[Ryokan]

Just watched this video and wanted to post it on Sound Science but it doesn't fit with any other thread so this seems the obvious place to post it. @71 dB you are very mathematically minded, I found this video very interesting (you are probably already familiar with this concept).
The mathematical techniques of working out answers, mentioned here, weren't taught in my schools and they make equations much simpler and quicker to do.
I would've enjoyed mathematics much more if I'd have been taught differently. I achieved a top grade none the less in basic mathematics but it was aimed at pupils who were less academically inclined.

 

[71 dB]

@71 dB you are very mathematically minded,

Thanks for the compliment! I do consider myself mathematically minded, but how much the word "very" holds up in this context is up to who you compare me with. Compared to someone who struggles to calculate 3*3*3 in his/her head my skills in math are massive, but compared to Terence Tao my skills are almost non-existing.

My mind is very intuitive which helps me understand mathematical concepts and see logical connections, but I am bad at "memorizing" stuff that are needed to do math in a rigorous manner. My MBTI is INTJ while the optimal type for mathematicians is INTP because of Introverted thinking being the dominant cognitive function.

I found this video very interesting (you are probably already familiar with this concept).

I do watch Mathologer's videos because they are excellent recreational math. I did watch this video around the time it come out. However, I don't think it's one if his best. This one is actually somewhat anti-climactic in nature because he demystifies Nikola Tesla's mystic theories. Not that it is a bad video, on the contrary, but not on the lever of Burkard Polsters' better video which can be quite mindblowing at best.

The mathematical techniques of working out answers, mentioned here, weren't taught in my schools and they make equations much simpler and quicker to do.

Do you mean you weren't taught modular arithmetic in school?

I would've enjoyed mathematics much more if I'd have been taught differently. I achieved a top grade none the less in basic mathematics but it was aimed at pupils who were less academically inclined.

Yes, the way math is taught can have a large impact on how interested and motivated the students are. Perhaps the best Youtuber math teacher (at college & highschool level) I have seen is the channel Prime Newtons .

 

[Ryokan]

Thanks for the compliment! I do consider myself mathematically minded, but how much the word "very" holds up in this context is up to who you compare me with. Compared to someone who struggles to calculate 3*3*3 in his/her head my skills in math are massive, but compared to Terence Tao my skills are almost non-existing.

My mind is very intuitive which helps me understand mathematical concepts and see logical connections, but I am bad at "memorizing" stuff that are needed to do math in a rigorous manner. My MBTI is INTJ while the optimal type for mathematicians is INTP because of Introverted thinking being the dominant cognitive function.


I do watch Mathologer's videos because they are excellent recreational math. I did watch this video around the time it come out. However, I don't think it's one if his best. This one is actually somewhat anti-climactic in nature because he demystifies Nikola Tesla's mystic theories. Not that it is a bad video, on the contrary, but not on the lever of Burkard Polsters' better video which can be quite mindblowing at best.


Do you mean you weren't taught modular arithmetic in school?


Yes, the way math is taught can have a large impact on how interested and motivated the students are. Perhaps the best Youtuber math teacher (at college & highschool level) I have seen is the channel Prime Newtons .

I struggled in a class environment, I wasn't a good pupil. Because one maths teacher was very large and scary (he used his size and feigned severity to make people learn) I learnt my times tables in just one week, and I loved logarithm's but every else was a huge struggle. What helped enormously was having a private tutor in my last year, she made the subject much more accessible and easier to understand, it is thanks to her I passed my school exam.
I passed my chemistry exam by writing down the whole periodic table on the back of an answer sheet to refer to. But I kinda cheated by making up rhymes for each column.

I read somewhere that maths geniuses are introverts and many at silicone valley fit this type.

 

[71 dB]

I struggled in a class environment, I wasn't a good pupil. Because one maths teacher was very large and scary (he used his size and feigned severity to make people learn)

Was he Matt Striker?

I read somewhere that maths geniuses are introverts and many at silicone valley fit this type.

