Operations on polynomials and series

In this article we will cover common operations that you will probably have to do if you deal with polynomials.

Basic Notion and Facts

Consider a polynomial $A(x) = a_0 + a_1 x + \dots + a_n x^n$ such that $a_n \neq 0$.

Basic implementation

Here you can find the basic implementation of polynomial algebra.

It supports all trivial operations and some other useful methods. The main class is poly<T> for polynomials with coefficients of class T.

All arithmetic operation +, -, *, % and / are supported, % and / standing for remainder and quotient in integer division.

There is also the class modular<m> for performing arithmetic operations on remainders modulo a prime number m.

Other useful functions:



The very core operation is the multiplication of two polynomials, that is, given polynomial $A$ and $B$: $$A = a_0 + a_1 x + \dots + a_n x^n$$ $$B = b_0 + b_1 x + \dots + b_m x^m$$ You have to compute polynomial $C = A \cdot B$: $$\boxed{C = \sum\limits_{i=0}^n \sum\limits_{j=0}^m a_i b_j x^{i+j}} = c_0 + c_1 x + \dots + c_{n+m} x^{n+m}$$ It can be computed in $O(n \log n)$ via the Fast Fourier transform and almost all methods here will use it as subroutine.

Inverse series

If $A(0) \neq 0$ there always exists an infinite series $A^{-1}(x) = \sum\limits_{i=0}^\infty a_i'x^i$ such that $A^{-1} A = 1$.

It may be reasonable for us to calculate first $k$ coefficients of $A^{-1}$:

  1. Let's say that $A^{-1} \equiv B_k \pmod{x^{a}}$. That means that $A B_k \equiv 1 \pmod {x^{a}}$.
  2. We want to find $B_{k+1} \equiv B_k + x^{a}C \pmod{x^{2a}}$ such that $A B_{k+1} \equiv 1 \pmod{x^{2a}}$: $$A(B_k + x^{a}C) \equiv 1 \pmod{x^{2a}}$$
  3. Note that since $A B_k \equiv 1 \pmod{x^{a}}$ it also holds that $A B_k \equiv 1 + x^a D \pmod{x^{2a}}$. Thus: $$x^a(D+AC) \equiv 0 \pmod{x^{2a}} \implies D \equiv -AC \pmod{x^a} \implies C \equiv -B_k D \pmod{x^a}$$
  4. From this we obtain that: $$x^a C \equiv -B_k x^a D \equiv B_k(1-AB_k) \pmod{x^{2a}} \implies \boxed{B_{k+1} \equiv B_k(2-AB_k) \pmod{x^{2a}}}$$

Thus starting with $B_0 \equiv a_0^{-1} \pmod x$ we will compute the sequence $B_k$ such that $AB_k \equiv 1 \pmod{x^{2^k}}$ with the complexity: $$T(n) = T(n/2) + O(n \log n) = O(n \log n)$$

Division with remainder

Consider two polynomials $A(x)$ and $B(x)$ of degrees $n$ and $m$. As it was said earlier you can rewrite $A(x)$ as:

$$A(x) = B(x) D(x) + R(x), \deg R < \deg B$$

Let $n \geq m$, then you can immediately find out that $\deg D = n - m$ and that leading $n-m+1$ coefficients of $A$ don't influence $R$.

That means that you can recover $D(x)$ from the largest $n-m+1$ coefficients of $A(x)$ and $B(x)$ if you consider it as a system of equations.

The formal way to do it is to consider the reversed polynomials: $$A^R(x) = x^nA(x^{-1})= a_n + a_{n-1} x + \dots + a_0 x^n$$ $$B^R(x) = x^m B(x^{-1}) = b_m + b_{m-1} x + \dots + b_0 x^m$$ $$D^R(x) = x^{n-m}D(x^{-1}) = d_{n-m} + d_{n-m-1} x + \dots + d_0 x^{n-m}$$

Using these terms you can rewrite that statement about the largest $n-m+1$ coefficients as:

$$A^R(x) \equiv B^R(x) D^R(x) \pmod{x^{n-m+1}}$$

From which you can unambiguously recover all coefficients of $D(x)$:

$$\boxed{D^R(x) \equiv A^R(x) (B^R(x))^{-1} \pmod{x^{n-m+1}}}$$

And from this in turn you can easily recover $R(x)$ as $R(x) = A(x) - B(x)D(x)$.

Calculating functions of polynomial

Newton's method

Let's generalize the inverse series approach. You want to find a polynomial $P(x)$ satisfying $F(P) = 0$ where $F(x)$ is some function represented as: $$F(x) = \sum\limits_{i=0}^\infty \alpha_i (x-\beta)^k$$

Where $\beta$ is some constant. It can be proven that if we introduce a new formal variable $y$, we can express $F(x)$ as: $$F(x) = F(y) + (x-y)F'(y) + (x-y)^2 G(x,y)$$

Where $F'(x)$ is the derivative formal power series defined as $F'(x) = \sum\limits_{i=0}^\infty (k+1)\alpha_{i+1}(x-\beta)^k$ and $G(x, y)$ is some formal power series of $x$ and $y$.

