a formula for the "circumference" of an ellipse.
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One last sanity check is to plug a=b. The only term that survives from the sum is the n=0 term as the rest have a power of (a-b), and the "factorial" term is trivial. We are left with 2 pi * (a+b)/2 = 2 pi a, or in standard lingo 2pi r
As people stated, the first term is 0^0 which is normally ill defined, but we can think instead of the limit of a->b since we want continuity in a,b and we have x^0 with x->0 which is 1.
One can make sure that this definition is consistent
@@tcoren1 Whenever we're writing series using summation notation, especially for power series, it is traditionally assumed you mean that a^0=1 for the 0th term regardless of the value of a just so that you can write the summation notation nicely. In this case, the first term was already 1 from when he brought it into the series at 18:19 and then got multiplied by pi(a+b) when taking the integral. So the first term of the ellipse series is always pi(a+b) which becomes the proper 2pi r for the circle without having to consider limits at all.
Another sanity check: Let b go to 0. L should be 4a. Looks like the sum should converge to 4/pi, which I don't see at the casual glance, but suppose should be provable.
@@user-gs6lp9ko1chow did you get rid of the full sum in that case?
@@tcoren1 As b goes to 0, the length of each side of the ellipse goes to 2*a, and L is two times that. From Michael's final equation, as b goes to 0 each term in the sum has an a multiplier, so that can be brought out front. At that point, we have L = 4a = pi*a* the sum. Therefore the sum should equal 4/pi. If Michael's work is correct, that in itself is a proof of the sum converging to that value. It would be a good check if there is another way to show the sum converges to that value, however.
Instead of using Cauchy's formula for the product of two series, I would've noted that the integral of exp(2i(n-m)) from 0 to 2pi is 0 if n != m, and otherwise is 2pi if n=m. I.e 2pi*kroneckerDelta(n,m).
I also want to point out that asserting sqrt[(1-A)(1-B)] = sqrt(1-A)sqrt(1-B) is sketchy when working with complex numbers
That's very roughly what he does at 17:20 but you still need to expand Cauchy's formula to get exp(2i(4k-2n)θ) (which is nonzero only for even n and k=n/2). Keeping the exponentials and later applying the Kronecker delta would have been clearer but perhaps a bit notation-heavy
@@bonzinipcan you explain me the kroneckerDelta please. I never heard about it. And what does it improve or make worse than his variant?
Matt parker did a video about this topic and he discussed about this infinite series to calculate the perimeter as well as a lot of different approximations for the formula. I still find it strange that the area of an ellipse is just a simple generalisatiom of the area of a circle but the perimeter is so much crazier than a circle.
For a given area, ellipses are families of shapes and have parametrised forms, so you can include a given area in many ways. Somehow you need to specify exactly which ellipse covers the same area, which is why the perimeter is more complicated. There's only one circle with a given area.
I think that Matts point in the end that circles are just as crazy, but we have hidden the infinite series inside of pi is a really nice way of looking at it.
Thanks. I now have done my morning math calisthenics. 😀
Interesting: no exact formula for the circumference of an ellipse, yet there is an exact formula for the area of an ellipse.
Depends on what you mean by exact. Exact in terms of polynomials? Define a sufficient number of “special” functions (like roots, sines, exponential, Bessel…) and many things become exact. More useful to think of computable (get arbitrarily close in a finite number of operations that you know how to complete) than exact.
@@glennjohnson4919 I think he was referring to elementary formula.
@@DrR0BERT Anything that depends on pi is only "exact" if you consider pi as exact. But there's no exact expression for pi - only things like infinite sum expressions. So when you say "C = 2 pi r" that's shorthand for an infinite sum expression, its just that the sum that comes to pi is very useful so we give it a name. The sum in the ellipse perimeter expression could have a special name, and then the perimeter of an ellipse would be "exact" too. See StandUpMaths th-cam.com/video/5nW3nJhBHL0/w-d-xo.html
@@glennjohnson4919 'exact' is a commonly understood term in mathematics. You just described computable things and not an exact equation
and it's so simple too :D
I had no idea you could just take the binomial coefficient of a fraction like that
The absurdity! I have half a chance, and you choose three of it.
