Is this integral too complex for Feynman's technique???

Рет қаралды 6,518

8 ай бұрын

Advanced MathWear:
my-store-ef6c0f.creator-sprin...
Complex analysis lectures:
• Complex Analysis Lectures
If you like the videos and would like to support the channel:
/ maths505
You can follow me on Instagram for write ups that come in handy for my videos:
maths.505?igshi...
My LinkedIn:
/ kamaal-mirza-86b380252

Пікірлер: 41

@KaRim-fc1sd 8 ай бұрын

Bro i noticed u reaaaaallly love Feynman's technique

@maths_505 4 ай бұрын

@SayedHamidFatimi 😂😂😂

@mcalkis5771 8 ай бұрын

Feynman's technique is literally never boring to watch.

@gesucristo0 8 ай бұрын

Could you do something about Maxwell’s equations? Both integral and differential form

@maths_505 8 ай бұрын

Hell yeah 🔥

@kaanetsu1623 8 ай бұрын

Same integral was given in my class under feynmann tech just instead of cos^2(x) there was x^2. So no complex was needed and its not in our syllabus also . BTW the golden ratio at the end was too goood!! 🤯🤯

@uggupuggu 8 ай бұрын

so cos^2(x^2) or just x^2

@kaanetsu1623 8 ай бұрын

@@uggupuggujust x2

@manstuckinabox3679 8 ай бұрын

2:39 This integral form has become something that has piqued my curiosity for a while now. I remember in last time's case, that is when alfa was equal to zero, I wondered if invoking summation of 1/1+sin^4(x) (cos(x) as well by king's property) would work, based of the equation we can prove using contour integration or the beta function. However, it seems we relied on the same method as last time, which still is mind-blowing to me! 5:01 oh yeah LOL! multi-valued functions go brr! Amazing vid as always my man! keep pumping them up!

@MrWael1970 8 ай бұрын

Very interesting integral. Thank you for your smart plan to solve this type of integrals.

@polpotify 6 ай бұрын

I think this is the first time I have heard swearing on this channel, and the answer was so satisfying and beautiful ❤

@alessandropacco2958 8 ай бұрын

Hey! Nice videos, I was wondering if you wanted to do also some cool probability stuff at some point. I guess there are a lot of really nice problems and tricks to do there too.

@krisbrandenberger544 8 ай бұрын

@ 3:30 The cos x terms in the denominators should both be squared.

@NikitaMelik-Marutov 8 ай бұрын

Hi, incredible youtuber!! Could you tell me what blackboard are you using in your videos?

@TheArtOfBeingANerd 8 ай бұрын

I love how you used phi in the process, and then the golden ratio popped out

@maths_505 8 ай бұрын

I avoid θ at all costs 😂

@robertsandy3794 8 ай бұрын

Only issue for me was that you used phi as an angle and then used phi as golden ratio. Confused me for a second or 2

@jieyuenlee1758 5 ай бұрын

14:47:me memorise that (sqrt5-1)/2=1/y (y=golfen ratio)

@Mayk_thegoat 5 ай бұрын

sir i got a nice way to evaluate this monster , it took me 2 min.

@PopPhyzzle 8 ай бұрын

Sweet maths dude I'm curious what app do you use for your videos? I wanna take my notes in that.

@maths_505 8 ай бұрын

It's Samsung notes. I use an S6 tab.

@PopPhyzzle 8 ай бұрын

@@maths_505 Thanks man!

@pandavroomvroom 8 ай бұрын

all in all, very cool!

@maths_505 8 ай бұрын

Thanks bro

@grumpyparsnip 2 ай бұрын

At 3:28, you are missing squares on the cosines in the partial fraction decomposition. This confused the heck out of me, so just a heads up.

@maths_505 2 ай бұрын

Oh sorry about that mate. My bad.

@biscuit_6081 8 ай бұрын

why did you add +C when it's a definite integral? does that have something to do eith fynman's technique?

@amritlohia8240 5 ай бұрын

He calculated the *indefinite* integral of I'(alpha) - this is separate from the fact that I(alpha) is defined as a definite integral.

@joelchristophr3741 8 ай бұрын

brother where do you get these integrals???? you even create them???

@maths_505 8 ай бұрын

This one's from Micheal Penn and Rizzy Math (Instagram)

@maths_505 8 ай бұрын

The one I'm about to upload is one I made up.

@aryaghahremani9304 8 ай бұрын

why is it necessary to take the limit for the tangent structure? isnt arctan infinity just pi/2

@maths_505 8 ай бұрын

1. Those are "limits" of integration. 2. Infinity isn't a number.

@aryaghahremani9304 8 ай бұрын

@@maths_505 thanks but i meant why turn arctan into the log form when the variable is just being divided by a constant

@maths_505 8 ай бұрын

@@aryaghahremani9304 its a good teaching practice to show things like how such results farrt forward into the complex realm.

@giuseppemalaguti435 8 ай бұрын

I=π/√2(√(√5-1))

@shivanshnigam4015 8 ай бұрын

Amaze Balls

@amitadeshpande8474 8 ай бұрын

This integral is complex enough for Feynman's technique

@Hobbitangle 4 ай бұрын

Terrible solution. I mean the calculation of (√(1+2i)-√(1-2i))/i Definitely the result should be the real number, so squaring the expression , then simplifying it and taking the square root gives us the result (1+2i+1-2i-2√(1+2i)•√(1-2i)))/i²= -2+2√5= 4•(√5-1)/2=4•2/(√5+1)=4/phi Where phi is the golden ratio The answer is π/2•√(4/phi)= =π/√phi