Given this we can find the coefficients of the solution iteratively:

  1. Assume that $F(Q_k) \equiv 0 \pmod{x^{a}}$, we want to find $Q_{k+1} \equiv Q_k + x^a C \pmod{x^{2a}}$ such that $F(Q_{k+1}) \equiv 0 \pmod{x^{2a}}$.
  2. Pasting $x = Q_{k+1}$ and $y=Q_k$ in the formula above we get: $$F(Q_{k+1}) \equiv F(Q_k) + (Q_{k+1} - Q_k) F'(Q_k) + (Q_{k+1} - Q_k)^2 G(x, y) \pmod x^{2a}$$
  3. Since $Q_{k+1} - Q_k \equiv 0 \pmod{x^a}$ we can say that $(Q_{k+1} - Q_k)^2 \equiv 0 \pmod{x^{2a}}$, thus: $$0 \equiv F(Q_{k+1}) \equiv F(Q_k) + (Q_{k+1} - Q_k) F'(Q_k) \pmod{x^{2a}}$$
  4. From the last formula we derive the value of $Q_{k+1}$: $$\boxed{Q_{k+1} = Q_k - \dfrac{F(Q_k)}{F'(Q_k)} \pmod{x^{2a}}}$$

Thus knowing how to invert arbitrary polynomial and how to compute $F(Q_k)$ quickly, we can find $n$ coefficients of $P$ with the complexity: $$T(n) = T(n/2) + f(n)$$ Where $f(n)$ is the maximum of $O(n \log n)$ needed to invert series and time needed to compute $F(Q_k)$ which is usually also $O(n \log n)$.


For the function $\ln P(x)$ it's known that: $$\boxed{(\ln P(x))' = \dfrac{P'(x)}{P(x)}}$$ Thus we can calculate $n$ coefficients of $\ln P(x)$ in $O(n \log n)$.

Inverse series

Turns out, we can get the formula for $A^{-1}$ using Newton's method. For this we take the equation $A=Q^{-1}$, thus: $$F(Q) = Q^{-1} - A$$ $$F'(Q) = -Q^{-2}$$ $$\boxed{Q_{k+1} \equiv Q_k(2-AQ_k) \pmod{x^{2^{k+1}}}}$$


Let's learn to calculate $e^{P(x)}=Q(x)$. It should hold that $\ln Q = P$, thus: $$F(Q) = \ln Q - P$$ $$F'(Q) = Q^{-1}$$ $$\boxed{Q_{k+1} \equiv Q_k(1 + P - \ln Q_k) \pmod{x^{2^{k+1}}}}$$

$k$-th power

Now we need to calculate $P^k(x)=Q$. This may be done via the following formula: $$Q = \exp\left[k \ln P(x)\right]$$ Note though, that you can calculate the logarithms and the exponents correctly only if you can find some initial $Q_0$.

To find it, you should calculate the logarithm or the exponent of the constant coefficient of the polynomial.

But the only reasonable way to do it is if $P(0)=1$ for $Q = \ln P$ so $Q(0)=0$ and if $P(0)=0$ for $Q = e^P$ so $Q(0)=1$.

Thus you can use formula above only if $P(0) = 1$. Otherwise if $P(x) = \alpha x^t T(x)$ where $T(0)=1$ you can write that: $$\boxed{P^k(x) = \alpha^kx^{kt} \exp[k \ln T(x)]}$$ Note that you also can calculate some $k$-th root of a polynomial if you can calculate $\sqrt[k]{\alpha}$, for example for $\alpha=1$.

Evaluation and Interpolation

Chirp-z Transform

For the particular case when you need to evaluate a polynomial in the points $x_r = z^{2r}$ you can do the following:

$$A(z^{2r}) = \sum\limits_{k=0}^n a_k z^{2kr}$$

Let's substitute $2kr = r^2+k^2-(r-k)^2$. Then this sum rewrites as:

$$\boxed{A(z^{2r}) = z^{r^2}\sum\limits_{k=0}^n (a_k z^{k^2}) z^{-(r-k)^2}}$$

Which is up to the factor $z^{r^2}$ equal to the convolution of the sequences $u_k = a_k z^{k^2}$ and $v_k = z^{-k^2}$.

Note that $u_k$ has indexes from $0$ to $n$ here and $v_k$ has indexes from $-n$ to $m$ where $m$ is the maximum power of $z$ which you need.

Now if you need to evaluate a polynomial in the points $x_r = z^{2r+1}$ you can reduce it to the previous task by the transformation $a_k \to a_k z^k$.

It gives us an $O(n \log n)$ algorithm when you need to compute values in powers of $z$, thus you may compute the DFT for non-powers of two.

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:

  1. Compute a segment tree such that in the segment $[l;r)$ stands the product $P_{l, r}(x) = (x-x_l)(x-x_{l+1})\dots(x-x_{r-1})$.
  2. Starting with $l=1$ and $r=n$ at the root node. Let $m=\lfloor(l+r)/2\rfloor$. Let's move down to $[l;m)$ with the polynomial $A(x) \pmod{P_{l,m}(x)}$.
  3. This will recursively compute $A(x_l), \dots, A(x_{m-1})$, now do the same for $[m;r)$ with $A(x) \pmod{P_{m,r}(x)}$.
  4. Concatenate the results from the first and second recursive call and return them.

