This is a cool example for those who know the technique already, but for someone who doesn't know it, I can't say I would recommend this video to learn it. For that, it would need to be longer, more motivated, and better explained
This explanation is... honestly a little too simplified. It might have been a good idea to go into a _lot_ more detail with more steps, more examples, and more reasoning for when this technique can and can't be done.
First of all F(x,t) must be definite in a set A measurable. Then in A F(.,t) must be integrable. And last F(x, .) must be derivable and the derivate and < g(x)
Thank you everybody for all of your support and constructive criticism! Honestly, I didn't expect the video to get so many views. I made it for an audience of 100 or so people who already had plenty of mathematical knowledge as more of a refresher (which is why the video is WAY TOO concise). My content from now on will certainly be more explanatory! :D I'll try and answer some commonly asked questions here: - Q: 'Why is k=7?' - A: 'I am answering the integral in the thumbnail and you should try k=2,3,4,... as it is just an arbitrary constant.' - Q: 'Why didn't you refer to Leibniz?' - A: 'Feynman popularised the Leibniz integral rule in physics (pilot-wave theory is my honours topic, so I'm biased). Since Leibniz generalises for non-constant limits of integration, if I ever want to fix this video in the future, I will probably do the more general Leibniz integral rule.' - Q: 'When can I swap the differential and integral signs? There is no rigour!' - A: 'To differentiate under the integral sign, the integrand must be continuously differentiable (luckily most functions in physics are analytic and so this isn't much of a worry). To prove this, some real analysis is involved which I'm happy to go through in the future.' - Q: Why didn't you take the indefinite integral instead of an integral from 0 to 7. Why haven't you mentioned +C? - A: The infamous +C has tripped many a mathematician up on their final exam (including me lol). To be honest, it was a shortcut. I(k)=ln(k+1)+C, but by observation, I(0)=0 (using the definition of I(k)) and therefore C=0 -> I(7)=ln(8). - Q: Why is a partial derivative of the LHS? - A: As a matter of convenience, since both notations are equivalent for a one-variable function (Wolfram does it at mathworld.wolfram.com/LeibnizIntegralRule.html but some other sources will use d/dk for the left hand side and partial derivatives on the right for the same equation). I prefer using partial derivatives because it highlights that we're dealing with a multivariable function f(x,k) (after swapping the integral). Anyways, if you are at the end of this, I hope this helps.
I thought your video was fine, man. Some people just get real analytical and can't Leibniz alone. Most of their complaints aren't integral to your pedagogy. Clairaut your mind, and don't let them test your limits or derive you crazy.
A bit more explanation and more examples would be nice. I mean, I know the trick/technique and just clicked out of curiosity, but yeah just a bit more in general would be helpful. Like if your channel is more for beginners, show the motivation, (which is simple, a lot of integrals are nasty buggers. Especially ones with denominators. And Feynam's technique is often godsent in taking those out) and a word of warning that you're not actually always allowed to switch the integral operator and the differential operator would be nice.
@@shmerox7683 in a very simplified explanation he looked for a way to eliminate the ln(x) in the denominator without changing the actual integral. So in the first step he defined a new function in terms of that integral and named it I and a new variable k and made the new function depend on k. He turned the original x^7+1/lnx into x^k+1/lnx. So you can see, I(7) is the original integral. What he did next is the actual Feynman technique. He took the derivative with respect to k! You would normally have to prove you can switch the integral sign and the derivative operator, but Feynman was a physicist and physicists don't care about details like that. Everything that has to do with x is now constant. Meaning he only has to take the derivative of x^k. Remember x is a constant so the derivative structure is identical to let's say 2^k. That would be ln(2)*2^k So the derivative of x^k is lnx*x^k which is why we did the shenanigans in the first place. We now have integral from 0 to 1 lnx/lnx x^k+1 dx so the LNs vanish. You know evaluate that integral You should get 1/k+1 as the result And now comes the tricky part. Remember this is I'=1/k+1 we need I(7) How do we get there? Well the fundamental theorem of calculus states F(x)=int from 0 t x f(t) dt Let x=7, t=k and f=I' and we just integrate 1/k+1 from 0 to 7 and you get the result ln8. Hopefully this unreadable textwall helped a little with whats going on? The actual Feynman trick is just introducing a second parameter to eliminate a difficult part of your integral with a derivative with respect to your second parameter. Its a powerful tool that can solve many nonelementary integrals like this rather easily. Imho it's so useful it SHOULD be taught in calc 2 courses
@@Metalhammer1993 Great explanation. Do you have any reference where someone could check the conditions under which swapping integral and derivative operators is allowed? I'd say Tom Apostol Mathematical Analysis could do the job but perhaps you know a more enlightening source.
You need to mention the fact there's a +C . This is only the answer because I(0) = 0 so theres no added constant. This is a very important detail that you missed out
Well pointed out. From the 1/(k+1) line you can either take the indefinite integral (which is ln|k+1|+C), realise C is 0, and then substitute for k=7, or if by inspection C is clearly 0 then the definite integral from 0 to 7 can be taken instead (but I can see that this 'shortcut' I used in the video might be confusing for some and doesn't give a more general statement).
@@jamesexplainsmath You say that "by inspection" c is "clearly 0". But in an educational video where people may not have much experience on it, no one would notice you skipped all of that.
This video let me know, in 67 seconds, that the technique exists. To actually learn it, I would need something much more extensive, but that's fine, knowing it exists is a great start, and if the video were 20 minutes long and included practice problems (like needed to learn the method) I wouldn't have watched it.
There is a slight addition: When evaluating I(7) you have set the lower limit to be 0, but that may not always be the case. In general we set the lower limit to be that value where we know the value of the integral. In this case luckily I(0) = 0 and we are saved.
This is a TERRIBLE explanation. Your second step don't follow from your first without several unstated assumptions. You didn't explain any general principle.
All good. I skimmed over the details on when and how to swap the integral and derivative in line 2. I chose to approach it from a non-rigorous standpoint (i.e. I didn't want to justify through any real analysis as I thought that would be confusing), so I just applied it and assumed the viewer knew partial differentiation. My aim in setting up a maths channel is certainly not to confuse anyone about a topic more and I'm sorry if I did!
I think the first question you need to ask yourself is "Who is my intended audience?" If it is first year calculus students or just people with an interest in math but limited coursework then you would need to provide more explanation and details. If it is someone who has more advanced math skills but may not be familiar with this technique then they might appreciate your keeping it short as you did.
Yes, the intended audience is someone who knows at least first year calculus (with partial differentiation). I kept it brief and didn't go into real analysis (i.e. when can we rigorously differentiate under the integral sign) for that reason.
@@jamesexplainsmath it's amazing! I studied calculus several years ago in college yet i wasn't aware of this. Video length is just great for the purpose of touching on some cool tricks. 👍 Keep it up! And happy new year everyone 🎉
Substitute ln(x) = -t, then the general integral will be from 0 to +∞ of (e^(-t) - e^(-(k+1)t))/t - a Frullani integral which is equal to ln((k+1)/1) = ln(k+1)
My approach was a bit different, I multiplied the expression inside the integral by x^a, which I think to be a lot more intuitive especially for me who is a beginner at this. It clearly differentiates to x^a*lnx, which cancels out the lnx in the denominator and the integral becomes trivial. Taking the limit as a approaches negative infinity (even though I don't quite understand how limits can be used to solve for C, but that's how it went for integrating sinx/x) you can see C is 0 and get to the same answer in the end
There is an integral that I saw in a mathematics Olympiad, where to solve it I had to transform it into a double integral and then it was solved that way, it was a very interesting trick. Someone knows the name of that technique, or some integral that is solved like this. The thing is that I saw it many years ago when I was studying at university, and I don't remember it, but I didn't forget that trick, transforming a simple integral into a double, it would never have occurred to me to do that.
