Items tagged with solve

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updated:
P := evalm(p2 + c*vector([cos(q1+q2+q3), sin(q1+q2+q3)]));
 
restart:
with(Groebner):
p1 := vector([a*cos(q1), a*sin(q1)]);
p2 := evalm(p1 + b*vector([cos(q1+q2), sin(q1+q2)]));
P := evalm(p2 + c*vector([cos(q1+q2+q3), sin(q1+q2+q3)]));
Pe := map(expand, P);
A := {cos(q1) = c1, sin(q1) =s1, cos(q2)=c2, sin(q2)=s2, cos(q3)=c3, sin(q3)=s3};
P := subs(A, op(Pe));
F1 := [x - P[1], y - P[2], s1^2+c1^2-1, s2^2+c2^2-1, s3^2+c3^2-1 ];
F2 := subs({a=1, b=1, c=1}, F1);
 
g2 := Basis(F2, plex(c3, s3, c2, s2, c1, s1));
LeadingTerm(g2[1], plex(c3, s3, c2, s2, c1, s1));
LeadingTerm(g2[2], plex(c3, s3, c2, s2, c1, s1));
LeadingTerm(g2[3], plex(c3, s3, c2, s2, c1, s1));
LeadingTerm(g2[4], plex(c3, s3, c2, s2, c1, s1));
LeadingTerm(g2[5], plex(c3, s3, c2, s2, c1, s1));
LeadingTerm(g2[6], plex(c3, s3, c2, s2, c1, s1));
LeadingTerm(g2[7], plex(c3, s3, c2, s2, c1, s1));
LeadingTerm(g2[8], plex(c3, s3, c2, s2, c1, s1));
LeadingTerm(g2[9], plex(c3, s3, c2, s2, c1, s1));
 
                                   1, c1
                               2       2    2   2
                           16 y  + 16 x , s1  s2
                                           2
                                 8 x, c1 s2
                                2      2    2  
                             2 y  + 2 x , s1  c2
                                 2 x, c1 c2
                            3            2        
                         2 x  - 2 x + 2 y  x, s2 c2
                                        2
                                   1, c2
                                   2 x, s3
                                    2, c3
originally i think
g2[1], g2[7], g2[9] have single variables c1, c2, c3 respectively
can be used to solve system
 
but without x and y, these equations can not be used
if choose leading term has x and y , but there is no single variable s1 or c1.
 
originally expect solve as follows
g2spec := subs({x=1, y=1/2}, [g2[3],g2[5],g2[6]]);
S1 := [solve([g2spec[1]])];
q1a := evalf(arccos(S1[1]));
q1b := evalf(arccos(S1[2]));
S2 := [solve(subs(s1=S1[1], g2spec[2])), solve(subs(s1=S1[2], g2spec[2])) ];
q2a := evalf(arccos(S2[1]));
q2b := evalf(arccos(S2[2]));
S3 := [solve(subs(s1=S2[1], g2spec[2])), solve(subs(s1=S2[2], g2spec[2])) ];
q2a := evalf(arccos(S3[1]));
q2b := evalf(arccos(S3[2]));
 

I am trying to solve a system of equations (I'm using MapleTA< but I'm pretty sure that this applies to any Maple product).  I have successfully solved the system, and obtain a set of solutions, which has name Soln.  I can access the element Soln[1], which is an expression:

vn2 = 12/7

Now, I just want that 12/7, as a decimal.  I try evalf(Soln[1]), but again I end up with vn2 = 12/7.  How do I get the decimal number out, without it being an expression?

Does Maple have build-in function, which when given an expression that depends on x and y, will separate it to a product of two functions, one that depedns on x only and the other that depends on y only?

The input mathematical expression is known to be seperable.

For example, If the input is

((3*y + y^2)*3*x)/(x + sin(x))

Then I'd the Maple function to take the above and return list or set of two parts  {(3*y + y^2)   ,     3*x/(x + sin(x) } (if it can't separate it, it can return null).

The API can be something as  

 f,g = find_product_functions(expression,[x,y])

Something like this is used on determining for example if RHS of first order ODE is separable in order to solve it more easily.  collect() does not really work for this. So Maple allready does this internally in its ODE solver when it checks if ODE is separable or not. But is the function available for users?

 

sys1:=-.736349402144656384 = -1.332282598*10^12*(-.99999999999999966)^po1-1.332282598*10^12*(-.99999999999999966)^po2-.735533633151605248*Resid;

sys2:=.326676717828940144 = 1.331567176*10^12*(-.99999999999999966)^po1+1.331567176*10^12*(-.99999999999999966)^po2+.325144093024965720*Resid;

sys3:=.590327283775080036 = -1.072184073*10^9*(-.99999999999999966)^po1-1.072184073*10^9*(-.99999999999999966)^po2+.589610307487437146*Resid;

Minimize(sys1, {sys2,sys3},assume = nonnegative);

complex value encountered;

I found this error extremely confusing when using the solve function:

 

Error, (in Engine:-Dispatch) cannot determine if this expression is true or false: 1000 < 5^(1/2)
 

Hi, i encountered this, error, and the link to the help page was broken.

