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Question Number 162530 by mr W last updated on 30/Dec/21

Commented by mr W last updated on 30/Dec/21

find the sum of areas of all (infinite)  red circles

$${find}\:{the}\:{sum}\:{of}\:{areas}\:{of}\:{all}\:\left({infinite}\right) \\ $$$${red}\:{circles} \\ $$

Commented by aleks041103 last updated on 30/Dec/21

This is a very difficult question.  Up untill now I could only prove that   the centers lie on a parabola.

$${This}\:{is}\:{a}\:{very}\:{difficult}\:{question}. \\ $$$${Up}\:{untill}\:{now}\:{I}\:{could}\:{only}\:{prove}\:{that}\: \\ $$$${the}\:{centers}\:{lie}\:{on}\:{a}\:{parabola}. \\ $$

Commented by mr W last updated on 30/Dec/21

indeed it′s very hard, perhaps not  solvable in an exact way.

$${indeed}\:{it}'{s}\:{very}\:{hard},\:{perhaps}\:{not} \\ $$$${solvable}\:{in}\:{an}\:{exact}\:{way}. \\ $$

Commented by aleks041103 last updated on 30/Dec/21

I found this recursive relation:  r_(i+1) =(r_i /((2r_i +1)^2 ))(3−2r_i −2(√(2−4r_i )))

$${I}\:{found}\:{this}\:{recursive}\:{relation}: \\ $$$${r}_{{i}+\mathrm{1}} =\frac{{r}_{{i}} }{\left(\mathrm{2}{r}_{{i}} +\mathrm{1}\right)^{\mathrm{2}} }\left(\mathrm{3}−\mathrm{2}{r}_{{i}} −\mathrm{2}\sqrt{\mathrm{2}−\mathrm{4}{r}_{{i}} }\right) \\ $$

Commented by mr W last updated on 30/Dec/21

i got this too.  (λ_(n+1) /λ_n )=((3−2λ_n −2(√(2(1−2λ_n ))))/((2λ_n +1)^2 ))  with λ_n =(r_n /R), R=radius of semicircle=1  r_0 =(1/2)  r_1 =(1/4)  r_2 =(1/(18))  r_3 =(1/(100))  r_4 =(1/(578))  r_5 =(1/(3 364))  r_6 =(1/(19 602))  r_7 =(1/(114 244))  r_8 =(1/(665 858))  r_9 =(1/(3 880 900))  r_(10) =(1/(22 619 538))  r_(11) =(1/(131 836 324))  r_(12) =(1/(768 398 402))  r_(13) =(1/(4 478 554 084))  r_(14) =(1/(26 102 926 098))  r_(15) =(1/(152 139 002 500))  r_(16) =(1/(886 731 088 898))  ...

$${i}\:{got}\:{this}\:{too}. \\ $$$$\frac{\lambda_{{n}+\mathrm{1}} }{\lambda_{{n}} }=\frac{\mathrm{3}−\mathrm{2}\lambda_{{n}} −\mathrm{2}\sqrt{\mathrm{2}\left(\mathrm{1}−\mathrm{2}\lambda_{{n}} \right)}}{\left(\mathrm{2}\lambda_{{n}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${with}\:\lambda_{{n}} =\frac{{r}_{{n}} }{{R}},\:{R}={radius}\:{of}\:{semicircle}=\mathrm{1} \\ $$$${r}_{\mathrm{0}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${r}_{\mathrm{1}} =\frac{\mathrm{1}}{\mathrm{4}} \\ $$$${r}_{\mathrm{2}} =\frac{\mathrm{1}}{\mathrm{18}} \\ $$$${r}_{\mathrm{3}} =\frac{\mathrm{1}}{\mathrm{100}} \\ $$$${r}_{\mathrm{4}} =\frac{\mathrm{1}}{\mathrm{578}} \\ $$$${r}_{\mathrm{5}} =\frac{\mathrm{1}}{\mathrm{3}\:\mathrm{364}} \\ $$$${r}_{\mathrm{6}} =\frac{\mathrm{1}}{\mathrm{19}\:\mathrm{602}} \\ $$$${r}_{\mathrm{7}} =\frac{\mathrm{1}}{\mathrm{114}\:\mathrm{244}} \\ $$$${r}_{\mathrm{8}} =\frac{\mathrm{1}}{\mathrm{665}\:\mathrm{858}} \\ $$$${r}_{\mathrm{9}} =\frac{\mathrm{1}}{\mathrm{3}\:\mathrm{880}\:\mathrm{900}} \\ $$$${r}_{\mathrm{10}} =\frac{\mathrm{1}}{\mathrm{22}\:\mathrm{619}\:\mathrm{538}} \\ $$$${r}_{\mathrm{11}} =\frac{\mathrm{1}}{\mathrm{131}\:\mathrm{836}\:\mathrm{324}} \\ $$$${r}_{\mathrm{12}} =\frac{\mathrm{1}}{\mathrm{768}\:\mathrm{398}\:\mathrm{402}} \\ $$$${r}_{\mathrm{13}} =\frac{\mathrm{1}}{\mathrm{4}\:\mathrm{478}\:\mathrm{554}\:\mathrm{084}} \\ $$$${r}_{\mathrm{14}} =\frac{\mathrm{1}}{\mathrm{26}\:\mathrm{102}\:\mathrm{926}\:\mathrm{098}} \\ $$$${r}_{\mathrm{15}} =\frac{\mathrm{1}}{\mathrm{152}\:\mathrm{139}\:\mathrm{002}\:\mathrm{500}} \\ $$$${r}_{\mathrm{16}} =\frac{\mathrm{1}}{\mathrm{886}\:\mathrm{731}\:\mathrm{088}\:\mathrm{898}} \\ $$$$... \\ $$

Commented by Tawa11 last updated on 30/Dec/21

Great sir. Please show us the prove sir.

