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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.
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.
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.
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.
@@Inndjkaawed2922-- 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.
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.
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.
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 :)
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!… 😂
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 . . . .
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
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.
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.
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.
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..
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.
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...
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.
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.
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
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.
2:00 shouldn't we need to precise that the expression tends to 1 when x near 0 so the integral is defined for every x and then swtich the integral and the derivative by a
That's pretty trivial so I never felt the need to mention that. Anyone with even a basic understanding of calculus can see that. But I like that you're paying so much attention to detail.
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.
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.
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.
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.
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 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.
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.
Wonderful solution! Can someone explain me why is a=1 our target case? I feel like I missed something. I understood that a=0 is essential to find the constant of integration but couldn't understand why a=1 is our target case? Thank you for your explanation! :)
@@kushal_2oo4 Thanks a lot man! The problem discussed so much about having'a' in it that I totally forgot that the original problem didn't have 'a' in it. My bad! Thanks a lot :)
@@yashsethi1110 any question is always a good question. Now that you've come to a realization of your original problem, the solution is now engrained in the structure of your brain and you will most likely never forget it. This is the invaluable part of realizing things you never understood before.
Well, I'm a CS student so I do not have this level at Calculus, so I lack all the theory that supports this integration method behind, so I really find it odd. I manage to understand just the basics and since I haven't studied formal Math since january I'm a little rusty, and we didn't get to see complex integration. But I do love Math, so some day I'll be able to fully understand this. However, I'm currently studying basic Electromagnetic Physics right now and I've discovered an integration method quite elegant if you ask me. It's pretty basic, but I came across it just now. Specifically it's the trigonometric substitution method, when trying to reach an expression to find the magnitude of the magnetic field created by a large, straight conductor in which current flows. I'm talking about the situation described at Sears-Zemansky vol.2 page 928. This conductor is parallel to the y axis, "y" is the distance between the center of the conductor and the coordinate axis, and "x" is the distance in the x axis between the center of the coordinate axis and the field point. 2a is the longitude of the conductor, so "a" will be the half. In the coordinate axis, the top half will be at +a and the bottom half at -a. ϴ is in the angle formed between the conductor and the position vector. So, the supplement angle is (π-ϴ), so sen(ϴ)=sen(π-ϴ)=(x)/(sqrt(x²+y²)) Given the fact that the Biot-Savart law for a current element determines that the magnitude of the magnetic field is: (µ0)/(4π)*((I*dl*sen(ϴ)/r²)), and given the fact that the distance between the field point and the source point of magnetic field is sqrt(x²+y²) according to Pythagoras' theorem, and that in this context the vector dl only has a component in the y asis, we can say that dl=dy; so the magnetic field is the definite integral from y=-a to y=a. Since µ0, and 4π are constant when integrating with respect to "y" and that (π-ϴ) is supplement to ϴ, the problem is reduced to find the antiderivate that when differentiate with respect to y equals to (x)/((x²+y²)^(3/2)). I love this integration method because it involves a lot of elementary trigonometric stuff. You just need to assume that a right triangle exist of a certain angle, where "y" is the opposite side and where "x" is the adjacent side. You could assume otherwise, where "x" is the opposite side and where "y" is the adjacent side and it should work too, but when I was resolving this I find the first to be easier. According to the first option, the tangent of the angle equals to y/x, so y = tg(angle) * x, so if you differenciate in the left of the equality to respect of y and in the right in respect to the angle, dy=x*sec²(angle)d(angle), this way, the indefinite integral transforms to: (µ0*I)/(4π)*∫[(x²*sec²(angle)/(x²*sec²(angle)^(3/2)))d(angle)], since x²+x²*tg²(angle) = x²*sec²(angle), this way we went from an integral with a "rather complicated" expression to a relative really easy expression just using elementary trigonometric stuff! I don't know, it just blows my mind that Math expressions can often get so simplified using stuff every person knows from school, and it's sometimes quite intuitive. I can't help but think that the world is a complex well-tight and connected machine that makes sense as a whole. Sometimes, a kid could ask why does he need to study trigonometrics or prime numbers, but it's hard to justify their usefulness and reason of existence without having a more advanced knowledge of the world and some abstract theory. I can't get enough of the beauty of linkage in Math, and how the nature seems to be written in "Math language", if that makes sense. I'm pretty sure this is the case too, I just find to be so ignorant of advanced/medium (or even just a bit more advanced basic) Calculus that I do not appreciate as much this beauty too, at least for now. I'm pretty sure Feynman or another physician can share this emotion somehow at some extent. I just wanted to share my thoughts, and the great opportunity and luck that I have to be alive and so ignorant that I will never ran out of things to understand things about Math, Physics or overrall Engineering. Well, maybe I won't need it as much as a Computer Scientist but I sure love Math and Physics as long as Computer Science related stuff.
