Zihan

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6 years, 166 days

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These are questions asked by Zihan

Hi my dear friends, sorry for boring you. I am hauted by a problem about 'mtalyor'.

I failed to expand the following equation by using the following command:

u:=x(t)/sqrt(x(t)^2+y(t)^2-2*x(t)+1)*diff(y(t),t);
indets(u,name);
w:=evalindets(u,function(identical~({x(t),y(t)})),s->op(0,s)(freeze(op(s))));
var:=indets(w,name) minus {t};
mtaylor(w,var,4);
thaw(%)

 Freeze is applied to the part in which the expansion will take place, obviously it didn't work well. 

Thank you in advance for taking a look. 

Sorry for boring you my friends. I am haunted by a problem of how to omit the undesired term.

For example, in the following equation, the a(t) , b(t), c(t), u(t), v(t), w(t), psi(t), phi(t), theta(t), varsigma(t), tau(t) and upsilon(t) and their first and second direvative to time t are considered as first order small variables. How could I omit the term greater than second order of small variables?

If we omit the undesired by hand, the omitted equation takes the form of:

R^2*rho*h*(diff(w(t), t, t))*Pi+R^2*rho*h*(diff(c(t), t, t))*Pi = 0;

The original equation is given as: 

-R^2*rho*h*cos(Omega*t)*(diff(tau(t), t, t))*a(t)*Pi+tau(t)*R^2*rho*h*(diff(tau(t), t))^2*a(t)*cos(Omega*t)*Pi-tau(t)*R^2*rho*h*(diff(tau(t), t))^2*b(t)*sin(Omega*t)*Pi+tau(t)*a(t)*Pi*cos(Omega*t)*(diff(varsigma(t), t))^2*R^2*h*rho+tau(t)*R^2*rho*h*a(t)*Omega^2*cos(Omega*t)*Pi-tau(t)*sin(Omega*t)*Pi*(diff(varsigma(t), t))^2*b(t)*R^2*h*rho-tau(t)*R^2*rho*h*b(t)*Omega^2*sin(Omega*t)*Pi+2*tau(t)*R^2*rho*h*(diff(a(t), t))*Omega*sin(Omega*t)*Pi+2*tau(t)*R^2*rho*h*(diff(b(t), t))*Omega*cos(Omega*t)*Pi-varsigma(t)*a(t)*sin(Omega*t)*Pi*(diff(varsigma(t), t))^2*R^2*h*rho-varsigma(t)*a(t)*sin(Omega*t)*Pi*Omega^2*R^2*h*rho-varsigma(t)*Pi*cos(Omega*t)*(diff(varsigma(t), t))^2*b(t)*R^2*h*rho-varsigma(t)*Pi*cos(Omega*t)*b(t)*Omega^2*R^2*h*rho-2*varsigma(t)*sin(Omega*t)*Pi*(diff(b(t), t))*Omega*R^2*h*rho+2*varsigma(t)*Pi*cos(Omega*t)*(diff(a(t), t))*Omega*R^2*h*rho+2*a(t)*Pi*cos(Omega*t)*(diff(varsigma(t), t))*Omega*R^2*h*rho-2*sin(Omega*t)*Pi*(diff(varsigma(t), t))*b(t)*Omega*R^2*h*rho+R^2*rho*h*(diff(tau(t), t, t))*sin(Omega*t)*b(t)*Pi+R^2*rho*h*(diff(w(t), t, t))*Pi+R^2*rho*h*(diff(c(t), t, t))*Pi-R^2*rho*h*(diff(tau(t), t))^2*c(t)*Pi+2*varsigma(t)*(diff(tau(t), t))*a(t)*Pi*cos(Omega*t)*(diff(varsigma(t), t))*R^2*h*rho-2*varsigma(t)*(diff(tau(t), t))*sin(Omega*t)*Pi*(diff(varsigma(t), t))*b(t)*R^2*h*rho+a(t)*sin(Omega*t)*Pi*(diff(varsigma(t), t, t))*R^2*h*rho+Pi*cos(Omega*t)*b(t)*(diff(varsigma(t), t, t))*R^2*h*rho-varsigma(t)*Pi*c(t)*(diff(varsigma(t), t, t))*R^2*h*rho-2*tau(t)*R^2*rho*h*(diff(tau(t), t))*(diff(c(t), t))*Pi-2*varsigma(t)*Pi*(diff(varsigma(t), t))*(diff(c(t), t))*R^2*h*rho+2*sin(Omega*t)*Pi*(diff(varsigma(t), t))*(diff(a(t), t))*R^2*h*rho+2*Pi*cos(Omega*t)*(diff(varsigma(t), t))*(diff(b(t), t))*R^2*h*rho-Pi*(diff(varsigma(t), t))^2*c(t)*R^2*h*rho-2*R^2*rho*h*cos(Omega*t)*(diff(tau(t), t))*(diff(a(t), t))*Pi+2*R^2*rho*h*(diff(tau(t), t))*sin(Omega*t)*(diff(b(t), t))*Pi+tau(t)*R^2*rho*h*(diff(b(t), t, t))*sin(Omega*t)*Pi-tau(t)*R^2*rho*h*(diff(a(t), t, t))*cos(Omega*t)*Pi-tau(t)*R^2*rho*h*(diff(tau(t), t, t))*c(t)*Pi+2*R^2*rho*h*Omega*sin(Omega*t)*(diff(tau(t), t))*a(t)*Pi+2*R^2*rho*h*(diff(tau(t), t))*Omega*cos(Omega*t)*b(t)*Pi+varsigma(t)*sin(Omega*t)*Pi*(diff(a(t), t, t))*R^2*h*rho+varsigma(t)*Pi*cos(Omega*t)*(diff(b(t), t, t))*R^2*h*rho = 0;

