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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?

I was trying to make a function (or procedure) that uses the simplify command and outputed all the different types of simplify that are in the right click menu.  I never know which one to chose so until I get the hang of what they are, I wanted to see all of them at once.  Can someone set me off on the right path?


I would like to simplify a trigonometric equation that I obtain with a vectorial closure (in mechanics)

Here the equation that I would like to simplify 

eq_liaison := (-sin(p(t)+g(t))*cos(a(t))-sin(b(t))*sin(a(t))*cos(p(t)+g(t)))*l2[1]+((-sin(p(t)+g(t))*cos(a(t))-sin(b(t))*sin(a(t))*cos(p(t)+g(t)))*cos(th(t))+(-cos(p(t)+g(t))*cos(a(t))+sin(a(t))*sin(b(t))*sin(p(t)+g(t)))*sin(th(t)))*l3[1]+(-sin(a(t))*sin(g(t))*sin(b(t))+cos(a(t))*cos(g(t)))*xb[1]+sin(a(t))*cos(b(t))*yb[1]+(sin(a(t))*sin(b(t))*cos(g(t))+cos(a(t))*sin(g(t)))*zb[1]+x(t)-xp(t) = 0

Do you have ideas so as to simplify again this expression ?

This expression can still be simplified. You can find here the result expected :


I find surprising that I have so many difficulties to make trigonometric simplications with the trigonometric functions.

I attached the code 


Thank you for your help


I have a complex equation EQ, it gives me 4 answers - two complex, one negative and one positive. Which assumptions do i need to use to automatically get one answer - real and positive one?

I tried similary to what I did with real equations:
simplify(solve(EQ)) assuming real, positive
But that didn't work


is there a way to substitute funktion combinations when they have the same arguments? I want to substitute abs(exp)*abs(1,exp) with exp. Algsubs works as long as I know the excat expressions. However, I want to do this substitution for any exp. Simplify doen't work here ether.

I also tried to write my own simplification rule with no success:




abs(a)*abs(1, a)



{abs(x)*abs(1, x) = x}


simplify(f,siderel);    #expected result: a

abs(a)*abs(1, a)








Thanks in advance!

Dear maple users,

I have a lengthy formulation of function f(x) which contains some constant coefficients (A1, A2, A3 ...). I would like to simplify f(x) for functions of same above mentioned coefficients as following:

f(x) = f1(x) A1 + f2(x) A2 + ... + fn(x) An

I tried the following command:

collect(simplify(f(x)), [A1, A2, A3 ...])

Maple returns the expected form of functions but the problem is that Maple did not simplify f1(x), f2(x)... Obviously, I do not want to simplify manually : simplify(f1(x)) ... again

How can I solve this problem?


Can Maple simplify these DE's by eliminating the d/dt VL(t) by taking the derrivative of the bottom equation and substituting in the first one? 

In this question, I asked for a way to simplify an expression containing radicals. The discussion led us to that as default field for simplicfication is the Complex number system we should use assume or assuming command to simplify the radicals. However, the mothod suggested there seems to not work in this new case that I have. For details please see the attached file. The terms sqrt{u} and sqrt{u-1} should cancel in denominator.

 What Maple Does


`ϕ` := (1+sqrt(5))*(1/2)



f := (1/2)*sqrt(-(u-1)*(u+1)*(u^2-u-1))*u*(4*u-3)/sqrt(u*(u-1))



`assuming`([combine(f)], [1 < u and u < `&varphi;`])



`assuming`([simplify(f)], [1 < u and u < `&varphi;`])



`assuming`([combine(f, radical)], [1 < u and u < `&varphi;`])



`assuming`([simplify(f, radical)], [1 < u and u < `&varphi;`])




 Remark by Markiyan Hirnyk. The below content is added by the questionner on 08.02.2016 .

What Mathematica Does


I assume that I'm not providing the correct input to the simplify command to get the simplification that I want.  In particular, for the following code:

assume(n, positive);

The expression should evaluate to 0.  However, the first expression does not simplify to 0 (it does not simplify at all in Maple) while the second expression simplifies to 0.

The simplification is fairly easy for the first expression by factoring 6 and combining terms; it seems like I'm not entering the command to simplify in this way.

I have the following expression


and I want to simplify it. Eventhough that I tell Maple that u is real and greater than 1 but it does not simplify the expression. What is wrong? Please see the attached file.

a1:= f(x) :
> T1 :=simplify((taylor(a1,x=alpha,N+3))):
> E1:=subs([seq(((D@@i)(f))(alpha) = 0,i=1..m-1),f(alpha)=0,x=e[n]+alpha],T1):
> g1 :=(convert(simplify(series((E1,e[n]=0,N))),polynom));


This is a stripped down example of something I've been doing. Basically I'm building matrices which I then, using unapply, convert into functions of some variables of t.
.... but found that simplify seems to often not work as i'd wish.

mm:=Matrix([[cos(sqrt(g__1^2)*t), (-I*g__1*sin(sqrt(g__1^2)*t))*(1/sqrt(g__1^2))], [(-I*g__1*sin(sqrt(g__1^2)*t))*(1/sqrt(g__1^2)) ,cos(sqrt(g__1^2)*t)]]);

#great - simplifies as i'd expect:
simplify(mm) assuming g__1::positive;

Do the same thing but when matrix is a function of t
mmFun:=unapply(mm, t);

#the function works - gives what i'd expect
mmFun(3); mmFun(t);

#but now the simplification does not work - why the g__1 in the argument of cos does not get properly simplified?
simplify(mmFun(t)) assuming g__1::positive;

Any ideas if this is a bug? I'm using maple 2015.2 on linux 64-bit.

here is the worksheet:



as a side note once can sometimes overcome this with mapping simplify  as in :

map(simplify, resultMatrix ) assuming g__1::positive;

but this is not optimal, and sometimes does not work when i first multiply the matrix by say a vector.




Dear all,

Thank you for helping me  to generate a table of values of f(x) starting with x=0 to 100 in steps of 1, that is for x=0,1,2,3,...,100.


I tried:

f:=x->2*sqrt(3)*a1*a2*(sum(pochhammer(1/3,k)*3^k*x^(3*k)/(3*k)! ,k=0..infinity)*sum(pochhammer(2/3,k)*3^k*x^(3*k+2)/(3*k+2)!  ,k=0..infinity)-sum(pochhammer(2/3,k)*3^k*x^(3*k+1)/(3*k+1)!  ,k=0..infinity)*sum(pochhammer(1/3,k)*3^k*x^(3*k+1)/(3*k+1)!  ,k=0..infinity));


But I the result is amazing.... I don't understand the problem.



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