$$\mathrm{Great}\:\mathrm{sir}.\:\mathrm{Please}\:\mathrm{show}\:\mathrm{us}\:\mathrm{the}\:\mathrm{prove}\:\mathrm{sir}. \\ $$

Commented by mr W last updated on 30/Dec/21

it seems that (1/r_n ) is always an integer.

$${it}\:{seems}\:{that}\:\frac{\mathrm{1}}{{r}_{{n}} }\:{is}\:{always}\:{an}\:{integer}. \\ $$

Commented by aleks041103 last updated on 30/Dec/21

I tried up to n=100, and indeed (1/r_n ) remains  an integer.

$${I}\:{tried}\:{up}\:{to}\:{n}=\mathrm{100},\:{and}\:{indeed}\:\frac{\mathrm{1}}{{r}_{{n}} }\:{remains} \\ $$$${an}\:{integer}. \\ $$

Commented by peter frank last updated on 31/Dec/21

great mind think a like

$$\mathrm{great}\:\mathrm{mind}\:\mathrm{think}\:\mathrm{a}\:\mathrm{like} \\ $$

Commented by mr W last updated on 01/Jan/22

as we can see later,  (1/r_n )=Σ_(k=0) ^n 2^k C_(2k) ^(2n) +1   which is indeed always integer.

$${as}\:{we}\:{can}\:{see}\:{later}, \\ $$$$\frac{\mathrm{1}}{{r}_{{n}} }=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\mathrm{2}^{{k}} {C}_{\mathrm{2}{k}} ^{\mathrm{2}{n}} +\mathrm{1}\: \\ $$$${which}\:{is}\:{indeed}\:{always}\:{integer}. \\ $$

Answered by mr W last updated on 30/Dec/21

Commented by mr W last updated on 02/Jan/22

say the radius of the semicircle is R.  say the biggest inscribed circle in the  middle of semicircle has the radus r_0   which is (R/2).  say the center of n^(th)  inscribed circle  is A and its radius is r_n .  say the center of (n+1)^(th)  inscribed   circle is B and its radius is r_(n+1) .    OA=R−r_n   OB=R−r_(n+1)   OC=(√((R−r_n )^2 −r_n ^2 ))=(√(R(R−2r_n )))  OD=(√((R−r_(n+1) )^2 −r_(n+1) ^2 ))=(√(R(R−2r_(n+1) )))  AB=r_n +r_(n+1)   (r_n +r_(n+1) )^2 −(r_n −r_(n+1) )^2 =((√(R(R−2r_n )))+(√(R(R−2r_(n+1) ))))^2   2r_n r_(n+1) =R^2 −R(r_n +r_(n+1) )+R(√((R−2r_n )(R−2r_(n+1) )))  let λ_n =(r_n /R)  2λ_n λ_(n+1) +λ_n +λ_(n+1) −1=(√((1−2λ_n )(1−2λ_(n+1) )))   determinant (((4𝛌_n ^2 𝛌_(n+1) ^2 +(𝛌_n +𝛌_(n+1) )^2 +4𝛌_n 𝛌_(n+1) (𝛌_n +𝛌_(n+1) −2)=0)))  ...(I)  (2λ_n +1)^2 ((λ_(n+1) /λ_n ))^2 −2(3−2λ_n )((λ_(n+1) /λ_n ))+1=0   (λ_(n+1) /λ_n )=((3−2λ_n −(√((3−2λ_n )^2 −(2λ_n +1)^2 )))/((2λ_n +1)^2 ))  (λ_(n+1) /λ_n )=((3−2λ_n −2(√(2(1−2λ_n ))))/((2λ_n +1)^2 ))  (λ_(n+1) /λ_n )=((2+1−2λ_n −2(√(2(1−2λ_n ))))/((2−(1−2λ_n ))^2 ))  (λ_(n+1) /λ_n )=((((√2)−(√(1−2λ_n )))^2 )/((((√2))−((√(1−2λ_n )))^2 )^2 ))  (λ_(n+1) /λ_n )=(1/(((√2)+(√(1−2λ_n )))^2 ))   (1/λ_(n+1) )=((((√2)+(√(1−2λ_n )))^2 )/λ_n )=((3−2λ_n +2(√(2(1−2λ_n ))))/λ_n )  ⇒(1/λ_(n+1) )=((3+2(√(2(1−2λ_n ))))/λ_n )−2   ...(i)    we can transform eqn. (I) also into  (2λ_(n+1) +1)^2 ((λ_n /λ_(n+1) ))^2 −2(3−2λ_(n+1) )((λ_n /λ_(n+1) ))+1=0  (λ_n /λ_(n+1) )=((3−2λ_(n+1) +(√((3−2λ_(n+1) )^2 −(2λ_(n+1) +1)^2 )))/((2λ_(n+1) +1)^2 ))  similarly to above,  (λ_n /λ_(n+1) )=(1/(((√2)−(√(1−2λ_(n+1) )))^2 ))  (1/λ_n )=((((√2)−(√(1−2λ_(n+1) )))^2 )/λ_(n+1) )=((3−2λ_(n+1) −2(√(2(1−2λ_(n+1) ))))/λ_(n+1) )  (1/λ_n )=((3−2(√(2(1−2λ_(n+1) ))))/λ_(n+1) )−2  or  ⇒(1/λ_(n−1) )=((3−2(√(2(1−2λ_n ))))/λ_n )−2   ...(ii)    (i)+(ii):  (1/λ_(n+1) )+(1/λ_(n−1) )=(6/λ_n )−4   determinant ((((1/λ_(n+1) )−(6/λ_n )+(1/λ_(n−1) )+4=0)))  let (1/λ_n )=ξ_n +c  (ξ_(n+1) +c)−6(ξ_n +c)+(ξ_(n−1) +c)+4=0  ξ_(n+1) −6ξ_n +ξ_(n−1) +4−4c=0  set 4−4c=0, i.e. c=1  ⇒ξ_(n+1) −6ξ_n +ξ_(n−1) =0  x^2 −6x+1=0  (characteristic eqn.)  ⇒x=3±2(√2)=((√2)±1)^2   ⇒ξ_n =A((√2)+1)^(2n) +B((√2)−1)^(2n)   (1/λ_n )=A((√2)+1)^(2n) +B((√2)−1)^(2n) +1  λ_0 =(1/2)  2=A+B+1 ⇒A+B=1  λ_1 =λ_(−1)  due to symmetry  A((√2)+1)^2 +B((√2)−1)^2 +1=A((√2)+1)^(−2) +B((√2)−1)^(−2) +1  A((√2)+1)^2 +B((√2)−1)^2 =A((√2)−1)^2 +B((√2)+1)^2   (A−B)((√2)+1)^2 =(A−B)((√2)−1)^2   ⇒A=B ⇒A=B=(1/2)   determinant ((((1/λ_n )=((((√2)+1)^(2n) +((√2)−1)^(2n) )/2)+1)))  ⇒(1/λ_n )=((((√2)+1)^(2n) +((√2)+1)^(−2n) +2)/2)  ⇒(1/λ_n )=2×[((((√2)+1)^n +((√2)+1)^(−n) )/2)]^2   ⇒(1/λ_n )=2×[((e^(nln ((√2)+1)) +e^(−nln ((√2)+1)) )/2)]^2   ⇒(1/λ_n )=2 cosh^2  [nln ((√2)+1)]   determinant (((r_n =(1/(2 cosh^2  [nln ((√2)+1)])))))    total area of all red circles:  A=πr_0 ^2 +2Σ_(n=1) ^∞ πr_n ^2   A=(π/4)+(π/2)Σ_(n=1) ^∞ (1/(cosh^4  [nln ((√2)+1)]))      ≈1.198 137 490 (76% of semicircle)    additional note:  we can also proceed as following:  (1/r_n )=((((√2)+1)^(2n) +((√2)−1)^(2n) )/2)+1  (1/r_n )=((Σ_(k=0) ^(2n) C_k ^(2n) ((√2))^k +Σ_(k=0) ^(2n) C_k ^(2n) ((√2))^k (−1)^(2n−k) )/2)+1  (1/r_n )=Σ_(k=0) ^n 2^k C_(2k) ^(2n) +1  this is clearly always an even integer.    [completed on 01/01/2022, 2:00h]

