Question-8073

Question Number 8073 by 314159 last updated on 29/Sep/16

Commented by 123456 last updated on 30/Sep/16

\cosh (x) = \frac{e^{x} + e^{- x}}{2}

x^{4} + x^{- 4} = 47

x = e^{t}

e^{4 t} + e^{- 4 t} = 47

2 \cdot \frac{e^{4 t} + e^{- 4 t}}{2} = 47

2 \cosh (4 t) = 47

t = \frac{argcosh (\frac{47}{2})}{4}

x^{3} + x^{- 3} = 2 \cdot \frac{e^{3 t} + e^{- 3 t}}{2} = 2 \cosh (3 t)

= 2 \cosh (\frac{3 argcosh (\frac{47}{2})}{4})

Commented by Rasheed Soomro last updated on 30/Sep/16

A p p r e c i a t i o n s f o r t h i s n o v e l a p p r o a c h!

C a n w e o b t a i n t h e a n s w e r i n m o r e s i m p l i f i e d f o r m ?

Commented by prakash jain last updated on 30/Sep/16

This gives only one solution .

x^{4} + \frac{1}{x^{4}} gives 8 solutions which are be

α, β, γ, θ, \frac{1}{α}, \frac{1}{β}, \frac{1}{γ}, \frac{1}{θ}

effectively we get 4 solutions for x^{3} + \frac{1}{x^{3}}

as given in {Rasheed}^{'} s Answer ? How do

we get same 4 answer using above approach ?

Actual calculation of the value gives 18 .

Commented by 123456 last updated on 30/Sep/16

y = \cosh (t) = \frac{e^{t} + e^{- t}}{2}

x = e^{t}

2 y = x + \frac{1}{x}

2 x y = x^{2} + 1

x^{2} - 2 x y + 1 = 0

Δ = 4 y^{2} - 4 = 4 (y^{2} - 1)

x = y \pm \sqrt{y^{2} - 1} (x ⩾ 0)

e^{t} = y + \sqrt{y^{2} - 1}

t = \ln (y + \sqrt{y^{2} - 1})

argcosh (t) = \ln (t + \sqrt{t^{2} - 1})

Commented by 123456 last updated on 30/Sep/16

we lost some roots

y = x^{2} \Rightarrow x = \pm \sqrt{y}

also

\cos (x + 2 π k) = \cos (x)

\cosh (x i) = \cos (x)

so \cosh are periodic too, but the period

is 2 π i

Commented by prakash jain last updated on 30/Sep/16

Thanks . I understand now .

2 \cosh (4 t) = 47

Will g i v e 4 t = 2 i n π \pm \cosh^{- 1} (47 / 2)

So we get 8 solution for t for n = 0, 1, 2, 3

Answered by Rasheed Soomro last updated on 29/Sep/16

x^{4} + \frac{1}{x^{4}} = 47, x^{3} + \frac{1}{x^{3}} = ?

x^{4} + \frac{1}{x^{4}} = 47

x^{4} + 2 + \frac{1}{x^{4}} = 47 + 2

{(x^{2})}^{2} + 2 (x^{2}) (\frac{1}{x^{2}}) + (\frac{1}{x^{2}}) = 49

{(x^{2} + \frac{1}{x^{2}})}^{2} = {(\pm 7)}^{2}

x^{2} + \frac{1}{x^{2}} = 7 ∣ x^{2} + \frac{1}{x^{2}} = - 7

x^{2} + 2 + \frac{1}{x^{2}} = 7 + 2 ∣ x^{2} + 2 + \frac{1}{x^{2}} = - 7 + 2

{(x)}^{2} + 2 (x) (\frac{1}{x}) + {(\frac{1}{x})}^{2} = 9 ∣ {(x)}^{2} + 2 (x) (\frac{1}{x}) + {(\frac{1}{x})}^{2} = - 5

{(x + \frac{1}{x})}^{2} = {(\pm 3)}^{2} ∣ {(x + \frac{1}{x})}^{2} = {(\pm \sqrt{5} i)}^{2}

x + \frac{1}{x} = \pm 3 ∣ x + \frac{1}{x} = \pm \sqrt{5} i

{(x + \frac{1}{x})}^{3} = {(\pm 3)}^{3} ∣ {(x + \frac{1}{x})}^{3} = {(\pm \sqrt{5} i)}^{3}

x^{3} + \frac{1}{x^{3}} + 3 (x + \frac{1}{x}) = {(\pm 3)}^{2} (\pm 3) ∣ x^{3} + \frac{1}{x^{3}} + 3 (x + \frac{1}{x}) = {(\pm \sqrt{5} i)}^{2} (\pm \sqrt{5} i)

