How to compute the cartesian coordinates of each vertices of a Spiral of Theodorus?

Question

I'd like to plot integers from 0 to N on a spiral and I have the intuition the easiest way to do this would be to use the Spiral of Theodorus as an approximation.

So, for 0 i would have (0,0), for 1 (1,0), for 2 (1,1) and here begins the difficult part for me.

Can someone help me to figure out a formula or a closure so that I could easily retrieve the coordinates of each positive integer?

Answer 1

The coordinates of the $n$th point are $$\alpha=\sum_{i=1}^n \arctan\left(\frac{1}{\sqrt{i}}\right)$$ $$r=\sqrt{n+1}$$ $$(x,y)=(r\cos(\alpha),r\sin(\alpha))$$ For $n=0$ we get $(1,0)$, for $n=1$ we get $(1,1)$, etc.

Answer 2

Let us start with $$ (x_k, y_k) = r_k (\cos \phi_k, \sin \phi_k) $$ you gave $$ r_k = \sqrt{k} $$ so we need $\phi_k$. For the increase we have $$ \tan \Delta \phi_k = 1 / r_k $$ and thus \begin{align} \phi_1 &= 0 \\ \phi_{k+1} &= \phi_k + \arctan 1/r_k \end{align} Here is a construction for your example:

You can fiddle with a live version here: link

Answer 3

Approximating $\boldsymbol{\theta_n}$ Using The Euler-Maclaurin Sum Formula

I was not able to compute a closed form for $\theta_n$, but using $$ \arctan\left(\frac1{\sqrt k}\right)=\frac1{\sqrt{k}}-\frac1{3\sqrt{k}^3}+\frac1{5\sqrt{k}^5}-\frac1{7\sqrt{k}^7}+\frac1{9\sqrt{k}^9}+O\!\left(\frac1{\sqrt{k}^{11}}\right)\tag1 $$ the Euler-Maclaurin Sum Formula says $$ \begin{align} \theta_n &=\sum_{k=2}^n\arctan\left(\frac1{\sqrt k}\right)\\ &=2\sqrt{n}+C+\frac7{6\sqrt{n}}-\frac{41}{120\sqrt{n}^3}+\frac{167}{840\sqrt{n}^5}-\frac{1147}{8064\sqrt{n}^7}+O\!\left(\frac1{\sqrt{n}^9}\right)\tag2 \end{align} $$ where $$ C=-2.943181160056894531\tag3 $$ was computed by extending the formula for $\theta_n$ out to $O\!\left(\frac1{\sqrt{n}^{23}}\right)$ and evaluating at $n=100$. The next term in the formula is $\frac{1411}{12672\sqrt{n}^9}\approx\frac1{9\sqrt{n}^9}$ which is a good estimate of the error of the approximation given in $(2)$.

For $n=10$, using the approximation in $(2)$ gives an error in $\theta_n$ of approximately $3.5\times10^{-6}$ radians or $2\times10^{-4}$ degrees. For larger $n$, the error gets smaller.

Then as in the previous answers, we have $$ (x_n,y_n)=\left(\sqrt{n}\cos(\theta_n),\sqrt{n}\sin(\theta_n)\right)\tag4 $$

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Approximating The Curve That Passes Through The Points

Using the first $3$ terms of the approximation, we get $$ \theta=2r+C+\frac7{6r}\tag5 $$ and solving for $r$, we have $$ \begin{align} r &=\frac{\theta-C+\sqrt{(\theta-C)^2-\frac{28}3}}4\\ &=\frac{\theta-C}2-\frac7{6(\theta-C)}+O\!\left(\frac1{(\theta-C)^3}\right)\tag6 \end{align} $$ This means that each revolution will increase the radius by $\pi$.