when Feynman's integration trick doesn't work...
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Yep! Feynman credited Fredrick Wood's text book "Advanced Calculus" for where he learned this, and, sure enough, Woods shows the function must be continuous to switch differentiation and integration.
"the function must be continuous to switch differentiation and integration."
Much simpler to understand than trying to explain with the solution of an integral with some obscure trick like the video at 1:00
@physnoct that trick is the feynman technique and to use it the function has to be continuous
I love Feyman too! but please don't say this is his integration rule. He only popularized it among engineers. This rule was well known by mathematicians since Leibniz time.
In FPS's other players quickly made grenade launchers "mine" because I just had soooo much fun with them. I'd like to think it's more like: "oh, here we go, Feynman's excited about this again" and -- as regularly happens -- context is not carried forward because if you know, it's not necessary.
I believe technically speaking Leibniz's integral rule just states that you can switch integration and a derivative, whereas Feynman's technique specifically refers to introducing a new variable into an integral, taking a derivative using Leibniz's rule, forming a differential equation, and solving for the original integral.
I'm sure this was done before Feynman, but it's not like 17th century math has a great history of crediting people anyways...
IMO it's fine to use an already well known name for it to make a distinction between the rule and the trick.
A quick Google search will show that "Feynman's Technique" or "Feynman's Trick" (add "for integration") is a common enough name that it's totally standard at this point. There are a bunch of misnomers in language (As Vsauce mentions in his "Misnomers" video, Singapore means "city of lions" although there have never been lions in Singapore). Unfortunate, maybe, but we're stuck with them anyway.
I agree: this isn't Feynman's technique. Feynman invented Feynman diagrams and discovered many interesting things in physics, but he didn't come up with this one! In fact, Feynman credits a 1920's calculus text, Fredrick Wood's "Advanced Calculus." The text was being used at RPI in the late 50's to teach calculus, by my dad, and I have his copy.
Also, if you look closely at many table of integrals, even those preceeding Feynman, or from the Soviet Union, you'll notice many adjacent integrals were obviously solved using this technique, as well as integrating under the integral, to find the solutions to the "child" integrals once the "parent" had been solved. (Okay, you mathematicians don't believe in tables of integrals, I get that, but I was an engineer and not paid to solve integrals!) 🙂
In front of the integral it would be better to write d/dx rather than the partial derivative because we are indeed dealing with a standard derivative in x of a function of the variable x
Partial is identical to derivative in single variable situation, there is no harm in using partial, as everybody would understand it.
Great video ! Altough I've seen several videos using Feyman's trick to solve the first integral, I've never seen one that was correct, because there is an really important step you forgot to mention to solve the first integral. With this reasoning, which uses the dominated convergence theorem and is equivalent to derivate under the integral, you can't directly prove this formula for all s greater or equal to zero, but the formula is only true for s stricly bigger than zero, meaning you can't directly plug in 0 at the end (this is because you can only dominate the first function you want to derive under the integral for all s>0). But the formula still holds for all s>0. So the correct way to prove it is to show that the limit as s goes to zero of F(s) is indeed the integral of sin(x)/x from 0 to infinity, and you can use the fact that the right side of the equation is continuous at zero to give the final result. However, showing that you can interchange the limit and the integral as s goes to zero is not that trivial, since you can't dominate the function properly. To do that, you first need to integrate by parts and only after that you can dominate the function properly and do an interchange of limits and integral that is valid, and get the final result.
But doesn't the dominated convergence theorem only hold for lebesque integrable functions? Sin(y)/y is classic example of being riemann integrable on (0,infty) but not lebesque integrable
The Leibniz rule for Riemann integration and for Lebesque integration are not strictly the same. You can show that there are conditions for swapping the integral and derivative for both definitions. For Riemann integration you do still need to show you can swap the limit and the integral . Technically that's a special case of the dominated convergence theorem, but practically for Riemann integration you just need uniform convergence.
IIRC when you build the Leibniz rule for Riemann integration, if you have the partial derivative continuous on the rectangle, you also have uniform convergence.
But also this gives that the Leibniz rule for Lebesque integration is more relaxed than for Riemann integration since you're not dependent on the special case of the dominated convergence theorem. I cbf to go through it but I suspect you'll find that your issue stems from sin(x)/x not being Lebesque integrable in the first place.
I always thought of Feynman's technique as introducing the extra variable so you have something to switch with Leibnitz's Rule, rather than any other application of that rule.
yeah it should be that way. feynan saw this proof in a textbook and "just" applied it.
At 6:22:
Professor Pen: - Here we have our "DU" ear muffs.
Me: - I understood that reference!
