Show that $x^3-\lambda x+1=0$ has an approximate root $x=\frac{1}{\lambda}$ when $\lambda$ is positive and large

Question

I'm currently trying to figure out a way to show that $\frac{1}{\lambda}$ is an approximate root for $$x^3-\lambda x +1=0$$

when $\lambda$ is positive and large.

My idea is to rearrange the equation as $x=(\lambda x-1)^\frac{1}{3}$ and use a binomial expansion, but I'm kinda unsure how to approach this question.

Answer 1

Rescale to get a proper perturbation problem. Set $z=λx$, then the equation transforms to $$ \frac1{λ^3}z^3-z+1=0. $$ Substitute $\epsilon = \frac1{λ^3}$ then $$ z=1+ϵz^3. $$ For $ϵ=0$ the solution is $z=1$, for $ϵ\approx 0$ one can obtain the solution as a perturbation series $z=1+c_1ϵ+c_2ϵ^2+...$ where $$ c_1ϵ+c_2ϵ^2+...=ϵ(1+c_1ϵ+c_2ϵ^2+...)^3=ϵ+3c_1ϵ^2+3(c_1^2+c_2)ϵ^3+... $$ Comparing coefficients one obtains $c_1=1$, $c_2=3$, $c_3=12$, ...

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This gives in the original scaling $$ x=\frac1λ+\frac1{λ^4}+\frac3{λ^7}+\frac{12}{λ^{10}}+... $$ Up to now we can not be sure if this series converges for any finite $λ$.

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Using the intermediate value theorem one can show that

there is indeed a root between $\frac1λ$ and $\frac1λ+\frac2{λ^4}$ or between $\frac1λ+\frac1{λ^4}$ and $\frac1λ+\frac1{λ^4}+\frac6{λ^7}$.

For the first claim, observe first that $x^3-λx+1=\frac1{λ^3}>0$ at $x=\frac1λ$, and at the other point $x=\frac1λ+\frac2{λ^4}$ the polynomial value is $$ (\frac1λ+\frac2{λ^4})^3-(1+\frac2{λ^3})+1=\frac6{λ^6}+\frac{12}{λ^9}+\frac8{λ^{12}}-\frac1{λ^3}<-\frac3{8λ^3}<0 $$ for $λ>2$.

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Using the Rouché theorem one can more exactly show that

there is exactly one root of $f(z)=1-z+ϵz^3$ with $|z-1|<2ϵ$.

On the circle $|z-1|=2ϵ$ the difference to $g(z)=1-z$ is bounded via $$|z^3|\le(1+2ϵ)^3=1+6ϵ+12ϵ^2+8ϵ^3<1+(6+\frac32+\frac18)ϵ<1+8ϵ<2$$ for $ϵ<\frac18$. Thus the condition $|f(z)-g(z)|<2ϵ=|g(z)|$ for Rouché is satisfied and there is exactly one root inside this cirlce.

That is, for $λ>2$ there is exactly one root of the original equation with $$\left|x-\frac1λ\right|<\frac2{λ^4}.$$

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Going one step further, one could also show that

there is a root of $f(z)=1-z+ϵz^3$ with $|z-1-ϵ|<6ϵ^2$.

Compare $f$ on the circle $|z-1-ϵ|=6ϵ^2$ with $g(z)=1+ϵ-z$ so that $|f(z)-g(z)|=ϵ|z^3-1|=ϵ|z-1|\,|z^2+z+1|$. Now $$|z-1|=ϵ+|z-1-ϵ|<ϵ+6ϵ^2<\frac32ϵ$$ for $ϵ<\frac1{12}$ and then $$|z^2+z+1|\le 3+|z-1|\,|z+2|<3+\frac32ϵ(3+\frac32ϵ)<\frac72$$ so that indeed $$ |f(z)-g(z)|<ϵ\cdot\frac32ϵ\cdot\frac72=\frac{21}4ϵ^2<6ϵ^2=|g(z)|. $$

In conclusion, for $λ>\frac52$ there is exactly one root of the original equation satisfying $$ \left|x-\frac1λ-\frac1{λ^4}\right|<\frac6{λ^7} $$

Answer 2

$x^3$ has zero derivation at origin, so it can be approximated by a constant zero function. Then you have: $$0 - \lambda x + 1 = 0$$ So the root is approximately $\frac1{\lambda}$.