New answers tagged parametric-functions
6
There is unlikely to be a "nice" closed-form solution for this equation, since it involves both transcendental and algebraic expressions for $v$. However, it can be solved in terms of special functions with a bit of work.
We can make the substitution $v = \alpha x + b$ to rewrite the equation (after some algebra) as
$$
x e^x < - e^{-b/\alpha }.
...
1
You can do either of the following
result1 = 16 (y - x);
result2 = x (45 - z) - y;
result3 = x*y - 4 z;
F[{x_, y_, z_}] := Evaluate[{result1, result2, result3}];
x = Array[a, 3];
F[x]
Or
result1 = 16 (y - x);
result2 = x (45 - z) - y;
result3 = x*y - 4 z;
F[{x_, y_, z_}] = {result1, result2, result3};
x = Array[a, 3];
F[x]
Both give same result you showed.
...
3
We can include derivative in solution as follows
sol = ParametricNDSolveValue[{e1'[t] == s e2[t], e2'[t] == -s e1[t],
e1[0] == 1, e2[0] == 0}, {e1, e2, e1', e2'}, {t, 0, 1}, s]
fun[t_, s_] := Through[sol[s][t]][[1]];
Plot[fun[t, 1], {t, 0, 10}]
dfun[t_, s_] := Through[sol[s][t]][[3]];
Plot[dfun[t, 10], {t, 0, 1}]
4
Considering the lagrangian L which follows,
Qk[p_] := 1 - (p - f - theta (alpha + beta))/theta/(1 - beta)
Qr[p_] := (p - f - theta (alpha + beta))/
theta/(1 - beta) - (p + (1 - m)/m f - (alpha + beta))/(1 - beta)
R[p_] := (1 - m) Qr[p]
Q[p_] := Qr[p] + Qk[p]
obj = Q[p] (p - k) + R[p] (f + r - p)
L = obj + l1 (p - f mu/m/(1 - theta) - s1^2)
+ l2 (p ...
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