现在chatgpt 4o版本 解封随便用了，我今天试了一下
How to solve a algebraically integrable system by Abel-Jaboci theory? show me an explicit example.

The **Abel-Jacobi theory** is a classical tool used to solve algebraically integrable systems, particularly those that can be reduced to equations on Riemann surfaces, often elliptic or hyperelliptic curves. These systems can be solved using integrals of algebraic functions, which leads to solutions in terms of elliptic or hyperelliptic functions.

### Steps of Solving an Algebraically Integrable System Using Abel-Jacobi Theory:

1. **Reduction to a Curve**: The first step in solving an algebraically integrable system is to reduce the system to an algebraic curve. This curve is typically an elliptic or hyperelliptic curve, and its form depends on the first integrals of the system.

2. **Abelian Integrals**: Once the system is reduced to a curve, the next step is to compute the integrals of differential forms on this curve. These integrals, called **Abelian integrals**, encode the solution of the system.

3. **Jacobian Inversion**: The Abel-Jacobi map connects the integrals on the curve (Abelian integrals) to coordinates in the Jacobian variety of the curve. The inversion of the Abel-Jacobi map gives the solution to the original dynamical system. This typically involves finding the coordinates in terms of elliptic or hyperelliptic functions.

### Example: The Simple Pendulum

Let’s illustrate the procedure using a classical example: the **simple pendulum**. The equation of motion for a pendulum of length \( l \) and mass \( m \) in the presence of gravity \( g \) is:

\[
\frac{d^2 \theta}{dt^2} + \frac{g}{l} \sin \theta = 0,
\]
where \( \theta(t) \) is the angular displacement.

#### Step 1: Reduction to a First Integral (Energy)
The system has a conserved quantity (the total energy), which simplifies the dynamics. The total energy is given by:

\[
E = \frac{1}{2} m l^2 \left( \frac{d \theta}{dt} \right)^2 - m g l \cos \theta.
\]

For convenience, we normalize this equation by dividing by \( m l^2 \):

\[
E = \frac{1}{2} \left( \frac{d \theta}{dt} \right)^2 - \frac{g}{l} \cos \theta.
\]

Let \( \frac{g}{l} = \omega^2 \), the equation simplifies to:

\[
E = \frac{1}{2} \left( \frac{d \theta}{dt} \right)^2 - \omega^2 \cos \theta.
\]

This can be solved for \( \frac{d\theta}{dt} \):

\[
\left( \frac{d \theta}{dt} \right)^2 = 2E + 2\omega^2 \cos \theta.
\]

#### Step 2: Change of Variables
Now, we make a change of variables to express this as an integral on an elliptic curve. Set:

\[
u = \sin \frac{\theta}{2}.
\]

Then, the differential equation transforms into:

\[
\left( \frac{du}{dt} \right)^2 = \omega^2 (1 - u^2)(k^2 - u^2),
\]
where \( k^2 \) is related to the energy \( E \). This equation describes the motion on an elliptic curve.

#### Step 3: Solve via Elliptic Integrals
The next step is to integrate this equation to find \( u(t) \). This leads to an elliptic integral of the form:

\[
t = \int \frac{du}{\omega \sqrt{(1 - u^2)(k^2 - u^2)}}.
\]

This is the standard form of an elliptic integral of the first kind. The solution \( u(t) \) is given in terms of the **Jacobi elliptic function** \( \mathrm{sn}(t) \), which is the inverse of the elliptic integral.

Thus, the solution for \( u(t) \), and consequently \( \theta(t) \), is expressed in terms of elliptic functions:

\[
\sin \frac{\theta(t)}{2} = \mathrm{sn}(\omega t; k).
\]

#### Step 4: Jacobi Inversion
Finally, using Jacobi inversion, we can express \( \theta(t) \) in terms of time explicitly using the elliptic function \( \mathrm{sn} \).

### Conclusion:
By following the Abel-Jacobi theory, the solution to the pendulum’s equation of motion can be expressed in terms of elliptic integrals and their inverses, the Jacobi elliptic functions. This method can be generalized to more complicated algebraically integrable systems, where the solutions are given by more general Abelian functions (for example, in the case of hyperelliptic curves).

答案无误。有点厉害了。
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修改:vinbo FROM 202.120.11.*
FROM 202.120.11.*

gpt 4o推荐的参考文献有个特点：文献都是经典文献没问题，但是书名有出入。。。所以书名不是抄的，是记住的，记得有偏差而已
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FROM 202.120.11.*

第三次回答如下
Apologies for misunderstanding your earlier point. Upon reviewing the example in **Section 3**, I found the actual typo:

> "If \( E = \mathbb{Q}(\sqrt{2}) \), the Galois group \( \text{Gal}(E/\mathbb{Q}) \) consists of two automorphisms: the identity automorphism \( \sigma(\alpha) = \alpha \) for all \( \alpha \in E \), and the automorphism that swaps \( \sqrt{2} \) with \( -\sqrt{2} \), i.e., \( \sigma(\sqrt{2}) = -\sqrt{2} \)."

Here, the **typo** is that \( \sigma(\alpha) = \alpha \) should say **\( \sigma(x) = x \) for all \( x \in \mathbb{Q} \)**, not \( \alpha \in E \). The identity automorphism only fixes elements of \( \mathbb{Q} \), not all elements of \( E \).

### Corrected Example:
> "If \( E = \mathbb{Q}(\sqrt{2}) \), the Galois group \( \text{Gal}(E/\mathbb{Q}) \) consists of two automorphisms:
> - The identity automorphism \( \sigma(x) = x \) for all \( x \in \mathbb{Q} \),
> - The automorphism that swaps \( \sqrt{2} \) with \( -\sqrt{2} \), i.e., \( \sigma(\sqrt{2}) = -\sqrt{2} \)."

Thank you for your patience!
※ 修改:·vinbo 于 Oct 10 21:38:28 2024 修改本文·[FROM: 202.120.11.*]
※ 来源:·水木社区 http://www.mysmth.net·[FROM: 202.120.11.*]

修改:vinbo FROM 202.120.11.*
FROM 202.120.11.*

我去，这用的着学？打开网址就直接用了啊

【 在 Z5boy 的大作中提到: 】
: 羡慕你，我也想学用chatgpt
:
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FROM 202.120.11.*

那是很久以前，现在美国突然发现越封中国的ai发展得越快，所以最近chatgpt对中国解封了，只要你能上chatgpt，他现在不检查你的ip了

【 在 Z5boy 的大作中提到: 】
: 要注册啊
: 听说要用海外手机号，我就放弃了
:
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FROM 202.120.11.*