Honestly this is the most intuitive and clear explanation I've found of the process of finding bounds with triangle inequality etc - thanks Dr. Peyam, this was very useful!
Hello ,Dr. πm) I would like to say thank you that you invited blackpenredpen to your lecture because they asked very useful questions for novices. As for me, I was graduated from the engineering faculty of the russian university and surprisingly I had all these stuff with proofs (surprisingly because even mathematicians have less mathematics than us- engineers ). Anyway , thank you that you reminded me of the best subject - Complex Analysis) . By the way ,Residue in Russian means "Вычет" 😁
Hokusai's Great Wave and Van Gogh's Starry Night : Peyam is definitely the Artist of mathematics ! Well, about contour integral, I have to work a bit more :-(
Complex analysis is *really* cool but for people to understand it from the ground and then people don’t have to assume it’s magic :) Basically the reason why the residue is important can be explained by the Taylor series of the (analytic, meaning it has a complex derivative) function, breaking it up into a sum of different powers of z. Every term has a nice primitive function [remember integral of x^n=(x^(n+1))/(n+1) ] *except* the 1/z term! (What’s kind of interesting in real analysis is really important in complex analysis here.) It’s the derivative of the complex logarithm which is multi valued. If we integrate on a closed curve, are start and endpoint is the same and so those terms with a primitive function integrate to zero! The exception is 1/z and famously the (complex) logarithm increases by 2*pi*i in one lap around the origin. So the only thing we care about when integrating on a closed curve is where the function has any singularities (else the 1/z term coefficient is necessarily 0 as the function is “nice”) and in those singularities, what the coefficient of 1/z is, and we call that the residue! (Note: some singularities have residue 0, like 1/z^2 in z=0, and do not contribute to the integral. It also does not have to be a simple “pole”, you can also integrate e^(1/z) around z=0 for example. This is an essential singularity where the function approaches infinity and 0 in the same point but it does not matter!)
To compute the residue just notice that 1/(z^2+1) = 1/(z-i) * 1/(z+i) and the second factor has no poles at z=i, so its Laurent expansion coincides to its Taylor expansion. The -1-th coefficient of 1/(z^2+1) now must be 1 (i.e., 1/(z-i)'s -1-th coefficient) times the 0-th coefficient of 1/(z+i), which is its value at z=i, namely 1/2i. So the residue is 1/2i (or -i/2).
Calcualtion of the residuum is much easier than demonstrated. As the pole is of first order, you can use the following identity for residuum calculation: \textstyle \operatorname {Res}_{a}f=\lim _{{z ightarrow a}}(z-a)f(z) The term (z-a) cancels out and the remaining function value is equal to the residuum. I have integrated some even more complex functions. This method is pretty easy.
A question about determined integrals that are equal to pi**N As this video shows Integral of 1/(x^2+1) from -inf to inf= pi ?Is there any function integrated on an interval that is equal to pi**2 or pi**N (where N is an integer N=2, 3.....). Obviously the limit of the integral do not contain explicit pi nor the integrated function. Nor the function contains trigonometric (or inverse of) functions.
Dr. Peyam is stylin!! Look at the sweet watch!! Steady pimpin' these lil Math tricks! Get Σ playa!! (I don't really talk like this; its just fun and I'm just very enthused about your attire.) But yo shirt be lookin' ill af; is dat silk???
Muhammad Qasim Dilawari It cannot be done. f(x) = x^^x provided f:R -> R is not integrable, since it is discontinuous almost everywhere. In fact, it is defined almost nowhere.
Excelent video dr peyam. I wonder what would have happened ...if when we were about to integrate 1/(z^2+1) wich goes to zero....instead we had put Arctan(z) what would be the result? Again 0? From z=R+0i to z=-R+0i the integrand satisfies Cauchy Riemann so...why not....
(gamma(t)) is just enclosing it in brackets so it's easier to square without confusion gamma(t) is just some function of real t that outputs complex values it sorta represents how one might draw the curve in 2D using a pen and t is the time it takes to get to the point gamma(t) while drawing the curve (but the time can be any interval over the reals, it doesn't literally mean real world time since starting the curve, just an analogy) there might be some constraints like maybe it has to be continuous or not undifferentiable at measures greater than 0 or something like that but anyway since it's a real input function it can be differentiated using bounds of integration notation (if it is an integrable function) just like any real functions you would deal with in calculus or real analysis, just that the output space would may complex (in this case since gamma(t) = t which is always real)
By the way I think you can do paramatrization this way too. Say z=x+i*y, from -R to R there is only real numbers so z=x if z=x then dz=dx. After that just put in place.