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.
When you say that there are no problems with switching differentiation and integration around, it is because you apply either Lebesgue's dominated convergence theorem or his monotone convergence theorem?
At 3:00 when he does chain rule for derivative of sin(ax^2) how come the chain ruled term is ax^2, i was under the impression it would be 2ax, is this Feyman's technique? Just slightly confused.
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
2:02 Yeaaaaah... You might want to take a look at the conditions of the theorem of inversion derivative/integral (not sure about the name, I'm French...), cause that's far from all you have to verify.
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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😎
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.
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
@@Inndjkaawed2922-- 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
@@Inndjkaawed2922 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
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
epic , thank you for making this technique so clear
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
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 :)
Wow yes this is so intuitive and elegant and beautiful and I totally followed you the whole way along
Thanks so much 😊
This was amazing, really gotta use it instead of by parts. Thanks a lot !
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
This is the most satisfying and exciting solution development ive seen on the interent so far.
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 . . . .
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
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.
Great technique, i also tried solving it by parts and using the gamma function, that worked too!
Been waiting for an explanation of my favorite’s, Feynman, noble prize topic.
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.
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
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.
My AP calculus BC brain has expanded… glad i’m pursuing a stem major 😃
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..
Wow. This technique is amazing. Maybe not even among the top 10 achievements of Richard Feynman but still fantastic!
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.
I came up with this myself in college. I hadn't known until now that this Feynman guy stole it.
😂😂😂
I completely believe you
don't worry, it'll be the dennishwhite4446 technique for me now
You're doing really good content. Please, moreeeeee Feynman Integrals!!
Once upon a time I would have been able to reproduce this. Now I am just watching and thinking wow.
The derivative of x squared is 2X
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)
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!
Ugh. The only thing I know after watching this is that there is one more thing in the world that I don’t understand, shrinking my Rumsfeld unknowns.
What a beautiful integral! You might also be able to solve this same integral using residues/contour integration.
The square root in complex numbers has two solutions. You also have e^7pi/8 as solution
Absolutely beautiful. Thank you for sharing!!
We used to study similar integrals using the residue theory in the complex field and the polar coordinates.
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.
Why impossible? The function is sometimes positive and sometimes negative
to the guy above me, no, the integer is a positive series, and can never be negative because of 0 to the positive infinity.
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.
at 2:00, what would be a problem regarding convergence or boundedness which would forbid the switchup?
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
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.
3 months ago I understood none of these.Now I finally understand it
Hell yeah 🔥
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.
Great video, primers are so much better than triggers
2:00 shouldn't we need to precise that the expression tends to 1 when x near 0 so the integral is defined for every x and then swtich the integral and the derivative by a
That's pretty trivial so I never felt the need to mention that. Anyone with even a basic understanding of calculus can see that. But I like that you're paying so much attention to detail.
Very nice presentation.
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. 😢
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.
Discarding makes it simpler
Integral calculus is already difficult do not invent new obstacles for yourself :)
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.
@@CeRz I guess I'm good with dropping the imaginary part at the very last step, but not before that.