$${say}\:{the}\:{radius}\:{of}\:{the}\:{semicircle}\:{is}\:{R}. \\ $$$${say}\:{the}\:{biggest}\:{inscribed}\:{circle}\:{in}\:{the} \\ $$$${middle}\:{of}\:{semicircle}\:{has}\:{the}\:{radus}\:{r}_{\mathrm{0}} \\ $$$${which}\:{is}\:\frac{{R}}{\mathrm{2}}. \\ $$$${say}\:{the}\:{center}\:{of}\:{n}^{{th}} \:{inscribed}\:{circle} \\ $$$${is}\:{A}\:{and}\:{its}\:{radius}\:{is}\:{r}_{{n}} . \\ $$$${say}\:{the}\:{center}\:{of}\:\left({n}+\mathrm{1}\right)^{{th}} \:{inscribed}\: \\ $$$${circle}\:{is}\:{B}\:{and}\:{its}\:{radius}\:{is}\:{r}_{{n}+\mathrm{1}} . \\ $$$$ \\ $$$${OA}={R}−{r}_{{n}} \\ $$$${OB}={R}−{r}_{{n}+\mathrm{1}} \\ $$$${OC}=\sqrt{\left({R}−{r}_{{n}} \right)^{\mathrm{2}} −{r}_{{n}} ^{\mathrm{2}} }=\sqrt{{R}\left({R}−\mathrm{2}{r}_{{n}} \right)} \\ $$$${OD}=\sqrt{\left({R}−{r}_{{n}+\mathrm{1}} \right)^{\mathrm{2}} −{r}_{{n}+\mathrm{1}} ^{\mathrm{2}} }=\sqrt{{R}\left({R}−\mathrm{2}{r}_{{n}+\mathrm{1}} \right)} \\ $$$${AB}={r}_{{n}} +{r}_{{n}+\mathrm{1}} \\ $$$$\left({r}_{{n}} +{r}_{{n}+\mathrm{1}} \right)^{\mathrm{2}} −\left({r}_{{n}} −{r}_{{n}+\mathrm{1}} \right)^{\mathrm{2}} =\left(\sqrt{{R}\left({R}−\mathrm{2}{r}_{{n}} \right)}+\sqrt{{R}\left({R}−\mathrm{2}{r}_{{n}+\mathrm{1}} \right)}\right)^{\mathrm{2}} \\ $$$$\mathrm{2}{r}_{{n}} {r}_{{n}+\mathrm{1}} ={R}^{\mathrm{2}} −{R}\left({r}_{{n}} +{r}_{{n}+\mathrm{1}} \right)+{R}\sqrt{\left({R}−\mathrm{2}{r}_{{n}} \right)\left({R}−\mathrm{2}{r}_{{n}+\mathrm{1}} \right)} \\ $$$${let}\:\lambda_{{n}} =\frac{{r}_{{n}} }{{R}} \\ $$$$\mathrm{2}\lambda_{{n}} \lambda_{{n}+\mathrm{1}} +\lambda_{{n}} +\lambda_{{n}+\mathrm{1}} −\mathrm{1}=\sqrt{\left(\mathrm{1}−\mathrm{2}\lambda_{{n}} \right)\left(\mathrm{1}−\mathrm{2}\lambda_{{n}+\mathrm{1}} \right)} \\ $$$$\begin{array}{|c|}{\mathrm{4}\boldsymbol{\lambda}_{\boldsymbol{{n}}} ^{\mathrm{2}} \boldsymbol{\lambda}_{\boldsymbol{{n}}+\mathrm{1}} ^{\mathrm{2}} +\left(\boldsymbol{\lambda}_{\boldsymbol{{n}}} +\boldsymbol{\lambda}_{\boldsymbol{{n}}+\mathrm{1}} \right)^{\mathrm{2}} +\mathrm{4}\boldsymbol{\lambda}_{\boldsymbol{{n}}} \boldsymbol{\lambda}_{\boldsymbol{{n}}+\mathrm{1}} \left(\boldsymbol{\lambda}_{\boldsymbol{{n}}} +\boldsymbol{\lambda}_{\boldsymbol{{n}}+\mathrm{1}} −\mathrm{2}\right)=\mathrm{0}}\\\hline\end{array}\:\:...\left({I}\right) \\ $$$$\left(\mathrm{2}\lambda_{{n}} +\mathrm{1}\right)^{\mathrm{2}} \left(\frac{\lambda_{{n}+\mathrm{1}} }{\lambda_{{n}} }\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{3}−\mathrm{2}\lambda_{{n}} \right)\left(\frac{\lambda_{{n}+\mathrm{1}} }{\lambda_{{n}} }\right)+\mathrm{1}=\mathrm{0}\: \\ $$$$\frac{\lambda_{{n}+\mathrm{1}} }{\lambda_{{n}} }=\frac{\mathrm{3}−\mathrm{2}\lambda_{{n}} −\sqrt{\left(\mathrm{3}−\mathrm{2}\lambda_{{n}} \right)^{\mathrm{2}} −\left(\mathrm{2}\lambda_{{n}} +\mathrm{1}\right)^{\mathrm{2}} }}{\left(\mathrm{2}\lambda_{{n}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\frac{\lambda_{{n}+\mathrm{1}} }{\lambda_{{n}} }=\frac{\mathrm{3}−\mathrm{2}\lambda_{{n}} −\mathrm{2}\sqrt{\mathrm{2}\left(\mathrm{1}−\mathrm{2}\lambda_{{n}} \right)}}{\left(\mathrm{2}\lambda_{{n}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$$$\frac{\lambda_{{n}+\mathrm{1}} }{\lambda_{{n}} }=\frac{\mathrm{2}+\mathrm{1}−\mathrm{2}\lambda_{{n}} −\mathrm{2}\sqrt{\mathrm{2}\left(\mathrm{1}−\mathrm{2}\lambda_{{n}} \right)}}{\left(\mathrm{2}−\left(\mathrm{1}−\mathrm{2}\lambda_{{n}} \right)\right)^{\mathrm{2}} } \\ $$$$\frac{\lambda_{{n}+\mathrm{1}} }{\lambda_{{n}} }=\frac{\left(\sqrt{\mathrm{2}}−\sqrt{\mathrm{1}−\mathrm{2}\lambda_{{n}} }\right)^{\mathrm{2}} }{\left(\left(\sqrt{\mathrm{2}}\right)−\left(\sqrt{\mathrm{1}−\mathrm{2}\lambda_{{n}} }\right)^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$$\frac{\lambda_{{n}+\mathrm{1}} }{\lambda_{{n}} }=\frac{\mathrm{1}}{\left(\sqrt{\mathrm{2}}+\sqrt{\mathrm{1}−\mathrm{2}\lambda_{{n}} }\right)^{\mathrm{2}} }\: \\ $$$$\frac{\mathrm{1}}{\lambda_{{n}+\mathrm{1}} }=\frac{\left(\sqrt{\mathrm{2}}+\sqrt{\mathrm{1}−\mathrm{2}\lambda_{{n}} }\right)^{\mathrm{2}} }{\lambda_{{n}} }=\frac{\mathrm{3}−\mathrm{2}\lambda_{{n}} +\mathrm{2}\sqrt{\mathrm{2}\left(\mathrm{1}−\mathrm{2}\lambda_{{n}} \right)}}{\lambda_{{n}} } \\ $$$$\Rightarrow\frac{\mathrm{1}}{\lambda_{{n}+\mathrm{1}} }=\frac{\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}\left(\mathrm{1}−\mathrm{2}\lambda_{{n}} \right)}}{\lambda_{{n}} }−\mathrm{2}\:\:\:...