Solving a DIFFICULT integral from My Hero Academia
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- เผยแพร่เมื่อ 14 พ.ค. 2024
- In this video, I go through the solution to a challenging definite integral featured in the manga My Hero Academia. I originally filmed this almost 2 years ago but got distracted by college, so I'm glad that I had some free time to edit and publish it. If you're confused by the solution, please feel free to ask questions in the comments.
Based on this Redditor's solution: / self_math_problem_from...
"Vibing Over Venus" Kevin MacLeod (incompetech.com)
Licensed under Creative Commons: By Attribution 4.0 License
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Instead of developping (1+u^2)^5, I recommend a second variable change t = 1 + u^2, or choosing directly u = cosh(x) to start with. Ifyou do so, you end up with int/ (t-1)*t^5 from t=1 to 2, which is much less daunting. Thanks for the video!
Yeah I think when I was filming this I was trying to figure out how to do something like that but didn’t get around to it, that’s definitely a better approach.
@@emachine003 Also, you could've just used Newton's binomial formula lol
don't forget the 1/2 from the derivative.
I swear you have to be a math Olympian to come up with that!
@@Sunkem1Not6Hacks trust me, being able to a calculus substitution is unfortunately not enough to become a math olymipian.
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)
Yea I was thinking bout this
fr lol, i thought leaving 1 sinh and expressing the sinh^2 with the pythagorean identity was like basic calc 2
yeah I did it exactly like this
Wow, it is a really simpler method.
just use the exponential definitions of sinh and cosh (which is in the original problem) and work it out with fairly ease by using logarithm properties to calculate cosh(ln(1+sqrt(2))) = sqrt(2)
Wow man as a CS student who is also interested in math (weather applied in cs or not), I really enjoy these videos where you tackle hard problems across various media. Plz do it more!
Good integral to do in my head while exhausted and sleepy. Took me 20 minutes and I some how didn't mess up. I went the route of substituting u=coshx because the resulting integrand had less terms. But in the end, I don't think that saved much time because I still had to plug in both bounds and not forget negatives.
The algorithm has done it's job and found an intersection between math, anime and coding. I'm a not-so-good coder so I look forward to watching your other videos.
Great! Yeah, I’ve done some other stuff about math as well if you want to check that out.
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
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.
Anyone who is aware of the math behind gojo’s divergence and convergence will know that the function inside the bracket is hollow purple
An imaginary technique that doesn't use any imaginary numbers... so satisfying.
At my university's calculus course we used log for the natural logarithm, so maybe that's the reason it's that way.
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.
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)
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.
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.
Yeah in higher math log is generally used to mean ln
Isnt it base 10 by default if the base isnt shown?
@algirdasltu1389 only if you are an engineer lol
being serious the more you go into advanced subjects the more you see the notation log reserved for the natural logarithm.
@@pl412Ln is just easier
Yea it’s kind of a dumb notation to just have log(x) in the first place and assume base just because the base you assume varies depending on who you talk to.
Which it shouldn’t be because it just introduces unneeded confusion and ambiguity
you can do integ (sinh(x))^3 * (cosh(x))^11 dx
= integ sinh(x) * ((cosh(x))^2 - 1) * (cosh(x))^11 dx
= (integ sinh(x) * (cosh(x))^13 dx) - (integ sinh(x) * (cosh(x))^11 dx)
then just do cosh(x) = u and you're good
That's the actual way to solve
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@@farragoprismproductions3337 ?
You don't even need to remember all these convoluted formulae. Just use the binomial expansion and everything cancels out near the end.
Looking forward to the day he will make a video about the dots exercise from assassination classroom
Haven’t seen that, might have to check it out
From cosh²x - sinh²x = 1 one gets sinh²x = cosh²x - 1, so the integrand becomes sinh³x · cosh¹¹x = sinh(x) (cosh²x - 1) cosh¹¹x = cosh¹³x sinh(x) - cosh¹¹x sinh(x).
Since the derivative of cosh(x) is sinh(X), the indefinite integral becomes:
integral(cosh¹³x sinh(x) - cosh¹¹x sinh(x)) dx = cosh^14(x)/14 - cosh^12(x)/12 + constant
The fun part is evaluating the definite integral. lol
This seemed like the clearest path to victory to me too.