I just always forgot why the semi circle is sufficient. Why not both poles have to be included. Because you can integrate using any curve? but you have to include at least 1 pole, clearly.. can't rememberrr
Can't this integral (i.e. from minus infinity to plus infinity of 1/(1+x^2) ) also be solved like this? 1: L=int(-inf,+inf, 1/(x^2+1) ) 1/(x^2+1) is even, so L can also be derived as 2 times the integral from 0 to infinity => 2: L=2*int(0,+inf, 1/(x^2+1) )=2*int(0,+inf, (1/(x+i))*(1/(x-i)) ) 3: f(x)=1/(x^2+1)=1/((x+i)*(x-i))=A/(x+i)+B/(x-i) => A=-B and -2Ai=1 => A=i/2=-B => 4: L=2*int(0,+inf, i/(2(x+i))-i/(2(x-i)) )=i*int(0,+inf, 1/(x+i)-1/(x-i) ) 5: F'(x)=2f(x)/i=1/(x+i)-1/(x-i) and so F(x)=ln|x+i|-ln|x-i|+C L=i*(F(+inf)-F(0) )=i*(lim( x->+inf, ln|(x+i)/(x-i)| ) - (ln(i)-ln(-i)) )=i*(0-i*pi/2-i*pi/2)=-i^2*pi=pi I wonder: how much am I doing (in)correct at step (4 to) 5 or maybe somewhere else?
@@blackpenredpen Serious question: Did you really not know complex integration? You know so damn much about integration and maths in general, it's weird to see you struggle at sth. I can actually do^^
@@JamesLaFleur 1. This doesn't have to do anything 2. How is education brainwashing? By that logic I could say that it's brainwashing if kids only lern about traditional families and relationships. 3. What's exactly your problem with that picture? You can't really see what those books are about, or is this about the outfit? If you want to have some sort of conversation you need to present your problem in some meaningfull way.
Woohoo!!!! 😄
You actually didn't forget the "i" at 14:38 but it got erased at 13:28 haha
@@hiy9846 it's not.
It's from C1
Pi- m
Hoowoo!!!!
Doctor Pi M (Peyam) :0
Best explanation of contour integral I have seen. Hopefully, y'all continue with a series of more complex integrals.
The One and Only Peyam has many more videos on his channel. You can check the link in description. This is meant to be a starter. : )
Essentially I can be treated as a constant becaus it is one
We can use logarithm complexe: 1/(x²+1)=1/2i((1/(x-i)-1/(x+i))
Int(1/(x²+1))=Ln((x-i)/(x+i))
Please make a video on complex analysis :)
This video summarizes my calc 4 exam for next month on residue theorem.
Thank you very much!
Marvel: Infinity War is the most ambitious crossover event!
Bprp and Dr. Peyam: Hold my markers
Why It? Hahaha thanks!! And also Chester!
This is the first time I know that a lot of Peyam’s videos were taken in his office. Always liked the picture beside the board.
Finally I found a worth watching video about contour integrals
Is it me or is this a lot more interesting than the usual videos? It gives an insight to how you'd work in university.
Man, how beautiful is it to see a culture that appreciates learning. So wonderful.
Honestly this is the most intuitive and clear explanation I've found of the process of finding bounds with triangle inequality etc - thanks Dr. Peyam, this was very useful!
Peyam's enthusiasm for math is infectious
Hello ,Dr. πm) I would like to say thank you that you invited blackpenredpen to your lecture because they asked very useful questions for novices. As for me, I was graduated from the engineering faculty of the russian university and surprisingly I had all these stuff with proofs (surprisingly because even mathematicians have less mathematics than us- engineers ). Anyway , thank you that you reminded me of the best subject - Complex Analysis) . By the way ,Residue in Russian means "Вычет" 😁
Wow this is amazing, he made some amazing points throughout the video
I just had complex analysis last semester and saw the thumbnail. I immediately thought residue theorem and it was super fun to see you use it
Worth every minute
C. D. Chester totallly!!!
This video brought back some memories from graduate school... I really miss those times...
atrath : )))
Dr Peyam is genius dude
This is so wholesome haha I love complex analysis as well!
