Le mercredi 26 février 2025 à 13:02:00 UTC+1, [email protected] a
écrit :
Dear Emmanuel,
The results of your approach might not be pretty. Unfortunatly I think they
seem to be wrong too.
See:
sage: r, t, alpha, delta, r = var('r, t, alpha, delta, r')
sage: k, x, lambda_ = function("k, x, lambda_")
sage: h1(t) = x(t)*e^(-r*t)
sage: D[0](h1)(x(t))
-r*e^(-r*x(t))*x(x(t)) + e^(-r*x(t))*D[0](x)(x(t))
Kindly note the "x(x(t))" in the first term. :-/
Indeed. I won't pretend to understand what exactly `D` does, a,d I may be
wrong. The (pseudo-)equality :
[image: \displaystyle{\frac{\partial h}{\partial
x}}\,=\,\frac{\frac{\partial h}{\partial t}}{\frac{\partial x}{\partial t}}]
might be more useful.
HTH,
Greetings,
Bernd
Bernd Breitschaedel schrieb am Mittwoch, 26. Februar 2025 um 12:43:34 UTC+1:
Thank you Eric & Emmanuel for your solution proposals.
Regards,
Bernd
Emmanuel Charpentier schrieb am Dienstag, 25. Februar 2025 um 18:08:15
UTC+1:
Complement : Sage’s notation for derivatives is somewhat baroque, and this
does not help in the present case.
Here, both k and x are functions of t ; writing h as a function of x and k
is just an intricate way to write a function of t, unique independent
variable.
As for notations : what we’d “manually” write in this case would be :
[image: \displaystyle{\frac{\partial h}{\partial
t}}\,=\,\displaystyle{\frac{\partial h}{\partial x}\,\frac{\partial
x}{\partial t}}]
and we could deduce
[image: \displaystyle{\frac{\partial h}{\partial
x}}\,=\,\frac{\frac{\partial h}{\partial t}}{\frac{\partial x}{\partial t}}]
.
Sage doesn’t have a direct notation for denoting derivative with respect to
another function ; instead, it uses the Doperator, which creates derivative
operators :
sage: t=var("t") sage: h, x=function("h, x") sage: diff(h(x(t)), t)
D[0](h)(x(t))*diff(x(t), t) sage: diff(h(x(t)), t)/diff(x(t), t)
D[0](h)(x(t))
Eric’s notation is an elegant way to work around this problem.
The use of standard notations leads to :
sage: r, t, alpha, delta, r = var('r, t, alpha, delta, r') sage: k, x,
lambda_ = function("k, x, lambda_") sage: h(t) = k(t)*(1-x(t))*e^(-r*t) +
lambda_(t)*(k(t)^alpha*x(t)-delta*k(t)) sage: D[0](h)(x(t)) r*(x(x(t)) -
1)*e^(-r*x(t))*k(x(t)) - (x(x(t)) - 1)*e^(-r*x(t))*D[0](k)(x(t)) -
e^(-r*x(t))*k(x(t))*D[0](x)(x(t)) + (alpha*k(x(t))^(alpha -
1)*x(x(t))*D[0](k)(x(t)) - delta*D[0](k)(x(t)) +
k(x(t))^alpha*D[0](x)(x(t)))*lambda_(x(t)) - (delta*k(x(t)) -
k(x(t))^alpha*x(x(t)))*D[0](lambda_)(x(t))
[image: \frac{\partial h(t)}{\partial x(t)}\,=\,r
{\left(x\left(x\left(t\right)\right) - 1\right)} e^{\left(-r
x\left(t\right)\right)} k\left(x\left(t\right)\right) -
{\left(x\left(x\left(t\right)\right) - 1\right)} e^{\left(-r
x\left(t\right)\right)}
\mathrm{D}_{0}\left(k\right)\left(x\left(t\right)\right) - e^{\left(-r
x\left(t\right)\right)} k\left(x\left(t\right)\right)
\mathrm{D}_{0}\left(x\right)\left(x\left(t\right)\right) + {\left(\alpha
k\left(x\left(t\right)\right)^{\alpha - 1} x\left(x\left(t\right)\right)
\mathrm{D}_{0}\left(k\right)\left(x\left(t\right)\right) - \delta
\mathrm{D}_{0}\left(k\right)\left(x\left(t\right)\right) +
k\left(x\left(t\right)\right)^{\alpha}
\mathrm{D}_{0}\left(x\right)\left(x\left(t\right)\right)\right)}
\lambda\left(x\left(t\right)\right) - {\left(\delta
k\left(x\left(t\right)\right) - k\left(x\left(t\right)\right)^{\alpha}
x\left(x\left(t\right)\right)\right)}
\mathrm{D}_{0}\left(\lambda\right)\left(x\left(t\right)\right).]
