Complex Taylor Series 2: Some Examples

[@MelThree-tb7kt]

I love you man. You've made this so much more digestible than my book

[@mdnia6592]

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

[@gokuvegeta9500]

Engineers*

[@amir.r1639]

Good for you , your now free from engineering courses Im in term 3 of mechanical engineering and Im going to die 😂

[@adetunjidaniel5064]

following your complex variables playlist. It has been awesome. Thanks

[@evionlast]

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

[@mattgiron2090]

great variety of examples, cramming for AP right now so this was great!

[@1112634]

On the third example the disk of convergence is the disk centered at the origin with radius 1 (|z|

[@gokuvegeta9500]

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

[@neutrino5695]

Thanks, man. This was amazing

[@ruinex7718]

Just awesome explaining. Thank you, it helped me alot

[@fernxndesofc]

Thanks so much. You helped me a lot.

[@gDrLemon]

At 8:01. Can someone please tell me why does the derivative of the series start at k=1 and not k=0? Thanks

[@Anonymous-od3bl]

when k=0, the value of that expression is 0. To nullify the constant we started from k=1.

[@Qits13]

Thank you so much! Helped me a lot!

[@bobogao6328]

very clear explanation! thank you

[@ottodvalishvili7601]

great videos, thanks so much

[@hagarsalem4996]

Could i know name of the book to solve similar problem?

[@kattynl]

very helpful... thank you

[@edgarkloster6017]

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!

[@s0mthingsmells]

multiply the numerator and denominator by i. for example 1/-i = 1/-i *i/i = (1*i)/(i*-i) = i/1 = i

[@zaimelkalai256]

@@s0mthingsmells im still confused, in your example you multiply the numerators while adding the denominators...

[@gokuvegeta9500]

@@zaimelkalai256 He rationalised the numerator

[@Aziz-sh3wq]

god bless ya shawty xoxo

[@okan3028]

Awesome thank you sir

[@kalki9924]

thank you sir

[@PhuongLe-ki7cb]

Great

[@zoezhuang8514]

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??

[@atulitsrivastava9199]

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

[@cameronspalding9792]

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)

[@96Willers]

Thanks

[@beophobic9653]

Dan Just don’t comment if you don’t have anything positive to say, man. Go away.

[@edgarcalochcampos1930]

@@beophobic9653 ?

[@gokuvegeta9500]

@@beophobic9653 🤡