Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

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?

share|cite|improve this question
You mean $e^{\pi\sqrt{163}}$. Is there a reason for your guess? – anon Feb 5 '14 at 4:17
see: [ ] 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. – janmarqz Feb 5 '14 at 4:21
Relevant: Why is $e^{\pi \sqrt{163}}$ almost an integer? – Ben Feb 5 '14 at 5:44
Your identity may be rewritten as $$\pi\approx\frac{392-175\sqrt{2}}{46}\approx3.1415(7)$$ – Jaume Oliver Lafont Jan 22 at 1:01
You can get more correct decimals using less digits: $$\pi\approx\frac{192-98\sqrt{2}}{17}\approx3.141592(4)$$ – Jaume Oliver Lafont Jan 22 at 1:10

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 $$\pi\approx\frac{392-175\sqrt{2}}{46},$$ so, at least for this case, some theory relating $\pi$ to algebraic integers would suffice.

A possible 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 $a\times(a+1)\times...\times (a+n-1)$.


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 $1$.

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

share|cite|improve this answer
I don't see how this is an answer to the question. – Gerry Myerson Mar 29 at 8:42
@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. – Jaume Oliver Lafont Mar 29 at 11:47
How is $\pi\approx a+b\sqrt2$ of the same form as $\pi/(a-\pi)\approx b+\sqrt2$? – Gerry Myerson Mar 29 at 21:46
@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](…) $a'=\frac{392}{46}=\frac{196}{23}$ and $b'=\frac{175}{46}$ from the approximation by the OP. – Jaume Oliver Lafont Mar 30 at 7:41

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.