Maple 2017 Questions and Posts

These are Posts and Questions associated with the product, Maple 2017

I have a trigonometric equation that outputs with a solution in terms of _B1 which I want to remove.

restart: solve({7*cos(2*t)=7*cos(t)^2-5, t>=0, t<=2*Pi}, t, allsolutions, explicit);


{t = arccos((1/7)*sqrt(14))},

{t = 2*Pi-arccos((1/7)*sqrt(14))},

{t = 2*arccos((1/7)*sqrt(14))*_B1-2*_B1*Pi+2*Pi*_Z1-arccos((1/7)*sqrt(14))+Pi}

Is there anyway to get rid of the _B1, or somehow evaluate it by a substitution?


Even numerically the answer still retains the _B1.

{t = 1.006853685}, {t = 5.276331623}, {t = -4.269477938*_B1+6.283185308*_Z1+2.134738969}


Also it would be nice to remove the _Z1 subscript too, as the domain of the equation is [0, 2pi].

I tried removing the 'AllSolutions' command , but then I am missing two solutions:

solve({7*cos(2*t)=7*cos(t)^2-5., t>=0 and t<=2*Pi}, t, Explicit);

 {t = 1.006853685}, {t = 2.134738969}

There should be 4 solutions in the domain [0, 2pi].

Hello people in mapleprimes,

I am reading Introduction to Maple. There, the following input and output appear.

> convert((-8)^(1/3), RootOf);
1+RootOf(_Z^2 + 3, index = 1)

I think, which implies



with the solution alpha of alpha^2+3=0, beta above is expressed as 1+alpha.

But, probably from my understanding being wrong, I can't understand why this becomes so.

So, though this is about mathematics, not about maple, if possible, can I ask you to teach me why

convert((-8)^(1/3), RootOf) brings 1+RootOf(_Z^2 + 3, index = 1)?

Thanks in advance.


Maple easily solves multi-points problems for an ODE, if the equation can be integrated analytically.

For example, the following 3-points problem  is solvable:

dsolve([diff(y(x), x$3)+diff(y(x), x$2)+y(x)=1, y(0)=0, y(1)=0, y(2)=1], [y(x)]);

But a similar problem cannot be solved numerically.

For example,

dsolve([diff(y(x), x$3)+diff(y(x), x$2)+y(x)=1, y(0)=0, y(1)=0, y(2)=1], [y(x)], type = numeric, 'output' = Array([seq(k/5, k=0..5)]));

generates: Error, (in dsolve/numeric/process_input) boundary conditions specified at too many points: {0, 1, 2}, can only solve two-point boundary value problems

I need to solve a certain number of multi-points problems for ODE systems numerically. Maybe, for this there are some workaround?

Hello together,

I´m a new member and I used in Maple 2017 the function "is prime" for the largest known Mersenne prime (277,232,917 − 1) with the command "isprime (277,232,917 − 1)" to test how much time the programm needs that it returns "true".

1.) Has anybody experiences with the function "is prime" in relation with such a large Mersenne prime ( 277,232,917 − 1) or similar Mersenne primes like ( 257,885,161 − 1), ( 274,207,281 − 1), etc... ? How many days it takes to get the confirmation in my case ?

My processor: Intel i5-4590 CPU @ 3.30 GHz

System type: 64 Bit  

I started the function isprime(277,232,917 − 1) before 5 days. The programm is still evaluating, next to "evaluating" the point changes his colors (black and white) continuous, so I think the programm is still working, but the memory stopped at 2440.33 M and the time stopped at 12326.65 s

2.) Has anybody an idea why Maple has stopped to count the time and the memory in this case ? 


I hope that somebody can answer my 2 questions....thx... 


I have a question about the Import of .txt file:

I can import the following .txt, but everytime I want to convert in a matrix or something else, maple chrashes. 

I need some parts (smaller matrices) of this data (for example: all values from coloumn 200-400 and row 100-600)

I hope somebody can help me




I created the following plot

plot([BesselJ(0, x), BesselJ(1, x)], x = 0 .. 10, color = [red, blue]);

What I want to do mext is calculate all x values on this interval wherethe two paths intersect

 Calculate all x values on this interval where

Is it possible that the two-variable arctan function arctan(a(x),b(x)) could be could be equal to arctan(a(x)/b(x)) as long as x>0? And why?

arctan(a(x),b(x))=arctan(a(x)/b(x) as long as x>0 ?

Thank's for any replies :)

Dear all,
I have made a code that solves a nonlinear system of equations by using Newton's Method. I am facing a problem in printing the Jacobian at each iteration. It only prints its lable not the Jacobian matrix itself. Please help me in this regard.

