Question

I am going through the above topic from CLRS(CORMEN) (page 834) and I got stuck at this point.

Can anybody please explain how the following expression,

A(x)=A^{[0]}(x^2) +xA^{[1]}(x^2)

follows from,

n-1                       `
 Σ  a_j x^j
j=0

Where,

A^{[0]} = a_0 + a_2x + a_4a^x ... a_{n-2}x^{\frac{n}{2-1}}  
A^{[1]} = a_1 + a_3x + a_5a^x ... a_{n-1}x^{\frac{n}{2-1}}

Answer 1

The polynomial A(x) is defined as

A(x) = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + ...

To start the divide-and-conquer strategy of polynomial multiplication by the FFT, CLRS introduces two new polynomials: one of the coefficients of the even-powers of x called A[0] and one of the coefficients of the odd-powers of x called A[1]

A[0](x) = a_0 + a_2 x + a_4 x^2 + ...
A[1](x) = a_1 + a_3 x + a_5 x^2 + ...

Now if we substitute x^2 into A[0] and A[1], we have

A[0](x^2) = a_0 + a_2 x^2 + a_4 x^4 + ...
A[1](x^2) = a_1 + a_3 x^2 + a_5 x^4 + ...

and if we multiply A[1](x^2) by x, we have

x A[1](x^2) = a_1 x + a_3 x^3 + a_5 x^5 + ...

Now if we add A[0](x^2) and x A[1](x^2), we have

A[0](x^2) + x A[1](x^2) = (a_0 + a_2 x^2 + a_4 x^4 + ...) + (a_1 x + a_3 x^3 + a_5 x^5 + ...)
                        = a_0 + a_1 x + a_2 x^2 + a_3 x^3 + ...
                        = A(x)

Q.E.D.