# modular forms and their fourier coefficients question

I was recently listening to Don Zagiers fourth lecture at ICTP (posted Feb 5, 2015) on mock modular forms. At roughly 22:00 in the lecture he makes two statements: 1. the product of the weight and the level of some modular form divided by 12 is roughly how the fourier coefficients behave early on (if I understand him correctly), and 2. to rigorously relate some two functions that happen to have coefficients of the same modular form then computing the weight, level, and first 33 fourier coefficients is a brute force way to do this.

My question is two fold: why does the number in 1. happen to be roughly 1/12, and why does the the number in 2. happen to be roughly 33? Are there any introductory texts that go over this?

Thank!

The Sturm bound for $$M_k(\Gamma_1(N))$$ is $$k m/12$$ where $$m$$ is the index of $$\Gamma_1(N)$$ in $$SL_2(\Bbb{Z})$$, that we can replace by $$N^2$$. Comparing the first $$k N^2/12$$ coefficients of $$f,g\in M_k(\Gamma_1(N))$$ is a proof that they are equal.
The proof is not difficult, let $$h=\prod_{\alpha \in \Gamma_1(N)\setminus SL_2(\Bbb{Z})} (f-g)|_k \alpha\in M_{mk}(SL_2(\Bbb{Z}))$$ if the constant term and the $$\lfloor km/12\rfloor$$ first coefficients of $$f-g$$ vanish then $$h/\Delta^{\lfloor km/12\rfloor}$$ vanishes at $$i\infty$$ ie. it is in $$S_{mk - 12\lfloor km/12\rfloor}(SL_2(\Bbb{Z}))$$ which contains only $$0$$ and hence $$h=0,f-g=0$$.