"restart;  iter:=5;  f[1](x,y):=3 x^(2)-y^(2);  f[2](x,y):=3 x^()*y^(2)-x^(3)-1;"



proc (x, y) options operator, arrow; 3*x^2-y^2 end proc


proc (x, y) options operator, arrow; 3*x*y^2-x^3-1 end proc


var := x, y

x, y


pointt := [x[n], y[n]]

[x[n], y[n]]





x[0] := 1; y[0] := 1





for n from 0 to iter do print('f1' = f[1](x[n], y[n]), 'f[2]' = f[2](x[n], y[n])); print('J'*[n] = J[n]); J[n] := Student[MultivariateCalculus][Jacobian]([f[1](x, y), f[2](x, y)], [var] = pointt, output = matrix); sol[n] := eval((Vector(2, {(1) = x[n], (2) = y[n]}))-1/J[n].(Vector(2, {(1) = f[1](x[n], y[n]), (2) = f[2](x[n], y[n])}))); x[n+1] := evalf(sol[n][1]); y[n+1] := evalf(sol[n][2]); print(x[n+1], y[n+1]) end do

f1 = 2, f[2] = 1


J*[0] = J[0]


.6111111111, .8333333333


f1 = .4259259256, f[2] = 0.44924554e-1


J*[1] = J[1]


HFloat(0.5036590808700434), HFloat(0.8524944221287727)


f1 = HFloat(0.03427066946790058), f[2] = HFloat(-0.029666658033242088)


J*[2] = J[2]


HFloat(0.4999641210723523), HFloat(0.8660456363859079)


f1 = HFloat(-1.4267722412308892e-4), f[2] = HFloat(-1.2576398193964167e-6)


J*[3] = J[3]


HFloat(0.50000000001492), HFloat(0.8660254018170033)


f1 = HFloat(3.45245787514159e-9), f[2] = HFloat(-5.089167087746205e-9)


J*[4] = J[4]


HFloat(0.5), HFloat(0.8660254037844386)


f1 = HFloat(1.1102230246251565e-16), f[2] = HFloat(-2.220446049250313e-16)


J*[5] = J[5]


HFloat(0.5), HFloat(0.8660254037844387)






the print output is too ugly, any help in making it more elegant would also be appreciated .

Am trying to integrat f(x)=x3/2 from 1-2 but can't seem to set it up properly.  

I am trying to make a contour plot of the function f(x,y):=exp(-(x2+y2)/3)cos(2xy)  

Hi all.

Hope the best for all

In the following program(written code), why i can't achieve the proper combined function in HybrFunc(2, 3, 1)?

thanks for any help

Is there any way to convert a complex number in to a phasor angle (radians or degrees) like the one marked below?

In the following workbook, I defined two equivalent one variable functions and plot their difference.

I'm surprised to see that the difference is not 0 between 55 and 60. But when I computed the respective function's values at these points, the difference is 0.

assume(n, integer, n > 0)


f := proc (n) options operator, arrow; 2^(n+10) end proc

proc (n) options operator, arrow; 2^(n+10) end proc


g := proc (n) options operator, arrow; 1024*2^n end proc

proc (n) options operator, arrow; 1024*2^n end proc








map(f, [seq(i, i = 55 .. 60)])

[36893488147419103232, 73786976294838206464, 147573952589676412928, 295147905179352825856, 590295810358705651712, 1180591620717411303424]


map(g, [seq(i, i = 55 .. 60)])

[36893488147419103232, 73786976294838206464, 147573952589676412928, 295147905179352825856, 590295810358705651712, 1180591620717411303424]


[36893488147419103232, 73786976294838206464, 147573952589676412928, 295147905179352825856, 590295810358705651712, 1180591620717411303424]+[-36893488147419103232, -73786976294838206464, -147573952589676412928, -295147905179352825856, -590295810358705651712, -1180591620717411303424]

[0, 0, 0, 0, 0, 0]





Hi everybody!

I am generating MATLAB code for a matrix created in Maple 2017 using the following code:


My question now: How can I specify that the generated code should not include zero elements of the matrix.

Since the matrices are sparse, there are hundrets of lines indicating zeros elements.

This is not necessary since I initialise the matrix with zeros in MATLAB anyway.

Many thanks for your help!


The following quiz doesn't react in the rigth way, when the respond is a fraction with nominator 1. When the fraction 4/20 is the answer, the respond 1/5 is incorrect

Grading:-Quiz("Reduce this fraction",
proc (Resp, Ans) evalb([op(InertForm:-Value(Ans))] = [op(Resp)]) end proc,
proc () local a, b, c; a := rand(1 .. 5); b := rand(1 .. 5); c := rand(2 .. 5)( ); `%/`(a( )*c, b( )*c) end proc, 
'inertform' = true)

Any proposals?

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