The Hardest Putnam Integral Ever….
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- Опубликовано: 25 окт 2024
- In this video, we tackle one of the hardest integrals ever featured in a Math Olympiad competition. Follow along as we break down the complex steps needed to solve this challenging problem. Whether you're preparing for math competitions, enhancing your calculus skills, or just enjoy solving difficult integrals, this video will walk you through every part of the solution with detailed explanations.
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![The most beautiful equation in math, explained visually [Euler’s Formula]](/img/1.gif)
how is this the moment where I learn how chappell roans name is actually pronouced
i think trig sub starting with x²=tan(θ) is way easier
Mistake spotted: 7:09 the factorisation of u⁴ - u², should become u²(u+1)(u-1) instead of what we see. The solution to this integral is exactly what we have already but with the negative sign swapped between the 2 ln terms :)
I'm a little bit confused with the part where it was u³/x.
Isn't that supposed to be 1/xu³ that is u^-3/x instead??
You're absolutely right. I've just realised the mistake thank you.
Okay so, correcting the mistake, in the denominator we would need to times by u^2. This leads to an integral which is identical to what we have, but with a negative on the front. Leading to the final answer being the same as what we got, however the ln(u+1) term is positive, and the ln(u-1) term is negative, with the arctan being the same
Thanks for pointing it out, in my original solution I didn't make this mistake but in the video I did ...
@@Jagoalexander Understood. And if the u-substitution results in u²/(1-u⁴), perhaps the natural logarithms will also be in the form of ln|1-u| or ln|1+u|??
Which would make the ln|1-u| term be positive and ln|1+u| be negative. Or maybe I didn't compute it right from up top. If you don't mind, please do let me know my mistake.
Either way, yours is a beautiful technique. Loved the clever substitution. You're a godsend brother ❤️❤️❤️ Glad to have subscribed here.
Could you solve the integral of e^-x^3 zero to infinity?
If I plug your answer into sympy differentiate and simplify I get a mess. If I plug the following answer into sympy -
I(x) = (1/2)*atanh(Abs(x)/(x**4+1)**(1/4))+(1/2)*atan(Abs(x)/(x**4+1)**(1/4))
I get -
dI/dx = x/((x**4 + 1)**(1/4)*Abs(x))
did you check your answer by differentiating?
I started with the substitution x**2 = sinh(y) and then used sinh(y)**2+1 = cosh(y)**2. Then used z = sqrt(tanh(y)). Then used partial fraction decomposition -
1/(1-z**4) = (1/2)*(1/(1-z**2)+1/(1+z**2)).
I've got the same result
Thank you for watching the video! If you’re not already, please subscribe. Also RIP Euler, you would have loved Chappell Roan 😔
5:33 What you set to be u^3/x should be 1/(xu^3). In fact, in your definition the radical stays at the numerator, while in the fraction you are considering, it stays at the denominator. Or maybe are we missing some minus sign?
This is just a simple integral. Considering int of x^n ≈ (1/n+1)*x^(n+1), we can write the displayed equation as int of (x^4 + 1)^(-1/4), note -(1/4) + (1/1) = (-1 + 4)/ 4 = 3/4 and x^4 ≈ (1/5)*x^5, hence giving us (3/4)*(1/5)*x^5*(x^4+1)^(-1/4) ≈ || (3/20)*x^5*(x^4+1)^(-1/4) : voila, my answer may differ from your answer, we call it relativity.
This is how I see this equation, let us consider the whole integral of (x^4+1)^(-1/4) as our starting static universe and the integral of (x^4+1) as a small change or perturbation in our universe, so after rigorous derivation, observation and analysis, we came to the conclusion that, we have keep our starting equation, which represents the universe as constant and then multiply it with the equation, which represents the small changes in the universe. The equation can then considered as the effect that a changing variable has on a static universe, therefore making it a dynamic change.
Using x^n ≈ (1/n+1)*x^(n+1) || x^4 ≈ (1/5)*x^5 , 1 ≈ x , -(1/4) + 1 ≈ 3/4 || therefore our previous answer becomes (4/3)*((1/5)*x^5 + x)*(x^4 + 1)^(3/4) . This is an indication that our universe is expanding.
another solution with x² = sh(t), then u = sqrt(tanh(t)). this leads directly to (1/2)(Atanh(u) + Atan(u)) which is quite fun...
I assume you mean x**2 = sinh(t). That is what I did too. Note you need to note that sqrt(x**2) = Abs(x).
I(x) = (1/2)*atanh(Abs(x)/(x**4+1)**(1/4))+(1/2)*atan(Abs(x)/(x**4+1)**(1/4))
If I use sympy to differentiate I(x) I get
dI/dx = x/((x**4 + 1)**(1/4)*Abs(x))
Did he somehow correct the error at 6:51 ?
I.e. that (u^4-u^2) actually equals (u^2+u)(u^2-u) and not what was in the video.
If that has not been corrected, surely the final answer is wrong?
7:09 Is it just me or does u^4-u^2 not factor into (u^2+1)(u^2-1)
No, as stated by OP, the factorisation is u^2(u+1)(u-1)
No, that would be u^4 - 1
1/(u^4-u^2)= A/(u^2-1)+B/u^2 ( Subtitute u^2=v). Now A/(u^2-1) is easy A=a*(u-1)+b*(u+1) and u-1=0 and a+1=0 give a and b, you can use L'hopital's rule, L'hopital's has a algebraic proof.
this is great!!!
AMAZING!!!!
And here I was thinking the answer was non-elementary!
This is a little bit below ln x, an evaluation before starting anything meaningful towards finding solution.
What do you mean sorry?
@@Jagoalexander I meaan that the initial function os approximately 1/x when x->infinity, consequently, the integral of 1/x is ln x.
6:59??