An Indian mathematician, Bhaskara I, gave the following amazing approximation of the sine (I checked the graph and some values, and the approximation is truly impressive.)

$$\sin x \approx \frac{{16x\left( {\pi - x} \right)}}{{5{\pi ^2} - 4x\left( {\pi - x} \right)}}$$

for $(0,\pi)$

Here's an image. Cyan for the sine and blue for the approximation.

¿Is there any way of proving such rational approximation? ¿Is there any theory similar to Taylor's or Power Series for rational approximations?

-
Off topic: May be you could say Indian Mathematician rather than a Hindu-Mathematician which attaches a religion to Mathematics. (Note: I am Hindu and not against any religion!) – user21436 Feb 6 '12 at 18:59
@KannappanSampath: Edited accordingly :) – user9413 Feb 6 '12 at 19:05
@KannappanSampath That is product of my own ignorance. I translated from my language, which uses "Hindú" for "Indian" ("Hinduista" is "Hindu", whereas "Indio" rather means native indian, and would be used for the people in cowboy movies). – Pedro Tamaroff Feb 6 '12 at 19:06
There is the theory of Padé approximants, but I don't think this particular function is one. – Rahul Feb 6 '12 at 19:06
@Peter: Well, i thought including Bhaskara's name in the question would be better because, if some unknown person wants to search something attributed to Bhaskara he may see this question in future :) – user9413 Feb 6 '12 at 19:45

• Here is an article which is written by Shailesh Shirali. Unfotunately my university doesn't have access to it.

• Here is one more article.

-

This is very close to a Padé approximant, and in this case the formula is simple enough that it's easy to derive. Firstly, we know that $\sin(x)$ is $0$ at $x=0, x=\pi$; this suggests recasting in terms of the variable $y=x(\pi-x)$. What we're after is a first-order rational approximation $\sin(x) = f(y) = \frac{ay+b}{cy+d}$; since we know that $f(y) = 0$ at $y=0$ (i.e., as $x$ approaches $0$ or $\pi$) then the constant term in the numerator is $0$, and after dividing out the approximation takes the form $\frac{y}{a+by}$.

Now, we certainly want our approximation to give $\sin(\pi/2) = f(\pi^2/4) = 1$; this means $\displaystyle{\frac{\pi^2/4}{a+b\pi^2/4}} = 1$, or $4a+b\pi^2 = \pi^2$, or $a=\frac{1-b}{4}\pi^2$. The other relation between $a$ and $b$ presumably comes from trying to match the derivative at $0$, $\left.d(\sin(x))/dx\right|_{x=0} = 1$; the condition for this this can easily be written out in terms of $df/dy$ at $y=0$. I'll spare the arithmetic (unless someone's really curious), but the result works out to be that $a=\pi$; this would give $b=(\pi-4)/\pi$ and the overall approximant $f(y) = \frac{\pi y}{\pi^2+(\pi-4)y}$, but instead the formula uses a second approximation by setting $a=5\pi^2/16$, which (thanks to the first relation) gives a rational value of $b$ (and in fact, the 'nice' value $1/4$). This approximation is equivalent to saying that $5\pi^2/16\approx\pi$, or in other words that $\pi\approx 16/5 = 3.2$; it means a slight error in the slope of the approximation at $x=0$, but that's a fair tradeoff for the ease of calculation gained.

-
Thank you for the answer. Upvoted. I chose the other since he provided me a full article with an explanation. – Pedro Tamaroff Feb 6 '12 at 20:24
Wholly understandable! (Note that the article gives a different derivation of the approximation $5\pi^2/16$ for $a$; it satisfies the relation that $\sin(\pi/6) = 1/2$, which keeps $b$ and $a/\pi^2$ rational at all steps). – Steven Stadnicki Feb 6 '12 at 20:49
(Since that's the only other relation of the form $\sin(p\cdot\pi)=q$ for rational $p,q$ this makes quite a bit of sense.) – Steven Stadnicki Feb 6 '12 at 21:04
How did you get the article? – Pedro Tamaroff Feb 6 '12 at 21:29
Analogous to taking $\pi \approx 16/5$ to get the correct gradient at $x=0$, you can take $\pi \approx \sqrt{10}$ to get the correct second differential at $x=\pi/2$. – Peter Taylor Feb 6 '12 at 22:34

Writing $x = \pi/2 + \pi t$, the approximation becomes $\cos(\pi t) \approx \frac{1-4t^2}{1+t^2} = 1 - 5 t^2 + O(t^4)$. In fact $\cos(\pi t) = 1 - \frac{\pi^2}{2} t^2 + O(t^4)$, but $\pi^2/2 \approx 4.9348$ is not far from $5$. In terms of uniform approximation to $\cos(\pi t)$ for $t \in [-1/2, 1/2]$, $\frac{1 - 4 t^2}{1+1.0043 t^2}$ would be somewhat better.

-
Actually, putting $\cos \frac{\pi x}{2}$ gives what in the article is stated as the "Bhaskara function":$$\cos \frac{\pi x}{2} \approx 4 \frac{1-x^2}{4+x^2}$$ which is an outstoundingly good approximation of that function. Thanks! – Pedro Tamaroff Feb 10 '12 at 3:17