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Aug 8, 2025
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Update polynomial.md #1491

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2 changes: 1 addition & 1 deletion src/algebra/polynomial.md
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Expand Up @@ -192,7 +192,7 @@ This algorithm was mentioned in [Schönhage's article](http://algo.inria.fr/semi

$$A^{-1}(x) \equiv \frac{1}{A(x)} \equiv \frac{A(-x)}{A(x)A(-x)} \equiv \frac{A(-x)}{T(x^2)} \pmod{x^k}$$

Note that $T(x)$ can be computed with a single multiplication, after which we're only interested in the first half of coefficients of its inverse series. This effectively reduces the initial problem of computing $A^{-1} \pmod{x^k}$ to computing $T^{-1} \pmod{x^{\lfloor k / 2 \rfloor}}$.
Note that $T(x)$ can be computed with a single multiplication, after which we're only interested in the first half of coefficients of its inverse series. This effectively reduces the initial problem of computing $A^{-1} \pmod{x^k}$ to computing $T^{-1} \pmod{x^{\lceil k / 2 \rceil}}$.

The complexity of this method can be estimated as

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