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owl3
United States
Registrace 31. 10. 2023
Welcome! This is a math channel dedicated to mini courses, playlists and RANDOM stuff from various math subjects. Focus will be Calculus, Algebra, Number Theory and more.
Feynman's trick is the MOST POWERFUL FORCE in the universe BY FAR
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Integral practice:
owlsmath.neocities.org/integrals.html
Website:
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#math
#integrals
#integrationtechniques
czcams.com/channels/artWliFdki6px-57oVh6VA.html
Check out my other channel OWLS SCHOOL OF MATH!
czcams.com/channels/AJaLg-yEcvhC_ggEr0Hl6g.html
Integral practice:
owlsmath.neocities.org/integrals.html
Website:
owlsmath.neocities.org
#math
#integrals
#integrationtechniques
zhlédnutí: 476
Video
Forcing the Weierstrass to happen
zhlédnutí 552Před 4 hodinami
Weierstrass Substitution Intro video: czcams.com/video/0D3VLR8Fg6g/video.html&ab_channel=OwlsSchoolofMath Intro video: czcams.com/video/quMwAUJzLTY/video.html&ab_channel=OwlsMath Example problem: czcams.com/video/rZkcobgdWFM/video.html&ab_channel=OwlsSchoolofMath Quiz: owlsmath.neocities.org/Weierstrass Substitution/integral Formulas: owlsmath.neocities.org/Trig Identities and Formulas/trig ide...
Hint: the answer might contain 2024
zhlédnutí 367Před 12 hodinami
Here's the related video I mentioned at the beginning: czcams.com/video/D5kwCC_Pa9A/video.html Check out my other channel OWLS MATH! czcams.com/channels/artWliFdki6px-57oVh6VA.html Check out my other channel OWLS SCHOOL OF MATH! czcams.com/channels/AJaLg-yEcvhC_ggEr0Hl6g.html Integral practice: owlsmath.neocities.org/integrals.html Website: owlsmath.neocities.org #math #integrals #integrationte...
I wanted to use the Standup Maths method but I was delusional
zhlédnutí 1,2KPřed 16 hodinami
Derivation of the formula for reducing the bounds: czcams.com/video/MZw5CKWbx_0/video.html I have no idea if pi^pi^pi^pi is an integer I have no idea how I could even relate that to this integral but it's fun to wish I could somehow use it on this problem! czcams.com/video/BdHFLfv-ThQ/video.htmlsi=mj1J7aEy-rkSsHw_ Check out my other channel OWLS MATH! czcams.com/channels/artWliFdki6px-57oVh6VA....
Can I turn this into an integral???
zhlédnutí 3,1KPřed 21 hodinou
Website: owlsmath.neocities.org Check out my other channel OWLS MATH! czcams.com/channels/artWliFdki6px-57oVh6VA.html Check out my other channel OWLS SCHOOL OF MATH! czcams.com/channels/AJaLg-yEcvhC_ggEr0Hl6g.html #math #mathcompetition
The most time I EVER spent on a thumbnail and it still looks weird!
zhlédnutí 2KPřed dnem
Check out my other channel OWLS MATH! czcams.com/channels/artWliFdki6px-57oVh6VA.html Check out my other channel OWLS SCHOOL OF MATH! czcams.com/channels/AJaLg-yEcvhC_ggEr0Hl6g.html Practice problems: owlsmath.neocities.org/integrals.html Website: owlsmath.neocities.org #math #integrals #integrationtechniques
Short Method vs Long Method
zhlédnutí 1,8KPřed 14 dny
Check out my other channel OWLS MATH! czcams.com/channels/artWliFdki6px-57oVh6VA.html Check out my other channel OWLS SCHOOL OF MATH! czcams.com/channels/AJaLg-yEcvhC_ggEr0Hl6g.html Practice integrals with King's Principle: owlsmath.neocities.org/King's Rule Integrals/king Website: owlsmath.neocities.org #math #integrals #integrationtechniques
Can we do this with a Laplace Transform???
zhlédnutí 563Před 14 dny
Here's the original video without Laplace Transforms: czcams.com/video/Lf4sAJfO8eA/video.html Laplace Transforms playlist: czcams.com/play/PLOvxeHw2nLaySIKdV-QTjiGHHkOx0M_2-.html Practice problems: owlsmath.neocities.org/Laplace 101/laplace owlsmath.neocities.org/Inverse Laplace 101/laplace owlsmath.neocities.org/Laplace Cheat Sheet/laplace cheat sheet Website: owlsmath.neocities.org Check out ...
