1-2-

Question Number 4268 by Momeen last updated on 06/Jan/16

1 + 2 =

Answered by Yozzii last updated on 06/Jan/16

\frac{1}{11} \times \frac{\partial^{4}}{\partial^{2} x \partial^{2} y} [\frac{11}{4} x^{2} y^{2}] (e x p (l n [\frac{24}{π} {\lim_{n \to + \infty} (\underset{m = 1}{\sum^{n}} \frac{n}{n^{2} + m^{2}})} (\frac{\sqrt{1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + \dots}}}}}{1 + \sqrt{5}})]))

Commented by prakash jain last updated on 06/Jan/16

\frac{1}{11} \times \frac{\partial^{4}}{\partial^{2} x \partial^{2} y} [\frac{11}{4} x^{2} y^{2}] = 1

(\frac{\sqrt{1 + \sqrt{1 + \sqrt{1 + \sqrt{1 + \dots}}}}}{1 + \sqrt{5}}) = 1

e x p (l n [\frac{24}{π} {\lim (\underset{m = 1}{\sum^{n}} \frac{n}{n^{2} + m^{2}})}]) = \frac{24}{π} {\lim (\underset{m = 1}{\sum^{n}} \frac{n}{n^{2} + m^{2}})}

\frac{24}{π} {\lim_{n \to \infty} (\underset{m = 1}{\sum^{n}} \frac{n}{n^{2} + m^{2}})} = 3

\lim_{n \to \infty} (\underset{m = 1}{\sum^{n}} \frac{n}{n^{2} + m^{2}}) = \frac{π}{8} ?

Commented by Yozzii last updated on 06/Jan/16

u = \sqrt{1 + \sqrt{1 + \sqrt{1 + \dots}}} = \sqrt{1 + u}

u^{2} - u - 1 = 0

u = \frac{1 \pm \sqrt{5}}{2} . u > 0 \Rightarrow u = \frac{1 + \sqrt{5}}{2}

∴ \sqrt{1 + \sqrt{1 + \sqrt{1 + \dots}}} = \frac{1 + \sqrt{5}}{2}

\Rightarrow \frac{\sqrt{1 + \sqrt{1 + \sqrt{1 + \dots}}}}{1 + \sqrt{5}} = \frac{1}{2}

B y R i e m a n n s u m f o r i n t e g r a l,

u s i n g r i g h t e n d - p o i n t s x = 1 + \frac{m}{n}

w h e r e 1 ⩽ m ⩽ n a n d t h e i n t e r v a l

i s 1 ⩽ x ⩽ 2,

I = \int_{1}^{2} f (x) d x = \int_{1}^{2} \frac{1}{1 + {(x - 1)}^{2}} d x = \lim_{n \to \infty} {\underset{m = 1}{\sum^{n}} f (1 + \frac{m}{n}) \times \frac{1}{n}}

I = {t a n}^{- 1} (x - 1) ∣_{1}^{2} = \frac{π}{4}

\frac{1}{n} f (1 + \frac{m}{n}) = \frac{1}{n} \times \frac{1}{1 + {(\frac{m}{n} + 1 - 1)}^{2}} = \frac{n}{n^{2} + m^{2}}

Commented by prakash jain last updated on 06/Jan/16

Thanks .

Answered by 123456 last updated on 06/Jan/16

\frac{3}{π} {(\underset{- \infty}{\int^{+ \infty}} e^{- t^{2}} d t)}^{2}

Commented by prakash jain last updated on 06/Jan/16

\underset{- \infty}{\int^{+ \infty}} e^{- t^{2}} d t = \sqrt{π}

\frac{3}{π} {(\underset{- \infty}{\int^{+ \infty}} e^{- t^{2}} d t)}^{2} = \frac{3}{π} {(\sqrt{π})}^{2} = \frac{3}{π} \times π = 3

1 + 2 = \frac{3}{π} {(\underset{- \infty}{\int^{+ \infty}} e^{- t^{2}} d t)}^{2}

Answered by sj121524 last updated on 08/Jan/16

1 + 1 + 1 = 3

Facebook Tweet Pin