It's perfectly clear where everything comes from and one can always just skip past the partial fractions if one wants. I think the example video was very well done and I think most people (should) understand that this is not in any way a lecture on laurent/taylor series but just a (nicely) worked out example. Thank you!
Dude, this was a great video, dont acknowledge that 1/8 factor guy. Not only was this a good starting example but it was very straightforward and crystal clear
The difference in art of manipulating alzebra of Z for Laurent and Taylor series was easily observed .This video accelerated my pace of study about Laurent series. Thanku dgory.
from indrada 3290609: "Taylor series converges within a certain radius, and a Laurent series converges outside a certain radius, he wanted it only to converge within a radius of 4, but outside a radius of two."
In this example, the function has singularities at z=-1 and z=-3. So we will get two circles one of radius 2 and the other of radius 4, both centred at the point we are expanding about, z=1. The next thing to do is to look at the question given and see which area in the complex plane they want the expansion. If it is between the two circles, as in this case, then we will need one Laurent series expansion, and one Taylor series expansion. Hope this helps. Please feel free to ask me more!
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It's easier to think about these as geometric series of the form 1/(1-r). For 1/(z+3)= 1/4*[1/(1+(z-1)/4)], we want |z-1|2 (use the expansion formula -1/r-1/r^1-1/r^3-...)
1. We need to expand with z-1 since we are finding a series about z=1. For example, if we were finding a series about z=-i we would need z+i. 2. Without going into too much detail we divide by 4 for the first series since this is a taylor series in the required region, and we divide by z-1 for the second series since this is a laurent series in the required region. This is the technique you should use and I encourage you to follow it closely. Best of luck for your exam!
I'm not entirely sure of your issue, but let me try clear it up. A Laurent and Taylor Series are always expanded about a point. This point is INDEPENDENT of the function and must be given to you in the question. In the example in this video the point we are expanding about is z=1. The next thing you need to do is visualise circles expanding from this point. We keep expanding the circles about the point z=1 (in this example) until we hit singularities of the function.
For the part ii of this particular question, you would do the same process to expand it to the outer part of the bigger circle? Thank you for your time and the video, helped me a lot to understand this.
I was fascinated of your explanation, it was great! I understood it well. Thumps up! However, just one favor though, can you please find the Laurent Series of the following: 1. sinz/z 2. cosz/z^2 3. exp(1/z^2) Thanks alot. More power sir!
why at -1, we need a laurent series and taylor series at -3? What is the idea that the Laurent series "keep's it out" and taylor "keeps it in"? What I only know is that we use taylor when it is analytic at that point and laurent when it is not analytic at that point.
Why divide by 4 (the constant) for first series and then divide by z - 1 for the second series (the function) instead of dividing by 2 (the constant)? Why do 'we *need* to expand with z - 1 in the denominator'?
Bunk Moreland Taylor series converges within a certain radius, and a Laurent series converges outside a certain radius, he wanted it only to converge within a radius of 4, but outside a radius of two.
this was in a comment below, i hope it helps "1. We need to expand with z-1 since we are finding a series about z=1. For example, if we were finding a series about z=-i we would need z+i. 2. Without going into too much detail we divide by 4 for the first series since this is a taylor series in the required region, and we divide by z-1 for the second series since this is a laurent series in the required region. This is the technique you should use and I encourage you to follow it closely."
You use the one which is bigger to divide, in order for the sum to converge -> the absolute to be less than 1. So for the first fraction, 1/(z+3) you're focusing on the outside circumference, |z-1| < 4. You have 1/((z-1) + 4) and you want the sum to converge, so you'll divide by the one which is greates, which is 4. You get the sum 1/(1+(z-1)/4), which will converge because |(z-1)/4| < 1. In the second fraction, 1/(z+1), it's the same principle. You're focusing on the 2 < |z-1| circumference. So |z+1| will be bigger, and that's why you have to divide by it. You get the sum 1/(1+ 2/(z-1)), which will converge because |2/(z-1)| < 1. I know this is a late response, but oh well.
