salim-barzani

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These are questions asked by salim-barzani

How do true evaluation for finding unkn...

Asked: salim-barzani 1560 Product: Maple 2024

March 21 2025

1 1

in this equation we have a list of a lot paramter which i have to find it but in prgress to find it some issue are appear i don't know how many term i have to replacing by algsubs there is any way for showing that something like lpring just for factoring and replacing if we did something like that in one step we can replacing all and then find our parameter there is any way for finding parameter like that?

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How remove sign of integral by substitut...

Asked: salim-barzani 1560 Product: Maple 2024

March 21 2025

1 3

in this integral PDE author did a substitution and the integral is simplify and removing how i can do that as mention in picture i did try but i think need a technique

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Why when do transform the equation to la...

Asked: salim-barzani 1560 Product: Maple 2024

March 06 2025

1 2

when i use change maple to latex most of that equation when i want to change the place of term are change how i can fix that for example in (R) if watch in exponential the x is first term but after changing to latex are change which i have to change by hand how i can fix this issue?

Warning, A new binding for the name `gamma` has been created. The global instance of this name is still accessible using the :- prefix, :-`gamma`. See ?protect for details.

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\Lambda = k_{i} \left(w_{i} t +y l_{i}+r_{i} z +x \right)+\eta_{i}

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f =
1+{\mathrm e}^{k_{i} \left(w_{i} t +y l_{i}+r_{i} z +x \right)+\eta_{i}}

eq15 := w[i] = (-1+sqrt(-4*beta*mu*l[i]-4*delta*mu*r[i]-4*mu*k[i]^2-4*alpha*mu+1))/(2*mu)

w[i] = (1/2)*(-1+(-4*beta*mu*l[i]-4*delta*mu*r[i]-4*mu*k[i]^2-4*alpha*mu+1)^(1/2))/mu

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w_{i} =
\frac{-1+\sqrt{-4 \beta \mu l_{i}-4 \delta \mu r_{i}-4 \mu k_{i}^{2}-4 \alpha \mu +1}}{2 \mu}

R := f(x, y, z, t) = 1+exp(k[i]*(x+l[i]*y+r[i]*z+(-1+sqrt(-4*beta*mu*l[i]-4*delta*mu*r[i]-4*mu*k[i]^2-4*alpha*mu+1))*t/(2*mu))+eta[i])

f(x, y, z, t) = 1+exp(k[i]*((1/2)*(-1+(-4*beta*mu*l[i]-4*delta*mu*r[i]-4*mu*k[i]^2-4*alpha*mu+1)^(1/2))*t/mu+y*l[i]+r[i]*z+x)+eta[i])

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f =
1+{\mathrm e}^{k_{i} \left(\frac{\left(-1+\sqrt{-4 \beta \mu l_{i}-4 \delta \mu r_{i}-4 \mu k_{i}^{2}-4 \alpha \mu +1}\right) t}{2 \mu}+y l_{i}+r_{i} z +x \right)+\eta_{i}}

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what is problem why i can't get result?...

Asked: salim-barzani 1560 Product: Maple 2024

March 05 2025

0 2

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eqw := w[i] = (-1+sqrt(-4*beta*mu*l[i]-4*delta*mu*r[i]-4*mu*k[i]^2-4*alpha*mu+1))/(2*mu)

Bij := proc (i, j) options operator, arrow; -24*mu/(sqrt(1+(-4*beta*l[j]-4*delta*r[j]-4*alpha)*mu)*sqrt(1+(-4*beta*l[i]-4*delta*r[i]-4*alpha)*mu)-1+((2*r[i]+2*r[j])*delta+(2*l[i]+2*l[j])*beta+4*alpha)*mu) end proc

eq17 := u(x, y, z, t) = 2*(diff(diff(f(x, y, z, t), x), x))/f(x, y, z, t)-2*(diff(f(x, y, z, t), x))^2/f(x, y, z, t)^2; equ := simplify(eval(eq17, eqfcomplex))

So we want to find a substitution that removes the time dependence from u. One way is to find the maximum and see how it moves. Here, the first solution gives what we want.

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There is any way for see the outcome of ...

Asked: salim-barzani 1560 Product: Maple 2024

March 05 2025

0 0

i need the result for (eqt33) but i can reach the result there is any other way for finding? i need to plot 3D of that function but without have the function how i can do explore on it

w1.mw

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