# Approximation for $\pi$

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?

-

– Kirthi Raman May 18 '12 at 21:01
@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. – ja72 May 18 '12 at 21:03
What are you expecting, validation of this approximation and some sort of variation of this approximation? – Kirthi Raman May 18 '12 at 21:16
I look for someone to say, I have seen this and it was used by x, or this is related to y approximation. – ja72 May 18 '12 at 21:21

## 2 Answers

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

-

Perfect! Exactly what I was looking for. – ja72 May 18 '12 at 21:29

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

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

-
{"extraction_info": {"found_math": true, "script_math_tex": 0, "script_math_asciimath": 0, "math_annotations": 0, "math_alttext": 0, "mathml": 0, "mathjax_tag": 0, "mathjax_inline_tex": 17, "mathjax_display_tex": 6, "mathjax_asciimath": 0, "img_math": 0, "codecogs_latex": 0, "wp_latex": 0, "mimetex.cgi": 0, "/images/math/codecogs": 0, "mathtex.cgi": 0}, "config": {"markdown_headings": true, "markdown_code": true, "boilerplate_config": {"ratio_threshold": 0.18, "absolute_threshold": 10, "end_threshold": 15, "enable": true}, "remove_buttons": true, "remove_image_figures": true, "remove_link_clusters": true, "table_config": {"min_rows": 2, "min_cols": 3, "format": "github"}, "remove_chinese": true, "remove_edit_buttons": true, "extract_latex": true}, "english_confidence": 0.9707184433937073, "perplexity_flag": "head"}