Question and Answers Forum

All Questions      Topic List

Geometry Questions

Previous in All Question      Next in All Question      

Previous in Geometry      Next in Geometry      

Question Number 162368 by mr W last updated on 29/Dec/21

Commented by mr W last updated on 29/Dec/21

$$findthesumofareaofallredcircles.$$

Commented by mr W last updated on 29/Dec/21

$$finalresult$$$$S=\pi {(\sqrt{2}-1)}^4+\frac{\pi {\psi }^{\left(3\right)}(2+\sqrt{2})}{3}$$$$\approx 0.172564$$

Answered by aleks041103 last updated on 29/Dec/21

$$Lettheradiusofthebiggestcirclebe{r}_0.$$$$Then,usingthetheoreminQ.161764,$$$$\frac{1}{\sqrt{r_1}}=\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{r_0}}$$$$Byanalogy:$$$$\frac{1}{\sqrt{r_{n+1}}}=1+\frac{1}{\sqrt{r_n}}$$$$Let{a}_n=\frac{1}{\sqrt{r_n}}.$$$$\Rightarrow a_{n+1}=1+a_n$$$$\Rightarrow a_n=n+a_0=n+\frac{1}{\sqrt{r_0}}=\frac{1}{\sqrt{r_n}}$$$$\Rightarrow r_n=\frac{1}{{(n+\frac{1}{\sqrt{r_0}})}^2}.$$$$WeseekS=\pi r_0^2+2\underset{i=1}{\sum ^{\mathrm{\infty }}}\pi r_i^2$$$$S=\pi r_0^2+2\pi \underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{(n+1/\sqrt{r_0})}^4}$$$$Nowwefind{r}_0.$$$${(r_0+1)}^2=2{(1-r_0)}^2$$$$r_0+1=\sqrt{2}-\sqrt{2}{r}_0$$$$(1+\sqrt{2})r_0=\sqrt{2}-1$$$$\Rightarrow r_0=\frac{\sqrt{2}-1}{\sqrt{2}+1}=\frac{{(\sqrt{2}-1)}^2}{2-1}={(\sqrt{2}-1)}^2=r_0$$$$\Rightarrow \frac{1}{\sqrt{r_0}}=\frac{1}{\sqrt{2}-1}=\sqrt{2}+1$$$$\Rightarrow S=\pi {(\sqrt{2}-1)}^4+2\pi \underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{(n+1+\sqrt{2})}^4}$$

Commented by mr W last updated on 29/Dec/21

$$alsothisisperfect!congratulations$$$$fortheperfectsolutionsir!$$

Commented by mr W last updated on 29/Dec/21

$${it}^{\prime }sageniusacttoapplytheformula$$$$fromQ161764!$$

Commented by aleks041103 last updated on 29/Dec/21

$$Letsfindthesum$$$$\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{(n+k)}^4}=s$$$${\psi }_0\left(z\right)=\frac{1}{\mathrm{\Gamma }\left(z\right)}\frac{d\mathrm{\Gamma }\left(z\right)}{dz}-digammafunction$$$$define$$$${\psi }_n\left(z\right)=\frac{d^n{\psi }_0\left(z\right)}{{dz}^n}$$$$Itiskownthat$$$${\psi }_n\left(z\right)={(-1)}^{n+1}n!\underset{k=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(z+k)}^{n+1}}$$$$\Rightarrow {\psi }_3\left(a\right)=6\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(n+a)}^4}$$$$\Rightarrow \underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{(n+a)}^4}=\frac{{\psi }_3\left(a\right)}{6}-\frac{1}{a^4}$$

Commented by aleks041103 last updated on 29/Dec/21

$$tobecontinued$$

Commented by mr W last updated on 29/Dec/21

$$thisisperfectsir!$$$$icreatedthisquestionwithoutknowing$$$$thatthesolutioncouldbesosimple$$$$asyouhavepresented.$$

Commented by aleks041103 last updated on 29/Dec/21

$$Also$$$${\psi }_3(a+1)=6\underset{n=0}{\sum ^{\mathrm{\infty }}}\frac{1}{{(n+1+a)}^4}=6\underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{(n+a)}^4}$$$$\Rightarrow \underset{n=1}{\sum ^{\mathrm{\infty }}}\frac{1}{{(n+a)}^4}=\frac{{\psi }_3(a+1)}{6}$$

Terms of Service

Privacy Policy

Contact: info@tinkutara.com