\left({i}\right) \\ $$$$ \\ $$$${we}\:{can}\:{transform}\:{eqn}.\:\left({I}\right)\:{also}\:{into} \\ $$$$\left(\mathrm{2}\lambda_{{n}+\mathrm{1}} +\mathrm{1}\right)^{\mathrm{2}} \left(\frac{\lambda_{{n}} }{\lambda_{{n}+\mathrm{1}} }\right)^{\mathrm{2}} −\mathrm{2}\left(\mathrm{3}−\mathrm{2}\lambda_{{n}+\mathrm{1}} \right)\left(\frac{\lambda_{{n}} }{\lambda_{{n}+\mathrm{1}} }\right)+\mathrm{1}=\mathrm{0} \\ $$$$\frac{\lambda_{{n}} }{\lambda_{{n}+\mathrm{1}} }=\frac{\mathrm{3}−\mathrm{2}\lambda_{{n}+\mathrm{1}} +\sqrt{\left(\mathrm{3}−\mathrm{2}\lambda_{{n}+\mathrm{1}} \right)^{\mathrm{2}} −\left(\mathrm{2}\lambda_{{n}+\mathrm{1}} +\mathrm{1}\right)^{\mathrm{2}} }}{\left(\mathrm{2}\lambda_{{n}+\mathrm{1}} +\mathrm{1}\right)^{\mathrm{2}} } \\ $$$${similarly}\:{to}\:{above}, \\ $$$$\frac{\lambda_{{n}} }{\lambda_{{n}+\mathrm{1}} }=\frac{\mathrm{1}}{\left(\sqrt{\mathrm{2}}−\sqrt{\mathrm{1}−\mathrm{2}\lambda_{{n}+\mathrm{1}} }\right)^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{\lambda_{{n}} }=\frac{\left(\sqrt{\mathrm{2}}−\sqrt{\mathrm{1}−\mathrm{2}\lambda_{{n}+\mathrm{1}} }\right)^{\mathrm{2}} }{\lambda_{{n}+\mathrm{1}} }=\frac{\mathrm{3}−\mathrm{2}\lambda_{{n}+\mathrm{1}} −\mathrm{2}\sqrt{\mathrm{2}\left(\mathrm{1}−\mathrm{2}\lambda_{{n}+\mathrm{1}} \right)}}{\lambda_{{n}+\mathrm{1}} } \\ $$$$\frac{\mathrm{1}}{\lambda_{{n}} }=\frac{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}\left(\mathrm{1}−\mathrm{2}\lambda_{{n}+\mathrm{1}} \right)}}{\lambda_{{n}+\mathrm{1}} }−\mathrm{2} \\ $$$${or} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\lambda_{{n}−\mathrm{1}} }=\frac{\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}\left(\mathrm{1}−\mathrm{2}\lambda_{{n}} \right)}}{\lambda_{{n}} }−\mathrm{2}\:\:\:...\left({ii}\right) \\ $$$$ \\ $$$$\left({i}\right)+\left({ii}\right): \\ $$$$\frac{\mathrm{1}}{\lambda_{{n}+\mathrm{1}} }+\frac{\mathrm{1}}{\lambda_{{n}−\mathrm{1}} }=\frac{\mathrm{6}}{\lambda_{{n}} }−\mathrm{4} \\ $$$$\begin{array}{|c|}{\frac{\mathrm{1}}{\lambda_{{n}+\mathrm{1}} }−\frac{\mathrm{6}}{\lambda_{{n}} }+\frac{\mathrm{1}}{\lambda_{{n}−\mathrm{1}} }+\mathrm{4}=\mathrm{0}}\\\hline\end{array} \\ $$$${let}\:\frac{\mathrm{1}}{\lambda_{{n}} }=\xi_{{n}} +{c} \\ $$$$\left(\xi_{{n}+\mathrm{1}} +{c}\right)−\mathrm{6}\left(\xi_{{n}} +{c}\right)+\left(\xi_{{n}−\mathrm{1}} +{c}\right)+\mathrm{4}=\mathrm{0} \\ $$$$\xi_{{n}+\mathrm{1}} −\mathrm{6}\xi_{{n}} +\xi_{{n}−\mathrm{1}} +\mathrm{4}−\mathrm{4}{c}=\mathrm{0} \\ $$$${set}\:\mathrm{4}−\mathrm{4}{c}=\mathrm{0},\:{i}.{e}.\:{c}=\mathrm{1} \\ $$$$\Rightarrow\xi_{{n}+\mathrm{1}} −\mathrm{6}\xi_{{n}} +\xi_{{n}−\mathrm{1}} =\mathrm{0} \\ $$$${x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{1}=\mathrm{0}\:\:\left({characteristic}\:{eqn}.