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f := x^4-c1*x^3+c2*x^2-c3*x+c4;

restart;
ferrai := -x1^3*x2*x3*x4-x1^2*x2^2*x3^2-x1^2*x2^2*x4^2-x1^2*x3^2*x4^2-x1*x2^3*x3*x4-x1*x2*x3^3*x4-x1*x2*x3*x4^3-x2^2*x3^2*x4^2+x1^2*x2*x3*y+x1^2*x2*x4*y+x1^2*x3*x4*y+x1*x2^2*x3*y+x1*x2^2*x4*y+x1*x2*x3^2*y+x1*x2*x4^2*y+x1*x3^2*x4*y+x1*x3*x4^2*y+x2^2*x3*x4*y+x2*x3^2*x4*y+x2*x3*x4^2*y-x1*x2*y^2-x1*x3*y^2-x1*x4*y^2-x2*x3*y^2-x2*x4*y^2-x3*x4*y^2+y^3;
coeff(ferrai, y^3);
coeff(ferrai, y^2);
coeff(ferrai, y);
res := ferrai - coeff(ferrai, y^3)*y^3 - coeff(ferrai, y^2)*y^2 - coeff(ferrai, y)*y;

c2 := -coeff(ferrai, y^2)/coeff(ferrai, y^3);
sys1 := c1*c3 - 4*c4 = coeff(ferrai, y)/coeff(ferrai, y^3);
sys2 := -c3^2-(c1^2)*c4+4*c2*c4 = res;
solve([sys1, sys2],[c1,c3,c4]);

number of equations is not enough, is it possible to find back c1,c3,c4?

though c2 is easy to know

How to solve the inequality

,

assuming a::real ?

Of course, with Maple. I'd like to demonstrate the difficulties, solving

>solve(log[2*abs(x-a)](abs(x+a)+abs(x-a)) < 1, x) assuming a > 0, a < 1/2

.

The correct answer under the above restrictions is

{x  > 0, x  <  a} union {a  <  x, x <  a + 1/2} union { -infinity < x, x < a - 1/2}.

This is a problem from Lviv math school olympiad '2016.

 

 

Hello people in mapleprime

 

maxi:=proc(obj,expre,x,y)
local eq1,eq2,eq3,lagrangean;
lagrangean:=obj+lambda*expre;
eq1:=diff(lagrangean,x)=0:
eq2:=diff(lagrangean,y)=0:
eq3:=diff(lagrangean,lambda)=0:
solve({eq1,eq2,eq3},{x,y});
end proc;

Using the above procedure, next code will not return the values of a and b, or a[2] and b[2].

Can you teach me the reason why?

maxi(a*b,z-a-b,a,b);

maxi(a[2]*b[2],z-a[2]-b[2],a[2],b[2])

Thanks in advance.

 

taro

To saving my time i create a function like this:

LTTS:=proc(ff)

local ll,r,r1,r2,r3;

ll:=rhs(ff)-lhs(ff);
solve01(ll):
solve02(ll):
solve03(ll):
solve04(ll):
solve05(ll):
end:

 But when i run it, it is only give me the result of solve05(ll). when i run alone from solve01 to solve05 it still give me result.

I also try:

myfunction:solve01(ll):solve02(ll):solve03(ll):solve04(ll):solve05(ll):

It still the same :(

 of equations from a set of solution after solve ?

what are these ways?

 

[a = s/RootOf(_Z^2-s^2+s), b = -RootOf(_Z^2-s^2+s)/s, c = RootOf(_Z^2-s^2+s)]

hi.after calculate Determinant of matrix  and gain value omega'' ω'' by fsolve rule ,when substuting result (ω) in matrix (q) and calculate Determinant again, this value is not zero!!!! may i use LUDecomposition?determinan.mw

Hello,

I would like to obtain all the solutions of a equation.

Here an extract of my code:

v:=unapply(H*sin(w*t),t);
L:=0.080;
H:=0.020;
Vf:=0.3;
w:=10;
fsolve(v(t)=0,t=0.5..2);

How can I do to obtain all the solutions of the equation in the wanted interval ?

I'm not fixed to use fsolve function.

Thank you for your help

See attached file and code

0. This is the differential equation I'm trying to do:

http://www.intmath.com/differential-equations/6-rc-circuits.php

https://i.imgur.com/zlVIisR.png

 

1. After you look at the above imgur link, could you help me with this error 

Error, (in Units:-Standard:-+) the units ``&Omega;`` and `1/F` have incompatible dimensions

  

2. Why does my solve ODE fail? 

 

See my code: 

test_maple.mw

I am able to get unlimeted numbers of equations describing my system. These equations are generally relate quotients of multivariate polynomials. Each additional equation I get is generally less than twice the length of the last, and it is not always the case that an equation is independant of the previous equations. Although I can get unlimited numbers of equations describing the system, it is not overdetermined.

I am interested in solving these equations for their variables. There are about 30 cases I am working on, the smallest number of evariables is six, the largest would be twenty.

I want to be able to solve these equations in the minimal time possible. But I don't understand the function solve well enough to do so.

How do I choose the equations to minimise the time taken for the command solve to proccess them?
How does the command solve work?

particularly:

  1. if I process the command solve([Eq1,Eq2,Eq3...Eqn],variables) would the command solve([Eq[1],Eq[2],Eq[3]...Eq[n],Eq[n+1]],variables) take longer if Eq[n+1] is not indipendant of the previous equations? 
  2. Is there a way of checking whether Eq[n+1] is independant of the previous vequations, fast enough for it to be useful to check the equations before they are processed?
  3. Does the ordering of the equations affect the speed of solve?
  4. Is there a way of pre processing the equations before they are put into solve that will save it time? (for example factorising them, simplifying them etc...)