A very famous example is the integral from -inf to inf of e^(-x^2). Transforming it into a double integral and changing coordinates to polar coordinates does the trick.
Hi, thanks for the video! A clear description and easy to follow along. Would love to see a physics example were this technique is applied. Keep it up!
@@jamesexplainsmath well, youtube would soon remove API access to the dislike count so the dislike count extension would stop working in the near future. Anyway, I guess your video is a great introduction feynmenn's method, but since these are some advanced stuff, I suggest you put more effort in explaining.
@@otaku-qi4dj All good, I'm pretty frustrated at youtube's removal of dislikes as well. Since I'm only just starting a channel, constructive criticism is always welcome :D
Regarding this example, a few comments to avoid confusion. 0. As with many similar examples, hand-waving the assumptions and conditions in which such technique can be applied. But that does not matter in terms of calculation and the targeted audience. 1. In the second equation, it is not partial differentiation but ordinary with respect to k. From the way it is set up, I is just a function of k. Of course it is implicitly assumed that the integral in Eqn (1) converges. 2. While I(7) = ln 8, the way presented in the middle of the last line, especially the upper limit 7, should be made clear. In fact, one simply goes from integrating dI/dt = 1/(k+1) and applies the initial condition I(0) = 0 from Eqn (1) to obtain I(k) = ln(k+1).
Some comments say "Leibniz single handed invented calculus". Yeah, no, he didn't single handled invented it. Leibniz and Newton both took several works from that time, like J. Kepler's, and both reached the same conclusion. While Newton focused on a more practical use of calculus, Leibniz gave it its proper foundation, and solid math definition. You can say, both discovered calculus. There are several works, from ancient Greece to Arabic mathematicians that got a good idea of what we know as integrals. Kinda sad people still couldn't get over the old discussion of who invented/discovered calculus. The actual conclusion is that both Leibniz and Newton did it.
Still the tecnique is useful even in that case, since the integral to calculate C is typically much easier (e.g. in this case, if you take -1 out at the numerator, everything stays the same but C=I(0) becomes integral from 0 to 1 of 1/log(x)dx, and the method is still much shorter than applying part integration 7 times to integral from 0 to 1 of x^7/log(x)dx 😅
@@Gna-rn7zx You should think better. Originally there is only one definite integral that depends on the parameter k. We define a function I(k) that is equal to this integral and we can find it's derivative, it equals to 1/( k+1). Now we have now right to conclude that if some abstract function F(x) has derivative f(x) then F(k) = int_0^k f(x)dx, because so far we only know that F is antiderivative of f, so it belongs to the family of functions int f(x)dx any two functions of which differs on a constant. But in our case with additional information that I(0) = int 0/ln(x) dx = 0 we can easily get that I(k) = int_0^k 1/(k+1)dx + I(0) = int_0^k 1/(k+1)dx = ln(k+1). If you have any doubts then what is the value of int_0^1 (x^7 - x) / ln(x) dx ?
Just as a sidequestion: doesn't proving the possobility to take the limit under the integral sign risk to be even more difficult than solving the integral itself?
I think it would have been a more meaningful example if you integrated over k the 1/(k+1) to get the indefinite integral with c constant, then solved for c by letting k=0 to find out c=0, then you had the general fomula I(k) = ln(k+1) and use k=7 to get the solution ln(8)
I’m sorry, I don’t understand where everything else after letting k = 7 come from. Why is the integral with respect to k? Why are the bounds of integration different and why from 0 to 7
Perhaps in my opinion I am a influenced by my "Bourbakist" approach, but I think that the description of example was a bit too concise, and therefore not oriented for grounding a general method: as a matter of fact, one has to define a the definite integral as a function I of some parameter k inside the integrand x function; then the derivative dI/dk has to be evaluated by partiallly k-derivating inside the definite integral (obviously specific conditions have to be verified!); this results useful provided that the last x-integral is analytically evaluable as a closed-form dI/dk=G(k) ; if so, as a last step we shall solve a differential equation in order to obtain I
Yes, it was intended as a short refresher rather than a complete overview of the method. Most physicists deal with analytic functions (which are infinitely differentiable) and don't often worry about the conditions on when to swap the differentiation and integration signs. Btw, I love the work of the Bourbaki group (especially Éléments de mathématique)
Sometimes life seems unfair, tough and cruel. It feels like I can't break this impassable wall of problems and misfortune. And then I come across such equations and I get so happy I don't need to be solving them... ever... It's a huge relief. My problems seem smaller after that. Everything becomes possible. And my life is bright and complete as long as I can say "no, I won't even attempt, it's not my competence". Like eating ice-cream after a long hot summer day in the desert. Like scratching your leg you couldn't touch for 3 months. Like first time kissing the girl you love. Lemme look at it again... Yes! God, yes, I'm so glad I don't care why it's power 7. And why -1. And what can I possibly do with it. Math is great, math is very useful and curious, it's beautiful and so deep. But I remember I can read 50 books, watch 200 films, build 3 houses instead of learning what this one formula means. Let those who love math more than I do deal with it. Peace.
The k=7 is but one example. It's about calculating this definite integral on a family of integrands of that form. k could be any constant. Also it won't work if the terms containing x won't cancel out. (But if they WILL, woo hoo.) There's such a thing as being a bit TOO brief.
The last integral is really confusing. I don't think that I(7) is equal to the last integral. Also why is I(k) taken as a partial derivative on the left hand side of line 2? I(k) is a function of 1 variable, i.e. the variable k.
The reason the partial derivative is on both sides is more a matter of convenience since both notations are equivalent for a one-variable function (Wolfram does it at mathworld.wolfram.com/LeibnizIntegralRule.html but some other sources will use d/dk for the left hand side and partial derivatives on the right for the same equation). I prefer using partial derivatives because it highlights that we're dealing with a multivariable function f(x,k) (after swapping the integral) For the last integral, the full working is int 1/(k+1) from 0 to 7 = int 1/u du from 1 to 8, = ln|8| - ln|1|, = ln|8|, such that u = k + 1 and du = dk. When k = 0, u = 1 and when k = 7, u = 8, which is why the limits of integration change. I hope this helps.
It is everything okay with numbers in the video though pretty much lacks thorough explanation. Firstly, (more generally speaking) I(k)=ln(k+1)+C, but you can clearly see that I(0)=0 (by definition of I(k)) and therefore C=0 -> I(7)=ln8 Your second question is probably not so much of a problem, more like about definitions. Yes, technically you can take a partial derivative there (which is totally equivalent to common derivative), but yes, there is no reason to do one instead of another.
"Oh, a one minute video on maths! This should be quick..." [pauses video half way through and stares blankly at a single frame for 5 minutes] [rewinds video]
The video omitted the value of I(0) since it used definite integration. I would prefer to use indefinite integration with initial condition I(0) to determine the constant.
The objective integral (k=7) doesn't appear anywhere in the video except in the thumbnail, so when you actually evaluate the function I(k) at k=7 it appears to be for totally no reason.
the last integral is an indefinite integral, so the function I(k) should be equal to ln(k+1) + C, where C is some constant independent of k. So the answer may include some constant ln(8)+C. C can be nonzero. You cannot just ignore C. For example, if you do the same steps as in the video for the function under the integral x^k/ln(x) instead of (x^k-1)/ln(x), you will get the same answer ln(8) which is not correct.
