The "s = var('s')" is not necessary (the argument s inside the functions
shadows it).
As for the original question, IMHO there is a learning opportunity here.
Numerical integration is powerful, but it doesn't give you symbolic
answers. Even if you make the integration bound a symbolic variable.
On Wednesday, September 10, 2014 6:09:06 AM UTC+1, Hal Snyder wrote:
>
> This works on my sage-6.1.1:
>
> s = var('s')
>
> def g(s):
> return numerical_integral(cos(pi*x^2/2), 0, s, max_points=100)[0]
>
> def h(s):
> return numerical_integral(sin(pi*x^2/2), 0, s, max_points=100)[0]
>
> p = plot((g,h),(-pi,pi),parametric=True)
> show(p)
>
> On Tuesday, September 9, 2014 5:17:14 PM UTC-5, Jotace wrote:
>>
>> Hi all,
>>
>> I want (my students) to plot Cornu's spiral, givent in parametric form by
>>
>> x(t) = integral cos(pi/u^2/2), u going from 0 to t , and y(t) defined
>> analogously using the sine function. The integral connot be evaluated
>> symbolically, I guess.
>>
>> The first attempt would be
>>
>> parametric_plot([integrate(cos(pi*u^2/2),u,0,t),integrate(sin(pi*u^2/2),u,0,t)],(t,-pi,pi))
>> which failw (coercion)
>>
>> The second attempt would be:
>>
>> parametric_plot([integral_numerical(cos(pi*u^2/2),0,t),integral_numerical(sin(pi*u^2/2),0,t)],(t,-pi,pi))
>> which also fails.
>>
>> I finally did:
>> def x(t):
>> return integral_numerical(cos(pi*u^2/2),0,t)[0]
>>
>> def y(t):
>> return integral_numerical(sin(pi*u^2/2),0,t)[0]
>>
>> Points = [(x(t),y(t)) for t in sxrange(-pi,pi,2*pi/200)]
>> line(Points).show(figsize=[5, 5],aspect_ratio=1)
>>
>> This works, but it looks highly inelegant. Also, i cannot expect my
>> students to come up with something like this in a first year undergrad
>> course.
>>
>> Is there a way to fix one of the first two options?
>>
>> Regards,
>> JC
>>
>>
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