1. Polynomials and the FFT(UNIT-3)
ADVANCED ALGORITHMS
Polynomials and the FFT(UNIT-3)

2. Polynomials :
A polynomial in the variable x over an algebraic field F represents a function A(x) as a formal sum :
A(x) =  aj xj
j = 0 to n-1
Here, we call the values a0, a1, a2, a3,….,an-1 as the
coefficients of the polynomial.
A polynomial has degree k, if its highest non-zero coefficient is ak.
Any integer strictly greater than the degree of a
polynomial ‘k’, is degree-bound ‘n’. i.e., n > k.

3. The representation of a Polynomial :
Ex-1 : A(x) = 𝑥3 − 2𝑥 − 1
Here, A(x) has degree 3.
A(x) has degree bounds 4,5,6,..
A(x) has coefficients (-1,-2,0,1)
Ex-2 : 𝑥4 + 𝑥2 − 1
Here, A(x) has degree 4.
A(x) has degree bounds 5,6,7,..
A(x) has coefficients (-1,0,1,0,1)

4. Coefficient Representation : (CR)
A CR of a polynomial of degree bound ‘n’ is a
vector of coefficients
a =
Adding Polynomials :
Let a = (a0,a1,a2,… , an-1)
b = (b0,b1,b2,… , bn-1)
Sum is the coefficient vector C, where
c = (c0,c1,c2,… , cn-1)
Here, cj = aj + bj  j = 0,1,2,..,n-1.

5. Multiplying Polynomials :
Multiplying of two degree-bound ‘n’ polynomials
A(x) and B(x) :
C(x) =
Ex-3 : Add & Multiply the following two polynomials :
A(x) = x x + 9
B(x) = x x - 5
A(x) + B(x) = 2x x + 4
A(x) X B(x) = x4 – 6 x3 – 36 x x - 45

6. Horner’s Rule : Given the Polynomial :
p(x) =  ai xi = a0 + a1 x + a2 x2 + … + an xn
i = 0 to n
Here,  i = 0,1,2,..,n-1,n ai is a real number.
The above polynomial can be written in the form :
p(x) = a0 + x (a1 + x (a2 + x (a3 + …. +
x (an-1 + an x )……))
Let x = x0
Now, assume bn := an
bn-1 := an-1 + bn xo
…………..
b0 := a0 + b1 xo

7. Here let us substitute b values into expression of p(x0).
So, p(x0) = a0 + x0 (a1 + x0 (a2 + x0 (a3 + …. +
x0 (an-1 + bn x0 )……))
= a0 + x0 (a1 + x0 (a2 + x0 (a3 + …. +
x0 bn-1 )……))
= …………
= a0 + x0 (a1 + x0 (a2 + x0 b3 ))
= a0 + x0 (a1 + x0 b2)
= a0 + x0 b1 = b0
Hence p(x0) = b0

8. Ex-4 : Evaluate 2x3 - 6x2 + 2x - 1 for x = 3
Synthetic Division : (Horner’s Method)
x0 | x x x x0
3 | |
i.e., The Value of p(x0) is 5.

9. Ex-5 : Evaluate
x x x for x = 2
Synthetic Division : (Horner’s Method)
x0 | x x x x0
2 | |
i.e., The Value of p(x0) is 0.

10. 3. Point-Value Representation :
A Point-Value Representation of a polynomial A(x) of
degree-bound n is a set of n point-value pairs :
{(x0 ,y0), (x1 ,y1), (x2 ,y2),…, (xn-1 ,yn-1),}
such that all of the xk are distinct and yk = A(xk)
for k = 0,1,2,…,n-1.
Ex-6 : A(x) = x x + 1
Here, xk : 0, 1, 2, 3
A(xk ) : 1, 0, 5, 22
 { (0,1), (1,0), (2,5), (3,22) }

11. Interpolation of PVR : The inverse of evaluation is Interpolation.
- determines coefficient form of polynomial from
point value representation.
Theorem : (Uniqueness of an interpolating polynomial)
“For any set {(x0 ,y0), (x1 ,y1), (x2 ,y2),…, (xn-1 ,yn-1),}
of n point-value pairs  all the xk values are distinct,
there is a unique polynomial A(x) of degree-bound n
such that yk = A(xk) for k = 0, 1, 2, …, n-1.”

12. The above theorem can be written as :
The matrix on the left is denoted V(x0 , x1 , x2 , …, xn-1).
It is known as : Vander-monde Matrix.
i.e., a = V -1 y.

13. Z =  j  k (x – xj) /  j  k (xk – xj)
Lagrange’s Formula :
A(x) =  yk Z, where
k = 0 to n-1
Z =  j  k (x – xj) /  j  k (xk – xj)
Ex-7 : Consider the PVR :
[ (0,1), (1,0), (2,5), (3,22) ]
Here, 1.[(x-1)(x-2)(x-3)] / [(0-1) (0-2) (0-3)]
= (x x x - 6) / - 6
= (-x x x + 6) / ….(1)
And 0.[(x-0)(x-2)(x-3)] / [(1-0) (1-2) (1-3)]
= ….(2)

14. And 5.[(x-0)(x-1)(x-3)] / [(2-0) (2-1) (2-3)]
= x x …Answer
The above is the polynomial for the given PVR.

15. Adding Polynomials using PVR :
Let C(x) = A(x) + B(x)
A : [(x0 ,y0), (x1 ,y1), (x2 ,y2),…, (xn-1 ,yn-1)]
B : [(x0 ,z0), (x1 ,z1), (x2 ,z2),…, (xn-1 ,zn-1)]
C : [ (x0 , y0+ z0), (x1 , y1+ z1), ….,
(xn-1 , yn-1+ zn-1) ]
Ex-8 : Find the value of C(x) in PVR, where
A(x) = x x + 1
B(x) = x3 + x2 + 1
Let x = (0,1,2,3)
Then A : [(0,1), (1,0), (2, 5), (3,22) ]
B : [ (0,1), (1,3), (2, 13), (3, 37) ]
Hence C : [(0,2), (1, 3), (2, 18), (3, 59) ]

