Відео

dude are you okay? you sounded like you were on the verge of tears throughout this 😭

lol, hearing how would a middle schooler solve this problem while being a middle schooler who can solve this problem 💀, anyways, GREAT VIDEO, you made it sound ez, keep up the great work

7:40 a simple argument would be to let P be the set of points within this volume, P1 and P2 are the subsets colsest to point A and B respectively. with P being the unione of P1 and P2. |P1|<= |P2| w.l.o.g. and because of A and B not being the same |P1|>= |P2| as well. Therefore |P1| = |P2| holds. might be nicer to proove by contradiction but whatever

this is a long-solved problem thats common to see in early x-ray crystallography coursework. cool to see an anime mention stuff i did in grad school! and good job coming up with the 3D distance formula solution!

In the anime, the dude didn't lose points because of his proposed solution but rather because he didnt finish the calculations and hence didnt get the final answer. He got partial credit for the calculations he managed to finish til then. Bro was just solving the polyhedron calculation you mentioned in his math exam, and didnt manage to finish it

A problem I have with the first solution is that it assumes no empty space... which looking at the diagram even in the anime there seems to be some. If this space were accounted for the solution would look closer to around 0.34*a^3 instead of 0.5*a^3. (sqrt(3)*pi/16)a^3. Great video regardless, thanks

nice, unneeded info at 3am when i need to finish homework. Thanks for the interesting video! I basically skipped the explaining when reading it.

If you guys don’t want spoilers, check out the spoiler-free version here: ua-cam.com/video/ZbwZmbnc-UU/v-deo.htmlsi=zAygGV2-kppyFKFQ

Great video! Your animations and models made the problems so much easier to understand

Comment for the algorithm 👍

the problem is very unpleasant. the most elegant exercises are those in which some trick or realization is what’s most challenging and vital. the problem here are the pesky calculations, I always find those cumbersome. the hyperbolic trig subs here were a great example of that which I’m talking about. I suggest you try to solve similar exercises (try it with less ridiculous exponents tho) using complex integration techniques.

You don't even need to remember all these convoluted formulae. Just use the binomial expansion and everything cancels out near the end.

just expand them out 🗿

I just used a power series😅

people in the comments are popping off, so i'm just gonna take a time to explain what log is for anyone wondering, since he just said "it's ln" definition of a logarithm is as follows: log_a(b) = c -> a^c = b. (where a is called the "base" and b is the "argument") and that's it, it's basically just the inverse function of exponentials. ln is just a special log, which is base e. ln(x) = log_e(x). it's good/special because of calculus stuff.

C'mom doge dog is funny man.

Anime helping students to better understand the mathematics, physics, science 😮

My resolution: 1)(e^x-e^-x)/2=sinh(x); 2)(e^x+e^-x)/2=cosh(x); 3)cosh²(x)=1+sinh²(2); 4)d/dx(sinh(x))= cosh(x); 5)d/dx(cosh(x))= sinh(x); After some substitutions: ʃ[sinh³(x)cosh¹¹(x)]dx = ʃ[sinh³(x)cosh(x)(cosh²(x))⁵]dx = ʃ[sinh³(x)cosh(x)(1+sinh²(x))⁵]dx. Let u=1+sinh²(x); u-1=sinh²(x); du= 2sinh(x)cosh(x); ʃ[sinh³(x)cosh(x)(1+sinh²(x))⁵]dx = ʃ[sinh²(x)sinh(x)cosh(x)(1+sinh²(x))⁵]dx = ʃ[(u-1)(u)⁵]/2du = ½ʃ(u⁶-u⁵)du =u⁷/14-u⁶/12 = [1+sinh²(x)]⁷/14 -[1+sinh²(x)]⁶/12 from 0 to 1+√2; sinh(0)=0: [1+sinh²(0)]⁷/14 -[1+sinh²(0)]⁶/12 = 1/14-1/12=6/84-7/84=-1/84; (e^x-e^-x)/2=sinh(x) -> sin(x)= [e^log(1+√2)-e^-log(1+√2)]/2 = [e^log(1+√2)-1/(e^log(1+√2))]/2 = [(1+√2)-1/(1+√2)]/2; 1/(1+√2) ×(1-√2)/(1-√2)=-(1-√2). [(1+√2)-1/(1+√2)]/2 = [(1+√2)--(1-√2)]/2 = [(1+√2)+(1-√2)]/2 = 1=sinh(1+√2); [1+sinh²(log(1+√2))]⁷/14 -[1+sinh²(log(1+√2))]⁶/12 = [1+1²]⁷/14 -[1+1²]⁶/12 = 2⁷/14-2⁶/12 = 2⁶/7-2⁵/6= (3×2⁷-7×2⁵)/42= 2⁵(3×2²-7)/2×7×3 = 2⁴×5/7×3 =80/21 Finally: 80/21--1/84= 80/21+1/84= 320/84+1/84= 321/84 = 107/28

Looking forward to the day he will make a video about the dots exercise from assassination classroom

dont diss doge like that, its up there i assume as a honorable image as one of the godfathers of memes.

