Question Number 191205 by ajfour last updated on 20/Apr/23
Commented by ajfour last updated on 20/Apr/23
$${Find}\:\underset{{i}=\mathrm{1}} {\overset{\infty} {\sum}}{r}_{{i}} \:\:.\:{Take}\:{r}_{\mathrm{1}} ={a},\:\theta=\alpha \\ $$
Commented by mr W last updated on 20/Apr/23
$${very}\:{nice}\:{question}! \\ $$
Answered by mr W last updated on 21/Apr/23
![θ_1 =α θ_2 =(θ_1 /2)=(α/2) ... θ_(n−1) =(α/2^(n−2) ) from Q191192: (r_(n−1) /(tan θ_(n−1) ))+2(√(r_n r_(n−1) ))=(r_n /(tan (θ_(n−1) /2))) let λ=(√(r_n /r_(n−1) )), t=tan (θ_(n−1) /2)=tan (α/2^(n−1) ) λ^2 −2tλ−((1−t^2 )/2)=0 ⇒λ=t+(√((1+t^2 )/2))=((1+(√2) sin (α/2^(n−1) ))/( (√2) cos (α/2^(n−1) ))) (r_n /r_(n−1) )=λ^2 =((((1/( (√2)))+ sin (α/2^(n−1) ))/( cos (α/2^(n−1) ))))^2 ⇒r_n =aΠ_(k=1) ^n ((((1/( (√2)))+sin (α/2^(k−1) ))/(cos (α/2^(k−1) ))))^2 or r_n =[((((1/( (√2)))+sin α)((1/( (√2)))+sin (α/2))...((1/( (√2)))+sin (α/2^(n−1) )))/(cos α cos (α/2) ... cos (α/2^(n−1) )))]^2 a −−−−−−−−−−− cos α cos (α/2) cos (α/2^2 ) ... cos (α/2^(n−1) ) =(1/(2 sin (α/2^(n−1) ))) cos α cos (α/2) cos (α/2^2 ) ... cos (α/2^(n−1) ) 2 sin (α/2^(n−1) ) =(1/(2^2 sin (α/2^(n−1) ))) cos α cos (α/2) cos (α/2^2 ) ... cos (α/2^(n−2) ) 2 sin (α/2^(n−2) ) ... =(1/(2^n sin (α/2^(n−1) ))) cos α 2 sin α =((sin 2α)/(2^n sin (α/2^(n−1) ))) −−−−−−−−−−−−−− S=Σ_(n=1) ^∞ r_n =aΣ_(n=1) ^∞ [((((1/( (√2)))+sin α)((1/( (√2)))+sin (α/2))...((1/( (√2)))+sin (α/2^(n−1) )))/(cos α cos (α/2) ... cos (α/2^(n−1) )))]^2 examples: α=60°: S≈79.619654909509a α=45°: S≈21.061652923501a α=30°: S≈7.043218497988a α=10°: S≈1.907032259467a](https://www.tinkutara.com/question/Q191214.png)
$$\theta_{\mathrm{1}} =\alpha \\ $$$$\theta_{\mathrm{2}} =\frac{\theta_{\mathrm{1}} }{\mathrm{2}}=\frac{\alpha}{\mathrm{2}} \\ $$$$… \\ $$$$\theta_{{n}−\mathrm{1}} =\frac{\alpha}{\mathrm{2}^{{n}−\mathrm{2}} } \\ $$$${from}\:{Q}\mathrm{191192}: \\ $$$$\frac{{r}_{{n}−\mathrm{1}} }{\mathrm{tan}\:\theta_{{n}−\mathrm{1}} }+\mathrm{2}\sqrt{{r}_{{n}} {r}_{{n}−\mathrm{1}} }=\frac{{r}_{{n}} }{\mathrm{tan}\:\frac{\theta_{{n}−\mathrm{1}} }{\mathrm{2}}} \\ $$$${let}\:\lambda=\sqrt{\frac{{r}_{{n}} }{{r}_{{n}−\mathrm{1}} }},\:{t}=\mathrm{tan}\:\frac{\theta_{{n}−\mathrm{1}} }{\mathrm{2}}=\mathrm{tan}\:\frac{\alpha}{\mathrm{2}^{{n}−\mathrm{1}} } \\ $$$$\lambda^{\mathrm{2}} −\mathrm{2}{t}\lambda−\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{2}}=\mathrm{0} \\ $$$$\Rightarrow\lambda={t}+\sqrt{\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{2}}}=\frac{\mathrm{1}+\sqrt{\mathrm{2}}\:\mathrm{sin}\:\frac{\alpha}{\mathrm{2}^{{n}−\mathrm{1}} }}{\:\sqrt{\mathrm{2}}\:\mathrm{cos}\:\frac{\alpha}{\mathrm{2}^{{n}−\mathrm{1}} }} \\ $$$$\frac{{r}_{{n}} }{{r}_{{n}−\mathrm{1}} }=\lambda^{\mathrm{2}} =\left(\frac{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\:\mathrm{sin}\:\frac{\alpha}{\mathrm{2}^{{n}−\mathrm{1}} }}{\:\mathrm{cos}\:\frac{\alpha}{\mathrm{2}^{{n}−\mathrm{1}} }}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{r}_{{n}} ={a}\underset{{k}=\mathrm{1}} {\overset{{n}} {\prod}}\left(\frac{\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\mathrm{sin}\:\frac{\alpha}{\mathrm{2}^{{k}−\mathrm{1}} }}{\mathrm{cos}\:\frac{\alpha}{\mathrm{2}^{{k}−\mathrm{1}} }}\right)^{\mathrm{2}} \\ $$$${or}\:{r}_{{n}} =\left[\frac{\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\mathrm{sin}\:\alpha\right)\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}\right)…\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\mathrm{sin}\:\frac{\alpha}{\mathrm{2}^{{n}−\mathrm{1}} }\right)}{\mathrm{cos}\:\alpha\:\mathrm{cos}\:\frac{\alpha}{\mathrm{2}}\:…\:\mathrm{cos}\:\frac{\alpha}{\mathrm{2}^{{n}−\mathrm{1}} }}\right]^{\mathrm{2}} {a} \\ $$$$−−−−−−−−−−− \\ $$$$\mathrm{cos}\:\alpha\:\mathrm{cos}\:\frac{\alpha}{\mathrm{2}}\:\mathrm{cos}\:\frac{\alpha}{\mathrm{2}^{\mathrm{2}} }\:…\:\mathrm{cos}\:\frac{\alpha}{\mathrm{2}^{{n}−\mathrm{1}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\:\mathrm{sin}\:\frac{\alpha}{\mathrm{2}^{{n}−\mathrm{1}} }}\:\mathrm{cos}\:\alpha\:\mathrm{cos}\:\frac{\alpha}{\mathrm{2}}\:\mathrm{cos}\:\frac{\alpha}{\mathrm{2}^{\mathrm{2}} }\:…\:\mathrm{cos}\:\frac{\alpha}{\mathrm{2}^{{n}−\mathrm{1}} }\:\mathrm{2}\:\mathrm{sin}\:\frac{\alpha}{\mathrm{2}^{{n}−\mathrm{1}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{\mathrm{2}} \:\mathrm{sin}\:\frac{\alpha}{\mathrm{2}^{{n}−\mathrm{1}} }}\:\mathrm{cos}\:\alpha\:\mathrm{cos}\:\frac{\alpha}{\mathrm{2}}\:\mathrm{cos}\:\frac{\alpha}{\mathrm{2}^{\mathrm{2}} }\:…\:\mathrm{cos}\:\frac{\alpha}{\mathrm{2}^{{n}−\mathrm{2}} }\:\mathrm{2}\:\mathrm{sin}\:\frac{\alpha}{\mathrm{2}^{{n}−\mathrm{2}} } \\ $$$$… \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}^{{n}} \:\mathrm{sin}\:\frac{\alpha}{\mathrm{2}^{{n}−\mathrm{1}} }}\:\mathrm{cos}\:\alpha\:\mathrm{2}\:\mathrm{sin}\:\alpha \\ $$$$=\frac{\mathrm{sin}\:\mathrm{2}\alpha}{\mathrm{2}^{{n}} \:\mathrm{sin}\:\frac{\alpha}{\mathrm{2}^{{n}−\mathrm{1}} }} \\ $$$$−−−−−−−−−−−−−− \\ $$$${S}=\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{r}_{{n}} ={a}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\left[\frac{\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\mathrm{sin}\:\alpha\right)\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\mathrm{sin}\:\frac{\alpha}{\mathrm{2}}\right)…\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}+\mathrm{sin}\:\frac{\alpha}{\mathrm{2}^{{n}−\mathrm{1}} }\right)}{\mathrm{cos}\:\alpha\:\mathrm{cos}\:\frac{\alpha}{\mathrm{2}}\:…\:\mathrm{cos}\:\frac{\alpha}{\mathrm{2}^{{n}−\mathrm{1}} }}\right]^{\mathrm{2}} \\ $$$${examples}: \\ $$$$\alpha=\mathrm{60}°:\:{S}\approx\mathrm{79}.\mathrm{619654909509}{a} \\ $$$$\alpha=\mathrm{45}°:\:{S}\approx\mathrm{21}.\mathrm{061652923501}{a} \\ $$$$\alpha=\mathrm{30}°:\:{S}\approx\mathrm{7}.\mathrm{043218497988}{a} \\ $$$$\alpha=\mathrm{10}°:\:{S}\approx\mathrm{1}.\mathrm{907032259467}{a} \\ $$
Commented by mr W last updated on 21/Apr/23
$${i}\:{can}'{t}\:{find}\:{a}\:{closed}\:{formula}\:{for} \\ $$$${r}_{{n}} \:{and}\:\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}{r}_{{n}} . \\ $$