From the book "Inside Interesting Integrals"
zhlédnutí 559Před 14 dny
Check out my other channel OWLS MATH! czcams.com/channels/artWliFdki6px-57oVh6VA.html Check out my other channel OWLS SCHOOL OF MATH! czcams.com/channels/AJaLg-yEcvhC_ggEr0Hl6g.html Practice integrals with King's Principle: owlsmath.neocities.org/King's Rule Integrals/king Website: owlsmath.neocities.org #math #integrals #integrationtechniques
King's Principle is less than ideal on this one
zhlédnutí 724Před 14 dny
Check out my other channel OWLS MATH! czcams.com/channels/artWliFdki6px-57oVh6VA.html Check out my other channel OWLS SCHOOL OF MATH! czcams.com/channels/AJaLg-yEcvhC_ggEr0Hl6g.html practice integrals with King's principle: owlsmath.neocities.org/King's Rule Integrals/king Website: owlsmath.neocities.org #math #integrals #integrationtechniques
General form of Serret's integral
zhlédnutí 861Před 21 dnem
Check out my other channel OWLS MATH! czcams.com/channels/artWliFdki6px-57oVh6VA.html Check out my other channel OWLS SCHOOL OF MATH! czcams.com/channels/AJaLg-yEcvhC_ggEr0Hl6g.html More on Gamma function: czcams.com/play/PLZza5ZCDgQWxWCjhPQ-rvTgS0LGJYBpRQ.html Practice problems: owlsmath.neocities.org/Gamma Quiz/gamma Website: owlsmath.neocities.org #math #integrals #integrationtechniques
Is this IMPOSSIBLE???
zhlédnutí 2,8KPřed 21 dnem
Check out my other channel OWLS MATH! czcams.com/channels/artWliFdki6px-57oVh6VA.html Check out my other channel OWLS SCHOOL OF MATH! czcams.com/channels/AJaLg-yEcvhC_ggEr0Hl6g.html Practice integrals with King's Principle: owlsmath.neocities.org/King's Rule Integrals/king Website: owlsmath.neocities.org #math #integrals #integrationtechniques
You don't need a dx when evaluating limits :)
zhlédnutí 598Před 21 dnem
Some practice problems for YOU: owlsmath.neocities.org/Limits Quiz/limits owlsmath.neocities.org/Limits Quiz 2/limit owlsmath.neocities.org/L'Hopitals Rule/lhopitals owlsmath.neocities.org/Delta Epsilon Limits/deltaEpsilon Website: owlsmath.neocities.org Check out my other channel OWLS MATH! czcams.com/channels/artWliFdki6px-57oVh6VA.html Check out my other channel OWLS SCHOOL OF MATH! czcams.c...
Why not use 2024 on the exponent???
zhlédnutí 1,3KPřed 28 dny
UK integration Bee: integration.soc.srcf.net/ Check out my other channel OWLS MATH! czcams.com/channels/artWliFdki6px-57oVh6VA.html Check out my other channel OWLS SCHOOL OF MATH! czcams.com/channels/AJaLg-yEcvhC_ggEr0Hl6g.html Practice problems: owlsmath.neocities.org/integrals.html Website: owlsmath.neocities.org #math #integrals #integrationtechniques
Revisiting this variation on the Dirichlet Kernel
zhlédnutí 555Před 28 dny
Original video: czcams.com/video/nNio8dX4UlY/video.html Another one using this method: czcams.com/video/k2mv9O8qprg/video.html Check out my other channel OWLS MATH! czcams.com/channels/artWliFdki6px-57oVh6VA.html Check out my other channel OWLS SCHOOL OF MATH! czcams.com/channels/AJaLg-yEcvhC_ggEr0Hl6g.html Practice problems: owlsmath.neocities.org/Dirichlet Kernel/dirichlet Website: owlsmath.n...
Grueling problem. Satisfying solution.
zhlédnutí 3,6KPřed měsícem
Grueling problem. Satisfying solution.
