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.
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 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.
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.
@@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.
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.
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.
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!
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.
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 . . . .
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.
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.
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...
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
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!… 😂
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.
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.
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 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.
@@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.
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 :)
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.
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."
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 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.
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
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!
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.
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
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.
At <a href="#" class="seekto" data-time="300">5:00</a>. This integral can be determined easily by switching to a 2D integral in polar coordinates. No need to use formulas from books.
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...
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..
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.
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 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.
Please tell me why we take just real part in <a href="#" class="seekto" data-time="223">3:43</a>. I see that we need just cos but I do not undersfand how can we ingore sin part of Eular formula.
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.
There are two points at which the technique used here needs further explanation: where the derivative of the integral becomes the integral of the derivative of the integrand, and the reason given is because the integrand is clearly bounded; the more crucial point is where part of the integrand is replaced by the real part of a complex term, and it is then assumed that integrating the integrand with the full complex term and then, when the integration is done, taking the real part, so discarding the imaginary part, is an equivalent result to integrating without the complex term replacement - that is quite an assumption since throughout the subsequent manipulations of the complex terms some real terms become imaginary and some imaginary terms become real, so some imaginary terms contribute to the real result, but the technique seems to rely on the imaginary part of the original complex replacement having no effect on the real part.
<a href="#" class="seekto" data-time="624">10:24</a>, 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.
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.
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.
Just assume (sin x^2)/x^2 = 1 And then integrate just the exp(-x^2). (sin x^2)/x^2 is actually less than 1, in general. But the value of exp(-x^2) will sharply fall and go to zero, before the value of (sin x^2)/x^2 moves away from one.
