor x \right\rfloor return different results (let's call this difference dd for further use). From graphing, it appears that d0.75d\approx 0.75 returns the quickest convergence, but I'm still investigating this. I believe d=0d=0 is safest, but any value 0d<10\le d<1 appears to converge to π\pi, just at different rates.

At 20002000 sums with d=0d=0, this method converges about 13341334 times more quickly than the pp-series of 22. At 20002000 sums with d=0.75d=0.75, this method converges about 2244530022445300 times more swiftly than the pp-series of 22. However, as the number of terms increases, dd must decrease or it loses accuracy. There seems to be some relationship between number of terms and optimal dd value, which I'm still looking into, but I suggest (in the meantime, at least) the use of d=0d=0 for any practicality. These values were calculated in Desmos (see my test graph here) as the following:

f(x)=16(2(x+d)n=1x(2n+1)(2(x+d)n)n2(n+1)2)πf\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<1d<1

g(x)=16n=1x1n2πg\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 x\left\lfloor x \right\rfloor as the upper bound of the pp series in order to compare the two sums with equal number of terms. So, the number of times quicker is f(x)g(x)\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 π\pi, but Desmos puts it above π\pi relatively quickly.

Note by Austin Antonacci
3 years, 11 months ago

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wow this is insane :o good job :D

Romeo Gomez - 3 years, 10 months ago

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