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Using Maple 18, I solved for minimum and maximum price. Instead of using fsolve I wanna use procedure programming structure in order to get the same results. How can I do it?

min_sol := fsolve([bc_cond, slope_cond, x[G, 1] = w[aggr, 1]], {p = 0 .. 1, x[G, 1] = 0 .. w[aggr, 1], x[G, 2] = 0 .. w[aggr, 2]}); p_min := subs(min_sol, p); max_sol := fsolve([bc_cond, slope_cond, x[G, 2] = w[aggr, 2]], {p = 0 .. 1, x[G, 1] = 0 .. w[aggr, 1], x[G, 2] = 0 .. w[aggr, 2]}); p_max := subs(max_sol, p);
{p = 0.3857139820, x[G, 1] = 127.8000000, x[G, 2] = 38.99045418}
0.3857139820
{p = 0.8841007104, x[G, 1] = 44.30160890, x[G, 2] = 164.2000000}
0.8841007104

Hello every one,

Is any one knows how to solve the following inequality with assumptions that all parameters are real positive and k<1 and delta > c*alpha

(1/2)*((alpha*k^2-3*alpha*k-2*beta)*sqrt(delta^2*(k-1)*(k-2)*(c*alpha-delta)^2)-k*delta*(alpha*k^2-3*alpha*k+2*alpha-2*beta)*(c*alpha-delta))/(delta^2*(alpha*k^2-3*alpha*k-2*beta))<0

I tried the following code but it  dosn't make sense:

u:=(1/2)*((alpha*k^2-3*alpha*k-2*beta)*sqrt(delta^2*(k-1)*(k-2)*(c*alpha-delta)^2)-k*delta*(alpha*k^2-3*alpha*k+2*alpha-2*beta)*(c*alpha-delta))/(delta^2*(alpha*k^2-3*alpha*k-2*beta))

solve({u < 0,alpha > 0, beta > 0, c > 0, delta > 0, delta > c*alpha, k > 0, k < 1, })

In fact I want to know under which circumastances the above inequality is negative.

THX

What is the best way to solve for the simple equation X^2+y^2=1[m]^2 symbolically for either x or y? I actually have a huge list of equations and want to solve the group but my problem boils down to the issue here where I get two possible solutions though using the assumption one is clearly negative and the assumption used should exclude negative results (see attempt below). Also solve doesn't seem to work with units either...  any ideas? Can I give the variables units in a meaningful way?

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restart;

with(RealDomain);
f := x^2+y^2 = 1;

                            x^2+y^2 = 1

assume(y > 0)

a := y > 0

y1 = solve(f, y, useassumptions = true)

                          y1 = (sqrt(-x^2+1), -sqrt(-x^2+1))

 

y2 = solve({a, f}, y)

                          y2 = ({y = sqrt(-x^2+1)}, {y = -sqrt(-x^2+1)})

-------------------------------------------------------------------------------------------

Why is y = -sqrt(-x^2+1) a solution?

Also, how do I use units when trying to solve 

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restart;
f := x^2+y^2 = Unit('m')^2;
                           x^2+y^2 = Unit('m')^2

assume(x > 0);
assume(y > 0);
d = solve(f, y, useassumptions = true);

Error, (in Units:-Standard:-+) the units `m^2` and `1` have incompatible dimensions

---------------------------------------------------------------------------------------------

 

THANKS!

Can anyone coax Maple to solve this reccurence relation? It seems harmless enough but Maple is struggng a bit with "hypergeomsols."

f := c -> (2*n-c)*f(c-1) - (c-1)*n*f(c-2);

f(0) := 1;

f(1) := 2*n-1;

 

Hi, I'm new to Maple and was trying to use it to solve 3 equations with 3 unknowns, in terms of another 2 parameters. This is what I put in, and the error that came up:

solve({(1-x)/(1+b) = b*y, (1-y)(1-a)/(-a*b+1) = b*z, (1-z)(1-b)/(-a*b+1) = a*x}, {x, y, z}); 

Error, (in SolveTools:-LinearSolvers:-Algebraic) unable to compute coeff

I want to solve the equations to get x,y,z in terms of a,b but I don't understand the error coming up - have I done something wrong or is it because not all the variables appear in all of the equations? As there are 3 equations and 3 unknowns there is a solution, but I want to check the answer I found on paper with something (the algebra got a bit messy!)

Any help greatly appreciated! :)

So I am trying to solve a given ODE using calculated christoffel symbols found by maple, and in order to get the correct christoffel symbols, I need my function to be r=x-x_s, where x is not a function of t. However, I then have to solve the ODE where x is a function of t. Maple used the r value to find the christoffel symbols which has x in it, and now I want to find the origonal function of x(t), but I can't have x and x(t) in the same ODE. If I change r=x-x_s to r=x(t)-x_s, I get the wrong christoffel symbols. How can I solve my ODE?

Hi,

I'm pretty new into MAPLE andI'm trying to get into it.