Thanks for the video. Does anyone know what this slideshow theme is? Is it a PowerPoint theme? I have studied Mathematics at two universities and so many of my lecturers from both had slides that looked just like this. Any help would be greatly appreciated.
This peed my off because I could not follow the steps (in my defense it was late at night and my brain was a bit fried.) Before going to bed I thought just use the definition of the derivative we were taught in calculus class. This morning, with a fresher brain, I got the problem to the point where I needed to find limit (x^a -1)/a as a -> 0, I did not know this but Wolfram Alpha did. Is there a "trick" to finding the above limit? Thanks.
Sorry that my video annoyed you. But hopefully I can help with your limit! You can apply L'Hopital's rule (since the limit is of an indeterminate form 0/0). So, differentiating both the numerator and denominator lim (x^a-1)/a = lim d/da(x^a-1)/d/da(a) = lim x^a*ln(x) as a -> 0 = x^0*ln(x) = ln(x). If you would like to know more about L'Hopital's rule, I'd suggest going through first year uni/high school calculus textbooks (i.e. Schaum's or Stewart Calculus) or having a look at www.mathsisfun.com/calculus/l-hopitals-rule.html for a brief overview. If you have any more questions, feel free to let me know.
@@jamesexplainsmath It annoyed me in a good way as I am now off to relearn L'Hopital's rule. Thank you. Edit, the proof in the Wikipedia article on L'Hopital's rule, I follow that.
Hs student here who has just started with integration (I know bits and pieces of the topic till the usage of partial fractions) . What prerequisites are needed to understand this? I've looked it up and they have mentioned things like the fundamental theorem of calculus, indefinite integrals, etc, in general terms that I can't understand. Is there any specific part of integrations I should read up on before attempting to understand this video? I tried looking up the wiki article for the Leibniz Integral Rule which only made my head spun
Hi and thanks for your question! If you're looking for a textbook, I would recommend a calculus textbook which is used by universities/colleges in first year (for me, I learnt with Stewart Calculus Early Transcendentals but I've also heard that Schaum's Calculus has good explanations if you want something shorter). If you want to learn each prerequisite, I'll list most that would give you a grounding in calculus off the top of my head: - Limits - Definition of the derivative (from first principles, using limits) - Differentiation rules - quotient, power, product, chain rules (and specific ones: log, trig, ...) - Implicit differentiation - Integration rules (log, trig, ...) - Integration (u-substitution, by parts, partial fractions) - Partial differentiation (similar to implicit differentiation, but all other variables are considered constant) It might look daunting at first! But I'm sure it will only take half a year to a year, if you have enough time to spare. Also, if you intend to do maths in the future, it will certainly be useful. Hope this helps :D P.S. I haven't gone into real analysis here, since that's what makes everyone's head spin on wikipedia and it's only useful if you want to prove the technique (and when it holds).
@@jamesexplainsmath Thank you for all the resources! Ive learnt all the topics you've listed save for partial differentiation (which isnt listed in my textbooks at all). Can I still understand this video and learn the underlying stuff without learning partial derivatives and stuff?
@@Sh-hg8kf Oh yes, I'm sure that would be enough to apply the technique to different integrals (note that when picking 'k' you're looking to simplify the integral after the differentiation - in my video, the logs on either side of the fraction cancel). If you want an overview of partial derivatives I'd suggest at least the first few minutes of ua-cam.com/video/JAf_aSIJryg/v-deo.html. Something I forgot to mention in the video (see my pinned comment) is that you can only swap the differential and integral signs if the integrand (function being integrated) is continuously differentiable (i.e. the derivative exists at each point on its domain). Hope this helps.
@@jamesexplainsmath Thank you once more ! I'll try going through the video again, hopefully being capable of understanding it this time. Second reads always help!
the method gets even better. Even if your original integral does not contain the "k", you can plug some k "somewhere in" and solve it with this method (in the end putting k=0 or k=1, depending on the modification you did). The "art" however is to find a good spot where to modify.
You can get trapped if you use this method without checking first that it can be legally applied (i.e., that within the given integration range the integrand is continuous wrt x and k, and a C^1 function wrt k).
Oh yeah but this only works when the integrated function and its derivative are dominated by an integrable function + some continuity hypothesis Nah I'm just joking, cool trick. To think how many super powerful theorems for simplificating integrals were found thanks to Lebesgue integral (this one, dominated convergence theorem, Beppo-Levi theorem...)
why is this called feynman's technique, differentiation under the integral sign was first popularized by Leibniz, infact Feynman learned it from an old text while in highschool
7 minuti fa A more Easy example is sheldon cooper problem in the big beng theory F(x)= x^2 * e^(-x) This function can be obtained from the double differentiation of I(b)=e^(-bx) and and then using b=1
I(0)=0 Look at the expression for I(k) on top of the page: the numerator inside the integral becomes (x^0-1)=0, and integrating 0 from 0 to 1 obviously gives 0
WHY FROM 0? … Because you want to take the integral of the derivative of k to get back to the original function. But that leaves you with a constant C from the indefinite integration. So you look for a point to start where the indefinite integral is zero and start from there. (If you think this is just me being anal about it…. just replace k by k+1 in the equation from the video and see how it all crumbles away if you start from 0.)
From the 1/(k+1) line you can either take the indefinite integral (which is ln|k+1|+C), realise C is 0, and then substitute for k=7, or if by inspection C is clearly 0 then the definite integral from 0 to 7 can be taken instead
@@jamesexplainsmath There are two cases: you either can know C or you can’t. If you do, then you can just take the indefinite integral, set it at k=7 and subtract C. If you can’t, then you can take the definite integral (so C cancels-out) from x_0 to 7. The result for this will be the desired one for k=7 with a new constant as the indefinite integral at x_0. So, why it did work out in your video is simply because in that case, at x_0=0, this new constant is zero.
@@jamesexplainsmath Exactly, but my point is that it is not clear that it is by pure coincidence that taking the lower bound to be 0 gives the desires answer.
@@vlad2718 Oh, all good. Honestly, I didn't expect the video to get so many views and I made it for an audience of 100 or so people who already had plenty of mathematical knowledge (which is why the video is VERY concise). My content from now on will certainly be more explanatory!
Wow, so the -1 in the numerator could be any constant ... same answer! Cool! EDIT: No wait, nevermind, it can't be anything because it changes the value of I(0). My bad! But at least the antiderivative form doesn't change.
k=7 was chosen as an example (you might enjoy trying other values of k=3,4,... and seeing what it reduces to). Since you can't differentiate with respect to 7, we use the placeholder variable k to simplify the integral before substituting it back in at the end. Hope this helps.
Thanks for asking. So, line 2 is the derivative of line 1 with respect to k. The full working for the numerator is d/dk(x^k-1)=d/dk(x^k)-d/dk(1), =x^k*ln(x).
Because in the thumbnail the original integral had x^7 in it instead of x^k. The whole idea of the trick is that you can treat the "7" (or whatever other number) as a variable to differentiate and then reintegrate on, then you set to "7" (or whatever the number was) once you've solved the easier integral. The video montage was done very badly, not adding it there too makes the video confusing and to some degree pointless.
Depends. Bear in mind that this solved the whole family of integrals for any value of k. This can also be used to solve problems like the Dirichlet integral.
Maths and especially integration is often like that, just a load of ad hoc tricks. It's like irregular French verbs, you just have to know them if you want to speak the language.
@@flyingpenandpaper6119 But in this context, a whole "family" of integrals barely scratches the surface of the infinite variety of problems which could actually arise. It's like if a chess beginner learns about pins and then thinks they can win every chess position with that idea. My point stands. For example, in this case, I'm pretty sure you could just add 1 in the denominator and all of a sudden, the trick doesn't work. It only really works for this exact denominator. It's a useful shortcut... in the specific situations where it happens to work.
