Items tagged with simplify

Hello people in mapleprimes,

I want to simplify the next expression which has 1/k as its exponent,

especially, I want to collect for T. I hope you will teach me how to do it.


If I do as


the output is


But, as all variables has 1/k as its exponent, I want to collect it to (...)^(1/k).

Is this possible?


The following product


(product(mu^x[i]/factorial(x[i]), i = 1 .. n))


does not simplify to the most obvious form whatever I try


mu^(sum(x[i], i = 1 .. n))/(product(factorial(x[i]), i = 1 .. n))


What can it be?



hi every one...

how i can simplify this result (R_arm_F2 $  Twflex) via tringular relations.

where Ixflex & tetadot and other... are constants




R := (Matrix(3, 3, {(1, 1) = 1, (1, 2) = 0, (1, 3) = 0, (2, 1) = 0, (2, 2) = cos(teta), (2, 3) = -sin(teta), (3, 1) = 0, (3, 2) = sin(teta), (3, 3) = cos(teta)})).(Matrix(3, 3, {(1, 1) = cos(phi), (1, 2) = 0, (1, 3) = sin(phi), (2, 1) = 0, (2, 2) = 1, (2, 3) = 0, (3, 1) = -sin(phi), (3, 2) = 0, (3, 3) = cos(phi)})).(Matrix(3, 3, {(1, 1) = cos(si), (1, 2) = -sin(si), (1, 3) = 0, (2, 1) = sin(si), (2, 2) = cos(si), (2, 3) = 0, (3, 1) = 0, (3, 2) = 0, (3, 3) = 1}))

R := Matrix(3, 3, {(1, 1) = cos(phi)*cos(si), (1, 2) = -cos(phi)*sin(si), (1, 3) = sin(phi), (2, 1) = sin(teta)*sin(phi)*cos(si)+cos(teta)*sin(si), (2, 2) = -sin(teta)*sin(phi)*sin(si)+cos(teta)*cos(si), (2, 3) = -sin(teta)*cos(phi), (3, 1) = -cos(teta)*sin(phi)*cos(si)+sin(teta)*sin(si), (3, 2) = cos(teta)*sin(phi)*sin(si)+sin(teta)*cos(si), (3, 3) = cos(teta)*cos(phi)})



RT := simplify(1/R)

RT := Matrix(3, 3, {(1, 1) = cos(phi)*cos(si), (1, 2) = sin(teta)*sin(phi)*cos(si)+cos(teta)*sin(si), (1, 3) = -cos(teta)*sin(phi)*cos(si)+sin(teta)*sin(si), (2, 1) = -cos(phi)*sin(si), (2, 2) = -sin(teta)*sin(phi)*sin(si)+cos(teta)*cos(si), (2, 3) = cos(teta)*sin(phi)*sin(si)+sin(teta)*cos(si), (3, 1) = sin(phi), (3, 2) = -sin(teta)*cos(phi), (3, 3) = cos(teta)*cos(phi)})


R_I_F2 := Matrix(3, 3, {(1, 1) = sin(phi)^2.(1-cos(si))+cos(si), (1, 2) = -(sin(phi).cos(phi).sin(teta))*(1-cos(si))-cos(phi).cos(teta).sin(si), (1, 3) = (sin(phi).cos(phi).cos(teta))*(1-cos(si))-sin(teta)*cos(phi).sin(si), (2, 1) = -(2*sin(phi).cos(phi).sin(teta).cos(teta))*(1-cos(si))+(cos(phi).sin(si))*(cos(teta)^2-sin(teta)^2), (2, 2) = (2*cos(phi)^2.(sin(teta)^2).cos(teta))*(1-cos(si))+cos(teta).cos(si)-sin(teta).sin(phi).sin(si), (2, 3) = -(2*cos(phi)^2.sin(teta))*cos(teta)^2*(1-cos(si))-sin(phi).cos(teta).sin(si)-sin(teta).cos(si), (3, 1) = (sin(phi).cos(phi))*(1-cos(si))*(cos(teta)^2-sin(teta)^2)+2*cos(phi).cos(teta).sin(teta).sin(si), (3, 2) = (cos(phi)^2.sin(teta))*(sin(teta)^2-cos(teta)^2)*(1-cos(si))+cos(si).sin(teta)+sin(phi).cos(teta).sin(si), (3, 3) = (cos(phi)^2.cos(teta))*(cos(teta)^2-sin(teta)^2)*(1-cos(si))-sin(phi).sin(teta).sin(si)+cos(teta).cos(si)})

