I read somewhere that


is almost an integer and strangely enough this isn't just a random coincidence but rather there exists some general theory


behind the occurences of these almost integers (and their relation to other areas of number theory)

Surely there are many other strange identities such as:

$$\sqrt{2} \approx \frac{3}{5} + \frac{\pi}{7 -\pi}$$

I'm guessing that this "coincidence" is probably similar to the earlier example a special case of some general theory that relates rational expressions of pi to algebraic integers.

Can someone point me in the right direction if not explain it here itself?

  • $\begingroup$ You mean $e^{\pi\sqrt{163}}$. Is there a reason for your guess? $\endgroup$
    – anon
    Feb 5, 2014 at 4:17
  • $\begingroup$ see: [ wolframalpha.com/input/?i=Heegner+number ] verbatim: The Heegner numbers have a number of fascinating connections with amazing results in prime number theory. In particular, the j-function provides stunning connections between e, pi, and the algebraic integers. They also explain why Euler's prime-generating polynomial n^2-n+41 is so surprisingly good at producing primes. $\endgroup$
    – janmarqz
    Feb 5, 2014 at 4:21
  • 2
    $\begingroup$ Relevant: Why is $e^{\pi \sqrt{163}}$ almost an integer? $\endgroup$
    – Ben
    Feb 5, 2014 at 5:44
  • $\begingroup$ Your identity may be rewritten as $$\pi\approx\frac{392-175\sqrt{2}}{46}\approx3.1415(7)$$ $\endgroup$ Jan 22, 2016 at 1:01
  • $\begingroup$ You can get more correct decimals using less digits: $$\pi\approx\frac{192-98\sqrt{2}}{17}\approx3.141592(4)$$ $\endgroup$ Jan 22, 2016 at 1:10

1 Answer 1


The approximation $$\sqrt{2} \approx \frac{3}{5} + \frac{\pi}{7 -\pi}$$

may be rewritten as $$\sqrt{2}-\frac{3}{5} \approx \frac{1}{\frac{7}{\pi} -1}$$

After some manipulation, this is found to be equivalent to $$\pi\approx\frac{392-175\sqrt{2}}{46}=\frac{7}{46}\left(56-25\sqrt{2}\right)$$ so, at least for this case, some theory relating $\pi$ to algebraic integers would suffice.

A useful direction is shown by the following series, which is related to a similar approximation to $\pi$:

$$\sum_{k=0}^{\infty} \frac{15!(k+1)}{(8k+1)_{15}}=\frac{15}{8}\left(1716-7\left(99\sqrt{2}-62\right)\pi\right)\approx 1$$

where $(a)_n$ is a rising factorial or Pochhammer symbol $a\times(a+1)\times...\times (a+n-1)$.

(see https://math.stackexchange.com/a/1657416/134791)

This gives the approximation

$$\pi \approx \frac{3676}{15(99\sqrt{2}-62)}=\frac{1838(62+99\sqrt{2})}{118185}$$

with eight correct decimal digits.

A general series that might provide an explanation for this approximation, as well as others of the form $a+b\sqrt{2}$ for rational $a$ and $b$, is given by

$$\sum_{k=0}^\infty \frac{c}{\prod_{i=1}^{7}((8k+i)(8k+16-i))^{w_i}} \approx 1,$$ with constant $c$ and binary weighting exponents $w_i$ taking values either $0$ or $1$. The example provided above is the particular case $w_i=1$ for all $i$ from $1$ to $7$.

