This is in continuation to my earlier post at : How many times tangent to a cubic curve $y = x^3$ from a point on it, meets again at another point.
A brief summary of the previous post :
If take a suitable start point as tangent manually to the curve; then a question arose : will the tangent from new point will again cut the curve. If yes, then how many times this will occur. The selected answer by @Siong Thye Goh, showed that there will be a geometric ratio between such points, with accompanying python code & graphs stated below by him to substantiate infinite sequence of such points.
Also, there are still lingering doubts: I wanted some proof in order to show an infinite sequence of such points, and there came two options that could be could to prove :
(i) convergence of slopes, i.e. decreasing slopes of successive tangents; else
(ii) to prove a bound on the slopes, that is never violated.
In order to prove it tried to see the first $50$ tangents' slopes, but the slopes are approaching $90^0$ very fast, and all are positive (i.e., the ones made from above to below, & vice-versa), while the points on the cubic curve alternated in each turn from far-left-bottom to far-right-top. Anyway, the alternation of sides and points going still further was a direct result of the ratio $-2$ chosen for the G.P. of such points (as in answer by @Siong Thye Goh).
Could not ascertain why the angles of slopes are nearly $90^0$, & hence could not use the idea stated at (i) or (ii) to prove infinitude of such points; but have stated the reason for angles being all positive in this post, as shown below.
Associated code (by @Siong Thye Goh) is at : http://py3.codeskulptor.org/#user301_3iiEyt0dXlou231_10.py
http://py3.codeskulptor.org/#user301_QKlzKmDtjSmgRMn.py
Associated graph (by @Siong Thye Goh) is at : https://www.desmos.com/calculator/j0frxb5gwt
Request vetting of the below statement :
The problem's (i.e., tangent from any point on cubic to another point on cubic has positive slope) reason seems that $tan(270^0+\theta)= \tan(\theta)$, & $\tan(90^0-\theta) = \tan(\theta)$ and any angle (of tangent) made from 3rd quad. to first one is having slope of $\tan(90^0 -\theta)$, and any angle (of tangent) made from 1st quad. to third one is having slope of $\tan(270^0 +\theta)$.
*Update * Need specify the angle from upper half(1st quad.) to lower half (3rd quad.) of cubic curve, as lying in the range of angle of slope $\theta' = 180^0+\theta$ to $270^0 -\theta$. This leads to positive $\tan(\theta')$ values; as $\tan(180^0+\theta)=\tan(\theta)$, so getting answer of such slopes as $\approx 90^0$, as $270^0=180^0+90^0$.
If $\theta$ is an acute angle, then $270+\theta$ is in the $4$-th quadrant and has negative tangent and cotangent value.
$\tan(90-\theta)=\cot(\theta)$
The slope of $x^3$ is $3x^2$, in fact, at any non-zero point, the slope is positive.
The sequence follows a geometric sequence $(-2)^na$ where $a$ is the initial point.
The slope of the sequence is $3\cdot 4^n\cdot a^2$, this sequence is positive and increasing if the first point $a\ne 0$ and tends to $\infty$.
To see that it is increasing, the graph of $4^x$ might help.
$\lim_{n \to \infty}\arctan(3\cdot 4^n\cdot a^2)=\lim_{x \to \infty} \arctan(x)=\frac{\pi}{2}.$
This is a graph of $\arctan{x}$.
$\frac{\pi}2 radians = 90^\circ$.
If you start with $a>0$, then the slope is $3a^2$. Hence whenever you move to the left by $1$ unit, the tangent value dropped by $3a^2$. Whenever you move to the left by $a$ units, the value dropped by $3a^3$ units. How many times do I have to move by $a$ unit step size such that the tangent meet the curve again?
$$a^3-3ka^3=(a-ka)^3$$
$$1-3k=(1-k)^3$$
$$3k^2-k^3=0$$
$$k^2(3-k)=0$$
It is independent of $a$ as long as they are positive, hence to understand the problem, just focus on small number, say $a=1$.