Calculate-the-area-of-the-green-shaded-portions-

Question Number 205161 by York12 last updated on 11/Mar/24

$$\mathrm{Calculate}\:\mathrm{the}\:\mathrm{area}\:\mathrm{of}\:\mathrm{the}\:\mathrm{green}\:\mathrm{shaded}\:\mathrm{portions} \\ $$

Commented by York12 last updated on 11/Mar/24

Answered by mr W last updated on 12/Mar/24

Commented by mr W last updated on 13/Mar/24

(−(1/a)+(1/b)+(1/r_n )+(1/r_(n+1) ))^2 =2((1/a^2 )+(1/b^2 )+(1/r_n ^2 )+(1/r_(n+1) ^2 )) (−(1/a)+(1/b))^2 +(1/r_n ^2 )+(1/r_(n+1) ^2 )+2(−(1/a)+(1/b))((1/r_n )+(1/r_(n+) ))+(2/(r_n r_(n+1) ))=2((1/a^2 )+(1/b^2 )+(1/r_n ^2 )+(1/r_(n+1) ^2 )) (1/r_(n+1) ^2 )−2((1/b)−(1/a)+(1/r_n ))(1/r_(n+1) )+(1/r_(n+1) ^2 )−2((1/b)−(1/a))(1/r_n )+((1/b)−(1/a))^2 +(4/(ab))=0 (1/r_(n+1) ^2 )−2((1/r_n )+(1/b)−(1/a))(1/r_(n+1) )+((1/r_n )−(1/b)+(1/a))^2 +(4/(ab))=0 ⇒(1/r_(n+1) )=(1/r_n )+(1/b)−(1/a)+2(√((1/r_n )((1/b)−(1/a))−(1/(ab)))) ⇒(1/r_(n−1) )=(1/r_n )+(1/b)−(1/a)−2(√((1/r_n )((1/b)−(1/a))−(1/(ab)))) ⇒(1/r_(n+1) )+(1/r_(n−1) )=2((1/r_n )+(1/b)−(1/a)) say k_n =(1/r_n ), α=(1/a), β=(1/b) ⇒k_(n+1) −2k_n +k_(n−1) +2(α−β)=0 k_n =A+Bn+(β−α)n^2 in current example: a=2, b=1, r_0 =a−b=1 k_0 =A=(1/r_0 )=(1/(a−b)) k_1 =A+B+(β−α) k_(−1) =A−B+(β−α)=k_1 ⇒B=0 ⇒k_n =(1/(a−b))+((1/b)−(1/a))n^2 =(((a−b)/(ab)))[((ab)/((a−b)^2 ))+n^2 ] r_n =((ab)/((a−b)[((ab)/((a−b)^2 ))+n^2 ]))=((ab)/((a−b)(λ^2 +n^2 ))) with λ=((√(ab))/(a−b)) S=πr_0 ^2 +2πΣ_(n=1) ^∞ r_n ^2 S=π(a−b)^2 +((2a^2 b^2 π)/((a−b)^2 ))Σ_(n=1) ^∞ (1/((λ^2 +n^2 )^2 )) we have (see Q205173) Σ_(n=1) ^∞ (1/((λ^2 +n^2 )^2 ))=(π^2 /(4λ^2 sinh^2 (λπ)))+(π/(4λ^3 tanh (λπ)))−(1/(2λ^4 )) S=π(a−b)^2 +((2a^2 b^2 π)/((a−b)^2 ))[(π^2 /(4λ^2 sinh^2 (λπ)))+(π/(4λ^3 tanh (λπ)))−(1/(2λ^4 ))] ⇒S=(((a−b)(√(ab))π^2 )/2)[((λπ)/(sinh^2 (λπ)))+(1/(tanh (λπ)))] with a=2, b=1, λ=(√2) S=(π^2 /( (√2)))[(((√2)π)/(sinh^2 ((√2)π)))+(1/(tanh ((√2)π)))] =((π^2 (e^((√2)π) +e^(−(√2)π) ))/( (√2)(e^((√2)π) −e^(−(√2)π) )))+((4π^3 )/(e^(2(√2)π) +e^(−2(√2)π) −2)) ≈6.997 958 338

