Question-178522

Question Number 178522 by peter frank last updated on 17/Oct/22

Answered by MJS_new last updated on 17/Oct/22

y=e^x >0 4y^4 +y^3 −6y^2 +y+1=0 this has no “nice” solution y≈1.09783726106 ⇒ x≈.0933421181361

$${y}=\mathrm{e}^{{x}} >\mathrm{0} \\ $$$$\mathrm{4}{y}^{\mathrm{4}} +{y}^{\mathrm{3}} −\mathrm{6}{y}^{\mathrm{2}} +{y}+\mathrm{1}=\mathrm{0} \\ $$$$\mathrm{this}\:\mathrm{has}\:\mathrm{no}\:“\mathrm{nice}''\:\mathrm{solution} \\ $$$${y}\approx\mathrm{1}.\mathrm{09783726106} \\ $$$$\Rightarrow\:{x}\approx.\mathrm{0933421181361} \\ $$

Commented by peter frank last updated on 18/Oct/22

$$\mathrm{more}\:\mathrm{clarification}\:\mathrm{please}\:\mathrm{step}\:\mathrm{3} \\ $$

Commented by MJS_new last updated on 18/Oct/22

4y^4 +y^3 −6y^2 +y+1=0 y^4 +(1/4)y^3 −(3/2)y^2 +(1/4)y+(1/4)=0 y=z−(1/(16)) z^4 −((195)/(128))z^2 +((225)/(512))z+((14973)/(65536))=0 we can solve this finding α, β, γ (z^2 −αz−β)(z^2 +αz−γ)=0 z^4 −(α^2 +β+γ)z^2 −α(β−γ)z+βγ=0 ⇔ −(α^2 +β+γ)=−((195)/(128)) −α(β−γ)=((225)/(512)) βγ=((14973)/(65536)) solve the 1^(st) for γ, then the 2^(nd) for β, the 3^(rd) becomes α^6 −((195)/(64))α^4 +((5763)/(4096))α^2 −((50625)/(262144))=0 α=(√(u+((65)/(64)))) u^3 −((27)/(16))u−((55)/(64))=0 this has no “nice” solution ⇒ we cannot use the exact solution for y

$$\mathrm{4}{y}^{\mathrm{4}} +{y}^{\mathrm{3}} −\mathrm{6}{y}^{\mathrm{2}} +{y}+\mathrm{1}=\mathrm{0} \\ $$$${y}^{\mathrm{4}} +\frac{\mathrm{1}}{\mathrm{4}}{y}^{\mathrm{3}} −\frac{\mathrm{3}}{\mathrm{2}}{y}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{4}}{y}+\frac{\mathrm{1}}{\mathrm{4}}=\mathrm{0} \\ $$$${y}={z}−\frac{\mathrm{1}}{\mathrm{16}} \\ $$$${z}^{\mathrm{4}} −\frac{\mathrm{195}}{\mathrm{128}}{z}^{\mathrm{2}} +\frac{\mathrm{225}}{\mathrm{512}}{z}+\frac{\mathrm{14973}}{\mathrm{65536}}=\mathrm{0} \\ $$$$\mathrm{we}\:\mathrm{can}\:\mathrm{solve}\:\mathrm{this}\:\mathrm{finding}\:\alpha,\:\beta,\:\gamma \\ $$$$\left({z}^{\mathrm{2}} −\alpha{z}−\beta\right)\left({z}^{\mathrm{2}} +\alpha{z}−\gamma\right)=\mathrm{0} \\ $$$${z}^{\mathrm{4}} −\left(\alpha^{\mathrm{2}} +\beta+\gamma\right){z}^{\mathrm{2}} −\alpha\left(\beta−\gamma\right){z}+\beta\gamma=\mathrm{0} \\ $$$$\Leftrightarrow \\ $$$$−\left(\alpha^{\mathrm{2}} +\beta+\gamma\right)=−\frac{\mathrm{195}}{\mathrm{128}} \\ $$$$−\alpha\left(\beta−\gamma\right)=\frac{\mathrm{225}}{\mathrm{512}} \\ $$$$\beta\gamma=\frac{\mathrm{14973}}{\mathrm{65536}} \\ $$$$\mathrm{solve}\:\mathrm{the}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{for}\:\gamma,\:\mathrm{then}\:\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{for}\:\beta,\:\mathrm{the}\:\mathrm{3}^{\mathrm{rd}} \\ $$$$\mathrm{becomes} \\ $$$$\alpha^{\mathrm{6}} −\frac{\mathrm{195}}{\mathrm{64}}\alpha^{\mathrm{4}} +\frac{\mathrm{5763}}{\mathrm{4096}}\alpha^{\mathrm{2}} −\frac{\mathrm{50625}}{\mathrm{262144}}=\mathrm{0} \\ $$$$\alpha=\sqrt{{u}+\frac{\mathrm{65}}{\mathrm{64}}} \\ $$$${u}^{\mathrm{3}} −\frac{\mathrm{27}}{\mathrm{16}}{u}−\frac{\mathrm{55}}{\mathrm{64}}=\mathrm{0} \\ $$$$\mathrm{this}\:\mathrm{has}\:\mathrm{no}\:“\mathrm{nice}''\:\mathrm{solution}\:\Rightarrow\:\mathrm{we}\:\mathrm{cannot}\:\mathrm{use} \\ $$$$\mathrm{the}\:\mathrm{exact}\:\mathrm{solution}\:\mathrm{for}\:{y} \\ $$

Commented by peter frank last updated on 22/Oct/22

$$\mathrm{thanks} \\ $$

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