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Multiplication of polynomials by Fast Fourier Transform (FFT):

Fast Fourier Transform (FFT):

The $4脳4$matrix of FFT is,

localid="1659009189962" $M_4\left({蝇}^{\text{'}}\right)=\left[\begin{array}{cccc}1& 1& 1& 1\\ 1& 蝇& {蝇}^2& {蝇}^3\\ 1& {蝇}^2& {蝇}^3& {蝇}^6\\ 1& {蝇}^3& {蝇}^6& {蝇}^9\\ & & & \end{array}\right].....\left(1\right)$

The difficult result of a nth fundamental plus unity is,

${蝇}^{\text{'}}=e^{\frac{2\mathrm{蟺颈}}{n}}$ 鈥︹�� (2)

Alternative in Equations, the value of " n " is " 4 " (2). As a result, the value of is

$$\begin{gathered} {蝇}^{\text{'}}=e^{\frac{2\mathrm{蟺颈}}{4}} \\ =e^{\frac{蟺i}{2}} \end{gathered}$$

Note: $e^{ix}=\cos \left(x\right)+i\sin \left(x\right)$

Thus, $e^{\frac{\mathrm{蟺颈}}{2}}$has already been prepared in accordance with the information provided:

$$\begin{gathered} 蝇=\cos \left(\frac{\mathrm{蟺}}{2}\right)+i\sin \left(\frac{\mathrm{蟺}}{2}\right) \\ =\cos \left(90\right)+i\sin \left(90\right) \\ =0+i\left(1\right) \\ 蝇=i...........\left(3\right) \end{gathered}$$

Alternative Equation (3) in Equation (1).

$M_4\left({蝇}^{\text{'}}\right)=\left[\begin{array}{cccc}1& 1& 1& 1\\ 1& i& i^2& i^3\\ 1& i^2& i^3& i^6\\ 1& i^3& i^6& i^9\\ & & & \end{array}\right].....\left(4\right)$

In general, the value of $i,i^2,i^3,i^4$are,

$$\begin{gathered} i=i \\ {i}^2=-1 \\ {i}^3=-i \\ {i}^4=1 \end{gathered}$$ ...........(5)

The counting value of $i^6,i^9$is,

$$\begin{gathered} i^6=i^4脳i^2=-1 \\ {i}^9=i^6脳i^3=i \end{gathered}$$ 鈥︹�� (6)

Alternative counting values are $i,i^2,i^3,i^4,i^6,i^9$in Equation (4). The FFT matrix is,

$M_4\left({蝇}^{\text{'}}\right)=\left[\begin{array}{cccc}1& 1& 1& 1\\ 1& i& -1& -i\\ 1& -1& 1& -1\\ 1& -i& -1& i\\ & & & \end{array}\right].....\left(7\right)$

Polynomial multiplication:

Question now been written in accordance with the information provided.

$A\left(x\right)=a_0+a_{1}x+....+a_d{x}^d.....\left(8\right)$

The first polynomial is x + 1 鈥︹�� (9)

Similarity equations (8) and (9),

$a_0=1,a_1=1,a_2=0\mathrm{and}{a}_3=0.....\left(10\right)$

Considering the polynomials into matrix transformation formula..

localid="1659009567989" $A=\left[\begin{array}{c}{a}_0\\ a_1\\ a_2\\ a_3\\ \end{array}\right]$ 鈥︹�� (11)

2^nd part of the above values of $a_0,a_1,a_2,a_3$ in Equation (11).

$A=\left[\begin{array}{c}\begin{array}{c}1\\ 1\\ 0\\ 0\end{array}\end{array}\right]...\left(12\right)$

The FFT of A ( 1, 1, 0, 0 )is calculated as follows:

FFT of $A=M_4\left(蝇\right)脳A..........\left(13\right)$

Substitute the values $M_4\left(蝇\right)$from Equation (7) and A from Equation (12) in Equation (13).

