1

I am studying trigonometry and I know that $\pi$ can be approximated using Gregory series, Rutherford series, etc. Also it's strange and mysterious that $\pi$ is just ratio of circumference to diameter. This profoundly shows integration in math between different branches.

I would like to know other ways of $\pi$ approximation in different fields of math. Please share your views.

Jam
  • 10,325
godonichia
  • 2,235

4 Answers4

0

$\pi/2$ is the least positive solution of $0=f(x)=(\sin x) -1.$ Since $f'(x)>0 >f(x) $ for $x\in (0,\pi/2)$ we can use Newton's method to approximate $\pi/2:$ Take any $x_1\in (0,\pi/2)$ and let $$x_{n+1}=x_n-f(x_n)/f'(x_n). $$ We can use the angle-sum formulas together with the power series for $\sin$ and $\cos, $ to more easily compute $\sin x_{n+1}$ and $\cos x_{n+1}$ from $\sin x_n$ and $\cos x_n.$ Of course $x_n\to \pi/2$ as $n\to \infty.$

Since $f'(\pi/2)=0,$ a faster convergence is obtained if we let $x_{n+1}=x_n-2f(x_n)/f'(x_n).$

There are many variations on this theme.

DanielWainfleet
  • 57,985
  • Why not use $f(x)=\cos(x)$ and say, $x_0=1$. Then $x_{n+1}=x_n+\cot(x_n)$. To floating point precision I have full convergence in 4 steps. (Also, this is a two year old question.) – Ian Jul 15 '16 at 18:31
  • (Three steps with $x_0=1.5$, in fact.) – Ian Jul 15 '16 at 18:43
0

The roots of $\frac{\sin{x}}{x}$ are the non-zero roots of its numerator $\sin{x}$: $\pm\pi$, $\pm 2\pi$, $\pm 3\pi$... and so on.

Using these roots, Euler factored the series for $\frac{\sin{x}}{x}$ as the infinite product: $$\frac{\sin{x}}{x}=\prod_{n=1}^\infty \left(1-\frac{x^2}{n^2\pi^2}\right)$$ If you use Euler's formula and set $x=\frac{\pi}{2}$, you can easily get $$\pi=\frac{2}{\prod_{n=1}^\infty \left(1-\frac{1}{4 n^2}\right)}.$$

alans
  • 6,475
  • This is pretty slow, unfortunately. With a product of $1000$ terms, you get just three correct digits. It's just slightly better than Leibniz's formula. – Ian Jul 15 '16 at 18:39
  • 1
    Actually, here's an interesting little problem (though it might be meaningless): that formula gives $\ln(\pi)=\ln(2)-\sum_{n=1}^\infty \ln(1-1/4n^2)$. One might approximate the latter sum by $\sum_{n=1}^\infty -\frac{1}{4n^2}=-\frac{\pi^2}{24}$, just by Taylor expansion. Doing that makes the sum itself bigger, so it makes the whole expression smaller. So $\ln(\pi)=\ln(2)+\delta+\frac{\pi^2}{24}$. What's $\delta$? Perhaps something to do with Euler-Mascheroni $\gamma$? – Ian Jul 15 '16 at 18:53
0

The half circumference of the unit circle can be computed from the implicit equation $x^2+y^2=1$, which expresses a constant distance to the origin.

$$H=\int_{-1}^1\sqrt{y'^2(x)+1}\ dx=\int_{-1}^1\frac{dx}{\sqrt{1-x^2}}=\arcsin x\Big|_{-1}^1=\pi.$$

Similarly, for the area. $$A=2\int_{-1}^1y\ dx=2\int_{-1}^1\sqrt{1-x^2}\ dx=\left(x\sqrt{1-x^2}-\arcsin x\right)\Big|_{-1}^1=\pi.$$ When you accept the analytic formula for the Euclidean distance, you make a connection between geometry and calculus. And from there the link with trigonometric functions and the power series that allow to compute the value of $\pi$.

By the way, numerical estimates of these integrals provide approximations for $\pi$.

0

The number of reduced fractions $p/q$ with $1\le p\le q\le n$ is approximately $(3/\pi^2)n^2$ --- approximately, in the sense of asymptotically equal (the ratio of the two quantities approaches 1 as $n\to\infty$). So $\pi$ can be approximated by counting the number of reduced fractions $p/q$ with $1\le p\le q\le n$, dividing by $n^2$, taking the reciprocal, multiplying by 3, and taking the square root. For example, for $n=7$, the reduced fractions are $1/1,1/2,1/3,2/3,1/4,3/4,1/5,2/5,3/5,4/5,1/6,5/6,1/7,2/7,3/7,4/7,5/7,6/7$, 18 of them; what approximation does that give you for $\pi$?

Also, the number of squarefree naturals up to $n$ (meaning, naturals not divisible by the square of any prime) is asymptotically $(6/\pi^2)n^2$. So $\pi$ can be approximated by counting the number of squarefree naturals up to $n$, dividing by $n^2$, taking the reciprocal, multiplying by 6, and taking the square root. For example, for $n=30$, the squarefree numbers are $1,2,3,5,6,7,10,11,13,14,15,17,19,21,22,23,26,29,30$, 19 of them. What approximation do you get for $\pi$?

Gerry Myerson
  • 179,216
  • Also in number theory, let $x$ and $y$ be chosen independently and randomly from $ {1,...,n} $ with each $j$ from $1$ to $n$ equally likely in each choice. Let $P_n$ be the chance that $x,y$ are co-prime. Then $\lim_{n\to \infty}P_n=6/\pi^2.$ – DanielWainfleet Jul 15 '16 at 17:02
  • 1
    @user, that's pretty much the same thing as the reduced fractions $p/q$ from the first sentence of the answer. – Gerry Myerson Jul 16 '16 at 06:00