@@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.
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 technique is elegant but can it be solved using complex integration involving cauchy residue theorem?
And a lot more easily
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.
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.
This makes me want to learn complex analysis. Great video considering I still understood most of it
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.
The reason why you can introduce the derivative into the integral is because the integration limits aren’t functions (Leibniz theorem)
Yess....precisely
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.
I am reviewing calculus and differential equations and this discouraged me max. However I will never ever give up.
You are mad man indeed ... You mad a great Difference. So clever...❤❤❤❤❤
I love Feynman Integration! Why isn't this taught in undergraduate?
because it's hard to predict what the parameter a is, and where you should put it? That's the Satan's level mate !
@@yassinetiaret505 so you are saying it's too hard to be taught for college students 🙃
it is in my program
It is in upper level Physics classes
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.
Beautifully done video!
Thank you Sir for your best explanation and working out of the problem🥰😍🤩
Thank you for the nice comment
The most difficult part always lies in the first inspiration
Can't wait to learn all this it seems interesting enough 🙂
This is basically a special case of Leibniz rule
I understood it but it still made my head spin!
that passion about maths =) I could feel it
I have no idea what's going on
I feel that im evolving after watching this!!
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.
I just realized that Jimmy Fallon could probably play him in a documentary.
Just infinitely beautiful!
SUIIIIIIIIIIIIIIII
10:35
Wouldn't it be better if you also took the euler form of 'i' and multiply it and later extract to polar form?
Wonderful solution! Can someone explain me why is a=1 our target case? I feel like I missed something. I understood that a=0 is essential to find the constant of integration but couldn't understand why a=1 is our target case? Thank you for your explanation! :)
putting a=1 gives us the original integral we were supposed to find. sin(ax^2) becomes sin(x^2) again
@@kushal_2oo4 Thanks a lot man! The problem discussed so much about having'a' in it that I totally forgot that the original problem didn't have 'a' in it. My bad! Thanks a lot :)
@@yashsethi1110 any question is always a good question. Now that you've come to a realization of your original problem, the solution is now engrained in the structure of your brain and you will most likely never forget it. This is the invaluable part of realizing things you never understood before.
@@CeRz Thank you for your kind words. I really appreciate them. I definitely won't forget the solution anytime soon :)
Thanks you , greeting from Argentina.
Can u explain feynman technique
very nice effort. good luck
Please share the software details for presentation. I like the presentation.
It's the default software on my samsung note phone
i kinda dont understanded why we look a=1 after a=0 and why we didn't investigate what was going on in infinity
a = 0 gives the constant of integration
a = 1 gives the original integrand
An integral of a complex function equates to a real number.
Well, I'm a CS student so I do not have this level at Calculus, so I lack all the theory that supports this integration method behind, so I really find it odd. I manage to understand just the basics and since I haven't studied formal Math since january I'm a little rusty, and we didn't get to see complex integration. But I do love Math, so some day I'll be able to fully understand this.
However, I'm currently studying basic Electromagnetic Physics right now and I've discovered an integration method quite elegant if you ask me. It's pretty basic, but I came across it just now.
Specifically it's the trigonometric substitution method, when trying to reach an expression to find the magnitude of the magnetic field created by a large, straight conductor in which current flows. I'm talking about the situation described at Sears-Zemansky vol.2 page 928.
This conductor is parallel to the y axis, "y" is the distance between the center of the conductor and the coordinate axis, and "x" is the distance in the x axis between the center of the coordinate axis and the field point.
2a is the longitude of the conductor, so "a" will be the half. In the coordinate axis, the top half will be at +a and the bottom half at -a.
