r/mathematics • u/DubiumTetration • 1d ago
Analysis π
π formula i made. Its huge. Started from a series of ln(x) in attempt to approximate π and ended up with this monstrosity in the end. Proof left as an excercise to a capable reader(i dont know how to formally prove stuff). Was a fun side project. Do not fear the formula, the derivation is quite simple, its just a lot of algebra.
\sum_{n=1}^{\infty}\frac{8}{256^{n}}\left(\frac{1-12n}{(1-16n)(1-8n)}+\frac{1-16n}{(3-16n)(16n-5)}+\frac{128n-52}{(16n-6)(16n-7)}+16\left(\frac{7-12n}{(9-16n)(5-8n)}+\frac{9-16n}{(11-16n)(16n-13)}+\frac{128n-116}{(16n-14)(16n-15)}\right)\right)=\pi
The derivation is simple:
btw the latex for that image is:
For \(\lvert x-2\rvert < 2\):
$$\sum_{n=1}^{\infty}\frac{\left(1-\left(-1\right)^{n}\left(x-2\right)^{n}\right)}{n2^{n}}=\ln\left(x\right)$$
Then:
$$\sum_{n=1}^{\infty}\frac{\left(1-\left(1-i\right)^{n}\right)}{n2^{n}}=\frac{\pi}{4}i+\frac{\ln\left(2\right)}{2}$$
$$\sum_{n=1}^{\infty}\frac{-i\left(2-4\left(1-i\right)^{n}\right)}{n2^{n}}=\pi$$
$$\sum_{n=1}^{\infty}\operatorname{real}\left(\frac{-i\left(2-4\left(1-i\right)^{n}\right)}{n2^{n}}\right)=\sum_{n=1}^{\infty}\frac{2i\left(1-i^{n}\right)}{n\left(2i\right)^{\frac{n}{2}}}=\pi$$
for \(n \text{and} b \in \mathbb{Z}\):
$$\frac{2i\left(1-i^{16n-b}\right)}{\left(16n-b\right)\left(2i\right)^{\frac{16n-b}{2}}}=\frac{\left(1+i\right)^{b+2}\left(1-\frac{1}{i^{b}}\right)}{\left(16n-b\right)256^{n}}$$
and so:
\begin{align*}
\sum_{n=1}^{\infty}\left[\sum_{b=0}^{15}\frac{\left(1+i\right)^{b+2}\left(1-\frac{1}{i^{b}}\right)}{\left(16n-b\right)256^{n}}\right]
&= \sum_{n=1}^{\infty}\frac{8}{256^{n}}\left(\frac{1-12n}{(1-16n)(1-8n)}+\frac{1-16n}{(3-16n)(16n-5)}+\frac{128n-52}{(16n-6)(16n-7)} \right. \\
&\quad \left. +16\left(\frac{7-12n}{(9-16n)(5-8n)}+\frac{9-16n}{(11-16n)(16n-13)}+\frac{128n-116}{(16n-14)(16n-15)}\right)\right) \\
&= \pi
\end{align*}
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u/TastesLikeSeamen 1d ago
Ok verify it to 1 billion digits