Question Number 4268 by Momeen last updated on 06/Jan/16
$$\mathrm{1}+\mathrm{2}= \\ $$
Answered by Yozzii last updated on 06/Jan/16
/π){lim_(n→+∞) (Σ_(m=1) ^n (n/(n^2 +m^2 )))}(((√(1+(√(1+(√(1+(√(1+...))))))))/(1+(√5))))]))](https://www.tinkutara.com/question/Q4270.png)
$$\frac{\mathrm{1}}{\mathrm{11}}×\frac{\partial^{\mathrm{4}} }{\partial^{\mathrm{2}} {x}\partial^{\mathrm{2}} {y}}\left[\frac{\mathrm{11}}{\mathrm{4}}{x}^{\mathrm{2}} {y}^{\mathrm{2}} \right]\left({exp}\left({ln}\left[\frac{\mathrm{24}}{\pi}\left\{\underset{{n}\rightarrow+\infty} {\mathrm{lim}}\left(\underset{{m}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{n}}{{n}^{\mathrm{2}} +{m}^{\mathrm{2}} }\right)\right\}\left(\frac{\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+…}}}}}{\mathrm{1}+\sqrt{\mathrm{5}}}\right)\right]\right)\right) \\ $$$$ \\ $$$$ \\ $$
Commented by prakash jain last updated on 06/Jan/16
![(1/(11))×(∂^4 /(∂^2 x∂^2 y))[((11)/4)x^2 y^2 ]=1 (((√(1+(√(1+(√(1+(√(1+...))))))))/(1+(√5))))=1 exp(ln[((24)/π){lim(Σ_(m=1) ^n (n/(n^2 +m^2 )))}])=((24)/π){lim(Σ_(m=1) ^n (n/(n^2 +m^2 )))} ((24)/π){lim_(n→∞) (Σ_(m=1) ^n (n/(n^2 +m^2 )))}=3 lim_(n→∞) (Σ_(m=1) ^n (n/(n^2 +m^2 )))=(π/8)?](https://www.tinkutara.com/question/Q4273.png)
$$\frac{\mathrm{1}}{\mathrm{11}}×\frac{\partial^{\mathrm{4}} }{\partial^{\mathrm{2}} {x}\partial^{\mathrm{2}} {y}}\left[\frac{\mathrm{11}}{\mathrm{4}}{x}^{\mathrm{2}} {y}^{\mathrm{2}} \right]=\mathrm{1} \\ $$$$\left(\frac{\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+…}}}}}{\mathrm{1}+\sqrt{\mathrm{5}}}\right)=\mathrm{1} \\ $$$${exp}\left({ln}\left[\frac{\mathrm{24}}{\pi}\left\{\mathrm{lim}\left(\underset{{m}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{n}}{{n}^{\mathrm{2}} +{m}^{\mathrm{2}} }\right)\right\}\right]\right)=\frac{\mathrm{24}}{\pi}\left\{\mathrm{lim}\left(\underset{{m}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{n}}{{n}^{\mathrm{2}} +{m}^{\mathrm{2}} }\right)\right\} \\ $$$$\frac{\mathrm{24}}{\pi}\left\{\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\underset{{m}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{n}}{{n}^{\mathrm{2}} +{m}^{\mathrm{2}} }\right)\right\}=\mathrm{3} \\ $$$$\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left(\underset{{m}=\mathrm{1}} {\overset{{n}} {\sum}}\frac{{n}}{{n}^{\mathrm{2}} +{m}^{\mathrm{2}} }\right)=\frac{\pi}{\mathrm{8}}? \\ $$
Commented by Yozzii last updated on 06/Jan/16
$${u}=\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+…}}}=\sqrt{\mathrm{1}+{u}} \\ $$$${u}^{\mathrm{2}} −{u}−\mathrm{1}=\mathrm{0} \\ $$$${u}=\frac{\mathrm{1}\pm\sqrt{\mathrm{5}}}{\mathrm{2}}.\:{u}>\mathrm{0}\Rightarrow{u}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\therefore\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+…}}}=\frac{\mathrm{1}+\sqrt{\mathrm{5}}}{\mathrm{2}} \\ $$$$\Rightarrow\frac{\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+…}}}}{\mathrm{1}+\sqrt{\mathrm{5}}}=\frac{\mathrm{1}}{\mathrm{2}} \\ $$$${By}\:{Riemann}\:{sum}\:{for}\:{integral}, \\ $$$${using}\:{right}\:{end}−{points}\:{x}=\mathrm{1}+\frac{{m}}{{n}} \\ $$$${where}\:\mathrm{1}\leqslant{m}\leqslant{n}\:{and}\:{the}\:{interval} \\ $$$${is}\:\mathrm{1}\leqslant{x}\leqslant\mathrm{2}, \\ $$$${I}=\int_{\mathrm{1}} ^{\mathrm{2}} {f}\left({x}\right){dx}=\int_{\mathrm{1}} ^{\mathrm{2}} \frac{\mathrm{1}}{\mathrm{1}+\left({x}−\mathrm{1}\right)^{\mathrm{2}} }{dx}=\underset{{n}\rightarrow\infty} {\mathrm{lim}}\left\{\underset{{m}=\mathrm{1}} {\overset{{n}} {\sum}}{f}\left(\mathrm{1}+\frac{{m}}{{n}}\right)×\frac{\mathrm{1}}{{n}}\right\} \\ $$$${I}={tan}^{−\mathrm{1}} \left({x}−\mathrm{1}\right)\mid_{\mathrm{1}} ^{\mathrm{2}} =\frac{\pi}{\mathrm{4}} \\ $$$$\frac{\mathrm{1}}{{n}}{f}\left(\mathrm{1}+\frac{{m}}{{n}}\right)=\frac{\mathrm{1}}{{n}}×\frac{\mathrm{1}}{\mathrm{1}+\left(\frac{{m}}{{n}}+\mathrm{1}−\mathrm{1}\right)^{\mathrm{2}} }=\frac{{n}}{{n}^{\mathrm{2}} +{m}^{\mathrm{2}} } \\ $$$$ \\ $$
Commented by prakash jain last updated on 06/Jan/16
$$\mathrm{Thanks}. \\ $$
Answered by 123456 last updated on 06/Jan/16
$$\frac{\mathrm{3}}{\pi}\left(\underset{−\infty} {\overset{+\infty} {\int}}{e}^{−{t}^{\mathrm{2}} } {dt}\right)^{\mathrm{2}} \\ $$
Commented by prakash jain last updated on 06/Jan/16
$$\underset{−\infty} {\overset{+\infty} {\int}}{e}^{−{t}^{\mathrm{2}} } {dt}=\sqrt{\pi} \\ $$$$\frac{\mathrm{3}}{\pi}\left(\underset{−\infty} {\overset{+\infty} {\int}}{e}^{−{t}^{\mathrm{2}} } {dt}\right)^{\mathrm{2}} =\frac{\mathrm{3}}{\pi}\left(\sqrt{\pi}\right)^{\mathrm{2}} =\frac{\mathrm{3}}{\pi}×\pi=\mathrm{3} \\ $$$$\mathrm{1}+\mathrm{2}=\frac{\mathrm{3}}{\pi}\left(\underset{−\infty} {\overset{+\infty} {\int}}{e}^{−{t}^{\mathrm{2}} } {dt}\right)^{\mathrm{2}} \\ $$
Answered by sj121524 last updated on 08/Jan/16
$$\mathrm{1}+\mathrm{1}+\mathrm{1}=\mathrm{3} \\ $$$$ \\ $$