\right) \\ $$$$\Rightarrow{x}=\mathrm{3}\pm\mathrm{2}\sqrt{\mathrm{2}}=\left(\sqrt{\mathrm{2}}\pm\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\Rightarrow\xi_{{n}} ={A}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}{n}} +{B}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}{n}} \\ $$$$\frac{\mathrm{1}}{\lambda_{{n}} }={A}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}{n}} +{B}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}{n}} +\mathrm{1} \\ $$$$\lambda_{\mathrm{0}} =\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\mathrm{2}={A}+{B}+\mathrm{1}\:\Rightarrow{A}+{B}=\mathrm{1} \\ $$$$\lambda_{\mathrm{1}} =\lambda_{−\mathrm{1}} \:{due}\:{to}\:{symmetry} \\ $$$${A}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} +{B}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}={A}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{−\mathrm{2}} +{B}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{−\mathrm{2}} +\mathrm{1} \\ $$$${A}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} +{B}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} ={A}\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} +{B}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\left({A}−{B}\right)\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}} =\left({A}−{B}\right)\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{A}={B}\:\Rightarrow{A}={B}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\begin{array}{|c|}{\frac{\mathrm{1}}{\lambda_{{n}} }=\frac{\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}{n}} +\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}{n}} }{\mathrm{2}}+\mathrm{1}}\\\hline\end{array} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\lambda_{{n}} }=\frac{\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}{n}} +\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{−\mathrm{2}{n}} +\mathrm{2}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\lambda_{{n}} }=\mathrm{2}×\left[\frac{\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{{n}} +\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{−{n}} }{\mathrm{2}}\right]^{\mathrm{2}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\lambda_{{n}} }=\mathrm{2}×\left[\frac{{e}^{{n}\mathrm{ln}\:\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)} +{e}^{−{n}\mathrm{ln}\:\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)} }{\mathrm{2}}\right]^{\mathrm{2}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{\lambda_{{n}} }=\mathrm{2}\:\mathrm{cosh}^{\mathrm{2}} \:\left[{n}\mathrm{ln}\:\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)\right] \\ $$$$\begin{array}{|c|}{{r}_{{n}} =\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{cosh}^{\mathrm{2}} \:\left[{n}\mathrm{ln}\:\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)\right]}}\\\hline\end{array} \\ $$$$ \\ $$$${total}\:{area}\:{of}\:{all}\:{red}\:{circles}: \\ $$$${A}=\pi{r}_{\mathrm{0}} ^{\mathrm{2}} +\mathrm{2}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\pi{r}_{{n}} ^{\mathrm{2}} \\ $$$${A}=\frac{\pi}{\mathrm{4}}+\frac{\pi}{\mathrm{2}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\mathrm{cosh}^{\mathrm{4}} \:\left[{n}\mathrm{ln}\:\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)\right]} \\ $$$$\:\:\:\:\approx\mathrm{1}.\mathrm{198}\:\mathrm{137}\:\mathrm{490}\:\left(\mathrm{76\%}\:{of}\:{semicircle}\right) \\ $$$$ \\ $$$$\underline{\boldsymbol{{additional}}\:\boldsymbol{{note}}:} \\ $$$${we}\:{can}\:{also}\:{proceed}\:{as}\:{following}: \\ $$$$\frac{\mathrm{1}}{{r}_{{n}} }=\frac{\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)^{\mathrm{2}{n}} +\left(\sqrt{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}{n}} }{\mathrm{2}}+\mathrm{1} \\ $$$$\frac{\mathrm{1}}{{r}_{{n}} }=\frac{\underset{{k}=\mathrm{0}} {\overset{\mathrm{2}{n}} {\sum}}{C}_{{k}} ^{\mathrm{2}{n}} \left(\sqrt{\mathrm{2}}\right)^{{k}} +\underset{{k}=\mathrm{0}} {\overset{\mathrm{2}{n}} {\sum}}{C}_{{k}} ^{\mathrm{2}{n}} \left(\sqrt{\mathrm{2}}\right)^{{k}} \left(−\mathrm{1}\right)^{\mathrm{2}{n}−{k}} }{\mathrm{2}}+\mathrm{1} \\ $$$$\frac{\mathrm{1}}{{r}_{{n}} }=\underset{{k}=\mathrm{0}} {\overset{{n}} {\sum}}\mathrm{2}^{{k}} {C}_{\mathrm{2}{k}} ^{\mathrm{2}{n}} +\mathrm{1} \\ $$$${this}\:{is}\:{clearly}\:{always}\:{an}\:{even}\:{integer}. \\ $$$$ \\ $$$$\left[{completed}\:{on}\:\mathrm{01}/\mathrm{01}/\mathrm{2022},\:\mathrm{2}:\mathrm{00}{h}\right] \\ $$

