Update: There's a small typo in this video: With the concavity part, it shouldn't be that cos is below its tangent line (which is still true), but rather that cos is above the secant line connecting the points (0,1) and (1,0), which is y = 1-t so the correct estimate is that cos(pi/2 t) is greater than or equal to 1-t. The rest of the video still proceeds the same. Thanks Tobias for catching that mistake :)
I am not convince that Int 0 to infinite of sin(x**2) actually converges with the usual definition. For all epsilon > 0, there exist M so x > M => abs(f(x)-c) < epsilon, where c is the proposed limit.
@@michellauzon4640 I think tho that because the period of sin(x²) decreases as x increases, the area between the curve and the x axis becomes infinitesimally thinner
I love how all of Dr. Peyam's methods are so clearly organized and all the stages like idea, strategy and so on are so easy to follow!! I just envy the students that get to study under him first hand!!!
This is creative math in its purest form. Coming up with the complex integrand -(exp(z2)) to integrate over the bottom half circle of the first quadrant so cleverly takes immense practice and insight. Trial and error alone, even with persistence, would cause anyone to give up. Thank you Dr Peyam!
*on the whiteboard: 0* Peyam:Ohhahahahahah BPRP:Finally Peyam: *falls onto the floor* Together:HAHAHAHAHAHHA Peyam: you won't see this in the video but it took us an hour to fix this mistake Together:HAHAHAHHHAHAHAH Oon Han: ... Lol
Good video. Your videos cover much more advanced stuff than your bprp's videos do. 2 months ago, I could not understand anything you said, but now, I find your videos very helpful as I move on to learning more advanced stuff.
I swear I forgot the question at 11:42, I asked myself, what am I trying to do, what's going on, did I really wished to watch this before? What did I searched for? These questions popped out of my mind
around 15:50 the reasoning is that if cos() < -πt/2+π/2, this implies that e^-R^2 cos() < e^-R^2(-πt/2+π/2). How? I would think the opposite, because of the -R^2 in the exponent.
This can be fixed by setting cos(2t)>=1-4t/π when t in [0, π/4] With z(t)=Re^it, consider the norm of the integrand |e^-z^2 dz/dt| = |iRe^it e^(-R^2 e^2it)| = |iRe^it| |e^(-R^2 ( cos(2t)+isin(2t)))| = R e^(-R^2 cos(2t))
We did this in 1st year at UWaterloo, no compex integrals. I can`t remeber how we did it and don`t have mynotes. How is it done without contour integrals, tia.
9:55 i think that is not because it is purely imaginary, but because the radius of e^(i.pi.t/4) is 1 and |e^(i.pi.t/4)| means the modulus or the radius of e^(i.pi.t/4) extra: e^(i.pi.t/4) is not purely imaginary because it has the real part which is cos(i.pi.t/4) correct me if i am wrong
I can not wrap my had around, why a sine wave with quadratic growing frequency ends up at sqrt(pi/8). Great work. Loved it :-) edit: I just had a closer look at the graph and looking at the part
I enjoyed it very much, thank you Peyam! The way you cope with the B-part is instructive, using techniques typical of real analysis (cosine vs. affine function). But the "Standardabschätzung" taken from my book on complex analysis (by Reinhold Remmert) is more effective: The absolute value of a complex path integral is smaller than or equal to the length of the path times the maximum of the absolute value of the function on the path. Complex analysis is magical somehow.
I remember long ago I was reading about the Fresnel integrals, but I always wondered why they evaluated to sqrt(pi/8) from 0 to inf. This was a good video, and it answered my question!
You don't need complex analysis to do this integral, but it's standard if you know it. There's a wonderful video on fapable maths channel which does it without complex, you can check it out if you want!
Just went thru this in some detail and when your looking at the integral along path C you substitute t' = R-t, dt'=dt and then the t' mysteriously disappears further down and becomes t. This puzzles me a little so can you please explain this step ?
alex zorba I didn't want to keep writing t' all the time, so basically for convenience I replaced t' by t. If you'd like, think of it as another substitution t = t', dt = dt'
this could be a little sum-of-infinitives flawed logic (like 1-1/2+1/3-1/4.... can be anything you want) but since sine starts out positive, and it is of x^2, doesn't that mean that for each period, more of the period is spent during the positive portion every time, and thus it must add to infinity?
