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Yes, extremely well explained. Well done.
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)
Thank you from the bottom of my heart. You are really a great teacher. Thank you
Thank you for your kind words!
Sir, for e^-i*n*theta= cos(n*theta)+isin(n*theta), It should bee^-i*n*theta = cos(n*theta) - i sin(n*theta)right?
you've led me astray here😂😂😂
Thank you!!
alot of mistakes man e^-inpi = cos(-npi) + isin(-npi) = cos(npi) + 0 = (-1)^n
good stuff.
Am struggling to understand this 😢
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?
It will be (-i)^n
@@DR_VIV ok thank you sir, I will try again
@@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,-1isnt that i^n?
@@ritesha8050 you should have a minus sign because exp(-i x) = cos x - i sin x
@@DR_VIV aah ok thanks sir
Can you please explain me how you got the negative sign in e^(-i*n*pi*x)/l?
The negative sign comes when you isolate the coefficient c_n using orthogonal properties of the complex exponential
Boss the limits of the integral have to be 0 and 2 right; you've written 0,1. Otherwise great video!😄Thank you
Yea, but the function is zero from 1 to 2, so the only nonzero contribution is from 0 to 1.
@@DR_VIV Ah yes I see. Thank you for the clarification!
@@DR_VIV what if instead of from1 to 2, the function was 4 instead of 0. Do you add the integrals?
@@Abdiedits_ yes. That would be the correct procedure.
what pen is that ?!
I use many types of fountain pens
Yes, extremely well explained. Well done.
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)
Thank you from the bottom of my heart. You are really a great teacher. Thank you
Thank you for your kind words!
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?
you've led me astray here😂😂😂
Thank you!!
alot of mistakes man e^-inpi = cos(-npi) + isin(-npi) = cos(npi) + 0 = (-1)^n
good stuff.
Am struggling to understand this 😢
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?
It will be (-i)^n
@@DR_VIV ok thank you sir, I will try again
@@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?
@@ritesha8050 you should have a minus sign because exp(-i x) = cos x - i sin x
@@DR_VIV aah ok thanks sir
Can you please explain me how you got the negative sign in e^(-i*n*pi*x)/l?
The negative sign comes when you isolate the coefficient c_n using orthogonal properties of the complex exponential
Boss the limits of the integral have to be 0 and 2 right; you've written 0,1. Otherwise great video!😄Thank you
Yea, but the function is zero from 1 to 2, so the only nonzero contribution is from 0 to 1.
@@DR_VIV Ah yes I see. Thank you for the clarification!
@@DR_VIV what if instead of from1 to 2, the function was 4 instead of 0. Do you add the integrals?
@@Abdiedits_ yes. That would be the correct procedure.
what pen is that ?!
I use many types of fountain pens