Size of bitwise OR between vectors represented as polynomials I am using polynomials to represent bit vectors, such that the value of the d-th coefficient is $1$ if the d-th bit in the vector is $1$, and $0$ otherwise.
For example, the vector $A = [1, 1, 0, 1, 0]$ is represented by the polynomial $P_A(x) = 1 + 1x + 0x^2 + 1x^3 + 0x^4 = 1 + x + x^3$.

I understand that I can find the number of bits $1$ in the bitwise AND of two vectors, $|A \wedge B|$, by multiplying $P_A(x)$ and the "inverted" $P_B(x)$ and checking the (d-1)-th coefficient in the resulting polynomial, as explained in Scalar product of vectors over polynomial rings.

Now I want a similar way to find the number of bits $1$ in the bitwise OR: $|A \vee B|$.
Considering $|A \vee B| = |A| + |B| - |A \wedge B|$, so far I thought about of $|A|$ by multiplying $P_A(x) \cdot P_1(x)$, where $P_1(x)$ is the polynomial having the value $1$ for all coefficients. Then, I would do the same to find $|B|$. 

This would require three polynomial multiplications and two additions: $P_A(x) \cdot P_1(x) + P_B(x) \cdot P_1(x) + P_A(x) \cdot P_B(x)$. 
Since the first two have $P_1(x)$ in common, it is possible to simplify it a bit: $(P_A(x) + P_B(x)) \cdot P_1(x) + P_A(x) \cdot P_B(x)$, using a total of two polynomial multiplications.

My question is: Is there a way to do this using fewer polynomial multiplications?
It holds that $$\begin{align}|A \vee B|&amp;=d-1-|\overline{A \vee B}|\\
&amp;=d-1-|\overline A \wedge \overline B|
\end{align}$$

Define $K(x)$ as the polynomial of degree $d-1$ with all 1 coefficients. $\overline A$ can be represented as the polynomial $K(x)-A(x)$.

Thus the desired $|A \vee B|$ is obtained by subtracting from $d-1$ the coefficient for $x^{d-1}$ of the polynomial $(K(x)-A(x))\cdot(K(x)-\tilde B(x))$, where $\tilde B(x)$ is $B(x)$ with "inversion" of coefficients, that is the coefficient of degree $i$ becomes coefficient of degree $d-1-i$.

We can use polynomial addition rather than subtraction: $d-1$ minus the coefficient for $x^{d-1}$ of the polynomial $(A(x)+L(x))\cdot(\tilde B(x)+L(x))$ also works (where $L(x)$ is the polynomial of degree $d-1$ with all $-1$ coefficients).