The elliptic integral is tremendously important in classical mechanics; it appears in the exact solution to the equations of motion of the undamped pendulum, when the swing angle is not small enough to linearize the equations. By extension, it is part of the mathematical toolbox used in studying all kinds of coupled oscillations - again, when linearization is not a good approximation.
4:34 - should be a b² - (b²-a²)sin²(theta) instead of 1-(b²-a²)sin²(theta). Divide by b² under the root and pull a b out in front of the integral and so long as a
Fun fact: these functions pop up quite often when you try to calculate scattering amplitudes for elementary particles using perturbative techniques, when Feynman diagrams with multiple loops are involved in the calculation.
I'm really fond of elliptic integrals, lets have more! Did you know: the Euler-Poinsot problem (the "tennis-racquet" problem) has a solution in terms of an incomplete elliptic integral of the third kind? It was only found in the 1970s, by Kinoshita and others, more than 200 years after Euler started work on the problem. Also: the bicylinder (intersection of crossed cylinders), which we had a few days ago, has a solution in terms of second kind elliptic integrals: V(volume) = 8*a*b^2*I where I=(E(pi/2,k)+(1/k)*E(arcsin(k),1/k))/3 and k=b/a (b=1).
Awesome video, need more advanced math videos! Btw, I'm still waiting for proof of Generalized Stokes' Theorem, so hope you'll do a video about it soon.
4:45 should be a² instead of 1 Edit: a²-b² is usually noted as c² which is the half distance between the foci of the elips, and c/a is noted as e (for eccentricity), so over all we get an integrand of the form: a*sqrt(1+e²sin(theta))
You made two small errors. Actually, if b > a as in the video, the factor in front of the square root is b, not a. Additionally, it should be a minus sign before e² instead of a plus sign.
A few years back I figured out an easy calculation to use in a pinch that is about 98-99% accurate. Assume "a" is the longer axis, "b" is the shorter axis. Axis being the radius distance from the center. For ellipses with an a/b ratio up to around 5/1, use Perimeter = [(3.7a/b)+2.4]b. If you get into more extreme ellipses, for an a/b ratio up to 10/1, use P = [(3.84a/b)+2]b. For a/b ratio up to 15/1, use P = [(3.9a/b)+1.7]b. For up to 20/1, P = [(3.93a/b)+1.55]b. You can also use a polynomial function (for whatever crazy reason). For a/b ratios up to around 5/1 try, P = [(0.072a/b)^2 + 3.26a/b + 2.93]b. For a/b ratios up to 20/1, try P = [(0.007a/b)^2 + 3.78a/b + 2.1]b.
The derivation simplifies considerably with three preliminary facts/results. 1. Gamma(a+n)/Gamma(a) = (a)_n where (a)_n is the rising factorial. 2. The binomial expansion can be written in terms of the rising factorial: (1-x)^a = Sum from n=0 to infinity of (-a)_n x^n/n! 3. B(m, n) = Gamma(m) Gamma(n)/Gamma(m + n) where B(m, n) is the usual Beta function. With these it's a two or three line derivation at most.
Take the familiar expression m choose n = m!/(n!(m-n)!) and rewrite m!/(m-n)! as the rising factorial (m-n+1)_n In this form we see that m choose n is perfectly well-defined when n is a nonnegative integer and m is any complex number.
I remember I needed the length of a sine wave. So, I tried to get an explicite antiderivative of sqrt(1+cos(x)^2) and then apply it between 0 and 2pi. But, I've never found it. I used some series to get an approximation. Now, I know why, it's a kind of elliptic integral of the second kind with a cosine instead of a sine and an imaginary k. That's weird and considering symmetries, my solution should be equivalent to 4*E(i).
As non native english speaker I have hard time distinguish between Cee and Zee in US accent. It would be much better if US people pronounce Z as Zat like British
Yup. Actually the only fact we need is that the force is radial i.e. always pointing toward the origin. Let r, v and _a_ be position, velocity and acceleration. Since acceleration is in the direction of force (F = ma) and the force is radial r and _a_ are parallel. Now, the rate at which area is being swept out is the area of triangle (origin, r, r+v) which is r×v/2. To show that this is constant we take the derivative and get 0. (r×v/2)' = (r×v' + r'×v)/2 by the product rule = (r×a + v×v)/2 = (0 + 0)/2 = 0 since the cross product of parallel vectors is 0. QED.
The elliptic integral is tremendously important in classical mechanics; it appears in the exact solution to the equations of motion of the undamped pendulum, when the swing angle is not small enough to linearize the equations. By extension, it is part of the mathematical toolbox used in studying all kinds of coupled oscillations - again, when linearization is not a good approximation.
And in cryptography too!
4:34 - should be a b² - (b²-a²)sin²(theta) instead of 1-(b²-a²)sin²(theta). Divide by b² under the root and pull a b out in front of the integral and so long as a
Right
Shouldn't it be 1 - a^2/b^2 = k^2 ?
@@xCorvus7x (b²-a²)/b²
@@ZetaGirlPower ah, it appears you forgot the brackets
One should mention thart k is the numerical eccentricity of the ellipse.
Fun fact: these functions pop up quite often when you try to calculate scattering amplitudes for elementary particles using perturbative techniques, when Feynman diagrams with multiple loops are involved in the calculation.
Except for the mistake that’s been pointed out repeatedly, this was very enlightening.
Thank you, professor.