Definite integral isn’t hard either since if you use the definition cosh(x)=(e^x+e^(-x))/2 you can easily see that cosh(ln(1+sqrt(2))=(1+sqrt(2)+(1/(1+sqrt(2)))/2 multiply the top and bottom by 1+sqrt(2) to get
(2sqrt(2)+4)/(2(sqrt(2)+1))
= (sqrt(2)+2)/(sqrt(2)+1) multiply top and bottom by 1-sqrt(2) to get
-sqrt(2)/-1 = sqrt(2)
Then just plug sqrt(2) into the equation and it simplifies very easily of course got to also remember to subtract the cosh(0) terms but cosh(0) is 1 so that’s easy enough
that was a fun watch :)
pretty simple u-sub problem with hyperbolic trig identities before sub
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.
"um actually its pronounced hyperbolic sine and hyperbolic cosine" 🤓
Couldnt you just use integration by parts? Is there something that would block that from happening?
The hyperbolic trig functions i can recognize but those exponents gmfu icl
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
What do you mean? It's still a "dx" at the end. The substitution hasn't occurred yet.
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?
Anime helping students to better understand the mathematics, physics, science 😮
Why tf does bro sound like he about to cry any minute now
Had to do too much calculus
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.
It’s way easier to expand (1+u^2)^5 using the binomial theorem, you can just go directly to the answer that way.
Also if you do the substitution u=cosh(x) instead it’s even easier because the power of (1+u^2) will just be 1
@@deananderson7714 true but can you do cosh(arcsinh(1)) in your head?
@@NStripleseven it’s actually not too hard if you just use the original ln(1+sqrt(2)) because if you look at the definition of cosh(x) which is (e^x+e^(-x))/2 you can see if you plug it in you have cosh(ln(1+sqrt(2))=(e^(ln(1+sqrt(2))+e^(-ln(1+sqrt(2)))/2 and since
e^ln(x) = x and -ln(x) = ln(x^-1)
this simplifies easily to
(1+sqrt(2)+1/(1+sqrt(2))/2 multiply by 1+sqrt(2) on top and bottom to get rid of the fraction to get
(2sqrt(2)+4)/(2(1+sqrt(2)) cancel a 2 and we have (sqrt(2)+2)/(sqrt(2)+1) multiply by 1-sqrt(2) on the top and bottom and this will simply simplify to -sqrt(2)/-1=sqrt(2)
So the answer is just sqrt(2)
@@NStripleseven its pretty easy if you use the given value of ln(1+sqrt(2)) just plug that into the definition of cosh(x) and it will simplify pretty easily. Notice how the definition contains e^x and e^(-x) and those will cancel the ln. You should get a value of sqrt(2) which will be very easy to plug into the antiderivative
@@deananderson7714 fair enough ig but not as easy as 1.
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
dont diss doge like that, its up there i assume as a honorable image as one of the godfathers of memes.
10:08 Is the sentiment still the same currently?
RIP Kabosu :(
You can actually write (sinhx )^3 as (sinhx)((coshx)^2-1) . I solved this in the anime itself .
Hey, Doge just died, show some respect 😢
If you are for real, I feel very sorry. His ability to explain is just gorgeous. He made a lot of afford, I know, he was liked
Solved it in like 3 minutes. Let u = cosh(x). What the hell are you doing lmao
arrogant af
Use sinh^2 = cosh^2-1 or something, and sinh is something like the derivative of cosh. Use sobstitution and solve
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
Wow that’s a lot to type. Good job!
C'mom doge dog is funny man.
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
I just used a power series😅
How would that work?
@@emachine003 easier to work with a few polynomials to approximate it since I had no idea they were actual functions and didnt want to try intergration by parts and it worked good enough
nais video 👍
So the answer is 107/28, right?
Yup!
just expand them out 🗿
Ans = 0.00000005
Theres an easier way using trig substitution
High school level it’s easy
you can do binomial expansion much faster using pascal's triangle!
5th row of Pascal's triangle: 1 5 10 10 5 1, so u have 1 + 5*u2 + 10*u4 + 10*u6 + 5*u8 + 1*u10.
Reasoning: for each product term in the expansion of (1+u^2)^5, we choose either 1 or u^2. Therefore each term is (n choose k)*1^k*(u^2)^(n-k), with k from 0 to 5.
Pascals triangle gives (n choose k) where n is the row and k is the column
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.
When writing logx, it can be in base 10 or base e. Some people just write ln as log, and then log as log10. Since the inside function uses exponential functions in base e, it would be silly to suggest that the log function is in base 10 for this function.
In a math context, you _always_ assume base _e_ unless otherwise stated.