The numerator $c$ may be set by letting the first term of the series equal $1$.

$$\sum_{k=0}^\infty \prod_{i=1}^7 \left(\frac{i(16-i)}{(8k+i)(8k+16-i)}\right)^{w_i} \approx 1,$$

A simple Dalzell-type integral that relates $\pi$ to approximations using $\sqrt{2}$ is given by

$$\pi=\frac{20\sqrt{2}}{9} - \frac{2\sqrt{2}}{3} \int_0^1 \frac{x^4(1 - x)^4}{1 + x^2 + x^4 + x^6}dx$$

The following example combines a rational approximation from below and an irrational approximation from above

$$\pi \approx \frac{1}{4}·\frac{25}{8}+\frac{3}{4}·\frac{20\sqrt{2}}{9}$$ to obtain an irrational one from below that improves over the rational one.

$$\pi= \frac{25}{32} + \frac{5\sqrt{2}}{3} + \int_0^1 \frac{x \left(1 - x\right)^4\left(\left(1 + x^4\right) \left(1 + 4 x + x^2\right) - 8 \sqrt{2} x^3\right)}{16 \left(1 + x^2 + x^4 + x^6\right)} dx$$

WA link

Since the integrand is non-negative in $\left(0,1\right)$, this integral is a proof that $$\pi > \frac{25}{32}+\frac{5\sqrt{2}}{3}$$

Hopefully a similar integral exists for your approximation. For instance, the integrand in

$$\frac{1}{184} \int_0^1 \frac{x^4(1 - x)^4}{1 + x^2 + x^4 + x^6} (210\sqrt{2}- (259 + 120 x (1 - x)^2) (1+x^4) ) dx \\= \pi-\frac{392-175\sqrt{2}}{46}$$

is small, so your approximation is justified, but unfortunately there is a sign change in $(0,1)$, so we do not know in advance whether it lies above or below $\pi$. This integral has been built combining linearly three Dalzell-type expressions for constants $\frac{22}{7}-\pi$, $\frac{377}{120}-\pi$ and $\frac{20\sqrt{2}}{9}-\pi$ in order to match your number.

The relationship between the above families of series and integrals can be established as in the proofs found in these answers: [1],[2].

Finally, the following integral evaluates to the error of your approximation and has small nonnegative integrand.

$$\frac{3+ 10 \sqrt{2}}{4393} \int_0^1 \frac{x^3(1 - x)^6}{1 + x^2 + x^4 + x^6} \left(3111662 - 2200275\sqrt{2} + 3465 \left(-898 + 635 \sqrt{2}\right) x^8\right) dx = \pi-\frac{7}{46}\left(56-25\sqrt{2}\right)$$

It is a linear combination of

$$\frac{4}{10\sqrt{2}-3}\int_0^1 \frac{x^3(1-x)^6}{1+x^2+x^4+x^6}dx = \pi-\frac{35}{10\sqrt{2}-3}$$


$$\frac{4}{10\sqrt{2}-3}\int_0^1 \frac{x^{11}(1-x)^6}{1+x^2+x^4+x^6}dx = \pi-\frac{17327}{495\left(10\sqrt{2}-3\right)}$$

  • $\begingroup$ I don't see how this is an answer to the question. $\endgroup$ Mar 29, 2016 at 8:42
  • $\begingroup$ @GerryMyerson This explains an approximation of the same form as the one in the question, so hopefully it points in some useful direction, as it was asked. I cannot discard that a similar series with a denominator of the form $\prod_{i=1}^{15} (8k+i)^{w_i}$, where coefficients $w_i$ take the value $0$ or $1$, directly explains the approximation. $\endgroup$ Mar 29, 2016 at 11:47
  • $\begingroup$ How is $\pi\approx a+b\sqrt2$ of the same form as $\pi/(a-\pi)\approx b+\sqrt2$? $\endgroup$ Mar 29, 2016 at 21:46
  • 1
    $\begingroup$ @GerryMyerson They are of the same form, but $a$ and $b$ are not the same. Solving for $\pi$ in $$\frac{\pi}{a-\pi} \approx b+\sqrt{2}$$ leads to $$\pi\approx a'+b'\sqrt{2}$$. If I made no mistake [here](math.stackexchange.com/questions/664175/…) $a'=\frac{392}{46}=\frac{196}{23}$ and $b'=\frac{175}{46}$ from the approximation by the OP. $\endgroup$ Mar 30, 2016 at 7:41
  • $\begingroup$ @GerryMyerson Updated the answer with a parallel of Dalzell's integral using $\sqrt{2}$ $\endgroup$ Apr 28, 2017 at 10:09

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