R_I_F2 := Matrix(3, 3, {(1, 1) = sin(phi)^2.(1-cos(si))+cos(si), (1, 2) = -(`.`(sin(phi), cos(phi), sin(teta)))*(1-cos(si))-`.`(cos(phi), cos(teta), sin(si)), (1, 3) = (`.`(sin(phi), cos(phi), cos(teta)))*(1-cos(si))-sin(teta)*cos(phi).sin(si), (2, 1) = -2*(sin(phi).cos(phi).sin(teta).cos(teta))*(1-cos(si))+(cos(phi).sin(si))*(cos(teta)^2-sin(teta)^2), (2, 2) = 2*(cos(phi)^2.(sin(teta)^2).cos(teta))*(1-cos(si))+cos(teta).cos(si)-`.`(sin(teta), sin(phi), sin(si)), (2, 3) = -2*(cos(phi)^2.sin(teta))*cos(teta)^2*(1-cos(si))-`.`(sin(phi), cos(teta), sin(si))-sin(teta).cos(si), (3, 1) = (sin(phi).cos(phi))*(1-cos(si))*(cos(teta)^2-sin(teta)^2)+2*(cos(phi).cos(teta).sin(teta).sin(si)), (3, 2) = (cos(phi)^2.sin(teta))*(sin(teta)^2-cos(teta)^2)*(1-cos(si))+cos(si).sin(teta)+`.`(sin(phi), cos(teta), sin(si)), (3, 3) = (cos(phi)^2.cos(teta))*(cos(teta)^2-sin(teta)^2)*(1-cos(si))-`.`(sin(phi), sin(teta), sin(si))+cos(teta).cos(si)})