I couldn't decide which Integration Bee I was working on!
zhlédnutí 1,8KPřed měsícem
I couldn't decide which Integration Bee I was working on!
fun random integrals & I can't think of a title
zhlédnutí 745Před měsícem
fun random integrals & I can't think of a title
UK Integration Bee Sample #11 (Alternative Method from adandap)
zhlédnutí 1,8KPřed měsícem
UK Integration Bee Sample #11 (Alternative Method from adandap)
Bernoulli's integral has a few tricky things going on
zhlédnutí 16KPřed měsícem
Bernoulli's integral has a few tricky things going on
floor(-2024pi)=-floor(2024pi)-1 I=int[-2024pi,ceil(-2024pi)](floor(x))dx (everything else cancels) width=ceil(-2024pi)+2024pi height=floor(-2024pi) I=floor(-2024pi)•(ceil(-2024pi)+2024pi) I=({2024pi}-1)•floor(-2024pi)
I=int[0,♾️](int[1,pi](1/(1+a^2•x^2))da)dx I=int[1,pi](int[0,♾️](1/(1+a^2•x^2))dx)da I=int[1,pi](pi/2a)da I=pi/2•ln(pi)
This is another of those problems where looking at it as two integrals you end up with the difference of two logarithmically divergent expressions. You can see that there's a problem if you put u = pi x in the first one, which transforms it into the second, giving zero. Since arctan(pi x) > arctan (x) for all x in [0, pi/2] that clearly can't be right. The problem is that the terms in the integrand both tend to pi/(2x) for large x, hence the divergence. So I tried writing y = arctan(a x) - arctan(x), so tan(y) = x(1-a)/(1 + a x^2) so the integrand becomes 1/x arctan(x(1-a)/(1 + a x^2) ). I couldn't do much with that, but it's clear that it's convergent, since the large x behaviour ~ 1/x arctan( (1-a)/(a x) ) ~ 1/x^2 * (1-a)/a. Then I gave up and used the Feynman thingy as the title suggested. ☺
😂 Feynman thingy. Interesting stuff!
Fastest way: (i think) use king's symmetry property to switch sin to cos then use identity 1+cos(x)=2cos²(x/2) so then we get 1/2 (0 to π/2) ∫ sec²(x/2) dx which easily integrates to tan(x/2) then applying bounds gives us tan(π/4) - tan(0) = 1
Like it! Yeah good method. 👍
dude, i was just visualizing the thing and approximated it to be between 0.4 and 0.5 as a wild guess. what a lucky shot
cool. I think we should approximate integrals more often!
This is actually an integral first evaluated by Euler about 275 years ago, if not more. All you have to do is a substitution x=2t and no other principles are necessary. For people who suggested any other methods (such as by parts), I have to disappoint you, this function does not have elementary antiderivative.
I learned another approach using a double integral. Use x<=y<=πx as the interval and tan(y)/x as the integrand.
That makes sense. Good way 👍
I see a hyperbolic claim about the effectiveness of Feynman's trick, I click
🤣🤣🤣 Yeah i think all previous claims for the power of Feynman trick have been understated
@@owl3mathtrue
@@txikitofandangoThe entire universe hinges on Feynman's trick. Without it we are toast.
@@MikeT10101Feynman: 🤫🧏♂️
It's also the same as the integral of sinx / (1 + cosx) And you can do that in your head to get ln( 1 + cos(π/3)) = ln( 3/2) Still with you almost daily, keep up the great work, you're nailing it! 😊👍
PS - as always, the alternative is just to take up space so the algorithm sees engagement and is not a suggestion for doing it that way. 😉😊
Hey @Ni999! It's been a while and I was wondering if you were still watching :) Glad to hear you are. Thanks as always for the support!
@@owl3math You're welcome!
Just finished vector calculus and linear algebra. What in the world is the floor function
basically it rounds down to the next greatest integer. Floor(3.87232) = 3 Floor(3) = 3 Floor(-2.1) = -3
@@owl3math what are it’s applications?
I enjoyed that. I've liked Weierstrasse substitution since I first learned about it.
Hey Mike :) Thanks! Yeah I like doing these. I've probably done too many Weierstrass videos but I like it 🤣 Anyway probably going for something different in the next video.
