## Some Nārāyaṇa-like convergents and their geometric and trigonometric connections

While playing with an iterative geometric construction in our youth we discovered for ourselves a particular right triangle whose sides are in the proportion $1: \sqrt{\phi} : \phi$, where $\phi= \tfrac{1+\sqrt{5}}{2}$ is the Golden Ratio. This triangle is of course famous as being the basis of the largest triangles of the Śrīyantra and the Tripuraśekhara- (Bālā-) yantra of the Śrīkula tradition and also the famous Great Pyramid of the ancient Egyptians. Revisiting it, we were able to provide the trigonometric proof for our construction, which in turn led several sequences of interest. The construction goes thus (Figure 1):

Figure 1

1) Start with an isosceles right triangle $\triangle AOB$ or $\triangle A'OB$, whose equal sides are 1 each and hypotenuse is $\sqrt{2}$.
2) Slide the hypotenuse down to point $O$ create a segment equal to it in length $\overline{OC_1}$. Join point $C_1$ to $B$.
3) Draw a perpendicular to $\overline{C_1B}$ to meet the base line $\overleftrightarrow{AC_1}$ to meet it at $C_1'$. Joint $C_1'$ to $B$ to create a right triangle $\triangle C_1'OB$.
4) Draw a segment equal to the hypotenuse $\overline{C_1'B}$ of this new triangle along the baseline $\overleftrightarrow{AC_1}$ to get a point at the distance equal to $\overline{C_1'B}$ from $O$.
5) Iterate the above steps with this point.

What will happen if you keep iterating thus? We learned to our surprise in our youth that it converges to a right triangle $\triangle K_1BK_2$ (shaded yellow in Figure 1), whose sides will be $(\sqrt{\phi}, \phi, \phi^{3/2})$. Further the altitude $\overline{OB}$ will divide the hypotenuse into segments $\overline{OK_1}=\sqrt{\tfrac{1}{\phi}}$ and $\overline{OK_1}=\sqrt{\phi}$.

The trigonometric proof we obtained for this is rather simple. It essential comprises of finding the attractor of the following map:

$x_{n+1}=\cos\left(\arctan(x_n)\right)$; where $x_0=1$

$\cos\left(\arctan(x)\right) = \dfrac{1}{\sqrt{1+x^2}}$

$\therefore x_1=\sqrt{\dfrac{1}{2}}; \; x_2= \sqrt{\dfrac{2}{3}}; \; x_3= \sqrt{\dfrac{3}{5}}; \; x_4=\sqrt{\dfrac{5}{8}}...$

Given the mātrā-meru-paṅkti: $f[1]=1, f[2]=2, f[n]=f[n-1]+f[n-2]; x_n=\sqrt{\dfrac{f[n]}{f[n+1]}}$

$\displaystyle \therefore \lim_{n \to \infty} x_n= \sqrt{\dfrac{1}{\phi}}$

One can see that the $x_n$ of this map corresponds to the bases of the right-side triangles in successive steps (Figure 1), starting with $\overline{OA}, \overline{OC_1'}...\overline{OK_1}$. Thus, $\overline{OK_1}=\tfrac{1}{\sqrt{\phi}} \;\;\; _{...\blacksquare}$

Now, we specify a similar map of the form:

$y_{n+1}=\sin\left(\arctan(y_n)\right)$; where $y_1=1$

Given that $\sin\left(\arctan(y)\right) =\tfrac{y}{\sqrt{1+y^2}}$, we get the sequence:

$y_2=\dfrac{1}{\sqrt{2}}; \; y_3=\dfrac{1}{\sqrt{3}}; \; y_4=\dfrac{1}{\sqrt{4}}... \; y_n=\dfrac{1}{\sqrt{n}}$

Thus, $\displaystyle \lim_{n \to \infty} x_n=0$

Inspired by the above, we define the below intertwined maps that generate interesting sequences and convergents:

Map 1
$x_{n+1}=\cos\left(\arctan\left( x_n\right) \right); \; x_0=1$

$y_{n+1}=\sin\left(\arctan\left( x_{n+1}\right)\right)$

Here, as $n \to \infty\;\; x_{n} \to \dfrac{1}{\sqrt{\phi}}, \; y_{n} \to \dfrac{1}{\phi}$

We see that $x_n \rightarrow \sqrt{\dfrac{1}{2}}; \sqrt{\dfrac{2}{3}}; \sqrt{\dfrac{3}{5}}; \sqrt{\dfrac{5}{8}}...$ and $y_n \rightarrow \sqrt{\dfrac{1}{3}}; \sqrt{\dfrac{2}{5}}; \sqrt{\dfrac{3}{8}}; \sqrt{\dfrac{5}{13}}...$

It is apparent that these sequences are constructed from consecutive or every second term of the mātrā-meru-paṅkti.

Map 2
$x_{n+1}=\cos\left(\arctan\left( y_n\right) \right); \; y_0=1$

$y_{n+1}=\sin\left(\arctan\left( x_{n+1}\right)\right)$

Here, as $n \to \infty\;\; x_{n} \to \dfrac{1}{\sqrt[4]{2}}, \; y_{n} \to \dfrac{1}{\sqrt{1+\sqrt{2}}}$

We see that $x_n \rightarrow \sqrt{\dfrac{1}{2}}, \sqrt{\dfrac{3}{4}}, \sqrt{\dfrac{7}{10}}, \sqrt{\dfrac{17}{24}}...$

This sequence can be derived from the sequence $\textbf{1, 2, 3, 4, 7, 10, 17, 24...}$ which has $f[1]=1, f[2]=2$ and alternating terms of the form $f[n+1]=f[n]+f[n-1], f[n+1]=f[n]+f[n-2]$

Further, we have $y_n \rightarrow \sqrt{\dfrac{1}{3}}, \sqrt{\dfrac{3}{7}}, \sqrt{\dfrac{7}{17}}....$

One notices that it is constructed from the above integer sequence by taking the square root of the ratio of ever second term.

