mehran rajabi

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8 years, 104 days

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These are questions asked by mehran rajabi

Hi everyone:

How can I solve numerically the system of nonlinear algebraic equations by Newton’s method?

eq1:= (1/2)*x[0]*sqrt(3)-(1/2)*x[1]*sqrt(3) = ((1/2)*x[0]*(t+(1/3)*sqrt(3))*sqrt(3)-(1/2)*x[1]*(t-(1/3)*sqrt(3))*sqrt(3))*(1-(1/6)*y[0]*(t+(1/3)*sqrt(3))*sqrt(3)+(1/6)*y[1]*(t-(1/3)*sqrt(3))*sqrt(3)-(1/8)*y[0]*(5*sqrt(3)*(1/12)-1/4+t)*sqrt(3)+(1/8)*y[1]*(-(1/4)*sqrt(3)-1/4+t)*sqrt(3)-(1/8)*y[0]*((1/4)*sqrt(3)-1/4+t)*sqrt(3)+(1/8)*y[1]*(-5*sqrt(3)*(1/12)-1/4+t)*sqrt(3))-5*t^3*(1/2)+49*t^2*(1/12)+17*t*(1/12)-23/6;
eq2:= (1/2)*y[0]*sqrt(3)-(1/2)*y[1]*sqrt(3) = ((1/2)*y[0]*(t+(1/3)*sqrt(3))*sqrt(3)-(1/2)*y[1]*(t-(1/3)*sqrt(3))*sqrt(3))*(-2+(1/2)*x[0]*(t+(1/3)*sqrt(3))*sqrt(3)-(1/2)*x[1]*(t-(1/3)*sqrt(3))*sqrt(3)+(1/4)*(-(1/12)*sqrt(3)-3/4)*((1/2)*x[0]*(5*sqrt(3)*(1/12)-1/4+t)*sqrt(3)-(1/2)*x[1]*(-(1/4)*sqrt(3)-1/4+t)*sqrt(3))+(1/4)*((1/12)*sqrt(3)-3/4)*((1/2)*x[0]*((1/4)*sqrt(3)-1/4+t)*sqrt(3)-(1/2)*x[1]*(-5*sqrt(3)*(1/12)-1/4+t)*sqrt(3)))+15*t^3*(1/8)-(1/4)*t^2+3*t*(1/8)-1;
eq3:=(1/2)*x[0]+(1/2)*x[1] = 1;
eq4:=(1/2)*y[0]+(1/2)*y[1] = 0;

 

Hi everyone:

How can I re-write the EQ with transformation       s=1+2*((tau-t)/T0)    ?

EQ:=int(f1(t-tau)*(Sum(y[k]*F[k](tau), k = 0 .. M)), tau = t-T0 .. t)

tnx...

Hi everyone:

I want to earn f(zeta) and zeta=x/a while the f(x) is: 

f:=(x)->A1*sin(k*x)+A2*cos(k*x)+A3*sinh(k*x)+A4*cosh(k*x)

zeta=x/a and a, k, A1..A4 are constants. 

f(zeta)=? 

 

 

Hi everybody:

I have an equation that attached with this question and my goal is to solve it, how can do it?

Note: in this equation must be considered 49.32883964 <= x and x is real.

solve_equation.mw

 

 

Hi everyone:

I'm going to write code to give me the following expression, how can I do it?
n is Natural and k=0..n. 

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