@@Gretchaninov It's better than nothing, and chess masters use pins frequently, so I don't even see what your point is. There is no single mathematical technique that solves everything, and you have to use a combination of techniques, as with chess. Is that surprising? This technique makes quite a few tough integrals tractable, so it's worth learning. Let me respond to your last remark with a question. Suppose we have an algorithm where given a starting number n, we multiply by 3 if it's odd, and divide by 2 if it's even, then repeat. Does every number eventually reach 1? Elementary number theory makes this very easy, right? Now let's "just add 1". Make it so that if n is odd, our next term is 3n+1, with the same rule for n even. Does every number converge to 1? :-) Indeed, the integrand in the video does not have an elementary anti-derivative for most values of k. I don't know any other way I could solve it. Do you?
@Flying Pen and Paper Perhaps you've missed my point. Pins are useful in chess, but knowing about pins hardly makes you a master of chess. No doubt Feynman's trick is useful with some integrals - I hadn't seen it before but it looks cool. However, it barely scratches the surface of the world of integrals and differential equations in general. I just think that should be emphasised more. It's useful in very specific situations only, meanwhile most integrals have no closed solution or easy method. Regarding alternative methods, these integrals can usually be solved using induction. Feynman's method can also be thought of in different ways and generalised. There are all kinds of patterns and sequences of progressively more complicated examples in maths, especially if they're designed to work out. Your reference to the x3+1 problem just reinforces my point - number theory is immensely complicated. I just get annoyed when a very neat, cute, specific problem and specific method to solve it get presented as if the method works in general. I prefer to get to the crux of why it works and when it works - to actually understand it more deeply.
Thanks for pointing this out. The main requirements are that the function being integrated is continuously differentiable and that the limits of integration are constant (the Leibniz integral rule is a generalisation with non-constant limits). To prove that these requirements are met, a bit of real analysis would be needed (see math.stackexchange.com/questions/2566783/differentiating-under-the-integral-sign-and-c1-class for example).
This is Leibniz's technique NOT Feynman's. Feynman took this from a book on Advanced Calculus. The real name of the technique is "Leibniz Integral Rule". To attribute this to Feynman is complete BS. He didn't come up with it. The book he got this method from as well a several other gems is: Advanced Calculus by Woods, 1926.
This video is putting under carpet so many details that I’d not advise to blindly apply it. Just make sure that the derivative exists, is systematically integrable and it’s derivative module is upper bounded by an integrable function. Theorem of derivation under integration (due to Leibniz) applies then. I don’t know why the author of this very unrigorous video says it’s Feynman.
A good example of how explaining too much can work against the explainer's end goal, and now I've just realized that I've just explained too much already....! Oh well too bad. Here's my dummy spit of the day. All this reminds me of Pierre de Fermat's Last Theorem - the world's greatest exercise in mathematical click-bait. Uh, except of course, that Fermat died without publishing anything, and his theories were found only by chance, and he was in any case not always right .... but hey, look how many genius mathematicians took the bait! Even Euler had no luck with FLT. Thank God for Andrew Wiles. Now I, and mathematicians all over the world for all time to come, can sleep peacefully at night knowing that X^N+Y^N=Z^N has no positive integer solutions for N>2. 😖 I really needed to know that.
This is a cool example for those who know the technique already, but for someone who doesn't know it, I can't say I would recommend this video to learn it. For that, it would need to be longer, more motivated, and better explained
Any tipps on where to find such a video?
Thanks.
-Someone who doesn't know the technique already.
@@yama123numbercauseytdemand4 blackpenredpen has a few videos using the technique, and I think he explains it fairly well.
The conceit of mathematics is to take away the scaffolding. If you explained everything, anyone could understand it.
@@MrKnivan Ah I see, thank you very much.
I HOPE EVERYBODY THAT WATCHES THIS VIDEO SEES THIS COMMENT
Gottfried Liebniz knew about this technique 300 years before Feynman
Very true and the Leibniz rule is generalised to non-constant limits of integration. Feynman really just popularised its use in Physics.
Yeah we were taught this in school
To be fair though, Feynman didn't exist 300 years ago.
@@13utt3r5cotch Iam not a feynam hater but to be really fair leibniz single handedly invented calculus.
@@bernhardriemann3821 the same time that Newton
This explanation is... honestly a little too simplified. It might have been a good idea to go into a _lot_ more detail with more steps, more examples, and more reasoning for when this technique can and can't be done.
A little? Worst shit ever.
@@Mazzphysics no u
It should work with any linear transformation, then integration and the anti-transform. Fourier transform, Laplace transform, Differentiation, etc.
First of all F(x,t) must be definite in a set A measurable.
Then in A F(.,t) must be integrable.
And last F(x, .) must be derivable and the derivate and < g(x)
Thank you everybody for all of your support and constructive criticism! Honestly, I didn't expect the video to get so many views. I made it for an audience of 100 or so people who already had plenty of mathematical knowledge as more of a refresher (which is why the video is WAY TOO concise). My content from now on will certainly be more explanatory! :D
I'll try and answer some commonly asked questions here:
- Q: 'Why is k=7?'
- A: 'I am answering the integral in the thumbnail and you should try k=2,3,4,... as it is just an arbitrary constant.'
- Q: 'Why didn't you refer to Leibniz?'
- A: 'Feynman popularised the Leibniz integral rule in physics (pilot-wave theory is my honours topic, so I'm biased). Since Leibniz generalises for non-constant limits of integration, if I ever want to fix this video in the future, I will probably do the more general Leibniz integral rule.'
- Q: 'When can I swap the differential and integral signs? There is no rigour!'
- A: 'To differentiate under the integral sign, the integrand must be continuously differentiable (luckily most functions in physics are analytic and so this isn't much of a worry). To prove this, some real analysis is involved which I'm happy to go through in the future.'
- Q: Why didn't you take the indefinite integral instead of an integral from 0 to 7. Why haven't you mentioned +C?
- A: The infamous +C has tripped many a mathematician up on their final exam (including me lol). To be honest, it was a shortcut. I(k)=ln(k+1)+C, but by observation, I(0)=0 (using the definition of I(k)) and therefore C=0 -> I(7)=ln(8).
- Q: Why is a partial derivative of the LHS?
- A: As a matter of convenience, since both notations are equivalent for a one-variable function (Wolfram does it at mathworld.wolfram.com/LeibnizIntegralRule.html but some other sources will use d/dk for the left hand side and partial derivatives on the right for the same equation). I prefer using partial derivatives because it highlights that we're dealing with a multivariable function f(x,k) (after swapping the integral).
Anyways, if you are at the end of this, I hope this helps.
I thought your video was fine, man. Some people just get real analytical and can't Leibniz alone. Most of their complaints aren't integral to your pedagogy. Clairaut your mind, and don't let them test your limits or derive you crazy.
A bit more explanation and more examples would be nice. I mean, I know the trick/technique and just clicked out of curiosity, but yeah just a bit more in general would be helpful. Like if your channel is more for beginners, show the motivation, (which is simple, a lot of integrals are nasty buggers. Especially ones with denominators. And Feynam's technique is often godsent in taking those out) and a word of warning that you're not actually always allowed to switch the integral operator and the differential operator would be nice.
Youre right. I mean. Ive no clue what he did there.
@@shmerox7683 in a very simplified explanation he looked for a way to eliminate the ln(x) in the denominator without changing the actual integral.
So in the first step he defined a new function in terms of that integral and named it I and a new variable k and made the new function depend on k.
He turned the original x^7+1/lnx into x^k+1/lnx.