R_arm_F2 := RT.R_I_F2

R_arm_F2 := Matrix(3, 3, {(1, 1) = cos(phi)*cos(si)*(sin(phi)^2.(1-cos(si))+cos(si))+(sin(teta)*sin(phi)*cos(si)+cos(teta)*sin(si))*(-2*(`.`(sin(phi), cos(phi), sin(teta), cos(teta)))*(1-cos(si))+(cos(phi).sin(si))*(cos(teta)^2-sin(teta)^2))+(-cos(teta)*sin(phi)*cos(si)+sin(teta)*sin(si))*((sin(phi).cos(phi))*(1-cos(si))*(cos(teta)^2-sin(teta)^2)+2*(`.`(cos(phi), cos(teta), sin(teta), sin(si)))), (1, 2) = cos(phi)*cos(si)*(-(`.`(sin(phi), cos(phi), sin(teta)))*(1-cos(si))-`.`(cos(phi), cos(teta), sin(si)))+(sin(teta)*sin(phi)*cos(si)+cos(teta)*sin(si))*(2*(`.`(cos(phi)^2, sin(teta)^2, cos(teta)))*(1-cos(si))+cos(teta).cos(si)-`.`(sin(teta), sin(phi), sin(si)))+(-cos(teta)*sin(phi)*cos(si)+sin(teta)*sin(si))*((cos(phi)^2.sin(teta))*(sin(teta)^2-cos(teta)^2)*(1-cos(si))+cos(si).sin(teta)+`.`(sin(phi), cos(teta), sin(si))), (1, 3) = cos(phi)*cos(si)*((`.`(sin(phi), cos(phi), cos(teta)))*(1-cos(si))-sin(teta)*cos(phi).sin(si))+(sin(teta)*sin(phi)*cos(si)+cos(teta)*sin(si))*(-2*(cos(phi)^2.sin(teta))*cos(teta)^2*(1-cos(si))-`.`(sin(phi), cos(teta), sin(si))-sin(teta).cos(si))+(-cos(teta)*sin(phi)*cos(si)+sin(teta)*sin(si))*((cos(phi)^2.cos(teta))*(cos(teta)^2-sin(teta)^2)*(1-cos(si))-`.`(sin(phi), sin(teta), sin(si))+cos(teta).cos(si)), (2, 1) = -cos(phi)*sin(si)*(sin(phi)^2.(1-cos(si))+cos(si))+(-sin(teta)*sin(phi)*sin(si)+cos(teta)*cos(si))*(-2*(`.`(sin(phi), cos(phi), sin(teta), cos(teta)))*(1-cos(si))+(cos(phi).sin(si))*(cos(teta)^2-sin(teta)^2))+(cos(teta)*sin(phi)*sin(si)+sin(teta)*cos(si))*((sin(phi).cos(phi))*(1-cos(si))*(cos(teta)^2-sin(teta)^2)+2*(`.`(cos(phi), cos(teta), sin(teta), sin(si)))), (2, 2) = -cos(phi)*sin(si)*(-(`.`(sin(phi), cos(phi), sin(teta)))*(1-cos(si))-`.`(cos(phi), cos(teta), sin(si)))+(-sin(teta)*sin(phi)*sin(si)+cos(teta)*cos(si))*(2*(`.`(cos(phi)^2, sin(teta)^2, cos(teta)))*(1-cos(si))+cos(teta).cos(si)-`.`(sin(teta), sin(phi), sin(si)))+(cos(teta)*sin(phi)*sin(si)+sin(teta)*cos(si))*((cos(phi)^2.sin(teta))*(sin(teta)^2-cos(teta)^2)*(1-cos(si))+cos(si).sin(teta)+`.`(sin(phi), cos(teta), sin(si))), (2, 3) = -cos(phi)*sin(si)*((`.`(sin(phi), cos(phi), cos(teta)))*(1-cos(si))-sin(teta)*cos(phi).sin(si))+(-sin(teta)*sin(phi)*sin(si)+cos(teta)*cos(si))*(-2*(cos(phi)^2.sin(teta))*cos(teta)^2*(1-cos(si))-`.`(sin(phi), cos(teta), sin(si))-sin(teta).cos(si))+(cos(teta)*sin(phi)*sin(si)+sin(teta)*cos(si))*((cos(phi)^2.cos(teta))*(cos(teta)^2-sin(teta)^2)*(1-cos(si))-`.`(sin(phi), sin(teta), sin(si))+cos(teta).cos(si)), (3, 1) = sin(phi)*(sin(phi)^2.(1-cos(si))+cos(si))-sin(teta)*cos(phi)*(-2*(`.`(sin(phi), cos(phi), sin(teta), cos(teta)))*(1-cos(si))+(cos(phi).sin(si))*(cos(teta)^2-sin(teta)^2))+cos(teta)*cos(phi)*((sin(phi).cos(phi))*(1-cos(si))*(cos(teta)^2-sin(teta)^2)+2*(`.`(cos(phi), cos(teta), sin(teta), sin(si)))), (3, 2) = sin(phi)*(-(`.`(sin(phi), cos(phi), sin(teta)))*(1-cos(si))-`.`(cos(phi), cos(teta), sin(si)))-sin(teta)*cos(phi)*(2*(`.`(cos(phi)^2, sin(teta)^2, cos(teta)))*(1-cos(si))+cos(teta).cos(si)-`.`(sin(teta), sin(phi), sin(si)))+cos(teta)*cos(phi)*((cos(phi)^2.sin(teta))*(sin(teta)^2-cos(teta)^2)*(1-cos(si))+cos(si).sin(teta)+`.`(sin(phi), cos(teta), sin(si))), (3, 3) = sin(phi)*((`.`(sin(phi), cos(phi), cos(teta)))*(1-cos(si))-sin(teta)*cos(phi).sin(si))-sin(teta)*cos(phi)*(-2*(cos(phi)^2.sin(teta))*cos(teta)^2*(1-cos(si))-`.`(sin(phi), cos(teta), sin(si))-sin(teta).cos(si))+cos(teta)*cos(phi)*((cos(phi)^2.cos(teta))*(cos(teta)^2-sin(teta)^2)*(1-cos(si))-`.`(sin(phi), sin(teta), sin(si))+cos(teta).cos(si))})