ϴ is in the angle formed between the conductor and the position vector. So, the supplement angle is (π-ϴ), so sen(ϴ)=sen(π-ϴ)=(x)/(sqrt(x²+y²))
Given the fact that the Biot-Savart law for a current element determines that the magnitude of the magnetic field is: (µ0)/(4π)*((I*dl*sen(ϴ)/r²)), and given the fact that the distance between the field point and the source point of magnetic field is sqrt(x²+y²) according to Pythagoras' theorem, and that in this context the vector dl only has a component in the y asis, we can say that dl=dy; so the magnetic field is the definite integral from y=-a to y=a.
Since µ0, and 4π are constant when integrating with respect to "y" and that (π-ϴ) is supplement to ϴ, the problem is reduced to find the antiderivate that when differentiate with respect to y equals to (x)/((x²+y²)^(3/2)).
I love this integration method because it involves a lot of elementary trigonometric stuff. You just need to assume that a right triangle exist of a certain angle, where "y" is the opposite side and where "x" is the adjacent side. You could assume otherwise, where "x" is the opposite side and where "y" is the adjacent side and it should work too, but when I was resolving this I find the first to be easier.
According to the first option, the tangent of the angle equals to y/x, so y = tg(angle) * x, so if you differenciate in the left of the equality to respect of y and in the right in respect to the angle, dy=x*sec²(angle)d(angle), this way, the indefinite integral transforms to:
(µ0*I)/(4π)*∫[(x²*sec²(angle)/(x²*sec²(angle)^(3/2)))d(angle)], since x²+x²*tg²(angle) = x²*sec²(angle), this way we went from an integral with a "rather complicated" expression to a relative really easy expression just using elementary trigonometric stuff!
I don't know, it just blows my mind that Math expressions can often get so simplified using stuff every person knows from school, and it's sometimes quite intuitive. I can't help but think that the world is a complex well-tight and connected machine that makes sense as a whole. Sometimes, a kid could ask why does he need to study trigonometrics or prime numbers, but it's hard to justify their usefulness and reason of existence without having a more advanced knowledge of the world and some abstract theory. I can't get enough of the beauty of linkage in Math, and how the nature seems to be written in "Math language", if that makes sense.
I'm pretty sure this is the case too, I just find to be so ignorant of advanced/medium (or even just a bit more advanced basic) Calculus that I do not appreciate as much this beauty too, at least for now. I'm pretty sure Feynman or another physician can share this emotion somehow at some extent.
I just wanted to share my thoughts, and the great opportunity and luck that I have to be alive and so ignorant that I will never ran out of things to understand things about Math, Physics or overrall Engineering. Well, maybe I won't need it as much as a Computer Scientist but I sure love Math and Physics as long as Computer Science related stuff.
Thanks for sharing, don't ever lose those feelings.
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.
How to know when to apply feynman technique?
When you say that there are no problems with switching differentiation and integration around, it is because you apply either Lebesgue's dominated convergence theorem or his monotone convergence theorem?
Why Am i loving watching integration videos
U r from India bro ?
You have been chosen.....
Just another example of why we should be having people take complex analysis
At 3:00 when he does chain rule for derivative of sin(ax^2) how come the chain ruled term is ax^2, i was under the impression it would be 2ax, is this Feyman's technique? Just slightly confused.
Can i solve any integral by this method or just the ones with limits as infinity especelly asking as a highschool student
I'm gonna upload one today that has finite limits.
At 6:32 you shouldn't think of i as the derivative of ai. It is really the coefficient of a in the context of the integration step..
Can it be done using Taylor series expansion? Expanding sin x² and e-x²
It would be easy for me to love mathematics if my teachers were like you!
What is the ñame of the program tiene récord the video From your mobile ?
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
Beautiful solution
How u define time?
2:02 Yeaaaaah... You might want to take a look at the conditions of the theorem of inversion derivative/integral (not sure about the name, I'm French...), cause that's far from all you have to verify.
Another beast of an integral laid to rest by the sword of Feynman!!! this really satisfy the problem
Which app you use for writing please tell me
What is the use in daily life?
When you set up the steps wrong and start trying to solve this on your calc test
yep, you got that right. Feynmann is the GOAT since Einstein