As we discussed in class a very important relation in Fourie

As we discussed in class, a very important relation in Fourier analysis is 1/N sigma exp [I (m -n)theta_p] = delta_m, n where theta_p = 2 pi p/N (p = 1, 2, ... N) are N angles equally spread out between 0 and 2pi. delta_m n = 0 If in and n are different integers. Take N = 5,m = 2,n = 4 and show that the relation above holds

Solution

Solution:Given \\frac{1}{N}\\sigma_{p}exp[i(m-n)\\theta_{p}] =\\delta_{m,n}.

\\theta_{p}=\\frac{2p\\pi/N}(p=1,2,3,...., N) and N=5, m=2, n=4

So \\theta_{1}=\\frac{2\\pi/5}, \\theta_{2}=\\frac{4\\pi/5}, \\theta_{3}=\\frac{6\\pi/5}, \\theta_{4}=\\frac{8\\pi/5},

\\theta_{5}=2\\pi.

Now \\frac{1}{5}\\sigma_{p}exp[i(2-4)\\theta_{p}]

=\\frac{1}{5}[exp[-2i\\theta_{1} + exp[-2i\\theta_{2}+exp[-2i\\theta_{3}+exp[-2i\\theta_{4}+exp[-2i\\theta_{5} ]

= \\frac{1}{5}[exp[-2i(\\frac{2\\pi/5})] + exp[-2i(\\frac{4\\pi/5})]+exp[-2i(\\frac{6\\pi/5})]+exp[-2i(\\frac{8\\pi/5})]

+exp[-2i(2\\pi) ]

=\\frac{1}{5}[exp[-(\\frac{4i\\pi/5})] + exp[-(\\frac{8i\\pi/5})]+exp[-(\\frac{12i\\pi/5})]+exp[-(\\frac{16i\\pi/5})]

+1]

= \\frac{1}{5}[1 +{\\cos (4\\pi)/5+\\cos (8\\pi)/5+\\cos (12\\pi)/5+\\cos (16\\pi)/5}

+ i{\\sin (4\\pi)/5+\\sin (8\\pi)/5+\\sin (12\\pi)/5+\\sin (16\\pi)/5} ]

=\\frac{1}{5}[1+{\\sin \\frac{(4.4\\pi)/5}{2}/\\sin \\frac{(4\\pi)/5}{2}}.\\cos((4\\pi/5) +(3/2).(4\\pi/5)) -

i[{\\sin \\frac{(4.4\\pi)/5}{2}/\\sin \\frac{(4\\pi)/5}{2}}.\\sin((4\\pi/5) +(3/2).(4\\pi/5))]

=\\frac{1}{5}[1+{\\sin \\frac{(8\\pi)}{5}/\\sin \\frac{(2\\pi)/5}{5}}.\\cos(20\\pi/10)) +

+i[{\\sin \\frac{(8\\pi)}{5}/\\sin \\frac{(2\\pi)/5}{5}}.\\sin(20\\pi/10)) ]

=\\frac{1}{5}[1+{\\sin \\frac{(8\\pi)}{5}/\\sin \\frac{(2\\pi)/5}{5}}.\\cos(2\\pi)) +
+i[{\\sin \\frac{(8\\pi)}{5}/\\sin \\frac{(2\\pi)/5}{5}}.\\sin(2\\pi/) ]

=\\frac{1}{5}[1+{-1.1}
+i[0]=\\frac{1}{5[0]

=0 = \\delta_{m,n}(proved)