About the second part of this particular question: how to write 1/(z+1) as a laurent series that is convergent when |z-1| is GREATER than 4? I'm struggling with it Edit: On the other hand if 1/(z+1) is written as a series that is convergent when |z-1| is GREATER than 2 and... 1/(z+3) is written as a series that is convergent when |z-1| is GREATER than 4 then the LOGICAL CONJUCTION of both convergence conditions means that f(z) can be written as a sum of series that's convergent when |z-1| > 4
Thank you for this video!! It helped me a lot!! But what can I do when my function has 3 singularities. Maybe like the function you've used in video with the singularities 1,3 and -3. Then it's still the same circle in which we are looking for our series isn't it? But do I have to subtract another term or will the taylor series be the same for both 3 and -3?
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1/(1-r) is the solution to an infinite sum of a geometric series with first term 1 and multiplier r, which corresponds also to its taylor series. He turned the equation into the form 1/4*(1/(1+(z-1)/4)), so (-1)(z-1)/4=r. The taylor series follows from the corresponding geometric sum.
What would you do for the second part where |z-1|>4 Just Laurent for for the outer circumference plus the other fraction? Laurent for both fractions? Please help
Sorry for asking but, since you reply to comments in the Laurent series why did you put 1/(z-1) infront ( i didnt understand what you said)? why not write it as this: (1/2) * 1/( 1 +(z-1)/2 ) ? thanks
+Jason K Wondering the same thing. Think it's because z-1/2 > 1 for |z-1|>2 but 2/z-1 2. And that term has to be smaller than 1 for the geometric series. Not 100% sure though, confirmation would be appreciated.
Do we say both expansions are around z=1? Since the Laurent series has infinitely number of terms do we say it is a essential singularity at x=-1? Thanks.
Correct! Both expansions are around z=1. However the singularity at z=-1 is not essential. To determine if a singularity is essential or not you will need to find a Laurent expansion about the point of singularity. In this example we looked at an expansion about another point, namely z=1. If you do however find the expansion about z=-1 (the singularity) you will find there is only one negative power. Hence it is a singularity of order 1. If there were infinitely many negative powers of z then it would be an essential singularity. Does this explanation help you?
dgory Yes it definitely helps! I guess the easiest way to see the point of expansion is to consider the standard form of laurent series c_j(z-z0)^j, for all integer j, identify z0 and treat everything else to be c_j, which is the coefficient.
dgory I was responding to this comment, sorry "In this example, the function has singularities at z=-1 and z=-3. So we will get two circles one of radius 2 and the other of radius 4, both centred at the point we are expanding about, z=1. The next thing to do is to look at the question given and see which area in the complex plane they want the expansion. If it is between the two circles, as in this case, then we will need one Laurent series expansion, and one Taylor series expansion. Hope this helps. Please feel free to ask me more!"
I've got an urgent question! I get that he had to write 1/4 in the front while developing the taylor series, to get 1/(1+(z-1)/4). BUT why didn't he do the same thing with the laurent series, except it would have been 1/2 this time. Instead he took out 1/(z-1) out the front. Why? Can anyone explain? You guys know what I mean? That 1/((z-1)+2) = 1/(2+(z-1)) = 1/2 * 1/(1+(z-1)/2)
I don't understand. Anyone explain it to me please. which one that expanded by Taylor series or Laurent series? Can we choose that 1/(z-1) expanded by Taylor series?..forgive me, i can't speek English well. I hope u understand what i Mean. Thankyou
Good explanation. But I have an opinion here. Since the objective of this video is to show how to work out with Laurent series, you don't need to show the steps for partial fraction in this video. You can directly show us the result of partial fraction without going through the steps. Then you can do another video for PARTIAL FRAction for those who wish to understand it. Thank you! :)
Unfortunately I never made one. I am currently making a mini course on Laurent series expansions for complex analysis, but that probably doesn't help you right now.
You should have left out the extraneous simple math. I imagine anyone interested in the Laurent series should already know how to find partial fractions.
i would normally agree with this but some students such as myself end up doing maths courses every 2nd semester and tend to forget all the teeny tiny bits... i thought it was really nice that he showed that.
It's perfectly clear where everything comes from and one can always just skip past the partial fractions if one wants. I think the example video was very well done and I think most people (should) understand that this is not in any way a lecture on laurent/taylor series but just a (nicely) worked out example. Thank you!
Dude, this was a great video, dont acknowledge that 1/8 factor guy. Not only was this a good starting example but it was very straightforward and crystal clear
I was troubled with the problem and I have an important exam . SO HELPFUL THANK YOU!