Why not use the half angle formulas? 1-cos(x) = 2(sin(x/2))^2 sin(x) = 2sin(x/2)cos(x/2) The integrand becomes tan(x/2).
You can and it's faster that way! I think I said that in the video? Anyway multiple ways will work for this one :) thanks
Yep that's exactly what I did
I kept getting myself confused with the bounds until I drew a number line and marked various points, setting floor(2024 pi) = N. Then it was easy enough to calculate the three contributions.
Hi adandap. Yep kind of some basic book keeping here. Kind of cool how everything cancels! Not sure I’ll remember this as a formula however that you always get the lower bound as the answer.
Here is the link to that previous video where I looked at the case with integer bounds and bounds going to infinity: czcams.com/video/D5kwCC_Pa9A/video.html
The sum can be represented in terms of the digamma function: S = -1/3Σ1/k - 1/(k+1/3) = -γ/3 - 1/3(-γ + Σ1/k - 1/(k+4/3-1)) = -1/3(ѱ(4/3) + γ) = -1/3(ѱ(1/3) + γ) - 1 By multiplication formula (z=1/3, k=3): ѱ(1/3) + ѱ(2/3) + ѱ(1) = 3(ѱ(1) - ln3) ѱ(1/3) + ѱ(2/3) = -2γ - 3ln(3) By reflection formula (z=1/3): ѱ(2/3) - ѱ(1/3) = πcot(π/3) = π/√3 So we have the linear relation: ѱ(1/3) + ѱ(2/3) = -2γ - 3ln(3) ѱ(2/3) - ѱ(1/3) = π/√3 ѱ(1/3) = -γ - 3ln(3)/2 - π/2√3 Hence S = ln(3)/2 + π/6√3 - 1 Alternatively you can use the Gauss digamma theorem
Integration by parts with u=arctan(x) , dv = 1/x^2 dx if we want to use numerical integration use substitution u = 1/x and equality arctan(x)+arctan(1/x) = pi/2 , for x > 0
Nice!
俺やったら出されても解けんな…
和分で解けた!
Beautiful!
Thanks!
A nice integral Thank you so much
Thank you! 🙏
I did not see the limits in the thumbnail😢 and was solving the indefinite integral, I finally came to a point where I got the integral of (xcosx)^(-1), I am pretty sure this is unsolvable. If anyone knows how to solve please help😅
😮 this one seems difficult as an indefinite integral!
@owl3math yeah, that's why I expressed sinx in polar form.
It equals approximately -0.14839396162690884587003100526504 .
😭😭WOW, THAT WAS BRILLIANT SIR❤❤
Hey Clifford. Thanks! appreciate it :) 🤩😍
I've been working on an integral similar to this one but in the sum reverse the order of the terms and substitute a natural number k instead of 3. So far it doesn't seem to have a nice solution but I'll still try to genralise because I have nothing else to do with my life.
It seems to be related to the Digamma function, so you might want to check that out.
This is easy af bro
It's a great video ! The only thing I would like to see more is (even if it is simple and short) an explanation of why we can switch the sum and integral. It makes the video more satisfying for me, It kid of "legitimates" the result. thank you :)
Thanks DJ! And it’s a good point too 👍
That’s the hardest part, the rest is easy af
@@owl3mathdo real maths bro, please man. It pains me to see someone that isn’t rigorous
Well done! 👌 (Maybe you would like to work a bit on your intonation as it sounds a little bit monotonous. But that may be just an interpretation by my own ears. ✌️)
Thanks. And I think you’re right. I could use a better tone but it’s not my tendency
That's fascinating ❤❤❤
Thanks!
Nice solution. I would do the following: Define three series, A,B,C defined as the sums of respectively 1/(3n-1), 1/(3n), 1/(3n+1), each from n=1 to n=N. Then A+B+C=H_(3N)-1 and B=H_N/3 where H denotes the harmonic numbers. Now the series A-C is nice and can be easily computed to be 1-π/(3√3) using the cotangent residues trick from complex analysis. Then we can find the series we want, C-B, as ((A+B+C)-3B-(A-C))/2. From the approximation H_N≈ln N+γ we find lim H_(3N)-H_N=ln 3. So in total we get (ln 3-1-(1-π/(3√3)))/2 which agrees with your answer.
