I love how, on one hand he glosses over the intricacies of trig substitution in integrals, but at the same time he carefully explains how we know that pi/4 is less than one. sidenote: For those in the future, the legendre duplication video was yesterday's: ruclips.net/video/BIR4jDur_Ik/видео.html
Dude... I just want to say this out loud. As a Physicist(and teacher myself) I cannot say how happy I am to have found your video series. As you might expect I'm more of a PDE kinda guy than a Number Theory kinda guy. But you help keep me sharp. Much obliged.
@5:10 while that inequality looks obvious, one has to be careful for x values between 0 and 1 where a higher exponent actually leads to a smaller number
5:30 the obvious inequality is not clear to me 🤔 If x < 1, i can't see the inequality at first glance because 1+x^a gets smaller. E. G. For a = 3 and x = 0.1, the left integrand is bigger.
You've explained it pretty well, despite saying that you don't understand it. Keep in mind he's not talking about the integrands, but the integrals. (The 'smaller' curve (larger values of alpha) has less area underneath it.)
@@xizar0rgyes, so it looks like the integral from 0 to 1 behaves different than the integral from 1 to infinity. How the sum of both parts behave 🤷♂️
Great video! For those of us lucky enough to guess the u-substitution u = 1/(1+x^a) with differential dx = -1/a u^(-1/a-1) (1-u)^(1/a-1) du, this gives the normally recognizable form of the Beta function directly without the trig substitutions. I got stuck on this problem trying to find a satisfying x for the equation with the Gammas.
Wow! It starts as "plain" calculus, then it take a long steep route through the Gamma function, and finally an elementary quadratic formula, that yields the Golden Ratio!
That was beautiful! Compliments to approaching the problem tying it in with gammas and betas and bringing it down to earth with some simple quadratics. And all in < 17 minutes
I'm pretty sure that evaluates to (1/a)Gamma(1/a)Gamma(a-1/a)/Gamma(a) for a=phi, phi-1/phi=1, 1+1/phi=phi, tGamma(t)=Gamma(1+t), (1/phi)Gamma(1/phi)Gamma(phi-1/phi)/Gamma(phi)=Gamma(1+1/phi)Gamma(1)/Gamma(phi)=Gamma(phi)/Gamma(phi)=1 This is based on making a u-substitution x^a=u, which gets it to the form (1/a)Int_0^Inf(u^(1/a-1)du/(1+u)^a), which is (1/a)Beta(1/a, a-1/a). One Euler's reflection formula later, and we're at the point where I started.
You get alpha to be 1.618. The limit of the divisor would then be x power 2.61, and integrated x power 1.61. This is about 1 divide by x power 3/2 (three halves). It is Kepler's law! The square root of time power 3 is an axis of an ellipse. So we get 1 / axis of an ellipse. This may be the eccentricity or the 1 / eccentricity if it has a limit. In other words, if it has a limit, it is an elliptical (and not hyperbolic orbit). It is not exactly 3/2 because 1. the Moon is moving away from the Earth and the planets are moving away from the Sun 2. the universe is expanding 3. one does not count in the Earth's motion against the fixed stars.
7:20 i can’t get why the upper integration bound after theta substitution is pi/2. It was obvious in case of substitution x=tan(theta) but in this case we’re doing substitution x=tan(theta)^(2/alpha) so shouldn’t the upper bound be like (pi/2)^(2/alpha) ?
The thing is, tan(pi/2)^(2/alpha) continues to be infinity, since (infinity)^(2/alpha) = infinity. So you should consider the angle that is inside the tan(theta)^(2/alpha) which is, in this case, only pi/2, if that makes sense.
A lot of math problems have "clever" solutions that make things work. That's what he's talking about when he mentions "wishful thinking". (This is similar to when Blackpenredpen uses "wouldn't it be nice".) He chooses a solution that works (golden ratio) as a "clever guess", and then leaves it as a homework problem to show there are no other solutions (while sketching out an explanation).
Upon checking numerically, the inequality appears rather spurious to me, since the integral has a unique minimum on (1,infinity) after which it approaches 1 from below as alpha->infinity.
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I love how, on one hand he glosses over the intricacies of trig substitution in integrals, but at the same time he carefully explains how we know that pi/4 is less than one.
sidenote: For those in the future, the legendre duplication video was yesterday's: ruclips.net/video/BIR4jDur_Ik/видео.html
I know right 😂
I think that's because that's how he sees the beauty in math: by logical implications, rather than long equations
Dude... I just want to say this out loud. As a Physicist(and teacher myself) I cannot say how happy I am to have found your video series. As you might expect I'm more of a PDE kinda guy than a Number Theory kinda guy. But you help keep me sharp. Much obliged.
10:30 in the definition of the beta function, the exponents should be v-1 and w-1. The second definition with sin and cos is fine though.
@5:10 while that inequality looks obvious, one has to be careful for x values between 0 and 1 where a higher exponent actually leads to a smaller number
The general case of the integral parametrized by alpha is pretty cool too
Somehow, once he said that 1 < a < 2, I suspected that phi would be a solution. Nice problem.
Thanks!
5:30 the obvious inequality is not clear to me 🤔
If x < 1, i can't see the inequality at first glance because 1+x^a gets smaller.