So you can see, I(7) is the original integral. What he did next is the actual Feynman technique.
He took the derivative with respect to k! You would normally have to prove you can switch the integral sign and the derivative operator, but Feynman was a physicist and physicists don't care about details like that.
Everything that has to do with x is now constant.
Meaning he only has to take the derivative of x^k. Remember x is a constant so the derivative structure is identical to let's say 2^k. That would be ln(2)*2^k
So the derivative of x^k is lnx*x^k which is why we did the shenanigans in the first place.
We now have integral from 0 to 1
lnx/lnx x^k+1 dx so the LNs vanish.
You know evaluate that integral
You should get 1/k+1 as the result
And now comes the tricky part.
Remember this is I'=1/k+1 we need I(7)
How do we get there? Well the fundamental theorem of calculus states F(x)=int from 0 t x f(t) dt
Let x=7, t=k and f=I' and we just integrate 1/k+1 from 0 to 7 and you get the result ln8.
Hopefully this unreadable textwall helped a little with whats going on?
The actual Feynman trick is just introducing a second parameter to eliminate a difficult part of your integral with a derivative with respect to your second parameter. Its a powerful tool that can solve many nonelementary integrals like this rather easily.
Imho it's so useful it SHOULD be taught in calc 2 courses
@@Metalhammer1993 Great explanation. Do you have any reference where someone could check the conditions under which swapping integral and derivative operators is allowed? I'd say Tom Apostol Mathematical Analysis could do the job but perhaps you know a more enlightening source.
@@anuman99ful honestly no. I only know it's got to do with Fubini's theorem. I do believe Flammable Maths has something on that topic
Dude, I saw you in the comments in ua-cam.com/video/AWA-1rsSSh4/v-deo.html , which is basically the same video in longer. Strange things can happen.
You need to mention the fact there's a +C . This is only the answer because I(0) = 0 so theres no added constant. This is a very important detail that you missed out
Well pointed out. From the 1/(k+1) line you can either take the indefinite integral (which is ln|k+1|+C), realise C is 0, and then substitute for k=7, or if by inspection C is clearly 0 then the definite integral from 0 to 7 can be taken instead (but I can see that this 'shortcut' I used in the video might be confusing for some and doesn't give a more general statement).
Lol
@@jamesexplainsmath You say that "by inspection" c is "clearly 0". But in an educational video where people may not have much experience on it, no one would notice you skipped all of that.
I'm a Physics major. Never added that C in definite integrals
Everything is working good till now
@@danielyuan9862 dI/dk = 1/(k+1) directly implies I(7)-I(0)=integral of dk/(k+1) from 0 to 7, there isn't really a need to introduce a constant here
This video let me know, in 67 seconds, that the technique exists. To actually learn it, I would need something much more extensive, but that's fine, knowing it exists is a great start, and if the video were 20 minutes long and included practice problems (like needed to learn the method) I wouldn't have watched it.
1:06 for Feynman's Technique? Not even Feynman would have explained it that fast.
There is a slight addition: When evaluating I(7) you have set the lower limit to be 0, but that may not always be the case. In general we set the lower limit to be that value where we know the value of the integral. In this case luckily I(0) = 0 and we are saved.
honestly, the most helpful video ive found on this. short and simple
This is a TERRIBLE explanation. Your second step don't follow from your first without several unstated assumptions.
You didn't explain any general principle.
All good. I skimmed over the details on when and how to swap the integral and derivative in line 2. I chose to approach it from a non-rigorous standpoint (i.e. I didn't want to justify through any real analysis as I thought that would be confusing), so I just applied it and assumed the viewer knew partial differentiation. My aim in setting up a maths channel is certainly not to confuse anyone about a topic more and I'm sorry if I did!
The intermediate steps are left as exercise for the reader.
@@martijnb5887 most dreadful line in all math books.
I think the first question you need to ask yourself is "Who is my intended audience?"
If it is first year calculus students or just people with an interest in math but limited coursework then you would need to provide more explanation and details.
If it is someone who has more advanced math skills but may not be familiar with this technique then they might appreciate your keeping it short as you did.
Yes, the intended audience is someone who knows at least first year calculus (with partial differentiation). I kept it brief and didn't go into real analysis (i.e. when can we rigorously differentiate under the integral sign) for that reason.
@@jamesexplainsmath it's amazing! I studied calculus several years ago in college yet i wasn't aware of this. Video length is just great for the purpose of touching on some cool tricks. 👍 Keep it up! And happy new year everyone 🎉
@@Alex-gk8ik Happy new year :D
Substitute ln(x) = -t, then the general integral will be from 0 to +∞ of (e^(-t) - e^(-(k+1)t))/t - a Frullani integral which is equal to ln((k+1)/1) = ln(k+1)
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OK but then you need to have memorized the Frullani integral, while the Feynman technique does everything from scratch.
@@Aramil4 yeah, I mean finding another approach is interesting even though it requires to know more
My approach was a bit different, I multiplied the expression inside the integral by x^a, which I think to be a lot more intuitive especially for me who is a beginner at this. It clearly differentiates to x^a*lnx, which cancels out the lnx in the denominator and the integral becomes trivial. Taking the limit as a approaches negative infinity (even though I don't quite understand how limits can be used to solve for C, but that's how it went for integrating sinx/x) you can see C is 0 and get to the same answer in the end
This technique goes back to Leibniz
en.wikipedia.org/wiki/Leibniz_integral_rule
There is an integral that I saw in a mathematics Olympiad, where to solve it I had to transform it into a double integral and then it was solved that way, it was a very interesting trick. Someone knows the name of that technique, or some integral that is solved like this. The thing is that I saw it many years ago when I was studying at university, and I don't remember it, but I didn't forget that trick, transforming a simple integral into a double, it would never have occurred to me to do that.
A very famous example is the integral from -inf to inf of e^(-x^2). Transforming it into a double integral and changing coordinates to polar coordinates does the trick.
@@ElIrracional
That's a fun one!
Very powerful. Almost shocking how simple the integral becomes which seemed hopeless on first sight!
I really appreciated this short but precious piece of knowledge about calculus, thanks!
I did know the Feynman's tecnique, but I didn't notice that we could use it here. Nice one
Hi, thanks for the video! A clear description and easy to follow along. Would love to see a physics example were this technique is applied. Keep it up!
This video is the reason why UA-cam should not remove the dislike button
Just download the extension at chrome.google.com/webstore/detail/return-youtube-dislike/gebbhagfogifgggkldgodflihgfeippi lmao
@@jamesexplainsmath well, youtube would soon remove API access to the dislike count so the dislike count extension would stop working in the near future.
Anyway, I guess your video is a great introduction feynmenn's method, but since these are some advanced stuff, I suggest you put more effort in explaining.
@@otaku-qi4dj All good, I'm pretty frustrated at youtube's removal of dislikes as well. Since I'm only just starting a channel, constructive criticism is always welcome :D
Regarding this example, a few comments to avoid confusion.
0. As with many similar examples, hand-waving the assumptions and conditions in which such technique can be applied. But that does not matter in terms of calculation and the targeted audience.
1. In the second equation, it is not partial differentiation but ordinary with respect to k. From the way it is set up, I is just a function of k. Of course it is implicitly assumed that the integral in Eqn (1) converges.
2. While I(7) = ln 8, the way presented in the middle of the last line, especially the upper limit 7, should be made clear. In fact, one simply goes from integrating dI/dt = 1/(k+1) and applies the initial condition I(0) = 0 from Eqn (1) to obtain I(k) = ln(k+1).
So this is what Howard meant in big bang theory lol! I didn't understand until this video. Thank you!!
