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Dear Maple Users,

I'm solving quite a complicated task, so I tried to simplified an example.

There is an equation:

SOL := fsolve(Nz+int(int(StrssCctXY(x, y), x = -(1/2)*b .. (1/2)*b), y = -(1/2)*h .. (1/2)*h) = 0, {C1 = -(1/2)*h .. (1/2)*h})

 StrssCctXY(x,y) is piecewise function containing C1 variable, to solve an equation I had to use assumptions on C1 via assume(C1<num1, C1>-num2) command, after that C1 becomes C1~;

Dear all

I tried to intgrate this problem

P:=(b8*Omega^8+b6*Omega^6)/(a10*Omega^10+a8*Omega^8+a6*Omega^6+a4*Omega^4+a2*Omega^2+a0);

using these limits

Omega_r1:=beta*Omega_c;
Omega_r2:=(2-beta)*Omega_c;

assume((2-2*beta)*Omega_c>0);
assume(a0>0,a2>0,a4>0,a6>0,a8>0,a10>0,b6>0,b8>0,beta>0);
P_I:=evalf(int(P,Omega=Omega_r1..Omega_r2)):

but the results was not useful I got (RootOf)

restart;
a:='a';
                               a
assume(a>0);
H:=a;
                               a
subs(a=b, H);
       ...

Dear,

I have problem that after using assume(...) and then use label such as (1.1)... then the assumptions are lost. Is it a bug?

Below is my attached worksheet (I used Maple 13 , I don't know if later version is better?)

label_and_assume.mw

Actually , the lost of assumptions will cause some problems such as make integration become unevaluated.... and will take you a hard time to debug...

After I excute : x^2+2*x+1;   and right click on the output then I saw the CompleteSquare option in the Context Menu but after I write  assume(x>0); and then right click on the output of x^2+2*x+1; then I didn't see this option (and there are only a few options appeared in the Context Menu compared to the case before using assume()). 

Sometimes It cause some inconvenience for me since I used Context Menu on outputs from previous lines in my worksheet...

How may I program the following with Maple?

(1) Define the function:

H(p) := p * c1 + p^2 * c2 + p^3 * c3  + ......

(2) Now construct the following expression:

(1-p) [ v''' + 1 ]  = H(p) [ v''' - 25 * v' + 1]

Here, v is some funciton of x.

(3) Now assume: v = u0 + u1 * p + u2 * p^2 + ...

and substitute this into the expression defined at the point (2).

Hello,

I would like to write some functions and algebra that work with dual numbers.

http://en.wikipedia.org/wiki/Dual_number 

http://en.wikipedia.org/wiki/Dual_quaternion 

I have not found a library that supports this.

The basis of dual numbers it is epsilon^2=0, similar to i,j,k in complex algebra where i^2=-1,i*j=-1 etc.

I would like to be able to enter an assumption into maple indicating one variable is much greater than another.  I would then like maple to eliminate the smaller variable term when it is added together with the much greater term.  However when I use the >>> symbol in maple along with assuming I get an error. 

> eqn1 := y = a*t+b*t;
                        ...

How I can say that an object,say k, is a constant(it may be positive or negative) inside assume()?

assume(k::??should I say)

pls also tell me what kind of help pages I should see for this kind of "type inside assume()" problems. 

Thanks.

Hi,

I encountered the following problem in Maple 14

expr1:=sqrt(a)

assume(a>0)

expr2:=sqrt(a)

additionally(a>1)

subs(a=2,expr1) #gives correct answer of sqrt(2)

subs(a=2,expr2) #does not substitute a=2 in expr2

It seems the last statement with expr2 does not work as expected, I wondered why this behaviour is so, because it does work with expr1. The added assumption does not invalidate the first...

Hi,

I have this problem when I use assume and assign a value to an assumed parameter.

Example:

restart:assume(beta>0):

g:=int(Sum(5*sin(n*beta*x),n=1..5),x) ;

beta:=1;

g;

Normaly beta:=1 is replaced into g. The result obtained is function of beta and beta:=1 is not replaced.

Why beta:=1 is not replaced in g ?

Thanks for suggestion

Hello-

I am trying to find an analytic solution to a cubic equation of the form ax^3 + cx + d, where a, c, and d are rather complicated coefficients.  If I simply use the solve command, it gives the solutions assuming that the discriminant is negative (that is, one real and two imaginary solutions).  However, I want the solutions in the case that the discriminant is positive (the solutions have a different functional form; it's not the case that the imaginary solutions simply become real...

It still seems that the original post won't accept new replies, so I'm starting a new post.

 It seems I can't add a response to this message, so I added some detail to it.

Dear All!

Could you explain how the coulditbe() function works?

> assume(q, integer, w, integer); about(q, w); coulditbe(q/w < 1); coulditbe(q < w);
Originally q, renamed q~:
  is assumed to be: integer

Originally w, renamed w~:
  is assumed to be: integer

                              FAIL
                              true

BR,

Zoltán Faigl