Find dimension of the intriguing vector space

Question

We are given vector space of polynomials over $\mathbb R$ of two variables with powers not higher than 2013.
Let's consider subspace $V$ which contains such polynomials $f$, so following holds for them:
$$\oint_{C} f(x,y)ds =0$$ Where $C = \{x^2+y^2 = R^2\}$.
Task is to find dim $V$.

First I must note that I felt undereducated for that problem when I first approached it, so I went through some books and materials and came up with solution.
Solution
I moved to polar coords and with use of the following formula I was able to write the required equation

$$R\int_0^{2\pi}\sum_{k_1=0; k_2=0}^{2013}{a_{k_1 k_2}R^{k_1k_2}cos^{k_1}t \cdot sin^{k_2} t\space dt} = 0$$ By the way, I have doubts that this formula applies there. But if it is, $$cos^{p}(0)=cos^p(0);\space sin^{p}(2\pi)=sin^p(2\pi)$$
Hence, it's a trivial subspace and any polynomial fits.
Than, power spans from $0$ to $2013$ for both variables and basis should have all of them, than $dim V = 2013^2$
I have doubts about it's correctness, but my main concern is understanding. So I would appreciate hints and corrections.

Answer 1

Sketch:

• Let $X$ be the vector space of all polynomials with powers up to 2013. Verify that $X$ has dimension $2014^2$ (remember to count the 0 power).

• Let $T : X \to \mathbb{R}$ be the map $Tf = \oint_C f$. Verify that $T$ is linear and surjective (i.e. its range is all of $\mathbb{R}$). (For surjectivity, it suffices to find a single $f$ such that $Tf \ne 0$).

• Observe that $V$ is the kernel (null space) of $T$.

• Recall the rank-nullity theorem.

Answer 2

Well, the dimension is certainly not $2013*2013$. For example, even-powered polynomials $x^{2n} y^{2m}$ certainly do not integrate to 0.

Hint: what do you know about $\int_0^{2\pi} \sin^p\!\theta \cos^q\!\theta \, d\theta$?

Hint2: apply integration by parts twice to reduce $\int_0^{2\pi} \sin^p\!\theta \cos^q\!\theta \, d\theta$ to $\int_0^{2\pi} \sin^{p-2}\!\theta \cos^{q+2}\!\theta \, d\theta$

Answer 3

The given integral equation imposes a single linear condition. You just need to check that the relation is not trivial by finding a polynomial whose integral is nonzero.