Jason Grout wrote:
> Jim Clark wrote:
>> The reference manual shows the following example for the gradient()
>> function:
>>
>> sage: x,y = var('x y')
>> sage: f = x^2+y^2
>> sage: f.gradient()
>> (2*x, 2*y)
>>
>> However, if instead I enter:
>>
>> sage: x,y,n = var('x y n')
>> sage: f = x^n+y^n
>> sage: f.gradient()
>> (y^n*log(y) + x^n*log(x), n*x^(n - 1), n*y^(n - 1))
>>
>> (not what I wanted, but I can understand what happened.)
>> So I tried:
>>
>> sage: f(x,y) = x^n+y^n
>> sage: f.gradient()
>> ((x, y) |--> y^n*log(y) + x^n*log(x), (x, y) |--> n*x^(n - 1), (x, y)
>> |--> n*y^(n - 1))
>>
>> So even if I specify that f is a function of x and y,
>> gradient() still insists on also differentiating w.r.t. n
>>
>> How do I tell gradient() that n is a constant?
>
>
> Good point. Right now, the gradient function looks like this:
>
> from sage.modules.free_module_element import vector
> l=[self.derivative(x) for x in self.variables()]
> return vector(l)
>
> That second line should probably be
> l=[self.derivative(x) for x in self.arguments()]
>
> and then your last example should work.
I've posted a patch to http://trac.sagemath.org/sage_trac/ticket/4343
Can you apply the patch and test it out?
Here is the new behavior:
sage: f(x,y) = x^n+y^n
sage: f.gradient()
((x, y) |--> n*x^(n - 1), (x, y) |--> n*y^(n - 1))
sage: f.gradient([y,x])
((x, y) |--> n*y^(n - 1), (x, y) |--> n*x^(n - 1))
Thanks,
Jason
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