Proportion of the area covered by circles

[Dhamnekar Win,odd]

On 1/12/2024 at 7:21 PM, KJW said:

The following is a solution to the second problem:

Let R be the radius of the large circle.
Let r be the radius of the small circles in the outermost layer.
Let αr be the radius of the small circles in the second outermost layer.
Let n be the number of small circles around the circumference of the large circle.
Let P be the proportion of the area of the large circle covered by small circles for n small circles around the circumference of the large circle.

Partitioning the large circle into n sectors, the area of each sector is:

=πR2n

The total area of small circles covering a sector:

=πr2(1+α2+α4+α6+⋯)=πr21−α2

Then the proportion of the area of a sector covered by small circles:

P=nr2(1−α2)R2

rR−r=sinπn

r=(R−r)sinπn

r(1+sinπn)=Rsinπn

rR=sinπn1+sinπn

P=nsin2πn(1−α2)(1+sinπn)2

Let (1−α2)=βπn

limn→∞P=n(πn)2βπn=πβ

Applying the law of cosines:

Note that (1+α)r is the distance between the centres of the outermost and nearest second outermost circles, and that (R−r) and α(R−r) are the distances from the centre of the large circle to the centres of the outermost circle and second outermost circle, respectively.

(1+α)2r2=(1+α2)(R−r)2−2α(R−r)2cosπn

(1+α)2r2(R−r)2=1+α2−2αcosπn

rR−r=sinπn

(1+α)2sin2πn=1+α2−2αcosπn

(1+α)2sin2πn=(1+α)2−2α(1+cosπn)

(1+α)2cos2πn−2α(1+cosπn)=0

(1+α)2cos2πn−2(1+α)(1+cosπn)+2(1+cosπn)=0

Let ϕ=πn

(1+α)2cos2ϕ−2(1+α)(1+cosϕ)+2(1+cosϕ)=0

Also let ϕ≈0

cosϕ≈1−12ϕ2

(1+α)2(1−ϕ2)−(1+α)(4−ϕ2)+(4−ϕ2)=0

This is a quadratic equation to be solved for (1+α) . However, only the lesser of the two solutions is of interest.

1+α=(4−ϕ2)−(4−ϕ2)2−4(1−ϕ2)(4−ϕ2)−−−−−−−−−−−−−−−−−−−−−−−√2(1−ϕ2)

1+α=(4−ϕ2)−(16−8ϕ2+ϕ4)−(16−20ϕ2+4ϕ4)−−−−−−−−−−−−−−−−−−−−−−−−−−−−−√2(1−ϕ2)

1+α=(4−ϕ2)−12ϕ2−3ϕ4−−−−−−−−−√2(1−ϕ2)

1+α=2−3–√ϕ

α=1−3–√ϕ

α2=1−23–√ϕ+3ϕ2

1−α2=βπn=23–√ϕ=23–√πn

β=23–√

Therefore:

limn→∞P=π23–√

 

 

 

Edited January 14 by Dhamnekar Win,odd
I answered my query myself

[Dhamnekar Win,odd]

KJW,

Would you show me this triangle graphically? 🤔☺️

[KJW]

3 hours ago, Dhamnekar Win,odd said:

I answered my query myself

That's ok, I was curious. How did you solve the second problem?

 

52 minutes ago, Dhamnekar Win,odd said:

Would you show me this triangle graphically? 🤔☺️

Edited January 14 by KJW

[Dhamnekar Win,odd]

KJW,

 How did you derive the distance from the center of circle lying in the second outermost circle to the center of the larger circle is \(\alpha(R-r) ?\) Can you give me a real world example?

Suppose R=10, r=2 and \(\alpha (R -r) = 1\) So, using simple mathematical reasoning, the distance from the center of the circle lying in the second outermost layer  to the center of the larger circle is  4 + 1 = 5. But using your formula \(\alpha (R-r)= \displaystyle\frac12(10- 2)= 4\). How is that? Where I am wrong?  

Edited January 15 by Dhamnekar Win,odd

[KJW]

I don't know where you got those numbers from, but have you interpreted [math]\alpha (R-r)[/math] as [math]\alpha[/math] as a function of [math]R-r[/math] instead of [math]\alpha[/math] multiplied by [math]R-r[/math]? [math]\alpha[/math] is the ratio of the radius of the second outermost circle to the radius of the outermost circle. It is also the ratio of the distance between the centre of the large circle and the center of the second outermost circle to the distance between the centre of the large circle and the center of the outermost circle. Thus, [math]\alpha r[/math] is the radius of the second outermost circles, and [math]\alpha (R-r)[/math] is the distance between the centre of the large circle and the center of the second outermost circle. I chose to define the ratio [math]\alpha[/math] rather than the radius and distance themselves. Much of the solution was about finding the value of [math]\alpha[/math], and even then, only for very large values of [math]n[/math]. It should be noted that the same value [math]\alpha[/math] describes both ratios because the second and subsequent outermost layers of circles are scaled down versions of the outermost layer of circles.

 

Edited January 15 by KJW

[Dhamnekar Win,odd]

@KJW,

How did you arrive at this equation pointed out by arrow? Would you show me its computations? What is mathematical reasoning (mathematical logic ) behind this computations?

Edited January 15 by Dhamnekar Win,odd

[KJW]

3 hours ago, Dhamnekar Win,odd said:

@KJW,

How did you arrive at this equation pointed out by arrow? Would you show me its computations? What is mathematical reasoning (mathematical logic ) behind this computations?