sage: D[0](h)(k(t)) r*(x(k(t)) - 1)*e^(-r*k(t))*k(k(t)) - (x(k(t)) -
1)*e^(-r*k(t))*D[0](k)(k(t)) - e^(-r*k(t))*k(k(t))*D[0](x)(k(t)) +
(alpha*k(k(t))^(alpha - 1)*x(k(t))*D[0](k)(k(t)) - delta*D[0](k)(k(t)) +
k(k(t))^alpha*D[0](x)(k(t)))*lambda_(k(t)) - (delta*k(k(t)) -
k(k(t))^alpha*x(k(t)))*D[0](lambda_)(k(t))
[image: \frac{\partial h(t)}{\partial k(t)}\,=\,r
{\left(x\left(k\left(t\right)\right) - 1\right)} e^{\left(-r
k\left(t\right)\right)} k\left(k\left(t\right)\right) -
{\left(x\left(k\left(t\right)\right) - 1\right)} e^{\left(-r
k\left(t\right)\right)}
\mathrm{D}_{0}\left(k\right)\left(k\left(t\right)\right) - e^{\left(-r
k\left(t\right)\right)} k\left(k\left(t\right)\right)
\mathrm{D}_{0}\left(x\right)\left(k\left(t\right)\right) + {\left(\alpha
k\left(k\left(t\right)\right)^{\alpha - 1} x\left(k\left(t\right)\right)
\mathrm{D}_{0}\left(k\right)\left(k\left(t\right)\right) - \delta
\mathrm{D}_{0}\left(k\right)\left(k\left(t\right)\right) +
k\left(k\left(t\right)\right)^{\alpha}
\mathrm{D}_{0}\left(x\right)\left(k\left(t\right)\right)\right)}
\lambda\left(k\left(t\right)\right) - {\left(\delta
k\left(k\left(t\right)\right) - k\left(k\left(t\right)\right)^{\alpha}
x\left(k\left(t\right)\right)\right)}
\mathrm{D}_{0}\left(\lambda\right)\left(k\left(t\right)\right).]
Not a pretty sight…
Le mardi 25 février 2025 à 11:01:27 UTC+1, [email protected] a écrit :
Hi,
Le lundi 24 février 2025 à 17:39:30 UTC+1, [email protected] a écrit :
So how should i define H in 'In [2]', so that 'In [4]' shows the
expected result.
Basically you should define H as a symbolic expression, not a function, and
make a distinction between functions and symbolic variables.
A solution is
t = var('t')
r = var('r')
delta = var('delta')
alpha = var('alpha')
x = function('x')
k = function('k')
lambda_ = function('lambda_')
X = var('X')
K = var('K')
L = var('L')
H = K*(1 - X)*exp(-r*t) + L*(K^alpha*X - delta*K)
to_functions = {X: x(t), K: k(t), L: lambda_(t)} # a dictionary to perform
substitutions
eq1 = (diff(H, X).subs(to_functions) == 0)
eq2 = (diff(H, K).subs(to_functions) == - diff(lambda_(t), t))
Best wishes,
Eric. .
--
You received this message because you are subscribed to the Google Groups
"sage-support" group.
To unsubscribe from this group and stop receiving emails from it, send an email
to [email protected].
To view this discussion visit
https://groups.google.com/d/msgid/sage-support/2f7e6fad-d453-4ec9-8d46-1ae186474986n%40googlegroups.com.
