Fix malformed LaTeX in polynomial.md.

algmyr · adamant-pwn · commit f1e8232c7060 · 2022-10-17T13:35:37.000+02:00
Seems like some LaTeX-like engine disagrees with LaTeX about what is an atom.

LaTeX parses `a^\cmd{x}{y}` as `{a^\cmd{x}}{y}`, rather than `a^{\cmd{x}{y}}` which is what's intended here.
diff --git a/src/algebra/polynomial.md b/src/algebra/polynomial.md
@@ -270,9 +270,9 @@ It gives us an $O(n \log n)$ algorithm when you need to compute values in powers
 
 Another observation is that $kr = \binom{k+r}{2} - \binom{k}{2} - \binom{r}{2}$. Then we have
 
-$$\boxed{A(z^r) = z^{-\binom{r}{2}}\sum\limits_{k=0}^n \left(a_k z^{-\binom{k}{2}}\right)z^\binom{k+r}{2}}$$
+$$\boxed{A(z^r) = z^{-\binom{r}{2}}\sum\limits_{k=0}^n \left(a_k z^{-\binom{k}{2}}\right)z^{\binom{k+r}{2}}}$$
 
-The coefficient of $x^{n+r}$ of the product of the polynomials $A_0(x) = \sum\limits_{k=0}^n a_{n-k}z^{-\binom{n-k}{2}}x^k$ and $A_1(x) = \sum\limits_{k\geq 0}z^{\binom{k}{2}}x^k$ equals $z^{\binom{r}{2}}A(z^r)$. You can use the formula $z^\binom{k+1}{2}=z^{\binom{k}{2}+k}$ to calculate the coefficients of $A_0(x)$ and $A_1(x)$.
+The coefficient of $x^{n+r}$ of the product of the polynomials $A_0(x) = \sum\limits_{k=0}^n a_{n-k}z^{-\binom{n-k}{2}}x^k$ and $A_1(x) = \sum\limits_{k\geq 0}z^{\binom{k}{2}}x^k$ equals $z^{\binom{r}{2}}A(z^r)$. You can use the formula $z^{\binom{k+1}{2}}=z^{\binom{k}{2}+k}$ to calculate the coefficients of $A_0(x)$ and $A_1(x)$.
 
 ### Multi-point Evaluation
 Assume you need to calculate $A(x_1), \dots, A(x_n)$. As mentioned earlier, $A(x) \equiv A(x_i) \pmod{x-x_i}$. Thus you may do the following:
