Question-223734

Question Number 223734 by behi834171 last updated on 03/Aug/25

Answered by Ghisom last updated on 03/Aug/25

we can only approximate all paths lead to polynomials of degree 8 I also tried to substitute x=(√3)cos 2θ which leads to the nice looking sin θ =((√6)/6)+((√6)/4)sin 4θ or sin θ =(1/( (√6)+6cos θ −12cos^3 θ)) but still we cannot get an exact solution x≈.328737896594

$$\mathrm{we}\:\mathrm{can}\:\mathrm{only}\:\mathrm{approximate} \\ $$$$\mathrm{all}\:\mathrm{paths}\:\mathrm{lead}\:\mathrm{to}\:\mathrm{polynomials}\:\mathrm{of}\:\mathrm{degree}\:\mathrm{8} \\ $$$$\mathrm{I}\:\mathrm{also}\:\mathrm{tried}\:\mathrm{to}\:\mathrm{substitute} \\ $$$${x}=\sqrt{\mathrm{3}}\mathrm{cos}\:\mathrm{2}\theta \\ $$$$\mathrm{which}\:\mathrm{leads}\:\mathrm{to}\:\mathrm{the}\:\mathrm{nice}\:\mathrm{looking} \\ $$$$\mathrm{sin}\:\theta\:=\frac{\sqrt{\mathrm{6}}}{\mathrm{6}}+\frac{\sqrt{\mathrm{6}}}{\mathrm{4}}\mathrm{sin}\:\mathrm{4}\theta \\ $$$$\mathrm{or} \\ $$$$\mathrm{sin}\:\theta\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}+\mathrm{6cos}\:\theta\:−\mathrm{12cos}^{\mathrm{3}} \:\theta} \\ $$$$\mathrm{but}\:\mathrm{still}\:\mathrm{we}\:\mathrm{cannot}\:\mathrm{get}\:\mathrm{an}\:\mathrm{exact}\:\mathrm{solution} \\ $$$$ \\ $$$${x}\approx.\mathrm{328737896594} \\ $$

Commented by fantastic last updated on 03/Aug/25

$${i}\:{tried}\:{substitution}. \\ $$$${same}\:{situation}\:{like}\:{you} \\ $$$$ \\ $$

Commented by behi834171 last updated on 03/Aug/25

thank you so much sir,for your time and attention. someone can sets: x=i(√3)sinh𝛉,and try. one root of : 8 degree equation may be z=1 ( i gusse and wish...)

$${thank}\:{you}\:{so}\:{much}\:{sir},{for}\:{your}\:{time} \\ $$$${and}\:{attention}. \\ $$$$\boldsymbol{{someone}}\:\boldsymbol{{can}}\:\boldsymbol{{sets}}:\:\boldsymbol{{x}}=\boldsymbol{{i}}\sqrt{\mathrm{3}}\boldsymbol{{sinh}\theta},\boldsymbol{{and}}\:\boldsymbol{{try}}. \\ $$$$\boldsymbol{{one}}\:\boldsymbol{{root}}\:\boldsymbol{{of}}\::\:\:\mathrm{8}\:\boldsymbol{{degree}}\:\boldsymbol{{equation}}\: \\ $$$$\boldsymbol{{may}}\:\boldsymbol{{be}}\:\boldsymbol{{z}}=\mathrm{1}\:\left(\:\boldsymbol{{i}}\:\boldsymbol{{gusse}}\:\boldsymbol{{and}}\:\boldsymbol{{wish}}…\right) \\ $$

Commented by fantastic last updated on 03/Aug/25

Commented by Ghisom last updated on 03/Aug/25

@fantastic there is a solution, plot it to see it

$$@\mathrm{fantastic} \\ $$$$\mathrm{there}\:\boldsymbol{{is}}\:\mathrm{a}\:\mathrm{solution},\:\mathrm{plot}\:\mathrm{it}\:\mathrm{to}\:\mathrm{see}\:\mathrm{it} \\ $$

