# How to linearize these terms in many different variables? [duplicate]

How to linearize these terms in q[t] and p[t,y]

term1=(covd[Derivative[1][q][t] + Derivative[1, 0][p][t, y], t] +
2*(1 + 2*\[Alpha][t, r, \[Theta], \[Phi]])*(Derivative[0, 1][a][t, y]*  (4*(p[t, y] + q[t])*Derivative[0, 1][a][t, y] -
a[t, y]*Derivative[0, 1][p][t, y]) +
2*a[t, y]*(p[t, y] + q[t])* Derivative[0, 2][a][t, y]))/(2*a[t, y]^2*(p[t, y] + q[t])^3* (1 + 2*\[Alpha][t, r, \[Theta], \[Phi]]))

term2=(Derivative[0, 1][a][t,y]*(4*(p[t, y] + q[t])*Derivative[0, 1][a][t, y] - a[t, y]*Derivative[0, 1][p][t, y]) + 2*a[t, y]*(p[t, y] + q[t])*
Derivative[0, 2][a][t, y])/(a[t, y]^2*(p[t, y] + q[t])^3)


the codes I use are:

Normal[Series[term1 /. {q -> (eps q[#] &), p -> (eps p[#] &)}, {eps, 0, 1}]]

The output is:

(2*(1 + 2*\[Alpha][t, r, \[Theta], \[Phi]])*(4*(p[t] + q[t])* Derivative[0, 1][a][t, y]^2 +
2*a[t, y]*(p[t] + q[t])*Derivative[0, 2][a][t, y]) +     (Derivative[1][p][t] + Derivative[1][q][t])* Derivative[1, 0][covd][0, t])/
(2*eps^2*a[t, y]^2*(p[t] + q[t])^3*(1 +
2*\[Alpha][t, r, \[Theta], \[Phi]])) +
((Derivative[1][p][t] + Derivative[1][q][t])^2*
Derivative[2, 0][covd][0, t])/
(4*eps*a[t, y]^2*(p[t] + q[t])^3*(1 +
2*\[Alpha][t, r, \[Theta], \[Phi]])) +
((Derivative[1][p][t] + Derivative[1][q][t])^3*
Derivative[3, 0][covd][0, t])/
(12*a[t, y]^2*(p[t] + q[t])^3*(1 +
2*\[Alpha][t, r, \[Theta], \[Phi]])) +
(eps*(Derivative[1][p][t] + Derivative[1][q][t])^4*    Derivative[4, 0][covd][0, t])/(48*a[t, y]^2*(p[t] + q[t])^3*
(1 + 2*\[Alpha][t, r, \[Theta], \[Phi]]))


So it did not linearized cause (q[t]+p[t,y])^3 still there

or:

Normal[Series[term2 /. {q -> Function[{t}, eps q[t]], p -> Function[{t, y}, eps p[t, y]]}, {eps, 0, 1}]] /. eps -> 1

The output returns the same function.

• Use the replacement term1 /. {q :> (eps q[##] &), p :> (eps p[##] &)}. Otherwise p[t, y] will be replaced by eps p[t]. Nov 26, 2023 at 16:57
• At which point do you want to linearize this? Currently, you seem to trying that at $q = 0$ and $p = 0$ where the expression has a pole of order 3 (so the expression is not linearizable there). Nov 26, 2023 at 16:58
• @HenrikSchumacher. The replacements works, thanks. q[t] and p[t] do not equal zero so the terms linearized and only (q[t]+p[t])^2 remains. Nov 26, 2023 at 17:18

DerivativeWithRespectToq = D[