Hi Dr. Penn!
I've never seen the integral of sin(x)/x evaluated in quite that way, at least not with those specifics.
gotta stand out from all the other videos he mentioned somehow
Could you make a video about the validity of changing either integration or differentiation with the summation sign!
In complex analysis, functions need to be holomorphic in the defined region to use certain techniques. But as a result, they're well-behaved and infinitely differentiable.
3:26 Is it not a problem that in order to have exp(y(-x+i)) approached zero when y approaches inf ,the x value must be strictly positive (if x =0 , the value of the limit will be undefined circling around origin of the complex plane)? Because eventually we're interested in x =0.
Technically, you can only differentiate while x > 0, and then take the limit as x -> 0. What needs to be done for a rigorous proof is showing that F(x) is continuous at 0 from the right
do we not have an indeterminate form when x is set to 0, so that a more detailed analysis is needed to show that the numerator equals 0?@@skit_inventor
@@barryzeeberg3672when we set x = 0, we get the original integral. I don't seem to get your question 🤔
f is interesting. In polar coordinates (for r not 0), f is independent of θ. Indeed, f(r,θ) = cos^3(θ) sin(θ)
At 9:25, importantly that is not a continuous function either just like the original f. Along the path x = y/2, it blows up to infinity and along x=y it equals 0. It shouldn't be surprising at all that a non-continuous function is also not differentiable and this is the concrete reason the trick does not work in this case.
Both f AND ∂f_x need to be continuous on a square for the theorem to hold, but f isn't required to be differentiable. The second condition doesn't imply the first.
@@shouligatv I am having trouble coming up with an example where f and ∂f_x are continuous, but f isn't differentiable. Can you give me an example?
Nevermind: I came up with a function. f(x,y) = |y| is obviously continuous and ∂f_x = 0, but ∂f_y does not exist at y=0.
Maybe you wanted to say: "...along x = 0 it equals 0" because it's false that along x=y it equals 0.
@@MarcoMate87 I missed the 3, which means I meant x=y*sqrt(3), and I didn't need x=y/2.
You can that integral via Feynman, but you need to express your integrals in terms of the correct distributions
3:25 as y approaches infinity e^ky will approach infinity not zero
Feynman's mathematics was a bag of tricks to solve problems.
I can't tell if this is a compliment or a critique XD
Feynman got the integration trick from the book "Advanced Calculus" by Woods.
Very nice and well explained. Thank you!
This happens because the integral has to be linear in the derivative variable to Leibniz Regel to appy
I need some help to understand.
At 1:00, he's adding an exponential in the integral. Is there some book or some video somewhere that explain this trick with more details and formalism?
Another video which use a similar trick:
ruclips.net/video/ZZccxuOpb4k/видео.html
Thank you, professor.
I dont see why exp(iy - xy) -> 0 as y -> inf
|exp(iy-xy)|=|exp(iy)|*|exp(-xy)|=exp(-xy), which approaches zero as y approaches infinity for positive x
the exp(iy) rotates around the unit circle in the complex plane as y->inf, whereas exp(-xy) goes to 0 for x>0. You can visualize it as a inward moving spiral in the complex plane, which starts on the unit circle and then collapses to the origin, like a sattelite moving from its orbit and falling to earth.
Exp(iy-xy)=exp(iy)/exp(xy)=(cos(y)+i*sin(y))/exp(xy). When y approaches infinity, denominator approaches infinity for any x>0 and numerator is any value on the circle on the complex plane with radius =1.
@@thatdude_93Is it not a problem that in order to have exp(y(-x+i)) approached zero when y approaches inf ,the x value must be strictly positive (if x =0 , the value of the limit will be undefined circling around origin of the complex plane)? Because eventually we're interested in x =0.
@@IoT_ x will never truly be set to zero. It will only be limited to zero.
Michael always chooses interesting content.
its cute but most, even engineers will simply choose to use transform methods.
Funny enough, so far no one noticed that Leibniz's Rule, as it is shown in the video, can be only applied when the domain of integration is compact. So we need some extension of this rule if we want to compute integrals on unlimited domains of integration. Probably, what is really involved is some sort of application of Lebesgue's theorem of the dominated convergence.
Is there a specific restriction to be imposed on ( not only f) the partial derivative , both ( fx and fy)continuous functions?