The difference in art of manipulating alzebra of Z for Laurent and Taylor series was easily observed .This video accelerated my pace of study about Laurent series. Thanku dgory.
Not only did this video explain exactly what I needed it to, but you sound exactly like a friend of mine which made it hilarious!
I'm glad it helped and that I sound exactly like a friend of yours!
Oh, this must be a reason why I said "Your are great, my friend" at the end of this perfect explanation! :)
I love you. After hours of reading through my book, I finally understand. Thank you so much.
This has saved my life, thank you for a detailed explanation of each little step!
How do u determine which one should be a laurent series and which one should be a taylor series?
from indrada 3290609:
"Taylor series converges within a certain radius, and a Laurent series converges outside a certain radius, he wanted it only to converge within a radius of 4, but outside a radius of two."
Best video I’ve found on Laurent series by far!
I as struggling so much w/ this material and now, finally its clear thanks to you! thank you so much
Thank you! After two long weeks I finally understood it! Thanks!
thank you...i totally never understood my lecturer.... but i have understood your explanation
This video clarifies to me some misunderstanding concepts ! Thank ..
I love you man! I thought I would never understand this subject !
Thanks, this is the one part of my upcoming exam I'm having trouble with.
up up.......helped my speed study 4 my exams
In this example, the function has singularities at z=-1 and z=-3. So we will get two circles one of radius 2 and the other of radius 4, both centred at the point we are expanding about, z=1.
The next thing to do is to look at the question given and see which area in the complex plane they want the expansion. If it is between the two circles, as in this case, then we will need one Laurent series expansion, and one Taylor series expansion.
Hope this helps.
Please feel free to ask me more!
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I stupidly forgot my account password. I love any assistance you can offer me.
@Everett Tobias instablaster :)
@Kieran Kylo Thanks so much for your reply. I found the site on google and Im trying it out now.
Takes quite some time so I will reply here later when my account password hopefully is recovered.
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Thanks so much, you really help me out :D
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Cheerio aussie brother. Very lucid explanations, excellent work :)
Best video in the category.
thanks so much! i have a midterm coming and ur vid helped a lot!! :)
thank you professor from korea
Nice and clear explanation. Thank you very much.
Would you be able to explain why we need a Laurent series expansion for the point at -1 and a Taylor series expansion for the point at -3 please?
It's easier to think about these as geometric series of the form 1/(1-r). For 1/(z+3)= 1/4*[1/(1+(z-1)/4)], we want |z-1|2 (use the expansion formula -1/r-1/r^1-1/r^3-...)
You have explain it soooo good. 👍
wow its actually THAT easy!? Thank you :))!
1. We need to expand with z-1 since we are finding a series about z=1. For example, if we were finding a series about z=-i we would need z+i.
2. Without going into too much detail we divide by 4 for the first series since this is a taylor series in the required region, and we divide by z-1 for the second series since this is a laurent series in the required region. This is the technique you should use and I encourage you to follow it closely.
Best of luck for your exam!
@6:47 how did you find the taylor series so eloquently like that??
Really helpful! Good job!
I'm not entirely sure of your issue, but let me try clear it up.
A Laurent and Taylor Series are always expanded about a point. This point is INDEPENDENT of the function and must be given to you in the question. In the example in this video the point we are expanding about is z=1.
The next thing you need to do is visualise circles expanding from this point. We keep expanding the circles about the point z=1 (in this example) until we hit singularities of the function.
Thanks a lot, it was helpful!
Could you tell me the name of the application you use to explain this?
Thank you so much for your help. It is much appreciated :)
Awesome video, thanks!
For the part ii of this particular question, you would do the same process to expand it to the outer part of the bigger circle? Thank you for your time and the video, helped me a lot to understand this.
I was fascinated of your explanation, it was great! I understood it well. Thumps up! However, just one favor though, can you please find the Laurent Series of the following: 1. sinz/z
2. cosz/z^2
3. exp(1/z^2)
Thanks alot. More power sir!