Hokusai's Great Wave and Van Gogh's Starry Night : Peyam is definitely the Artist of mathematics !
Well, about contour integral, I have to work a bit more :-(
also woke on the wall
Complex analysis is *really* cool but for people to understand it from the ground and then people don’t have to assume it’s magic :)
Basically the reason why the residue is important can be explained by the Taylor series of the (analytic, meaning it has a complex derivative) function, breaking it up into a sum of different powers of z.
Every term has a nice primitive function [remember integral of x^n=(x^(n+1))/(n+1) ] *except* the 1/z term! (What’s kind of interesting in real analysis is really important in complex analysis here.) It’s the derivative of the complex logarithm which is multi valued.
If we integrate on a closed curve, are start and endpoint is the same and so those terms with a primitive function integrate to zero! The exception is 1/z and famously the (complex) logarithm increases by 2*pi*i in one lap around the origin.
So the only thing we care about when integrating on a closed curve is where the function has any singularities (else the 1/z term coefficient is necessarily 0 as the function is “nice”) and in those singularities, what the coefficient of 1/z is, and we call that the residue!
(Note: some singularities have residue 0, like 1/z^2 in z=0, and do not contribute to the integral.
It also does not have to be a simple “pole”, you can also integrate e^(1/z) around z=0 for example. This is an essential singularity where the function approaches infinity and 0 in the same point but it does not matter!)
Love dr. P. Very approachable person and loves his subject.
To compute the residue just notice that 1/(z^2+1) = 1/(z-i) * 1/(z+i) and the second factor has no poles at z=i, so its Laurent expansion coincides to its Taylor expansion. The -1-th coefficient of 1/(z^2+1) now must be 1 (i.e., 1/(z-i)'s -1-th coefficient) times the 0-th coefficient of 1/(z+i), which is its value at z=i, namely 1/2i. So the residue is 1/2i (or -i/2).
This was cool.....Dr. P. rocks!
MIND BLOWN.
Nasty flashbacks to undergrad ... while complex analysis is amazing, I found it very challenging.
It's more fun without the exams!
@@douglasstrother6584 I should probably dig out my lecture notes, then!
@@flaviusclaudius7510 You'll be surprised by what you *did* understand.
It's a lot more fun w/o the time pressure.
wow!! Dr. Peyam. Brilliant try. Thank you. BlackPenRedPen!!
One of your best videos in all the time, I watched it for several times!
Thank you so much dear *bP🖋️rP🖍️*
Ohhh ... Dr Peyam is left-handed!!! :D
Calcualtion of the residuum is much easier than demonstrated. As the pole is of first order, you can use the following identity for residuum calculation:
\textstyle \operatorname {Res}_{a}f=\lim _{{z
ightarrow a}}(z-a)f(z)
The term (z-a) cancels out and the remaining function value is equal to the residuum. I have integrated some even more complex functions. This method is pretty easy.
A question about determined integrals that are equal to pi**N
As this video shows
Integral of 1/(x^2+1) from -inf to inf= pi
?Is there any function integrated on an interval that is equal to pi**2 or pi**N (where N is an integer N=2, 3.....).
Obviously the limit of the integral do not contain explicit pi nor the integrated function. Nor
the function contains trigonometric (or inverse of) functions.
Really enjoyed this one!
Nice paintings man .... Van Gogh, Hokusai , and something modernist with many colours resembling the Inca Empire flag ....
Loved it....!!!!
Uhh Dr.Peyam rocking the Rolex, okay, I see you xD
Happy Pride Peyam!!
Tom FW Happy pride 🙂
We can use logarithm complexe
1/(x²+1)=1/(x+i)(x-i)
............= (1/2i)[1/x-i +1/x+i]
You guys are fun. I want to see a Very Special Episode of "The Big Bang Theory" where you give them all swirlies.
i can picture this so vividly
Wooooowwwwwwwwwwwwwwwwwwwww I really appriciate the teacher , well done sir
Where was this channel when I was in college?!?! 😊
Awesome Video 😄
Dr. Peyam is stylin!! Look at the sweet watch!! Steady pimpin' these lil Math tricks! Get Σ playa!!