Commented by fantastic last updated on 03/Aug/25

$${I}\:{know}\:{sir}. \\ $$$${but}\:{i}\:{dont}\:{know}\:{why}\:{the}\:{calculator} \\ $$$${says}\:{no}\:{solution}\:{exists}. \\ $$

Commented by fantastic last updated on 03/Aug/25

Commented by Ghisom last updated on 03/Aug/25

@behi I just tried but I don′t get any useable solution how do you get z?

$$@\mathrm{behi} \\ $$$$\mathrm{I}\:\mathrm{just}\:\mathrm{tried}\:\mathrm{but}\:\mathrm{I}\:\mathrm{don}'\mathrm{t}\:\mathrm{get}\:\mathrm{any}\:\mathrm{useable} \\ $$$$\mathrm{solution} \\ $$$$\mathrm{how}\:\mathrm{do}\:\mathrm{you}\:\mathrm{get}\:{z}? \\ $$

Commented by behi834171 last updated on 03/Aug/25

z=±1 other part is: z^6 +z^4 +z^2 +1−((2z^3 )/( (√3)))=0

$${z}=\pm\mathrm{1} \\ $$$${other}\:{part}\:{is}: \\ $$$$\boldsymbol{{z}}^{\mathrm{6}} +\boldsymbol{{z}}^{\mathrm{4}} +\boldsymbol{{z}}^{\mathrm{2}} +\mathrm{1}−\frac{\mathrm{2}\boldsymbol{{z}}^{\mathrm{3}} }{\:\sqrt{\mathrm{3}}}=\mathrm{0} \\ $$

Commented by Ghisom last updated on 03/Aug/25

but how do you get from x=i(√3)sinh θ to z?

$$\mathrm{but}\:\mathrm{how}\:\mathrm{do}\:\mathrm{you}\:\mathrm{get}\:\mathrm{from}\:{x}=\mathrm{i}\sqrt{\mathrm{3}}\mathrm{sinh}\:\theta\:\mathrm{to}\:{z}? \\ $$

Commented by behi834171 last updated on 03/Aug/25

$$\boldsymbol{{z}}=\boldsymbol{{e}}^{−\frac{\boldsymbol{\theta}}{\mathrm{2}}} \\ $$

Commented by Ghisom last updated on 03/Aug/25

x=i(√3)sinh θ z=e^(−θ/2) ⇔ θ=−2ln z ⇒ x=i(((√3)(z^4 −1))/(2z^2 )) inserting & transforming gives ⇒ z^8 +((2(√3))/3)(1−i)z^5 +((4i)/3)z^4 −((2(√3))/3)(1+i)z^3 −1=0 ...either I made a mistake or you...

$${x}=\mathrm{i}\sqrt{\mathrm{3}}\mathrm{sinh}\:\theta \\ $$$${z}=\mathrm{e}^{−\theta/\mathrm{2}} \:\Leftrightarrow\:\theta=−\mathrm{2ln}\:{z} \\ $$$$\Rightarrow \\ $$$${x}=\mathrm{i}\frac{\sqrt{\mathrm{3}}\left({z}^{\mathrm{4}} −\mathrm{1}\right)}{\mathrm{2}{z}^{\mathrm{2}} } \\ $$$$\mathrm{inserting}\:\&\:\mathrm{transforming}\:\mathrm{gives} \\ $$$$\Rightarrow \\ $$$${z}^{\mathrm{8}} +\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\left(\mathrm{1}−\mathrm{i}\right){z}^{\mathrm{5}} +\frac{\mathrm{4i}}{\mathrm{3}}{z}^{\mathrm{4}} −\frac{\mathrm{2}\sqrt{\mathrm{3}}}{\mathrm{3}}\left(\mathrm{1}+\mathrm{i}\right){z}^{\mathrm{3}} −\mathrm{1}=\mathrm{0} \\ $$$$…\mathrm{either}\:\mathrm{I}\:\mathrm{made}\:\mathrm{a}\:\mathrm{mistake}\:\mathrm{or}\:\mathrm{you}… \\ $$