Ryan Roberson That's a great intuition, but it turns out that it spends *slightly* more time in the positive part which gets smaller and smaller so that in the end we have sqrt(pi/8). It's like in (1+1/n)^n, The 1/n in the parenthesis is just slightly enough bigger than 1 so that the resulting limit is e, kinda neat, huh?
I used u=x+2 , u-2=x then i found: integral cos ( u ^3/2 - 2 u ^ 1/2 ) du= integral ( cos (u ^ 3/2) ) ( cos 2 U ^ 1/2 ) + Integral ( sin u ^ 3/2 )( sin 2 u ^ 1/2 ) I am not sure if it is correct or not. it looks ugly.
![Blox Fruits Dragon Rework Update [Full Stream]](http://i.ytimg.com/vi/EqDAp8udhm0/mqdefault.jpg)
Update: There's a small typo in this video: With the concavity part, it shouldn't be that cos is below its tangent line (which is still true), but rather that cos is above the secant line connecting the points (0,1) and (1,0), which is y = 1-t so the correct estimate is that cos(pi/2 t) is greater than or equal to 1-t. The rest of the video still proceeds the same. Thanks Tobias for catching that mistake :)
Wills Zng Add to that: Concave = Second derivative is nonpositive, Convex = Second derivative nonnegative (if the function is smooth)
Can u plz do it for (sinx^2)^2 ??
I am not convince that Int 0 to infinite of sin(x**2) actually converges with the usual definition. For all epsilon > 0, there exist M so x > M => abs(f(x)-c) < epsilon, where c is the proposed limit.
ঙঙঙঙঙঙঙ
@@michellauzon4640 I think tho that because the period of sin(x²) decreases as x increases, the area between the curve and the x axis becomes infinitesimally thinner
I love how all of Dr. Peyam's methods are so clearly organized and all the stages like idea, strategy and so on are so easy to follow!! I just envy the students that get to study under him first hand!!!
"Thank you, Wikipedia" You have no idea how OFTEN I said that during my time in university....
@Dirac is Dune The Desert Planet I don't understand yours, though....
I didn't get a half of it but idc, I loved it.
This is creative math in its purest form. Coming up with the complex integrand -(exp(z2)) to integrate over the bottom half circle of the first quadrant so cleverly takes immense practice and insight. Trial and error alone, even with persistence, would cause anyone to give up.
Thank you Dr Peyam!
Sorry I gave up on words sin(x^2) = e^-z^2
Wow. What a mammoth effort. This would have to be the most complicated contour integration I've ever seen. Thanks for posting
25:22 I shouted "NOOOO" at my screen, and you never fixed your mistake... :'((
MrQwefty lololol, sorry! *hug* See the pinned comment, though :)
same. e^(-x) is a decreasing function!
Wow! What a beautiful result. 23 minutes in and we finally got to the integral in question. I love it!
Only Dr. Peyam does complicated yet elegant methods
correct me if i'm wrong, but i'm pretty sure you've made a mistake at around 15:30. You say cos(pi/2)
See the pinned comment! :)
bprp is the cameraman, did not expect this
*on the whiteboard: 0*
Peyam:Ohhahahahahah
BPRP:Finally
Peyam: *falls onto the floor*
Together:HAHAHAHAHAHHA
Peyam: you won't see this in the video but it took us an hour to fix this mistake
Together:HAHAHAHHHAHAHAH
Oon Han: ... Lol
Oon Han hahhahahaha. Life isnt easy
Oon Han isn't it?!
isnt
Thank you so much, I'm a colombian physics student and I didn't understand this exercice until I saw your video, Thank you a thousand times more
De nada!
Man you're amazing, really was thinking about it , was doing my homework about limits and continuity.
THANK YOU SO MUCH, Dr. Peyam! This helped me so much with my Complex Analysis homework! Keep doing amazing videos please! More Complex Analysis!
Lots of love from India ❤️
Thank you for such an amazing video🔥
I used to watch your videos just for fun, now I watch them to help me study for my tests and I still have a lot of fun doing so!
Good video. Your videos cover much more advanced stuff than your bprp's videos do. 2 months ago, I could not understand anything you said, but now, I find your videos very helpful as I move on to learning more advanced stuff.
This is a great video. You're fun and your passion comes through just watching you:) Keep up the great work!
