Asymptotic Riemann Roch I am reading Ravi Vakil's Foundation of Algebraic Geometry 20.1.I.
I want to show this:

Suppose $X$ is projective (over an infinite field), $\mathfrak{F}$ is a coherent sheaf on $X$ with support of dimension $= n$, and $\mathfrak{L}$ is a line bundle on $X$. Show that $\chi(\mathfrak{L}^{\otimes m}\otimes \mathfrak{F})$ is a polynomial with degree at most $n$, and the coefficient of $m^n$ is $(\mathfrak{L}^n\cdot \mathfrak{F})$
Hint: Expanding $(\mathfrak{L}^{n+1}\cdot (\mathfrak{L}^{\otimes i}\otimes \mathfrak{F}))=0$ to get a  recursion for $\chi (\mathfrak{L}^m \otimes \mathfrak{F})$.

For the case that $\mathfrak{L}$ is very ample, I can obtain the result by induction on $n$ as follows:
The case $n=1$ follows from Riemman Roch for nonreduced curves (Exercise 18.4 S). And if $n>1$, then I can find a global section $s$ of $\mathfrak{L}$, such that $s$ does not passes through the assocaited points of $\mathfrak{F}$.
By the short exact sequence
$$0\to \mathfrak{L}^{\vee}\otimes \mathfrak{F}\xrightarrow{\times s}\mathfrak{F}\to \mathfrak{G}\to 0$$ where $\mathfrak{G}$ has support with dimension $n-1$. I can obtain the result from the fact that Euler characteristic is additive.
What I want to know is how to do the general case, I know that every line bundle on $X$ can be expressed as the difference of two very ample line bundles, but I don't know how to obtain the result from this. I also don't know how to use the hint, from the hint, I can obtain
$$\chi(\mathfrak{L}^{\otimes m}\otimes \frak{F})=\sum_{i=1}^{n+1}(-1)^{j+1}\chi(\frak{L}^{\otimes (m-i)}\otimes \frak{F})$$
but I don't know why this formula is useful.
Any help or hints are appreciated, thank you.
 A: I am also not sure how the recursive formula $$\mathcal{L}^{n+1} \cdot (\mathcal{L}^{\otimes i} \otimes \mathcal{F})=0$$ is supposed to help.
To prove polynomiality one can combine your idea for very ample line bundles with some techniques from this section of stacks project.
Write $\mathcal{L}=\mathcal{A} \otimes \mathcal{B}^{\vee}$ as the product of very ample bundles, $\mathcal{A}=\mathcal{O}(D_1), \mathcal{B}=\mathcal{O}(D_2)$. (16.6.F Vakil).
Consider $$0 \to \mathcal{O}(-D_1) \to \mathcal{O}_X \to i_{*} \mathcal{O}_{D_1} \to 0,$$ tensor it with $\mathcal{A}^{\otimes n_1} \otimes (\mathcal{B}^{\vee})^{\otimes n_2} \otimes \mathcal{F}$ and apply additivity of $\chi$: $$\chi(\mathcal{A}^{\otimes n_1} \otimes (\mathcal{B}^{\vee})^{\otimes n_2} \otimes \mathcal{F})-\chi(\mathcal{A}^{\otimes n_1 - 1} \otimes (\mathcal{B}^{\vee})^{\otimes n_2} \otimes \mathcal{F}) = \chi(\mathcal{A}^{\otimes n_1} \otimes (\mathcal{B}^{\vee})^{\otimes n_2} \otimes \mathcal{F} \otimes i_{*} \mathcal{O}_{D_1} )$$where now $\mathcal{F} \otimes i_{*} \mathcal{O}_{D_1}$ has lower dimensional support.
Analagously for $D_2$; thus we can inductively assume that $$P(n_1,n_2)-P(n_1 -1, n_2)$$ $$P(n_1, n_2) - P(n_1, n_2 -1)$$ are polynomials in $(n_1,n_2)$. Then, as the Stacks Project says, "A simple arithmetic argument shows that $P$ is a numerical polynomial of total degree at most dim (Supp ($\mathcal{F}$))". Then set $p(m)=P(m,m)$.
As far as the leading coefficient is concerned, what we want to do is to "take $n$-th derivative" (where $n$ is  dim (Supp ($\mathcal{F}$))). Discretely, we could take "$n$-th order backward discrete difference" to obtain $$n! a_n = \sum_{i=0}^{n} (-1)^i{{n}\choose{i}} p(-i)=\sum_{i=0}^{n} (-1)^i{{n}\choose{i}} \chi((\mathcal{L}^{\vee})^{i} \otimes \mathcal{F})= \mathcal{L}^n \cdot \mathcal{F}.$$
A: Here is my approach:
Let $f(t) = \chi(X, L^{\otimes t} \otimes F)$. Then, $int_X(L^{int: n+1} , L^{\otimes i} \otimes F) = 0$.
Expanding $ L^{\otimes i} \otimes F) = 0$ using the definition, we note that there are $\binom{n+1}{k}$  subsets of cardinality $k$ of $\{1, ... n\}$. Therefore, $int_X(L^{int: n+1} , L^{\otimes i} \otimes F) = \sum_{k=0}^{n+1} (-1)^k \binom{n+1} {k}\chi(X, L^{\otimes t-k} \otimes F)$. Therefore, $\sum_{k=0}^{n+1} (-1)^k \binom{n+1} {k} f(t-k) = 0$.
Extracting the $k=0$ piece separately , we get $f(t) = \sum_{k=0}^{n+1} (-1)^{k+1} \binom{n+1} {k} f(t-k) = 0$.
This is a linear recurrence whose characteristic polynomial is $(x-1)^{n+1}$. Therefore, from the formula for linear recurrences, $f(t)$ is a polynomial of degree $n$.
