25
$\begingroup$

I just stumbled upon

$$ \pi \approx \sqrt{ \frac{9}{5} } + \frac{9}{5} = 3.141640786 $$

which is $\delta = 0.0000481330$ different from $\pi$. Although this is a rather crude approximation I wonder if it has been every used in past times (historically). Note that the above might also be related to the golden ratio $\Phi = \frac{\sqrt 5 + 1}{2} $ somehow (the $\sqrt5$ is common in both).

$$ \Phi = \frac{5}{6} \left( \sqrt{ \frac{9}{5} } + \frac{9}{5} \right) - 1 $$

or

$$ \Phi \approx \frac{5}{6} \pi - 1 $$

I would like to know if someone (known) has used this, or something similar, in their work. Is it at all familiar to any of you?

Related Question (link).

$\endgroup$
5
  • $\begingroup$ Another interesting discussion (related) here math.stackexchange.com/questions/108510/… $\endgroup$
    – Kirthi Raman
    Commented May 18, 2012 at 21:01
  • $\begingroup$ @Artin this first link does not apply because it is an identity formula and not an approximation, and the second link is already included in the original posting. $\endgroup$
    – John Alexiou
    Commented May 18, 2012 at 21:03
  • $\begingroup$ What are you expecting, validation of this approximation and some sort of variation of this approximation? $\endgroup$
    – Kirthi Raman
    Commented May 18, 2012 at 21:16
  • $\begingroup$ I look for someone to say, I have seen this and it was used by x, or this is related to y approximation. $\endgroup$
    – John Alexiou
    Commented May 18, 2012 at 21:21
  • 1
    $\begingroup$ The closest approximation for $\pi$ I have found is $$\pi \approx \frac{9}{5}\sqrt{3}$$, which comes from the first term of a series: math.stackexchange.com/a/1682189/134791 $\endgroup$
    – Jaume Oliver Lafont
    Commented Mar 5, 2016 at 15:21

3 Answers 3

Reset to default
28
$\begingroup$

I have not seen it before. Note that $\pi = \sqrt{a} + a$ where $a = (1+2\,\pi -\sqrt {1+4\,\pi })/2$, and what you're saying is that a rational approximation of $a$ is $9/5$. In fact, we have a continued fraction $$ a = 1 + \dfrac{1}{1 + \dfrac{1}{3+ \dfrac{1}{1+\dfrac{1}{1139 + \ldots}}}}$$ and $1+1/(1+1/(3+1/1)) = 9/5$. The fact that the first omitted element, $1139$, is so large makes this a very good approximation: the error in approximating $a$ by $9/5$ is only about $3.5 \times 10^{-5}$. Four elements later comes $7574$, so an even better approximation is $1+1/(1+1/(3+1/(1+1/(1139+1/(1+1/(15+1/1)))))) = 174530/96963$ with error about $1.4 \times 10^{-14}$.

EDIT: Perhaps even more remarkable are $$ \eqalign{\pi - \sqrt{1 + \dfrac{47}{35} \pi} &\approx \dfrac{6}{7}\cr \pi - \sqrt{\dfrac{3}{5} + \dfrac{5}{2} \pi } &\approx \dfrac{216}{923}\cr}$$

corresponding to the continued fractions

$$ \eqalign{\pi - \sqrt{1 + \dfrac{47}{35} \pi} &= \dfrac{1}{1+ \dfrac{1}{6 + \dfrac{1}{126402+ \ldots}}}\cr \pi - \sqrt{\dfrac{3}{5} + \dfrac{5}{2} \pi} &= \dfrac{1}{4+\dfrac{1}{3+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{1+\dfrac{1}{19+\dfrac{1}{133286+\ldots}}}}}}}\cr}$$

$\endgroup$
23
$\begingroup$

Ramanujan found this approximation, among many others, according to Wolfram MathWorld equation 21 in linked page.

$\endgroup$
1
  • $\begingroup$ Perfect! Exactly what I was looking for. $\endgroup$
    – John Alexiou
    Commented May 18, 2012 at 21:29
2
$\begingroup$

This is not a complete answer, but it may be useful.

The largest root of the simple polynomial $$x^2-3x+1$$

is $$\Phi^2=\frac{3+\sqrt{5}}{2}=\left(\frac{1+\sqrt{5}}{2}\right)^2=\Phi+1$$

Modifying the coefficients of the polynomial using $5$ and $6$ it becomes

$$5^2x^2-5\times6\times 3 x+6^2$$

and its largest root is this approximation to $\pi$.

$$\pi\approx \frac{6}{5}\left(\frac{3}{2}+\frac{\sqrt{5}}{2}\right)=\frac{9}{5}+\sqrt{\frac{9}{5}}$$

This procedure seems related to the one for another approximation by Ramanujan.

$\endgroup$
2
  • 1
    $\begingroup$ Where did the negative sign of $-3x$ go? $\endgroup$
    – John Alexiou
    Commented Apr 1, 2016 at 15:32
  • $\begingroup$ @ja72 Corrected, thank you! $\endgroup$
    – Jaume Oliver Lafont
    Commented Apr 1, 2016 at 15:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .