各位用你的AI牛刀试试这个问题:

[张旺教授]

Suppose that $f: C \to C$ is a function satisfying
(1) $f(x+y)=f(x)+f(y)$,
(2) $f(e^x)=e^{f(x)}$,
for all complex numbers $x$ and $y$.

(a) Prove $f(2^{1/2})=2^{1/2}$.
(b) Must it be the case that $f(2^{1/3})=2^{1/3}$? What about $f(2^{1/4})=2^{1/4}$ or $f(\ln 2)=\ln 2$?

[张旺教授]

注：
问题(a)证明不难。
问题(b)目前open, 无人能解。

[TheMatrix]

转发了三个版：
军版
书歌版
肚皮舞版 - 这个有人称术版

这三个版都有人对技术问题感兴趣。

[TheMatrix]

Suppose that $f: C \to C$ is a function satisfying
(1) $f(x+y)=f(x)+f(y)$,
(2) $f(e^x)=e^{f(x)}$,
for all complex numbers $x$ and $y$.

(a) Prove $f(2^{1/2})=2^{1/2}$.
(b) Must it be the case that $f(2^{1/3})=2^{1/3}$? What about $f(2^{1/4})=2^{1/4}$ or $f(\ln 2)=\ln 2$?

我的AI得到如下结论：

它第一段的分析很有章法：

[TheMatrix]

AI能分析出f是一个field automorphism of ℂ fixing ℚ已经很厉害了。

我觉得它是见过这个问题的。

[张旺教授]

(b)很难的，从2022年至今无人能解。

让AI好好做做。

[张旺教授]

To conclude f(2^{1/3})=2^{1/3}, one must prove 3|k, which is not known from these assumptions alone. It is true if one adds a regularity condition—for example, if f is continuous, measurable, has the Baire property, or maps ℝ into ℝ. Under such conditions f is either the identity or complex conjugation, and both fix 2^{1/3}.

此题的难点在于f无regularity condition。

如果加上regularity condition，则f只能是f(z)=z or f(z)=\bar{z} (conjugate of z). 那么(b)就容易了。但原题f无regularity condition，故难倒了n多数学家。

[TheMatrix]

To conclude f(2^{1/3})=2^{1/3}, one must prove 3|k, which is not known from these assumptions alone. It is true if one adds a regularity condition—for example, if f is continuous, measurable, has the Baire property, or maps ℝ into ℝ. Under such conditions f is either the identity or complex conjugation, and both fix 2^{1/3}.

此题的难点在于f无regularity condition。

如果加上regularity condition，则f只能是f(z)=z or f(z)=\bar{z} (conjugate of z). 那么(b)就容易了。但原题f无regularity condition，故难倒了n多数学家。

嗯。和雅可比问题类似，本来是充分条件，拿掉一个，问还能不能推出。这类问题确实很难。

这个问题有没有出处？

[TheMatrix]

To conclude f(2^{1/3})=2^{1/3}, one must prove 3|k, which is not known from these assumptions alone. It is true if one adds a regularity condition—for example, if f is continuous, measurable, has the Baire property, or maps ℝ into ℝ. Under such conditions f is either the identity or complex conjugation, and both fix 2^{1/3}.

此题的难点在于f无regularity condition。

如果加上regularity condition，则f只能是f(z)=z or f(z)=\bar{z} (conjugate of z). 那么(b)就容易了。但原题f无regularity condition，故难倒了n多数学家。

不对啊。3 | k 中，k是哪来的？

f既然是 C 的 automorphism，

f(2^{1/3})³
= f(2^{1/3})f(2^{1/3})f(2^{1/3})
= f(2^{1/3}2^{1/3}2^{1/3})
= f(2)
= 2

f(2) = 2 是之前证明过的。

那么 f(2^{1/3}) = 2^{1/3}

这样不行吗？差在哪？

[张旺教授]

再提示一下, 为什么不加a regularity condition 问题(b) 很难。

例：著名的柯西函数方程：f(x+y)=f(x)+f(y) (f: R->R). 求函数f(x).

显然，f(x)=cx 是一个解。如果加上一个合适的regularity condition，则这是柯西方程的唯一解。

但是地但是，“If no further conditions are imposed on ⁠f, then (assuming the axiom of choice) there are infinitely many other functions that satisfy the equation. This was proved in 1905 by Georg Hamel using Hamel bases. Such functions are sometimes called Hamel functions”。

[张旺教授]

不对啊。3 | k 中，k是哪来的？

f既然是 C 的 automorphism，

f(2^{1/3})³
= f(2^{1/3})f(2^{1/3})f(2^{1/3})
= f(2^{1/3}2^{1/3}2^{1/3})
= f(2)
= 2

f(2) = 2 是之前证明过的。

那么 f(2^{1/3}) = 2^{1/3}

这样不行吗？差在哪？

这是AI做的：

Under the stated assumptions alone, the desired conclusion does not follow by an elementary argument; it reduces to a difficult unresolved issue about endomorphisms of the complex exponential field.

Let L=ln⁡2, the usual real logarithm. First,

f(0)=0,f(1)=f(e⁰)=1,

so f(2)=2. Therefore

e^f(L)=f(e^L)=f(2)=2.

Every complex logarithm of 2 has the form L+2πik, so for some k∈ℤ,

f(L)=L+2πik.

Since 2¹/3=e^L/3, additivity gives f(L/3)=f(L)/3, and hence

f(2¹/3)=f(e^L/3)=e^f(L/3)=e(L+2πik)/3=2^{1/3}e^{2πik/3}.

Thus the assumptions prove only that

f(2¹/3)∈{2¹/3,2^{1/3}ω,2^{1/3}ω²},ω=e^{2πi/3}.

To conclude f(2¹/3)=2¹/3, one must prove 3|k, which is not known from these assumptions alone. It is true if one adds a regularity condition—for example, if f is continuous, measurable, has the Baire property, or maps ℝ into ℝ. Under such conditions f is either the identity or complex conjugation, and both fix 2¹/3.

[张旺教授]

Under the stated assumptions alone, the desired conclusion does not follow by an elementary argument; it reduces to a difficult unresolved issue about endomorphisms of the complex exponential field.

Let \(L=\ln 2\), the usual real logarithm. First,

\[
f(0)=0,\qquad f(1)=f(e⁰)=1,
\]

so \(f(2)=2\). Therefore

\[
e^{f(L)}=f(e^L)=f(2)=2.
\]

Every complex logarithm of \(2\) has the form \(L+2\pi i k\), so for some \(k\in\mathbb Z\),

\[
f(L)=L+2\pi i k.
\]

Since \(2^{1/3}=e^{L/3}\), additivity gives \(f(L/3)=f(L)/3\), and hence

\[
\begin{aligned}
f(2^{1/3})
&=f(e^{L/3})\\
&=e^{f(L/3)}\\
&=e^{(L+2\pi i k)/3}\\
&=2^{1/3}e^{2\pi i k/3}.
\end{aligned}
\]

Thus the assumptions prove only that

\[
\boxed{f(2^{1/3})\in
\left\{2^{1/3},\,2^{1/3}\omega,\,2^{1/3}\omega^2\right\}},
\qquad \omega=e^{2\pi i/3}.
\]

To conclude \(f(2^{1/3})=2^{1/3}\), one must prove \(3\mid k\), which is not known from these assumptions alone. It is true if one adds a regularity condition—for example, if \(f\) is continuous, measurable, has the Baire property, or maps \(\mathbb R\) into \(\mathbb R\). Under such conditions \(f\) is either the identity or complex conjugation, and both fix \(2^{1/3}\).

[张旺教授]

这个网站的LaTeX不给力。

[张旺教授]

嗯。和雅可比问题类似，本来是充分条件，拿掉一个，问还能不能推出。这类问题确实很难。

这个问题有没有出处？

是2022年老教授Jan Mycielski提出的问题征解题。问题(b)几年来无人能解。Mycielski passed away in 2025 at the age of 92.

[张旺教授]

怀疑问题(b)的解可能与Hamel基有关。但也不好说，毕竟Hamel基是在R上的。

也许AI能找到解法。

[TheMatrix]

怀疑问题(b)的解可能与Hamel基有关。但也不好说，毕竟Hamel基是在R上的。

也许AI能找到解法。

Hamel基是什么？能介绍一下吗？

[bigbendan]

This is a fascinating problem that combines additive functions (Cauchy's equation) with an exponential condition. Let me solve it step by step.

---

Solution

Given Conditions

For all complex numbers \(x, y \in \mathbb{C}\):

1. Additivity: \(f(x+y) = f(x) + f(y)\)
2. Exponential condition: \(f(e^x) = e^{f(x)}\) (where \(e\) is the base of natural logarithm)

---

(a) Prove \(f(\sqrt{2}) = \sqrt{2}\)

Step 1: Establish basic properties from additivity

From additivity alone, we can derive:

• \(f(0) = 0\): Since \(f(0) = f(0+0) = f(0) + f(0)\), so \(f(0) = 0\).
• \(f(n) = n \cdot f(1)\) for all integers \(n\).
• \(f(-x) = -f(x)\): Since \(f(x) + f(-x) = f(0) = 0\).

But importantly, additivity alone does not imply multiplicativity over reals without continuity assumptions. However, we have the exponential condition to help.

Step 2: Relate the two conditions

Set \(x = 0\) in condition (2):
\[
f(e⁰) = e^{f(0)} \implies f(1) = e⁰ = 1
\]

So \(f(1) = 1\).

Now, from additivity:
\[
f(n) = n \cdot f(1) = n
\]
for all integers \(n\). In fact, \(f(q) = q\) for all rational \(q\).

Step 3: Key insight — iterating the exponential condition

Apply condition (2) repeatedly:
\[
f(e^{e^x}) = f(e^{e^x}) = e^{f(e^x)} = e^{e^{f(x)}}
\]

This suggests that \(f\) commutes with exponentiation in a certain way.

Step 4: Find a specific value to get \(\sqrt{2}\)

We want to prove \(f(2^{1/2}) = 2^{1/2}\).

Let's see if we can use the fact that \(f(1) = 1\):

Since \(1 = e^0\), condition (2) gives:
\[
f(1) = e^{f(0)} \implies 1 = e⁰ \quad \text{(consistent, already used)}
\]

But here's the key: Take \(x = 0\) and \(y = \ln 2\) in condition (1):
\[
f(0 + \ln 2) = f(0) + f(\ln 2) \implies f(\ln 2) = f(\ln 2)
\]

...this doesn't give us anything new.

Step 5: The crucial step — using the exponential condition twice

Let's set \(x\) such that \(e^x = 2^{1/2}\). That is, \(x = \frac{1}{2}\ln 2\).

Then condition (2) gives:
\[
f(2^{1/2}) = e^{f(\frac{1}{2}\ln 2)}
\]

So if we can show \(f(\frac{1}{2}\ln 2) = \frac{1}{2}\ln 2\), we're done.

Step 6: Determine \(f(\ln 2)\)

From condition (2) with \(x = \ln 2\):
\[
f(e^{\ln 2}) = e^{f(\ln 2)} \implies f(2) = e^{f(\ln 2)}
\]

But we already know \(f(2) = 2\) (since \(f(n) = n\) for integers). Therefore:
\[
2 = e^{f(\ln 2)} \implies f(\ln 2) = \ln 2
\]

Step 7: Now find \(f(\frac{1}{2}\ln 2)\)

From additivity:
\[
f(\ln 2) = f\left(\frac{1}{2}\ln 2 + \frac{1}{2}\ln 2\right) = 2 \cdot f\left(\frac{1}{2}\ln 2\right)
\]

Since \(f(\ln 2) = \ln 2\), we get:
\[
\ln 2 = 2 \cdot f\left(\frac{1}{2}\ln 2\right) \implies f\left(\frac{1}{2}\ln 2\right) = \frac{1}{2}\ln 2
\]

Step 8: Apply condition (2)

Now, \(e^{\frac{1}{2}\ln 2} = 2^{1/2}\). Using condition (2):
\[
f(2^{1/2}) = f(e^{\frac{1}{2}\ln 2}) = e^{f(\frac{1}{2}\ln 2)} = e^{\frac{1}{2}\ln 2} = 2^{1/2}
\]

Thus, \(f(\sqrt{2}) = \sqrt{2}\).

---

(b) Analyze \(f(2^{1/3})\), \(f(2^{1/4})\), and \(f(\ln 2)\)

Result 1: \(f(\ln 2) = \ln 2\)

We already proved this in Step 6 above. So \(f(\ln 2) = \ln 2\).

Result 2: \(f(2^{1/4}) = 2^{1/4}\)

Following the same pattern:

1. From \(f(\ln 2) = \ln 2\) and additivity:
   \[
   f\left(\frac{1}{4}\ln 2\right) = \frac{1}{4}\ln 2
   \]

2. Apply condition (2):
   \[
   f(2^{1/4}) = f(e^{\frac{1}{4}\ln 2}) = e^{f(\frac{1}{4}\ln 2)} = e^{\frac{1}{4}\ln 2} = 2^{1/4}
   \]

So \(f(2^{1/4}) = 2^{1/4}\).

Result 3: \(f(2^{1/3}) = 2^{1/3}\)

Similarly:

1. \(f\left(\frac{1}{3}\ln 2\right) = \frac{1}{3}\ln 2\)
2. \(f(2^{1/3}) = f(e^{\frac{1}{3}\ln 2}) = e^{f(\frac{1}{3}\ln 2)} = e^{\frac{1}{3}\ln 2} = 2^{1/3}\)

So \(f(2^{1/3}) = 2^{1/3}\).

---

Summary of Results

Expression	Value of \(f\)
\(f(\sqrt{2}) = f(2^{1/2})\)	\(\sqrt{2}\)
\(f(2^{1/3})\)	\(2^{1/3}\)
\(f(2^{1/4})\)	\(2^{1/4}\)
\(f(\ln 2)\)	\(\ln 2\)

---

Key Insight

The two conditions together force \(f\) to be the identity on numbers that are powers of \(e\) with rational exponents, which includes \(\ln 2\) and \(2^{1/n}\). This is because:

• Additivity lets us move rational factors around.
• The exponential condition "maps" the argument of \(e^x\) through \(f\) in a consistent way.

Therefore: Yes, \(f(2^{1/3}) = 2^{1/3}\), \(f(2^{1/4}) = 2^{1/4}\), and \(f(\ln 2) = \ln 2\).

[张旺教授]

This is a fascinating problem that combines additive functions (Cauchy's equation) with an exponential condition. Let me solve it step by step.

---

Solution

Given Conditions

For all complex numbers \(x, y \in \mathbb{C}\):

1. Additivity: \(f(x+y) = f(x) + f(y)\)
2. Exponential condition: \(f(e^x) = e^{f(x)}\) (where \(e\) is the base of natural logarithm)

---

(a) Prove \(f(\sqrt{2}) = \sqrt{2}\)

Step 1: Establish basic properties from additivity

From additivity alone, we can derive:

• \(f(0) = 0\): Since \(f(0) = f(0+0) = f(0) + f(0)\), so \(f(0) = 0\).
• \(f(n) = n \cdot f(1)\) for all integers \(n\).
• \(f(-x) = -f(x)\): Since \(f(x) + f(-x) = f(0) = 0\).

But importantly, additivity alone does not imply multiplicativity over reals without continuity assumptions. However, we have the exponential condition to help.

Step 2: Relate the two conditions

Set \(x = 0\) in condition (2):
\[
f(e⁰) = e^{f(0)} \implies f(1) = e⁰ = 1
\]

So \(f(1) = 1\).

Now, from additivity:
\[
f(n) = n \cdot f(1) = n
\]
for all integers \(n\). In fact, \(f(q) = q\) for all rational \(q\).

Step 3: Key insight — iterating the exponential condition

Apply condition (2) repeatedly:
\[
f(e^{e^x}) = f(e^{e^x}) = e^{f(e^x)} = e^{e^{f(x)}}
\]

This suggests that \(f\) commutes with exponentiation in a certain way.

Step 4: Find a specific value to get \(\sqrt{2}\)

We want to prove \(f(2^{1/2}) = 2^{1/2}\).

Let's see if we can use the fact that \(f(1) = 1\):

Since \(1 = e^0\), condition (2) gives:
\[
f(1) = e^{f(0)} \implies 1 = e⁰ \quad \text{(consistent, already used)}
\]

But here's the key: Take \(x = 0\) and \(y = \ln 2\) in condition (1):
\[
f(0 + \ln 2) = f(0) + f(\ln 2) \implies f(\ln 2) = f(\ln 2)
\]

...this doesn't give us anything new.

Step 5: The crucial step — using the exponential condition twice

Let's set \(x\) such that \(e^x = 2^{1/2}\). That is, \(x = \frac{1}{2}\ln 2\).

Then condition (2) gives:
\[
f(2^{1/2}) = e^{f(\frac{1}{2}\ln 2)}
\]

So if we can show \(f(\frac{1}{2}\ln 2) = \frac{1}{2}\ln 2\), we're done.

Step 6: Determine \(f(\ln 2)\)

From condition (2) with \(x = \ln 2\):
\[
f(e^{\ln 2}) = e^{f(\ln 2)} \implies f(2) = e^{f(\ln 2)}
\]

But we already know \(f(2) = 2\) (since \(f(n) = n\) for integers). Therefore:
\[
2 = e^{f(\ln 2)} \implies f(\ln 2) = \ln 2
\]

Step 7: Now find \(f(\frac{1}{2}\ln 2)\)

From additivity:
\[
f(\ln 2) = f\left(\frac{1}{2}\ln 2 + \frac{1}{2}\ln 2\right) = 2 \cdot f\left(\frac{1}{2}\ln 2\right)
\]

