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I want to obtain the critical points or if you like a real solution for the first derivative of the function set to zero, and the second derivative also vanishes of the function :

P[v_] = t/v - (d - 3)/((d - 2) Pi v^2) + 32 t a/((d - 2)^2 v^3) - 
  16 a (d - 5)/((d - 2)^3 Pi v^4) + 256 t a^2/((d - 2)^4 v^5) - 
  256 a^2 (d - 7)/(3 (d - 2)^5 Pi v^6)

so I used the code

T[rh_] = FullSimplify[D[P[v], v]]
b[rh_] = FullSimplify[D[T[rh_], v]]
Solve[T[rh_] == 0, v, Reals]
Solve[b[rh_] == 0, v, Reals]

The true answer is that there are two solutions, which (obviously) I didn't obtain, the two solutions are enter image description here

eq 18 is just the first and second derivative of P with respect to v set to zero, these two solutions are real only if 2<d<12.

d and a are constants, t and v are variables, should I consider the first and second derivative with respect to v as a system of 2-equations with two variables t and v to be solved?

Thank you

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  • $\begingroup$ What's known about parameters a,d,t ? $\endgroup$ 2 days ago
  • $\begingroup$ d and a are constants, t is also a variable $\endgroup$ 2 days ago
  • $\begingroup$ That means you try to solve for t,v: Solve[{P'[v] == 0, P''[v] == 0}, {t, v}, Reals]? $\endgroup$ 2 days ago
  • $\begingroup$ yes that is true $\endgroup$ 2 days ago
  • $\begingroup$ You should check, how the solutions are derived in the paper $\endgroup$ 2 days ago

1 Answer 1

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Try

P = Function[{v, t},t/v - (d - 3)/((d - 2) Pi v^2) + 32 t a/((d - 2)^2 v^3) -16 a (d - 5)/((d - 2)^3 Pi v^4) + 256 t a^2/((d - 2)^4 v^5) -256 a^2 (d - 7)/(3 (d - 2)^5 Pi v^6)];

Solve[{Derivative[1, 0][P][v, t] == 0,Derivative[2, 0][P][v, t] == 0}, {t, v}, Reals]
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  • $\begingroup$ Thank you my man I only needed to edit your anwser so i add Fullsimplify and remove reals $\endgroup$ 2 days ago

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