Not stated but used in the solution is the series expansion of [math]\cos \phi[/math]:

[math]\cos \phi = 1 - \dfrac{\phi^2}{2} + \dfrac{\phi^4}{24} - \dfrac{\phi^6}{720} + \cdots = \displaystyle \sum_{j=0}^{\infty} \dfrac{(-1)^j}{(2j)!} x^{2j}[/math]

For [math]\phi \approx 0[/math], the terms after [math]\dfrac{\phi^2}{2}[/math] can be ignored, and the closer [math]\phi[/math] is to [math]0[/math], the better [math]1 - \dfrac{\phi^2}{2}[/math] is as an approximation of [math]\cos \phi[/math]. In the limit of [math]\phi \to 0[/math] ([math]n \to \infty[/math]), the approximation is effectively exact. But [math]\phi[/math] can't be exactly [math]0[/math] ([math]\cos 0 = 1[/math]) because then the ratio of the total area of the small circles to the area of the large circle in the limit of [math]n \to \infty[/math] becomes indeterminate.

Thus:

[math]\cos^2 \phi = (1 - \dfrac{\phi^2}{2})^2 = 1 - \phi^2 + \dfrac{\phi^4}{4} = 1 - \phi^2[/math]    for [math]\phi \to 0[/math]    (ignoring the [math]\dfrac{\phi^4}{4}[/math] term)

[math]2(1 + \cos \phi) = 2 + 2(1 - \dfrac{\phi^2}{2}) = 4 - \phi^2[/math]    for [math]\phi \to 0[/math]

Also, solving the quadratic equation involved carefully ignoring the higher order terms, leaving only the term that is linear in [math]\phi[/math].

 

 

Edited January 15 by KJW

[Dhamnekar Win,odd]

How did you get

 

 

I got \( 1 +\alpha = \displaystyle\frac{3\cdot |{\phi}| +\phi^2 -4}{2\phi^2 -2}\) How is that? Where I am wrong?

Edited January 16 by Dhamnekar Win,odd

[Dhamnekar Win,odd]

 KJW, I solved my query myself. ☺️

[KJW]

On 1/17/2024 at 2:15 AM, Dhamnekar Win,odd said:

How did you get

 

[math]1 + \alpha = \dfrac{(4 - \phi^2) - \sqrt{12 \phi^2 - 3 \phi^4}}{2 (1 - \phi^2)}[/math]

From here, I took a shortcut and simply ignored the two [math]\phi^2[/math] outside the square root and the [math]3 \phi^4[/math] under the square root because I could see that these are not going to be a part of the final result. Thus:

[math]1 + \alpha = \dfrac{4 - \sqrt{12 \phi^2}}{2}[/math]    for [math]x \to 0[/math]  (ignoring higher-order terms)

[math]1 + \alpha = 2 - \sqrt{3} \phi[/math]

 

However, one can verify that the shortcut leads to the same result as follows:

[math]1 + \alpha = \dfrac{(4 - \phi^2) - \sqrt{12 \phi^2 - 3 \phi^4}}{2 (1 - \phi^2)}[/math]

[math]1 + \alpha = \dfrac{(4 - \phi^2) - \sqrt{12} \phi \sqrt{1 - \dfrac{1}{4} \phi^2}}{2 (1 - \phi^2)}[/math]

 

The series expansion of [math]\dfrac{1}{1 + x}[/math] and [math]\sqrt{1 + x}[/math]:

[math]\dfrac{1}{1 + x} = 1 - x + x^2 - x^3 + \cdots = 1 - x[/math]    for [math]x \to 0[/math]  (ignoring higher-order terms)

[math]\sqrt{1 + x} = 1 + \dfrac{1}{2} x - \dfrac{1}{8} x^2 + \dfrac{1}{16} x^3 - \dfrac{5}{128} x^4 + \cdots = 1 + \dfrac{1}{2} x[/math]    for [math]x \to 0[/math]  (ignoring higher-order terms)

Thus:

[math]\dfrac{1}{1 - \phi^2} = 1 + \phi^2[/math]    for [math]x \to 0[/math]  (ignoring higher-order terms)

[math]\sqrt{1 - \dfrac{1}{4} \phi^2} = 1 - \dfrac{1}{8} \phi^2[/math]    for [math]x \to 0[/math]  (ignoring higher-order terms)

 

[math]1 + \alpha = \dfrac{1}{2} (4 - \phi^2) (1 + \phi^2) - \sqrt{3} \phi (1 - \dfrac{1}{8} \phi^2) (1 + \phi^2)[/math]

[math]1 + \alpha = \dfrac{1}{2} (4 + 3 \phi^2 - \phi^4) - \sqrt{3} \phi (1 + \dfrac{7}{8} \phi^2 - \dfrac{1}{8} \phi^4)[/math]

[math]1 + \alpha = 2 - \sqrt{3} \phi + \dfrac{3}{2} \phi^2 - \dfrac{7 \sqrt{3}}{8} \phi^3 - \dfrac{1}{2} \phi^4 + \dfrac{\sqrt{3}}{8} \phi^5[/math]

[math]1 + \alpha = 2 - \sqrt{3} \phi[/math]    for [math]x \to 0[/math]  (ignoring higher-order terms)

 

 

Edited January 19 by KJW