-2

Approximate value of the cotangent function we can find by the following formula, where $x$ represents the value of the semicircle minor arc angle

$ \sqrt [90]{\frac{{\pi}}{4\sin\left(\frac{{\pi}}{4}\right)}}^\left(2x-90\right)\cdot \left(\frac{90}{x}-1\right)\cdot \sqrt{2}+1$

For example if $x$= 60 we will get 1.73229..or for example $x$=30 we will get 3.7311 There is formula similare to mine, in literature

Srbin
  • 29
  • Are you asking "Is there a formula ... ?"? – Blue Mar 26 '21 at 12:08
  • Why write $4\sin(\pi/4)$ when you could write $2\sqrt{2}$? If nothing else, mixing radians there with degrees for $x$ is a little confusing. ... Come to think of it, I suggested further simplification in a comment to your recent question, so I guess I have nothing more to offer. :) – Blue Mar 26 '21 at 12:35
  • What is the question ? –  Mar 26 '21 at 13:26
  • I guess that your formula is wrong by a factor $2$, $x$ is not in degrees but in half degrees. –  Mar 26 '21 at 13:45

1 Answers1

0

More than $\large\color{red}{1,400}$ years ago were made the following approximations $$\sin(x) \simeq \frac{16 (\pi -x) x}{5 \pi ^2-4 (\pi -x) x}\qquad (0\leq x\leq\pi)$$ $$\cos(x) \simeq\frac{\pi ^2-4x^2}{\pi ^2+x^2}\qquad (-\frac \pi 2 \leq x\leq\frac \pi 2)$$ So, for $0 \lt x \leq \frac \pi 2$, it could have been possible to write $$\cot(x) \sim \frac{\left(5 \pi ^2-4 (\pi -x) x\right) \left(\pi ^2-4 x^2\right)}{16 (\pi -x) x \left(x^2+\pi ^2\right)}$$ Making $x=\frac \pi n$, this would give the simple $$\cot \left(\frac{\pi }{n}\right)\sim \frac{\left(n^2-4\right) \left(5 n^2-4 n+4\right)}{16 (n-1) \left(n^2+1\right)}$$

By the way, your values are not correct.

Why don't you try my simpler $$\cot(x)\sim \frac{15 \left(x^4-28 x^2+63\right)}{x \left(x^4-105 x^2+945\right)}$$

Claude Leibovici
  • 260,315
  • Not counting that raising to a fractional exponent can be as difficult as computing the cotangent itself ! –  Mar 26 '21 at 13:27
  • 1
    @YvesDaoust. For sure ! And I don't like at all the mixture of degrees and radians. – Claude Leibovici Mar 26 '21 at 13:29
  • I see that your formula matches very well two periods of the cotangent. There should be a similar one that matches a single period wit even better accuracy, or on the opposite, simpler with the same accuracy. –  Mar 26 '21 at 13:50
  • @YvesDaoust. There are so many things we can do for approximation. By the way, are you familiar with two points Taylor series expansions ? Cheers :-) – Claude Leibovici Mar 26 '21 at 13:55
  • Never heard of :) If I understand, they are polynomials such that their values and values of derivatives match a given function at two points rather than one ? –  Mar 26 '21 at 14:00
  • @YvesDaous. Have a look at https://mathematica.stackexchange.com/questions/171090/about-two-point-taylor-series-expansion . – Claude Leibovici Mar 26 '21 at 14:05
  • That's what I did ;-) I am familiar with Hermite interpolation (I am precisely working on cubic splines at the moment), but not with this generalization. –  Mar 26 '21 at 14:06