$=\left[\begin{array}{cccc}1& 1& 1& 1\\ 1& i& -1& -i\\ 1& -1& 1& -1\\ 1& -i& -1& i\end{array}\right]*\left[\begin{array}{c}1\\ 0\\ 0\\ 0\end{array}\right]$

$$\begin{gathered} =\left[\begin{array}{c}1+1+0-0\\ 1+\mathrm{i}+0+0\\ 1-1+0+0\\ 1-\mathrm{i}+0+0\end{array}\right] \\ \mathrm{FFT}\mathrm{of}\mathrm{A}=\left[\begin{array}{c}2\\ \mathrm{i}+1\\ 0\\ \mathrm{i}-1\end{array}\right]....\left(14\right) \end{gathered}$$

The 2^nd polynomial is $x^2+1$ 鈥︹�� (15)

equal equations (8) and (15)

$$\begin{gathered} a_0=1 \\ {a}_1=0 \\ {a}_2=1........\left(16\right) \\ {a}_3=0 \end{gathered}$$

Take a look there at formula on converting a polynomial to something like a matrix..

localid="1659011072851" $B=\left[\begin{array}{c}{a}_0\\ a_1\\ a_2\\ a_3\\ \end{array}\right]$ 鈥︹�� (17)

Alternative the above values of $a_0,a_1,a_2,a_3$in Equation (17).

$B=\left[\begin{array}{c}1\\ 0\\ 1\\ 0\end{array}\right]$ 鈥︹�� (18)

The FFT of B ( 1,0,1,0) is calculated as follows:

$\mathrm{FFT}ofB=M_4脳B$鈥︹�� (19)

Within Equations (7), replace all values of Equation (18) with M₄(w) the values from Equation (7). (19).

localid="1659067306314" $$\begin{gathered} \mathrm{The}\mathrm{FFT}\mathrm{of}B=\left[\begin{array}{cccc}1& 1& 1& 1\\ 1& i& -1& -i\\ 1& -1& 1& -1\\ 1& -i& -1& i\end{array}\right]脳\left[\begin{array}{c}1\\ 0\\ 1\\ 0\end{array}\right] \\ =\left[\begin{array}{c}1+0+1+0\\ 1+0-1+0\\ 1+0+1+0\\ 1+0+1+0\end{array}\right] \\ \mathrm{FFT}\mathrm{of}\mathrm{B}=\left[\begin{array}{c}2\\ 0\\ 2\\ 0\end{array}\right]............\left(20\right) \end{gathered}$$

This polynomial's combination is. X+1and x² +1 is

Product = (X+1) * (x² +1)

= x³ + x + x² +1

= 1 + x + x²+ x³ 鈥︹�� (21)

Let the product of FFT of A and B be P. The product is calculated as shown below:

P = A * B 鈥︹�� (22)

,

Substitute the value of 鈥� A鈥� from Equation (14) and 鈥�B鈥� from Equation (20) in Equation (22).

$$\begin{gathered} P=\left[\begin{array}{c}2\\ i+1\\ 0\\ i-1\end{array}\right]*\left[\begin{array}{c}2\\ 0\\ 2\\ 0\end{array}\right] \\ =\left[\begin{array}{c}4\\ 0\\ 0\\ 0\end{array}\right]鈥�鈥�\left(23\right) \end{gathered}$$

Inverse FFT:

Use the inverse FFT, which equals Inverse FFT, to get the final result. $FFT=\frac{1}{n}$ $\mathrm{Mn}\left({\mathrm{w}}^{-1}\right)$ 鈥︹�� (24)

Here, from Equation (3), the value of $蝇=i$Thus, the value of ${蝇}^{-1}$is:

$$\begin{gathered} {蝇}^{-1}={\left(i\right)}^{-1}={\left[\cos \frac{\mathrm{蟺}}{2}+i\sin \frac{\mathrm{蟺}}{2}\right]}^{-1} \\ =\cos \left(-\frac{\mathrm{蟺}}{2}\right)+i\sin \left(-\frac{\mathrm{蟺}}{2}\right) \\ =\cos \left(\frac{\mathrm{蟺}}{2}\right)-i\sin \left(\frac{\mathrm{蟺}}{2}\right) \\ =0-i\left(1\right) \\ {蝇}^{-1}=-i..........\left(25\right) \end{gathered}$$

