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Dear all,

I wold like to find the solution of the next system of two equations with three unknowns but we assume that the unknows are positive integers. How the following code can work. Many thanks

 

 

 

> restart;
> assume(J, integer, J >= 0);
> assume(A, integer, A >= 0);
> assume(T, integer, T >= 0);
> eq1 := J+10*A+50*T=500;
   eq2 := J+A+T = 100;
  solve( {eq1,eq2},{J,A,T});

 

With the following equation

eqn:=y=1/2+(1/2)*erf((1/2)*sqrt(2)*(x-mu)/sigma)-exp(-lambda*(x-mu)+(1/2)*lambda^2*sigma^2+ln(1/2-(1/2)*erf((1/2)*sqrt(2)*(lambda^2*sigma^2-lambda*(x-mu))/(lambda*sigma))));

and with

x:=solve(eqn,x) assuming sigma > 0, lambda > 0;

I got the following solution

x := -(1/2)*(-lambda^2*sigma^2-2*lambda*mu+2*RootOf(-exp(_Z)*erf((1/4)*sqrt(2)*(lambda^2*sigma^2+2*_Z)/(lambda*sigma))+exp(_Z)+erf((1/4)*sqrt(2)*(-lambda^2*sigma^2+2*_Z)/(lambda*sigma))+2*y-1))/lambda;

In order to get rid of RootOf I gave the command:

allvalues(%);

However, RootOf did not disappear. How should I proceed? 

 

hi...how i can gain result for solve three equations,in which term '' Root of'' dont appear?

thanks

root.mw

restart; Q1 := aa*(y-x)

aa*(y-x)

(1)

Q2 := -ll*x*z+bb*x

-ll*x*z+bb*x

(2)

 

-ll*x*z+bb*x

(3)

Q3 := -cc*z+hh*x*x+kk*y*y

hh*x^2+kk*y^2-cc*z

(4)

SOLL := solve({Q1, Q2, Q3}, {x, y, z})

{x = 0, y = 0, z = 0}, {x = RootOf((hh*ll+kk*ll)*_Z^2-bb*cc), y = RootOf((hh*ll+kk*ll)*_Z^2-bb*cc), z = bb/ll}

(5)

``

 

Download root.mw

Hi all,

I seem to be quite stuck on figuring out how to leave certain letters (e.g. planck's constant h) inside the equation without having to assign it as some particular number. 

What I am trying to do is find the value of a when the following equation is at a minimum:

E = (a*(h^2)/2m) + 0.3989422804/sqrt(a)

Here h and m are what I want to set as constants without actually setting them to h := 1 because I want a in terms of h and m. I have already found the derivative dE/da:

((h^2)/2m) - 0.1994711402/a^(3/2)

But I cannot use fsolve to find the value of a at the minimum because it keeps saying that h and m are variables and unsolved for.

Any help would be greatly appreciated.

How to solve following recurrence equation:

 

a(0)=2;

a(n+1)=a(n)+a(n)^2

 

I tried,but it doesn't work.

How to find the sequence an ?

Maple_worsheet.mw

Mariusz Iwaniuk

hi dear

i wright this formula for solving by two variable but it dont solve and i recicive this message (warning solutions may have been lost)

please help my

thanks

 

i have attcahed my ode with complex bvp

can anyone solved mine

NULL

restart

with(plots):

NULL

Eq1 := (11-10*d)*(diff(h(eta), eta))+2*f(eta) = 0;

(11-10*d)*(diff(h(eta), eta))+2*f(eta) = 0

 

(11-10*d)*(diff(diff(f(eta), eta), eta))-h(eta)*(diff(f(eta), eta))-f(eta)^2+g(eta)^2 = 0

 

diff(diff(g(eta), eta), eta)-h(eta)*(diff(g(eta), eta))-2*f(eta)*g(eta) = 0

 

diff(p(eta), eta)+2*(diff(f(eta), eta))-2*f(eta)*h(eta) = 0

(1)

NULL

NULL

`Vλ` := [0.5e-1, 1.5, 1.5]:

etainf := 3:

bcs := h(0) = 0, p(0) = 0, (D(f))(0) = lambda*f(0)^(4/3)/(f(0)^2+(1-g(0))^2)^(1/3), (D(g))(0) = -Typesetting:-delayDotProduct(lambda*f(0)^(1/3)*(1-g(0)), 1/(f(0)^2+(1-g(0))^2)^(1/3)), f(etainf) = 0, g(etainf) = 0;

h(0) = 0, p(0) = 0, (D(f))(0) = lambda*f(0)^(4/3)/(f(0)^2+(1-g(0))^2)^(1/3), (D(g))(0) = -f(0)^(1/3)*(1-g(0))*lambda/(f(0)^2+(1-g(0))^2)^(1/3), f(3) = 0, g(3) = 0

(2)

NULL

dsys := {Eq1, Eq2, Eq3, Eq4, bcs}:

for i to 3 do lambda := `Vλ`[i]; dsol[i] := dsolve(dsys, numeric, continuation = d); print(lambda); print(dsol[i](0)) end do

Error, (in dsolve/numeric/bvp) singularity encountered

 

NULL

NULL

NULL

 

Download compre1.mw

and attch back

I am trying to find the root of an equation. The problem is, I keep getting the error

"Error, (in fsolve) Can't handle expressions with typed procedures"

and 

Warning, solutions may have been lost


whenever I try to solve it. Anyone have any ideas? My worksheet is here:  1.mw

Hey all,

 

The title is probably very poorly explained and doesn't make much sense at all, but here goes nothing:

I define at the start of my .mw file that M:=1, but I need to be able to change it in order to run multiple different iterations.