E. G. For a = 3 and x = 0.1, the left integrand is bigger.
You've explained it pretty well, despite saying that you don't understand it. Keep in mind he's not talking about the integrands, but the integrals. (The 'smaller' curve (larger values of alpha) has less area underneath it.)
@@xizar0rgyes, so it looks like the integral from 0 to 1 behaves different than the integral from 1 to infinity.
How the sum of both parts behave 🤷♂️
@@thomashoffmann8857Right. One needs to prove that (1, infinity) overweighs (0,1) which is a homework @14:55 😂
Great video! For those of us lucky enough to guess the u-substitution u = 1/(1+x^a) with differential dx = -1/a u^(-1/a-1) (1-u)^(1/a-1) du, this gives the normally recognizable form of the Beta function directly without the trig substitutions. I got stuck on this problem trying to find a satisfying x for the equation with the Gammas.
Wow! It starts as "plain" calculus, then it take a long steep route through the Gamma function, and finally an elementary quadratic formula, that yields the Golden Ratio!
That was beautiful! Compliments to approaching the problem tying it in with gammas and betas and bringing it down to earth with some simple quadratics. And all in < 17 minutes
awesome result
Gee..that was fun❕ The general formula for the integral could come in handy too.
It's been scientifically proven that watching Michael Penn videos every day makes you more smarter.
I have never used calculus for years, but I found watching a math problem being solved satisfying.
As soon as he said it’s between 1 and 2, I said I bet it’s going to be phi. And it was phi. Glorious!
Shouldn't the inequality at the end be flipped?
ohh there is another video in internet of similar or exact same task where also it a power of golden ratio, using that a² = a+1, etc...
I'm pretty sure that evaluates to (1/a)Gamma(1/a)Gamma(a-1/a)/Gamma(a)
for a=phi, phi-1/phi=1, 1+1/phi=phi, tGamma(t)=Gamma(1+t),
(1/phi)Gamma(1/phi)Gamma(phi-1/phi)/Gamma(phi)=Gamma(1+1/phi)Gamma(1)/Gamma(phi)=Gamma(phi)/Gamma(phi)=1
This is based on making a u-substitution x^a=u, which gets it to the form (1/a)Int_0^Inf(u^(1/a-1)du/(1+u)^a), which is (1/a)Beta(1/a, a-1/a). One Euler's reflection formula later, and we're at the point where I started.
15:30
Fun fact, phi is the only value in (1,2) for which antiderivative has a nice form.
What form does it take?
Beta func def need z-1 and w-1?
You get alpha to be 1.618. The limit of the divisor would then be x power 2.61, and integrated x power 1.61. This is about 1 divide by x power 3/2 (three halves). It is Kepler's law! The square root of time power 3 is an axis of an ellipse. So we get 1 / axis of an ellipse. This may be the eccentricity or the 1 / eccentricity if it has a limit. In other words, if it has a limit, it is an elliptical (and not hyperbolic orbit). It is not exactly 3/2 because 1. the Moon is moving away from the Earth and the planets are moving away from the Sun 2. the universe is expanding 3. one does not count in the Earth's motion against the fixed stars.
The inequality in the HW should be the other way
love how a intimidating looking integral and quotients of beta and gamma functions boiled down to a simple quadratic equation
As soon as I saw that set up I had a feeling the golden ratio was coming
Haha, same - something about the repeated powers of alpha.
What if I press the like button non-gently?
Then a small black hole will appear in your basement and pull you towards it for your entire life
Just kidding just dont break your device you know
Daddy
I won't kink shame
Asking the real questions
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Hi Michael, thank you for the video. Appreciate if due credit be given to the author of the math problem.
To golden calculus, golden number!
Thank you very much
What a yet additional beautiful manifestation of the golden ratio!
golden ratio
*“It’s wabbit season!”*
7:20 i can’t get why the upper integration bound after theta substitution is pi/2. It was obvious in case of substitution x=tan(theta) but in this case we’re doing substitution x=tan(theta)^(2/alpha) so shouldn’t the upper bound be like (pi/2)^(2/alpha) ?
The thing is, tan(pi/2)^(2/alpha) continues to be infinity, since (infinity)^(2/alpha) = infinity. So you should consider the angle that is inside the tan(theta)^(2/alpha) which is, in this case, only pi/2, if that makes sense.
α = +log(10² )
why 1/alpha + 1 = alpha?
he didnt say that it's a necessary condition, just that if it were to be true then alpha would solve the problem.
A lot of math problems have "clever" solutions that make things work. That's what he's talking about when he mentions "wishful thinking". (This is similar to when Blackpenredpen uses "wouldn't it be nice".)
He chooses a solution that works (golden ratio) as a "clever guess", and then leaves it as a homework problem to show there are no other solutions (while sketching out an explanation).
Oh I see
I had the same question 😜
I think the HW should be clearly reversed, since for alpha->1 the integral diverges and must be greater than for any beta>alpha.
Upon checking numerically, the inequality appears rather spurious to me, since the integral has a unique minimum on (1,infinity) after which it approaches 1 from below as alpha->infinity.
Is it just me or the integral boils down to b(1/a, a-1/a)/a instead of twice that
This is exactly what maths 505 did.
Of course its the golden ratio. Why wouldn’t it be the golden ratio. Why would it make sense for it to be anything but the golden ratio? Lmao
Thoose who know
Take a look at problem 5 of IMO 2022
And Problem 1 of IMO 2023
Both are number theory problems :)