Maple 2019 Questions and Posts

These are Posts and Questions associated with the product, Maple 2019

Hi, I am trying to integrate a lengthy-expression but maple does not give a result and got hanged even after waiting 1 hour and more, pls help me to handle this or is this any other way to get a result?

Hi, how to write a loop and solve the algebraic expression? is the loop and do loop are the same thing? if different then pls mention how to solve the same question by using do loop?

Suppose I have the equation C := y^2*z + yz^2 = x^2, then I want to test for which triples (x,y,z) with x,y,z in {0,1} the equation is satisfied? Is there a quick way of doing this in Maple?

I am trying to rearrange the elements of an equation by the absolute value of their coefficients. eg -3y^2 x+2x z^2+6z^2 to

2x z^2 -3y^2 +6z^2 



1. How did Maple come up with this answer?

I've tried all the packages in SumTools and none of them give an answer except Hypergeometric.

2. Is there a way to trace the steps Maple is using so I can try and answer this myself?

3. Why did it sum the series when I didn't even ask - I deliberately used the inert form (no arguments though - I like what it did).

Thank you.


H1a := Sum(GAMMA(2*b - 1 + 2*n)*GAMMA(2*n - s)*(b - 1/2 + 2*n)/(GAMMA(2*b + 2*n + s)*GAMMA(2*n + 1)), n = 0 .. infinity);


The answer H1b it came up with is 

GAMMA(-s)*2^(1 + 2*b)*GAMMA(b)/(8*GAMMA(b + s)*2^(2*b));

which seems to be correct.

Hello there, 

Would please tell me how to pick up numerical vaules from answers given by 'solve()' command?

If you look at the worksheet (sorry for the error), one possible way is labeled by 'solution 1'. However, when I tried the expression in the 'attempt 1' label, I got an error. Therefore, I'm wondering if there is a way to extract the values from the answers, instead of using the 'rhs()' command. 

Maple Worksheet - Error

Failed to load the worksheet /maplenet/convert/ .



Best Regards, 

In Kwon Park

Hello all, 

When I enter the following expression:

eqaux_2 := l__bb = L__aa0 + L__aa2 * cos(2*theta - 4/3*Pi);

the cos function got converted an equivalent sin function automatically and the result was displayed like this:

l__bb = L__aa0 - L__aa2*sin(2*theta + Pi/6)

Is there any way to ask Maple not to do this automatic conversion?

Maple Worksheet - Error

Failed to load the worksheet /maplenet/convert/ .



Hello there, 

The equations described below are third-order equations. Therefore, the number of solutions would be 3. 

However, when I tried to solve them using the 'solve' command, only two solutions came out. 

Is there any chance to find the last one?

Maple Worksheet - Error

Failed to load the worksheet /maplenet/convert/ .


Hello there, 

When I tried to solve an equation using the 'solve' command in Maple, I got an answer. 

However, I could not understand how Maple got to the answer. Would you tell me what steps Maple might have gone through in order to come to the answer?

Here is my worksheet:

Maple Worksheet - Error

Failed to load the worksheet /maplenet/convert/ .


PS) Perhaps this editbox began to dislike Google Chrome.




I was able to get Procedure 2 to work using a for - do loop. I was wondering if it is possible to speed up the calculation by using map to find the number of roots? I do not fully understand map and passing data.

tgf := proc(a, b, c, d, t, m, n)

          local X;

          X := [solve(abs(a*x + b) + abs(c*x + d) - t*x^2 + m*x - n = 0)]; 

          return nops(X);

end proc;

res := CodeTools:-Usage(map(tgf, L));
Error, (in CodeTools:-Usage) invalid input: tgf uses a 2nd argument, b, which is missing

The L Array is tripping me up, here is a partial display of the array:

Array(1..262, 1..7, [[5,2,3,9,1,1,1],[5,2,3,9,2,1,1],[5,4,3,7,1,1,1],[5,4,3,7,2,1,1],[5,5,3,6,1,1,1],[5,5,3,6,2,1,1],[5,5,4,8,1,1,2],[5,5,4,8,2,1,2], ... ,[10,10,5,10,2,2,2]], datatype = integer[4]).

I made L Array into a list of list, R. Somewhat works.

Here is the script: 

Thanks for any help.

Can anyone correct me, what's wrong with it.




I'm experimenting the RSA encryption algorithm and in one example, I need to calculate (m^e) mod n where :




and I receive : Error, numeric exception: overflow

Even though these are large numbers, modular arithmetic algorithms should be able to deal with that.

Why is it not the case ?

NB : One of the mathApps in the category Computer Science is "RSA Encryption" and proves to be fairly comfortable with such large numbers, no overflow!

V1 := [1.666666667, 1.983050847, 2.372881356, 2.768361582, 
    3.380681818, 3.977272727, 4.767045455, 5.755681818, 
    6.937500000, 8.244318182, 9.801136364, 0.2971428571, 
    0.6914285714, 1.085714286, 0.3942857143]
dV1 := [0.03315280331, 0.03866282527, 0.04508264551, 
   0.05165406892, 0.06212371278, 0.07219635124, 0.08557994498, 
   0.1023815798, 0.1225076192, 0.1447933406, 0.1713677762, 
   0.01380612592, 0.01872473080, 0.02454205309, 0.01488350319]
V2 := [17/10, 2, 12/5, 14/5, 17/5, 4, 24/5, 29/5, 7, 42/5, 10, 3/10, 7/10, 11/10, 2/5]
dV2 := [0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0]


i have trouble adding error bars to my plots despite experimenting with it for hours. I tried using ErrorBars but i never get close to what i need. What i have is V1 (values for y) and V2 (values for x) to produce a ScatterPlot which works alright. What i need is something like this: .

For the errors in x and y-direction i have 2 lists dV1 (y errors) and dV2 (x errors).

This is my first post and i´m a bit in a rush so please excuse me for any mistakes i may have made.

Calculation of these integrals takes much time.

Is there a way to reduce the time?


I have this procedure to perform a Boole-Mobius Transform. I took me quite a while to figure out. Whereas it works, I wonder how it should be done efficiently? The document is also attached which shows the steps I went through to derive the procedure. I can't get the document to display.

BooleMobiusTransform := proc(V) 
local n, im, istep, jm, h, istart, i, j, k; n := ilog2(numelems(V)); im := 2^n/2; istep := im; jm := 1; h := 2^n; 
for k to n do
 istart := 1; 
for j to jm do 
for i from istart to im - 1 + istart do 
V(istep + i) := (V[istep + i] + V[i]) mod 2;
 end do;
 istart := istart + h; 
end do;
 im := 1/2*im; istep := 1/2*istep; jm := 2*jm; h := 1/2*h; 
end do; 
return V; 
end proc

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