They Won't Teach You This Trick
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- Опубликовано: 13 мар 2022
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How have you commented this 10 days ago ? You posted this video a day ago
brain left the chat lol
@@masternobody1896 time has left the chat
This video was first accessible to members , then to us
So, he had pinned it 10 days ago...........
@@SaurabhKumar-jo6dp oh
To be clear, Leibniz' rule and Feynman's trick refer to different things. Leibniz' rule, in reality, refers to a more general theorem, which you apply when you have an integral of a function g(x, y), and the bounds of the integral depend on x or y. Feynman's trick refers to a specific algorithm by which you evaluate the integral of a function f(x), by introducing a function h(x, y) and letting h(x, c) = f(x), for some c, using Leibniz' rule, and then finding the boundary conditions, and solving the corresponding differential equation. They are certainly related, but not the same thing. Also, there is a very good reason why this is not taught: you need to be acquainted with differential equations and various techniques on how to solve them. But you only study differential equations after learning integration techniques.
Also, I would say this method is not actually the best method to use for the integral in the video, and there are better integrals to solve using this method. The integral in the video can be more simply solved by using a substitution, specifically by letting ln(x) = -t. This just gives you something in terms of the Gamma function, which can be simplified to a factorial.
I solved it in a very similar way but first made the substitution x=e^u then we essentially we are solving the generalised integral for the gamma function :)
Mathematics mi solved this integral under 1 minute
You can also use substitutions y=ln x, -y=k,101k=a, there Is then gamma function And result Gamma(101)/101^101=100!/101^101
Yep, that's the method I went for. Any time I see the integral of a weird product of x and ln(x) between 0 and 1 I jump to this substitution in hopes of using the gamma function.
Honestly, that is the first thing I tried. A gamma function is always pretty at the end.
That's amazing, thank you a lot, the video is really well made!
I loved this trick. You're a genius !!
One of the best math videos ever
Hey dude really love your content can you please provide more such math learning youtube channels
I defined the function I(n,m)=int_0^1 x^n(ln(x))^mdx, and then integrated by parts with du=x^ndx and v=ln(x)^m. Using this you can construct a power reduction formula which related I(n,m) to I(n,m-1), and then a simple inductive argument allows you to compute I(100,100)
First one can be solved by Gamma integral. but your method is also too interesting. Thank you!
Congratulations brother
That was a really slick problem there.
totally agree, awesome ... solving integrals is so much fun
@@silentintegrals9104 Yup, agree. Although they are harder than differentiation, but that is the reason to love it.
@@gitgudnoobs7917 agree, each integral is like a challenge 🎁
they actually do teach the first integral here. It can be properly derived using by parts,
you can assume the integral to be I(m) = integral from 0 to 1 (x^n [lnx]^m) dx
use by parts, taking the first function as [lnx]^m and second as x^n, after simplification, you will get a telescopic product, which upon solving gives the same result.
Amazing!
I love Mathematics and Physics
nice approach
You can just substitute u = ln(x)
Then you get the integral from 0 to -inf of e^(101u)u^100 du
Then with the DI method you end up seeing that everything, except the later term, are just equal 0. So you end just needing to evaluate 100!e^(101u)/101^101 at u=0 and lim u->-inf. And you don't need to compute the 100 or so terms
That was amazing 👍👍
omg a fine addition to my collection
Awesome video as always brother!! 💯 😀
Appreciate it!
By substitution we can get Gamma function
By parts we can get reduction (recursive relation)
Yes in my native language it is taught as Leibniz rule for integration
or differentiation under the integral sign
Angelito can believe that this is different thing but it is simple consequence of Leibniz rule and people had used it before Feynman was born
just set u = ln(x) or u = x^n+1 and u get ur answer by integration by parts
Thanks some of the complex High school problems can be easily solved!
awesome, solving integrals is so much fun
Yes they are!
Nice!!😃. The wont teache us this but you and MI teaches us this!
Sir your all tricks are fabulous ..and it helps me lot .. But it really a talk of pleasure for me..if you make a video on " How to study Mathematics " ... b/c I am weak in mathematics ...and I hope it will help me to get good grades and to understand complicated concepts..
Watch The Math Sorcerer's videos.
i love feynman trick
yes, this is really nice
@@silentintegrals9104 but in my country, no youtube channel teaches about it :(
Another very interesting video! I also really appreciate how commenters are pointing out that you can make a substitution to get an integral close to that for the gamma function.
You can solve the original problem you suggested would be impressed if they could using an iterated integration by parts, integrate the X^100 and differentiate the (ln x)^100 101 times (d/dx (ln x)^n = n (ln x)^(n-1) * 1/x, and since you have a x^100 that keeps gaining powers, the 1/x you get will always have something to cancel out with so eventually you can differentiate away the ln x and solve the problem as a polynomial.
Hey can you tell me what software do u use to make these videos?
Thank You
Hey! Could you please do a video on solving for "n" in this equation. A=P*(1+r/n) ^nt
"Differentiating under the integral" feels similar to "deriving under the influence," I got my eyes on you Bri!
Also, love your editing, always beautiful!
Lololol
A point of contention, methinks. Just a historical appreciation...
I would call the property of integrals that allows you to move a "derivative under the integral sign" as the Leibniz rule, after its discoverer.
I would call the technique of introducing a second variable and using the Leibniz rule to evaluate a single-variable integral as the Feynman technique, after the man who popularized this approach to simplifying complicated integrals.
what is Calculus sir,
what are it's principles
and what are the objective and apparatus of Calculus
Hey BRITHEMATHGUY,
Here is my challenge to you. This integral is said to be the toughest integral out there.
