goli

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12 years, 80 days

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These are questions asked by goli

Daer all! I want to solve the following equation

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

where H(0) denotes the value of H(z) at z=0 and other constants are

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

If I want to use the fsolve command as

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

> Y := z-> fsolve(eq(z), H=1):  

what should I imply instead of H(0)? 

Dear guys! I have an equation as

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

where

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

At this equation H is an unknown function of z, h is the value of H at z=0 and I want to obtain H(z) actually(for example the graph of H(z) for z=-1..5). So I did as following

> Y := z-> fsolve(eq(z), H=h):    **

plot(eval(H, H='Y(z)'), z=-1 .. 5);

Dear guys!

I want to solve this system:

> c := 1: RC := 0.03: h := 1: m := 0.3:

> a := H(z)^2/(1+z)-h^2*(m*(1+z)^2+2*RC/(1+z));

> ode1 := diff(H(z),z) = (H(z)^4/((1+z)^2)+m*(1+z)*h^2*H(z)^2-2*m^2*h^4*(1+z)^4)/(2*H(z)*a) - (2*H(z)*h^2*RC)/(a*(1+z)^2) + (2*RC*h^2*(1+z)/(c*H(z)^2*a))*(H(z)^2/(1+z)^2-(H(z)^2/(1+z)-m*h^2*(1+z)^2)^2/(4*RC*h^2))^(3/2):

> ode2 := diff(M(z),z)=3*M(z)/(1+z)-2*M(z)*diff(H(z),z)/H(z):

> sys := {ode1, ode2}:

Dear all! I have 2 different lists:

ch1 := [[0.1e-1, -.56], [0.25e-1, -.57], [0.35e-1, -.555], [.115, -.43], [.16, -.3], [.195, -.24], [.28, -.155], [.315, -0.25e-1], [.32, -0.85e-1], [.41, .19], [.425, .11], [.425, .425], [.425, .43], [.425, .44], [.43, .12], [.43, .265], [.43, .405], [.43, .445], [.435, .295], [.435, .38], [.465, .44], [.5, .29], [.505, .31], [.505, .395], [.51, .37]];

ch2 := [[0.3e-1, -.5], [0.5e-1, .575], [.11, .495], [.15, -.295], [.19, -.33...

nonflat1-bestfit-H0-.mw  sn_union2.txt  Hi all! Can anyone tell me what the problem is in my worksheet? It works for first two initial conditions but then it does not work. The data file that is needed to run the program has attached, too. Thanks. 

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