Another beast of an integral laid to rest by the sword of Feynman!!! The solution development is absolutely gorgeous and the result is surprisingly satisfying.
If you like the videos and would like to support the channel: www.patreon.com/Maths505 You can follow me on Instagram for write ups that come in handy for my videos: instagram.com/maths.505?igshid=MzRlODBiNWFlZA==
@thermodynamics458
Жыл бұрын
Young mathematically talented kids these days are so lucky to have the internet as a resource to keep them stimulated. This kind of video is exactly what I needed as a young teenager.
@Targeted_1ndividual
Жыл бұрын
As a teenage self-proclaimed math goblin / Feynman acolyte, I concur.
@caspermadlener4191
Жыл бұрын
Most of the current IMO participants also watch a lot of math videos. As fourth of Europe at the IMO last year, I am surprised how much there is to learn on the internet.
@mayasudhakar9595
Жыл бұрын
I feel so jealous of them 😁
@slavinojunepri7648
Жыл бұрын
I wish I had access to resources of this king when I was young. I grew in a village with no books and libraries. I barely had a blackboard with some pieces of chalk and a kerosene lamp that hurt my eyes at night during homework. But somehow I took pleasure in math.
@Targeted_1ndividual
Жыл бұрын
@@slavinojunepri7648 where did you grow up?
@manstuckinabox3679
Жыл бұрын
The more I watch feynmann integration technique videos, the more powerful I become.
@azizbekurmonov6278
Жыл бұрын
Same!!
@Dagestanidude
Жыл бұрын
@@azizbekurmonov6278 азизбек.не русскоговорящий ты случайно?
@azizbekurmonov6278
Жыл бұрын
@@Dagestanidude Da ya panimayu
@onceuponfewtime
Жыл бұрын
Lol
@marcokonst4144
Жыл бұрын
Xp farming on this video
@jul8803
Жыл бұрын
So to sum it up and generalize: Craftily plug in a parameter a so the derivative of the integrand with respect to a is simpler, now you have I(a) and you're looking for I = I(a0) Derive the integral with respect to the parameter making sure swapping places between the integral and the derivative is allowed (check convergence) Make your way towards an explicit expression for I'(a) Integrate I'(a) yielding an extra constant in the I(a) expression Determine the constant by plugging in I(a) a nice value for a making it trivial to compute Replace a by a0 and voilà, I(a0) à-la-Feynman, serve hot with a light Chianti.
@brendawilliams8062
Жыл бұрын
No wonder they use a math sign language. What a ride!
@rondovk
10 ай бұрын
Hero
@TheChrisSimpson
9 ай бұрын
My summary: Find someone better at math than me and ask them for help. Maybe I'll find this guy's email somewhere...
@mq-r3apz291
7 ай бұрын
We makin it outa Cornell wit dis one😎
@kwgm8578
Жыл бұрын
It's been 50 years since I've solved a complex integral. This guy moves too fast for me! I'm reminded of my old teacher, and later friend, Wolfram Stadler. Rest in Peace, Wolf.
@blkcat6184
Жыл бұрын
Ditto. Learned how, then never had to use them again. Today, fugetaboutit!
@LetsbeHonest97
11 ай бұрын
sir, may I ask what you studied and what you did in your professional career? I'm planning to get back to grad school for math and computing
@kwgm8578
11 ай бұрын
@@LetsbeHonest97-- If you're asking me, I earned an undergrad in EE in 1980 and a master's in CS in 1984. Go and do it as soon as you can -- school gets more difficult as you age.
@LetsbeHonest97
11 ай бұрын
@@kwgm8578 absolutely ... Will do asap
@kwgm8578
11 ай бұрын
@@LetsbeHonest97 Good luck to you!
@smaari
10 ай бұрын
Excellent work, a good way to check the answer is by plotting the function (e^-x^2)*sin(x^2)/x^2 and estimating the area from 0 to infinity under the curve. The function is > 0 from x=(0 to 1.722), and the function is almost zero for x=(1.722 to 2.35) and then zero for all values of x>2.35. You can approximate the area under the curve as a right tringles with sides of 1 and 1.722. The area for that right triangle is (1x 1.722)/2=0.861. The exact answer per the video is 0.806626.