Commented by behi834171 last updated on 03/Aug/25

you are right sir.I have a typo.but,fixed. (the RED (2)).other terms are true.

$${you}\:{are}\:{right}\:{sir}.{I}\:{have}\:{a}\:{typo}.{but},{fixed}. \\ $$$$\left({the}\:{RED}\:\left(\mathrm{2}\right)\right).{other}\:{terms}\:{are}\:{true}. \\ $$

Commented by Ghisom last updated on 03/Aug/25

where′s the imaginary part gone? inserting x=ip cannot yield a real equation, no matter what p∈R is besides: z=±1 lead to x=0 which is wrong z^6 +z^4 +z^2 +1−((2z^3 )/( (√3)))=0 z_(1, 2) ≈−.447705530±.997921795i z_(2, 3) ≈−.374245561±.834181794i z_(4, 5) ≈.821951091±.569558078i inserted we get x_(1, 2, 3, 4) ≈±1.31456410±.207506738i [all 4 combinations of signs] x_(5, 6) ≈±1.62171530 none of these solve the given equation

$$\mathrm{where}'\mathrm{s}\:\mathrm{the}\:\mathrm{imaginary}\:\mathrm{part}\:\mathrm{gone}? \\ $$$$\mathrm{inserting}\:{x}=\mathrm{i}{p}\:\mathrm{cannot}\:\mathrm{yield}\:\mathrm{a}\:\mathrm{real}\:\mathrm{equation}, \\ $$$$\mathrm{no}\:\mathrm{matter}\:\mathrm{what}\:{p}\in\mathbb{R}\:\mathrm{is} \\ $$$$ \\ $$$$\mathrm{besides}: \\ $$$${z}=\pm\mathrm{1}\:\mathrm{lead}\:\mathrm{to}\:{x}=\mathrm{0}\:\mathrm{which}\:\mathrm{is}\:\mathrm{wrong} \\ $$$${z}^{\mathrm{6}} +{z}^{\mathrm{4}} +{z}^{\mathrm{2}} +\mathrm{1}−\frac{\mathrm{2}{z}^{\mathrm{3}} }{\:\sqrt{\mathrm{3}}}=\mathrm{0} \\ $$$${z}_{\mathrm{1},\:\mathrm{2}} \approx−.\mathrm{447705530}\pm.\mathrm{997921795i} \\ $$$${z}_{\mathrm{2},\:\mathrm{3}} \approx−.\mathrm{374245561}\pm.\mathrm{834181794i} \\ $$$${z}_{\mathrm{4},\:\mathrm{5}} \approx.\mathrm{821951091}\pm.\mathrm{569558078i} \\ $$$$\mathrm{inserted}\:\mathrm{we}\:\mathrm{get} \\ $$$${x}_{\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:\mathrm{4}} \approx\pm\mathrm{1}.\mathrm{31456410}\pm.\mathrm{207506738i} \\ $$$$\:\:\:\:\:\left[\mathrm{all}\:\mathrm{4}\:\mathrm{combinations}\:\mathrm{of}\:\mathrm{signs}\right] \\ $$$${x}_{\mathrm{5},\:\mathrm{6}} \approx\pm\mathrm{1}.\mathrm{62171530} \\ $$$$\mathrm{none}\:\mathrm{of}\:\mathrm{these}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{given}\:\mathrm{equation} \\ $$

Commented by Ghisom last updated on 03/Aug/25

...with x≈.328737896594 we get 4 values for z: z≈±.995445471±.0953326519i [all 4 combinations of signs]

$$…\mathrm{with}\:{x}\approx.\mathrm{328737896594}\:\mathrm{we}\:\mathrm{get}\:\mathrm{4}\:\mathrm{values} \\ $$$$\mathrm{for}\:{z}: \\ $$$${z}\approx\pm.\mathrm{995445471}\pm.\mathrm{0953326519i} \\ $$$$\:\:\:\:\:\left[\mathrm{all}\:\mathrm{4}\:\mathrm{combinations}\:\mathrm{of}\:\mathrm{signs}\right] \\ $$

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