To get the value of $M_4\left({蝇}^{-1}\right)$, substitute i =-iin Equation (7).

localid="1659067624148" $M_4\left({蝇}^{-1}\right)=\left[\begin{array}{cccc}1& 1& 1& 1\\ 1& -i& -1& i\\ 1& -1& 1& -1\\ 1& i& -1& -i\end{array}\right]$鈥︹�� (26)

The inverse FFT of P is calculated as shown below:

$\mathrm{Inverse}\mathrm{FFT}\left(P\right)=\frac{1}{4}P{M}_n\left({蝇}^{-1}\right)$鈥︹�� (27)

Substitute the value of $M_4\left({蝇}^{-1}\right)$from Equation (26), P from Equation (23) in Equation (27).

role="math" localid="1659067795192" $$\begin{gathered} \mathrm{Inverse}\mathrm{FFT}\left(P\right)=\frac{1}{4}\left[\begin{array}{cccc}1& 1& 1& 1\\ 1& -i& -1& i\\ 1& -1& 1& -1\\ 1& i& -1& -i\end{array}\right]脳\left[\begin{array}{c}4\\ 0\\ 0\\ 0\end{array}\right] \\ =\frac{1}{4}\left[\begin{array}{c}4\\ 4\\ 4\\ 4\end{array}\right] \\ \mathrm{Inverse}\mathrm{FFT}\left(P\right)=\left[\begin{array}{c}1\\ 1\\ 1\\ 1\end{array}\right]........\left(28\right) \end{gathered}$$

Formula (28)'s polynomial reflection of all the above matrix seems to be $1+x+x^2+x^3$.

As a result, this polynomial's composition is $1=x+x^2+x^3$and its FFT is (4,0,0,0) .

The first polynomial is $1+x+2x^2$ 鈥︹�� (29)

Compare equations (8) and (29)

a₀= 1

a₁= 1

a₂= 2

a₃= 0 鈥︹�� (30)

Consider the formula of polynomial to matrix transformation.

$A=\left[\begin{array}{c}{a}_0\\ a_1\\ a_2\\ a_3\end{array}\right]$鈥︹�� (31)

,

Substitute the above values of $a_0,a_1,a_2,a_3$in Equation (31).

$A=\left[\begin{array}{c}1\\ 1\\ 2\\ 0\end{array}\right]$鈥︹�� (32)

The FFT of A $(1,1,2,0)$is calculated as follows:

$\mathrm{FFT}\mathrm{of}A=M_4\left(蝇\right)脳A............\left(33\right)$

Substitute the values $M_4\left(蝇\right)$ from Equation (7) andA from Equation (32) in Equation (33).

$$\begin{gathered} \mathrm{FFT}\mathrm{of}A=\left[\begin{array}{cccc}1& 1& 1& 1\\ 1& i& -1& -i\\ 1& -1& 1& -1\\ 1& -i& -1& i\end{array}\right]脳\left[\begin{array}{c}1\\ 1\\ 2\\ 0\end{array}\right] \\ =\left[\begin{array}{c}1+1+2+0\\ 1+i-2+0\\ 1-1+2+0\\ 1-i-2+0\end{array}\right] \\ \mathrm{FFT}\mathrm{of}A=\left[\begin{array}{c}4\\ i-1\\ 2\\ -i-1\end{array}\right]....\left(34\right) \end{gathered}$$

The second polynomial is 2+3x鈥︹�� (35)

Compare equations (8) and (35)

localid="1659069567337" $$\begin{gathered} a_0=2 \\ {a}_1=3 \\ {a}_2=0 \\ {a}_3=0 \end{gathered}$$鈥︹�� (36)

Consider the formula of polynomial to matrix transformation.

$B=\left[\begin{array}{c}{a}_0\\ a_1\\ a_2\\ a_3\end{array}\right]$鈥︹赌�(37)

Substitute the above values of $a_0,a_1,a_2,a_3$in Equation (37).

