digital fourier transform and cubic equations.docx
- 1. Quadratic and Cubic equation as demonstration of Digital Fourier Transform Digital Fourier transform is in the very base of modern information technology. We are not going to discuss even the bare fundamentals of DFT. Just an example to solve cubic equations and thus an introduction to DFT through that. Besides that opens up many area of polynomial equations such as discriminant , resolvents etc. of polynomial equations, Such as Lagrange’s resolvents, Galois resolvents etc. which are entirely not discussed in this booklet. DFT of f (n) of f (0),f (1)….f (N−1)are defined by F(K)= 1 N ∑ m=0 N −1 f (m)e −2πinK N , K=0,1,… N−1 DFT transforms numbers to complex periodic functions in the manner shown. It may be shown that DFT F(K) ofF(0), F(1)….F (N−1) are f (n)=∑ K=0 N−1 F(K )e 2πinK N , n=0,1,…N −1 , Call them inverse transform? Does it recover original number ? Let us see. Substituting the values from the previous equation, f (n)=∑ K=0 N−1 (1 N ∑ m=0 N −1 f (m)e −2 πimK N )e 2πinK N ¿ ∑ K=0 N−1 1 N ∑ m=0 N −1 f (m)e −2πimK N e 2πinK N ¿ ∑ K=0 N−1 1 N ∑ m=0 N −1 f (m)e 2 πi(n−m)K N ¿ ∑ m=0 N −1 f (m) 1 N ∑ K =0 N −1 e 2 πi(n−m)K N
- 2. It can be shown that if N=1, from above, f (0)=∑ m=0 0 f (m) 1 1 ∑ K=0 0 e 2πi (n−m)K 1 =F (0) Let N=1 in f (n)=∑ K=0 0 F(K )e 2πinK 1 =f (0). Let N ≠1 andn=m , then 1 N ∑ K =0 N−1 e 2πi (n−m)K N = 1 N ∑ K=0 N −1 e0 = 1 N N=1. Let N ≠1 and ifn≠m , then, 1 N ∑ K =0 N−1 e 2πi (n−m)K N = 1 N ∑ K=0 N −1 e 2πipK N =0,p≠0, a positive integer. This is 0 as sum of n-th roots of unity is 0. Combining the two cases, 1 N ∑ K =0 N−1 e 2πi (n−m) N =δmn. δmn=0 if ≠m , and δmn=1 if n=m Kronekar delta function. At last ∑ m=0 N−1 f (m) 1 N ∑ K =0 N −1 e 2 πi(n−m) N =∑ m=0 N−1 f (m)δmn=f (n) Quadratic equation Let x0=f (0),x1=f (1) be roots of the quadratic equation x 2 +bx+c=0. We have f (n)= 1 N ∑ K =0 N−1 F(K )e 2πin N , n=0,1,N−1, For roots of quadratic equation,N=2 .
- 3. If we denoteX0=F(0), X1=F (1) , x0=X0e 2 π .0.0 2 +X1 e 2π.0 .1 2 =X0+ X1 , x1=X0 e 2π.1 .0 2 +X1 e 2 π.01.1 2 =X0−X1 , Now x 2 +bx+c=(x−x0)(x−x1) ¿(x−(X0+X1))(x−(X0−X1)) ¿ x 2 −2 X0 x+ X0 2 −X1 2 . Comparing coefficients ofx from both sides, b=−2 X0 and ¿ X0 2 −X1 2 . So X0= −b 2 and X1=√b2 −4 c 2 . Cubic equation In continuation of notations from above paragraph, along the same lines, the cubic equation taken is x3 +b x2 +cx+d=0. The roots assumed are x0=X0+ X1+X2 , x1=X0+ω X1+ω 2 X2, ω=e 2πi ,cube root of unity. x1=X0+ω2 X1+ω X2, Plugging these values in the cubic, x3 +b x2 +cx+d=(x−x0)(x−x1)(x−x2)
- 4. ¿ x3 −3 X0 x2 +(3 X0 2 −3 X1 X2)x−X0 3 +3 X0 X1 X2−(X1 3 +X2 3 ) Comparing coefficients from both sides, X0= −b 3 , X1 X2= b 2 −3c 9 , X1 3 +X2 3 = −27 d−9bc+2b 3 27 From this one quadratic equation in mayX1 , X2 may be set up and along with X0from above, the roots x0 ,x1 ,x2 may be easily found. There is no special advantage using DFT for solving polynomials of degree 2, 3 or even 4. For roots of polynomials of degree 5 or more, the DFT method must fail to give roots in closed forms, i.e., in radicals, but it would give approximate solutions employing a computer programming for DFT method. But iteration methods are also handy and may be easily built into computer programs. ut DFT plays crucial role in signal processing, information technology, and other fields. This article aimed at giving a simple introduction and natural motivation to DFT. Quartic and biquadratic equations A quartic equation is a polynomial equation of degree 4. It is different from biquadratic equation. The latter is also a polynomial equation of degree 4, but contains only terms of even degree in the variable. So it is only a quadratic equation in disguise and can be solved easily by substitutingy=x 2 , and let us not discuss that any further. Given below is a procedure for solving a quartic. A quartic equation may be written in the form x4 +4B x3 +C x2 + Dx+E=0, just to avoid fractions.
- 5. By a transformation x=u−Bwe remove the 2nd term and reduce the equation and convert it into a depressed quartic. We get the following. (−B is average of all its 4 roots) (u−B)4 +4 B(u−B)3 +C (u−B)2 +D (u−B)+E=0. By expanding and simplifying we get u 4 +au 2 +bu+c=0, Where ¿−6B 2 +2C , b=2B 3 −2C +D, c=−3B4 +B2 C−BD+E. We may write u 4 +cu 2 +du+e=0 Assume its factors as (u 2 + pu+q)(u 2 +ru+s), Or, u4 +(p+r)u3 +(q+s+ pr)u2 +( ps+qr )u+qs=u4 +au2 +bu+c , Equating coefficients of similar powers from both sides, p+r=0,q+s+ pr=c, ps+qr=d ,qs=e, These become the following using p=−r; q+s=p 2 +c ,s−q= d p ,qs=e, Now the identity (s+q)2 −(s−q)2 =4 sq becomes, ( p2 +c) 2 − d 2 p2 =4 e, If we setp2 =t , it becomes (t+c)2 − d 2 t =4e, This is a cubic equation int and can be solved, giving the values of q ,s which finally solves the quartic equation. For finding out roots of polynomials of degree 5 or higher Galois theory confirms that there is no method for computing the roots
- 6. employing an universal formula as in case of quadratic, cubic and quartic equations. So computation is done with algorithmic approach.