The whole procedure will run in $O(n \log^2 n)$.


There's a direct formula by Lagrange to interpolate a polynomial, given set of pairs $(x_i, y_i)$:

$$\boxed{A(x) = \sum\limits_{i=1}^n y_i \prod\limits_{j \neq i}\dfrac{x-x_j}{x_i - x_j}}$$

Computing it directly is a hard thing but turns out, we may compute it in $O(n \log^2 n)$ with a divide and conquer approach:

Consider $P(x) = (x-x_1)\dots(x-x_n)$. To know the coefficients of the denominators in $A(x)$ we should compute products like: $$P_i = \prod\limits_{j \neq i} (x_i-x_j)$$

But if you consider the derivative $P'(x)$ you'll find out that $P'(x_i) = P_i$. Thus you can compute $P_i$'s via evaluation in $O(n \log^2 n)$.

Now consider the recursive algorithm done on same segment tree as in the multi-point evaluation. It starts in the leaves with the value $\dfrac{y_i}{P_i}$ in each leaf.

When we return from the recursion we should merge the results from the left and the right vertices as $A_{l,r} = A_{l,m}P_{m,r} + P_{l,m} A_{m,r}$.

In this way when you return back to the root you'll have exactly $A(x)$ in it. The total procedure also works in $O(n \log^2 n)$.

GCD and Resultants

Assume you're given polynomials $A(x) = a_0 + a_1 x + \dots + a_n x^n$ and $B(x) = b_0 + b_1 x + \dots + b_m x^m$.

Let $\lambda_0, \dots, \lambda_n$ be the roots of $A(x)$ and let $\mu_0, \dots, \mu_m$ be the roots of $B(x)$ counted with their multiplicities.

You want to know if $A(x)$ and $B(x)$ have any roots in common. There are two interconnected ways to do that.

Euclidean algorithm

Well, we already have an article about it. For an arbitrary euclidean domain you can write the Euclidean algorithm as easy as:

template<typename T>
T gcd(const T &a, const T &b) {
    return b == T(0) ? a : gcd(b, a % b);

It can be proven that for polynomials $A(x)$ and $B(x)$ it will work in $O(nm)$.


Let's calculate the product $A(\mu_0)\cdots A(\mu_m)$. It will be equal to zero if and only if some $\mu_i$ is the root of $A(x)$.

For symmetry we can also multiply it with $b_m^n$ and rewrite the whole product in the following form: $$\boxed{\mathcal{R}(A, B) = b_m^n\prod\limits_{j=0}^m A(\mu_j) = b_m^n a_m^n \prod\limits_{i=0}^n \prod\limits_{j=0}^m (\mu_j - \lambda_i)= (-1)^{mn}a_n^m \prod\limits_{i=0}^n B(\lambda_i)}$$

The value defined above is called the resultant of the polynomials $A(x)$ and $B(x)$. From the definition you may find the following properties:

  1. $\mathcal R(A, B) = (-1)^{nm} \mathcal R(B, A)$.
  2. $\mathcal R(A, B)= a_n^m b_m^n$ when $n=0$ or $m=0$.
  3. If $b_m=1$ then $\mathcal R(A - CB, B) = \mathcal R(A, B)$ for an arbitrary polynomial $C(x)$ and $n,m \geq 1$.
  4. From this follows $\mathcal R(A, B) = b_m^{\deg(A) - \deg(A-CB)}\mathcal R(A - CB, B)$ for arbitrary $A(x)$, $B(x)$, $C(x)$.

Miraculously it means that resultant of two polynomials is actually always from the same ring as their coefficients!

Also these properties allow us to calculate the resultant alongside the Euclidean algorithm, which works in $O(nm)$.

template<typename T>
T resultant(poly<T> a, poly<T> b) {
    if(b.is_zero()) {
        return 0;
    } else if(b.deg() == 0) {
        return bpow(b.lead(), a.deg());
    } else {
        int pw = a.deg();
        a %= b;
        pw -= a.deg();
        base mul = bpow(b.lead(), pw) * base((b.deg() & a.deg() & 1) ? -1 : 1);
        base ans = resultant(b, a);
        return ans * mul;

Half-GCD algorithm

There is a way to calculate the GCD and resultants in $O(n \log^2 n)$. To do this you should note that if you consider $a(x) = a_0 + x^k a_1$ and $b(x) = b_0 + x^k b_1$ where $k=\min(\deg a, \deg b)/2$ then basically the first few operations of Euclidean algorithm on $a(x)$ and $b(x)$ are defined by the Euclidean algorithm on $a_1(x)$ and $b_1(x)$ for which you may also calculate GCD recursively and then somehow memorize linear transforms you made with them and apply it to $a(x)$ and $b(x)$ to lower the degrees of polynomials. Implementation of this algorithm seems pretty tedious and technical thus it's not considered in this article yet.