$$\left(−\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{r}_{{n}} }+\frac{\mathrm{1}}{{r}_{{n}+\mathrm{1}} }\right)^{\mathrm{2}} =\mathrm{2}\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}_{{n}} ^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}_{{n}+\mathrm{1}} ^{\mathrm{2}} }\right) \\ $$$$\left(−\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{{r}_{{n}} ^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}_{{n}+\mathrm{1}} ^{\mathrm{2}} }+\mathrm{2}\left(−\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{b}}\right)\left(\frac{\mathrm{1}}{{r}_{{n}} }+\frac{\mathrm{1}}{{r}_{{n}+} }\right)+\frac{\mathrm{2}}{{r}_{{n}} {r}_{{n}+\mathrm{1}} }=\mathrm{2}\left(\frac{\mathrm{1}}{{a}^{\mathrm{2}} }+\frac{\mathrm{1}}{{b}^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}_{{n}} ^{\mathrm{2}} }+\frac{\mathrm{1}}{{r}_{{n}+\mathrm{1}} ^{\mathrm{2}} }\right) \\ $$$$\frac{\mathrm{1}}{{r}_{{n}+\mathrm{1}} ^{\mathrm{2}} }−\mathrm{2}\left(\frac{\mathrm{1}}{{b}}−\frac{\mathrm{1}}{{a}}+\frac{\mathrm{1}}{{r}_{{n}} }\right)\frac{\mathrm{1}}{{r}_{{n}+\mathrm{1}} }+\frac{\mathrm{1}}{{r}_{{n}+\mathrm{1}} ^{\mathrm{2}} }−\mathrm{2}\left(\frac{\mathrm{1}}{{b}}−\frac{\mathrm{1}}{{a}}\right)\frac{\mathrm{1}}{{r}_{{n}} }+\left(\frac{\mathrm{1}}{{b}}−\frac{\mathrm{1}}{{a}}\right)^{\mathrm{2}} +\frac{\mathrm{4}}{{ab}}=\mathrm{0} \\ $$$$\frac{\mathrm{1}}{{r}_{{n}+\mathrm{1}} ^{\mathrm{2}} }−\mathrm{2}\left(\frac{\mathrm{1}}{{r}_{{n}} }+\frac{\mathrm{1}}{{b}}−\frac{\mathrm{1}}{{a}}\right)\frac{\mathrm{1}}{{r}_{{n}+\mathrm{1}} }+\left(\frac{\mathrm{1}}{{r}_{{n}} }−\frac{\mathrm{1}}{{b}}+\frac{\mathrm{1}}{{a}}\right)^{\mathrm{2}} +\frac{\mathrm{4}}{{ab}}=\mathrm{0} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{r}_{{n}+\mathrm{1}} }=\frac{\mathrm{1}}{{r}_{{n}} }+\frac{\mathrm{1}}{{b}}−\frac{\mathrm{1}}{{a}}+\mathrm{2}\sqrt{\frac{\mathrm{1}}{{r}_{{n}} }\left(\frac{\mathrm{1}}{{b}}−\frac{\mathrm{1}}{{a}}\right)−\frac{\mathrm{1}}{{ab}}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{r}_{{n}−\mathrm{1}} }=\frac{\mathrm{1}}{{r}_{{n}} }+\frac{\mathrm{1}}{{b}}−\frac{\mathrm{1}}{{a}}−\mathrm{2}\sqrt{\frac{\mathrm{1}}{{r}_{{n}} }\left(\frac{\mathrm{1}}{{b}}−\frac{\mathrm{1}}{{a}}\right)−\frac{\mathrm{1}}{{ab}}} \\ $$$$\Rightarrow\frac{\mathrm{1}}{{r}_{{n}+\mathrm{1}} }+\frac{\mathrm{1}}{{r}_{{n}−\mathrm{1}} }=\mathrm{2}\left(\frac{\mathrm{1}}{{r}_{{n}} }+\frac{\mathrm{1}}{{b}}−\frac{\mathrm{1}}{{a}}\right) \\ $$$${say}\:{k}_{{n}} =\frac{\mathrm{1}}{{r}_{{n}} },\:\alpha=\frac{\mathrm{1}}{{a}},\:\beta=\frac{\mathrm{1}}{{b}} \\ $$$$\Rightarrow{k}_{{n}+\mathrm{1}} −\mathrm{2}{k}_{{n}} +{k}_{{n}−\mathrm{1}} +\mathrm{2}\left(\alpha−\beta\right)=\mathrm{0} \\ $$$${k}_{{n}} ={A}+{Bn}+\left(\beta−\alpha\right){n}^{\mathrm{2}} \\ $$$${in}\:{current}\:{example}: \\ $$$${a}=\mathrm{2},\:{b}=\mathrm{1},\:{r}_{\mathrm{0}} ={a}−{b}=\mathrm{1} \\ $$$${k}_{\mathrm{0}} ={A}=\frac{\mathrm{1}}{{r}_{\mathrm{0}} }=\frac{\mathrm{1}}{{a}−{b}} \\ $$$${k}_{\mathrm{1}} ={A}+{B}+\left(\beta−\alpha\right) \\ $$$${k}_{−\mathrm{1}} ={A}−{B}+\left(\beta−\alpha\right)={k}_{\mathrm{1}} \\ $$$$\Rightarrow{B}=\mathrm{0} \\ $$$$\Rightarrow{k}_{{n}} =\frac{\mathrm{1}}{{a}−{b}}+\left(\frac{\mathrm{1}}{{b}}−\frac{\mathrm{1}}{{a}}\right){n}^{\mathrm{2}} =\left(\frac{{a}−{b}}{{ab}}\right)\left[\frac{{ab}}{\left({a}−{b}\right)^{\mathrm{2}} }+{n}^{\mathrm{2}} \right] \\ $$$${r}_{{n}} =\frac{{ab}}{\left({a}−{b}\right)\left[\frac{{ab}}{\left({a}−{b}\right)^{\mathrm{2}} }+{n}^{\mathrm{2}} \right]}=\frac{{ab}}{\left({a}−{b}\right)\left(\lambda^{\mathrm{2}} +{n}^{\mathrm{2}} \right)} \\ $$$${with}\:\lambda=\frac{\sqrt{{ab}}}{{a}−{b}} \\ $$$${S}=\pi{r}_{\mathrm{0}} ^{\mathrm{2}} +\mathrm{2}\pi\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{r}_{{n}} ^{\mathrm{2}} \\ $$$${S}=\pi\left({a}−{b}\right)^{\mathrm{2}} +\frac{\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} \pi}{\left({a}−{b}\right)^{\mathrm{2}} }\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\lambda^{\mathrm{2}} +{n}^{\mathrm{2}} \right)^{\mathrm{2}} } \\ $$$${we}\:{have}\:\left({see}\:{Q}\mathrm{205173}\right) \\ $$$$\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\mathrm{1}}{\left(\lambda^{\mathrm{2}} +{n}^{\mathrm{2}} \right)^{\mathrm{2}} }=\frac{\pi^{\mathrm{2}} }{\mathrm{4}\lambda^{\mathrm{2}} \:\mathrm{sinh}^{\mathrm{2}} \:\left(\lambda\pi\right)}+\frac{\pi}{\mathrm{4}\lambda^{\mathrm{3}} \:\mathrm{tanh}\:\left(\lambda\pi\right)}−\frac{\mathrm{1}}{\mathrm{2}\lambda^{\mathrm{4}} } \\ $$$${S}=\pi\left({a}−{b}\right)^{\mathrm{2}} +\frac{\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} \pi}{\left({a}−{b}\right)^{\mathrm{2}} }\left[\frac{\pi^{\mathrm{2}} }{\mathrm{4}\lambda^{\mathrm{2}} \:\mathrm{sinh}^{\mathrm{2}} \:\left(\lambda\pi\right)}+\frac{\pi}{\mathrm{4}\lambda^{\mathrm{3}} \:\mathrm{tanh}\:\left(\lambda\pi\right)}−\frac{\mathrm{1}}{\mathrm{2}\lambda^{\mathrm{4}} }\right] \\ $$$$\Rightarrow{S}=\frac{\left({a}−{b}\right)\sqrt{{ab}}\pi^{\mathrm{2}} }{\mathrm{2}}\left[\frac{\lambda\pi}{\mathrm{sinh}^{\mathrm{2}} \:\left(\lambda\pi\right)}+\frac{\mathrm{1}}{\mathrm{tanh}\:\left(\lambda\pi\right)}\right] \\ $$$${with}\:{a}=\mathrm{2},\:{b}=\mathrm{1},\:\lambda=\sqrt{\mathrm{2}} \\ $$$${S}=\frac{\pi^{\mathrm{2}} }{\:\sqrt{\mathrm{2}}}\left[\frac{\sqrt{\mathrm{2}}\pi}{\mathrm{sinh}^{\mathrm{2}} \:\left(\sqrt{\mathrm{2}}\pi\right)}+\frac{\mathrm{1}}{\mathrm{tanh}\:\left(\sqrt{\mathrm{2}}\pi\right)}\right] \\ $$$$\:\:\:=\frac{\pi^{\mathrm{2}} \left({e}^{\sqrt{\mathrm{2}}\pi} +{e}^{−\sqrt{\mathrm{2}}\pi} \right)}{\:\sqrt{\mathrm{2}}\left({e}^{\sqrt{\mathrm{2}}\pi} −{e}^{−\sqrt{\mathrm{2}}\pi} \right)}+\frac{\mathrm{4}\pi^{\mathrm{3}} }{{e}^{\mathrm{2}\sqrt{\mathrm{2}}\pi} +{e}^{−\mathrm{2}\sqrt{\mathrm{2}}\pi} −\mathrm{2}} \\ $$$$\:\:\:\approx\mathrm{6}.\mathrm{997}\:\mathrm{958}\:\mathrm{338} \\ $$

Commented by mr W last updated on 12/Mar/24