Omg thanks so much! The teacher we have at our university just grinds all this theory into us without showing us how to actually calcutlate and use the stuff we learn. And at an ingeneering university you don't need all that theory if you kinda understand it from a mathematical perspective and can use it. The most important thing for us ingeneers is to actually use it! Helped alot! Thanks again:D
Theorem 0:13 Useful facts 0:53 Example 1 Maclaurin series for f(z)=1/(3+6z)² 5:20 Example 2 expand f(z)=(z-1)/(3-z) around z_0=1 Example 3 obtain the Maclaurin series for f(z)= i/(z-i)(z+2i) 11:56 Example 4 expand ln(1+z) in a Maclaurin series 14:26
But when we use partial fractions the terms split into f(z)=(1/(z-2i))-(1/(z-1)) So the Taylor's series expansion for the first term doesn't have the singularity z=i in it Furthermore when we combine the Taylor series of both terms having radius of convergence R
At 13:09...why is 1/(2i) the same as i/2 ??? (same question for 1/i = i) I had a different solution because i solved it with 1/(2i) (and i in the second term). Is it wrong to calculate it with 1/(2i) (and i) times the series?? please help!
Hi, I have a question on the second example u did. when u say that the series are absolutely convergent inside the circle of convergence. Do u mean the circle with radius 2i or i centered at 0??
Qianqian Zhuang a circle with radius 2i will also include singularity point z=i so I don't think he is talking about circle with radius 2i cos for expansion the function must be analytic within the circle of radius 2i
I love you man. You've made this so much more digestible than my book
Omg thanks so much! The teacher we have at our university just grinds all this theory into us without showing us how to actually calcutlate and use the stuff we learn. And at an ingeneering university you don't need all that theory if you kinda understand it from a mathematical perspective and can use it. The most important thing for us ingeneers is to actually use it!
Helped alot! Thanks again:D
Engineers*
Good for you , your now free from engineering courses Im in term 3 of mechanical engineering and Im going to die 😂
following your complex variables playlist. It has been awesome. Thanks
Theorem 0:13
Useful facts 0:53
Example 1 Maclaurin series for f(z)=1/(3+6z)² 5:20
Example 2 expand f(z)=(z-1)/(3-z) around z_0=1
Example 3 obtain the Maclaurin series for f(z)= i/(z-i)(z+2i) 11:56
Example 4 expand ln(1+z) in a Maclaurin series 14:26
great variety of examples, cramming for AP right now so this was great!
On the third example the disk of convergence is the disk centered at the origin with radius 1 (|z|
But when we use partial fractions the terms split into f(z)=(1/(z-2i))-(1/(z-1))
So the Taylor's series expansion for the first term doesn't have the singularity z=i in it
Furthermore when we combine the Taylor series of both terms having radius of convergence R
Thanks, man. This was amazing
Just awesome explaining. Thank you, it helped me alot
Thanks so much. You helped me a lot.
At 8:01. Can someone please tell me why does the derivative of the series start at k=1 and not k=0? Thanks
when k=0, the value of that expression is 0. To nullify the constant we started from k=1.
Thank you so much! Helped me a lot!
very clear explanation! thank you
great videos, thanks so much
Could i know name of the book to solve similar problem?
very helpful... thank you
At 13:09...why is 1/(2i) the same as i/2 ??? (same question for 1/i = i)
I had a different solution because i solved it with 1/(2i) (and i in the second term).
Is it wrong to calculate it with 1/(2i) (and i) times the series??
please help!
multiply the numerator and denominator by i. for example 1/-i = 1/-i *i/i = (1*i)/(i*-i) = i/1 = i
@@s0mthingsmells im still confused, in your example you multiply the numerators while adding the denominators...
@@zaimelkalai256
He rationalised the numerator
god bless ya shawty xoxo
Awesome thank you sir
thank you sir
Great
Hi, I have a question on the second example u did. when u say that the series are absolutely convergent inside the circle of convergence. Do u mean the circle with radius 2i or i centered at 0??
Qianqian Zhuang a circle with radius 2i will also include singularity point z=i so I don't think he is talking about circle with radius 2i cos for expansion the function must be analytic within the circle of radius 2i
For the expansion about z0=1: could you not find a maccularin series of f(z+1) then adapt it to get back to f(z)
Thanks
Dan Just don’t comment if you don’t have anything positive to say, man. Go away.
@@beophobic9653 ?
@@beophobic9653
🤡