# Is there an integral or series for $\frac{\pi}{3}-1-\frac{1}{15\sqrt{2}}$?

The approximation

$$\pi\approx\frac{22}{7}=3+\frac{1}{7}$$

suggests that the closest integer to $\frac{1}{\left(\pi-3\right)}$ is $7$.

However, $$\frac{1}{\left(\pi-3\right)^2}\approx49.879$$ is closer to 50 than 49: $$\left({\frac{1}{\left(\pi-3\right)^2}}-50\right)^2< \left(\frac{1}{\left(\pi-3\right)^2}-7^2\right)^2$$ so $1+\frac{1}{15\sqrt{2}}$ is closer to $\frac{\pi}{3}$ than $1+\frac{1}{21}$ is.

$$\frac{\pi}{3}\approx 1.04719$$

$$1+\frac{1}{15\sqrt{2}}\approx 1.0471(4)$$

$$1+\frac{1}{21}\approx 1.047(6)$$

In other words,$$\left(\pi-3\right)^2\approx0.020048\approx0.02=\frac{1}{50}$$ so $$\pi-3\approx\frac{1}{\sqrt{50}}=\frac{1}{5\sqrt{2}}$$

Equivalently, $$\frac{\pi}{3}\approx1+\frac{1}{15\sqrt{2}}$$ An integral for $\frac{22}{7}-\pi$ is given by

$$\frac{22}{7}-\pi=\int_{0}^{1}\frac{x^4(1-x)^4}{1+x^2}dx$$

and a series for $\frac{\pi}{3}-1$ is given by

$$\frac{\pi}{3}-1=\sum_{k=1}^{\infty}\frac{1}{\left(1+k\right)\left(1+2k\right)\left(1+4k\right)}$$ Q: Is there a similar integral or series for $\frac{\pi}{3}-\left(1+\frac{1}{15\sqrt{2}}\right)$?

[EDIT 4/02/2016] Q2: Is there a variant of $$\pi=\frac{8}{3}\sum_{k=1}^{\infty}\frac{sin\left(\frac{k\pi}{4}\right)}{k}$$ that truncates to $3+\frac{1}{5\sqrt{2}}$?

For instance, there is $$\pi=\frac{1}{8944} \sum_{k=0}^\infty \left(\frac{77810+903\sqrt{2}}{8k+1}-\frac{131520}{8k+2}+\frac{9870+903\sqrt{2}}{8k+3}+\frac{43840}{8k+4}+\frac{77810-903\sqrt{2}}{8k+5}-\frac{131520}{8k+6}+\frac{9870-903\sqrt{2}}{8k+7}+\frac{43840}{8k+8}\right)$$ with first term $3+\frac{1}{5\sqrt{2}}$ but there may be a simpler one.

• Or in other words, $\pi-3 \approx \frac{1}{5\sqrt{2}}$. We only get three digits out of that, though -- which aligns roughly with the two nonzero digits of the right-hand side plus another digit to account for the wide choice in how one combines those two digits into a formula. Jan 23, 2016 at 10:26
• We only get two correct digits out of $log(3)\approx1$ but there is at least one series that truncates to it... Jan 23, 2016 at 10:28
• Why shouldn't it? You can approximate any constant arbitrarily close by other expressions. And yes, there is a series: $\frac \pi 3 = 1 + \frac{1}{15 \sqrt{2}} + \left(\frac \pi 3- 1 - \frac{1}{15\sqrt{2}} \right) +0 + 0 + 0 + 0 + \ldots$ Jan 23, 2016 at 10:30
• Note that, since ${\mathbb Q}+{\mathbb Q}\sqrt{2}$ in a dense subgroup of $\mathbb R$, you can get approximations of $\frac{\pi}{3}$ as precise as you want by numbers of the form $a+b\sqrt{2}$ (you can even choose $a$ and $b$ to be integers). Jan 23, 2016 at 10:41
• @JaumeOliverLafont: Since this question has been put on hold, can you ask another one on what is the Ramanujan formula responsible for $\pi \approx 2\sqrt{1+\sqrt{2}}\,$? That has more concrete answer and should not be closed. It can also provide a possible, though unlikely, solution to $\displaystyle 1+\frac{1}{15\sqrt{2}}$. Jan 23, 2016 at 16:33

