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Question Number 223256 by ajfour last updated on 19/Jul/25
Commented by ajfour last updated on 19/Jul/25
Commented by ajfour last updated on 19/Jul/25
x^3 −x−c=0
$$\:\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:{x}^{\mathrm{3}} −{x}−{c}=\mathrm{0} \\ $$$$ \\ $$
Commented by Frix last updated on 20/Jul/25
But x=(1+((100)/c^2 ))T−(c/5) is just a linear substitution ⇒ it doesn′t change the nature of the roots. So what′s the point?
$$\mathrm{But}\:{x}=\left(\mathrm{1}+\frac{\mathrm{100}}{{c}^{\mathrm{2}} }\right){T}−\frac{{c}}{\mathrm{5}}\:\mathrm{is}\:\mathrm{just}\:\mathrm{a}\:\mathrm{linear}\:\mathrm{substitution} \\ $$$$\Rightarrow\:\mathrm{it}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{change}\:\mathrm{the}\:\mathrm{nature}\:\mathrm{of}\:\mathrm{the}\:\mathrm{roots}. \\ $$$$\mathrm{So}\:\mathrm{what}'\mathrm{s}\:\mathrm{the}\:\mathrm{point}? \\ $$
Commented by ajfour last updated on 20/Jul/25
I hadnot started with this substitution. But for many cases it gives i think one root of T and hence one value of x without intermediaries going complex. You can check, may be.
$${I}\:{hadnot}\:{started}\:{with}\:{this}\:{substitution}. \\ $$$${But}\:{for}\:{many}\:{cases}\:{it}\:{gives}\:{i}\:{think} \\ $$$${one}\:{root}\:{of}\:{T}\:{and}\:{hence}\:{one}\:{value}\: \\ $$$${of}\:{x}\:{without}\:{intermediaries}\:{going}\: \\ $$$${complex}.\: \\ $$$${You}\:{can}\:{check},\:{may}\:{be}. \\ $$
Commented by Frix last updated on 20/Jul/25
f(x)=x^3 −x−c Substituting x=t+β, β∈R just shifts the function along the x−axis. Substituting x=αt, α∈R\{0} shrinks or stretches f(x) along the x−axis. You cannot change the nature of the zeros using x=αt+β. x^3 +px+q=0 D=(p^3 /(27))+(q^2 /4) You cannot change the sign of D or arrive at D=0 by substituting x=αt+β. You must change the values of p or q to achieve this but then you′re no longer solving your given equation...
$${f}\left({x}\right)={x}^{\mathrm{3}} −{x}−{c} \\ $$$$\mathrm{Substituting}\:{x}={t}+\beta,\:\beta\in\mathbb{R}\:\mathrm{just}\:\mathrm{shifts}\:\mathrm{the} \\ $$$$\mathrm{function}\:\mathrm{along}\:\mathrm{the}\:{x}−\mathrm{axis}. \\ $$$$\mathrm{Substituting}\:{x}=\alpha{t},\:\alpha\in\mathbb{R}\backslash\left\{\mathrm{0}\right\}\:\mathrm{shrinks}\:\mathrm{or} \\ $$$$\mathrm{stretches}\:{f}\left({x}\right)\:\mathrm{along}\:\mathrm{the}\:{x}−\mathrm{axis}. \\ $$$$\mathrm{You}\:\mathrm{cannot}\:\mathrm{change}\:\mathrm{the}\:\mathrm{nature}\:\mathrm{of}\:\mathrm{the}\:\mathrm{zeros} \\ $$$$\mathrm{using}\:{x}=\alpha{t}+\beta. \\ $$$$ \\ $$$${x}^{\mathrm{3}} +{px}+{q}=\mathrm{0} \\ $$$${D}=\frac{{p}^{\mathrm{3}} }{\mathrm{27}}+\frac{{q}^{\mathrm{2}} }{\mathrm{4}} \\ $$$$\mathrm{You}\:\mathrm{cannot}\:\mathrm{change}\:\mathrm{the}\:\mathrm{sign}\:\mathrm{of}\:{D}\:\mathrm{or}\:\mathrm{arrive} \\ $$$$\mathrm{at}\:{D}=\mathrm{0}\:\mathrm{by}\:\mathrm{substituting}\:{x}=\alpha{t}+\beta. \\ $$$$\mathrm{You}\:\mathrm{must}\:\mathrm{change}\:\mathrm{the}\:\mathrm{values}\:\mathrm{of}\:{p}\:\mathrm{or}\:{q}\:\mathrm{to} \\ $$$$\mathrm{achieve}\:\mathrm{this}\:\mathrm{but}\:\mathrm{then}\:\mathrm{you}'\mathrm{re}\:\mathrm{no}\:\mathrm{longer}\:\mathrm{solving} \\ $$$$\mathrm{your}\:\mathrm{given}\:\mathrm{equation}… \\ $$
Commented by Ghisom last updated on 21/Jul/25
you can try a Tschirnhaus Transformation but it has some issues... x^3 −x−c=0 y=x^2 +αx+β ⇒ x=−(α/2)±(√(y+(α^2 /4)−β)) inserting, transforming, squaring, ... y^3 +(3β+2)y^2 −(3αc+α^2 −(β+1)(3β+1))y −(c^2 +α(α^2 −3β−1)c−(α^2 −(β+1)^2 )β) =0 let β=−(2/3)∧α=−((3c)/2)+((√(3(27c^2 −4)))/6) ⇒ y^3 =(((√3)c(27c^2 −4)^(3/2) )/(18))−(((27c^2 −4)^2 )/(54)) try for yourself to see the problems...
$$\mathrm{you}\:\mathrm{can}\:\mathrm{try}\:\mathrm{a}\:\mathrm{Tschirnhaus}\:\mathrm{Transformation} \\ $$$$\mathrm{but}\:\mathrm{it}\:\mathrm{has}\:\mathrm{some}\:\mathrm{issues}… \\ $$$${x}^{\mathrm{3}} −{x}−{c}=\mathrm{0} \\ $$$${y}={x}^{\mathrm{2}} +\alpha{x}+\beta\:\Rightarrow\:{x}=−\frac{\alpha}{\mathrm{2}}\pm\sqrt{{y}+\frac{\alpha^{\mathrm{2}} }{\mathrm{4}}−\beta} \\ $$$$\mathrm{inserting},\:\mathrm{transforming},\:\mathrm{squaring},\:… \\ $$$${y}^{\mathrm{3}} \\ $$$$\:\:\:\:\:+\left(\mathrm{3}\beta+\mathrm{2}\right){y}^{\mathrm{2}} \\ $$$$\:\:\:\:\:−\left(\mathrm{3}\alpha{c}+\alpha^{\mathrm{2}} −\left(\beta+\mathrm{1}\right)\left(\mathrm{3}\beta+\mathrm{1}\right)\right){y} \\ $$$$\:\:\:\:\:−\left({c}^{\mathrm{2}} +\alpha\left(\alpha^{\mathrm{2}} −\mathrm{3}\beta−\mathrm{1}\right){c}−\left(\alpha^{\mathrm{2}} −\left(\beta+\mathrm{1}\right)^{\mathrm{2}} \right)\beta\right) \\ $$$$\:\:\:\:\:=\mathrm{0} \\ $$$$\mathrm{let}\:\beta=−\frac{\mathrm{2}}{\mathrm{3}}\wedge\alpha=−\frac{\mathrm{3}{c}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}\left(\mathrm{27}{c}^{\mathrm{2}} −\mathrm{4}\right)}}{\mathrm{6}} \\ $$$$\Rightarrow \\ $$$${y}^{\mathrm{3}} =\frac{\sqrt{\mathrm{3}}{c}\left(\mathrm{27}{c}^{\mathrm{2}} −\mathrm{4}\right)^{\mathrm{3}/\mathrm{2}} }{\mathrm{18}}−\frac{\left(\mathrm{27}{c}^{\mathrm{2}} −\mathrm{4}\right)^{\mathrm{2}} }{\mathrm{54}} \\ $$$$\mathrm{try}\:\mathrm{for}\:\mathrm{yourself}\:\mathrm{to}\:\mathrm{see}\:\mathrm{the}\:\mathrm{problems}… \\ $$
Answered by ajfour last updated on 19/Jul/25
Answered by ajfour last updated on 19/Jul/25

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