Feynman's technique is the greatest integration method of all time
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- čas přidán 20. 03. 2023
- 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.
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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.
As a teenage self-proclaimed math goblin / Feynman acolyte, I concur.
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.
I feel so jealous of them 😁
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.
@@slavinojunepri7648 where did you grow up?
The more I watch feynmann integration technique videos, the more powerful I become.
Same!!
@@azizbekurmonov6278 азизбек.не русскоговорящий ты случайно?
@@Dagestanidude Da ya panimayu
Lol
Xp farming on this video
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.
No wonder they use a math sign language. What a ride!
Hero
My summary:
Find someone better at math than me and ask them for help. Maybe I'll find this guy's email somewhere...
We makin it outa Cornell wit dis one😎
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.
Ditto. Learned how, then never had to use them again. Today, fugetaboutit!
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
@@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.
@@kwgm8578 absolutely ... Will do asap
@@LetsbeHonest97 Good luck to you!
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.
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.
Surely you're joking, Mr Feynman... ;)
JgHaverty, spoken like a true ignoramus.
@@TheSireverard, and also "What Do You Care What Other People Think?"
@jamesedwards6173 what the hell are you talking about? Hahaha
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.
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!
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
Love how you talk about mathematics with passion while solving :)
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.
This is advanced
epic , thank you for making this technique so clear
Been waiting for an explanation of my favorite’s, Feynman, noble prize topic.
Beautifully done video!
Wow yes this is so intuitive and elegant and beautiful and I totally followed you the whole way along
Thanks so much 😊
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.
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 . . . .
Absolutely beautiful. Thank you for sharing!!
You're doing really good content. Please, moreeeeee Feynman Integrals!!
This was amazing, really gotta use it instead of by parts. Thanks a lot !
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.
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
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.
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
@@patrick-kees8962 I believe it would still work if you consider the Imaginary part of the integral instead of the Real part
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.
Once upon a time I would have been able to reproduce this. Now I am just watching and thinking wow.
We used to study similar integrals using the residue theory in the complex field and the polar coordinates.
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.
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!
Wow. This technique is amazing. Maybe not even among the top 10 achievements of Richard Feynman but still fantastic!
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!… 😂
It's crazy
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 :)
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.
Very nice presentation.
I came up with this myself in college. I hadn't known until now that this Feynman guy stole it.
😂😂😂
I completely believe you
What a beautiful integral! You might also be able to solve this same integral using residues/contour integration.
Very cool! Thanks for sharing.
This makes me want to learn complex analysis. Great video considering I still understood most of it
Amazing content!
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...
Absolutely. I feel exactly the same!
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...
nice demonstration 👍
Why choose to throw alpha into the sine function as opposed to the x^{2} in the denominator or the exponent exp{-x^{2}} in the numerator?
very nice effort. good luck
Great video. Thank you
Brilliant! Thank you.
Beautiful solution
What is the ñame of the program tiene récord the video From your mobile ?
Thank you Sir for your best explanation and working out of the problem🥰😍🤩
Thank you for the nice comment
Just infinitely beautiful!
SUIIIIIIIIIIIIIIII
Where did the pi under the first radical come from in the last line? Shouldn't it just be root 2 of root 2 multiplied by sin pi/8?
Beautiful!
The derivative of x squared is 2X
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.
Which app you use for writing please tell me
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.
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."
Nah Leibnitz rule is different.
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
Can't wait to learn all this it seems interesting enough 🙂
Fascinating technique, are there applications of this integral?
I love this video!!
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
He covered that in the video, albeit somewhat handwavingly.
@@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.
@@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.
Mathematicians will worry about that, physicists not so much.
@@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.
Can feynman's methhod be used for all integrals? If not, what are the restrictions please?
I'd like to ask what's the device you record on? 👀
Great video, primers are so much better than triggers
I understood it but it still made my head spin!
You are mad man indeed ... You mad a great Difference. So clever...❤❤❤❤❤
3 months ago I understood none of these.Now I finally understand it
Hell yeah 🔥
Which software are you using for drawing?
can u make a video about the feynman technique itself ?
How come e^-x2 differentiates to (e^-x2)/x^2 shouldn’t it be (e^-x2)/-x^2 ? In other words where does the negative go
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.
Thanks you , greeting from Argentina.
Can this be done using laplace transform ?
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
quick stupid question from me: does power series work?
beautiful
Intégration by paramètre it is really powerful method.
But it isn't Feynman's method.
Nice video!
Wonderful!
How do we know that differentiating with respect to "a" wont change the value of the integral?
Nice video. What application and writing device(pen) are you using to write so nicely math?
I would like to know also.
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)
What software are you using Man?
Wouldn’t the constant of integration be part of the argument of Re() because the integral da was within Re()?
This is basically a special case of Leibniz rule
It would be easy for me to love mathematics if my teachers were like you!
I like the pace, you don't go at a snail's pace like some others. Great job!
I feel that im evolving after watching this!!
An integral of a complex function equates to a real number.
that passion about maths =) I could feel it
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.
I liked the sine term at the end but yeah the radicals are quite nice too
Radical!
Me too.
Awesome!
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.
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.
Instead of -pi/4 i used 2pi-pi4=7pi/4 which is the same but got different answer. 😢
There is no reason for this thumbnail to go so hard
Very awesome technique, I love it - great👌
The square root in complex numbers has two solutions. You also have e^7pi/8 as solution
What are the constrains on fx when using Feynmans method - you mentioned convergence and considered if the function was increasing or not
Also does it only work with fx with bounds 0-infinity?
How to know when to apply feynman technique?
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.
It is a minor omission, but you are right
@@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.
This technique is elegant but can it be solved using complex integration involving cauchy residue theorem?
And a lot more easily
Good job