I have these four equations:


eq1:=1.6*10^(-7)*R*sin(t)-4.4*10^(-14)*R^2*cos(t)*sin(t)-1.6*10^(-14)*R^2*cos(t)^2+4.2*10^(-14)*R^2-1.3+2.1*10^(-9)*R*cos(t)=0; eq2 :=8.3*10^(-8)*R*sin(t)-1.2*10^(-13)*R^2*cos(t)*sin(t)-2.9*10^(-44)*R^2*cos(t)^2+7.1*10^(-14)*R^2-1.3+8.3*10^(-8)*R*cos(t)=0; eq3 :=8.3*10^(-8)*R*sin(t)-1.2*10^(-13)*R^2*cos(t)*sin(t)-2.2*10^(-44)*R^2*cos(t)^2+7.1*10^(-14)*R^2-1.3+8.3*10^(-8)*R*cos(t)=0; eq4 :=2.1*10^(-9)*R*sin(t)-4.4*10^(-14)*R^2*cos(t)*sin(t)+1.6*10^(-14)*R^2*cos(t)^2+2.6*10^(-14)*R^2-1.3+1.6*10^(-7)*R*cos(t)=0;
 

restart:

Eq1:=x=(phi*(theta[m]-1/x)/theta[m]+(1-phi)/2);

solve({Eq1},x);

x := (1/4)*(phi*theta[m]+theta[m]+sqrt(phi^2*theta[m]^2+2*phi*theta[m]^2-16*phi*theta[m]+theta[m]^2))/theta[m];

Now plotting the first root

p1:=plot(subs(phi=0,x),theta[m]=0..14,color=red):
p2:=plot(subs(phi=0.1,x),theta[m]=0..14,color=green,linestyle=2):
p3:=plot(subs(phi=0.5,x),theta[m]=0..14,color=blue,linestyle=3):
p4:=plot(subs(phi=1,x),theta[m]=0..14,color=black):
display({p1,p2,p3,p4},axes=boxed,view=[0..14,0..1]);

Note: I'm unable to reproduce the following plot. Any idea?

 SA.mw

hello

I have a system of equation. to solve i using DirectSearch pakage. i think in answer DirectSearch use rounding number but in answer it's second and third digit after the decimal point is important. please help me. bbw.mw

if b and c increase 0.01 and -0.01 it's ok. another main question is why by increasing the intensity answer don't change.

Hello everybody,

 

I guess my problem is a very little one, but I don't have any idea how to solve the following equation:

The equation is defined as follows:

 

gl1 := a = (arctan(b^2/(z*sqrt(b^2+b^2+z^2)))+b^2*z*(1/(z^2+b^2)+1/(z^2+b^2))/sqrt(b^2+b^2+z^2))/(2*pi);
                                       
z := solve(gl1, z);

Warning, solutions may have been lost

Does anybody of you have an idea how to solve this problem?

 

Thanks so much in advance!

Here's my problem,

I'm this line of code

result:=solve({x+y=1,x+2y=4},{x,y})
and the value of result is
{x=-2,y=3} and the value of result[1] is x=-2.

I would like the value of result to be {-2,3} for I need to use those output later. Is there any way to put those output in a list?

Hi,

 

  Suppose I have the following code

 

***

f[1]:=x1^2+x2;
f[2]:=-x2^3+x1-2;

solve({f[1], f[2]}, {x1, x2}):
a,b:= eval([x1, x2], %[1])[]:
a, b;

***

 

It could work and give me "1 and -1". 

 

I would like to have a general subroutine, solving f[1]...f[n]. I cannot input  "solve({f[1], f[2],....,f[n]}, {x1, x2}):" just like that. Is ther any way to use "solve" or similar module to handle arbitary number of variables?

I am trying to use solve to determine 5 unknowns from 12 equations each with seperate data points); however, maple requires you to have equal number equations and variables. Is there any way around this?

There are two ways of expressing the solution to a cubic equation, one of them uses cos and arccos [1]. How do I / is there a way to ask Maple to get this form?

More generally, can Maple be instructucted to solve equations using trig identities?

[1] http://en.wikibooks.org/wiki/Trigonometry/The_solution_of_cubic_equations

Hi there,

I've got the following differential equation system:,

dU/dt = delta·dotD -lambda·U - kappa·U^2
dL/dt = (1-phi)·lambda·U + 1/4 ·kappa·U^2


being phi, delta, kappa, lambda, kappa some fixed parameters of the system, and where dotD (the derivative wrt time of a function D), which is defined a piecewise funtion:

dotD(t)=1/(3·T1)·DT for t in [0,T1]

dotD(t)=2/(3·(T2-T1-T))·DT for t in [T1+T,T2]

where T and DT are also known, and T1 approaches 0, and T2 approaches T1+T.

Setting the equation system in Maple and trying to solve it, gives a NULL result. However, trying to solve each piece separately seems to work fine.

Why is this?

 

Furthermore, taking limits for the [T1+T,T2] part (having solved each piece separately) yields an invalid limits point error. Ain't the possibility to take limits for both parameters at the same time?

Any ideas?

 

This is the Maple worksheet: MaplePrimes_LQ_model_solve.mw

Thank you.

jon

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