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

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

Calculate the area of the green shaded 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

{(- \frac{1}{a} + \frac{1}{b} + \frac{1}{r_{n}} + \frac{1}{r_{n + 1}})}^{2} = 2 (\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{r_{n}^{2}} + \frac{1}{r_{n + 1}^{2}})

{(- \frac{1}{a} + \frac{1}{b})}^{2} + \frac{1}{r_{n}^{2}} + \frac{1}{r_{n + 1}^{2}} + 2 (- \frac{1}{a} + \frac{1}{b}) (\frac{1}{r_{n}} + \frac{1}{r_{n +}}) + \frac{2}{r_{n} r_{n + 1}} = 2 (\frac{1}{a^{2}} + \frac{1}{b^{2}} + \frac{1}{r_{n}^{2}} + \frac{1}{r_{n + 1}^{2}})

\frac{1}{r_{n + 1}^{2}} - 2 (\frac{1}{b} - \frac{1}{a} + \frac{1}{r_{n}}) \frac{1}{r_{n + 1}} + \frac{1}{r_{n + 1}^{2}} - 2 (\frac{1}{b} - \frac{1}{a}) \frac{1}{r_{n}} + {(\frac{1}{b} - \frac{1}{a})}^{2} + \frac{4}{a b} = 0

\frac{1}{r_{n + 1}^{2}} - 2 (\frac{1}{r_{n}} + \frac{1}{b} - \frac{1}{a}) \frac{1}{r_{n + 1}} + {(\frac{1}{r_{n}} - \frac{1}{b} + \frac{1}{a})}^{2} + \frac{4}{a b} = 0

\Rightarrow \frac{1}{r_{n + 1}} = \frac{1}{r_{n}} + \frac{1}{b} - \frac{1}{a} + 2 \sqrt{\frac{1}{r_{n}} (\frac{1}{b} - \frac{1}{a}) - \frac{1}{a b}}

\Rightarrow \frac{1}{r_{n - 1}} = \frac{1}{r_{n}} + \frac{1}{b} - \frac{1}{a} - 2 \sqrt{\frac{1}{r_{n}} (\frac{1}{b} - \frac{1}{a}) - \frac{1}{a b}}

\Rightarrow \frac{1}{r_{n + 1}} + \frac{1}{r_{n - 1}} = 2 (\frac{1}{r_{n}} + \frac{1}{b} - \frac{1}{a})

s a y k_{n} = \frac{1}{r_{n}}, α = \frac{1}{a}, β = \frac{1}{b}

\Rightarrow k_{n + 1} - 2 k_{n} + k_{n - 1} + 2 (α - β) = 0

k_{n} = A + B n + (β - α) n^{2}

i n c u r r e n t e x a m p l e :

a = 2, b = 1, r_{0} = a - b = 1

k_{0} = A = \frac{1}{r_{0}} = \frac{1}{a - b}

k_{1} = A + B + (β - α)

k_{- 1} = A - B + (β - α) = k_{1}

\Rightarrow B = 0

\Rightarrow k_{n} = \frac{1}{a - b} + (\frac{1}{b} - \frac{1}{a}) n^{2} = (\frac{a - b}{a b}) [\frac{a b}{{(a - b)}^{2}} + n^{2}]

r_{n} = \frac{a b}{(a - b) [\frac{a b}{{(a - b)}^{2}} + n^{2}]} = \frac{a b}{(a - b) (λ^{2} + n^{2})}

w i t h λ = \frac{\sqrt{a b}}{a - b}

S = π r_{0}^{2} + 2 π \underset{n = 1}{\sum^{\infty}} r_{n}^{2}

S = π {(a - b)}^{2} + \frac{2 a^{2} b^{2} π}{{(a - b)}^{2}} \underset{n = 1}{\sum^{\infty}} \frac{1}{{(λ^{2} + n^{2})}^{2}}

w e h a v e (s e e Q 205173)

\underset{n = 1}{\sum^{\infty}} \frac{1}{{(λ^{2} + n^{2})}^{2}} = \frac{π^{2}}{4 λ^{2} \sinh^{2} (λ π)} + \frac{π}{4 λ^{3} \tanh (λ π)} - \frac{1}{2 λ^{4}}

S = π {(a - b)}^{2} + \frac{2 a^{2} b^{2} π}{{(a - b)}^{2}} [\frac{π^{2}}{4 λ^{2} \sinh^{2} (λ π)} + \frac{π}{4 λ^{3} \tanh (λ π)} - \frac{1}{2 λ^{4}}]

\Rightarrow S = \frac{(a - b) \sqrt{a b} π^{2}}{2} [\frac{λ π}{\sinh^{2} (λ π)} + \frac{1}{\tanh (λ π)}]

w i t h a = 2, b = 1, λ = \sqrt{2}

S = \frac{π^{2}}{\sqrt{2}} [\frac{\sqrt{2} π}{\sinh^{2} (\sqrt{2} π)} + \frac{1}{\tanh (\sqrt{2} π)}]

= \frac{π^{2} (e^{\sqrt{2} π} + e^{- \sqrt{2} π})}{\sqrt{2} (e^{\sqrt{2} π} - e^{- \sqrt{2} π})} + \frac{4 π^{3}}{e^{2 \sqrt{2} π} + e^{- 2 \sqrt{2} π} - 2}

\approx 6.997 958 338

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

Commented by York12 last updated on 12/Mar/24

thank you

Commented by York12 last updated on 12/Mar/24

very good solution

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