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, egourg...@gmail.com a écrit :
>
>> Hi,
>>
>> Le lundi 24 février 2025 à 17:39:30 UTC+1, bernd.bre...@gmail.com 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. .
>>
>
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