It's a Taylor series. The binomial expansion for positive integer powers is also a Taylor series, a special case where it terminates.
I remember in high school, after calculating the circumference of a circle by an integral, the teacher (improvising) tried to calculate the circumference of an ellipse. The rest of class was spent trying to calculate incomplete elliptic integrals in terms of elementary functions. It didn't go well. The next class, he apologized.
I did the same thing with data analysis teaching pre-algebra last Tuesday! Lately I’m focused on getting my students to ask good questions and just try things before I tell them about a new idea, method or approach. When everything is canned, we’re basically actors and salesmen. When things can be random we become clearly imperfect co-learners/explorers (i.e. mathematicians [instead of standards delivery bots]).
About 30 years ago at small, remote branch of a university in the southwest, their welding inspector asked me how to determine the amount of metal he needed to make an elliptical top on a horse trailer was.
Off of the top of my head, I couldn't remember any such formula and I tried to tell him that he was better off with a quick estimate of the distance than to try to get it exactly -- that it would be extremely difficult to actually form an exact ellipse by bending metal and cutting it with a cutting torch.
I thought that it would probably be better to have two quarter circles of metal, one on each side, and connect them with a piece of flag metal.
The guy didn't like my suggestions at all. He was bound and determined to measure out a perfect half ellipse.
Does he use the exact value of pi in his calculations as well?
When you derive a sum like this, it would be cool to spend a few minutes on the rate of convergence afterwards. Besides knowing how many terms you might need for reasonable accuracy it says something about the formula.
Yes, I want to know this *and* its relationship to 1 *and* what conditions there might be that make it greater or lesser.
Yes, I've been frustrated with ellipses for a while, why do you ask?
Ellipse is 4x (or 2x depend on how you see it) symmetrical hyperbolic that sewn/looping together, so yeah its definitely convergence.
Yes it's easy to see it converges, but perhaps not obvious how quickly. It's not easy to see, but the coefficients decrease at roughly a rate of 1/(pi*n^(5/2))
despite English being my second language, thus only being able to understand about half what you're saying using context clues, I still found this very interesting!
Instead of a and b a similar formula could be based on r and the "eccentricity" f. For any specific eccentricity the formula would then be r * . And I guess the main reason that we consider the circle circumference formula - i.e. the case f = 1 - "simple" and those for other ellipses "strangely complicate" is convention, since in this case we have a short name for the (actually also complicate) factor, we call it "2π". If the factor for e.g. all 2:1 ellipses would also have a name, e.g. "µ", the general circumference formula for those would be just as simple: µ * r
The eccentricity of the circle is 0. An eccentricity of 1 corresponds to the parabola. (Don't take this as me disregarding the idea. Writing the formula in terms of eccentricity makes a lot of sense.)
Alternatively, you could replace Pi with (-1/2)! ^2 seeing as we're already using factorials. (Or use gamma.)
This is really intriguing. I would never have thought that the calculation of the perimeter of an ellipse would be so complex. I realize that now I must complete my study of Calc I.
One of your best videos yet!
it was great to see how he simplified this calculation
Phew ! Thank you for your clear explanations.
Maybe you could make some videos on the arithmetic-geometric mean and elliptic integrals? It seems interesting, but I've never been able to get past the complicated notation and identities to figure out what's really going on.
He's already made one. th-cam.com/video/KI-S567giR4/w-d-xo.htmlsi=KHT6v01MAfq6wsh5
To me, it always appeared that the perimeter of the ellipse is similar to that of the circle, just with two parameters for the axes instead of the radius, and a constant that acts like pi, but is confined to exactly that ellipse, depending on a and b. You finally showed me the series to calculate that pi-replacement. Thank you!
20:08
For me, watching Michael Penn go through a derivation like this is like riding shotgun for a friend driving like a bat out of hell on a dangerous, winding mountain road. I know the way, but if I were behind the wheel, I'd be going a helluva lot slower, and still making wrong turns along the way. In the end, I just sit back & enjoy the scenery.
Can you do the version that depends on eccentricity too?