+JP PANTORILLA
1. Sin(z) = z - z^3/3! + z^5/5! + ....
so Sin(z)/z = 1 -z^2/3! + z^4/5! + ... = Sum(-1)^n(z^(2n)/(2n+1)!) (So no principal part)
2. Cos(z)/z^2 = 1/z^2 - 1/2 + z^2/4! + ... = Sum(-1)^n(z^(2n-2)/(2n!) (So principal part is 1/z^2 - 1/2)
3. exp(1/z^2) = Sum(1/(n!z^2))
why at -1, we need a laurent series and taylor series at -3? What is the idea that the Laurent series "keep's it out" and taylor "keeps it in"? What I only know is that we use taylor when it is analytic at that point and laurent when it is not analytic at that point.
very helpful! Thank you!
Why divide by 4 (the constant) for first series and then divide by z - 1 for the second series (the function) instead of dividing by 2 (the constant)?
Why do 'we *need* to expand with z - 1 in the denominator'?
thank for your video, it helps a lot :)
hi. can you tell me why you took laurent expansion at z=1 and taylor at z=3?
Think it's because the z+3 for the taylor and z+1 from the laurent
well, then why "z+3 for the taylor and z+1 from the laurent"?
Bunk Moreland Taylor series converges within a certain radius, and a Laurent series converges outside a certain radius, he wanted it only to converge within a radius of 4, but outside a radius of two.
Your voice is like a cowboy teaching math haha. Thanks!
this is great, thank you very much :)
I have an urgent question. Why to you divide 4 on the first fraction whereas you divide (z-1) on the second fraction??
I also want to know this....
and nobody yet commented..... :(
this was in a comment below, i hope it helps "1. We need to expand with z-1 since we are finding a series about z=1. For example, if we were finding a series about z=-i we would need z+i. 2. Without going into too much detail we divide by 4 for the first series since this is a taylor series in the required region, and we divide by z-1 for the second series since this is a laurent series in the required region. This is the technique you should use and I encourage you to follow it closely."
You use the one which is bigger to divide, in order for the sum to converge -> the absolute to be less than 1.
So for the first fraction, 1/(z+3) you're focusing on the outside circumference, |z-1| < 4. You have 1/((z-1) + 4) and you want the sum to converge, so you'll divide by the one which is greates, which is 4.
You get the sum 1/(1+(z-1)/4), which will converge because |(z-1)/4| < 1.
In the second fraction, 1/(z+1), it's the same principle. You're focusing on the 2 < |z-1| circumference. So |z+1| will be bigger, and that's why you have to divide by it.
You get the sum 1/(1+ 2/(z-1)), which will converge because |2/(z-1)| < 1.
I know this is a late response, but oh well.
awesome guide!
What is the region of validity of the expansion? And what is the principle part of the series? Thanks!
You sound like what I imagine King Tut would sound like.
About the second part of this particular question:
how to write 1/(z+1) as a laurent series that is convergent when |z-1| is GREATER than 4?
I'm struggling with it
Edit: On the other hand if 1/(z+1) is written as a series that is convergent when |z-1| is GREATER than 2 and...
1/(z+3) is written as a series that is convergent when |z-1| is GREATER than 4 then the LOGICAL CONJUCTION of both convergence conditions means that f(z) can be written as a sum of series that's convergent when |z-1| > 4
Thank you for this video!! It helped me a lot!!
But what can I do when my function has 3 singularities. Maybe like the function you've used in video with the singularities 1,3 and -3. Then it's still the same circle in which we are looking for our series isn't it? But do I have to subtract another term or will the taylor series be the same for both 3 and -3?
thank you sir
really helpful
Hi there, great video! It really helps! Quick question, what pen tablet are you using for writing? Your handwriting looks nice! Thanks
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In the part that you did the Taylor series what is the shortcut that you to find the Taylor series expansion?
1/(1-r) is the solution to an infinite sum of a geometric series with first term 1 and multiplier r, which corresponds also to its taylor series. He turned the equation into the form 1/4*(1/(1+(z-1)/4)), so (-1)(z-1)/4=r. The taylor series follows from the corresponding geometric sum.
It's a nice trick, but only works generally for fractional functions as in the video.
such a great video.
By the way, where's that accent from ?
What would you do for the second part where |z-1|>4 Just Laurent for for the outer circumference plus the other fraction? Laurent for both fractions? Please help
Laurent for both fractions, since both expansions are for the outside of their respective circles. Well done!