Map 3
$x_{n+1}=\cos\left(\arctan\left(y_n\right) \right); \; y_0=1$

$y_{n+1}=\sin\left(\arctan\left(\frac{x_{n+1}}{y_n}\right)\right)$

Here, as $n \to \infty\;\; x_{n} \to \dfrac{1}{\sqrt[6]{2\tau}}, \; y_{n} \to \dfrac{1}{\sqrt{\tau}}$

where $\tau$ is the convergent of Nārāyaṇa’s sāmāsika-paṅkti (known in the Occident as the tribonacci constant)

We see that $x_n \rightarrow 1, \sqrt{\dfrac{1}{2}}, \sqrt{\dfrac{3}{4}}, \sqrt{\dfrac{13}{22}}, \sqrt{\dfrac{367}{536}}, \sqrt{\dfrac{225273}{359962}}...$

This sequence defines an oscillatory convergence to $\tfrac{1}{\sqrt[6]{2\tau}}$ and can be derived from integer sequence: $\textbf{1, 1, 1, 2, 3, 4, 13, 22, 367, 536, 225273, 359962...}$. Here, $f[1]= 1, f[2]=1, f[3]=1$ and every alternate term is defined by the formulae:

$f[n]=f[n-1] + f[n-3]^2$

$f[n]=f[n-1] \cdot f[n-2]+f[n-4]^4$

Further, we see that $y_n \rightarrow \dfrac{1}{\sqrt{3}}, \dfrac{3}{\sqrt{13}}, \dfrac{13}{\sqrt{367}}...$

This sequence can again be derived from the above integer sequence by taking every other term, 1, 3, 13, 367 and results in an oscillatory convergence to $\tfrac{1}{\sqrt{\tau}}$

Map 4
$x_{n+1}=\cos\left(\arctan\left(x_n^2\right) \right); \; x_0=1$

$y_{n+1}=\sin\left(\arctan\left(x_{n+1}^2\right)\right)$

To understand the convergent of this map consider the unit circle $x^2+y^2=1$ and the cubic parabola $y=x^3$ (Figure 2).

Figure 2

Here, $\lambda_1$ is the positive intersect of the two curves with coordinates $(\lambda_x, \lambda_y)$. Further, $r=\tfrac{\lambda_y}{\lambda_x}$ is the only real root of the cubic equation $x^3+x-1=0$. Thus, $\lambda_x=\cos(\arctan(r)), \lambda_y=\sin(\arctan(r))$

Then, as $n \to \infty\;\; x_{n} \to \lambda_x, \; y_{n} \to \lambda_y$

We see that $x_n \rightarrow 1, \dfrac{1}{\sqrt{2}}, \dfrac{2}{\sqrt{5}}, \dfrac{5}{\sqrt{41}}, \dfrac{41}{\sqrt{2306}}...$

and $y_n \rightarrow \dfrac{1}{\sqrt{5}}, \dfrac{4}{\sqrt{41}}, \dfrac{25}{\sqrt{2306}}, \dfrac{1681}{\sqrt{8143397}}...$

These sequences are related to the integer sequence: $\textbf{1, 2, 5, 41, 2306, 8143397...}$ which can be described by the formula: $f[n]=f[n-1]^2+f[n-2]^4$ with $f[1]=1, f[2]=2$. Then $x_n$ is derived from successive terms with the square root of the large term. $y_n$ is derived similarly from every other term with the square of the first and the square root of the next.

Map 5

$x_{n+1}=\cos\left(\arctan\left( \frac{y_n}{x_n}\right) \right); \; x_0=1, y_0=1$

$y_{n+1}=\sin\left(\arctan\left( \frac{x_{n+1}}{y_n}\right)\right); \; y_0=1$

Here, as $n \to \infty\;\; x_{n}, \; y_{n} \to \dfrac{1}{\sqrt{2}}$

The $x_n$ is interesting in that it begins with $\tfrac{1}{\sqrt{2}}$ and converges back to it in an oscillatory fashion. This convergence takes the form:

$x_n \rightarrow \dfrac{1}{\sqrt{2}}, \dfrac{2}{\sqrt{7}}, \dfrac{7}{\sqrt{101}}, \dfrac{101}{\sqrt{19609}}, \dfrac{19609}{\sqrt{782362082}}...$

The convergence of $y_n$ to $\tfrac{1}{\sqrt{2}}$ takes the form:

$y_n \rightarrow 1, \sqrt{\dfrac{2}{3}}, \sqrt{\dfrac{7}{13}}, \sqrt{\dfrac{101}{192}}, \sqrt{\dfrac{19609}{39001}}...$

These sequences can be derived from the integer sequence $\textbf{1, 1, 2 ,3, 7, 13, 101, 192, 19609, 39001...}$. Given $f[1]=1, f[2]=1, f[3]=2$, the subsequent alternate terms of this sequence are defined the formulae

$f[n]=f[n-1]+f[n-2] \cdot f[n-3]$

$f[n]=f[n-2]^2+f[n-1] \cdot f[n-4]^2$

More generally the convergents obtained for any of the above maps will be reached from any other starting $(x_0,y_0)$ but using $x_0=y_0=1$ gives us a convergence via easy to understand explicit sequences.

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