Commented by Tawa11 last updated on 30/Dec/21

Wow, thanks sir. God bless you more.

$$\mathrm{Wow},\:\mathrm{thanks}\:\mathrm{sir}.\:\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{more}. \\ $$

Commented by Tawa11 last updated on 30/Dec/21

Thanks for your time sir

$$\mathrm{Thanks}\:\mathrm{for}\:\mathrm{your}\:\mathrm{time}\:\mathrm{sir} \\ $$

Commented by ajfour last updated on 22/Jan/22

r_(n+1) ^2 +x_(n+1) ^2 =(R−r_(n+1) )^2   (x_(n+1) −x_n )^2 =4r_n r_(n+1)   ⇒  ((√(R^2 −2Rr_(n+1) ))−(√(R^2 −2Rr_n )))^2                   =4r_n r_(n+1)   ⇒  2R^2 −2R(r_n +r_(n+1) )−4r_n r_(n+1)             =2(√((R^2 −2Rr_n )(R^2 −2Rr_(n+1) )))  say  R=1   ⇒  (1−r_n −r_(n+1) −2r_n r_(n+1) )^2         =1+4r_n r_(n+1) −2(r_n +r_(n+1) )  ⇒  (r_n +r_(n+1) )^2 +4r_n ^2 r_(n+1) ^2        +4r_n r_(n+1) (r_n +r_(n+1) )=4r_n r_(n+1)   ⇒  (1+4r_n +4r_n ^2 )r_(n+1) ^2   +(2r_n +4r_n ^2 −4r_n )r_(n+1) +r_n ^2 =0  ...

$${r}_{{n}+\mathrm{1}} ^{\mathrm{2}} +{x}_{{n}+\mathrm{1}} ^{\mathrm{2}} =\left({R}−{r}_{{n}+\mathrm{1}} \right)^{\mathrm{2}} \\ $$$$\left({x}_{{n}+\mathrm{1}} −{x}_{{n}} \right)^{\mathrm{2}} =\mathrm{4}{r}_{{n}} {r}_{{n}+\mathrm{1}} \\ $$$$\Rightarrow\:\:\left(\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Rr}_{{n}+\mathrm{1}} }−\sqrt{{R}^{\mathrm{2}} −\mathrm{2}{Rr}_{{n}} }\right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\mathrm{4}{r}_{{n}} {r}_{{n}+\mathrm{1}} \\ $$$$\Rightarrow\:\:\mathrm{2}{R}^{\mathrm{2}} −\mathrm{2}{R}\left({r}_{{n}} +{r}_{{n}+\mathrm{1}} \right)−\mathrm{4}{r}_{{n}} {r}_{{n}+\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:=\mathrm{2}\sqrt{\left({R}^{\mathrm{2}} −\mathrm{2}{Rr}_{{n}} \right)\left({R}^{\mathrm{2}} −\mathrm{2}{Rr}_{{n}+\mathrm{1}} \right)} \\ $$$${say}\:\:{R}=\mathrm{1}\:\:\:\Rightarrow \\ $$$$\left(\mathrm{1}−{r}_{{n}} −{r}_{{n}+\mathrm{1}} −\mathrm{2}{r}_{{n}} {r}_{{n}+\mathrm{1}} \right)^{\mathrm{2}} \\ $$$$\:\:\:\:\:\:=\mathrm{1}+\mathrm{4}{r}_{{n}} {r}_{{n}+\mathrm{1}} −\mathrm{2}\left({r}_{{n}} +{r}_{{n}+\mathrm{1}} \right) \\ $$$$\Rightarrow\:\:\left({r}_{{n}} +{r}_{{n}+\mathrm{1}} \right)^{\mathrm{2}} +\mathrm{4}{r}_{{n}} ^{\mathrm{2}} {r}_{{n}+\mathrm{1}} ^{\mathrm{2}} \\ $$$$\:\:\:\:\:+\mathrm{4}{r}_{{n}} {r}_{{n}+\mathrm{1}} \left({r}_{{n}} +{r}_{{n}+\mathrm{1}} \right)=\mathrm{4}{r}_{{n}} {r}_{{n}+\mathrm{1}} \\ $$$$\Rightarrow \\ $$$$\left(\mathrm{1}+\mathrm{4}{r}_{{n}} +\mathrm{4}{r}_{{n}} ^{\mathrm{2}} \right){r}_{{n}+\mathrm{1}} ^{\mathrm{2}} \\ $$$$+\left(\mathrm{2}{r}_{{n}} +\mathrm{4}{r}_{{n}} ^{\mathrm{2}} −\mathrm{4}{r}_{{n}} \right){r}_{{n}+\mathrm{1}} +{r}_{{n}} ^{\mathrm{2}} =\mathrm{0} \\ $$$$... \\ $$

