But I AM joking, Mr. Feynman!

Рет қаралды 114,345

Dr Peyam

Күн бұрын

Пікірлер: 195

@Galileosays 3 жыл бұрын

So the result is not normal, but nearly perfect (Peerfc(1)).

@drpeyam 3 жыл бұрын

😂😂😂

@puerulus 3 жыл бұрын

The differentiation under the integral sign technique is covered in many calculus textbooks from the 1800s and early 1900s. Richard Feynman himself pointed that out. He probably wouldn't have wanted the technique named after him.

@euva209 2 жыл бұрын

An example of watering down the curriculum? I'm sure it's an isolated case.😁

@UnforsakenXII 3 жыл бұрын

I remember being obessesed with these kind of integrals back in sophomore year of undergrad, needless to say, it paid off tremendously and I wish it was taught more standardly at least in physics curricula.

@jacobharris5894 3 жыл бұрын

I’m still obsessed with them now but I learned it pretty recently. Whenever I see a hard looking integral I try to find any excuse I can to use it.

@SimsHacks Жыл бұрын

it's a theorem usually covered in Measure theory. Sadly physicians don't often have this course.

@thelink9959 3 жыл бұрын

That french was spot-on and so was the math.

@AnkitSharma-ef7md 3 жыл бұрын

While others get to read early morning text from their sweet and loved ones, I get a notification from youtube that Dr. Peyam has uploaded another astonishing problem from mathematics. So good to see you professor, Amazing approach. I was puzzled at the step where you nicely adjusted the error function. Loved it. Seasonal greetings to you.

@ChariotduNord 3 жыл бұрын

nothing tickles my fancy more than math with fun jokes in other languages!

@hOREP245 3 жыл бұрын

One of my favourite integrals! This can be easily extended using the Gamma and Incomplete function for any n>0 (where this is the special case n = 2).

@robertl.crawford4369 2 жыл бұрын

Cool. thanks

@noonesperfect 3 жыл бұрын

new generation is so lucky to have these videos around. It really helps a lot you cant imagine , thx to prof. :)

@constantinototis4859 3 жыл бұрын

So true!

@77Chester77 3 жыл бұрын

13:37 😁 Sie gehen mir NIE auf die Nerven, Dr πm

@drpeyam 3 жыл бұрын

Hahahaha

@thomasborgsmidt9801 3 жыл бұрын

Ich glaube Du hast im Schwarzen getroffen! πm does not need to make the student feel like an idiot - but he points out and lets the student decide for himself, that maybe he has to reevaluate. Math is - like all sciences - not about finding answers, but finding the relevant question. It is like Kepler that just asked himself: Does planetary orbits HAVE to be circular? Well, they don't - the concept of circular orbits was made to make calculations simpler - and elipses do not have a rough and ready formula for the length of the circumfence. But then again the quest for making a simple circular equations valid led to a religious conviction that was much more complicated as an end result.

@mudkip_btw 3 жыл бұрын

Living dangerously pays off, that's why I like it :} Though I feel like it's getting time for me to learn the proper theorems at this point, like dominated convergence as you talked about. I did quite badly in my calculus 1, 2 classes but much better later, once we got to vector calculus etc. since I'm more of a visual learner. Glad to have people like you around to keep me interested in proper maths as a physics student :)

@GRBtutorials 3 жыл бұрын

I was so confused at the beginning... you’re probably the only KZfaqr who says “thanks for watching” at the beginning of the video instead of at the end!

@naveensagar2408 3 жыл бұрын

Integration under diff followed by lebini t' z rule feynman technique is very awsome ! being a masters in physics i can tell only 1 thing this technique is very useful in solving the famous dirichlet integrals . & Many problems in physics.

@6612770 2 жыл бұрын

Congratulations, you have pleasantly blown my mind again. Thank you! 🙂

@alexdemoura9972 3 жыл бұрын

In Portuguese: "Por que simplificar, se se pode complicar?" 😁

@TheTKPizza 3 жыл бұрын

Warum einfach, wenn es auch kompliziert geht? - German

@morgard211 3 жыл бұрын

Proč dělat něco jednoduše, když to jde složitě? - Czech

@agfd5659 3 жыл бұрын

@@morgard211 přesně jsem to teď chtěl napsat :D

@drpeyam 3 жыл бұрын

I love how we all have the same saying 😂

@souhilaoughlis5832 3 жыл бұрын

''ⴰⵢⵖⴻⵔ ⴰⴷ ⵜⵏⴻⵔ ⴷ ⵍⵉEⴰ ⵎⴰ ⵏⴻⵣⵎⴻⵔ ⴰⴷ ⵜⵏⴻⵔ ⴷ ⴰⵙⴰⵡⴻⵏ ''In Tamazight

@colonelburak2906 3 жыл бұрын

I'm a PhD student in applied maths, and I still learn so much from your videos, Dr. Peyam! The French saying we have in Sweden as well: Varför göra något enkelt när man kan göra det svårt?

@muratkaradag3703 3 жыл бұрын

11:16 I didnt know that you speak german ;) Lots of love From Germany Berlin!. THanks for showing your Magic to us

@plnmech 3 жыл бұрын

You went way over my head but one thing I learned is that you are afflicted with the same disease that I am, you are left handed.

@integralboi2900 3 жыл бұрын

Leibniz: Am I a joke to you?

@marinmaths3826 3 жыл бұрын

Lmaoooo

@itswakke 3 жыл бұрын

The most trouble I have with these is: how do I decide where the “a” term goes?

@drpeyam 3 жыл бұрын

It’s an art, there’s no reason why the a should be at a specific point

@vai_-cn9br 3 жыл бұрын

That was beautiful. Thanks for sharing this sir :)

@channalbert 3 жыл бұрын

By instinct, maybe biased by "Feynman" in the title, I tried to find the most direct way to apply Feynman's trick, and I came up with multiplying outside the integral by e and inside by e^-1. The latter, together with e^(-x^2) in the integrand form e^[-(1+x^2)] and you can now see an easy parametrization to do the trick: e^[-a(1+x^2)] with the limits from 1 to infty.

@spintwohalves 3 жыл бұрын

Another wonderful video. Thanks Dr.πm!

@alphaglucopyranose6928 3 жыл бұрын

Or just use residue integral and integrate on a semicircle on the upper half plane. The function has a simple pole at z=i. Plug in z=i into 2πi*e^(-z^2)/(z+i). We get πe.

@theelk801 Жыл бұрын

which is incorrect

@joaquingutierrez3072 3 жыл бұрын

Nice video!!! Thank you!! Amazing technique

@chriswinchell1570 3 жыл бұрын

Hi Dr. Peyam, I accept your challenge to make things worse: 1. Expand 1/(x^2) in a series and integrate against guassian function. You get an infinite sum of alternating even moments. (A sum of gamma functions, I think.) 2. Expand the guassian and integrate against the Cauchy distribution so you get a sum of moments of Cauchy. 3. Find the integral of the inverse Fourier transform of the convolution of guass and Fourier transform of Cauchy (It’s something like e^-abs(x)). I wonder if this works out better than it sounds. 4. Things aren’t always difficult, remember: toujours aimer le vent qui leve les joupons.

@marinmaths3826 3 жыл бұрын

Amazing Dr Peyam. I love Feynman’s technique. Very cool

@Lupercus-ht1xt 11 ай бұрын

You're an awesome professor. I laughed a lot and the problem was interesting. Thanks

@CoffeeTroll 3 жыл бұрын

1:13 I love this saying ❤️

@adritobiswas1982 2 жыл бұрын

That x square + 1 in the denominator was looking sooo delicious.

@triton62674 3 жыл бұрын

German, French *and* Spanish?! Surely you're joking Dr. Peyam??

@umerfarooq4831 3 жыл бұрын

The title of the video got me lol the equation was amazing as well

@ny6u 3 жыл бұрын

beautiful 👏🏻👏🏻👏🏻

@mertaliyigit3288 3 жыл бұрын

I love how bottom left of the board dances on 720p60

@drpeyam 3 жыл бұрын

I know, it’s so weird it does that 😂

@vivelesport8197 3 жыл бұрын

Best and fast... I remember a long bad proof in university....

@joeremus9039 2 жыл бұрын

Thanks. I really enjoy your videos.

@bernardlemaitre4701 3 жыл бұрын

jolie démonstration !

@thomasborgsmidt9801 3 жыл бұрын

Hmmm... the gauss'ian integral is so very close to being just a constant.

@maxsch.6555 3 жыл бұрын

I tried to solve this integral by myself. I introduced the factor e^(-a(1+x^2)) in the integrand and used feynmans trick. I defined f(a) to be the integral from 0 to inf of e^(-x^2)/(1+x^2) * e^(-a(1+x^2)) dx. Notice that the integral we want is 2f(0). For f'(a) i got -sqrt(pi)/2 * e^(-a)/sqrt(1+a). To get the original function f(a) I integrated from inf to a. I could express f(a) using the incomplete gamma function: f(a) = sqrt(pi)/2 * e * gamma(1/2, 1+a) So the original integral is 2f(a) = sqrt(pi)*e*gamma(1/2, 1) If we use the relation between the incomplete gamma function and the error function we get the same answer as in the video. :)

@MrRyanroberson1 3 жыл бұрын

Erf(1) is such an underrated constant and really puts things into perspective when working wuth statistics. Erf(1)*2 many people lie withing just one standard deviation of the average, and so on, as a fundamental definition.