I thought this was going to be a meme where he just puts it in wolfram alpha
Some comments say "Leibniz single handed invented calculus".
Yeah, no, he didn't single handled invented it. Leibniz and Newton both took several works from that time, like J. Kepler's, and both reached the same conclusion. While Newton focused on a more practical use of calculus, Leibniz gave it its proper foundation, and solid math definition.
You can say, both discovered calculus. There are several works, from ancient Greece to Arabic mathematicians that got a good idea of what we know as integrals.
Kinda sad people still couldn't get over the old discussion of who invented/discovered calculus. The actual conclusion is that both Leibniz and Newton did it.
This actually requires some hypothesis over the function that you are integrating so that you can differentiate under the integral sign
And you missed the most difficult part - you must proof that you can differentiate integral and everything converges
Do not forget about the condition that I(0) = 0, otherwise I(k) = ln(k+1) + C, but we do not know the value of C
Still the tecnique is useful even in that case, since the integral to calculate C is typically much easier (e.g. in this case, if you take -1 out at the numerator, everything stays the same but C=I(0) becomes integral from 0 to 1 of 1/log(x)dx, and the method is still much shorter than applying part integration 7 times to integral from 0 to 1 of x^7/log(x)dx 😅
I don't think the C is needed because everything here is definite integrals?
@@Gna-rn7zx You should think better.
Originally there is only one definite integral that depends on the parameter k.
We define a function I(k) that is equal to this integral and we can find it's derivative, it equals to 1/( k+1). Now we have now right to conclude that if some abstract function F(x) has derivative f(x) then F(k) = int_0^k f(x)dx, because so far we only know that F is antiderivative of f, so it belongs to the family of functions int f(x)dx any two functions of which differs on a constant. But in our case with additional information that I(0) = int 0/ln(x) dx = 0 we can easily get that I(k) = int_0^k 1/(k+1)dx + I(0) = int_0^k 1/(k+1)dx = ln(k+1).
If you have any doubts then what is the value of int_0^1 (x^7 - x) / ln(x) dx ?
why am i watching this? i really should be busy procrastinating
A typical video how math should not be explained i.e. leaving out all the necessary steps
ua-cam.com/video/YO38MCdj-GM/v-deo.html
Just as a sidequestion: doesn't proving the possobility to take the limit under the integral sign risk to be even more difficult than solving the integral itself?
This technique was known to the man who discovered calculus (Gottfried Wilhelm Leibniz)
If fact, Leibniz's integral rule generalises for non-constant limits of integration!
I think it would have been a more meaningful example if you integrated over k the 1/(k+1) to get the indefinite integral with c constant, then solved for c by letting k=0 to find out c=0, then you had the general fomula I(k) = ln(k+1) and use k=7 to get the solution ln(8)
I’m sorry, I don’t understand where everything else after letting k = 7 come from. Why is the integral with respect to k? Why are the bounds of integration different and why from 0 to 7
Perhaps in my opinion I am a influenced by my "Bourbakist" approach, but I think that the description of example was a bit too concise, and therefore not oriented for grounding a general method: as a matter of fact, one has to define a the definite integral as a function I of some parameter k inside the integrand x function; then the derivative dI/dk has to be evaluated by partiallly k-derivating inside the definite integral (obviously specific conditions have to be verified!); this results useful provided that the last x-integral is analytically evaluable as a closed-form dI/dk=G(k) ; if so, as a last step we shall solve a differential equation in order to obtain I
Yes, it was intended as a short refresher rather than a complete overview of the method. Most physicists deal with analytic functions (which are infinitely differentiable) and don't often worry about the conditions on when to swap the differentiation and integration signs. Btw, I love the work of the Bourbaki group (especially Éléments de mathématique)
@@jamesexplainsmath you said the truth... physicists (and engineers) simply use and don't worry... ;)
Sometimes life seems unfair, tough and cruel. It feels like I can't break this impassable wall of problems and misfortune. And then I come across such equations and I get so happy I don't need to be solving them... ever... It's a huge relief. My problems seem smaller after that. Everything becomes possible. And my life is bright and complete as long as I can say "no, I won't even attempt, it's not my competence". Like eating ice-cream after a long hot summer day in the desert. Like scratching your leg you couldn't touch for 3 months. Like first time kissing the girl you love.
Lemme look at it again... Yes! God, yes, I'm so glad I don't care why it's power 7. And why -1. And what can I possibly do with it. Math is great, math is very useful and curious, it's beautiful and so deep. But I remember I can read 50 books, watch 200 films, build 3 houses instead of learning what this one formula means. Let those who love math more than I do deal with it. Peace.
Thank you so much , it was very helpful and made me learn something new :)
The k=7 is but one example. It's about calculating this definite integral on a family of integrands of that form. k could be any constant. Also it won't work if the terms containing x won't cancel out. (But if they WILL, woo hoo.) There's such a thing as being a bit TOO brief.
please make a longer, more in depth video with explanation, if you could! Thank you
This method is not part of my course but i can surely use it for the mcqs !! haha amazing method and well explained :)
Calculus speed running strats
The last integral is really confusing. I don't think that I(7) is equal to the last integral. Also why is I(k) taken as a partial derivative on the left hand side of line 2? I(k) is a function of 1 variable, i.e. the variable k.
The reason the partial derivative is on both sides is more a matter of convenience since both notations are equivalent for a one-variable function (Wolfram does it at mathworld.wolfram.com/LeibnizIntegralRule.html but some other sources will use d/dk for the left hand side and partial derivatives on the right for the same equation). I prefer using partial derivatives because it highlights that we're dealing with a multivariable function f(x,k) (after swapping the integral)
For the last integral, the full working is
int 1/(k+1) from 0 to 7 = int 1/u du from 1 to 8,
= ln|8| - ln|1|,
= ln|8|,
such that u = k + 1 and du = dk. When k = 0, u = 1 and when k = 7, u = 8, which is why the limits of integration change. I hope this helps.
It is everything okay with numbers in the video though pretty much lacks thorough explanation.
Firstly, (more generally speaking) I(k)=ln(k+1)+C, but you can clearly see that I(0)=0 (by definition of I(k)) and therefore C=0 -> I(7)=ln8
Your second question is probably not so much of a problem, more like about definitions. Yes, technically you can take a partial derivative there (which is totally equivalent to common derivative), but yes, there is no reason to do one instead of another.
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This was an interesting video, but I’m tad confused on why we’re picking 7 to plug into our function. Where did that come from?
0:19 he sets it as the example.
"Oh, a one minute video on maths! This should be quick..."
[pauses video half way through and stares blankly at a single frame for 5 minutes]
[rewinds video]
Sorry lol
I have absolutely no idea what happened, I hope you will out already have made a more explanatory video in the future
The video omitted the value of I(0) since it used definite integration. I would prefer to use indefinite integration with initial condition I(0) to determine the constant.
The objective integral (k=7) doesn't appear anywhere in the video except in the thumbnail, so when you actually evaluate the function I(k) at k=7 it appears to be for totally no reason.
This was hard. But now it's harder.😄
lol
the last integral is an indefinite integral, so the function I(k) should be equal to ln(k+1) + C, where C is some constant independent of k. So the answer may include some constant ln(8)+C. C can be nonzero.
You cannot just ignore C. For example, if you do the same steps as in the video for the function under the integral x^k/ln(x) instead of (x^k-1)/ln(x), you will get the same answer ln(8) which is not correct.
Thanks for the video. Does anyone know what this slideshow theme is? Is it a PowerPoint theme? I have studied Mathematics at two universities and so many of my lecturers from both had slides that looked just like this. Any help would be greatly appreciated.