@pleasegivemeaciggy
Жыл бұрын
Love how you talk about mathematics with passion while solving :)
@Unidentifying
Жыл бұрын
epic , thank you for making this technique so clear
@chrisc4208
9 ай бұрын
Wow yes this is so intuitive and elegant and beautiful and I totally followed you the whole way along
@maths_505
9 ай бұрын
Thanks so much 😊
@JuhoKim-qg1tk
Ай бұрын
This is AMAZING!! Thank you for your great video. I think I lack some basic techniques regarding imaginary number but except that everything was super clear and easy.
@lawrencelawsen6824
25 күн бұрын
This is advanced
@user-lu5nj7yw5i
2 ай бұрын
Absolutely beautiful. Thank you for sharing!!
@vincentstrgar441
9 ай бұрын
Beautifully done video!
@markburnham7512
Жыл бұрын
My favorite aspect of Feynman is that, while he was certainly a genius, he has a big dose of ordinary guy that we can relate to. I'm not in his league by a long shot, but I bet it would have been a blast to hang out with him.
@JgHaverty
10 ай бұрын
With respect, what are you talking about lol? 😂 Feynman's brilliance was only matched by his ego and capability to be a complete asshole. His lecture series are engaging and make him out to be what youre trying to portray, but the reality of his personality was quite a bit more grim in both nature and circumstance of his life. He was a good teacher; as that tied into his work, but no you really wouldnt want to be "buds" with him and he most certainly is not a strong candidate for representing the "every man". Sorry to burst your bubble; but best to keep his legacy wrapped in his brilliance and contributions to science as a whole, not his personality.
@TheSireverard
10 ай бұрын
Surely you're joking, Mr Feynman... ;)
@jamesedwards6173
9 ай бұрын
JgHaverty, spoken like a true ignoramus.
@jamesedwards6173
9 ай бұрын
@@TheSireverard, and also "What Do You Care What Other People Think?"
@JgHaverty
9 ай бұрын
@jamesedwards6173 what the hell are you talking about? Hahaha
@user-dl8rb2do5s
Жыл бұрын
This was amazing, really gotta use it instead of by parts. Thanks a lot !
@scottlapierre1773
Жыл бұрын
Been waiting for an explanation of my favorite’s, Feynman, noble prize topic.
@pierre-adelinmercier1427
Жыл бұрын
Noticing that d/dx(-exp(-x^2)/x) = 2exp(-x^2) + exp(-x^2)/x^2, I went for an integration by parts, which also works nicely, but is less elegant I admit. I found amusing that in that case, the result appears in the form of sqrt(Pi/sqrt(2))(cos(Pi/8) - sin(Pi/8)). After multiple careful checks for mistakes, I eventually realized it is actually the same result as in the video!
@yogsothot
Жыл бұрын
In the video is =d/da[sin((ax²) dx =f of d/da X² ½-a The -exp =to its integral, but its sin8 and exp
@azizbekurmonov6278
Жыл бұрын
You're doing really good content. Please, moreeeeee Feynman Integrals!!
@ShimmerArc
10 ай бұрын
Very cool! Thanks for sharing.
@pesto484
Жыл бұрын
Very nice presentation.
@julianmldc
Жыл бұрын
Amazing content!
@zunaidparker
Жыл бұрын
Nice integral! I wonder if it's solvable putting the a parameter into the exponential instead? Seems like you should end up at the same place. To solve the constant of integration you would need to let a tend to Infinity instead of setting it to zero, and the rest should be the same.
@patrick-kees8962
Жыл бұрын
I'd imagine you'd get issues with the fact you'd still have the sin and therfore a complex exponential which makes things more complicated
@GilbertoCunha-tq2ct
Жыл бұрын
@@patrick-kees8962 I believe it would still work if you consider the Imaginary part of the integral instead of the Real part
@edmundwoolliams1240
Жыл бұрын
Amazing! I solved this by defining an I(a,b) equal to the integral with a parameter inside the e and the cos. Then differentiating partially and adding to get a first order PDE. Then conjugating and using partial integration to get the required result! Your method is much slicker, as you just took the real part rather than dealing with the whole complex function!… 😂
@zed_961
7 ай бұрын
It's crazy
@edcoad4930
Жыл бұрын
Did it (after seeing video) with the a on the exponential term.....follows pretty much the same route except using the Im operator as sin(x^2) is a constant. Other than proving Im(sin(x^2) = 0) over the range, pleasingly we get the same answer.