What a journey!
This is super intense!
When ever I see a derivation like this, which to me seems like a complex chain of steps that don't obviously follow one from the other I wish I knew the thought processes of the originator. So much could be learned that way! I am very grateful to M.P. for doing alot of videos in this mold.
While the intuition of the people who look for these derivations really help. A lot of the time it still boils down to trying a whole bunch of different things and stumbling across a useful solution.
@@ewanlee6337 Thanks. But its hard to believe they are "stumbling around completely in the dark," like an amateur like me would be.
@@r2k314 I did say their intuition helps and they are able to come up with and try possibilities much faster than normal people. Plus they’ll likely have standard techniques that’ll solve some derivatives. But there’s no algorithm to finding derivations so there’ll always be an aspect of just guessing and checking, even if they can make better guesses.
@@ewanlee6337 Your right. I misread you at first. Sorry. Thanks for your response.
I mean, you can't write the formula for a circles circumference without an infinite series either...we just happen to condense the entire series into pi...in theory i think you could do the same for ellipses, where youd end up with a different pi-like term for any selection of a and b
There is no pi like term for an ellipse.
Newton binomial for a fraction without any justification, I’m shook
Of course the power series representation can then be expressed in a fairly standard way using the hypergeometric 2F1.
Before I watch the video, can't you just find the length of the parametrized curve that represents an ellipse from t=0-2pi?
It could have been interesting to make a small detour into the work of Laplace and Bessel in this, and how it relates to the time of flight problem in astrodynamics.
Few improvements. In the final expressions should have kept (a+b) x PI which would be a naive formula for circumference in line with that of a circle. Then the series are corrections to this starting at 1 + etc. The 1x3x5x...x(2n-3) is usually denoted via double factorial so (2n-3)!!. No need to complicated with Cauchy, just multiplying put the two sums you get that unless the exponents in the indices don't offset, the terms are zero due to integration, so basically what remains are diagonal terms.
I am puzzled by the “1×3×5×…×(2n-3)” part of the formula, which Michael applies to both n=1 and n=0 at the end of the video. But in those cases the last term, (2n-3) will be negative!
How to interpret “1×3×5×…×(-1)” (for n=1) and “1×3×5×…×(-3)” (for n=0)?
Presumably you can think of them as the empty product, which would conventionally be taken to be 1
@@Alex_Deamso the numerator is ...
for n = 0 ... ( 1 )
for n = 1 ... ( 1 )
for n = 2 ... ( 1 )
for n = 3 ... ( 1 * 3 )
for n = 4 ... ( 1 * 3 * 5 )
for n = 5 ... ( 1 * 3 * 5 * 7 )
etc ???
@@MrSummitville I imagine so
this way of writing the series-product annoys me greatly; the version with decreasing half-integers is at least intelligible for n=1, but needlessly masking the first factor ruins it (it would have become -1)
and even then it's poorly defined for n=0, which itself needs a slightly better look at whats going on with these non-integer binomial coefficients from the start, to be sure which trivial value makes sense (like sure it's probably 1, but guessing isn't math, and it's not moving quickly in my head right now)
so there's almost certainly a much better way to write this expansion down, given these starting points
It would be nice if somebody could put up a webpage wherein one could input a and b, and then the answer for the perimeter would be spat out -- the website doing all the number-crunching for the user. These maths are above my head, and I would give just about anything to be able to intuitively 'grok' the inner meanings and concepts . . . but, alas, my math skills dead-end at a mental brick-wall, you might say, and I have to take the word of somebody like Michael Penn, who obviously knows what he's talking about!
In theory, one could draw an ellipse with a Long-to-Short axis ratio of, say, 3-to-2, and then use a flexible tape measure to actually physically measure the perimeter. That would give you an 'accurate' perimeter length, though how precise it might be is dependent on how precisely one does the measuring. Wrapping such a flexible tape measure around an elliptic cylinder should be a simple enough procedure, right? Wrapping a flexible tape measure around a circular cylinder which has a diameter of, say, 1 meter, should result in that tape measure ending up with a final reading of 3,142 millimeters (rounded to the nearest millimeter, of course) . . . right?