Hi, I use LATEX Beamer - Warsaw theme for presentations and my editor is overleaf. Hope this helps :D
@@jamesexplainsmath Thanks James! You have answered a question I've had for years! Yet another reason for me to work on my LATEX.
If y=(1/(k+1)) is antiderivative of I(K), then if we find a derivative of y, it wil not be equal to I(K).
This peed my off because I could not follow the steps (in my defense it was late at night and my brain was a bit fried.) Before going to bed I thought just use the definition of the derivative we were taught in calculus class. This morning, with a fresher brain, I got the problem to the point where I needed to find limit (x^a -1)/a as a -> 0, I did not know this but Wolfram Alpha did. Is there a "trick" to finding the above limit?
Thanks.
Sorry that my video annoyed you. But hopefully I can help with your limit! You can apply L'Hopital's rule (since the limit is of an indeterminate form 0/0). So, differentiating both the numerator and denominator
lim (x^a-1)/a = lim d/da(x^a-1)/d/da(a)
= lim x^a*ln(x) as a -> 0
= x^0*ln(x)
= ln(x).
If you would like to know more about L'Hopital's rule, I'd suggest going through first year uni/high school calculus textbooks (i.e. Schaum's or Stewart Calculus) or having a look at www.mathsisfun.com/calculus/l-hopitals-rule.html for a brief overview.
If you have any more questions, feel free to let me know.
@@jamesexplainsmath It annoyed me in a good way as I am now off to relearn L'Hopital's rule. Thank you. Edit, the proof in the Wikipedia article on L'Hopital's rule, I follow that.
@@andyeverett1957 Fantastic! I'm very glad that this has provoked you to have another look at L'Hopital's rule and that I could be of help.
Hs student here who has just started with integration (I know bits and pieces of the topic till the usage of partial fractions) . What prerequisites are needed to understand this?
I've looked it up and they have mentioned things like the fundamental theorem of calculus, indefinite integrals, etc, in general terms that I can't understand. Is there any specific part of integrations I should read up on before attempting to understand this video?
I tried looking up the wiki article for the Leibniz Integral Rule which only made my head spun
Hi and thanks for your question! If you're looking for a textbook, I would recommend a calculus textbook which is used by universities/colleges in first year (for me, I learnt with Stewart Calculus Early Transcendentals but I've also heard that Schaum's Calculus has good explanations if you want something shorter).
If you want to learn each prerequisite, I'll list most that would give you a grounding in calculus off the top of my head:
- Limits
- Definition of the derivative (from first principles, using limits)
- Differentiation rules - quotient, power, product, chain rules (and specific ones: log, trig, ...)
- Implicit differentiation
- Integration rules (log, trig, ...)
- Integration (u-substitution, by parts, partial fractions)
- Partial differentiation (similar to implicit differentiation, but all other variables are considered constant)
It might look daunting at first! But I'm sure it will only take half a year to a year, if you have enough time to spare. Also, if you intend to do maths in the future, it will certainly be useful. Hope this helps :D
P.S. I haven't gone into real analysis here, since that's what makes everyone's head spin on wikipedia and it's only useful if you want to prove the technique (and when it holds).
@@jamesexplainsmath Thank you for all the resources! Ive learnt all the topics you've listed save for partial differentiation (which isnt listed in my textbooks at all).
Can I still understand this video and learn the underlying stuff without learning partial derivatives and stuff?
@@Sh-hg8kf Oh yes, I'm sure that would be enough to apply the technique to different integrals (note that when picking 'k' you're looking to simplify the integral after the differentiation - in my video, the logs on either side of the fraction cancel). If you want an overview of partial derivatives I'd suggest at least the first few minutes of ua-cam.com/video/JAf_aSIJryg/v-deo.html.
Something I forgot to mention in the video (see my pinned comment) is that you can only swap the differential and integral signs if the integrand (function being integrated) is continuously differentiable (i.e. the derivative exists at each point on its domain).
Hope this helps.
@@jamesexplainsmath Thank you once more ! I'll try going through the video again, hopefully being capable of understanding it this time. Second reads always help!
the method gets even better. Even if your original integral does not contain the "k", you can plug some k "somewhere in" and solve it with this method (in the end putting k=0 or k=1, depending on the modification you did). The "art" however is to find a good spot where to modify.
I don't understand this yet, but I'll be back in time
Great!
You can get trapped if you use this method without checking first that it can be legally applied (i.e., that within the given integration range the integrand is continuous wrt x and k, and a C^1 function wrt k).
At the end, when you say let k = 7 ... I think you mean "As example, to find I(7), do int_0^7 __dk to both sides and note that I(0) =0" ?
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please, I don't understand why the differential of I (k) doesn't give integral 0_1 of kx^k-1/lnx dx.
The rule of differentiation by the parameter under the sign of the integral is the Leibniz formula. Why do you call it Feynman's Technique? .
Oh yeah but this only works when the integrated function and its derivative are dominated by an integrable function + some continuity hypothesis
Nah I'm just joking, cool trick. To think how many super powerful theorems for simplificating integrals were found thanks to Lebesgue integral (this one, dominated convergence theorem, Beppo-Levi theorem...)
just learned this in real analysis ii. love it
@@Marcus12813 well, I don't know what real analysis II is since I know nothing about how a university works but I trust you ^^
Can this method be applied to every integral? If not, how does one know when to use it?
very brief and not detailed so I had to rewatch like 3 times and ive taken 3 calculus courses. But yea still a better explanation than some I've seen
why is this called feynman's technique, differentiation under the integral sign was first popularized by Leibniz, infact Feynman learned it from an old text while in highschool
7 minuti fa
A more Easy example is sheldon cooper problem in the big beng theory
F(x)= x^2 * e^(-x)
This function can be obtained from the double differentiation of I(b)=e^(-bx) and and then using b=1
I(7) - I(0) = Integral { di(k)/dk }. What is the value of I(0)?
I(0)=0
Look at the expression for I(k) on top of the page: the numerator inside the integral becomes (x^0-1)=0, and integrating 0 from 0 to 1 obviously gives 0
If only I knew this method back then
WHY FROM 0?
…
Because you want to take the integral of the derivative of k to get back to the original function. But that leaves you with a constant C from the indefinite integration. So you look for a point to start where the indefinite integral is zero and start from there. (If you think this is just me being anal about it…. just replace k by k+1 in the equation from the video and see how it all crumbles away if you start from 0.)
From the 1/(k+1) line you can either take the indefinite integral (which is ln|k+1|+C), realise C is 0, and then substitute for k=7, or if by inspection C is clearly 0 then the definite integral from 0 to 7 can be taken instead
@@jamesexplainsmath There are two cases: you either can know C or you can’t. If you do, then you can just take the indefinite integral, set it at k=7 and subtract C. If you can’t, then you can take the definite integral (so C cancels-out) from x_0 to 7. The result for this will be the desired one for k=7 with a new constant as the indefinite integral at x_0. So, why it did work out in your video is simply because in that case, at x_0=0, this new constant is zero.
@@vlad2718 Yes, that's what I'm saying. Since x_0=0, I shortened the calculation
@@jamesexplainsmath Exactly, but my point is that it is not clear that it is by pure coincidence that taking the lower bound to be 0 gives the desires answer.
@@vlad2718 Oh, all good. Honestly, I didn't expect the video to get so many views and I made it for an audience of 100 or so people who already had plenty of mathematical knowledge (which is why the video is VERY concise). My content from now on will certainly be more explanatory!
Amazing insight 🌷
very funny critical reference, +10 points
Wow, so the -1 in the numerator could be any constant ... same answer! Cool!