You sound akward as fuck, but i really liked the video

10:08 Is the sentiment still the same currently?

High school level it’s easy

Why tf does bro sound like he about to cry any minute now

the actual integral in the show is much harder though?? maybe its a mistake, but on the show the expression that's cubed has a plus instead of a minus, making it cosh ^14... which is much harder to solve. i gave up after looking up some of the hyperbolic identities that i never learned in my calc 2 class and can we talk about the fact that they're supposed to be FRESHMEN in high school doing this?

Hey, Doge just died, show some respect 😢

bruh just substitute e^x+e^-x =t then dt=(e^x-e^-x)dx then it just becomes a simple polynomial with 2 terms

Great video all in all! Generally for the hyperbolic trig functions, they are pronounced: sinh = "sinch" cosh = "cosh" (exactly how it looks like you would say it) But they are weird functions anyways.

Originally I was gonna say u=coshx with a trig identity makea this integral a lot quicker but Ive seen a couple of comments already. I think for any integral involving exponentials and trig or hyperbolic trig you can make your life a lot easier by looking at identities. The most important Identity you can possibly take the time to understand with these 3 kinds of functions is e^iα=cosα+isinα and e^α=e^-i(iα)=cos(iα)+isin(iα)=coshα+isinhα. You can derive any trig identity from the euler identity so it makes integration a lot easier. But of course you dont touch some of these things until complex analysis.

Couldnt you just use integration by parts? Is there something that would block that from happening?

My solution without watching the video: Since the quadrilateral is cyclic, angle ADC is 120 degrees. Solve for AC using Law of Cosines on ABC: obtain AC=sqrt(21). Then solve for AD using Law of Cosines on ACD. The math works out nicely and we obtain AD=1. Using the triangle area formula (1/2)ab*sin(theta), we can compute the area of each triangle individually. They have areas 5*sqrt(3) and sqrt(3), which adds to 6*sqrt(3) for the total area. Seems too tough for a classroom setting, but it's not a hard problem at the competitive math level.

Use sinh^2 = cosh^2-1 or something, and sinh is something like the derivative of cosh. Use sobstitution and solve

If you do e^2x + e^-2x + 2 = u substitution and take out 2^-15 integral becomes so much more easier 2^-15 * ( integral of u^6 - 4u^5 from 4 to 8)

The hyperbolic trig functions i can recognize but those exponents gmfu icl

Theres an easier way using trig substitution

Here I am sitting with just integration by part (take u = e^-e^-x)^2 and v = (e^x-e^-x)(e^x+e^-x)^11 and you're good to go I just got outta high school not even in college yet ok I dunno hyperbolic trignometry and shid

Solved it in like 3 minutes. Let u = cosh(x). What the hell are you doing lmao

So the answer is 107/28, right?

nais video 👍

Anyone who is aware of the math behind gojo’s divergence and convergence will know that the function inside the bracket is hollow purple

Cool! I didn't know any hyperbolic trig so I just let x = i*θ to have ((e^x - e^-x)/2)^3 = -i*sin^3(θ), ((e^x + e^-x)/2)^11 = cos^11(θ) and then did some integration by parts.

At my university's calculus course we used log for the natural logarithm, so maybe that's the reason it's that way.

log is base 10. ln is base e. They're the same idea / concept, but different scales. log (10) = 1, log (100) = 2, etc ln (e) = 1, ln (e^2) = 2, etc.

6:22 gotta be careful here, you've done some u sub but still have the old bounds so this line is technically not equal to the rest, to fix this, writing "x=..." and "u=..." for a couple lines in the bounds avoids the ambiguity

that was a fun watch :)

There is a cleaner way, rewrite the integrand as : sinh(x) sinh^2(x) cosh^11(x) sinh (cosh^2 - 1) cosh^11 sinh (cosh ^13 - cosh^11) Then you can sub directly u = cosh(x), and you get no messy developments. The tricky part is calculating cosh(ln(1+sqrt(2))) = sqrt(2)

holy fuck this entire video boils my brain, i wish you stopped talking or didnt even make this video in the first place ngl

Since I don't know jack about hyperbolic function, I simply separated one (e^x - e^-x) and wrote the remaining (e^x - e^-x)²=(e^x + e^-x)² - 4 and substituted e^x+e^-x=t which yields 1/2^14 into integral 2 to 2√2 (t²-4)t¹¹dt.

Bro please use OneNote instead of Paint

It’s way easier to expand (1+u^2)^5 using the binomial theorem, you can just go directly to the answer that way.