@peterdecupis8296 2 жыл бұрын

This brilliant example paradoxically highlights that the success of Feynman method is critically dependent on a "fortunate" guess of the parametric generalization of the integrand function.

@uxueanderealdazabal128 2 жыл бұрын

Gracias por sus vídeos... Manos a la obra

@ahmedgaafar5369 3 жыл бұрын

i agree this was very beautiful integral and astonishing result.

@yaskynemma9220 Жыл бұрын

The answer almost says "perfect"

@shreyashegde4032 2 жыл бұрын

Your voice feels as if my friend is talking to me😄. Love your presentation

@quinktap 3 жыл бұрын

Belief has no function here

@AstroB7 3 жыл бұрын

As a French Canadian person, I found your accent wonderful !! Did you take any french class ?

@drpeyam 3 жыл бұрын

I went to a French Lycée :)

@AstroB7 3 жыл бұрын

@@drpeyam oh that’s fantastic, french is a very hard language ! It’s always nice to see your great content ! Keep up the good work !

@luna9200 3 жыл бұрын

Nice application of this that I thought of: If you do the same thing but with sin(x^2)/(1+x^2), you differentiate twice and use the Fresnel Integrals, and you get a second order ODE! A nice answer comes out if you're willing to use the Fresnel S and C functions.

@bprpfast 3 жыл бұрын

Love from SoCal

@drpeyam 3 жыл бұрын

🏝🏝🏝

@paulg444 3 жыл бұрын

The first thing I would do is bound that integral by putting x=0 in the denominator term (1+x^2)=1. .. that way I have a quick relation with the ERF function, proceed then to solve and then I double check that my solution does not exceed the upper bound. Else Ive got so many moving parts that I dont know if Im right at the end or if I made a few teen mistakes !

@LucPatry 3 жыл бұрын

mathematics done beautifully

@KarlMarX_93 3 жыл бұрын

Awesome! Great video before going to the job. 😂

@ericpham8205 3 жыл бұрын

If use i^2 as -1

@quinktap 3 жыл бұрын

Is the error function complimentary? What is important here? Understanding Euler's equations or perhaps his thoughts on expressing his understanding?

@chemistrychamp3369 Жыл бұрын

Thanks amazing integration. I Never really know using this method, when u integrate for da how to put the limits of the integrale and why…

@emiliomontes2043 3 жыл бұрын

¡Manos a la obra! , cordiales saludos Dr. U should make videos talking bout' Fourier transform !!! I love your work :)

@AndDiracisHisProphet 3 жыл бұрын

11:10 es geht eher an die Perven, oder? :D

@jesusalej1 3 жыл бұрын

There is always beauty in maths!

@g-smith4466 3 жыл бұрын

Actually what is the meaning of " why do something simply, when it can be done in a more complicated way?" It hits home somehow, but I am not sure I get it right.

@mlgswagman6002 Жыл бұрын

Nice, took a similar approach but it doesn't go through the differential equation route. Let I(a) = integral from -inf to inf of (e^(a(-x^2-1)) / (x^2 + 1)) Note that e*I(1) is the desired integral. Differentiating under the integral now completely destroys the denominator. I'(a) = -sqrt(pi)/(e^a * sqrt(a)) Now, just directly solve for I(a) using that lim as a -> inf I(a) = 0. You can do some trickery with integration by parts and a substitution to get the same result you did.

@peterchindove7146 2 жыл бұрын

Brilliant!

@the_nuwarrior 3 жыл бұрын

What criteria should be used to know where to enter the parameter?

@drpeyam 3 жыл бұрын

Pure guess

@justjacqueline2004 3 жыл бұрын

Ballet in a math tutu.

@YWanek-ft2fs 3 жыл бұрын

Es geht mir auf die πERF(en). Haha greetings from Germany 🇩🇪

@acac7353 3 жыл бұрын

Manos a la obra!! Jejeje... Saludos desde Andalucía (España)

@superkarnal13 3 жыл бұрын

Bellísimo

@Idtelos 3 жыл бұрын

And don't call me Shirley.

