Thank you from the bottom of my heart. You are really a great teacher. Thank you
@DR_VIV Жыл бұрын
Thank you for your kind words!
@shidzboutogodown323 Жыл бұрын
Thank you so much. I was confused as to how the constant came to be singular and one only 😅(I was comparing it to the standard condition)
@hongheng48019 ай бұрын
Sir, for e^-i*n*theta= cos(n*theta)+isin(n*theta), It should be e^-i*n*theta = cos(n*theta) - i sin(n*theta) right?
@clarencetkachikira284 Жыл бұрын
you've led me astray here😂😂😂
@pro_coder11 ай бұрын
alot of mistakes man e^-inpi = cos(-npi) + isin(-npi) = cos(npi) + 0 = (-1)^n
@joseEspinarPartal4 ай бұрын
Thank you!!
@dimas5187 Жыл бұрын
Am struggling to understand this 😢
@packrat32132 жыл бұрын
good stuff.
@ritesha8050 Жыл бұрын
hello sir thanks for the explanation. i am doing a question where p=4 and therefore r=2. is e^(-n pi i/2)=(i)^n?
@DR_VIV Жыл бұрын
It will be (-i)^n
@ritesha8050 Жыл бұрын
@@DR_VIV ok thank you sir, I will try again
@ritesha8050 Жыл бұрын
@@DR_VIV sir, i tried again but i didnt get the ans. e^(-n pi i/2) = cos(n pi/2) + i sin(n pi/2) when i wrote the terms from n=1 to 6, i got i,-1,-i,1,i,-1 isnt that i^n?
@DR_VIV Жыл бұрын
@@ritesha8050 you should have a minus sign because exp(-i x) = cos x - i sin x
@ritesha8050 Жыл бұрын
@@DR_VIV aah ok thanks sir
@tron68227 ай бұрын
Can you please explain me how you got the negative sign in e^(-i*n*pi*x)/l?
@DR_VIV7 ай бұрын
The negative sign comes when you isolate the coefficient c_n using orthogonal properties of the complex exponential
@the_boi_man68602 жыл бұрын
Boss the limits of the integral have to be 0 and 2 right; you've written 0,1. Otherwise great video!😄Thank you
@DR_VIV2 жыл бұрын
Yea, but the function is zero from 1 to 2, so the only nonzero contribution is from 0 to 1.
@the_boi_man68602 жыл бұрын
@@DR_VIV Ah yes I see. Thank you for the clarification!
@Abdiedits_6 ай бұрын
@@DR_VIV what if instead of from1 to 2, the function was 4 instead of 0. Do you add the integrals?
@DR_VIV6 ай бұрын
@@Abdiedits_ yes. That would be the correct procedure.