@MrWael1970
Жыл бұрын
very nice effort. good luck
@AJ-et3vf
9 ай бұрын
Great video. Thank you
@denniswhite4446
Жыл бұрын
I came up with this myself in college. I hadn't known until now that this Feynman guy stole it.
@maths_505
Жыл бұрын
😂😂😂
@RohanDhandr8
Жыл бұрын
I completely believe you
@ziggy6698
Жыл бұрын
Cool video. :D Another way I think you could do is using my #1 favorite method, ha ha. Once you've differentiated and the integrand is in the cosine form, use Euler's definition to re-write cos. Then you have a sum of integrals of exponentials. Then the trick is, make a u subsitution for the argument of the exponential, that puts the integrals into the form of a Euler's integral definition of gamma. The power of u allows you to determine each z.
@gheffz
Жыл бұрын
Brilliant! Thank you.
@aarohibhavsar1520
10 ай бұрын
This makes me want to learn complex analysis. Great video considering I still understood most of it
@Amb3rjack
Жыл бұрын
As someone who failed their A level maths almost forth years ago, I found this video utterly fascinating and understood (or rather, could follow) practically none of it . . . .
@cassianperera2426
10 ай бұрын
Thank you Sir for your best explanation and working out of the problem🥰😍🤩
@maths_505
10 ай бұрын
Thank you for the nice comment
@manfredgeilhaupt5070
10 ай бұрын
very perfect, I tried to do it myself and needed the video again and again. But now I got it all. See research gate if you are missing 2 or 5 steps in between.
@gauranshbansal
9 ай бұрын
Can't wait to learn all this it seems interesting enough 🙂
@jpaulc441
Жыл бұрын
I'm one of the very unlucky ones who are incapable of math beyond basic algebra but am fascinated by it. I watched the entire video despite understanding nothing. I'm not sure if this is just an elaborate form of self-harm...
@Amb3rjack
Жыл бұрын
Absolutely. I feel exactly the same!
@sirius1255-
Жыл бұрын
technically you also have to ensure that the differentiation and integration are interchangeable (which is not true in general for integrable functions) which can be quite tedious, especially when working with improper integrals
@thomasdalton1508
Жыл бұрын
He covered that in the video, albeit somewhat handwavingly.
@egdunne
Жыл бұрын
@@thomasdalton1508 Yes. The handwaving ignored the potential problem at the left-hand side, where x=0 and x^2 is in the denominator. It's fine, but should be addressed.
@thomasdalton1508
Жыл бұрын
@@egdunne It doesn't need to converge at x=0 does it? The integral is from 0 to infinity, so it needs to converge on the *open* interval (0, infinity). The boundary points don't matter.
@evertvanderhik5774
Жыл бұрын
Mathematicians will worry about that, physicists not so much.
@thomasdalton1508
Жыл бұрын
@@evertvanderhik5774 Physicists might not worry about proving rigorously that it converges appropriately, but they need to worry about whether it does or not otherwise they'll get the wrong answer. You can determine that using rules of thumb rather than a rigorous analysis, but you have to do it.
@dougr.2398
9 ай бұрын
This may be one of Feynman’s integration techniques (he has several and needed them to perform integrations necessary to compute Feynman diagram calculations) but it isnt the one he was most famous for…. Integrating by analogy with finite summations and vice versa. This particular technique, or parts of it (particularly integration by differentiating under the integral sign) is discussed in Engineering Mathematics Advanced texts such as Sokolnikoff & Sokolnikoff . This particular calculation is a bit more involved as complex variables are introduced
@Schlaousilein67
11 ай бұрын
I love this video!!
@annanemustaph
2 ай бұрын
nice demonstration 👍
@johnrosen7856
Жыл бұрын
Beautiful solution
@dipankarmondal7662
11 ай бұрын
Just infinitely beautiful!
@maths_505
11 ай бұрын
SUIIIIIIIIIIIIIIII
@aaabbb-lw3ob
10 ай бұрын
Beautiful!