I'm really fond of elliptic integrals, lets have more! Did you know: the Euler-Poinsot problem (the "tennis-racquet" problem) has a solution in terms of an incomplete elliptic integral of the third kind? It was only found in the 1970s, by Kinoshita and others, more than 200 years after Euler started work on the problem. Also: the bicylinder (intersection of crossed cylinders), which we had a few days ago, has a solution in terms of second kind elliptic integrals: V(volume) = 8*a*b^2*I where I=(E(pi/2,k)+(1/k)*E(arcsin(k),1/k))/3 and k=b/a (b=1).
Awesome video, need more advanced math videos!
Btw, I'm still waiting for proof of Generalized Stokes' Theorem, so hope you'll do a video about it soon.
I just wrote out some notes for this video the other day! It took me a while to get something I was happy with.
4:45 should be a² instead of 1
Edit:
a²-b² is usually noted as c² which is the half distance between the foci of the elips,
and c/a is noted as e (for eccentricity),
so over all we get an integrand of the form: a*sqrt(1+e²sin(theta))
not b^2?
You made two small errors. Actually, if b > a as in the video, the factor in front of the square root is b, not a. Additionally, it should be a minus sign before e² instead of a plus sign.
A few years back I figured out an easy calculation to use in a pinch that is about 98-99% accurate. Assume "a" is the longer axis, "b" is the shorter axis. Axis being the radius distance from the center. For ellipses with an a/b ratio up to around 5/1, use Perimeter = [(3.7a/b)+2.4]b.
If you get into more extreme ellipses, for an a/b ratio up to 10/1, use P = [(3.84a/b)+2]b. For a/b ratio up to 15/1, use P = [(3.9a/b)+1.7]b. For up to 20/1, P = [(3.93a/b)+1.55]b.
You can also use a polynomial function (for whatever crazy reason). For a/b ratios up to around 5/1 try, P = [(0.072a/b)^2 + 3.26a/b + 2.93]b. For a/b ratios up to 20/1, try P = [(0.007a/b)^2 + 3.78a/b + 2.1]b.
The derivation simplifies considerably with three preliminary facts/results.
1. Gamma(a+n)/Gamma(a) = (a)_n where (a)_n is the rising factorial.
2. The binomial expansion can be written in terms of the rising factorial:
(1-x)^a = Sum from n=0 to infinity of (-a)_n x^n/n!
3. B(m, n) = Gamma(m) Gamma(n)/Gamma(m + n) where B(m, n) is the usual Beta function.
With these it's a two or three line derivation at most.
Nicely simplified. Thank you. I noticed that you used recursion for sine integral, beta function also works I suppose.
thank you for this series Michael!
I just ran into these for the first time today while messing around in Mathematica , what a coincidence
b^2 - (b^2-a^2) sin(O)^2
1/2 choose n? Does that make any sense? What happens to the choose function when you evaluate it outside the usual range?
He gives the definition in the video
Take the familiar expression m choose n = m!/(n!(m-n)!) and rewrite m!/(m-n)! as the rising factorial (m-n+1)_n
In this form we see that m choose n is perfectly well-defined when n is a nonnegative integer and m is any complex number.
Are you going to get into Sturm-Liouville stuff in this series?
Very nice explanation 😀😂😀.
I remember I needed the length of a sine wave. So, I tried to get an explicite antiderivative of sqrt(1+cos(x)^2) and then apply it between 0 and 2pi. But, I've never found it. I used some series to get an approximation. Now, I know why, it's a kind of elliptic integral of the second kind with a cosine instead of a sine and an imaginary k. That's weird and considering symmetries, my solution should be equivalent to 4*E(i).
Don't need imaginary numbers, sqrt(1+c^2) = sqrt(2-s^2) = sqrt(2) sqrt(1 - (sin(x)/sqrt(2))^2), so the integral is sqrt(2)E(1/sqrt(2)).
@@aadfg0 Indeed. Just take the right identity
Did we lose an b^2 in the radical when we replaced it with a 1 term?
As non native english speaker I have hard time distinguish between Cee and Zee in US accent. It would be much better if US people pronounce Z as Zat like British
Wouldn't it be even better if the US people eschewed English altogether and spoke _your_ native language instead? 🤔
@@nHans My native language is totally horrible for learning and explaining STEM skill, english is the best language
@@ThainaYu I'm curious, which one is that?
@@xCorvus7x Thai
@@ThainaYu could you tell me how exactly Thai is horrible to speak about maths and sciences?
very cool video
13:06
Beat me to it
I don’t follow how the binomial expansion works, specifically the part about abs(x)
Ah never mind somebody else explained in another comment that the binomial expansion is extended to account for n>alpha
OK I've got this ellipse I need to measure the circumference of (have a very demanding boss!) - what do I do again? x
4:34
I guess, I should be sqrt(b^2 - (b^2 - a^2)*sin^2(theta))
Yes indeed.
Right
Nice
I was just preparing for my conc section test tomorrow and found this ...
👍
Is there any easy proof of Kepler's law that planets in their elliptic orbits sweep out equal areas in equal time?
Yup. Actually the only fact we need is that the force is radial i.e. always pointing toward the origin. Let r, v and _a_ be position, velocity and acceleration. Since acceleration is in the direction of force (F = ma) and the force is radial r and _a_ are parallel. Now, the rate at which area is being swept out is the area of triangle (origin, r, r+v) which is r×v/2. To show that this is constant we take the derivative and get 0.
(r×v/2)' = (r×v' + r'×v)/2 by the product rule
= (r×a + v×v)/2 = (0 + 0)/2 = 0 since the cross product of parallel vectors is 0. QED.
Stand Up maths also made a video about it