goli

160 Reputation

6 Badges

12 years, 76 days

MaplePrimes Activity


These are replies submitted by goli

GRBmatrix.txt @Markiyan Hirnyk ok here is the data

GRBmatrix.txt @Markiyan Hirnyk ok here is the data

@Markiyan Hirnyk Thanks. I did it and after changing I obtained this integral

int((x^2)*(1-x^4)^((3-2*f)/(4*f-4)),x);

This integral has a solution in terms of hypergeom, as well. I know what's the problem. I should consider the power of second term ( (3-2*f)/(4*f-4) ) as an integer. Am I right? But I don't know how should I do it. Please guide me.

@Markiyan Hirnyk Thanks. I did it and after changing I obtained this integral

int((x^2)*(1-x^4)^((3-2*f)/(4*f-4)),x);

This integral has a solution in terms of hypergeom, as well. I know what's the problem. I should consider the power of second term ( (3-2*f)/(4*f-4) ) as an integer. Am I right? But I don't know how should I do it. Please guide me.

Thank you. Can you explain more, please? Then waht is "x"? My integral is in terms of "t".

Thank you. Can you explain more, please? Then waht is "x"? My integral is in terms of "t".

I tried to simplify the most important term in the solution, i.e.,

hypergeom([1/4, (f-3/2)/(2*f-2)+1/(2*f-2)], [1+(f-3/2)/(2*f-2)+1/(2*f-2)], t^(2*f-2))

First I tried "simplify" command. Then I tried the MTM approach. But none of them can simplify the answer. Can anybody solve this problem and simplify the answer?

Thanks.

@Markiyan Hirnyk No, you missed h. I mean an approach like this: (if I replace H(0) in equation with h)

> restart:

> m := 0.211: h := 0.741: c := 0.80: r := 0.338: e := 1:

> eq := z-> H*(1-m*h^2*H^(-2)*(1+z)^(3))-2*e*sqrt(1-c^2)*sqrt(r)*h=0;

> yp := implicitdiff(eq(z), H, z);

> ode := diff(H(z), z) = subs(H=H(z), yp);

> sol := dsolve({ode, H(0) = h}, numeric, output = listprocedure, stiff = true);

It works well.

@Markiyan Hirnyk No, you missed h. I mean an approach like this: (if I replace H(0) in equation with h)

> restart:

> m := 0.211: h := 0.741: c := 0.80: r := 0.338: e := 1:

> eq := z-> H*(1-m*h^2*H^(-2)*(1+z)^(3))-2*e*sqrt(1-c^2)*sqrt(r)*h=0;

> yp := implicitdiff(eq(z), H, z);

> ode := diff(H(z), z) = subs(H=H(z), yp);

> sol := dsolve({ode, H(0) = h}, numeric, output = listprocedure, stiff = true);

It works well.

@Markiyan Hirnyk I want to best fit this parameter H(0) using some observational data. So, I need to solve this equation many times for different values of H(0). It should be about H(0)=0.7. But I was wrong a little. I had H(0) in my equation that I showed it by h and I gave it a value h=0.741 and then I wanted to solve the equation. Obviously, the H(z) I will obtain has a H(0) that differs from h. I want to find the H(0) where appears in my equation and I think your approach in http://www.mapleprimes.com/questions/129110-Fsolve-With-Unknown-Initial-Condition?submit=129159#comment129159 is not correct.    

@Markiyan Hirnyk I want to best fit this parameter H(0) using some observational data. So, I need to solve this equation many times for different values of H(0). It should be about H(0)=0.7. But I was wrong a little. I had H(0) in my equation that I showed it by h and I gave it a value h=0.741 and then I wanted to solve the equation. Obviously, the H(z) I will obtain has a H(0) that differs from h. I want to find the H(0) where appears in my equation and I think your approach in http://www.mapleprimes.com/questions/129110-Fsolve-With-Unknown-Initial-Condition?submit=129159#comment129159 is not correct.    

Thanks a lot. If I use a different approach

> yp := implicitdiff(eq(z), H, z); 

> ode := diff(H(z), z) = subs(H=H(z), yp);

then I can solve this differential equation by dsolve command for any initial condition that I want, not only H(0) = 0.

What does this mean? Does it mean that I can not choose any initial condition I want and I must find the right initial value and then use it? From your approach it seems to me that H(0) can not to be nothing but just zero.  

Thanks a lot. If I use a different approach

> yp := implicitdiff(eq(z), H, z); 

> ode := diff(H(z), z) = subs(H=H(z), yp);

then I can solve this differential equation by dsolve command for any initial condition that I want, not only H(0) = 0.

What does this mean? Does it mean that I can not choose any initial condition I want and I must find the right initial value and then use it? From your approach it seems to me that H(0) can not to be nothing but just zero.  

@Markiyan Hirnyk O.K. thanks.

@Markiyan Hirnyk O.K. thanks.

2 3 4 5 6 7 8 Last Page 4 of 12