@alexanderkolesnik9357
Жыл бұрын
Almost everything is cool, except for one. Complex numbers have two square roots. It would be nice to mention this and show that it does not affect the result.
@svetlanapodkolzina1081
Жыл бұрын
It is a minor omission, but you are right
@ZiqoR
Жыл бұрын
@@svetlanapodkolzina1081 It's not a minor omition, we don't have logarithm complex function because of monodromy. It's impossible to define square root on all of C.
@JimTDF
Жыл бұрын
Why did we stop? application of a formula for the cosine of double angle shows that sin(pi/8) equals sqrt(2-sqrt(2))/2 ... which allows us to simplify the entire answer to sqrt( pi (sqrt(2) - 1) / 2) ; that final formula does not use any trig functions (sin,cos,etc). Just a thought :)
@arioriabdulrafiu8773
Жыл бұрын
You are mad man indeed ... You mad a great Difference. So clever...❤❤❤❤❤
@EmpyreanLightASMR
Жыл бұрын
Been listening to the Feynman audiobook ("Surely...") and Feynman was a PLAYA wowwww. Dude got around! And then he talks about this, so I had to look it up. I've only taken Calc 1, so this is way beyond me but fun to watch. I'll have to watch more videos to understand it better.
@AbouTaim-Lille
10 ай бұрын
We used to study similar integrals using the residue theory in the complex field and the polar coordinates.
@peterzinya1
Жыл бұрын
The derivative of x squared is 2X
@nicolasgomezgimenez212
Жыл бұрын
Thanks you , greeting from Argentina.
@arctic_haze
9 ай бұрын
Wow. This technique is amazing. Maybe not even among the top 10 achievements of Richard Feynman but still fantastic!
@svetlanapodkolzina1081
Жыл бұрын
Nice video!
@gevodem
Жыл бұрын
What a beautiful integral! You might also be able to solve this same integral using residues/contour integration.
@probro9898
Жыл бұрын
I understood it but it still made my head spin!
@facurod1392
Жыл бұрын
I just want to know which drawing tablet do you use for mathematics and which app (on Android Tablet I suppose) ?. Thank you very much. And great content!
@davidbakker1170
Жыл бұрын
Once upon a time I would have been able to reproduce this. Now I am just watching and thinking wow.
@shibammanna7706
Жыл бұрын
This technique is elegant but can it be solved using complex integration involving cauchy residue theorem?
@larrymorley2579
Жыл бұрын
And a lot more easily
@michaelbaum6796
Жыл бұрын
Very awesome technique, I love it - great👌
@michaelmello42
9 ай бұрын
Wonderful!
@haydenbritt1915
Жыл бұрын
I love Feynman Integration! Why isn't this taught in undergraduate?
@yassinetiaret505
Жыл бұрын
because it's hard to predict what the parameter a is, and where you should put it? That's the Satan's level mate !
@akagami3
Жыл бұрын
@@yassinetiaret505 so you are saying it's too hard to be taught for college students 🙃
@sirius1255-
Жыл бұрын
it is in my program
@DLites151
Жыл бұрын
It is in upper level Physics classes
@roadchewerpe5759
Жыл бұрын
I think some of the math involved in this problem isn’t undergraduate level math, unless you’re a math major. For example, I don’t know much about a lot of the things he did with the imaginary numbers except from an identity we used in differential equations.
@beautyofmath6821
9 ай бұрын
beautiful
@albertolemosduran5685
Жыл бұрын
The reason why you can introduce the derivative into the integral is because the integration limits aren’t functions (Leibniz theorem)
@aritrakundu1464
8 ай бұрын
Yess....precisely
@maths_505
8 ай бұрын
Not exactly You have to make sure the integral function converges. For that you can apply tests like Dirichlet's test or just look at a graph.
@atmanh8372
4 ай бұрын
3 months ago I understood none of these.Now I finally understand it
@maths_505
4 ай бұрын
Hell yeah 🔥
@renerpho
Жыл бұрын
Why stop there? If you evaluate sin(pi/8) further, you can write the result as sqrt(pi*(sqrt(2)-1)/2), which I think is quite nice.
@maths_505
Жыл бұрын
I liked the sine term at the end but yeah the radicals are quite nice too
@tolkienfan1972
Жыл бұрын
Radical!
