@tsunamiBTP  See corrected file  untitled5_new1.mw 

@tsunamiBTP  You wrote "This seems to work like a nested FOR loop". No, these are the different things. Here is a simple example of use nested seq. Let's say you want to create a matrix 10x10 that has symbols  x1, x2, ... ,x10  in the first line, the second line contains the same symbols, which squared, then cubed and so on:

Matrix([seq([seq((x||j)^i, j=1..10)], i=1..10)]);

Another example. It is very convenient and effective to use  seq  command  to obtain the sequence or list of objects with specified properties. In the example, we generate the sequence of Pythagorean triplets with  hypotenuses <=100:

seq(seq(seq(`if`(a^2+b^2=c^2,[a,b,c],NULL), b=a+1..c-1), a=1..c-2), c=3..100);

[3, 4, 5], [6, 8, 10], [5, 12, 13], [9, 12, 15], [8, 15, 17], [12, 16, 20], [7, 24, 25], [15, 20, 25], [10, 24, 26], [20, 21, 29], [18, 24, 30], [16, 30, 34], [21, 28, 35], [12, 35, 37], [15, 36, 39], [24, 32, 40], [9, 40, 41], [27, 36, 45], [14, 48, 50], [30, 40, 50], [24, 45, 51], [20, 48, 52], [28, 45, 53], [33, 44, 55], [40, 42, 58], [36, 48, 60], [11, 60, 61], [16, 63, 65], [25, 60, 65], [33, 56, 65], [39, 52, 65], [32, 60, 68], [42, 56, 70], [48, 55, 73], [24, 70, 74], [21, 72, 75], [45, 60, 75], [30, 72, 78], [48, 64, 80], [18, 80, 82], [13, 84, 85], [36, 77, 85], [40, 75, 85], [51, 68, 85], [60, 63, 87], [39, 80, 89], [54, 72, 90], [35, 84, 91], [57, 76, 95], [65, 72, 97], [28, 96, 100], [60, 80, 100]

@Markiyan Hirnyk  Here is a check (I just added a few commands to your document):

Download Check.mw

@mehdibaghaee Sorry, I did not notice the second condition.
Should be:

seq(`if`(V[i]>0 and V[i]<>infinity, [i, V[i]], NULL), i=1..10);

@Preben Alsholm

For example, if we have a function of one variable, given explicitly  f:=x->f(x)  and we want to  calculate the derivative value at a point  x=x0 , then I think that  D(f)(x0)  is better than  eval(diff(f(x),x), x=x0)  or  unapply(diff(f(x),x), x)(x0) .

@Preben Alsholm 

D operator can be used in similar situations, only the function must be specified explicitly:

restart;
f:=(x,y)->(x^3+y^3)^(1/3);
r:=(s,t)->(s*sin(t), s+56);
D[1](unapply((f@r)(s,t), s,t));
 

@tsunamiBTP

1. Since you mentioned  plottools:-getdata  command, I understood your request as a desire to get data from the plot.

2. If you want to get data from the second your graph, write:

data := op([2, 1], P);  # data as a matrix
convert(data, listlist);   # data as a list of lists

@Carl Love  For greater clarity, you can add one row to the final matrix:

<< x(`in degrees`) | sin(x) | Error> ;  < d | S | Err >>;

@tomleslie   I understood this condition  0<= beta, delta <= 1, that both roots should be searched in the range 0..1. If we follow your understanding, then here are 4 plots (I used to believe my eyes). The first two graphs clearly show the roots that are not in the list of DirectSearch. The last 2 plots show that the last 2 solutions of DirectSearch are wrong.

plots:-implicitplot([focdeltapioptS2Tbeta_eg, focbetapioptS2Tbeta_eg], beta=0..2, delta=-20..1, color=[red,blue], thickness=2,  gridrefine=5, axes=normal);

plots:-implicitplot([focdeltapioptS2Tbeta_eg, focbetapioptS2Tbeta_eg], beta=0.99..1.01, delta=-50..1, color=[red,blue], thickness=2,  gridrefine=5, axes=normal);

plots:-implicitplot([focdeltapioptS2Tbeta_eg, focbetapioptS2Tbeta_eg], beta=10.8..11, delta=-2.2..-2, color=[red,blue], thickness=2,  gridrefine=5, axes=normal);

plots:-implicitplot([focdeltapioptS2Tbeta_eg, focbetapioptS2Tbeta_eg], beta=14.7..14.9, delta=0.9..1.1, color=[red,blue], thickness=2,  gridrefine=5, axes=normal);
 

@Wtolrud  You can use a simple procedure named  Sort  that sorts any polynomial in ascending order of degrees. Formal parameters of the procedure: P is a polynomial (possibly with symbolic coefficients), var is a polynomial variable (by default this is x):

Sort:=proc(P::polynom, var::name:=x)
sort(P, var, ascending);
end proc:

Examples of use:

Sort(a*x^2+b*x+c);
Sort(-3*t^3+t+2*t^2+5, t);

                                   c+b*x+a*x^2
                                 5+t+2*t^2-3*t^3
 

@omkardpd  Place your specific system here in a copyable form as a text (not as a picture).

You forgot the sign of multiplication. Should be:

5*x^3+7*x^2+2*x^3

@Ramakrishnan  I took frames = 121 so that one of the animation frames matched the value of the parameter a = -3, at which a qualitative change in the structure of the solution occurs. Therefore, the step for the parameter  a  can be taken from several variants: 0.5, 0.2, 0.1, 0.05, ... . I chose the latter option and then the number of frames will be  6 / 0.05 + 1 = 121

@Ramakrishnan  Substantially reduce the number of frames, for example to 60, or write so:

restart;
A := plots:-animate(plot, [x^3+a*x+2, x = -4 .. 4, -15 .. 15, color = blue, thickness = 2], a = -6 .. 0, frames = 120): 
A;

Also note that I slightly simplified the penultimate line of my code above (for C).
I advise you to think about why I took in the code  frames=121  instead of frames=120 .

@vv  Very interesting how did you get these results? My way for some reason fails for large n (even option remember does not help), and Markiyan's way takes a long time:

P(10^5);
   Error, (in P) too many levels of recursion

restart;
ts:=time():
x(1) := 1: N := 100000; for n to N do x(n+1) :=
sum(convert(x(n), base, 10)[j]*10^(nops(convert(x(n), base, 10))-j), 
j = 1 .. nops(convert(x(n), base, 10)))+n end do:

x(10^5);
time()-ts;

                          N := 100000
                           9966045232
                             84.281