Відео

Problem 1's solution doesnt make sense to me. Clearly we can see that for any t > 0 integral_{-t}^t f(x) dx, where f is defined by problem 1, is given by 2 * arctan( (t^3-4t) sec t ) , and as improper integral given in the question is simply taking the limit of t as t goes to infty, we get the answer here dne as lim_{t -> infty} sec t does not exist (dne, oscillates from -infty to infty). Thus by definition of improper integrals, this integral is undefined. b. I'm not sure about the adhoc graphical method you used to find the answer as -3pi. c. You can always make the substitution x = u(t), as long as u is continuously differentiable on the domain. It doesnt matter if 'u' is increasing or 'decreasing' or both.

Ur a legend, searching the perfect video for hours and really appreciate it.

you sound like you use ":3"

saw this on instagram tried doing trig identities first then partial fraction then tried doing summation gave up and ended up here because I have class at 8 am

This is one of the most enlightening math videos I have ever watched in line with fourier transforms and p-adic numbers.

I really love this video ❤❤❤❤❤

Thank You

my brain is gonna explode

Dude, I literally can't read your handwriting. I'm sorry, I'm not trying to be mean. Please, if you managed to improve your handwriting it would make this so much better.

At 9:15 I think you can alternately turn the sum into a product inside the log, then use euler's factorization of the sine function ->account for the lower bound being 2 instead of one, and take a limit to get the same answer

This video popped up again on my UA-cam, and I really enjoy seeing this monster integral destroyed by Feynman's technique. Good work once again!

Excellent

How do I check my answers?

There's a formula for this using just the x and y intercepts and extremities of the ellipse we use a lot in elliptic polarization so never really thought about this till now. I don't see anywhere else this could be used apart from solid state physics and transmission optics. Was surprised at the solution being so involved and requiring factorisation, but a nice approach nonetheless. I'd recommend this video to a few students but not my colleagues because of the handwriting(they would mock me to oblivion because of it). If you could make a typed version for the video it would be nice but going on the same topic again would be a bit too much to ask. Anyways good video and best wishes for your future👍

Just saying: integration by parts is pretty OP. Just ask Euler/Basel problem/Riemann hypothesis.

Some of these looked approchable while others just made me look and cry

The title is a lie, I didnt get the factorial of zeta(4) :(

I swear to god I HATE CHEMISTRY!!!!!!!!!!!

Can you tell me where we learn about the lapse transform? Harmonic Analysis? I've watched a few videos on it at khan academy but I want to go deeper.

1. In the Gamma(a+bi) equation, the convergence condition (a>0) must be mentioned. 2. Gamma(i+1) is not equal to i Gamma(i) in the original equation because of a=0. Gamma(x+1) = x Gamma(x) is valid only in the x>0 condition.

Your writtin' absolute trash, write properly, what you write ain't visible clearly

Sahih al-Bukhari 230 Narrated 'Aisha: I used to wash the semen off the clothes of the Prophet and even then I used to notice one or more spots on them. حَدَّثَنَا قُتَيْبَةُ، قَالَ حَدَّثَنَا يَزِيدُ، قَالَ حَدَّثَنَا عَمْرٌو، عَنْ سُلَيْمَانَ، قَالَ سَمِعْتُ عَائِشَةَ، ح وَحَدَّثَنَا مُسَدَّدٌ، قَالَ حَدَّثَنَا عَبْدُ الْوَاحِدِ، قَالَ حَدَّثَنَا عَمْرُو بْنُ مَيْمُونٍ، عَنْ سُلَيْمَانَ بْنِ يَسَارٍ، قَالَ سَأَلْتُ عَائِشَةَ عَنِ الْمَنِيِّ، يُصِيبُ الثَّوْبَ فَقَالَتْ كُنْتُ أَغْسِلُهُ مِنْ ثَوْبِ رَسُولِ اللَّهِ صلى الله عليه وسلم، فَيَخْرُجُ إِلَى الصَّلاَةِ وَأَثَرُ الْغَسْلِ فِي ثَوْبِهِ بُقَعُ الْمَاءِ‏

I actually saw this in an old "Higher Algebra" book by W.L. Ferrar. Its in the first chapter on Finite Series and is called the "Method of Differences" for calculating certain sums. I Didnt know it had developed into this broader theory.

anyone else struggling with #2?

Thank You

One way to speed up P5 is to just use the generating function of the fibonacci numbers, which is pretty well known. It is 1/(1-x-x^2) and subbing x=1/4 gives 1/(1-(1/4)-(1/16))=16/11, and multiplying by 1/4 gives 4/11 (as we have (1/4)^(n+1) in the sum, not (1/4)^n). I would guess this is what the contestants used to evaluate the sum

at 8:26 minutes, why are you dividing the two equations to eliminate the 2 and why do you then not consider the negative in the second equation after eliminating the 2. I thought that negative is meant to change the signs inside the bracket of that second equation

Thanks for your video. It helped me to getting a grasp of the idea of analytic continuation. I have one question though: for the complex function, you transform the gamma integral to integrate t^(a-1) e^(-t) cos(b ln t) dt + i integrate t^(a-1) e^(-t) sin(b ln t) dt (I assume you have a typo in your video, as you write x instead of t for the argument.) This equation, however, gives me some headache: how do you want to evaluate cos(ln t) or sin(ln t) as t -> 0 <=> ln t -> -infty?

Why going in imaginative numbers . Cant we just manipulate the numerator and keep factorising out the denominator?

gamma(1/12) gamma (5/12) can be written as sqrt2 3^(1/4) gamma^2 (1/4) Can you prove it?

Hi. Is there any version of the chain rule in discrete calculus?

very good！

Thanks to the subtitles feature.

I tried the xlnx/(x^5+1) integral using keyhole contour and now I need to solve the x/(x^5+1). Shall I use another contour integral for that?

Hello, are you using Obsidian to write?

Sorry, but your writing is abysmal.

maybe you implied it, but for the box contour, its purpose is to enclose just one or a small number of poles (minimizing number of residue calculations) of a function that has a large number of poles, that is, a function that an infinite semicircular half plane contour would require the calculation of a large number (possible infinite) of residues. In other words, a box contour is more suitable than an infinite semicircular half plane, for a function having many, many poles. Really like your videos and learn a lot.

two items: 1) it can be very confusing to use x as the variable for parameterization: recommend that variable t be used instead. 2) did not understand branch cut choice at end of video

too fast💀💀

How do I approach the sinx/x integral? I tried semicircle, but cannot show that inegral over Г is 0

Interesting. Today is the first time I've seen integration of an inverse function.

Please write clearly I could not able to understand

this is so cool!! Thanks a lot!

Is there a way to solve it using partial fractions?

personally, i'd put glasser's master theorem in b tier if not one more thing: it applies for infinite series too, as long as it converges, so with taking limits we can also change x+tan(x) to just x. i might've learned it from your channel tho lmao

Thanks

The logarithmic form of Stirling's approximation for large n is super useful and succinct, especial in statistical mechanics. log(n!) ~ n*log(n) - n

Any resources or reference books for this topic?

Tysm

I solved the first integral like this. S = Sum (-1)^k /(2k+1)^3 k from 0 to inf and I is the integral. Did the same thing as you to get I = 8S. Then I = 2 Int (x^2)/cosh(x) from 0 to inf, substitute x = arcosh(t) = ln(t+sqrt(t^2-1)), dx=(t+sqrt(t^2-1))/(t^2-1+t*sqrt(t^2-1)) dt. So I = 2 Int (ln(t+sqrt(t^2-1))^2)*(t+sqrt(t^2-1))/(t*(t^2-1+t*sqrt(t^2-1))) dt t from 1 to inf. Now t+sqrt(t^2-1)=u, (u^2-1)/2=t^2-1+t*sqrt(t^2-1), t=(u^2+1)/(2u), dt=(u^2-1)/(2u^2)du. So I = 4 Int (ln(u)^2)/(1+u^2) du u from 1 to inf. Split the integral like int from 0 to inf - int from 0 to 1. The Integral from 0 to 1 is 2S, power series for 1/(1+u^2) and Int x^m (ln(x)^n) formula. The Integral from 0 to inf: f(t)= Int (u^(t-1))/(1+u^2) du from 0 to inf. Write 1/(1+u^2) as Sum (-1)^k/k! * gamma(k+1) * (-1)^k * cos(pi*k/2) u^k, use ramanujans master theorem, f(t)=pi/(2sin(pi*t/2)) calculate f''(t=1)=pi^3/8, so I = pi^3/2-8S and I=8S, we get S=pi^3/32 and I = pi^3/4