@doctorb9264
Жыл бұрын
Me too.
@mus3equal
6 ай бұрын
Great video, primers are so much better than triggers
@TruthOfZ0
Жыл бұрын
I feel that im evolving after watching this!!
@Darakkis
10 ай бұрын
Awesome!
@Outcast_Raj
Жыл бұрын
can u make a video about the feynman technique itself ?
@stevenwilson5556
Жыл бұрын
Fascinating technique, are there applications of this integral?
@BederikStorm
9 ай бұрын
The square root in complex numbers has two solutions. You also have e^7pi/8 as solution
@teqnify63
Жыл бұрын
My AP calculus BC brain has expanded… glad i’m pursuing a stem major 😃
@JgHaverty
10 ай бұрын
Eh this is pretty entry level stuff on tbe grand scheme of things. If you really want to "expand your brain", go noodle around feynman diagrams; with regards to path integrals and quantization 😅. If you REALLLY wanna see where this rabbit hole can go, then go over neutron transport while youre there 😂 Recommend calming the hubris of your AP calculus class. The reality is if youre pursuing a degree in engineering, physics, or whatnot; your best interest is actually not using AP credits for anything other than humanities. Encumbent on what programs you narrow down and get accepted to of course [if your program only requires calc 1, then yes of course use your ap credit in that capacity]. Its a good path to be on; just take it in stride. That said, AP credits are kind of useless beyond gpa padding and i dont understand why highschools put so much weight on them in the first place..
@andrevadyaskin164
Жыл бұрын
Intégration by paramètre it is really powerful method. But it isn't Feynman's method.
@usernameisamyth
Жыл бұрын
amazing
@NimrodTargaryen
9 ай бұрын
Wonderful 🎉❤
@user-nw7qn8nm1t
Жыл бұрын
Please tell me why we take just real part in 3:43. I see that we need just cos but I do not undersfand how can we ingore sin part of Eular formula.
@matthiasehrhardt3469
Жыл бұрын
At 5:00. This integral can be determined easily by switching to a 2D integral in polar coordinates. No need to use formulas from books.
@mopcku77
10 ай бұрын
Nice video. What application and writing device(pen) are you using to write so nicely math?
@kenfrank2730
10 ай бұрын
I would like to know also.
@Bill0102
5 ай бұрын
This is sheer brilliance. I found something with a similar message, and it was beyond words. "The Art of Meaningful Relationships in the 21st Century" by Leo Flint
@illumexhisoka6181
Жыл бұрын
I have a great integral as an idea for a video The integral from 0 to ∞ of e^(A(x^B)) Where A and B are any complex numbers except the values of divergencey and to find what are they
@vidaripollen
Жыл бұрын
How u define time?
@wuhaochina
Жыл бұрын
10:24, I think we have two cases: -π/8 or 7π/8. But for case 7π/8, we can find that the final result of the intergration is negative which is impossible.
@user-sq8go3dg5n
Жыл бұрын
Why impossible? The function is sometimes positive and sometimes negative
@CeRz
Жыл бұрын
to the guy above me, no, the integer is a positive series, and can never be negative because of 0 to the positive infinity.
@jmcsquared18
28 күн бұрын
The one thing I dislike about the Feynman trick in everyday situations is that it's ad hoc. You need some level of foresight coupled with sufficient freetime, or just some serious courage, to use it in an actual scenario where you're trying to compute a new integral for the first time. For instance, if you put the parameter in the exponential, would it still work? In this case, it appears so based on the chain rule, but in a different situation, it might not be so clear. Or, how should the parameter be introduced? Can you tell ahead of time where it should go? I've used it on several insane integrals, it should be in everyone's toolbelt. But best method ever? I'd content it has a nice but isn't always the most useful thing to do. Cauchy and regularization could both be argued to be just as useful in many practical situations.
@pgress1867
Жыл бұрын
@maths_505 what is the name of the software you are using to write the stuff down?
@maths_505
Жыл бұрын
It's my galaxy note phone... Nothing else
@pgress1867
Жыл бұрын
@@maths_505 thanks
@JanPBtest
Жыл бұрын
Why is it called Feynman's technique? This is standard classic calculus fact, usually called simply the Leibniz-Newton formula (differentiation under the integral sign, it's used all the time in complex analysis). Weird.