Answered by aleks041103 last updated on 30/Dec/21

Commented by aleks041103 last updated on 30/Dec/21

As seen in the image PH=PN=r  ⇒OP+PH=OP+PN=ON=R=const.  ⇒OP+PH=const.  ⇒P lies on a parabola with focus at O

$${As}\:{seen}\:{in}\:{the}\:{image}\:{PH}={PN}={r} \\ $$$$\Rightarrow{OP}+{PH}={OP}+{PN}={ON}={R}={const}. \\ $$$$\Rightarrow{OP}+{PH}={const}. \\ $$$$\Rightarrow{P}\:{lies}\:{on}\:{a}\:{parabola}\:{with}\:{focus}\:{at}\:{O} \\ $$

Commented by aleks041103 last updated on 30/Dec/21

that′s just interesting to note.

$${that}'{s}\:{just}\:{interesting}\:{to}\:{note}. \\ $$

Commented by aleks041103 last updated on 31/Dec/21

Since it is a parabola with focus at O  f(x)=ax^2 +b  f(±1)=a+b=0  f(0)=r_0 =(1/2)=b  ⇒f(x)=((1−x^2 )/2)

$${Since}\:{it}\:{is}\:{a}\:{parabola}\:{with}\:{focus}\:{at}\:{O} \\ $$$${f}\left({x}\right)={ax}^{\mathrm{2}} +{b} \\ $$$${f}\left(\pm\mathrm{1}\right)={a}+{b}=\mathrm{0} \\ $$$${f}\left(\mathrm{0}\right)={r}_{\mathrm{0}} =\frac{\mathrm{1}}{\mathrm{2}}={b} \\ $$$$\Rightarrow{f}\left({x}\right)=\frac{\mathrm{1}−{x}^{\mathrm{2}} }{\mathrm{2}} \\ $$

Commented by mr W last updated on 31/Dec/21

alex sir:  yes, the locus of the center of a circle  inscribed in a circular segment is  a parabola.

$${alex}\:{sir}: \\ $$$${yes},\:{the}\:{locus}\:{of}\:{the}\:{center}\:{of}\:{a}\:{circle} \\ $$$${inscribed}\:{in}\:{a}\:{circular}\:{segment}\:{is} \\ $$$${a}\:{parabola}. \\ $$

Commented by mr W last updated on 31/Dec/21

Commented by aleks041103 last updated on 31/Dec/21

mr W, how are you making those drawings?

$${mr}\:{W},\:{how}\:{are}\:{you}\:{making}\:{those}\:{drawings}? \\ $$

Commented by mr W last updated on 22/Jan/22

the diagram is made with  the App LEHK DIAGRAM  (recommended by Ajfour sir)

$${the}\:{diagram}\:{is}\:{made}\:{with} \\ $$$${the}\:{App}\:{LEHK}\:{DIAGRAM} \\ $$$$\left({recommended}\:{by}\:{Ajfour}\:{sir}\right) \\ $$

Answered by aleks041103 last updated on 31/Dec/21

I found a recursive formula for (1/r_i )  (1/r_(i+2) )=(6/r_(i+1) )−(1/r_i )−4  the proof is waay to long, so I′m going  to make a pdf with the solution.  Anyone interested in the solution may  write me a mail at   alex_tinkutara_041103@abv.bg

$${I}\:{found}\:{a}\:{recursive}\:{formula}\:{for}\:\frac{\mathrm{1}}{{r}_{{i}} } \\ $$$$\frac{\mathrm{1}}{{r}_{{i}+\mathrm{2}} }=\frac{\mathrm{6}}{{r}_{{i}+\mathrm{1}} }−\frac{\mathrm{1}}{{r}_{{i}} }−\mathrm{4} \\ $$$${the}\:{proof}\:{is}\:{waay}\:{to}\:{long},\:{so}\:{I}'{m}\:{going} \\ $$$${to}\:{make}\:{a}\:{pdf}\:{with}\:{the}\:{solution}. \\ $$$${Anyone}\:{interested}\:{in}\:{the}\:{solution}\:{may} \\ $$$${write}\:{me}\:{a}\:{mail}\:{at}\: \\ $$$${alex\_tinkutara\_}\mathrm{041103}@{abv}.{bg} \\ $$

Commented by aleks041103 last updated on 31/Dec/21

After all the math we come to the answer:   determinant (((area=(π/2)((1/2)+Σ_(i=1) ^∞ sech^4 (ln(1+(√2)) i)))))

$${After}\:{all}\:{the}\:{math}\:{we}\:{come}\:{to}\:{the}\:{answer}: \\ $$$$\begin{array}{|c|}{{area}=\frac{\pi}{\mathrm{2}}\left(\frac{\mathrm{1}}{\mathrm{2}}+\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}{sech}^{\mathrm{4}} \left({ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)\:{i}\right)\right)}\\\hline\end{array} \\ $$

