dimension of vector space & polynomial in two variables Say that a polynomial with real coefficients in two variable, $ x, y $ , is balanced if the average value of the polynomial on each circle centred at the origin is $ 0 $ . The balanced polynomials of degree at most $2009$ form a
vector space $ V $ over  $\mathbb{R}$. Find the dimension of V .
I tried doing some calculations using graphs but didn’t find any proper approach to solve this problem 
 A: This solution is incomplete, as it has a gap.

First, we consider a monomial of the form $ x^n y^m$, for some integers $n,m$. It is clear that if either one is odd (say $n$), then by reflection about the line $y=0$, we have a balanced polynomial.
By symmetry about the line $y=x$, terms of the form $ x^{2n}y^{2m} - x^{2m} y^{2n}$ are also balanced.
It remains to consider polynomials which are made up solely of terms of the form $ x^{2n} y^{2m} $, where $ n \geq m$. It is not clear to me if these are all.
A: Supplementing Calvin Lin's answer by the following observations.


*

*Monomials of different degrees scale differently as a function of the radius, so a polynomial is in the space $V$, iff all its homogeneous parts are.

*It suffices to test a homogeneous polynomial on the unit circle.

*The expected value of a form of an odd degree is automatically zero: $(x,y)\mapsto (-x,-y)$ negates everything.

*The expected value (=the average) of a form $F(x,y):=\sum_{i=0}^{2k}a_ix^iy^{2k-i}$ of an even degree $2k$,  is 
$$
E(F)=\sum_{i=0}^{2k}a_i E(x^iy^{2k-i})=\sum_{i=0, 2\mid i}^{2k} a_iE(x^iy^{2k-i}).
$$
Here all the expected values $E(x^iy^{2k-i})$ are clearly positive ($i$ even), so the constraint $E(F)=0$ determines one (any one) of the coefficients $a_i$, $2\mid i$, as a linear combination of the others.

*Therefore we get a single linear constraint on the coefficients of forms of a given even degree to be balanced, and no other constraints.

*The space of (bivariate) forms of degree $m$ has dimension $m+1$, so the answer is
$$\dim V=\sum_{k=0}^{2009}(m+1)-\sum_{k=0,2\mid k}^{2009}1.$$ Leaving the calculation of this sum to you :-)