I swear I forgot the question at 11:42, I asked myself, what am I trying to do, what's going on, did I really wished to watch this before? What did I searched for? These questions popped out of my mind
What is the name of this integration method?
Hi! It was almost perfect
around 15:50 the reasoning is that if cos() < -πt/2+π/2, this implies that e^-R^2 cos() < e^-R^2(-πt/2+π/2). How? I would think the opposite, because of the -R^2 in the exponent.
This can be fixed by setting cos(2t)>=1-4t/π when t in [0, π/4]
With z(t)=Re^it, consider the norm of the integrand
|e^-z^2 dz/dt|
= |iRe^it e^(-R^2 e^2it)|
= |iRe^it| |e^(-R^2 ( cos(2t)+isin(2t)))|
= R e^(-R^2 cos(2t))
Oh yeah, Peyam said that too - sorry.
The result was pretty obvious, but demonstrating is much harder. Congrats :-)
Love contour integrals. This certainly reminds me of some best time learning complex analysis in college.
We did this in 1st year at UWaterloo, no compex integrals. I can`t remeber how we did it and don`t have mynotes. How is it done without contour integrals, tia.
You can check out the video on fapable math's channel; he does it without complex analysis :)
how do I get to said math channel,tia.
GetSmart008 There ya go: ruclips.net/video/iOBU4wEOyKw/видео.html
Series, yay
how about something notin German........new video idea for yourself
Can it be solved using 'semicircle+real axis' contour instead of pizza? If not why not? Something to think about.
Ein wunderbarer Substitutionsmechaniker.
when i can't solve a math problem, i always say "thank you wikipedia" to myself🤣
Peyam can also use the Riemann Lebesgue to show that the integral over the arc goes to 0?
9:55 i think that is not because it is purely imaginary, but because the radius of e^(i.pi.t/4) is 1 and |e^(i.pi.t/4)| means the modulus or the radius of e^(i.pi.t/4)
extra: e^(i.pi.t/4) is not purely imaginary because it has the real part which is cos(i.pi.t/4)
correct me if i am wrong
What I meant is that the exponent is purely imaginary, hence the modulus is 1
@@drpeyam ahhh i see, sorry i thought it is was the e^(i.pi.t/4)
How does e^(-i*R^2*sin(pi/2*t)) become 1?
EDIT: It's because of the minus sign .-.
not quite, it automatically has modulus 1 because e^(iw) is distance 1 from the origin for any complex number w
@@WhattheHectogon For any w *REAL*
@@theoleblanc9761 yes that's correct! my mistake
I am from India. Your videos are very nice. Thank you
A little variation:
Integral of sin(x^2)/(1+x^2), and cos(x^2)/(1+x^2), where x goes from minus infinity to plus infinity.
Ooooh, that one’s nice too!
beautiful problem. I solved the cosine version by using Feyman's trick
I can not wrap my had around, why a sine wave with quadratic growing frequency ends up at sqrt(pi/8). Great work. Loved it :-)
edit: I just had a closer look at the graph and looking at the part
I enjoyed it very much, thank you Peyam!
The way you cope with the B-part is instructive, using techniques typical of real analysis (cosine vs. affine function).
But the "Standardabschätzung" taken from my book on complex analysis (by Reinhold Remmert) is more effective:
The absolute value of a complex path integral is smaller than or equal to the length of the path times the maximum of the absolute value of the function on the path.
Complex analysis is magical somehow.
04:01 What theorem?
The composition of analytic functions is analytic (by the Chen Lou/chain rule), and z^2 is a power function, hence analytic.
Could you please spell the name of the theorem? I couldn't figure it out from hearing.
When you've taught or will teach complex analysis in the future. What is your advice for students to not make the wrong branch cut?
i have been thinking about how to integrate 1/[4-6(cosX)^6] , x/[4-6(cosX)^6] or x^2/[4-6(cosX)^6] ,or in general term x^n / a-bsin^6X as a whole.
I was very depressed when I started seeing this video, within first 20 seconds I started laughing.
I remember long ago I was reading about the Fresnel integrals, but I always wondered why they evaluated to sqrt(pi/8) from 0 to inf.
This was a good video, and it answered my question!
so evaluating the fresnel integral requires using complex analysis ? is there another way to do it ?
You don't need complex analysis to do this integral, but it's standard if you know it. There's a wonderful video on fapable maths channel which does it without complex, you can check it out if you want!
thank you !! amazing video too !!