@robj144
Жыл бұрын
He's talking about the entire technique. That's one part of the technique.
@frenchimp
Жыл бұрын
@@robj144 Nonsense. This "technique" was in use well before Feynmann's days.
@ushakiran8549
Жыл бұрын
It would be easy for me to love mathematics if my teachers were like you!
@kingbeauregard
Жыл бұрын
I'm never comfortable with just discarding the "i*sinx" part, especially when the cosine can be defined as (e^(ix) + e^(-ix))/2, no discarding of terms required. But the math would proceed much the same either way.
@maalikserebryakov
Жыл бұрын
Discarding makes it simpler Integral calculus is already difficult do not invent new obstacles for yourself :)
@CeRz
Жыл бұрын
with complex numbers this is totally okay because they have a real part and an imaginary part. If we're looking for the real part then there is a 0% probability to make any mistakes by leaving out the complex part in instances like this. You can obviously still make calculus errors etc. but that wasn't the issue here.
@kingbeauregard
Жыл бұрын
@@CeRz I guess I'm good with dropping the imaginary part at the very last step, but not before that.
@CeRz
Жыл бұрын
@@kingbeauregard and that is totally fine. However, if you ever change your mind for optimal efficiency you're still aware that it is possible to execute it like this aswell. To each their own. Good day.
@ivarorno
11 ай бұрын
Around minute 10, you can just use the fact that 1-i has angle -π/4 so the square root has half that, and multiplying by i rotates it by π/2 meaning that the new real part(cosine) is the old imaginary part(sine). Just seems slightly easier and more intuitive than the algebraic argument.
@georgemaclaurin3705
10 ай бұрын
Instead of -pi/4 i used 2pi-pi4=7pi/4 which is the same but got different answer. 😢
@nicolasgonzalez8482
Жыл бұрын
What is the ñame of the program tiene récord the video From your mobile ?
@user-hi8vb8rg5s
Жыл бұрын
Good job
@mauroariascontreras9284
29 күн бұрын
that passion about maths =) I could feel it
@choppa_the_cut
Жыл бұрын
shouldn't the -i be in the numerator after you solved int I'(a) da by substitution, hence providing the neg solution to that integral? sry if i am wrong, it has been some time...
@ashishraje5712
6 ай бұрын
Brilliant
@MSloCvideos
11 ай бұрын
Calling it Feynman's technique makes it appear as though it took centuries to develop it, when in reality this is also known as Leibniz's rule after one of the creators of integral calculus, so it was actually known pretty much since integration became a thing.
@csharpmusic9866
9 ай бұрын
Hey, just to add to your knowledge the lebinitz rule basically deals with differentiating a function under integration, whereas Feynman's techinque is a way to find definite integrals of non integrable functions by introduction of a parameter while 'using' the lebinitz rule as a smart tool and hence " lebinitz rule is different from Feynman's techinque, one helps the other."
@epikherolol8189
5 ай бұрын
Nah Leibnitz rule is different.
@user-oh2kt8lf6g
Жыл бұрын
sin(pi/8) is easy to calculate: sqrt((sqrt(2)-1)/sqrt(2))/sqrt(2). Hence, we can simplify the result: I = sqrt(pi/2) * sqrt(sqrt(2)-1)
@yuyo1948
Жыл бұрын
¡¡¡Brillante!!!
@suvosengupta4657
Жыл бұрын
feynman --->always😎💯
@ricardoruiz3248
10 ай бұрын
I'd like to ask what's the device you record on? 👀
@MathswithHiteshsir
10 ай бұрын
Which app you use for writing please tell me
@morgengabe1
9 ай бұрын
Honestly, using Re on euler's theorem that way is more impressive than feynman's technique, imo. That's precisely the sort of chicanery that i started to love these subjects for! edit: first time I saw that integral was statistical mechanics and the professor just gave the formula without proof or derivation. In numerical methods we got to see montecarlo integration, and that's probably my favourite integration method. Didn't see any of this in complex variables, which I went on to fail.
@thomaskeating7539
Жыл бұрын
I like the pace, you don't go at a snail's pace like some others. Great job!
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