2
$\begingroup$

I've never really understood how point-value multiplication of polynomials work, so I was wondering if somebody could talk me through it with an example.

Say if I was given the following two polynomials $A(x)$ and $B(x)$ in their point value representations: $((1,4), (i,0), (-1,0), (-i,0))$ and $((1,6),(i,0),(-2,0),(-i,0))$ respectively. How would I go about computing $C(x) = A(x) \cdot B(x)$?

I know that I have to multiply the values of $A$ and $B$ at each point, but that would only give 4 points which would not be enough to deduce the coefficient form of $C(x)$. I've read in some books that I would have to extend the point-value representation of $A(x)$ and $B(x)$ but I don't know how I would do this.

(I'm not sure if this will make much difference to the suggestions/explanations, but I need to know about this in the context of the Fast Fourier Transform and using it to multiply polynomials in $\Theta (n \cdot log(n))$ time)

$\endgroup$
1
  • $\begingroup$ well, first thing in order to do the multiplication A and B you need to know the degree bound of A and B. $\endgroup$ – dave Jun 13 '16 at 20:40
0
$\begingroup$

Suppose you have

A(x)=1+x

B(x)=1-x

degree of C(x) is 2

so I choose A=(1,1,0,0) B=(1,-1,0,0)

the point value representations are

A=( (i,1+i),(-1,0),(-i,1-i),(1,2) )

B=( (i,1-i),(-1,2),(-i,1+i),(1,0) )

Then

C=( (i,2),(-1,0),(-i,2),(1,0)} )

so now you can do the inverse of the Fast Fourier Transform and obtain

1-x^2=(1,0,-1,0)

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.