Regardless, it would be nice for somebody to set up an ellipse perimeter calculator website which allows a user to just plug in a and b and then get the answer for the perimeter length spat out, preferably to a number of significant digits that provides as much accuracy as one desires.
Here is another related video on the circumference of an ellipse by Michael Penn: th-cam.com/video/fYcUcj5kYqI/w-d-xo.html
you can use the double factorial to simplify the notation
How
I guess it's just not useful enough to introduce a new symbol that doesn't help solve the problem
So many steps required to get the solution. Here’s the key question … how do you know a step is getting you closer to the solution versus a dead end?
For example, a computer could grind through all possible chess moves out 20 steps and recognize the move sequence to get a checkmate; but algorithms can identify obvious bad moves, eliminate those possibilities and make finding the move sequence solution much more manageable.
Likewise, there is a standard cookbook of possible math steps; such as tables of derivative solutions or transforms between coordinate systems or euler’s conversions between trig functions and complex exponentials. With so many possible steps and dozens of steps required to get the solution, you must have a heuristic algorithm in your head (call it intuition or experience if it can’t be described, but it’s there)
One thing worth noting, when a=b, most of the terms of the sum vanish except for the 0th term, leaving pi*(2a). When a=b, it's typically called the radius of a circle; i.e. C=2pi*r
Excellent vid! Always seemed strange to me that you can deform a circle (maybe just by measuring x and y in different units!?) and a world of complexity opens up.
That is just an infinitesimal part of geometries. All this happens in an ideal 2d plane with an ellipse being a set of points where the ideal ellipse has two fixed parameters a, b. When a = b or b = a (there is relativity) it is case of a circle with radius r = a = b and pi becomes the thing that it is well known about being. Another poster commented value of pi is really an exceptional case - and in many way it is. But pi on the other hand can be very very helpful.
An ellipse may be traced out and this approximates an ideal ellipse in 2d space.
@@Alan-zf2tt I suppose what feels weird to me (disclaimer - physics not maths is my bg) is that if you measure x and y in different units then you cannot (easily) reconcile them into a circle. *Even if* they were measured in the same units - they would be. But I feel I need to ponder your answer a bit more :)
@@MacHooolahan I do not know if it will help but:
fixed pin at one position
fixed pin at another position
loop of fine twine or string
trace out a line (locus of points) with a pen/pencil constrained by string.
This where distances a and b are measured from to define an ideal ellipse
When the fixed pins coincide exactly at same position then this is the case of an ideal circle.
There are some interesting TH-cam videos on projective geometry or algebraic geometry.
Important? Ans: Yeh!
a 2d triangle has sum of internal angles at 180 degrees
A triangle traced out on a 3d sphere can have angle sum 270 degrees
For practical applications, it would be nice to see how approximate formulae can be derived from this one. Also, it would be nice to have an estimate of the error made if only the first k terms of the series are used to calculate the perimeter. For example, if k=0, the formula reduces to calculating the circumference of a circle having as radius the average of the two semi-axes.
Nice!
It would have been useful to compare a range of examples and compare answers to those derived from other estimates of the ellipse perimeter.
One thing that is problematic to me is that if we set a = b, we should expect the usual formula for the circumference of a circle 2πa, however, because we have (a-b)^(2*n) in the numerator, the whole formula evaluates to zero. So what, if anything, have I done wrong.
That expression shoud make sense when a=b, and it does!
How does all this stuff can be kept in a single man's head. Michael is a GOD
No he doesn't. He looks at his paper periodically
I always use a pin and a bit of string.
Man, I don't know any of this stuff. I just came to say that I love the platypus in the thumbnail.
It would be good to explain what's going on with the a=b case. It looks like in that one all the terms are zero except the first one which has a 0^0 term
brutal, just brutal, I guess this is why elliptic integral have the reputation they have
Math was beyond me but very interesting. Matt Parker did a video a while back exploring formulas that approximate the circumference and touched on the calculus approach but not to this extent.
"We're gonna derive this really quickly" Continues to assume without justification that the binomial formula holds for fractional exponents. Yeah, that will make quick work of the identity.