I haven't learnt Complex Analysis, can anyone tell me about why taking the imaginary part to the front works? Just some intuition would be enough
e^iy = cos(y) + i*sin(y)
So sin(y) = im [e^iy]
This is the first time I've seen this method, came to the comments for a better explanation 3:05
Statistics problem
Mrs. Smith lives in a rural area. Each day she brings produce into town at 4 a.m. There's one traffic light in town with red or green every 30 seconds. She turns a corner that reveals the traffic light within 30 seconds of travel at the posted speed limit. She wishes to calculate a strategy that will maximize her average velocity going through the light over time. There's almost never any other traffic at that time of the morning, so any strategy can be used.
I think Mrs. Smith Should settle in urban area
The best strategy would be go to travel at c
@@TheEternalVortex42 Mrs. Smith wouldn't dream of breaking the speed limit.
@@TheEternalVortex42 If you didn't mean the speed of light :), going too fast will require stopping and result in a lower long term average velocity. Going a constant velocity seems intuitive, but if so, which velocity?
Without some bounds on acceleration, she should always proceed at the speed limit, and only brake at the last moment for a red light. She can instantaneously resume at the speed limit as soon as it changes back. Her average progress (time to pass the light) will be greatest.
But as you are specifically concerned with her instantaneous velocity at the light, I assume you define any situation where she hits the red light to be a "zero". In which case upon turning the corner and seeing a red, she can always be sure to stay at the speed limit and hit a green by the time she hits the intersection. Upon turning and seeing a green, she can be sure she will hit a red light (I am of course ignoring the insignificant case of turning precisely on a change). So, she can stop any time, and resume with instant acceleration to the speed limit as soon as she see green. She will be sure to pass the light at the speed limit.
I think this only gets interesting if you impose bounds on the acceleration. Then we have the same strategy for red, but must plot the best decelerate, coast, accelerate trajectory for green. Which is fun and probabilistic. Is this what you had in mind?
3:57 how is the numerator 0 when you have e^i*inf?
@3:22 why does y approaching infinity make the numerator 0?
Exp(iy-xy)=exp(iy)/exp(xy)=(cos(y)+i*sin(y))/exp(xy). When y approaches infinity, denominator approaches infinity for any x>0 and numerator is any value on the circle on the complex plane with radius =1.
@@IoT_very nice, I think the simplification of yi - xy to y(-x+i) threw me off
@@IoT_ yes I understand that but why are we assuming x>0
Especially when it will end up being 0 in the end anyway
@@Happy_Abe there is a very nice comment explaining this stuff. But yeah, first you have to prove that such limit exists (x->0) which is not trivial thing to do.
@@IoT_ yeah so that’s the part that confuses me, I understood the rest just not the assuming x>0 and the limit holding
Will this work if f(x,y) is defined = 1/2 at (0,0)?
No. Consider the limit when y=0 is 0, and the limit when x=0 is positive or negative infinity. Also consider neither of the earlier derivations actually used the special value at zero.
No, because you can take the limit as (x,y)->(0,0) from a different direction (e.g., along the line y=-x) and get a different value (e.g. -1/2).
Does it always work when f is continuous
This method is NOT called "Feynman Integration" , IT'S CALLED *Leibniz Integral Rule* .
Gottfried Leibniz DISCOVERED THE RULE,
Feynman POPULARISED IT.
THIS IS Leibniz's technique, NOT FEYNMAN'S.
GIVE THE CREDIT TO THE RIGHT PERSON FOR GOODNESS SAKE 😡😡
anyone know if there is a connection between the schwarz lemma and lie algebras?
Yes, there is a connection between Lie algebras and the Schwarz lema. Lie algebras are mathematical structures studied in the theory of groups and algebras, which have applications in various Fields, including physics and geometry. Schwarz' s lemma is an important result in the theory of partial differential equations and concerns the symmetry of the order exchange of partial derivatives in an equation. This connection between Lie algebras and Schwarz's lemma arises in the context of harmonic analysis and the Fourier transform of partial differential equation.
Awesome, again!
What's the difference between the starting equation and expressing the fundamental theorem of analysis as a definite integral of Newton's heat equation?
I suspect the implications could be relevant to the recent, more entropy oriented formulations of mechanics; often leaning on Noether current conservations.
Nice video :)
UConn represent!
Feynman stole the method from an unknown person called leibniz and just say he read it somewhere. Everything is leibniz here
He didn't "steal" it, he just used it. Like every one else does. It's just that it has kind of been popularised in his name since he used it so much and also he's somewhat responsible for exposing a lot physicists and engineers.
"Unknown"???
"unknown" ikr ! Who's that random loser anyway ? ...
@@Noam_.Menashe I think it was an attempt at sarcasm
@@allanjmcpherson leibniz with a lowercase l is indeed unknown. Leibniz with a capital L is better known, but (somewhat unfortunately) some guy called Leibnitz with a t seems to be more famous, especially among English speakers.