So does the Laurent for 1/(z+1) remain the same and then you just find a Laurent for 1/(z+3)?
Edit* Thanks for the excellent video!!
What about singularities which lie within the region? For part 2 is it 2 Laurent series?
Sorry for asking but, since you reply to comments in the Laurent series why did you put 1/(z-1) infront ( i didnt understand what you said)? why not write it as this: (1/2) * 1/( 1 +(z-1)/2 ) ?
thanks
+Jason K Wondering the same thing. Think it's because z-1/2 > 1 for |z-1|>2 but 2/z-1 2. And that term has to be smaller than 1 for the geometric series. Not 100% sure though, confirmation would be appreciated.
what happens if u have two singularities in the exterior radius? (three in total with the next exercise) i.e: sin(z)/ z^2 + 1/(3-z^2) for 0
Really nice, thanks
Do we say both expansions are around z=1? Since the Laurent series has infinitely number of terms do we say it is a essential singularity at x=-1? Thanks.
Correct! Both expansions are around z=1. However the singularity at z=-1 is not essential. To determine if a singularity is essential or not you will need to find a Laurent expansion about the point of singularity. In this example we looked at an expansion about another point, namely z=1. If you do however find the expansion about z=-1 (the singularity) you will find there is only one negative power. Hence it is a singularity of order 1. If there were infinitely many negative powers of z then it would be an essential singularity. Does this explanation help you?
dgory
Yes it definitely helps! I guess the easiest way to see the point of expansion is to consider the standard form of laurent series c_j(z-z0)^j, for all integer j, identify z0 and treat everything else to be c_j, which is the coefficient.
Great job
Hi, i have quation when we use a partial fraction ?
what if there were more than two singularities inside the area? we'd still had to use Taylor and Laurent series?
I suspect you may be getting confused with Cauchy's Integral Formula. Can you provide an example?
dgory I was responding to this comment, sorry
"In this example, the function has singularities at z=-1 and z=-3. So we will get two circles one of radius 2 and the other of radius 4, both centred at the point we are expanding about, z=1. The next thing to do is to look at the question given and see which area in the complex plane they want the expansion. If it is between the two circles, as in this case, then we will need one Laurent series expansion, and one Taylor series expansion. Hope this helps. Please feel free to ask me more!"
But now how do you determine points of singularity from this?
You find the points of singularity by setting the denominator equal to zero. z+1=0 => z=-1 and z+3=0 => z=-3
I've got an urgent question!
I get that he had to write 1/4 in the front while developing the taylor series, to get 1/(1+(z-1)/4).
BUT why didn't he do the same thing with the laurent series, except it would have been 1/2 this time.
Instead he took out 1/(z-1) out the front. Why? Can anyone explain?
You guys know what I mean?
That 1/((z-1)+2) = 1/(2+(z-1)) = 1/2 * 1/(1+(z-1)/2)
Same question here.
I don't understand. Anyone explain it to me please. which one that expanded by Taylor series or Laurent series? Can we choose that 1/(z-1) expanded by Taylor series?..forgive me, i can't speek English well. I hope u understand what i Mean. Thankyou
good work
because use fraction partial?
I love you.
why laurentz at one and taylour series at other
why not apply laurentz at both
A Taylor series is a special case of a Laurent series.
It's not the answer for this question.
Good explanation. But I have an opinion here. Since the objective of this video is to show how to work out with Laurent series, you don't need to show the steps for partial fraction in this video.
You can directly show us the result of partial fraction without going through the steps. Then you can do another video for PARTIAL FRAction for those who wish to understand it.
Thank you! :)
where is example 2 video?
Unfortunately I never made one. I am currently making a mini course on Laurent series expansions for complex analysis, but that probably doesn't help you right now.
what tutorials?
Thanks
there is minor logical mistakes..
Please Help Exam tomorrow :)
How did you go?
You should have left out the extraneous simple math. I imagine anyone interested in the Laurent series should already know how to find partial fractions.
i would normally agree with this but some students such as myself end up doing maths courses every 2nd semester and tend to forget all the teeny tiny bits... i thought it was really nice that he showed that.
no
I totally agree with you. That partial fraction waste my 1 minute. LOL
You neglected a 1/(n!) in your Taylor Series.