AnnD

I see the link with the differential equation, but according to the question, I only need to find the value of a. [math] a_{k-1} = \frac{-ka_k}{2} [/math] Isn't right ? BTW thank you very much for the help

What I've done is; [math] y(x) = \sum_{k=0}^{\infty} a_k x^k [/math] [math] y'(x) = \sum_{k=1}^{\infty} ka_k x^{k-1} [/math] By developing both series and grouping I got; [math] (2a_0+a_1) + (2a_1+a_2)x + ... [/math] Because they must all equal 0 to satisfay the first equation; [math] a_{k-1} = \frac{-ka_k}{2} [/math] But I'm not sure, and I have no clue how to verify this kind of problem with Maple (the "solve" function does not allow "\sum" in it)

How can you show that [math] \sum_{k=1}^{\infty} ka_kx^{k-1}+2\sum_{k=0}^{\infty} a_kx^k = 0 [/math]

But the question is if the curve is algebraic or not. Like shadow said, f(x) = cos(arccos(x)) is algebraic, despite having cosinus in it. So a cercle should also be algebric, despite being written with a cosinus.

Yes but on wikipedia; "In mathematics, a transcendental curve is a curve that is not an algebraic curve. Here for a curve C what matters is the point set (typically in the plane) underlying C, not a given parametrisation." Clearly; x = cos(t), y = sin(t) can be translated to x^2+y^2-1=0, so it's an algebraic curve.

Hello I'm curious about something about the function f(x) = (cos(x),sin(x)). Is it a transcendental or an algebraic curve ? It looks like an algebraic curve, but I don't know if you can have cos and sin in an algebraic curve.