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- เผยแพร่เมื่อ 4 ก.พ. 2023
- Here's another awesome application of Feynman's trick to an aesthetically pleasing integral involving the golden ratio. The results are just as pleasing as the solution development.
φ^2(3φ-4)
=(φ+1)(3φ-4)
=3φ^2-φ-4
=3(φ+1)-φ-4
=2φ-1
=sqrt(5)
Please, more of the φ series! I'm loving this!
Side note: bro really missed two awesome chances while integrating the natural logs. Could've either evaluated them TOO using Feynman's, or could've used Dr. Peyam's life-changing trick and saved a couple steps, adding onto the elegance. Regardless, it's a really beautiful integral. :)
Final result can be simplified to sqrt(5)Ln(phi)
What I like about Feynman's technique is that it doesn't just offer you a numeric answer towards a specific computation, but an insight towards not only the general solution to the problem, but how functions of a curve "change" when adding a certain parameter, it's like finding how all the family of functions behave which is quite insightful when one thinks of it in a certain way, I also always remember complex line integrals when incountered with feynman.
Kind of like level curves from calc 3 but applied to integrals
5:43, when doing the integration by parts, you can multiply the integral by 1/2*2, and choose antiderivative 2t+1 for int(2), or choose 1/2*(2t+1) =t+1/2 instead of t for the antiderivative of 1 to make it easier. This way, you still get int(ln u)du=uln u - u, the integral will differ up to a constant, which is absorbed by the constant of integration.
Similar logic for I_2.
Another awesome integral and I love how you just smashed through everything in a very clear way that made it simple to follow.
One small comment: in your final answer I would write the term in parentheses before the ln, not after. Otherwise it can seem like you're taking the log of the entire expression. To your last point, I think it's already in simplest terms since all the phi terms are linear in your final answer.
Ln(2t+1) can be solved easily by taking 2t+1 = z and use lnx formula or if we don't remember we can solve once and reuse it for both I2 and I1
Or instead of substitution you can use the integral of f(ax+b) if you already know integral of f(x)
3φ³-4φ²=3(φ²+φ)-4φ²=-φ²+3φ=-(φ+1)+3φ=2φ-1. So I=(2φ-1)ln φ.
Wonderful!
= sqrt(5)*ln(φ)
Nice integration video! What is the app you are using?
Asume y is the golden ratio
it is the solution to the eqn x²-x-1=0
We can subsitutte back:y²-y-1=0
y²=y+1---(1)
y³=y²+y (1)×y
y³=y+1+y----from(1)
y³=2y+1---(2)
3y³-4y²=3(2y+1)-4(y+1)
=6y+3-4y-4
=2y-1
=sqrt5 +1-1
=sqrt5
4:15 if you just factor a 2 out of the first term and get a denominator of t+1/2 and a factor of 4/2 in the numerator, but then you are led to 2 ln(t+1/2). Where did I go wrong?
You can't just switch the parcial derivative and the integral, first you have see if the parcial derivative of the function is bounded.
Convergence criteria here is quite trivial; I have an entire playlist on Feynman's trick and I've discussed convergence everytime it seemed rather tricky.
i think using feynmans technique is the only way to solve that thing lol
Not the only way! If there's anything I've learned in integration, it's that there's ALWAYS another way to solve any problem. I'm too lazy to actually do the work, but I believe after the substitution t = -ln(x) --> x = e^-t and applying integration by parts, we get the form of a Frullani integral.
Edit: it worked.
Indeed it is!
I still would've applied Feynman's trick to it though just cuz I'm always looking for an excuse to use it😂😂😂
Frullani integral video coming up soon!
@@maths_505 yay
@@maths_505 awesome
A me risulta,derivando 2 volte una I(t) opportuna 2((@+1)ln(@+1)-@)...i passaggi erano semplici,se ho sbagliato è per distrazione