**3. On Numerical Methods and Rotation–Translation Equation**

*3.1. On Numerical Methods for Solving Nonlinear Equations*

Although the Newton's iterative method

$$z\_{k+1} = z\_k - \frac{f(z\_k)}{f'(z\_k)}, \ k = 0, 1, 2, \dots \tag{2}$$

is one of the most popular and commonly used methods, *numerical analysis* offers many iterative methods that can be used in the stage of solution of Equation (1). In general to calculate the roots of nonlinear equations (of the type (1)), we have to use approximate (iterative) methods. When studying an iterative method, two of the most important aspects to consider are:


Most of the known iterative algorithms for solving nonlinear equations are only locally convergent, i.e., before using such a method, we need to locate the unknown root at a sufficiently small interval. Even if the root sought is located at the appropriate interval, if we do not choose the initial approximation in a proper way, the process may not be convergent. Usually, iterative methods of this type require the following convergence conditions:


Then, the corresponding iterative process converges to the sought root for an initial approximation *z*<sup>0</sup> which is the end of the interval [*a*, *b*], where *f*(*z*0)*f* (*z*0) > 0 (or *f*(*z*0)*f* (*z*0) < 0).

For some examples of more computationally efficient and higher order iterative methods, we refer the reader to [28].

In the encryption algorithm that we will introduce later, we will use the following iterative function

$$z\_{k+1} = z\_k - \frac{h(z\_k)}{2} \left( \frac{3f'(u\_k) + f'(z\_k)}{3f'(u\_k) - f'(z\_k)} \right), \ k = 0, 1, 2, \dots \tag{3}$$

where *<sup>h</sup>*(*zk*) = *<sup>f</sup>*(*zk* ) *<sup>f</sup>* (*zk* ) and *uk* <sup>=</sup> *zk* <sup>−</sup> <sup>3</sup> <sup>2</sup> *h*(*zk*). This iterative algorithm is explored by Jarrat in [29], and it is known as *Jarrat's method* (see also [30]).

The reason we prefer iteration method (3) over method (2) is its faster convergence. The order of convergence of Jarrat's method is four, while the one of Newton's method is only two (see [30]). In addition, method (3) has higher computational efficiency, although at each step of the iteration one value of *f* and two values of *f* are calculated (while in the Newton's method, one value of *f* and one value of *f* are calculated). Thus, if the function *f* is a polynomial, then calculating the value of the function *f* is always more complex than calculating its derivative *f* .

## *3.2. Base of Rotation–Translation Equation*

In order to avoid the vulnerability to statistical attack, we include additional randomness by using the following space contraction formula based on rotation–translation equation of the form [31]

$$\begin{aligned} x\_{k+1} &= a + b(x\_k \cos \theta\_k - y\_k \sin \theta\_k), \\ y\_{k+1} &= b(x\_k \sin \theta\_k + y\_k \cos \theta\_k), \end{aligned} \tag{4}$$

where the angle of rotation is

$$
\theta\_k = c + \frac{d}{x\_k^2 + y\_k^2}.\tag{5}
$$

The translation value is *a* = 6, the space contraction value is *b* = 0.8 < 1, and rotation values are *c* = *a*/2 and *d* = *a*. The rotation–translation Equation (4) with initial conditions *x*<sup>0</sup> = 0.233, *y*<sup>0</sup> = −0.67 is presented in Figure 1.

**Figure 1.** Space contraction.
