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Approximating $\frac{\pi}{2}$ from above

Since $$\left(\frac{\pi}{2}\right)^9\approx 58.220897$$ the root $$58^\frac{1}{9}\approx 1.5701$$ is not far from $$\frac{\pi}{2}\approx 1.570796$$

This approximation may be improved by noticing that $$0.22\approx \frac{2}{9} = 0.22...\,$$

so $$\frac{\pi}{2} \approx \left(58+\frac{2}{9}\right)^\frac{1}{9}\approx 1.570800$$

Approximating $\frac{\pi}{2}$ from below

A similar approximation is given by $$\left(1+\frac{2}{9}\right)^\frac{9}{4} \approx 1.5707$$ which may be extended to

$$\frac{\pi}{2} \approx \left(1+\frac{2}{9}+\frac{1}{2^3·5^5}\right)^\frac{9}{4} \approx 1.5707963$$ to yield seven correct decimals.

Alternatively,

$$\frac{\pi}{2} \approx \left(58+\frac{2}{9}\right)^\frac{1}{9}-\frac{1}{2^4·5^6}\left(1-\frac{13}{3·5^4}\right) \approx 1.57079632679437$$

has twelve correct decimals.

Inequalities for $\frac{\pi}{2}$ and $\log(2)$

Combining both approximations we may write $$\left(58+\frac{2}{9}\right)^\frac{1}{9}-\frac{1}{2^4·5^6}\left(1-\frac{13}{3·5^4}\right)<\frac{\pi}{2}<\left(58+\frac{2}{9}\right)^\frac{1}{9}$$

following the pattern of the double inequality for $\log(2)$ $$\left(\frac{2}{5}\right)^\frac{2}{5}<\log(2)<\left(\frac{1}{3}\right)^\frac{1}{3}$$

taken from a comment to this question.

Q Is this 58 related to the almost-integer $e^{\sqrt{58}\pi} \approx24591257751.99999982$ ?

(see equation (68) here)

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  • $\begingroup$ I do not mean to be rude or anything, but why would anyone be interested in these ad hoc approximations? Like if you computed that $$\left(\frac{\pi}{2}\right)^9\approx 58.220897$$ then isn't it obvious that $58^\frac{1}{9}$ is gonna be an under approximation for $\frac{\pi}2$? $\endgroup$ – BigbearZzz Mar 13 '16 at 16:17
  • $\begingroup$ @BigbearZzz It is not so obvious that is that $58+\frac{2}{9}$ is a convergent of $\left(\frac{\pi}{2}\right)^9$, is it? Maybe I should add that to the question. Would you consider $$\pi\approx\left(97+\frac{2}{5}+\frac{1}{110}\right)^\frac{1}{4}$$ an ad hoc approximation? $\endgroup$ – Jaume Oliver Lafont Mar 13 '16 at 17:39
  • $\begingroup$ I still don't understand. How did you get $$\left(\frac{\pi}{2}\right)^9\approx 58.220897$$ in the first place? $\endgroup$ – BigbearZzz Mar 13 '16 at 17:40
  • $\begingroup$ It has no hidden steps, it is just an evaluation: wolframalpha.com/input/?i=(Pi%2F2)%5E9 I am not sure to understand your question. $\endgroup$ – Jaume Oliver Lafont Mar 13 '16 at 17:46
  • $\begingroup$ Of course, it is not obvious a priori that $58.222...$ is gonna be an approximation of $\left(\frac{\pi}{2}\right)^9$, but once you have computed $\left(\frac{\pi}{2}\right)^9$ using program, it's obvious since $$58.222...\approx 58.220897$$. $\endgroup$ – BigbearZzz Mar 13 '16 at 17:51

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