I am guessing this is a nice introduction to solving elliptic integrals and starts where most calculus text leave off after introducing power series. I have often wondered, what are they leaving out of the chapter on power series? Now, I know. Thank you! Feel free to reply if my understanding is not correct.
wondering if there are any "special" values of a and b that give simpler expressions through the use of identities
let b=ax, for example setting x=1 gives the standard circle perimeter, things like that
I believe there is an error in your final formula for the perimeter where the sum starts at n=0 but the last factor in the numerator is (2n-3) which is negative.
a much simpler approximation that's gonna be close enough for pretty much anyone's purposes is:
for eccentricity, k: p = 4 +(2pi -4) *cos(pi/2 *k)^(0.65 *k^(1/5))
the error is almost always less than 1%
as you can see it's basically just a single cosine scaled to fit between the circumference of a circle, 2pi, and the perimeter of an ellipse with eccentricity of one, 4. part of that scaling involves an awkward exponent, but that's it.
@@sumdumbmick David W. Cantrell developed an approximation that is said to be quite good, yet isn't horribly complicated. One web page mentions "an overall accuracy of 4.2 ppm." But it is an approximation. The result in this video provides as much precision as your computing power and patience will allow.
It must be x=a.cos @ ^ y=b.sin @ instead x=a.sin @ ^ y=b.cos @. Is not the same, for example, if @=0º, we´ll have A(0; b) instead A(a; 0).
Very nice indeed. We won't worry about the surface area of an ellipsoid in 3 dimensions which would be the next level of complexity.
@Micheal Penn, If a circle is a special ellipse, can you use the perimeter of an ellipse equation to derive the perimeter of a circle.
Yes. A circle of radius r is an ellipse with a=b=r, using the notation in this video. If you make this substitution into the sum formula for the ellipse perimeter, every term aside from the n=0 term disappears, as they are each connected to a power of a-b (which is 0 here). The remaining n=0 term gives us (pi)*(a+b)=2*(pi)*r, which is the circumference of the circle.
It is not too far ( but not easy) from here to estimate even the Spira mirabilis " circumference"
woah!
In theory, there should be a function such that f(0) = 4, f(1) = 2pi, and f(b) = the perimeter of any ellipse inscribed in the unit circle (i.e. a = 1 and b is a variable between 0 and 1). I wonder why it's so hard to find such a function.
I don't get what the function should look like. Why f(0)=4?
@@not_vinkami because as the lesser radius of the ellipse approaches 0, the perimeter of the ellipse approaches 4 times the greater radius (which, if the ellipse is inscribed in the unit circle, is 1).
As for what the function would look like, I'm not sure. It's probably not linear; I'm sure f(1/2) would not be 2 + pi. It's probably some sort of curve, and which way it curves probably depends on whether f(1/2) is greater than or less than 2 + pi.
@@paulchapman8023 the thing is there's a proof that that formula has not a form expresable in terms of usual functions. In particular it's not linear, but that can be seen in a simpler way just checking it's value in b = 0, b=1 and, divided by b (I mean f(b)/b) in b= +infty.
Very nice! Some steps were surprising to me. Ellipses seem to be tricky bastards...🤣
I think it's better to keep the formula in the penultimate stage, where you have (a+b)/2 * 2pi * stuff.
Because then you can compare to 2pi*r and say: the circumference of an ellipse is the same as the circumference of a circle with the radius equal to the average "radius" of ellipse, except modified by (stuff).
Then we can explore (stuff) in regimes like a = 0, a = b and a = 2b to see whether or not it's what we expect
nice. too bad we can't exploit the "conic slice viewpoint" of a "tilted circle" to make it work. i've tried ... it's still rough.
This is a problem that has haunted me for years. I work as an architect, so ellipses are very important for your designs.
Please don't use this to build something ugly 😂
Buildings shouldn't need ellipses, stick with straight lines. Love from a disgruntled builder 😂
What is the geometric interpretation of the elliptic integral of second kind?
Thanks!! Now if I can just pick my brain up off the floor....