Twflex := Typesetting:-delayDotProduct(Ixflex, (Typesetting:-delayDotProduct(tetadot, Typesetting:-delayDotProduct(sin(phi)^2, 1-cos(si))+cos(si))+Typesetting:-delayDotProduct(sidot, sin(phi)^3+Typesetting:-delayDotProduct(cos(phi)^2, Typesetting:-delayDotProduct(sin(phi), cos(si)+Typesetting:-delayDotProduct(cos(teta), 1-cos(si)))+Typesetting:-delayDotProduct(sin(teta), sin(si)))))^2)












Hello people in Mapleprimes,


I have an expression which I want to modify with another equation.

They are simple, and looks easy to simplify.


hh := (L*lambda*(k-sigma+1)*upsilon*tau*v)/(f__F*sigma*(-tau^2+upsilon^2)*k)=rho;


I want to express nb with hh as


With the next code, that modification can be done.


But, this isolates hh for f__F, which does not look intuitive.

On the other hand, the outcome of the substitution looks so simple, which you find with executing  the codes of

nb, and hh.

But, algsubs, and subs, and simplify/siderel won't work properly.


What I want to ask is this. Isn't there any nice way to substitute hh into nb other than isolating f__F, so that the result is expressed with rho?


I will be very glad if you will give me answers.


Best wishes.






I am trying to solve an equation using surd and I get a strange result.

    4, -4, 4 I, -4 I

These solutions are clearly wrong.

The equation (x^4)^(1/8) = -2 has no solution.

This problem is equivalent to asking the computer to solve sqrt(x) = -2

which has no solution in R or C.


However if I type

solve((x^4)^(1/8) = -2) , then I get no answer, which is what I expected.

Why does surd behave in this unexpected way.


Also another thing I am wondering, why doesn't Maple simplify (x^4)^(1/8) to x^(1/2).

I tried the simplify command it didn't work.


What is the Maple command to simplify  -x^(a)+x^n  to zero under the assumption that a=n?

I have attempted the following command

simplify(-x^(a)+x^n) assuming a=n;

However, it did not produce zero. The following command however produces zero


 How to do this operation using assumptions. How to inform Maple to assume that a=n.



Here is my question.


convert(diff(lambda(t-t1), t), diff)

eval(diff(lambda(t2), t2), {t2 = t-t1})


subs(lamda(t-t1) = sin(t-t1), %)

eval(diff(lambda(t2), t2), {t2 = t-t1})



I am trying to substitute value of lambda(t-t1) further but it is not taking the value.

Thanks in advance.




Hello people in the mapleprimes,

I have a question, so I hope someone give me answers to it.

I calculated for the solution of the follwing differential equation.

b:=diff(y(x),x)+a*y(x)=f(x);#where a and f(x) is not specified.

subs({f(x)=exp(x),a=2},%);where f(x) and a are specified.


The solution of the above was

y(x) = (1/3)*exp(x)+_C1/(exp(x))^2,  (A)

where please note that the second term takes

the form of fraction _C/(exp(x))^2.


On the other hand, next I calculated the following differential equation where f(x) and a are specified from the start.




y(x) = (1/3)*exp(x)+exp(-2*x)*_C1  (B)

was the obtained solution.


Each (A) and (B) are the same substantially mathematically. But, for Maple, the variable powered to minus brabra

is not the same as one over variable powered to brabra, so that (A) and (B) takes different forms, and maple will see them 

different with each other.


  Surely, with algsubs, algsubs(_C1/(exp(x))^2=exp(-2*x)*_C1,c) transforms (A) to (B).

But, I want to know whether there are some other ways than that  to modify (A) to (B).

If there are any good ways for it, I will be happy if you teach them to me.

