great video series, understandable explanations ! :)
@HadisMalekpour-zm6xy3 күн бұрын
You explain everything really well, even better than my teacher. 😁 Thanks 🙏
@nuclearrambo31674 күн бұрын
I would use fourier transform
@individual1st6485 күн бұрын
ive been trying for a few weeks to blindly get a line segment rotating at one endpoint, but i just couldn't figure it out (i knew it had to do with y=tan(a) but i didn't know what bounds i should insert) then i figured, maybe i just look up how to make a clock! and there we have it (thanks!)
@Halleluyah8313 күн бұрын
Wow😮
@KSM94K13 күн бұрын
Maybe I didn't understand it very well, why the Psi 1 and Psi 2 don't they cancel here? In the residue theorem proof something like that used to be cancelled
@Saqouro15 күн бұрын
This is really helpful. Thanks
@KSM94K15 күн бұрын
You're seriously underrated, nice explanation
@rashmipandey920117 күн бұрын
Nice
@physicstube812417 күн бұрын
I never thought the job could be done this easier 🎉
@marrom680817 күн бұрын
If i want to reflect relative to what a is, i.e. if y = x then a = 1 and then we have ((1, 0), (0, -1)), does this mean ((1,0),(0,-a)), so i.e. a = 2 means ((1,0), (0, -2)) ?
@txikitofandango19 күн бұрын
This is so wholesome
@luaiderar660026 күн бұрын
I think I saw you outside melbourne central. Apparently you're my friends calc 1 tutor at unimelb lmaoo.
@qncubed325 күн бұрын
:O
@kianheus248726 күн бұрын
Hi, loving the videos so far! I also have a question: at 15:08 you say to simply assume the existence of a C^k atlast, but how does that work in practice? Surely you can't just assume one to exist out of the blue.
@weselise248929 күн бұрын
you saved me thank you
@nicholasnick6499Ай бұрын
Great video
@meruem6022Ай бұрын
Hello @qncubed3 , I wanted to know if you have any bibliographical reference on solving the Basel problem using that method
@user-wu8yq1rb9tАй бұрын
I missed you..... Just tell me where are you??!!
@buddychumpalfriendhomiebud9242Ай бұрын
For a simpler method, substitute u=1/x , move around some things and replace u with x and you'll see that the integral is equal to its negative
@nizogosАй бұрын
Can't you say that 1/abs(z^4+1) <1/abs(z^4) ? Simply because the right fraction has a smaller denominator, without going into triangle inequality stuff
@qncubed3Ай бұрын
This only works if z^4 is positive. If z^4=-2 for example then you would get 1<1/2
@natnaelayele8388Ай бұрын
The most helpful video on the topic in the entire KZfaq
@anshuagrawal7364Ай бұрын
Amazing!!! Could you please continue posting more videos on Manifolds? 🙏🙏🙏
@MatvejPetrovicBronsteinАй бұрын
Great work!Ad maiora semper...
@smftrsddvjiou6443Ай бұрын
funny, but useless.
@iphilip1Ай бұрын
Can you still show the analyse program that you have in your CAS link in you description or comments section
@mai1906Ай бұрын
what a loser
@xxux1Ай бұрын
Seems the link is not working😅
@iphilip1Ай бұрын
yeah
@itisajem8645Ай бұрын
Interesting the result looks like the reflection formula for the gamma function but with 1/n
@ayandas8299Ай бұрын
Bro you are the goat
@nuclearrambo3167Ай бұрын
cOnSeRvAtIvE vEcToR fIElD
@user-qk5ce5rr7uАй бұрын
thank you, I ask you one more series expansion of tan(z) 31:17 I want to learn principle of this series of tan(z).
@Mouse-qm8wnАй бұрын
Thanks a lot for great videos ❤😊. Can you please give examples of when to use this in real life or is it only relevant in general relativity?
@Caller8194Ай бұрын
good video
@user-qk5ce5rr7uАй бұрын
Great!,thank you for your lecture, would you explain series of tan(z)(31:17)? I don't know result of this series.
@LatifaEssaАй бұрын
Vraiment c est super Je sais pas comment je peux remercier
@hyeonsseungsseungiАй бұрын
5:34 It's easier not to reduse (s+i) or (s-i) because (i+i)^4 or (-i-i)^4 is 16, so thier denominator becomes rational.
@williammartin4416Ай бұрын
Thanks!
@SarmadnessАй бұрын
Great explanation, thanks!
@NeptoidАй бұрын
The Jacob Collier of Physics
@GingerMathАй бұрын
After two years I found this video and it makes sense!! Thank you <3
@anandarunakumar6819Ай бұрын
Great explanation. Nice teaser to conformal technique.
@matejcataric2259Ай бұрын
This channel rocks!! Greetings from Croatia.
@PlayingcivilizationАй бұрын
Much easier with double integration I = \int_0^\infty \int_a^b sin(tx)/x dt dx = [using Foubini] = \int_a^b \int_0^\infty sin(tx) / x dx dt = \int_a^b pi/2 dt = pi/2 * (b - a)
@MuhammadYogavasАй бұрын
Is it possible to prové it with only gamma function ? Because our Professor didnt teach us the Beta function yet
@richard_larrainАй бұрын
0 is not a pole of logarithm
@PlayingcivilizationАй бұрын
Hi Could you please do a video about integral from 0 to infty of ln(x)/((x+1)*x^{3/4}). I was trying to solve it using complex integration but failed to find proper contour:(
@devanshgupta39352 ай бұрын
My final today had this exact integral. I did kind of figure out that we need to take a Gaussian integral to come to the result but used some QUESTIONABLE mathematics. Wish I'd watched this vid earlier. Anyways this channel is absolutely amazing! Please keep it up.