Commented by aleks041103 last updated on 31/Dec/21

turns out:  (1/r_i )=1+(((3+2(√2))^(i−1) +(3−2(√2))^(i−1) )/2)

$${turns}\:{out}: \\ $$$$\frac{\mathrm{1}}{{r}_{{i}} }=\mathrm{1}+\frac{\left(\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\right)^{{i}−\mathrm{1}} +\left(\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\right)^{{i}−\mathrm{1}} }{\mathrm{2}} \\ $$

Commented by mr W last updated on 31/Dec/21

i knew you will arrive at the final  solution. it′s perfect again sir. you  have solved a problem which seemed  unsolvable. really great job!

$${i}\:{knew}\:{you}\:{will}\:{arrive}\:{at}\:{the}\:{final} \\ $$$${solution}.\:{it}'{s}\:{perfect}\:{again}\:{sir}.\:{you} \\ $$$${have}\:{solved}\:{a}\:{problem}\:{which}\:{seemed} \\ $$$${unsolvable}.\:{really}\:{great}\:{job}! \\ $$

Commented by aleks041103 last updated on 31/Dec/21

mr W, I really want to share my solution  with you. If you want you can type your email   address here, or just write me an email  at alex_tinkutara_041103@abv.bg

$${mr}\:{W},\:{I}\:{really}\:{want}\:{to}\:{share}\:{my}\:{solution} \\ $$$${with}\:{you}.\:{If}\:{you}\:{want}\:{you}\:{can}\:{type}\:{your}\:{email}\: \\ $$$${address}\:{here},\:{or}\:{just}\:{write}\:{me}\:{an}\:{email} \\ $$$${at}\:{alex\_tinkutara\_}\mathrm{041103}@{abv}.{bg} \\ $$

Commented by mr W last updated on 31/Dec/21

thanks alot sir! i′m very interested.  i′ll also try if i could come to the  same result. maybe you can publish  your solution here so that more  people can benifit from it. you don′t  need to typewrite it, you can also  make a photo from your handwriting  or from your pdf, and then post the  image here.

$${thanks}\:{alot}\:{sir}!\:{i}'{m}\:{very}\:{interested}. \\ $$$${i}'{ll}\:{also}\:{try}\:{if}\:{i}\:{could}\:{come}\:{to}\:{the} \\ $$$${same}\:{result}.\:{maybe}\:{you}\:{can}\:{publish} \\ $$$${your}\:{solution}\:{here}\:{so}\:{that}\:{more} \\ $$$${people}\:{can}\:{benifit}\:{from}\:{it}.\:{you}\:{don}'{t} \\ $$$${need}\:{to}\:{typewrite}\:{it},\:{you}\:{can}\:{also} \\ $$$${make}\:{a}\:{photo}\:{from}\:{your}\:{handwriting} \\ $$$${or}\:{from}\:{your}\:{pdf},\:{and}\:{then}\:{post}\:{the} \\ $$$${image}\:{here}. \\ $$

Commented by mr W last updated on 01/Jan/22

alex sir:  i could come to the same result. please  review my completed working above.  thanks for your support!

$${alex}\:{sir}: \\ $$$${i}\:{could}\:{come}\:{to}\:{the}\:{same}\:{result}.\:{please} \\ $$$${review}\:{my}\:{completed}\:{working}\:{above}. \\ $$$${thanks}\:{for}\:{your}\:{support}! \\ $$

Answered by aleks041103 last updated on 01/Jan/22

Commented by aleks041103 last updated on 01/Jan/22

Commented by aleks041103 last updated on 01/Jan/22

Commented by aleks041103 last updated on 01/Jan/22

Commented by aleks041103 last updated on 01/Jan/22

Commented by aleks041103 last updated on 01/Jan/22

Commented by aleks041103 last updated on 01/Jan/22

Commented by aleks041103 last updated on 01/Jan/22

Commented by aleks041103 last updated on 01/Jan/22

Commented by aleks041103 last updated on 01/Jan/22

Commented by aleks041103 last updated on 01/Jan/22

Commented by aleks041103 last updated on 01/Jan/22

Commented by aleks041103 last updated on 01/Jan/22

Commented by aleks041103 last updated on 01/Jan/22

Commented by mr W last updated on 01/Jan/22

i have followed your working. it′s   amazing how you analysed the thing  and got the recursive relation.   i used a different way for my solution,  which even looks a little bit easier   than yours. but to be honest, without   your result, i even have stoped to go   further and have accepted that it′s   impossible to solve for r_n  in terms   of n.

$${i}\:{have}\:{followed}\:{your}\:{working}.\:{it}'{s}\: \\ $$$${amazing}\:{how}\:{you}\:{analysed}\:{the}\:{thing} \\ $$$${and}\:{got}\:{the}\:{recursive}\:{relation}.\: \\ $$$${i}\:{used}\:{a}\:{different}\:{way}\:{for}\:{my}\:{solution}, \\ $$$${which}\:{even}\:{looks}\:{a}\:{little}\:{bit}\:{easier}\: \\ $$$${than}\:{yours}.\:{but}\:{to}\:{be}\:{honest},\:{without}\: \\ $$$${your}\:{result},\:{i}\:{even}\:{have}\:{stoped}\:{to}\:{go}\: \\ $$$${further}\:{and}\:{have}\:{accepted}\:{that}\:{it}'{s}\: \\ $$$${impossible}\:{to}\:{solve}\:{for}\:{r}_{{n}} \:{in}\:{terms}\: \\ $$$${of}\:{n}. \\ $$

Commented by mr W last updated on 01/Jan/22

wow! what an effort you made!  you have really done a great job!  it looks like a scientific paper which  you can publish in magazines.

$${wow}!\:{what}\:{an}\:{effort}\:{you}\:{made}! \\ $$$${you}\:{have}\:{really}\:{done}\:{a}\:{great}\:{job}! \\ $$$${it}\:{looks}\:{like}\:{a}\:{scientific}\:{paper}\:{which} \\ $$$${you}\:{can}\:{publish}\:{in}\:{magazines}. \\ $$

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