Error, (in RootOf) expression independent of _S000100
 

Trying to solve:

solve (arctan((2*x^2-1)/(2*x^2+1)) = 0, x);

The answer I get is the original function:

 
            arctan((2*x^2-1)/(2*x^2+1))

 

This example is from the Maple book by Keck, and he shows the Maple V answer as

1/2 sqrt(2) -1/2 sqrt(2)     

Suggestions?

I want to find the first positive solution of the system of trigonometric equations inside the loop.

The solutions are in the form of "d=number*_Z +number" but I need one exact solution to use it for next run of the loop.
 

restart;
L[0]:=0:
for i from 1 by 1 to 3 do
assume(0<d[i], d[i]<1):
assume(-0.01<a[i], a[i]<0):
L[i]:= L[i-1]+ d[i]:
sys[i]:={Re((-80*Pi*I*a[i]/((a[i]+1)^3))*exp(4*Pi*I*L[i])) = -0.4, Im((-80*Pi*I*a[i]/((a[i]+1)^3))*exp(4*Pi*I*L[i])) = 0.8}:
solve(sys[i], {a[i],d[i]}, useassumptions = true,AllSolutions=true):
end do;
 

These are the solutions:

d[1] = 0.03689590440 + 0.5000000000 _Z1

d[2] = -1.000000000 d[1] + 0.03689590440 + 0.5000000000 _Z2

d[3] = -1.000000000 d[1] - 1.000000000 d[2] + 0.03689590440 + 0.5000000000 _Z3

 

Hey friends

I want to solve this relation with respect to "M"analytically but maple answer me: "Warning, solutions may have been lost"

How I can solve this problem and get to an analytical solution. It must be noted -1<w<-1/3. we can fix "w" with any value inn this interval. It be accepted any solution for any fixed "w".

Thank you

Analytically_solution.mw

Hi,

I want to solve this equation with respect to M, But Maple answer me: "solution may have been lost"

 

How I can solve this equation?

Thanks guyz

Solutionmayhavebeenlost.mw

Greetings Sirs,

I have recently aquired Maple for some mathematics, and being a new user, I basically google for everything at the moment.

While it has gone well so far, I seem to have hit a bump that I cannot figure out.

I have a function: f(x)=3.2+0.4sin(1.25x), 0<x<5

Trying to find the places where "f(x)=3.5" would normally be done with the equation "3.5=3.2+0.4sin(1.25x)", and when I solve for the equation in Maple I get a solution too.

Problem is though, I know there is supposed to be multiple solutions. Having used wolframalpha, and being capable of seeing the plot in Maple, I know there is two points within the period "x=0..5" that is the solution.

But when I try to solve the equation, I get only one solution per solve, and the second solve doesn't make much sense for me. These are what I use:

As you can see, in the first solve the entire function is being taking into consideration, yet I only get one solution... In the second solve I have tried specifying a period, but I still only get one solution.

Basically any help here is appreciated, because from what I understand, having read google, the solve command or fsolve command is supposed to give multiple results if they are there.

With appreciation,
Ciesi

(Edit: Image size changed)

I want to solve this equation with assumptions!!!

restart;
assume(d::real, d>0):
assume(a::real, -0.01 < a, a < 0):
sys:={-800*Pi*a*cos(6.557377048*Pi*(3.470797713+d))/(a+1)^3 = -.9396060697, 800*Pi*a*sin(6.557377048*Pi*(3.470797713+d))/(a+1)^3 = -.3238482794};
solve(sys, {a,d},useassumptions=true, AllSolutions=true);

one of the solutions has true "a" but "d" is wrong, I want one true solution!

I have the following equality:(N*(P-p[th])/K+p[th])/(2+2*N*Gamma/K+(1+N*Gamma/K)^2/(N*(P-p[th])/K+p[th])) = p[th]/(2+1/p[th]).

How can I get the solution step by step? I have read the forums around and tried few of the tutors, but most of them are either for single variable or I have to specify values for the variables.

can Maple solve bessel function by itself?

Or it seems i need to work out the solving process on myown?

my function is just like this, a simple one

please let me know if yo have any idea about it.

best regards

memdream

 

v=u+at                      (1)
s=u*t+1/2*a*t^2        (2)

below 3 equations, can substitute  (1)  into it to form (2)
s=1/2*(u+v)*t       (3)
v^2=u^2+2*a*s    (4)
s=v*t-1/2*a*t^2    (5)

can these 5 equations be considered as a solution set of solve function?

or

is only first 2 equations be a solution set?

if so, number of equations less than 5 variables, is there something missing?
 

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