Question-142725

Question Number 142725 by mathlove last updated on 04/Jun/21

Commented by Olaf_Thorendsen last updated on 04/Jun/21

h t t p s : / / p e t e r j a m e s t h o m a s . c o m / c a t e g o r y / g e n e r a l / s o c i a l - m e d i a / ? h c b = 1

Commented by MJS_new last updated on 04/Jun/21

there are about 7 900 000 000 people on this planet

and probably less than 79 000 000 can solve

this :

x^{4} - 8 x^{3} + 2 x^{2} + 8 x + 1 = 0

I can - so what ?!

Commented by 1549442205PVT last updated on 05/Jun/21

This equation is too easy!

x^{4} - {8 x}^{3} + {2 x}^{2} + 8 x + 1 = 0

\Leftrightarrow (x^{2} + \frac{1}{x^{2}}) - 8 (x - \frac{1}{x}) + 2 = 0 . put (x - \frac{1}{x}) = t

we get t^{2} + 2 - 8 t + 2 = 0 \Leftrightarrow t^{2} - 8 t + 4 = 0

Δ^{'} = 12 = {(2 \sqrt{3})}^{2} \Rightarrow t = 4 \pm 2 \sqrt{3}

i) x - \frac{1}{x} = 4 + 2 \sqrt{3} \Leftrightarrow x^{2} - (4 + 2 \sqrt{3}) x - 1 = 0

Δ^{'} = {(2 + \sqrt{3})}^{2} + 1 = 8 + 4 \sqrt{3} = {[\sqrt{2} (\sqrt{3} + 1)]}^{2}

x = 2 + \sqrt{3} \pm (\sqrt{6} + \sqrt{2})

ii) x - \frac{1}{x} = 4 - 2 \sqrt{3} \Leftrightarrow x^{2} - 2 (2 - \sqrt{3}) x - 1 = 0

Δ^{'} = {(2 - \sqrt{3})}^{2} + 1 = 8 - 4 \sqrt{3} = {(\sqrt{6} - \sqrt{2})}^{2}

x = 2 - \sqrt{3} \pm (\sqrt{6} - \sqrt{2})

Commented by MJS_new last updated on 05/Jun/21

Ok, now ask 1000 randomly chosen people

and tell us the percentage of right answers

Commented by Dwaipayan Shikari last updated on 05/Jun/21

A c t u a l l y n o o n e h a s d o n e i t y e t . I t i s R i e m a n n H y p o t h e s i s .

0 . 0000000001 % c a n d o t h i s

Commented by MJS_new last updated on 05/Jun/21

I know . but follow the link posted by Sir Olaf .

Commented by Tawa11 last updated on 06/Nov/21

Nice

Answered by Dwaipayan Shikari last updated on 05/Jun/21

z = w + b i

\underset{n = 1}{\sum^{\infty}} \frac{1}{n^{z}} = \underset{n = 1}{\sum^{\infty}} n^{- w} n^{- b i}

= \underset{n = 1}{\sum^{\infty}} n^{- w} e^{- b i l o g (n)} = \underset{n = 1}{\sum^{\infty}} n^{- w} c o s (b l o g (n)) - {i n}^{- w} s i n (b l o g (n))

ℜ (ζ (z)) = \underset{n = 1}{\sum^{\infty}} \frac{c o s (b l o g (n))}{n^{w}}

N o w \underset{n = 1}{\sum^{\infty}} \frac{c o s (b l o g (n))}{n^{w}} - i \frac{s i n (b l o g (n)}{n^{w}} = 0

P e o p l e b e l i e v e t h a t ℜ (z) = \frac{1}{2}

Commented by MJS_new last updated on 04/Jun/21

welcome among the chosen few!

Facebook Tweet Pin