In an exemplary expression
expr = -6 3^(1/3) c1 kL^4 L^2 (δ^3/L^3)^(1/3)
all parameters are real, especially L > 0
. FullSimplify[]
with the assumption L > 0
FullSimplify[ expr, L > 0]
cancels out the parameter L
.
How can I force Mathematica to do this simplification inside a function?
sol = {w ->Function[{x}, -6 3^(1/3) c1 kL^4 L^2 (δ^3/L^3)^(1/3)
BesselI[1/3, (2 (-kL^2 L)^(3/2))/(3 L^(3/2) δ^3)] Gamma[4/3] ]}
Thanks!
L^(3/2)
in the numerator and denominator should cancel out, then @Artes's suggestion works. $\endgroup$L > 0
, i.e., the second argument toFullSimplify
. AlthoughSimplify
would be sufficient. $\endgroup$