My problem: if a = b then we have a circle. But putting a = b inside this formula gives 0 ( because on a-b term).
For n = 0, it will be 0⁰. Which is widely defined as 1 for infinite series
Does it though? Isn't it the case that the formula can't be used in the case where a=b because the first term of the sum would then involve 0 to the power 0 and so the whole thing would be undefined?
@@mikenorman2525 If we take a look at power series, at n = 0 we will get something like ax⁰. It will be a problem if x = 0. So, we usually define 0⁰ = 1 for this case. I think it's the same reason for a = b (in this video).
The a-b exists on all but the n=0 term. This means, for a=b, the result is just \pi (a+b) = 2\pi a as required
Can we show that the lim(a->b) of the end expression equals 2pi*b?
I have a stupid question. Charaterize ellipse with pi(a+b) as circumference. Beside a circle a=b.
Very interesting. Such a simple shape ... yet quantifying its perimeter is much more difficult than it is for a circle. Nor is there a convenient way to get the length of an arc between two arbitrary angles -- again, unlike a circle. It is disappointing that solutions which are both simple and exact remain out of reach after millennia of study.
yes you can, if you do it with percent calculating a sirkle from a square (78.5%) and you know the rest. its 78.5% of the rectangle
The perimeter of circle is (pi/4) of the perimeter of the square. But that math fails for the perimeter of a very long & narrow ellipse vs the perimeter of the rectangle - they are nearly equal.
Nice derivation that avoids defining elliptic integrals, which are quite messy!
The perimeter of an ellipse is pretty simple: 4aE(e), where a is the semi-major axis, e is the eccentricity, and E() is the complete elliptic integral of the second kind.
If you think that isn't "simple", you need to ask yourself why you think that. What makes (say) trigonometric functions "simpler" than elliptic integrals? Reality doesn't care what we choose to teach in high school.
What is this descending product thing? Shouldn't you use the gamma function on 1/2! ? Г(1/2)/n!Г(1/2-n) ?
Spends 20mins calculating an integral relating to ellipses... not once mentions the phrase "elliptic integral".
So when b goes to 0, a can be taken out in front of the summation. Presumably, the summation then goes to 4/pi, although I don't see it at a casual glance....
Wouldn't the sum go to 4a, not 4/pi ?
@@Mark_Bridges The perimeter of the ellipse, as b goes to 0, is L = 4a. Then, in Michael's final equation, as b goes to 0, there is an a in each termof the sum that can be taken out front. At that point, we have L = 4a = pi*a* the sum (where I mean the sum after the a is taken out, of course). That sum must equal 4/pi where the pi in the denominator cancels the pi in front.
@@user-gs6lp9ko1c I see what you mean now. Luck figuring it out.
I wonder, when is a bad place to stop?
Isn't there some way to represent this with elliptic integrals or something?
Yes, the elliptic integrals were originally invented to work out the circumference of an ellipse.
There is, but there is no 'elliptic integral' button on my calculator. Nor does Excel have a function for them. AFIK, one needs a power series to get the length of an elliptical curve. This series allows writing a program to compute the length of the entire perimeter.
Isn't this Jacobi's Epsilon function?
I think it is better to leave the last line written in the form
π(a+b)*sum(...)
because when we replace a=b we get the circle length formula. Otherwise, if we replace a=b in the last line of the video, we would get that the length of the circle is π, and is independent of the radius, which is absurd.
I get perimeter = π * ( a + b ) which is correct when a = b.
This video doesn't include Exponential Numbers. When are there going to be exponential numbers?
Can someone explain the argument from 17:20 why the double sum collapses to the one sum?
It's Merten's theorem or Cauchy product, I've an article written for my students on it, but it's a bit technical. But the idea is simple, just multiply 2 polinommials of partial sums and extend it to series. The technical part is that effectively the limits work in a friendly way here. I speak spanish so sorry if Im not clear.
@@Ligatmarpingcan you send the article?
The formula should simplify to 2*pi*r when a=b but it appears that each term would be zero due to the a-b factor in the numerator so what gives??