16. Multiplying Polynomials using PVR :
Let C(x) = A(x) . B(x)
A : [(x0 ,y0), (x1 ,y1), (x2 ,y2),…, (x2n-1 ,y2n-1)]
B : [(x0 ,z0), (x1 ,z1), (x2 ,z2),…, (x2n-1 ,z2n-1)]
C : [ (x0 , y0 z0), (x1 , y1 z1),….,(x2n-1 , y2n-1+ z2n-1) ]
Ex-9 : Find the value of C(x) in PVR, where
A(x) = x x ; B(x) = x3 + x2 + 1
Let x = (-3,-2,-1,0,1,2,3)
A : [(-3,20), (-2,-3), (-1,2),(0,1), (1,0), (2, 5), (3,22) ]
B : [ (-3,-17),(-2,-3), (-1, 1),(0,1), (1,3), (2, 13), (3,17)]
C : [(-3,340),(-2,9),(-1,2),(0,1), (1, 0), (2, 65), (3,814) ]

17. Ex-8 Contd…
There are FOUR cordinates :
C : [(0,2), (1, 3), (2, 18), (3, 59) ]
Basing on First & Second Coordinates :
- 2 x x x …..(1)
9 x x x …..(2)
Basing on Third & Fourth Coordinates :
- 54 x x x …..(1)
59 x x x …..(2)
The denominator is 6.
Finally 2 x3 + x x + 2

18. 4. Complex Roots of Unity :
A complex nth root of unity (1) is a complex number
 such that n = 1.
There are exactly n complex nth roots of unity :
e2  i k / n for k = 0, 1, 2, …., n-1
where eiu = cos (u) + i sin (u)
Using e2  i k / n = cos(2 k / n) + i sin(2 k / n)
we can check that it is a root.
(e2  i k / n)n = e2  i k
= cos(2 k) + i sin (2k) = = 1

19. Principal nth Root of Unity : The Value n = e 2 ∏ i /n
Degrees :
SIN 0 1/ 2 1 0
COS 1 1/
Principal nth Root of Unity :
The Value n = e 2 ∏ i /n
is called the ‘principal nth root of unity’.
All of the other complex nth roots of unity are
powers of n .
The n complex nth roots of unity are :
n0, n1, n2,….., nn-1
Note : Here, nn = n0 = 1

20. Ex-10 : Find the 8th complex roots of unity.
The two roots of unity are : +1, - 1
The complex 4th roots of unity are :
1, - 1, i, - i where (-1) = i
In general, the n complex roots of unity are equally spaced around the circle of unit radius centered at the origin of the complex plane.
Note : nj . nk = nj+k = n(j + k) mod n

21. Visualizing 8 Complex 8th Roots of Unity :

22. First Root : 80 =  = 1
Here, k = 0, n = 8
So, 80 = e 2 ∏ i k / n = e 0 = 1
Second Root : Here, k = 1, n = 8
81 = e 2 ∏ i k / n = e 2 ∏ i / 8 = e ∏ i / 4
by using the formula
eiu = cos (u) + i sin (u)
So, 81 = cos (/4) + i sin (/4)
= 1 /  i / 2

23. So, 82 = cos (/2) + i sin (/2) = i
Third Root : Here, k = 2, n = 8
82 = e 2 ∏ i k / n = e 2 ∏ i 2 / 8 = e ∏ i / 2
So, 82 = cos (/2) + i sin (/2) = i
Fourth Root : Here, k = 3, n = 8
83 = e 2 ∏ i k / n = e 2 ∏ i 3 / 8 = e ∏ i 3 / 4
So, 83 = cos (3/4) + i sin (3/4)
= cos ( - /4) + i sin ( - /4)
= - cos /4 + i sin /4)
[since cos ( - x) = - cos x, sin ( - x) = sin x]
= / 2 + i /2

24. Fifth Root : Here, k = 4, n = 8
84 = e 2 ∏ i k / n = e 2 ∏ i 4 / 8 = e ∏ i
So, 84 = cos () + i sin () = - 1
Sixth Root : Here, k = 5, n = 8
85 = e 2 ∏ i k / n = e 2 ∏ i 5 / 8 = e ∏ i 5 / 4
So, 85 = cos (5/4) + i sin (5/4)
= cos ( + /4) + i sin ( + /4)
= - cos /4 + i (- sin /4)
[since cos ( + x) = - cos x, sin ( + x) = - sin x]
= /  i /2

25. Seventh Root : Here, k = 6, n = 8
86 = e 2 ∏ i k / n = e 2 ∏ i 6 / 8 = e 3∏ i / 2
So, 86 = cos ( + /2 ) + i sin ( + /2) = - cos (/2) + i (- sin (/2) )
= i ( - 1) = - i
Eighth Root : Here, k = 7, n = 8
87 = e 2 ∏ i k / n = e 2 ∏ i 7 / 8 = e ∏ i 7 / 4
So, 87 = cos (7/4) + i sin (7/4)
= cos ( + 3/4) + i sin ( + 3/4)
= - cos 3/4 + i (- sin 3/4)
[since cos ( + x) = - cos x, sin ( + x) = - sin x]
= cos /4 + i (- sin /4)
= /  i /2