Just went thru this in some detail and when your looking at the integral along path C you substitute t' = R-t, dt'=dt
and then the t' mysteriously disappears further down and becomes t. This puzzles me a little so can you please explain this step ?
alex zorba I didn't want to keep writing t' all the time, so basically for convenience I replaced t' by t. If you'd like, think of it as another substitution t = t', dt = dt'
Dr. Peyam. I see, that makes sense. Thanks for your reply, and also for taking the time to produce and post this very interesting integral.
you should make a video of solving the limit of the definition of the derivative for sin(x)
Paul Montes I've already made such a video! :)
You start off by assuming that the integral actually exists. Why is this justified?
this could be a little sum-of-infinitives flawed logic (like 1-1/2+1/3-1/4.... can be anything you want) but since sine starts out positive, and it is of x^2, doesn't that mean that for each period, more of the period is spent during the positive portion every time, and thus it must add to infinity?
Ryan Roberson That's a great intuition, but it turns out that it spends *slightly* more time in the positive part which gets smaller and smaller so that in the end we have sqrt(pi/8). It's like in (1+1/n)^n, The 1/n in the parenthesis is just slightly enough bigger than 1 so that the resulting limit is e, kinda neat, huh?
not that far off of what i thought, i just didn't know the decreasing trend was strong enough to cause a limit to occur.
I like your videos!!! Saludos desde México 😎
Can't this also be done with Taylor series and a little easier?... Ah, no, I tried it and the integration does not work out nicely.
Thank you man!... You save our lifes!
Your method becomes great because of your craziness😂👍
Isn’t this the Fresnal integral?
Yes
Dr. Peyam's Show
it’s super scary to integrate (lim n-> inf)!
Isn't it an eighth of a pie?
This integral really converges? İ think this integral converges more beatifull than the answer.
I love your content ! :D
I love how he loves maths.
what about abs(sin(x^2))?
I have a feeling that this is just going to be infinity!
doesn't this integral diverges anyway?
raziel Keren no, it is uniform convergent
Mi Les but doesn't it necessary for the function to approach zero at both limits?
Hello. Is this function no elementary?😮
It’s not
great video! i have been looking for integral of cos x sqrtx . I couldn't find it on google yet. any tip?
Do you mean cos(x sqrt(x)) or cos(x) sqrt(x) ?
I googled and worked on it almost 3 hours! I tried what ever I found. my knowledge is limited but I really like to know how I can solve it!!!
integral cos ( x ( sqrt (x+2) ) ) ,
sorry I didn't write it correctly first.
I used u=x+2 , u-2=x
then i found:
integral cos ( u ^3/2 - 2 u ^ 1/2 ) du=
integral ( cos (u ^ 3/2) ) ( cos 2 U ^ 1/2 ) + Integral ( sin u ^ 3/2 )( sin 2 u ^ 1/2 )
I am not sure if it is correct or not. it looks ugly.
I mean it’s correct so far, but it does look ugly :P
oh my god... i'm amazed at 26:28 thx
I just completed Cal 2. I thought I would be able to keep up... but I got lost 3 minutes after the video started 😭
You just apply u sub to the integral from 0 to infinity of e^-(it^2)...Then you are done
Dr. Peyam i solve the problem with the cone ,thanks for integrals :)))
oh my god this is amazing
Isn't it a Fresnel Integral? (I'll watch the video after work)
Yeah
U can solve this problem by LAPLACE transformation, which is easier then this complex analysis!
If I could suscribe again I would! hahaha Amazing work, love this channel.
Sir solve this integrate x2e-^x2 cosx dx from - infinity to + infinity
Wow! Great video. I got to part B and then I got stuck :)
Thank you so much. 👏👏👏
Great! Although I prefer doing it with Laplace transform
Mauricio Huicochea Toledo ? How do you do it that way
Me encanta como enseña
Actually a Nice vid
Awesome!
Thanks!
I repeated the intro like 100 times🤣🤣🤣🤣
thank u !
26:48 oooooooh ;J
From India 👌
You lost me almost right away
👏🏼👏🏼
Good
🤯🤯🤯
Hi :D
Hello! :)
/// hi!
Greetings.
f(z) dizzy
Hello sir
Subtítulos plis :´v
Wast of time.