This was earlier a bit hastily closed. However, one aspect of the question might have an interesting connection to the Tribonacci constant $T$. First, let $w = \frac{\sqrt{2}}{T^{-1}+1}$, so,

$$j\big(\tfrac{1+\sqrt{-11}}{2}\big) = \frac{(w^{24}-16)^3}{w^{24}} = -2^{15}$$

where $j(\tau)$ is the j-function. We then get the Ramanujan/Chudnovsky-type pi formula,

$$\frac{1}{\pi} = 4\sum_{n=0}^\infty (-1)^n\frac{(6n)!}{(3n)!\,n!^3}\frac{154n+15}{(2^{15})^{n+1/2}}\tag1$$

The first term of this just so happens to be,

$$\frac{1}{\pi} \approx \frac{4\times 15}{\sqrt{2^{15}}} = \frac{15\sqrt{2}}{64}$$

hence,

$$\frac{\pi}{64} \approx \frac{1}{15\sqrt{2}}\tag2$$

We wish to find a connection between $\frac{\pi}{64}$ and $\frac{\pi}{3}$. Looking at the continued fraction (WA link) of $\frac{\pi}{64}$, the fourth convergent is $\color{brown}{\frac{3}{61}}$ and it turns out that,

$$\frac{\pi}{64}+\color{brown}{\frac{\pi}{64}\frac{61}{3}} = \frac{\pi}{3}$$

However, since $\displaystyle \color{brown}{\frac{\pi}{64}\frac{61}{3}} \approx 1$, then,

$$\frac{\pi}{64}+1 \approx \frac{\pi}{3}\tag3$$

So $(2)$ and $(3)$ "explains" the relatively close approximation,

\begin{aligned} \frac{\pi}{3} &= 1.04719\dots\\ 1+\frac{1}{15\sqrt{2}}&= 1.04714\dots \end{aligned}

• The approximation (2) has unit ratio $$\frac{15\pi}{32\sqrt{2}}=1.041...$$ A similar approximation comes from the expansion of $${\left(1+x\right)}^\frac{1}{2} asin \left(\frac{x}{2}\right)=\frac{1}{2}x+\frac{1}{4}x^2-\frac{1}{24}x^3+\frac{1}{24}x^4+...$$ Setting $x=1$ and taking four terms yields $$\frac{9}{2\sqrt{2}\pi}=1.0128...$$ or $$\frac{2\pi}{135}\approx\frac{1}{15\sqrt{2}}$$ The remaining $\frac{135}{43}\approx\pi$ is the fifth convergent of $\frac{\pi}{15}$ wolframalpha.com/input/… Jan 27, 2016 at 5:51
• @Jaume: Using the $\arcsin$ at $x=1$ and taking four terms, one gets, $$\frac{\sqrt{2}\pi}{6} \approx \frac{3}{4}$$ In fact, the fourth convergent is indeed $3/4$. However, the relation, $$\frac{2\pi}{135} \approx \frac{1}{15\sqrt{2}}$$ is a bit contrived since this is just, $$\frac{2\pi}{9m} \approx \frac{1}{m\sqrt{2}}$$ where $m$ can be any number. (Note that your $1/43$ disappears without any explanation.) Jan 27, 2016 at 16:09
• From $$\pi\approx \frac{9}{2\sqrt{2}}$$ and $$\frac{43\pi}{135}\approx 1$$ we have $$\frac{\pi}{3}=\frac{43}{135}\pi+\frac{2}{135}\pi=\frac{43}{135}\pi+\frac{1}{15}\frac{2}{9}\pi\approx1+\frac{1}{15\sqrt{2}}$$ $m$ can only be $15$ if we take the irrational part of the approximation from the truncated series and the rational part from the continued fraction. Jan 27, 2016 at 21:09

A linear combination of integrals for approximations $3$, $\frac{333}{106}$ and $\frac{20\sqrt{2}}{9}$ from this collection gives

$$\int_0^1 \frac{\left[x(1 - x)^2 + 9x^4 (1 - x)^4 (7 - 30 x(1 - x)^2)\right](1+x^4) -15\sqrt{2}x^4(1-x)^4}{500\left(1+x^2+x^4+x^6\right)}dx =\\ \pi - \left(3+\frac{1}{5\sqrt{2}}\right)$$

with nonnegative integrand.