The numerator is = 1
Is there a similar video to derive the surface area of an ellipsoid?
pi*a*b that one is easy, the perimeter is not
oops you wanted area of an ellipsoid sorry! Exact formulae are available for spheroids i.e. ellipses rotated about an axis of symmetry so that say a=b etc. , but more general ellipsoids also yield elliptic integrals as does the ellipse perimeter. The Knud Thompson formula is 4*pi*(((a*b)1.6075 + (a*c)1.6075 + (b*c)1.6075)/3)(1/1.6075). but this approximate with an error of around 1% typically.
that was a bit unnecessarily long way to arrive at the √(1 - x) form, also I don't see why the 1/4 became 1/2
I need a pair of aspirin ...
half of the video is just algebraic flexing and can be skipped by integrating in rectangle coordinates over the positive part and multiplying by two the result(cuz symetry).
I must be getting more cynical as I always had a paper copy of the worked solution that I checked my on-board work with. This way I caught myself if I forgot a sign or denominator. It was a check of my work so students following would have the best chance to get it right upon their review for homework.
this right here is why I didn't go into math hehe... hope everyone is doing well!!
You should have shown the first 3 terms. I imagine the first was pi*ab.
The first one is pi*(a+b) as expected for the circle
WHY ON EARTH isn't EVERYONE else asking where that formula for something raised to the 1/2 power come from? You can't do 1/2 choose not it's not an integer..so unless you extend the formual.somehow but indont see a logical.wsy..tbis seems arbitrary and I don't see how or why anyone would think of it??
I'm confused. If a = b and it is a circle, won't every term be zero?
Every term expect the first
Take the limit a->b, then when n = 0, (a-b)^2n = 1, (a+b)^2n-1 = 1/2, so the sum = 2, you get 2pi as a result.
Wow can’t believe that an innocuous and pretty ellipse could get this ugly?
So how do planets figure it out? 😂
We are also forgetting that we need an infinite series to "calculate" - or rather represent - PI exactly in itself.
So we could introduce a new "constant" ( depending on a and b) for each type of ellipse.
A constant the depends upon the value of "a" and "b" is not a constant.
Hi,
If you set a=b in the formula, you get 0 instead of 2 pi R 😁
No you don't. What you get is 0 for every term of the infinite sum other than the first term where you get 0 to the power 0. That first term is then either simply undefined in which case the formula is not applicable or, as other commenters have pointed out, it actually has the value 1 in which case you do get 2 pi R
@@mikenorman2525 Ok, you are right, I wrote too quickly.
Has it been proven that a simple formula does not exist or have we not found one?
We just didn't bother to tell everyone about a symbol specifically made for making the formula simpler at and only at the first glance
Well, 2pi*r is simple, but this elliptic formula is significantly more complicated for me.
@@paulgillespie542and what is π ?
It seems that after more than 300 years we can finally apreciate a nice approach .
It is interesting to draw comparison between this approach and the way the perimeter can be expressed when the equation of a conic section is expressed in the polar coordinate system
Can you show a proof that there is no formula for the perimiter?
He gave you a formula ...
So obvious.
Glad I did a physics degree and not maths.
I tried searching the channel for proof of binomial expansion formula for non-integer powers, but I failed. :(
Edit: Wait, it's actually simply derived from the definition od Taylor series. Easily differentiable.
Edit2: In 13:00 it's important to note that because |u| < 1, we are in the interior of convergence circle of the Taylor series. If we were at the border, this could be troublesome, because we would need to prove that this formula still discribes the same function.
Edit3: Also later you use the fact that the integral of the series is the series of the integral, this also is not automatic. And here again we use the fact that the series are absolutely convergent inside the circle.
Unfortunate that you stopped at making it in terms of pi; i wonder, and may look into, is there a nice form of pi such that when you multiply gives you something a little cleaner in terms of just a and b?
Like pi is transcendental, its not exactly nice either
Didn't Matt Parker do a Vid on this very nutty problem several years ago on UTube? There's only so many Mathemagical problems under the Sun - or so it seems!
Yeah, but did he ever discuss the solutions that he asked viewers to send in?
I remember coming up with some approximation, but I never saw a followup video, afaik.