I am currently following this course I found on the internet about modular forms: http://www.few.vu.nl/~sdn249/modularforms16/ . My goal is to prove that the equality $$\sigma_9(n)=\frac{21}{11}\sigma_5(n)-\frac{10}{11}\sigma_3(n)+\frac{5040}{11}\sum_{j=1}^{n-1}\sigma_3(j)\sigma_5(n-j)$$ holds for all $n\in\mathbf{Z}_{>0}$.
So far I have noticed that the space of modular forms of weight $10$ has dimension 1, which tells me that if I want to check whether two modular forms of weight $10$ with $q$-expansions $\sum_{n=0}^\infty a_nq^n$ and $\sum_{n=0}^\infty b_nq^n$ coincide, then I only need to check that $a_0=b_0$ and $a_1=b_1$, which holds if we substitute $n=0$ and $n=1$ in the equality above. Is that it?
Because if that is the case, then my question is, who came up with that equality? Is there a straightforward way of deducing it? Because later on in that problem, I am asked to find similar expressions for $\sigma_{13}$ in terms of $\sigma_3$ and $\sigma_9$ and in terms of $\sigma_5$ and $\sigma_7$, and I do not see how to prove that. Could anyone help me?
Thank you for your time.
Thanks to Somos' answer I got the following solution:
SOLUTION: We want to show that $E_4E_6=E_{10}$. Observe that \begin{align*} E_4E_6&=\left(1+240\sum_{n=1}^{+\infty}\sigma_3(n)q^n\right)\left(1-504\sum_{n=1}^{+\infty}\sigma_5(n)q^n\right)\\ &=1+240\sum_{n=1}^{+\infty}\sigma_3(n)q^n-504\sum_{n=1}^{+\infty}\sigma_5(n)q^n-240\cdot 504\sum_{n=1}^{+\infty}\sum_{m=1}^{n-1}\sigma_3(m)\sigma_5(n-m)q^n\\ &=1+\sum_{n=1}^{+\infty}\left(240\sigma_3(n)-504\sigma_5(n)-120960\sum_{m=1}^{+\infty}\sigma_3(m)\sigma_5(n-m)\right)q^n. \end{align*} On the other hand, we have that $$E_{10}=1-264\sum_{n=1}^\infty\sigma_9(n)q^n.$$ It is clear that the $0$-th coefficient of $E_4E_6$ and $E_{10}$ coincides, and it is $1$. On the other hand, the first coefficients are, respectively $-264$ and $240-504=-264$, hence they are also equal. Since both $E_4E_6$ and $E_{10}$ are modular forms of weight $10$ and $\dim M_{10}=1$, by Cor.2.13 from the course notes, we have that the previous is enough to show that $E_4E_6=E_{10}$, hence in particular the coefficients of the $q$-expansion satisfy the desired equality.