 

 

Dear all
Please guide me how to convert system of expressions into system of equations, so as solve them using "solve command".

The following expressions are just coefficients extracted from certain equation.

16*a[2]^4*delta[1]^2-48*a[2]^3*a[3]*delta[1]*delta[2]-2*a[2]*b[2]*delta[1]^2, 48*a[2]^2*a[3]^2*delta[1]*delta[2]-80*a[3]^4*delta[1]*delta[2]-4*a[3]*b[3]*delta[1]*delta[2], 64*a[2]^3*a[3]*delta[1]*delta[2]-64*a[2]*a[3]^3*delta[1]*delta[2]-2*a[2]*b[3]*delta[1]*delta[2]-2*a[3]*b[2]*delta[1]*delta[2]

16*a[2]^4*delta[1]^2-48*a[2]^3*a[3]*delta[1]*delta[2]-2*a[2]*b[2]*delta[1]^2, 48*a[2]^2*a[3]^2*delta[1]*delta[2]-80*a[3]^4*delta[1]*delta[2]-4*a[3]*b[3]*delta[1]*delta[2], 64*a[2]^3*a[3]*delta[1]*delta[2]-64*a[2]*a[3]^3*delta[1]*delta[2]-2*a[2]*b[3]*delta[1]*delta[2]-2*a[3]*b[2]*delta[1]*delta[2]

(1)

It possible for me to write (1) in the following form

for EQ in 16*a[2]^4*delta[1]^2-48*a[2]^3*a[3]*delta[1]*delta[2]-2*a[2]*b[2]*delta[1]^2, 48*a[2]^2*a[3]^2*delta[1]*delta[2]-80*a[3]^4*delta[1]*delta[2]-4*a[3]*b[3]*delta[1]*delta[2], 64*a[2]^3*a[3]*delta[1]*delta[2]-64*a[2]*a[3]^3*delta[1]*delta[2]-2*a[2]*b[3]*delta[1]*delta[2]-2*a[3]*b[2]*delta[1]*delta[2] do EQ = 0 end do

16*a[2]^4*delta[1]^2-48*a[2]^3*a[3]*delta[1]*delta[2]-2*a[2]*b[2]*delta[1]^2 = 0

 

48*a[2]^2*a[3]^2*delta[1]*delta[2]-80*a[3]^4*delta[1]*delta[2]-4*a[3]*b[3]*delta[1]*delta[2] = 0

 

64*a[2]^3*a[3]*delta[1]*delta[2]-64*a[2]*a[3]^3*delta[1]*delta[2]-2*a[2]*b[3]*delta[1]*delta[2]-2*a[3]*b[2]*delta[1]*delta[2] = 0

(2)

But I want to write (1) in the following form

16*a[2]^4*delta[1]^2-48*a[2]^3*a[3]*delta[1]*delta[2]-2*a[2]*b[2]*delta[1]^2 = 0, 48*a[2]^2*a[3]^2*delta[1]*delta[2]-80*a[3]^4*delta[1]*delta[2]-4*a[3]*b[3]*delta[1]*delta[2] = 0, 64*a[2]^3*a[3]*delta[1]*delta[2]-64*a[2]*a[3]^3*delta[1]*delta[2]-2*a[2]*b[3]*delta[1]*delta[2]-2*a[3]*b[2]*delta[1]*delta[2] = 0

16*a[2]^4*delta[1]^2-48*a[2]^3*a[3]*delta[1]*delta[2]-2*a[2]*b[2]*delta[1]^2 = 0, 48*a[2]^2*a[3]^2*delta[1]*delta[2]-80*a[3]^4*delta[1]*delta[2]-4*a[3]*b[3]*delta[1]*delta[2] = 0, 64*a[2]^3*a[3]*delta[1]*delta[2]-64*a[2]*a[3]^3*delta[1]*delta[2]-2*a[2]*b[3]*delta[1]*delta[2]-2*a[3]*b[2]*delta[1]*delta[2] = 0

(3)

``

 

Download Maple_Query.mw

I downloaded Maple Player, and I don't know how can I use it resolve Maths problems. Kindly, could you explain to me the procedure I shall follow for the reason to use it?

Dear all;

Good morning everyone.

I solve a simple ode, i want how can I write this program as procedure with output the two coefficient involved in the solution, after solving with ics. i.e coef:=dsolve(ode);  in my example I want as output [4, -1] .

 

restart:
ode:=diff(y(x),x,x)=3*diff(y(x),x)-2*y(x);
coef:=dsolve(ode);
ics:=y(0)=3, D(y)(0)=2;
dsolve({ics,ode});

with best regards

 

In Maple 2015.1 we have

restart;

solve([sin(2*x)/cos(x+3*Pi/2)=1,  x>-4*Pi, x<-5*Pi/2], x, allsolutions, explicit);

solve([sin(2*x)/cos(x+3*Pi/2)=1, x>0, x<2*Pi], x, allsolutions, explicit);

 

 

In the first example, the error message is not clear (actually there exists a unique root  x=-11*Pi/3), in the second example, one root  (x=5*Pi/3) is lost.

 

assume a,b,c,d,B2,B3 are matrices and y is unknown

eq2 := a*b+c*d+a;
eq3 := a*c+c*d+c;
eq4 := a*b+c*a+b*c;
eq5 := a*b+a*d+b*c;
solve([eq2=B2,eq3=B3,eq4=B2,eq5=y],[a,b,c,d]);

which function can solve this kind of system of matrices?

how to solve a,b,c,d in terms of y?

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