(I don't really talk like this; its just fun and I'm just very enthused about your attire.)
But yo shirt be lookin' ill af; is dat silk???
Pls can you make a video about macclaurin series thx.
Bprp: because "i" don't like to be on the bottom.
Dr. P.: Yes, Nice.
Bprp: ... Someone don't watch my videos
Hahaha hahaha
Great video !
I would think that someone who could do 1000 integrals, would be a complex analysis jock.
Sometimes you do something just to see that it agrees with what you already know. Other times you do it because it is actually easier.
I enjoyed math for the first time
Do the integral of y=x^^x
Muhammad Qasim Dilawari It cannot be done. f(x) = x^^x provided f:R -> R is not integrable, since it is discontinuous almost everywhere. In fact, it is defined almost nowhere.
Ah shit,my bad
Man... I love your smile 😍😅😛
Chester has a great shirt😄
Rot definitely
Excelent video dr peyam. I wonder what would have happened ...if when we were about to integrate 1/(z^2+1) wich goes to zero....instead we had put
Arctan(z) what would be the result? Again 0?
From z=R+0i to z=-R+0i the integrand satisfies Cauchy Riemann so...why not....
I really liked the host
Hi Blackpenredpen, I have a question for you:
When you write(for example) sin^-1(x), you mean 1/sin(x) or arc sin (x)?
99% sure that its arcus. I struggle with the same problem everyday
Typically arcsin(x) because we call the other one csc(x)
Res. Integral [C] dF(z)
Why didn't you use the evaluation of arctan(x) between inf and -inf? It was much simpler... Did you just want to show an alternative method?
yes
Great to see this again. We used the limit rule. The Laurent series was for me an eye opener, or should I say an i opener.
Oke, it’s very interesting
Can't we use Euler's form for z??..
What course would you learn this in?
Integration 101
Just Joking. You would learn this in Complex Analysis usually.
When Dr.Peyam was parameterizing the line segment on the semi circle from (-R,R) he wrote that ( γ(t)=t) but what is the meaning of (γ(t)).
(gamma(t)) is just enclosing it in brackets so it's easier to square without confusion
gamma(t) is just some function of real t that outputs complex values
it sorta represents how one might draw the curve in 2D using a pen and t is the time it takes to get to the point gamma(t) while drawing the curve (but the time can be any interval over the reals, it doesn't literally mean real world time since starting the curve, just an analogy)
there might be some constraints like maybe it has to be continuous or not undifferentiable at measures greater than 0 or something like that
but anyway since it's a real input function it can be differentiated using bounds of integration notation (if it is an integrable function) just like any real functions you would deal with in calculus or real analysis, just that the output space would may complex (in this case since gamma(t) = t which is always real)
By the way I think you can do paramatrization this way too. Say z=x+i*y, from -R to R there is only real numbers so z=x if z=x then dz=dx. After that just put in place.
"ok" - blackpenredpen 5/1/19
Zachary Usher ?
Isn't it easier to use Cauchy's Integral Formula? with f(z)=(z+i)^-1 and w=i? you get the exact same result. Also this only works for R>1 right?
I just always forgot why the semi circle is sufficient. Why not both poles have to be included. Because you can integrate using any curve? but you have to include at least 1 pole, clearly.. can't rememberrr
please can help me to solve let a and b are matrices, if a*b=b*a prove that a*(b^-1)=(b^-1)*a
Wonderful 🌷
Why in that case the absolute value of z its iqual to R?
Nice Big Classroom He Has There
BPRP : "OK"
Me : master saitama?
it flew , just over the cereberum
Can't this integral (i.e. from minus infinity to plus infinity of 1/(1+x^2) ) also be solved like this?