Of course they are. INTP personality type which is the most optimal for mathematicians is the second most introverted while my own personality type, INTJ is the most introverted. People in Silicon Valley consist heavily of these two personality types.

 

[bigshot]

I was lousy in math. I was always conceptual, and memorization was painful to me. But for some reason I scored in the top 10% in math on my SATs. Either most other people sucked more at it, or what little I did learn stuck with me.

Now I use a calculator and a computer and I don't worry about it.

 

[The Jester]

Some people are just “wired” for maths, remember years ago one customer was a University Math Professor … gave him the account for his car repairs, and as I’d chatted with him several times earlier joked “what’s the square root of that” … in a matter of fact way gave me the correct answer to 3 decimal places .. checked it on a calculator after he left ….

 

[71 dB]

We all learned in school how to calculate the circumference of a circle with a radius r is 2𝝿r. The circumference of an ellipse is (much) harder to calculate. If we have an ellipse with major and minor axis a and b, the exact circumference C is calculated with the integral

C = 4a ∫ (1- (𝝴*sin 𝞱)^2)^½ d𝞱 from 𝞱 = 0 to 𝞱 = 𝝿/2,

where 𝝴 = eccentricity = (1 - (b^2/a^2))^½.

Unfortunately this elliptic integral of the 2nd kind doesn't have a closed form solution. An arbitrarily close value for C can be calculated by summing enough terms of the infinite sum:

C = 𝝿(a+b) [1+∑ ((2n-1)!!/2^n*n!) * (h^n/(2n-1)^2 ],

where h = (a-b)^2 / (a+b)^2. Calculating the first few term of this infinite sum we get

C = 𝝿(a+b) [1+ h/4 + h²/64 + h³/256 + 25h⁴/16384 + 49h⁵/65536 + 441h⁶/1048576 + ...].

Note that if a = b, h = 0 and C = 2𝝿a, the circumference of a circle of radius a as it should be.

Of course calculating the circumference of an ellipse using the method above is unnecessarily cumbersome, if an ballpark estimation is enough. There are several less and more accurate approximations for C. The simplest of these is considering the ellipse a circle with the radius of the average of the major and minor axis:

C ≈ 𝝿(a+b),

but this simple formula gives only about 10 % accuracy. Indian math genius Srinivasa Ramanujan (1887 - 1920) came up with much better approximations of which the formula:

C ≈ 𝝿(a+b)(1+3h/(10+√(a+3h)))

is extremely good.

I came up (intuitively and iteratively) with my own formula

C ≈ 4a*(1+k*(b/a)^𝜷 )^(1/𝜷 ),

where 𝜷 ≈ 1.55325525 (iterated to make the circumference curve match the accurate value at b = a/2) and k = (𝝿/2)^𝜷 - 1 ≈ 1.016620987 (to give the correct circumference when b = a). My formula gives automatically the correct value C = 4a when b = 0. Below is a comparison of Ramanujan's and my methods:

My methods hits better the circumference of 4 when b = 0 and a = 1 while Ramanujan's method gives 3.99839065. I think my method if worse (too large values) when b is around 0.1 - 0.3.

 

[Ryokan]

We all learned in school how to calculate the circumference of a circle with a radius r is 2𝝿r. The circumference of an ellipse is (much) harder to calculate. If we have an ellipse with major and minor axis a and b, the exact circumference C is calculated with the integral

C = 4a ∫ (1- (𝝴*sin 𝞱)^2)^½ d𝞱 from 𝞱 = 0 to 𝞱 = 𝝿/2,

where 𝝴 = eccentricity = (1 - (b^2/a^2))^½.

Unfortunately this elliptic integral of the 2nd kind doesn't have a closed form solution. An arbitrarily close value for C can be calculated by summing enough terms of the infinite sum:

C = 𝝿(a+b) [1+∑ ((2n-1)!!/2^n*n!) * (h^n/(2n-1)^2 ],

where h = (a-b)^2 / (a+b)^2. Calculating the first few term of this infinite sum we get

C = 𝝿(a+b) [1+ h/4 + h²/64 + h³/256 + 25h⁴/16384 + 49h⁵/65536 + 441h⁶/1048576 + ...].

Note that if a = b, h = 0 and C = 2𝝿a, the circumference of a circle of radius a as it should be.

Of course calculating the circumference of an ellipse using the method above is unnecessarily cumbersome, if an ballpark estimation is enough. There are several less and more accurate approximations for C. The simplest of these is considering the ellipse a circle with the radius of the average of the major and minor axis:

C ≈ 𝝿(a+b),

but this simple formula gives only about 10 % accuracy. Indian math genius Srinivasa Ramanujan (1887 - 1920) came up with much better approximations of which the formula:

C ≈ 𝝿(a+b)(1+3h/(10+√(a+3h)))

is extremely good.

I came up (intuitively and iteratively) with my own formula

C ≈ 4a*(1+k*(b/a)^𝜷 )^(1/𝜷 ),

where 𝜷 ≈ 1.55325525 (iterated to make the circumference curve match the accurate value at b = a/2) and k = (𝝿/2)^𝜷 - 1 ≈ 1.016620987 (to give the correct circumference when b = a). My formula gives automatically the correct value C = 4a when b = 0. Below is a comparison of Ramanujan's and my methods:



My methods hits better the circumference of 4 when b = 0 and a = 1 while Ramanujan's method gives 3.99839065. I think my method if worse (too large values) when b is around 0.1 - 0.3.

Would you mind breaking that down and explaining what each value and symbol represents in the equations? Thank you.

The story of Ramanujan is fascinating.

 

[71 dB]

Would you mind breaking that down and explaining what each value and symbol represents in the equations? Thank you.

That's quite a many equations to "break down." Which equation(s) are you interested the most?

The story of Ramanujan is fascinating.

Indeed!

 

[Ryokan]

That's quite a many equations to "break down." Which equation(s) are you interested the most?

How about the first one?

C = 4a ∫ (1- (𝝴*sin 𝞱)^2)^½ d𝞱 from 𝞱 = 0 to 𝞱 = 𝝿/2

Is a 'closed form solution' a number that doesn't have an infinity value?

 

[71 dB]

How about the first one?

C = 4a ∫ (1- (𝝴*sin 𝞱)^2)^½ d𝞱 from 𝞱 = 0 to 𝞱 = 𝝿/2

Is a 'closed form solution' a number that doesn't have an infinity value?

That's a good place to start if you feel that way. Closed form solution mean the integrated function can't be expressed with basic functions. For example function f(x) = 12x² can be integrated for a closed form solution:

F(x) = ∫ f(x) dx = ∫ 12x² dx = 4x³ + C,

where C is unknown integration constant. If you derivate 4x³ + C with respect to x, you get 12x² back, because the derivative of a constant C is zero. Note, this C is not the same thing as C, which is the circumference of the ellipse in question. Unfortunate, that they are both the same letter here.

In the case of elliptic integrals, the function E(𝞱, 𝝴 ) to be integrated is

E(𝞱, 𝝴 ) = (1- (𝝴*sin 𝞱)^2)^½

and it has been proven (beyond my skills) this function doesn't have closed form solution. We can't write a function whose derivative would be this particular function. However, if we integrate the function from 𝞱 = 0 to 𝞱 = 𝝿/2 as is the case when calculating the circumference of an ellipse, the value of the definitive integral can be calculated (approximated) with the infinite sum

C = 𝝿(a+b) [1+ h/4 + h²/64 + h³/256 + 25h⁴/16384 + 49h⁵/65536 + 441h⁶/1048576 + ... ]

The more terms you calculate, the more precise answer you'll get.

 

[Ryokan]

Still trying to understand, it's not a language I'm familiar with @71 dB.

Searching on-line I found this, and they mention our friend Ramanujan:


Ramanujan Formulas of Perimeter of Ellipse.
'Ramanujan, one of the famous mathematicians, came up with some formulas that would give better approximations of the perimeter of ellipse. Ramanujan's formulas for finding the perimeter of ellipse became famous as they are simple and easy to use. Though these formulas do not give the exact perimeter, they can give reasonably a very close answer. The formulas are':

1. P ≈ π [ 3 (a + b) - √[(3a + b) (a + 3b) ]]
2. P ≈ π (a + b) [ 1 + (3h) / (10 + √(4 - 3h) ) ], where h = (a - b)2/(a + b)2. From Cuemath.

Why do his equations look so different to yours?

 

[71 dB]

Still trying to understand, it's not a language I'm familiar with @71 dB.

That's okay. This is somewhat advanced math. For me this is very "familiar" language, because I have a university degree in engineering (M.Eng.). There is not shame in trying to understand something. The shame is not trying to understand! There has been time in the past when I struggled to understand these things and as I said there are some aspects I still don't understand myself.

Searching on-line I found this, and they mention our friend Ramanujan:

Ramanujan Formulas of Perimeter of Ellipse.
'Ramanujan, one of the famous mathematicians, came up with some formulas that would give better approximations of the perimeter of ellipse. Ramanujan's formulas for finding the perimeter of ellipse became famous as they are simple and easy to use. Though these formulas do not give the exact perimeter, they can give reasonably a very close answer. The formulas are':

1. P ≈ π [ 3 (a + b) - √[(3a + b) (a + 3b) ]]
2. P ≈ π (a + b) [ 1 + (3h) / (10 + √(4 - 3h) ) ], where h = (a - b)2/(a + b)2. From Cuemath.

Why do his equations look so different to yours?

Because Ramanujan's and my equations are just approximations. They don't need to "look" the same. What they need to do is give good approximations for the perimeter of ellipse. As you see from the graph I attached in my original post, the curves given by Ramanujan's 2nd equation (the better one) and my equation are almost identical, even when the equations themselves look very different. How is this possible? It's because we are using just a small region of the curves which is 0 ≤ b ≤ a. Outside this region the equations probably give quite different values, (even complex values!), but this doesn't matter, because the region 0 ≤ b ≤ a is all we need. It covers all the shapes ellipses can have.

I don't know how Ramanujan came up with his equations (he was a genius and his mind worked on another level). I came up with my equation with the help of intuition (as an INTJ introverted intuition is my dominant cognitive function) and the experience of approximating datasets in engineering. You could say Ramanujan's way is how mathematicians do it, while my way is how engineers do it. Different approaches, but both seems to give results of very similar accuracy in this case.

 

[Ryokan]

That's okay. This is somewhat advanced math. For me this is very "familiar" language, because I have a university degree in engineering (M.Eng.).

My subject was History, so a world apart from advanced mathematics. Would it be possible to explain the concept of working out the circumference of an ellipse without an equation?

I don't know how Ramanujan came up with his equations (he was a genius and his mind worked on another level).

Neither did Ramanujan, he saw divinity in mathematics and explained it as that.


This video is helping me understand an integral:

 

[71 dB]

My subject was History, so a world apart from advanced mathematics.

What's the application of ellipses in history? I wasn't expecting anyone without good knowledge of math be even interested of my post about the circumference of ellipses.

Would it be possible to explain the concept of working out the circumference of an ellipse without an equation?

I guess it is possible, but I don't think I can give such explanation without thinking about it a lot, if even then. Anyway, the idea is to break the ellipse into infinite pieces that are infinitely short and then put them all one after one. We get a line and the length of that line is the circumference of the ellipse. Of course we can't do that in real life, but math can do it in the abstract realm. That's the power of math!

This video is helping me understand an integral:

Well, that's a short introductory video, but of course it can be a good first step into integrals. I'm sorry I assumed these things to be more familiar to you, because you expressed interest into my post, in fact you are the only one here doing so. Thanks for that!

If you are interested of calculus, I suggest you start with derivation, the opposite operation of integration. Derivatives are often much easier to calculate and they give a good foundation for integrals.