[VA] Short & Sweet Math Challenge #25 "San Valentin's Special: Weird Math"

[C.Ret]

(02-19-2021 11:32 PM)Valentin Albillo Wrote:        
Hi, all:

Concoction the Fourth: Weird Graph

      This is mostly unaddressed. In particular no program or description of the operations required to produce the graph have been posted so far and worse, no one has produced and posted the graph itself, i.e.: an actual image of it. This is the kind of functionality for which graphics calculators are intended and I included this part specifically to give HP Prime's or RPL-models' users some opportunity to show off their calculator's worthiness for this challenge.

      Also, no attempts to factorize the polynomial have been reported, either successful or unsuccessful. This is the kind of functionality for which CAD is intended. Why don't you give it a try ? Hint: It would help to answer the main question.

      Finally, as for the main question proper, What's so weird about this graph?, it's still left unanswered. Apart from its so-described "funny" appearance, there's more to the graph than it seems at first sight. Some sleuthing would surely help.

Thanks and best regards.  
V.

I spent so much time on these wierd graph that my valentine leave me and she is actualy going out with a former best friend of mine.

The good thing is that I am now full time free of wasting plenty of time on my followed and trusty HP Color Graphing Calculator.

Another good fact is that this polynomial is perfectly even for both arguments x and y this simply greatly the investigation:

\( Po(x,y)=Po(-x,y)=Po(x,-y)=Po(-x,-y) \)

One may investigate it out in the first graph quadrant aka x>=0 and y>=0 only since other quadrants are exact symmetric reflections.

   

I was unable to directly factorize this wierd Polynomial. But by factorizing specific sub-expressions I get information about where to look for roots on specific axis:

\( Po(x,0)=9x^8-100x^6+182x^4-100x^2+9=(x-3)(x-1)^2(x+1)^2(x+3)(3x-1)(3x+1) \)

This clearly indicate that roots on the Ox axis are found at abscissas -3 -1 -1/3 +1 +1/3 +3.
But the squared factors trigger me, I was under alert, there is double root there. Nodes points. Are same time hard to trace on graphics !

\( Po(0,y)=9y^8-4y^6-10y^4-4y^2+9=(y-1)^2(y+1)^2(3y^2-2y+3)(3y^2+2y+3) \)

Fortunately, \( 3y^2 ? 2y+3 \) have no real roots , only pure complex zeros. So the only expected zero on the Oy axis are at -1 and +1. Again, double roots alert there .

I was on the way to investigate other specify factorization of \( Po(-3,y) \,Po(-1,y)\,Po(-\frac{1}{3},y)\,\cdots \) but I realize how powerful is a Graphing Calculator and his Advanced Graphics Application:

   

The position \(Po(x,y)=0 \) are marked by black traces. The graphic windows is from (-3.2,2.4) to (3.2,-2.4). Ticks are 1 unit for both x and y axis. Square zoom 9.
The red color indicate positive values of\( Po(x,y)\), the darker the highest.
The blue color indicate negative values of \( Po(x,y) \), the darker the deepest.

The Advanced Graphic App makes a decent job.
But I spent a few hours developing my own way of doing it:


   


The weirdest with this polynomial is to catch the two zeros at (-1,0) and (+1,0) since the polynomial reach zeros there but without any change of sign , all surrounding values stay negative except at the exact zeros points.



Thank a lot for this amazing pulzze. I really spend a good time playing on it !