1: L=int(-inf,+inf, 1/(x^2+1) )
1/(x^2+1) is even, so L can also be derived as 2 times the integral from 0 to infinity =>
2: L=2*int(0,+inf, 1/(x^2+1) )=2*int(0,+inf, (1/(x+i))*(1/(x-i)) )
3: f(x)=1/(x^2+1)=1/((x+i)*(x-i))=A/(x+i)+B/(x-i) =>
A=-B and -2Ai=1 => A=i/2=-B =>
4: L=2*int(0,+inf, i/(2(x+i))-i/(2(x-i)) )=i*int(0,+inf, 1/(x+i)-1/(x-i) )
5: F'(x)=2f(x)/i=1/(x+i)-1/(x-i) and so F(x)=ln|x+i|-ln|x-i|+C
L=i*(F(+inf)-F(0) )=i*(lim( x->+inf, ln|(x+i)/(x-i)| ) - (ln(i)-ln(-i)) )=i*(0-i*pi/2-i*pi/2)=-i^2*pi=pi
I wonder: how much am I doing (in)correct at step (4 to) 5 or maybe somewhere else?
3:10 shouldn't it be absolute value abs(γ'(t)) in general?
Yes
He forgot it
why is it x in the title, not z?
oh, i commented before watching
Okay
arctan(inf) - arctan (-inf) = π ??
I'd like an explanation please. You just glossed over that as if it's trivial
Alexis Mandelias
arctan(inf) is pi/2
And arctan(-inf) is -pi/2
@@blackpenredpen nvm I'm beyond stupid. I was thinking of a different function. Thanks for reply though
Are you also a doctor? , so do you have phd?
Super u and Dr πr
I'm the like number 1000........yeeeeeeeeeeeeeeeeeeeeee ;-)
8:40
Did you make a mistake? The integral was 1/z^2 + 1 and you used the triangle inequality with 1/z^2-1?
1/abs(z^2+1)=1/abs(z^2-(-1)) We needed a subtraction here, and when we broke it up, abs(-1) just became 1
🏳️🌈🏳️🌈🏳️🌈
I get it or should I say 2i get it.
No entendí una pija pero vi todo el video 📹 por que se cagaban de risa 😂.
Subtitules in spanish please!!!! Tranks.
Wow that starry night picture(wrong focus sorry)
Mak Vinci oh ok!! Hahaha
@@blackpenredpen Though that wont distract the great lesson by Dr. P 😁
Mak Vinci there’s another really cool picture, coming up soon
Talking about Starry Night, Please make a video chaos theory and turbulence.
Nice focus Mark, not a worthless one. There's a mathematical/physical backstory to it: accurate depiction of Turbulence.
Oke
I umm... I have no clue what just happened.
In physics we learn this in 2nd semester lol.
Ana Bang
👍
We learn this in Peyam’s office.
@@blackpenredpen
Serious question: Did you really not know complex integration?
You know so damn much about integration and maths in general, it's weird to see you struggle at sth. I can actually do^^
Ana Bang
I kinda remember it but kinda don’t. I learned this like over 12 years ago and hadn’t touch it since then.
You learn complex integration in a physics class? 🤔
i stan the pride flag in the background
Third
Wait wait... A rainbow flag 🤔🤔
Rura FS Ye
Fags use it
🌈
Your point is?
Well yeah, I mean I'm pretty sure Dr. Peyam is.... a wonderful human being and a gift from the heavens, duh. ;)
I got a little giddy when I saw that
Why is there this rainbow flag?
Why not the rainbow flag?
🏳️🌈🏳️🌈🏳️🌈🏳️🌈
June is pride month
@@sensei9767 Thank you!
@@fuckyou1640 Everyone should live how he wants to. But this is brainwashing. lbpost.com/wp-content/uploads/2017/10/DMN9oKgUIAAls6N.jpg
@@JamesLaFleur 1. This doesn't have to do anything
2. How is education brainwashing? By that logic I could say that it's brainwashing if kids only lern about traditional families and relationships.
3. What's exactly your problem with that picture? You can't really see what those books are about, or is this about the outfit? If you want to have some sort of conversation you need to present your problem in some meaningfull way.
We stan with an LGBTQ+ math professors!
I get so lonely lonely lonely lonely lonely
LGBTQ flag ? 😡
周知院学院 周知のじーじつ
Show the teacher some respect
Sydney Carton ?
What?
Second
Anyone watch while playing minecraft?
@@mipmip4575 same trying to get into 2b2t rn
I was just wondering... Is Dr Peyam gay?
I mean, the " 'i' don't like to be in the bottom" misunderstanding and the huge pride flag might mean something who knows
First!