x^{3} + \frac{1}{x^{3}} + 3 (\pm 3) = (9) (\pm 3) ∣ x^{3} + \frac{1}{x^{3}} + 3 (\pm \sqrt{5} i) = (- 5) (\pm \sqrt{5} i)

x^{3} + \frac{1}{x^{3}} = (9) (\pm 3) - 3 (\pm 3) ∣ x^{3} + \frac{1}{x^{3}} = (- 5) (\pm \sqrt{5} i) - 3 (\pm \sqrt{5} i)

x^{3} + \frac{1}{x^{3}} = (9 - 3) (\pm 3) ∣ x^{3} + \frac{1}{x^{3}} = (- 5 - 3) (\pm \sqrt{5} i)

x^{3} + \frac{1}{x^{3}} = 6 (\pm 3) ∣ x^{3} + \frac{1}{x^{3}} = - 8 (\pm \sqrt{5} i)

x^{3} + \frac{1}{x^{3}} = \pm 18 ∣ x^{3} + \frac{1}{x^{3}} = \mp 8 \sqrt{5} i

x^{3} + \frac{1}{x^{3}} = 18, - 18, - 8 \sqrt{5} i, 8 \sqrt{5} i

Answered by prof.kerdus last updated on 29/Sep/16

i t s 18

∙ ⌣ ∙

Answered by Rasheed Soomro last updated on 30/Sep/16

\underline{Another approach}

x^{4} + \frac{1}{x^{4}} = 47, x^{3} + \frac{1}{x^{3}} = ?

{(x^{4} + \frac{1}{x^{4}})}^{3} = {(47)}^{3}

{(a + \frac{1}{a})}^{3} = a^{3} + \frac{1}{a^{3}} + 3 (a + \frac{1}{a})

{(x^{4})}^{3} + {(\frac{1}{x^{4}})}^{3} + 3 (x^{4} + \frac{1}{x^{4}}) = 103823

x^{12} + \frac{1}{x^{12}} + 3 (47) = 103823

x^{12} + \frac{1}{x^{12}} = 103823 - 141 = 103682

{(x^{6})}^{2} + {(\frac{1}{x^{6}})}^{2} = 103682

{(x^{6})}^{2} + 2 + {(\frac{1}{x^{6}})}^{2} = 103682 + 2

{(x^{6})}^{2} + 2 (x^{6}) (\frac{1}{x^{6}}) + {(\frac{1}{x^{6}})}^{2} = 103684

{(x^{6} + \frac{1}{x^{6}})}^{2} = {(\pm 322)}^{2}

x^{6} + \frac{1}{x^{6}} = \pm 322

x^{6} + 2 + \frac{1}{x^{6}} = \pm 322 + 2

{(x^{3})}^{2} + 2 (x^{3}) (\frac{1}{x^{3}}) + {(\frac{1}{x^{3}})}^{2} = 324, - 320

{(x^{3} + \frac{1}{x^{3}})}^{2} = {(\pm 18)}^{2} ∣ {(x^{3} + \frac{1}{x^{3}})}^{2} = {(\pm 8 \sqrt{5} i)}^{2}

x^{3} + \frac{1}{x^{3}} = \pm 18 ∣ x^{3} + \frac{1}{x^{3}} = \pm 8 \sqrt{5} i

x^{3} + \frac{1}{x^{3}} = 18, - 18, 8 \sqrt{5} i, - 8 \sqrt{5} i

Answered by Rasheed Soomro last updated on 30/Sep/16

S l i g h t l y d e f f e r e n t v e r s i o n

o f m y a n s w e r d a t e d 29 / 9 / 16

x^{4} + \frac{1}{x^{4}} = 47, x^{3} + \frac{1}{x^{3}} = ?

L e t x + \frac{1}{x} = y

{(x + \frac{1}{x})}^{2} = y^{2}

x^{2} + \frac{1}{x^{2}} = y^{2} - 2

{(x^{2} + \frac{1}{x^{2}})}^{2} = {(y^{2} - 2)}^{2}

x^{4} + \frac{1}{x^{4}} = {(y^{2} - 2)}^{2} - 2 = 47

{(y^{2} - 2)}^{2} = 47 + 2 = 49

{(y^{2} - 2)}^{2} = {(\pm 7)}^{2}

y^{2} - 2 = \pm 7

y^{2} = \pm 7 + 2

y^{2} = 7 + 2, - 7 + 2

y^{2} = {(\pm 3)}^{2} ∣ y^{2} = {(\pm \sqrt{- 5})}^{2}

y = 3, - 3 ∣ y = \sqrt{5} i, - \sqrt{5} i

x + \frac{1}{x} = 3, - 3, \sqrt{5} i, - \sqrt{5} i

F o r r e m a i n i n g p r o c e s s s e e m y a n s w e r d a t e d 29 / 09 / 16