$B=\left[\begin{array}{c}2\\ 3\\ 0\\ 0\end{array}\right]$ .........(38)

The FFT of B ( 2, 3, 0, 0 ) is calculated as follows:

$\mathrm{FFT}\mathrm{of}B=M_4\left(蝇\right)脳B$鈥︹�� (39)

Substitute the values $M_4\left(蝇\right)$from Equation (7) and B from Equation (38) in Equation (39).

localid="1659070329001" $$\begin{gathered} \mathrm{FFT}\mathrm{of}B=\left[\begin{array}{cccc}1& 1& 1& 1\\ 1& i& -1& -i\\ 1& -1& 1& -1\\ 1& -i& -1& i\end{array}\right]脳\left[\begin{array}{c}2\\ 3\\ 0\\ 0\end{array}\right] \\ =\left[\begin{array}{c}2+3+0+0\\ 2+3i+0+0\\ 2-3+0+0\\ 2-3i+0+0\end{array}\right] \\ \mathrm{FFT}\mathrm{of}B=\left[\begin{array}{c}5\\ 2+3i\\ -1\\ 2-3i\end{array}\right] \end{gathered}$$

The product of the polynomial is $1+x+x^2$and $2+3x$is

$$\begin{gathered} Product=(1+x+2x^2)脳(2+3x) \\ =2+3x+2x+3x^2+4x^2+6x^3 \\ =2+5x+7x^2+6x^3......\left(41\right) \end{gathered}$$

Let the product of FFT of A and FFT of B be P. The product is calculated as shown below:

$\mathrm{P}=\mathrm{FFT}\mathrm{of}\mathrm{A}脳\mathrm{FFT}\mathrm{of}\mathrm{B}$鈥︹�� (42)

Substitute the values of FFT of 鈥淎鈥� from Equation (34) and FFT of 鈥淏鈥� from Equation (40) in Equation (42).

$$\begin{gathered} P=\left[\begin{array}{c}4\\ i-1\\ 2\\ -i-1\\ \end{array}\right]脳\left[\begin{array}{c}5\\ 2+3i\\ -1\\ 2-3i\\ \end{array}\right] \\ =\left[\begin{array}{c}20\\ 2+3i^2+-2-3i\\ -2\\ -2i-3i^2-2-3i\\ \end{array}\right] \\ =\left[\begin{array}{c}20\\ -2-i-3\\ -2\\ -2+i-3\\ \end{array}\right] \\ =\left[\begin{array}{c}20\\ -5-i\\ -2\\ -5+i\\ \end{array}\right] \end{gathered}$$

The inverse FFT of P is calculated as shown below:

Inverse FFT(P) $\frac{1}{4}P{M}_n\left({蝇}^{-1}\right)$ 鈥︹�� (44)

Substitute the value of $M_4\left({蝇}^{-1}\right)$from Equation (26), P from Equation (43) in Equation (44).

Inverse of FFT (P) role="math" localid="1659073083704" $$\begin{gathered} \mathrm{Inverse}\mathrm{FFT}\left(P\right)=\frac{1}{4}\left[\begin{array}{cccc}1& 1& 1& 1\\ 1& -i& -1& i\\ 1& -1& 1& -1\\ 1& i& -1& -i\\ & & & \end{array}\right]脳\left[\begin{array}{c}20\\ -5-i\\ -2\\ -5+i\\ \end{array}\right] \\ =\frac{1}{4}\left[\begin{array}{c}20-5-i-2-5-i\\ 20-i(-5-i)-2-i(-5+i)\\ 20-(-5-i)-2-1(-5+i)\\ 20+i(-5-i)-(-2)-i(-5+i)\\ \end{array}\right] \\ =\frac{1}{4}\left[\begin{array}{c}8\\ 20\\ 28\\ 24\\ \end{array}\right] \\ \mathrm{Inverse}\mathrm{FFT}\left(P\right)=\left[\begin{array}{c}2\\ 5\\ 7\\ 6\\ \end{array}\right].....\left(45\right) \end{gathered}$$

The polynomial representation of the above matrix from Equation (45) is

$2+5x+7x^2+6x^3$.

Therefore, the product of the polynomial is $2+5x+7x^2+6x^3$and its FFT is

$(20,-5-i,-2,-5+i)$.