This is a very nice analysis, Ilias. Thanks for bringing rigor to my laziness! This connection between polygonal dissections (and trees, Dyck paths, etc?) and integer partitions is very interesting. I’ve never understood why partitions were so important. But this may be a path to some fundamental understanding.

I come back to solving equations. Because in fact, the name of this thread is “Solving the general polynomial equation”. Do you remember ?

Let a first example : solve the equation (E):x ^ 3 + x² + x + c_0 = 0.

With all these calculations, we know that a root r of (E) is : r=\sum_{i=0}^{\infty} a_i \times c_0^{i}

where a_i are calculated with a mix of bitri coefficients and c_1, c_2,c_3 powers. As we have computers, programs, and formulas, we are able to calculate the a_i terms. Starting with a_0 up to a_{20} we have :

0, -1, -1, -1, 0, 4, 14, 30, 33, -55, -429, -1365, -2652, -1428, 12920, 64600, 178296, 277932, -152950, -2770350,-10785390. These numbers are obtained with a small modification of “my” python program (See above in this thread). It’s not easy to calculates the convergence radius of this serie, but, i’m sure that it is less than 1 ! Despite this, it is possible, in some cases, to calculate the sum of the serie outside the radius of convergence. On this point, I advise you to watch this excellent video series Mathematical Physics 01 - Carl Bender - YouTube

It work with Padé approximants. These are rational fractions which most closely approximate the serie \sum_{i=0}^{\infty} a_i \times c_0^{i} and therefore r.

The principle is not complicated, we have to calculate the Padé approximants

P_{p,q}(x)=\frac{\sum_{i=0}^{p} p_i \times x^{i}}{\sum_{j=0}^{q} q_j \times x^{j}}=\sum_{i=0}^{p+q+1} a_i \times x^{i} (In the following I use x instead of c_0)

To calculate an approximant of Padé, it is therefore necessary to calculate the p_i and q_j according to the a_i, p and q. For that, it is necessary to solve a system of linear equations. But nothing very complicated. For example :

P_{5,5}(x)=\frac{-x+8x^2-24x^3+31x^4-11x^5}{1-9x+32x^2-54x^3+37x^4-5x^5}

P_{6,5}(x)=\frac{-x+7.2x^2-21.6x^3+29.8*x^4-16.2x^5+1.2x^6)}{1-8.2x+28.8x^2-50.4x^3+41.8x^4-11.4*x^5}

In the picture below, in red y=\sum_{i=0}^{\infty} a_i x^{i} and 4 Padé approximants

As y(-2) =\sum_{i=0}^{\infty} a_i \times (-2)^{i} diverges, it’s not possible to use this method to solve the x ^ 3 + x² + x -2 = 0 but with Padé approximants, it’s easy:

P_{5,5}(-2)=0.807

P_{6,5}(-2)=0.813

These are good approximations of the root r = 0.8105.

Thanks Chris!!Let’s see how it goes!

I watched the first lecture. Very interesting. Wildberger’s technique seems very closely allied with this perturbation theory technique, yes?

Yes exactly !

In the case of the equation of degree 3 for example, Norman propose to solve ax^3 + bx^2 +cx+\epsilon=0. He uses t instead \epsilon, but it’s the same method.

Carl Bender introduce the Padé approximants in the lecture 6.

By the way, do you have seen the last video of Norman ? At the end, he proposes a new problem.

I have already wrote a Python program to explore the roots of the graph below.

The root’s set seems endless with a pseudo cycle of 12 generations …

It will be a good idea to create a new thread on this topic. no ?

Yes, excited about the new series. Although I hope we make some good progress here before he starts!

I’ll start a new topic for it.

Did you watch guys the last lecture of Norman where he connects the division of a polygon into triangles and quadrilaterals with the binary and ternary trees.

Thanks for letting us know. I haven’t been getting notifications about these videos – perhaps because they are posted a month ago.

I’ve spent some time with the tree interpretation. It’s quite straight-forward. Now, I’m trying to connect the polys and trees to Dyck paths. This is a bit harder. This paper has been of some help:

http://pages.stat.wisc.edu/~callan/papers/polygon_dissections/polygon_dissections.pdf

yeap, no worries man! I have a look on the paper! thanks! you are right about trees, it is straight forward!