$ϵx^3-x+1=0$渐近解

hbghlyj

1. AsymptoticSolve[ϵ x^3-x+1==0,{x},{ϵ,0,2}]

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\begin{array}{l}
1+\epsilon +3 \epsilon ^2 \\
-\frac{1}{\sqrt{\epsilon }}-\frac{1}{2}+\frac{3 \sqrt{\epsilon }}{8}-\frac{\epsilon }{2} \\
\frac{1}{\sqrt{\epsilon }}-\frac{1}{2}-\frac{3 \sqrt{\epsilon }}{8}-\frac{\epsilon }{2} \\
\end{array}按照MSE加了相同的标签

hbghlyj

Algebraic closure nlab
Theorem 3.2. If $K$ is algebraically closed and has characteristic $0$, then the algebraic closure of the field of (restricted) Laurent series $K((x))$ over $K$ is the field of Puiseux series over $K$.

Here “restricted” refers to Remark 2.4. See Puiseux series for more details on this result.

hbghlyj

math.stackexchange.com/questions/704309

I'm trying to find the expansion for the roots of this equation. I've found one root as $x\sim 1+\epsilon $. Now considering the dominant balance I want to rescale so that   
$\epsilon x^3\sim O(x) \Rightarrow x=O(1/\sqrt\epsilon )$  
Setting $x=y(1/\sqrt\epsilon )$ where $y=O(1)$ I get the new equation
$$y^3-y+\sqrt\epsilon=0$$

发帖者找到了一个根 $x\sim 1+\epsilon $. 现在考虑剩下的两个遥远的根.
$y\sim y_0+\epsilon y_1+\epsilon ^2 y_2+...$ 不起作用(因为洛朗级数不是代数闭的)
我们必须考虑分数指数(Puiseux series)
$y\sim y_0+\sqrt\epsilon y_1+\epsilon  y_2+...$

Czhang271828

考虑 $\epsilon X^3=X-1$ 在 $X_0=1$ 附近的单根 $X_\epsilon$. 显然 $X_\epsilon-1\sim \epsilon$, 观察知
\[
X_\epsilon=\sum_{n\geq 0}\dfrac{\binom{3n}{n}}{2n+1}\cdot \epsilon^n.
\]

hbghlyj

$P(x)ϵ+Q(x)=0$
如果$P(x)$的次数高于$Q(x)$ 则$x$的渐近解 是分数次幂的?
如果$P(x)$的次数低于$Q(x)$ 则$x$的渐近解 是整数次幂的?

hbghlyj

Puiseux series
It is the algebraic closure of the field of formal Laurent series, which itself is the field of fractions of the ring of formal power series.

hbghlyj

Algebraic Closure of Laurent Series
Apr 21, 2006 · 7 posts · 1 author
Incidentally, these are called Puiseux series. Anyways, we're talking formal Laurent series, etc, right? It's easy to reduce the problem to ...

hbghlyj

Czhang271828 发表于 2023-5-29 15:36
\[
X_\epsilon=\sum_{n\geq 0}\dfrac{\binom{3n}{n}}{2n+1}\cdot \epsilon^n.
\]

sum_(n=0)^∞ (binomial(3 n, n) ϵ^n)/(2 n + 1) converges when $\abs ϵ\le\frac4{27}$

hbghlyj

And from this I get 3 solutions if I'm not mistaken, one of them being the one you listed. – Katie Mar 9, 2014 at 10:12

应该有 3 个根

Czhang271828 发表于 2023-5-29 15:36
考虑 $\epsilon X^3=X-1$ 在 $X_0=1$ 附近的单根 $X_\epsilon$. 显然 $X_\epsilon-1\sim \epsilon$, 观察知 ...

4#已经写出1附近的根的级数系数的一般表达式，但我们能否写出其余2个遥远的根的级数系数的一般表达式
WolframAlpha $\epsilon=0.0001$:
$x_1 = -100.496$
$x_2 = 1.0001$
$x_3 = 99.4962$