## 165 Reputation

12 years, 311 days

## inverse of a matrix...

GRBmatrix.txt @Markiyan Hirnyk ok here is the data

## inverse of a matrix...

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.

## What do you mean ?...

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

## What do you mean ?...

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

## simplifying hypergeom...

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. ...

@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. ...

@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...

@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...

@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 diffe...

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 diffe...

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.

## @Markiyan Hirnyk O.K. thanks....

@Markiyan Hirnyk O.K. thanks.

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