Thanks in advance.



How can I simplify $\sqrt{1−r^2\exp(2i\theta)}$ in Maple. I could do it by hand but I need this type of simplification later for far more complicated expressions.  I allready tried to enter this as a complex number using II, but simplify(...,'symbolic') didn't simplify this expression. Any suggestion?

Hi everybody

In the following attached file that is a simple code, 2 problems occur:

1- The values of Lc1 and Lc2 are specified in the third and fourth lines of the code. when I execute Pcl10, Maple does not replace Lc1 with its value that is 0.5*L. This problem does not happen for Tcl10 and Tcl21. Why Maple does not replace Lc1 with its value that is specified at the start of the worksheet?

2- In the last line, when I try to simplify the Tcl10, Maple returns an error, while for example the first element i.e. Tcl20(1,1) can be simplified as: cos(thet1+thet2). What is the source of error?


Thanks in advance

I am curious, can simplify/siderels be executed in mod p by some equivalent Maple function call?

I am trying to simplify the square of a parameterized polynomial mod 2. My parameters are intended to be either 0 or 1. How do I accomplish this?

For example:


alias(alpha = RootOf(x^4+x+1))



z := alpha^3*a[3]+alpha^2*a[2]+alpha*a[1]+a[0]``



z2 := collect(`mod`(Expand(z^2), 2), alpha)








I would like to simplify the squared parameters modulo 2. a[3]^2=a[3], etc.

Any help would be appreciated. Elegant methods even more so!






i use the pdsolve to find the solutions of a system of partial differential equations,

but the result contains some indefinite integrals, how to simplify it further?

thank you


eq1 := {6*(diff(_xi[t](x, t, u), u))-3*(diff(_xi[x](x, t, u), u)), 12*(diff(_xi[t](x, t, u), u, u))-6*(diff(_xi[x](x, t, u), u, u)), 2*(diff(_xi[t](x, t, u), u, u, u))-(diff(_xi[x](x, t, u), u, u, u)), diff(_eta[u](x, t, u), t)+diff(_eta[u](x, t, u), x, x, x)+(diff(_eta[u](x, t, u), x))*u, 18*(diff(_xi[t](x, t, u), x, u))+3*(diff(_eta[u](x, t, u), u, u))-9*(diff(_xi[x](x, t, u), x, u)), 6*(diff(_xi[t](x, t, u), x, x))+3*(diff(_eta[u](x, t, u), x, u))-3*(diff(_xi[x](x, t, u), x, x)), 6*(diff(_xi[t](x, t, u), x, u, u))+diff(_eta[u](x, t, u), u, u, u)-3*(diff(_xi[x](x, t, u), x, u, u)), 12*(diff(_xi[t](x, t, u), u))-6*(diff(_xi[x](x, t, u), u))+6*(diff(_xi[t](x, t, u), x, x, u))-6*(diff(_xi[t](x, t, u), u))*u+3*u*(diff(_xi[x](x, t, u), u))-3*(diff(_xi[x](x, t, u), x, x, u))+3*(diff(_eta[u](x, t, u), x, u, u)), 12*(diff(_xi[t](x, t, u), x))-6*(diff(_xi[x](x, t, u), x))+2*(diff(_xi[t](x, t, u), t))+2*(diff(_xi[t](x, t, u), x, x, x))-4*(diff(_xi[t](x, t, u), x))*u+2*(diff(_xi[x](x, t, u), x))*u+_eta[u](x, t, u)-(diff(_xi[x](x, t, u), t))+3*(diff(_eta[u](x, t, u), x, x, u))-(diff(_xi[x](x, t, u), x, x, x))};



Dear all

I have such an expression in the file ( Obviously it is not a simple form. 10should be cancelled both in denominator and nominator. I have applied the function simplify(). However, it doesn't work.  I hope someone can do me a favor. That will help me a lot. Thanks

I know that the following expression (a:=6*s*sqrt(9*s^2+32)+18*s^2+32)

can be rewritten as

However, none of the following maple functions is abble to give the factored result:




Someone could help me to understand what is going on, please?

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