@karabodibakoane3202 3 жыл бұрын

Is Dr. Peyam really doing calculations in the kitchen? I thought the kitchen was meant for cooking, but no, I was wrong.

@h4z4rd28 3 жыл бұрын

In Slovak: Prečo to urobiť jednoducho, keď sa to dá urobiť zložito? :)

@shibhanlalpandita6975 3 жыл бұрын

Explain the problem. Y can't you put it integral where numerator is 1?

@bruwayn2158 2 жыл бұрын

we can also solve the 1st order differential equation : f'(a)-2a f(a) = -2square(Pi)

@yashagrahari 3 жыл бұрын

Nice integral

@petervinella5545 2 жыл бұрын

Feyman’s idea is very similar to the calculus of variations as discussed in Evan’s PDE textbook

@joniiithan 2 жыл бұрын

your jokes are hilarious

@DeanCalhoun 2 жыл бұрын

this problem just screams residue theorem to me

@dihinamarasinghe9278 Жыл бұрын

Can this be solved using complex analysis? If so can you upload a video on it too

@drpeyam Жыл бұрын

Maybe, you come up with a solution :)

@alexcaesar5377 3 жыл бұрын

What a easy equation! It is a piece of cake.

@BaoNguyen6742 3 жыл бұрын

If you substitute a as -1 can it still give the same answer like this

@stochasticxalid9853 Жыл бұрын

Pourquoi faire ça si on peut faire compliqué ? Vamos a la obra ! Delicious Pi ! Es geht mir auf die Nerven! ....Love u doc.❤

@ilovephysics350 3 жыл бұрын

Set 1 to "t". That would be perfect.

@PhilesArt 3 жыл бұрын

I'm french and I was confused when u started talking french xD

@DynestiGTI 3 жыл бұрын

Can all integrals be turned into differential equations like this?

@brendanlawlor2214 Жыл бұрын

fantastic integral wish Peyam was my college teacher 🔥😃

@meiwinspoi5080 3 жыл бұрын

may post a video on numerical calculation og error function

@tgx3529 3 жыл бұрын

Dr Peyam, please , what is the result for this integral.We are not able get primitiv function for integral exp-x^2 from -infinity till a. I am not sure, mabye I can take origin integral as e* integral from (exp -(1+x^2))/(1+x^2), and then calculate integral f(a) =exp (-a(1+x^2))/(1+x^2), there will be cancel (1+x^2) for f'(a).... , but I am not sure, if this way is OK

@guill3978 3 жыл бұрын

Is that constant, the result of the integral, a transcendental number?

@adityaekbote8498 3 жыл бұрын

Hey what whiteboard do you use?

@andresvivas2814 3 жыл бұрын

Me, whatching this when I can Hardy Solve a simple integral using U substitution...

@alecapin 3 жыл бұрын

Dicho por el Negro Olmedo: “¡Si lo vamo’ a hacer, lo vamo’ a hacer bien!” :D

@guill3978 3 жыл бұрын

One question, is the number Ei(1) a transcendental number?

@shanmugasundaram9688 3 жыл бұрын

Very nice presentation.

@ismailaitabdelkarim7164 Жыл бұрын

6:13 you made things complicated. It is a differential equation of order 1 which is very simple to solve.

@drpeyam Жыл бұрын

@arupabinash2263 3 жыл бұрын

I love this method

@alireza-vq6ul 3 жыл бұрын

It is useful. thanks.

@physicsboy3108 3 жыл бұрын

Can be solved by using and Gamma or Beta function!!!🤩

@alwysrite 3 жыл бұрын

when you did f(a) why did you only put 'a' in numerator and not denominator? Denominator also had an 'x' term?

@drpeyam 3 жыл бұрын

It would make it more complicated! Putting a once is simple yet already solves our problem

@waleedalzamil2228 3 жыл бұрын

Can you please recommend a good reference to study math...?and to understand more please...

@aneeshsrinivas9088 7 ай бұрын

I’m surprised nobody brings up int_{-infty}^{infty} e^(ix)/(x^2+1)dx=pi/e as beautiful identities involving fundamental cosntants.

@psjt9230 3 жыл бұрын

What an idea

@harrisonford4103 3 жыл бұрын

In Chinese it is translated as following: 如果你可以做更难的事情，那么为什么还要做简单的呢？

@thomasfritz8174 2 жыл бұрын

And the function f(a) actually has a name: it is the Faddeeva function w(iz) , well exactly it is f(a) = w(ia). I deal with the Faddeeva fct. very often.

@drpeyam 2 жыл бұрын

Interesting!