TI-30X Pro MathPrint - How does num-solve work?

[wb.c]

I've been messing around with the TI-30X Pro MathPrint for a while now, and I was trying to understand how num-solve works exactly. It is well understood that the initial guess has an impact on the solution found, when there is more than one solution. Initially I thought that the solver must do a check of sorts to determine which direction it will step towards to find the solution.
   
This lead me to think that the solver must always do a validation of sorts to the right of the initial guess to determine if the results are getting smaller (closer to zero or smaller than the result from the initial guess). But it seems this is not the case, and I'm curious how this solver actually works.

Take for example now f(x)=0.6sin(x)-0.2x+1.
The results are quite interesting and not exactly what I expected.
The most interesting results are highlighted.

   
   

It seems that the upper and lower bounds did not have an impact on the result, other than excluding certain results and where the Error appeared. Where the Error appears, I would have expected to get the first result greater than the lower bound, as was seen with the first tabulated outcome. In this first outcome, any value greater than 10.8303 with the bounds outside of the solutions always results in the first solution (3.62). However, the same setup, but with the lower bound higher than the first solution but lower than the second one, did not produce the second solution as the result, it still gave the third solution (7.98).

Does anyone know how the TI-30X Pro MathPrint numeric solver functions?

[rawi]

On the website education.ti.com there is under materials a text about the solver.
https://ti-unterrichtsmaterialien.net/ma...e973fa5152
But this is not very speficific and it is in German (perhaps there is as well an English version).

First it is explained how the equation sin(x)=x-1 can be solved.
Change the equation so that x is on one side: x=1+sin(x). Take a starting value for x, e.g. x(0) = 1. i=0
Loop: Compute x(i+1) = 1+sin(x(i)) : i=i+1
Until there is no change.

The text continues: The numeric solver applies a similar procedure as shown above.
But the text is not specific in what respect there is a difference to this procedure.
Sorry, but that is all I could find.

[wb.c]

I’m actually quite familiar with the documents from the German TI site. They are made by a third party group called T3. I’m working on an updated English version of that very document right now, when it’s done I will share it. In any case. The TI doesn’t quite use the technique shown in their example. As I understand, most calculators use Newton’s method for approximating solutions. I did a lot of reading about this last night, basically a crash course on numerical analysis, but I think I figured out the behavior of the solver in this particular instance. It has to do with how the Newton method uses tangent lines to move closer to the point of intersection. Newton’s method had strange behavior around local extremum, which I believe is the reason for the outcomes tabulated. I need to manually solve this equation using Newton’s method to prove this, but it makes sense now.

https://computingskillset.com/solving-eq...ns-method/

This article does a good job explaining it I think.

[wb.c]

So here is an example of Newton's Method of approximation for a starting value of 10.83.
As previously tabulated, the result should be 7.9772, and it indeed is.
The path to that answer is just wild!

   

This really shows how important the initial guess is for more complicated equations. Having an idea of what the function looks like and then using a table to isolate sign changes is really important.

For comparison, the Casio fx-991ex (latest edition) has these results.
x0= >10.84 : Cannot Solve
x0= >10.8 : Cannot Solve
x0= >10.83 : x=3.62

Makes me wonder which approximation method the Casio is using.