Integral (sin^2 (x)) / (1- 2sin^2 (x)cos^2 (x))
Can you solve the integral of :
ln(sinx+cosx)/(cosx-sinx) dx
There. Is a closed form of integral 0 to 1
Of xⁿ(lnx)^m
What is it? It is left as a exercise to reader.
Let y = ln(x), hence the interval of integration is (-♾, 0] and the integrand is y^m·exp[(n + 1)·y]. Let z = (n + 1)·y. The interval of integration remains unchanged, and the integrand is now 1/(n + 1)^(m + 1)·z^m·exp(z). Finally, let z = -a, the interval of integration is now [0, ♾), and the integrand is simply [-1/(n + 1)]^(m + 1)·a^m·exp(-a). This integral is equal to m!·[-1/(n + 1)]^(m + 1).
Super cool
That was smart
I'm failing my math classes and here I am watching a math video
There is a beautiful like beetween this and the basel problem... int_1^0 x^n * ln x dx = 1/(x+1)^2...
I solved it using integration by parts simply the result is x^(m+1)(Σ(((-1)^(k+2))((lnx)^(n-k))(n!)/(((m+1)^(k+1))(n-k)) where k=0 and finishes at k=n
Watching this at 10:52 PM, in a train, on my way home😄
Cool trick by the way.
That first integral can be easily done using some substitution and gamma function integral definition! Did you get very impressed! Lol
∫1/x dx = ln(x)+C
The absolute value sign is only there because not all calculators support imaginary numbers and not everyone knows how to take the log of a negative number.
Differentiating under the integral sign is a standard result in today's analysis courses and we use it extensively. One of its main uses is precisely what you showed - repeated applications and then a quick induction proof. Are you saying "they won't teach you this in high school"? If so, then that is expected and reasonable, I'd say.
FIS-IDFIS rule
Do you think there's a point in this?
Pls solve this question : 20^200{20x200^20} -that question is my secondary 2 question
10^100 is Googol number
Infinity isnt the maxuim
xM to the power of xXXM
xXXM
xXXM
Is maximum like infinity x infinity
Hint : Utter Oblivious is the number before infinity
The feyman trick
I’m genuinely confused. It looks like you can do basic calc things to solve them😭
now present the proof that integral and differential operators are interchangable
I used to be really good at math. But I still can’t remember the purpose when all problems are mathematically theorized.
I've been re-teaching myself Calculus after graduating 3 years ago with a physics B.Sc.
What I've discovered is that if you define and derive everything from first principles, you come to a better understanding of *why* these formulas and operations solve specific problems, rather than the reverse of finding a problem and then applying a formula or operation to solve the problem.
I found that the latter method is how physics majors are taught to perform Calculus, but the former method, which I would assume math majors are familiar with, is the *best* way to learn and perform Calculus: derive everything from fundamentals, which *implies* deriving general theorems and then specifying them to whatever problem it is you're working on.
This is why higher mathematics is so much more abstracted and theoretical than your standard Calc I-III courses, because by necessity the study *must* become more rigorous if the concepts are to be applied to, say, n-dimensional spaces where n > 3 (i.e., x, y, z, t, ...) which would require more than mere triple integrals if you wanted to understand such concepts in terms of Calc III knowledge. Hence why generalized theories are so useful once you understand the notation.
If you use this First Principles Method, you automatically need to do less problems to ingrain the material in your brain, because you'll always know *HOW* to develop the model *such that* a simple application of formulas can solve your problem. Too often people throw a puzzle of theorems at a problem hoping it gets solved when understanding the problem from a generalized perspective would help them to better visualize the solution for when numbers are substituted-in.
This is NOT Feynmann’s trick!!
Couldn't I have learnt this before I had the advanced integration test?
Mathematics mi solved this integral under 1 minute
Why the FUCK i am watching this at 2 am.. without any reason FML.. i am bored
they do teach this because they want to feel superior
why are use uptalk in your voice speech so much?
Happy Pi Day
Don't click on read more
Never gonna give you up
I warned you ;)
Hey BRITHEMATHGUY,
Here is my challenge to you. This integral is said to be the toughest integral out there.
Integral (sin^2 (x)) / (1- 2sin^2 (x)cos^2 (x))
Hey BRITHEMATHGUY,
Here is my challenge to you. This integral is said to be the toughest integral out there.
Integral (sin^2 (x)) / (1- 2sin^2 (x)cos^2 (x))
Hey BRITHEMATHGUY,
Here is my challenge to you. This integral is said to be the toughest integral out there.
Integral (sin^2 (x)) / (1- 2sin^2 (x)cos^2 (x))
Hey BRITHEMATHGUY,
Here is my challenge to you. This integral is said to be the toughest integral out there.
Integral (sin^2 (x)) / (1- 2sin^2 (x)cos^2 (x))
Use 1-cos2x/2 , 1+cos2x/2 then proceed with substitution
Hey BRITHEMATHGUY,
Here is my challenge to you. This integral is said to be the toughest integral out there.
Integral (sin^2 (x)) / (1- 2sin^2 (x)cos^2 (x))
Hey BRITHEMATHGUY,
Here is my challenge to you. This integral is said to be the toughest integral out there.
Integral (sin^2 (x)) / (1- 2sin^2 (x)cos^2 (x))
Hey BRITHEMATHGUY,
Here is my challenge to you. This integral is said to be the toughest integral out there.
Integral (sin^2 (x)) / (1- 2sin^2 (x)cos^2 (x))
-tanx+0.5ln[sec2x+tan2x] +c. ??
Hey BRITHEMATHGUY,
Here is my challenge to you. This integral is said to be the toughest integral out there.
Integral (sin^2 (x)) / (1- 2sin^2 (x)cos^2 (x))
You're missing a dx