I have some lecture notes and examples that suddenly appears to be corrupted. I am getting the following error code when trying to open the file: "There was a problem in the loading process, you worksheet may be incomplete."

I've tried the following trouvleshooting:

- Close and reopen Maple

- Restarting computer

- Tried different computer

note: I have no backup files as I do not have autosave switched on.

I have also tried looking into the solution posted on a previous question on this site (https://www.mapleprimes.com/questions/235363-How-To-Fix-Corrupted-Maple-File), i.e.  changing the file format to.xml and making changes but with no luck :(

Is anyone able to help me fix it and un-corrupt this file:

Lecture_2_-_notes.mw

I cannot find an explanation why a boolean evaluation with higher Digits is not the same

RealRange seems to be evaluated differently.
Any explanation for that?

(expression extracted from an error message)

Download eval_of_range_in_ln.mw

Dear Maple user

I want to add caption for all four graphs with a,b,c,d at the bottom centre of each graph that should be arrange in 2X2 array and the array bottom caption as Figure 6, also i want to save the output of the array graphs in desktop with jpeg format. sample graph i have uploaded.

restart:
with(plots):
A1:= plot(sin(x), x = -Pi .. Pi,title = "Graph at M^2=1",caption = "Fig a"):# This caption is at the centre bottom of first graph and the title of A1 is M*M (M^2)should be display
A2:= plot(cos(x), x = -Pi .. Pi,caption = "Fig b"):# This caption is at the centre bottom of second graph
A3 := plot3d(1.5^x*sin(y), x = -3 .. 3, y = -3 .. 3,caption = "Fig c"):# This caption is at the centre bottom of Third graph
A4 := plot3d(1.3^x*sin(y), x = -1 .. 2*Pi, y = 0 .. Pi,caption = "Fig d"):# This caption is at the centre bottom of forth graph
display(Array(1 .. 2, 1 .. 2, [[A1, A2], [A3, A4]]), aligncolumns = [1],caption = "Fig 6"):# This caption at the bottom centre of array graphs 

I have and input parameter to a procedure called shift:=05. It sets the position of test between to points.

this particular procedure has two potential sets of text one in Q and the other in vec.
Can Q and vec have the own individual values for shift?

I have shown the input to the procedure and what it outputs. The procedure itself to too long and akward to the post.

Also given is the input definiton of the procedure and how I could possibly find a shift in a list.

Download 2024-09-18_Select_name_and_Value_from_List.mw

The first time they are evaluated, some aborts can occur; the second time they are evaluated, no exception is thrown: 
 

Download run__twice.mw

Why does (the outer) RootOf have to be evaluated twice? 
Note. There are other similar examples, but they are less concise (so they are omitted here).

I have a differential equation which I am looking for a series solution in inverse powers of r. I am doing this by matching inverse powers between the RHS and LHS of the differential equation and then finding the coefficient in the series solution that would solve this. I have written a little procedure which returns the coefficients, I am just curious if there is a command that will do this for me or if i can improve on this technique? I tried the Solve command in PDETools but it just kept evaluating and did not return an answer. 

Thanks in advance. 

EDIT: More detail provided in the first reply. 

coefficient_question.mw

Hello, 

I have a list of data {a,b,c,d} for which I need to create permutations without inbult function in Maple. The list of inputs is not fixed and it can contain aither one value or n-values. I was thinking about use of recursion in loops but it doesn't work as I was expected. Do you know please how to set a such recursive calculation with loops?
 

Result should be as follows for the list whole list range:

a+b

a+c

a+d

a+b+c

a+b+d
a+c+b
a+c+d

a+d+b

a+d+c

a+b+c+d

a+c+d+b

a+d+b+c

b+a

b+c

b+d

b+a+c

b+a+d

b+c+a

b+c+d

b+d+a

b+d+c

b+a+c+d

b+c+d+a

b+d+a+c

etc...

Comb1:=proc(S)
local G2::DataFrame,x,xs,i,j,T,k;

G2:=DataFrame([[]]);

if numelems(S)=0 then 
		return {}
		elif numelems(S)=1 then
			return {S}
		else 
			for i in seq(1..numelems(S))do
				for j in seq(1..numelems(S))do
					if i=j then
						next j
					end if;
			x:=S[j];
			xs:=S[i]+S[j];	

			#collect data into a dataframe
			G2:=Append(G2,DataSeries(<xs>),mode=row);

			#recursion
			for T in Comb1([xs]) do 
				for k in seq(1..numelems(S)) do
					if k=i or k=j then 
						next k
						end if;
				G2:=Append(G2,DataSeries(<[S[k]]+T>),mode=row);
				
					end do;
				end do;
			end do;
		end do;
	end if;
		
return G2;

end proc:

Comb1({a,b,c,d});

I cannot remember that this was an issue in Maple 6/9/11, so it must have changed later on up to 22.

The following limit fails to execute

restart;
1/(r_S^5*(_C2*r_S^4 + _C2*r_S^3 + _C2*r_S^2 + f[3](0)*r_S + f[4](0)));
limit(%, f[4](0) = infinity);

It fails to calculate the limit with result as zero.

Thanks

I want to solve a pde numerically, where one of the initial conditions involves an integral expression. In essence, I would like to get the following code to run. 

restart;
with(PDEtools);

# the heat equation
pde := diff(v(x, t), t) = diff(v(x, t), x, x);

f := x -> exp(-x^2);

a := -1.68858; b := 1.68858; delx := 1/100; t_final := 5;

pds := pdsolve(pde, {v(a, t) = 0, v(b, t) = 0, v(x, 0) = 1 - int(f(f(s)), s = 0 .. x)}, numeric, range = a .. b, time = t, spacestep = delx);

Error, (in pdsolve/numeric/process_IBCs) improper specification of initial/boundary condition: additional unknown(s) {s} were found in v(x,0) = 1-int(exp(-exp(-s^2)^2),s = 0 .. x)

pds:-animate(frames = 120, t = t_final);

Maple doesn't have any built-in function to evaluate the integral expression, so the initial condition doesn't evaluate to a simple expression involving x. I hope that because pdsolve is outputting numerical results, it should be acceptable for the initial condition to output floats as well.

Where can I find my purchase code to complete the installation for Maple Calculator? I am not able to use it without one.

If you draw a chord in any curve, when the latter becomes infinitely small, the ratio of the surface segment to the triangle formed by the chord and the associated tangents is 2:3.

(Source: Archive of Mathematics and Physics, editor Johann August Grunert, 31st part of 1858, pp. 449-453, "On a remarkable general theorem on curves",
Author: Andreas Völler)

(The curve may be assumed to be sufficiently differentiable.)