EDIT: No wait, nevermind, it can't be anything because it changes the value of I(0). My bad! But at least the antiderivative form doesn't change.
Where is the 7 coming from?
Just an example (I am solving the integral in the thumbnail). Try k=2,3,4,... for fun if you would like :D
Math majors: Hey, that's just a fancy way of stating the Newton-Leibnitz rule!
Isn’t the integral evaluated from 0 to 1 of x^k with respect to x equal to (x^(k+1))/(k+1) evaluated from 0 to 1? Confused how you got 1/(k+1)
That's the indefinite integral. The limits are substituted into it and 1/(k+1) is the answer
@@mattp4805, oh duh. Thanks!
@@Andyg2g No worries. I do that kind of shit all the time lol
I was confused too thanks!
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can someone explain why the upper limit is now our value of k rather than the 1 in the original integral?
Why k=7? And why not set k to 7 everywhere (on both sides of equal sign set *ALL* instances of k to 7)?
k=7 was chosen as an example (you might enjoy trying other values of k=3,4,... and seeing what it reduces to). Since you can't differentiate with respect to 7, we use the placeholder variable k to simplify the integral before substituting it back in at the end. Hope this helps.
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@@jamesexplainsmath thank you
Is this from Alfonso?
Why are the numerators on line 1 and 2 different? I don't understand.
Thanks for asking. So, line 2 is the derivative of line 1 with respect to k. The full working for the numerator is
d/dk(x^k-1)=d/dk(x^k)-d/dk(1),
=x^k*ln(x).
Why did you equal k to 7?
Because in the thumbnail the original integral had x^7 in it instead of x^k. The whole idea of the trick is that you can treat the "7" (or whatever other number) as a variable to differentiate and then reintegrate on, then you set to "7" (or whatever the number was) once you've solved the easier integral. The video montage was done very badly, not adding it there too makes the video confusing and to some degree pointless.
If you think thats good you should try wolfram alpha
No explanation why integral from 0 to 7 (7 is clear but 0 why), so, to me, I(k)=ln(k+1)+c
but when k=1 l(k)=0 thats why c=0 and I(k)=ln(k+1).k=7 l=ln8
I was about to write EXACTLY what you wrote about the lack of explanation of the lower limit, and that c can be worked out by an initial condition.
Why put k=7 ? Is that arbitrary ? Wouldn't the answer be better put as I(k) = ln(k+1) ? I got confused by the specificity of putting k=7.
thumbnail.
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How x^k-1 became x^klnx?
d/dk(x^k-1)=d/dk(x^k)-d/dk(1)
=x^k*ln(x)
Hope this helps.
@@jamesexplainsmath oh yeah thanks
Can i solve any definite integral by this
some but not others
I don't get it. Why is C=0? Why is the definition of I(0) =0?
I(k)=ln(k+1)+C, but by observation, I(0)=0. We us the definition of I(k) that
(x^0-1)/ln(x) = (1-1)/ln(x) = 0
Therefore, C=0 -> I(7)=ln(8).
I got it already but thanks anyway
@@jamesexplainsmath lol
progress of Leibniz integral rule
Great refresher! Thanks
This is cool but seems like it will only really work in specific cases designed to work nicely.
Depends. Bear in mind that this solved the whole family of integrals for any value of k. This can also be used to solve problems like the Dirichlet integral.
Maths and especially integration is often like that, just a load of ad hoc tricks. It's like irregular French verbs, you just have to know them if you want to speak the language.
@@flyingpenandpaper6119 But in this context, a whole "family" of integrals barely scratches the surface of the infinite variety of problems which could actually arise. It's like if a chess beginner learns about pins and then thinks they can win every chess position with that idea. My point stands.
For example, in this case, I'm pretty sure you could just add 1 in the denominator and all of a sudden, the trick doesn't work. It only really works for this exact denominator. It's a useful shortcut... in the specific situations where it happens to work.
@@Gretchaninov It's better than nothing, and chess masters use pins frequently, so I don't even see what your point is. There is no single mathematical technique that solves everything, and you have to use a combination of techniques, as with chess. Is that surprising? This technique makes quite a few tough integrals tractable, so it's worth learning.
Let me respond to your last remark with a question. Suppose we have an algorithm where given a starting number n, we multiply by 3 if it's odd, and divide by 2 if it's even, then repeat. Does every number eventually reach 1? Elementary number theory makes this very easy, right? Now let's "just add 1". Make it so that if n is odd, our next term is 3n+1, with the same rule for n even. Does every number converge to 1? :-)
Indeed, the integrand in the video does not have an elementary anti-derivative for most values of k. I don't know any other way I could solve it. Do you?
@Flying Pen and Paper Perhaps you've missed my point. Pins are useful in chess, but knowing about pins hardly makes you a master of chess.
No doubt Feynman's trick is useful with some integrals - I hadn't seen it before but it looks cool. However, it barely scratches the surface of the world of integrals and differential equations in general. I just think that should be emphasised more. It's useful in very specific situations only, meanwhile most integrals have no closed solution or easy method.
Regarding alternative methods, these integrals can usually be solved using induction. Feynman's method can also be thought of in different ways and generalised. There are all kinds of patterns and sequences of progressively more complicated examples in maths, especially if they're designed to work out.
Your reference to the x3+1 problem just reinforces my point - number theory is immensely complicated. I just get annoyed when a very neat, cute, specific problem and specific method to solve it get presented as if the method works in general. I prefer to get to the crux of why it works and when it works - to actually understand it more deeply.
👍🏾
You must show uniformity to switch the derivative sign inside the integral. Otherwise, this will not be true.
Thks someone who notices
Thanks for pointing this out. The main requirements are that the function being integrated is continuously differentiable and that the limits of integration are constant (the Leibniz integral rule is a generalisation with non-constant limits). To prove that these requirements are met, a bit of real analysis would be needed (see math.stackexchange.com/questions/2566783/differentiating-under-the-integral-sign-and-c1-class for example).
This is Leibniz's technique NOT Feynman's. Feynman took this from a book on Advanced Calculus. The real name of the technique is "Leibniz Integral Rule". To attribute this to Feynman is complete BS. He didn't come up with it.
The book he got this method from as well a several other gems is: Advanced Calculus by Woods, 1926.
I finished diffy eq last semester why the FUCK is this in my recommended NOW
This video is putting under carpet so many details that I’d not advise to blindly apply it. Just make sure that the derivative exists, is systematically integrable and it’s derivative module is upper bounded by an integrable function. Theorem of derivation under integration (due to Leibniz) applies then.
I don’t know why the author of this very unrigorous video says it’s Feynman.
Why set k = 7?
Just an example, try k=2,3,4,5,.... etc if you would like (since it will still work). I.e. I am answering the thumbnail integral.
why choose seven?
Just for an example. You might like to see what k=2,3,4,... reduces to.
This trick was invented by leibniz. Why do you say it is Feynman's
Nice, but this technique has very little to do with Feynman. The idea goes
back to Euler, at least.
A good example of how explaining too much can work against the explainer's end goal, and now I've just realized that I've just explained too much already....!
Oh well too bad. Here's my dummy spit of the day. All this reminds me of Pierre de Fermat's Last Theorem - the world's greatest exercise in mathematical click-bait. Uh, except of course, that Fermat died without publishing anything, and his theories were found only by chance, and he was in any case not always right .... but hey, look how many genius mathematicians took the bait! Even Euler had no luck with FLT.
Thank God for Andrew Wiles. Now I, and mathematicians all over the world for all time to come, can sleep peacefully at night knowing that X^N+Y^N=Z^N has no positive integer solutions for N>2. 😖 I really needed to know that.