Nice 👍 thanks
Nice job!
Hey Mike. Thank you!
Technically you should have left the -n at the last step instead of putting in -Infinity, then gotten to e^0+e^(-n), and then take the limit to get e^0=1. Convergence on limits if you take it sooner has very stingy rules, and I'm sure they don't apply automatically to bounds of integration
Makes sense. I glossed over it
was about to comment this but i think it's ok since we have continuous functions of n in the end
Ugh. I didnt care for that - and i can vouch that having arctan in the floor function makes it impossible! (The thumbnail is kinda hard to read so I started with your first handwritten version). One other comment is that ln(1/u) = - ln(u) so you didnt really need the y sub.
Yep thumbnail is a mess!!! But don’t you still need the substitution to clean up the u^2 in the denominator?
@@owl3math No, just use parts. Differentiate ln(u) and integrate 1/u^2
@@adandap oh yeah. That works :)
I gave this problem to my 10 year old cousin and he got it right. "1?!" Bro's gonna win the bee this year frfr
Are you serious????
@@owl3math he really did guess it right. It seems to me that sometimes the best course of action is to just guess the answer haha
Ha! Good point!
@@owl3math bro I was baffled too 😅😅
Imagine being a calc 1 student who just got into trig integrals and on the exam next day, there's a question that needs the long method to prove why the integral of 1/(x^2+1) is arctan(x) 💀
that student would be very very lucky and well prepared :)
One could write the integral as I = 4 \int^(2pi)_(0) dx 1/[3 + cos 4x ]. Defining 4x=t, I = \int^(8pi )_(0) dt 1/(3 + cos t) = 4 \int^(2pi)_(0) dt 1/(3+cos t). If z=e^it, This can be converted to a contour integral over the unit circle C. I = 8/i \int_{C} dz/(z^2+6z+1). Using the residue theorem, I = 8/i (2 pi i) 1/(2√8) = 2 √2 pi.
The given integral, I = (d/dt)^n|t=a-1 [ \int^1_0 dx x^t], as d/dt x^t = x^t ln x. Now \int^1_0 dx x^t = 1/(t+1). So, I = [n! (-1)^n]/a^(n+1).
The constant value with ln(-1) is not completly trivial in the sense you are working with the complex logarithm and to make it properly defined you have to remove a half line from the domain, usually from 0 to minus infinity, so its not completly obvious (still its constant I know, but it should be mentioned that it is not completely trivial)
Right. Its more involved if we need to get a value for that part 👍
POV: Me trying to solve the 2 mark question in my exam
It’s easier to understand things intuitively if you know the underlying logic
you could also do complex factoring (1+i)(1-i) and do partial fractions. BPRP did it this way for fun. It's also a cool way of doing it!
Nice! Yes, I'm familiar with that way and I think it's a good way to get that "complex" definition for arctan I show in the video. Thanks :)
Can we get Gamma function ?
Hey Jacek. What do you mean? Thats how i did it right?
@@owl3math This integral remids me Gamma function after substitution but Gamma functon can be avoided assuming that n is positive integer
@@holyshit922 yep good point! Thats funny because I was just working on basically that exact thing. I think that's an upcoming video :)
Love it
Thanks!
I usually prefer the longer methods, and trying diffetent techniques, etc. It can be a great learning experience.
Totally feel the same way. And sometimes trying some long methods on easy problems makes it easier to use later when you really need it. 👍
Hello how can I contact with you because I have some questions about learn mathematics
Sure. You can use my email address if you have some questions: owlsmathcontact@gmail.com
@@owl3math I sent you the message to your email. I hope you can help me.
I sent you a message hoping you can help me @@owl3math
Nice approach.
Thanks but What!?!? No alternative methods?
getting used to the amount of variables
Hey Felipe. Good point. It does make things confusing on this one!
It is important to state that this is an improper integral at 0
Yep good point. True 👍
1 integration by parts with du = 1/xdx , v = ln(1-x) or substitution u=1-x 2. Geometric series expansion 3. Change order of summation and integration 4. Integration by parts once again
Another integral from -a to a of even(x)/(1+b^odd(x)). very common
Right. Pretty easy to spot. I think I need to do this less often. 🤣