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Show that all Eisenstein series $G_k$ can be expressed as polynomials in $G_4$ and $G_6$, e.g. express $G_8$ and $G_{10}$ in this way. Hint: Setting $a_n := (2n+1)G_{2n+2}$, show that $2n(2n-1)a_n = 6(2a_n + \sum\limits_{k=1}^{n-2} a_{k}a_{n-1-k})$ , $n>2$

Obviously, what I have trouble proving, is the Hint. It seems, like I should use the second derivative of Weierstrass function representation, using the Eisenstein series (namely $\wp = \displaystyle\frac{1}{z^2} + \sum\limits_{m=1}^{\infty} (2m+1)G_{2m+2}z^{2m}$), but I'm not sure where the sum on the right comes from.

Any help will be appreciated.

EDIT: Can someone help some more?

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up vote 3 down vote accepted

Use the differential equaltion satisfied by $\wp$: $$ \wp'^2 = 4\wp^3 -g_2 \wp -g_3 $$ from this you get easily (as $g_2 = 60 G_4$) $$ \wp'' = 6 \wp^2 -10 a_1 $$ which translates directly into the hint if you substitute $\wp$ with the series you give.

EDIT: TO clarify further, squaring $$ \wp(z) = \frac{1}{z^2} + \sum_{m>=1} a_m z^{2m} $$ gives $$\begin{align*} \wp(z)^2 &= \frac{1}{z^4} + \frac{2}{z^2}\sum_{m\ge1} a_m z^{2m} + \left(\sum_{m>=1} a_m z^{2m}\right)^2 =\\ &= \frac{1}{z^4} + 2\sum_{m\ge 1} a_m z^{2m-2} + \sum_{m\ge 1} z^{2m} \sum_{k,l \ge 1, k+l = m} a_ka_l\\ &= \frac{1}{z^4} + 2\sum_{m\ge 1} a_m z^{2m-2} + \sum_{m\ge 2} z^{2m-2} \sum_{k=1}^{m-2} a_ka_{m-1-k} \end{align*} $$ I hope this helps.

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Thanks for your answer, but I still can't see it. As I thought, the left-wise side of the equation in Hint is the coefficient of $z^{2n-2}$ in $\wp''$. As I see it, I should look for the coefficient of the same in $\wp^2$, meaning that I'm looking for the coefficient of $z^{n-1}$ (as squared) and $z^{2n}$ (as multiplication by $\frac{1}{z^2}$). Is it the wrong way? – Pavel Mar 7 '11 at 16:51
Thanks! Never known a good way to work with sums like those. – Pavel Mar 7 '11 at 20:40

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