So what I've come up with so far is a way to get a variable ammount of equations named "eqc(1,3,5,...)" The number of equations I get is equal to the M defined in the beggining. How would I go about solving this?
To give you an idea of something to work with:

So basically I'd need to solve as many of these eqc equations as I get. If I change M to, lets say 30, I'd need to solve 30 equations. This solve option above doesn't work and I've messed around with Vectors and Matrixes but I honesly have no idea what I'm doing there, so I thought best to seek out help.

 

Thanks in advance, Rafael.

i need this equation solve by maple

eql := alap*[(la(a)^4+6*la(a)+1+alap*(la(a)^4-1)/difl1)*la(a)*j(a)-((la(a)^2+1)*(1+2/difl)+la(a)^2-1)*intdj/a^4+[(4*(la(a)^4-1))/difl-8*la(a)^2]*intdj/a^2]-(4*(la(a)^2+1))*intdj/a^4+(4*(la(a)^4-1))*intdj1/a^2 = 0;

I used SOLVE to solve an inequality. The result shows things like this: 

How can I read the upper bound to a variable? 

THanks!

 For solving polynomial systems I used RootFinding[Isolate]. But after discussing the question http://www.mapleprimes.com/questions/211774-Roots-Of--Expz--1
I decided to compare Isolate and evalf(solve ([...], [...])). It seemed to me that solve some convenient. The only if in the equation there are integers as a real, they should be recorded with a decimal point. (For real solutions of this procedure should be used with (RealDomain).)  Examples:

SOLVE_ISOLATE.mw

I wonder why then the need Root Finding [Isolate]?

Maple 15.

I have a set of equations I can solve manually, but, solve fails.

restart;
eq1 := tgtX[1] = 0;
eq2 := tgtY[2] - y[2]    = m[2]*(tgtX[2]-x[2]);
eq3 := tgtY[3] - y[3]    = m[3]*(tgtX[3]-x[3]);
eq4 := tgtY[4] - y[4]    = m[4]*(tgtX[4]-x[4]);
eq5 := tgtY[2] - tgtY[1] = vy*t[2];
eq6 := tgtY[3] - tgtY[1] = vy*t[3];
eq7 := tgtY[4] - tgtY[1] = vy*t[4];
eq8 := tgtX[2]           = vx*t[2];
eq9 := tgtX[3]           = vx*t[3];
eq10:= tgtX[4]           = vx*t[4];
#
# solve the equations
eqs  := {eq1,eq2,eq3,eq4,eq5,eq6,eq7,eq8,eq9,eq10};

solvx := solve(eq10,vx);
solvy := solve(eq7,vy);
sol1  := subs(vx=solvx,{eq8,eq9});
sol2  := subs(vy=solvy,{eq5,eq6});

soln  := subs(sol1,{eq2,eq3});

soln  := subs(tgtY[2]=solve(sol2[1],tgtY[2]),soln);
soln  := subs(tgtY[3]=solve(sol2[2],tgtY[3]),soln);
soln  := subs(tgtY[1]=solve(soln[1],tgtY[1]),soln[2]);
soln  := solve({eq4,soln},{tgtX[4],tgtY[4]});

# this returns empty solution
solve(eqs,{tgtX[4],tgtY[4]});

Any ideas?

Tom Dean

Dear all,

I have a question: how to compute the roots of exp(z) = -1 with z in C? 

I tried: 

fsolve( exp(z) = -1, z, complex );

But it only gives one root (0.1671148658e-3+4.934802220*10^9*I) which does not even seem to be correct. I would prefere smth like z_n = I*(2*n-1)*pi or at least multiple roots...

By using

solve(exp(x) = -1, x);

it returns I*Pi.

 

MATLAB MuPAD gives the desired result:


solve(exp(x) = -1, x)

(PI*I + 2*PI*k*I, k in Z)

 

 

Thanks!

Hi,

I have been trying to solve the following equation with respect to y, but I have not been successful. In fact, I always get answer RootOf(...). I should mention that all variables and parameters are real non-negative. I have also tested with "assume", but it did not help. Any suggestion would be appreciated. 

with(RealDomain):

eq := -((y-b)*mu-y)*x^beta*alpha+y^beta*varepsilon*(x-a) = 0

-((y-b)*mu-y)*x^beta*alpha+y^beta*varepsilon*(x-a) = 0

(1)

solve(eq, y)

RootOf(-x^beta*alpha*b*mu+x^beta*alpha*mu*_Z-x^beta*alpha*_Z+_Z^beta*varepsilon*a-_Z^beta*varepsilon*x)

(2)

remove_RootOf(%)

-x^beta*alpha*b*mu = 0

(3)

``

``

Download Equation.mw

 

Thanks.

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