 

Thank you in advance for taking a look ;)

Sorry for disturbing you again. I am haunted by a problem about Taylor expansion in Maple.

For example, when the angle alpha is small, thus we could use taylor(sin(x),x=0,1) to get a approximated function of sin(x) with different order of precision.

But, if the sin(x) meets diff(x(t),t), (x is a time-dependant variable), could we also apply the Tayolr expansion in Maple to obtain a approximation?

For example, the function: sin(x(t))*diff(x(t),t), x(t) is given with a small value, but diff(x(t),t) could not be neglected, because even x(t) is with a small value , diff(x(t),t) could be great. We could conduct the Talyor expansion with hands, and sin(x(t))*diff(x(t),t) becomes x(t)*diff(x(t),t). But I wonder if we could use Maple to realize this procedure. And if the Maple could make it, I would like to know if the Maple could handle the multi-variable Taylor expansion accompanied with the time derivative. For example, the x(t) and y(t) are considered as small value variables, could we simplified the following function sin(x(t))*cos(y(t))*diff(x(t),t)*diff(y(t),t).

Thank you very much in advance for taking a look. 

Sorry for disturbing you. I am wondering if there is an easier approach in Maple that could convert a system of second order differential equations into matrix form. Of course, we could do it by hand easily if the degrees of freedom is small. I would like to know if we could use Maple to do so. 

Here is an example with 6 degrees of freedom: the variables are u, v, w, alpha, beta and gamma. And, this is a uncoupled system.

Vector(6, {(1) = 2*R^2*(diff(w(t), t))*Pi*Omega*h*rho+R^2*(diff(u(t), t, t))*Pi*h*rho-R^2*u(t)*Pi*Omega^2*h*rho = 0, (2) = R^2*(diff(v(t), t, t))*Pi*h*rho = 0, (3) = -2*R^2*(diff(u(t), t))*Pi*Omega*h*rho+R^2*(diff(w(t), t, t))*Pi*h*rho-R^2*w(t)*Pi*Omega^2*h*rho = 0, (4) = (1/4)*R^4*Pi*(diff(alpha(t), t, t))*h*rho+(1/12)*R^2*Pi*(diff(alpha(t), t, t))*h^3*rho+(1/6)*R^2*Pi*(diff(gamma(t), t))*Omega*h^3*rho-(1/12)*R^2*Pi*alpha(t)*Omega^2*h^3*rho = 0, (5) = (1/2)*R^4*Pi*(diff(beta(t), t, t))*h*rho-(1/2)*R^4*Pi*beta(t)*Omega^2*h*rho = 0, (6) = (1/4)*R^4*Pi*(diff(gamma(t), t, t))*h*rho+(1/12)*R^2*Pi*(diff(gamma(t), t, t))*h^3*rho-(1/6)*R^2*Pi*(diff(alpha(t), t))*Omega*h^3*rho-(1/12)*R^2*Pi*gamma(t)*Omega^2*h^3*rho = 0});

The objective is to reform it into matrix form : M*diff(X(t), t, t)+C*diff(X(t), t)+K*X(t)=F.

Thank you in advance for taking a look. 

Hi my dear friends, I am haunted by a problem of how to convert a very very very long equation to Latex or Word properly?

Thank you for taking a look at the following 'long and boring' equation and sharing your brilliant idea.

 

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))*cos(gamma(t))*sin(alpha(t))+1/6*R^2*h^3*rho*diff(theta(t),t)*Pi*cos(beta(t))^2*diff(psi(t),t)*sin(phi(t))*cos(phi(t))*cos(alpha(t))*cos(theta(t))*cos(gamma(t))^2*sin(alpha(t)) = 0"

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