It seems Mathematica doesn’t recognize a[1] + 1 like the same variable a[1] plus one:
a[1] = \[Omega]
a[2] = l
V[x_, a[1], a[2]] := 1/4*a[1]^2*x^2 + a[2]*(a[2] + 1)/x^2
V[x, a[1] + 1, a[2]]
V[x, a[1] + 1, a[2]]
simply does not match the pattern:
V[x_, a[1], a[2]]
I think you want:
ClearAll[V]
V[x_, a1_, a2_] := 1/4*a1^2*x^2 + a2*(a2 + 1)/x^2
I defined a[1] instead of a1 cause later i have to call it with a[i] where i=1 or 2 and it seems it doesnt work with ai
V[x_, a1_, a2_] := 1/4a1^2x^2 + a2*(a2 + 1)/x^2
this function can still ‘receive’ a[1] or a[2] as an argument, a1 and a2 are just temporary names
If that doesn’t work for some reason; you can also make definitions with more elaborate patterns:
V[x_, a_[y_], b_[z_]] := ...
Thank you but it seems it doesn’t work. What i would like to do is this:
a[1][\[Omega]_] = \[Omega]
a[2][l_] = l
V[x_, a_[1], a_[2]] := 1/4*a[1]^2*x^2 + a[2]*(a[2] + 1)/x^2
V[x, a[1] + 1, a[2]]
This works, but after i found a k value (like k=1) i would be able to eliminate a[2] dependence from V, defining a new function.
k := 1
G[x_, a_[k]] := V[x, a[1], a[2]]
G[x, a[k] + 1]
This doesn’t work
That first example does NOT work. What is k?
To make this question/topic useful explain what you want to do, your input, and the expected output as simple as possible.
You’re right. So i have 2 problems. Do you have an idea?
k is a value i found with an algorithm that gives me the right parameter a[1] or a[2]
Ok sorry… i try to explain it.
I have a function V depending from x and some parameters a[i].
After an algorithm i found the parameter that needs to be modified and so i found a value for k.
Next i have to eventually substitute a[k] with a[k] + c for any value of k and so i thought to define a new function G depending only of x and a[k]
That is a bit more clear. Still not very have a look at Replace and ReplaceAll.
I tried that but i can’t do something like this
Q[x_,a[k]_]:=V[x,a[k]+c]-V[x,a[k]]