The charisma and humour and the way you explained the subject makes it not only fun and interesting but also clear to absorb and understand the concept. I am happy l discovered your lectures. Thanks
ah i used a similar but less rigorous argument for I_2. i just taylor expanded f(z) around a (because f(z) is analytic there) and said that for a sufficiently small circle f(z) \approx f(a). then the numerator is 0 and the whole integral is 0. it's pretty much exactly what you did but not as rigorous as the epsilon delta stuff. i like what you did better. thanks for the videos they have been great for helping me prepare for my exam.
Oh I remember this topic when taking Mathematical Methods for Physicists, by Arfken back in the mid 90s. I wish I had RUclips videos like this back then! So jealous 😁
These are great! Thank you for sharing and... is this okay? integral over closed curve of [f(z) / (z-a)]dz = I = I₁+ I₂ as shown about 7:09 By linearity of integral operator on I₂ the numerator f(z) - f(a) over common denominator (z-a) can be split into two parts wlog f(z)/(z-a) and -f(a)/(z-a) and this is negative result to I₁ so sum to zero And so integral I = I₁+ I₂₍₁₎ - I₂₍₂₎ becomes integral over closed curve of [f(z)/(z-a)]dz =f(a)2πi from Cauchy Integral Formula since second part of integral in I₂ is negative value in I₁ Equivalently I₂ = I₂₍₁₎ + I₂₍₂₎ and I₂₍₂₎ = -I₁ I cannot help but wonder at some Newtonian things happening here such as the center of gravity of a sphere equals sum of all its 'parts'. When these are summed up the value is zero.
For I2 didn't you just prove that the integral will be zero for the smaller circle. How can we be sure that the integral along the curve will be zero from that???
For I2 didn't you just prove that the integral will be zero for the smaller circle. How can we be sure that the integral along the curve will be zero from that???
For I2 didn't you just prove that the integral will be zero for the smaller circle. How can we be sure that the integral along the curve will be zero from that???
The charisma and humour and the way you explained the subject makes it not only fun and interesting but also clear to absorb and understand the concept. I am happy l discovered your lectures. Thanks
Wow, thank you!
You are a very gifted teacher
Been searching for a video like this for a while. Great job
Glad you found it helpful!
ruclips.net/video/NM76WpwA1LA/видео.htmlsi=CsE78qYZyX7gWx0J
Such video also uploaded.😊
Many thanks for these videos, they are truly entertaining and brilliant.
ah i used a similar but less rigorous argument for I_2. i just taylor expanded f(z) around a (because f(z) is analytic there) and said that for a sufficiently small circle f(z) \approx f(a). then the numerator is 0 and the whole integral is 0. it's pretty much exactly what you did but not as rigorous as the epsilon delta stuff. i like what you did better. thanks for the videos they have been great for helping me prepare for my exam.
Thank you so much! I can understand complex numbers thanks to you!
Oh I remember this topic when taking Mathematical Methods for Physicists, by Arfken back in the mid 90s. I wish I had RUclips videos like this back then! So jealous 😁
You can also write f as a Taylor series to prove the theorem.
2:59! That’s why! I never knew that but have always been wondering. Cool.
Whats this board used to write called?
Dear sir, as you have established the inequality I2 is less than or equal to 2pi*iota*epsilon. But we know that "iota" does not obey order relations .
These are great! Thank you for sharing and... is this okay?
integral over closed curve of [f(z) / (z-a)]dz = I = I₁+ I₂ as shown about 7:09
By linearity of integral operator on I₂ the numerator f(z) - f(a) over common denominator (z-a)
can be split into two parts wlog f(z)/(z-a) and -f(a)/(z-a) and this is negative result to I₁ so sum to zero
And so integral I = I₁+ I₂₍₁₎ - I₂₍₂₎ becomes integral over closed curve of [f(z)/(z-a)]dz =f(a)2πi from Cauchy Integral Formula since second part of integral in I₂ is negative value in I₁
Equivalently I₂ = I₂₍₁₎ + I₂₍₂₎ and I₂₍₂₎ = -I₁
I cannot help but wonder at some Newtonian things happening here such as the center of gravity of a sphere equals sum of all its 'parts'.
When these are summed up the value is zero.
Thank you for this video
Remember people, "Don't drink and derive"😂
The link you solved the exercise on Matlab (I didn't find any code in python):
Started at time 40:22 on ruclips.net/video/NZ25MQ5W2eA/видео.html.
“Don’t drink and derive”. Seriously? 😂😂
😂
"Analytic functions are continuous on steroids" man😂🔥
Gracias 👍
Great🎉🎉🎉
Axel Walter in Lund
😹😹
Shoutout to Seif, also in Lund 💣
don't derink and derive 5:08 💀
For I2 didn't you just prove that the integral will be zero for the smaller circle. How can we be sure that the integral along the curve will be zero from that???
In time 5:08: Don't drink in drive 😂😂😂.
“It’s koh-shee, not 🛋y” 👁👄👁🪄
For I2 didn't you just prove that the integral will be zero for the smaller circle. How can we be sure that the integral along the curve will be zero from that???
For I2 didn't you just prove that the integral will be zero for the smaller circle. How can we be sure that the integral along the curve will be zero from that???