Could you help me with what Mathematica is doing here, in the last lines? (The rest is here for context.) I removed the output to let you simply run this. Thanks.
k=10/3
u[c_,l_]:=Log[c]-Log[1+l^(1+k)/(1+k)]
T[z_]:=(1-0.84/1.3) * z
lType[n_]:=ArgMax[{u[n l-T[n l],l],l>=0},l]
zType[n_] := n lType[n]
Type := InverseFunction[zType]
Type[5345]
Type[2]
Type[4324424]
With a much longer version with some context, you can also switch to https://stackoverflow.com/questions/10641713/mathematica-define-inverse-only-where-it-exists
Or please reconsider the same code with some output to see what my problem is with the inverse:
k=10/3
10/3
Utility function as in Saez 2001 allowing for income effects:
u[c_,l_]:=Log[c]-Log[1+l^(1+k)/(1+k)]
Actual tax schedule (approximate)
T[z_]:=(1-0.84/1.3) * z
Equation 14:
lType[n_]:=ArgMax[{u[n l-T[n l],l],l>=0},l]
zType[n_] := n lType[n]
Type := InverseFunction[zType] does not seem to work.
Type := InverseFunction[zType]
Type[5345]
5030.985810589009`
Type[2]
(zType^(-1))[2]
Type[4324424]
(zType^(-1))[4324424]
Merge Cells
before the copy you won't have all the In/Out tags. $\endgroup$