Since \(f(\ln 2) = \ln 2\), we get:
\[
\ln 2 = 2 \cdot f\left(\frac{1}{2}\ln 2\right) \implies f\left(\frac{1}{2}\ln 2\right) = \frac{1}{2}\ln 2
\]

Step 8: Apply condition (2)

Now, \(e^{\frac{1}{2}\ln 2} = 2^{1/2}\). Using condition (2):
\[
f(2^{1/2}) = f(e^{\frac{1}{2}\ln 2}) = e^{f(\frac{1}{2}\ln 2)} = e^{\frac{1}{2}\ln 2} = 2^{1/2}
\]

Thus, \(f(\sqrt{2}) = \sqrt{2}\).

---

(b) Analyze \(f(2^{1/3})\), \(f(2^{1/4})\), and \(f(\ln 2)\)

Result 1: \(f(\ln 2) = \ln 2\)

We already proved this in Step 6 above. So \(f(\ln 2) = \ln 2\).

Result 2: \(f(2^{1/4}) = 2^{1/4}\)

Following the same pattern:

1. From \(f(\ln 2) = \ln 2\) and additivity:
   \[
   f\left(\frac{1}{4}\ln 2\right) = \frac{1}{4}\ln 2
   \]

2. Apply condition (2):
   \[
   f(2^{1/4}) = f(e^{\frac{1}{4}\ln 2}) = e^{f(\frac{1}{4}\ln 2)} = e^{\frac{1}{4}\ln 2} = 2^{1/4}
   \]

So \(f(2^{1/4}) = 2^{1/4}\).

Result 3: \(f(2^{1/3}) = 2^{1/3}\)

Similarly:

1. \(f\left(\frac{1}{3}\ln 2\right) = \frac{1}{3}\ln 2\)
2. \(f(2^{1/3}) = f(e^{\frac{1}{3}\ln 2}) = e^{f(\frac{1}{3}\ln 2)} = e^{\frac{1}{3}\ln 2} = 2^{1/3}\)

So \(f(2^{1/3}) = 2^{1/3}\).

---

Summary of Results

Expression	Value of \(f\)
\(f(\sqrt{2}) = f(2^{1/2})\)	\(\sqrt{2}\)
\(f(2^{1/3})\)	\(2^{1/3}\)
\(f(2^{1/4})\)	\(2^{1/4}\)
\(f(\ln 2)\)	\(\ln 2\)

---

Key Insight

The two conditions together force \(f\) to be the identity on numbers that are powers of \(e\) with rational exponents, which includes \(\ln 2\) and \(2^{1/n}\). This is because:

• Additivity lets us move rational factors around.
• The exponential condition "maps" the argument of \(e^x\) through \(f\) in a consistent way.

Therefore: Yes, \(f(2^{1/3}) = 2^{1/3}\), \(f(2^{1/4}) = 2^{1/4}\), and \(f(\ln 2) = \ln 2\).

The solution is not correct for \(f:\mathbb C\to\mathbb C\). The key error occurs in Step 6.

From

\[
e^{f(\ln 2)}=2
\]

the solution concludes

\[
f(\ln 2)=\ln 2.
\]

That implication is valid over \(\mathbb R\), but not over \(\mathbb C\), because the complex exponential is periodic:

\[
e^z=2
\quad\Longleftrightarrow\quad
z=\ln 2+2\pi i k,\qquad k\in\mathbb Z.
\]

Thus all that can be concluded is

\[
\boxed{f(\ln 2)=\ln 2+2\pi i k}
\]

for some integer \(k\).

Consequences

By additivity, \(f(qx)=qf(x)\) for every \(q\in\mathbb Q\). Therefore, for any positive integer \(n\),

\[
f\left(\frac{\ln 2}{n}\right)
=\frac{f(\ln 2)}n
=\frac{\ln 2+2\pi i k}{n}.
\]

Applying the exponential condition gives

\[
\begin{aligned}
f(2^{1/n})
&=f\left(e^{(\ln 2)/n}\right)\\
&=e^{f((\ln 2)/n)}\\
&=e^{(\ln 2+2\pi i k)/n}\\
&=2^{1/n}e^{2\pi i k/n}.
\end{aligned}
\]

Hence the correct conclusions from this argument are:

\[
\boxed{f(\sqrt2)=(-1)^k\sqrt2},
\]

so

\[
f(\sqrt2)\in\{\sqrt2,-\sqrt2\};
\]

\[
\boxed{f(2^{1/3})=2^{1/3}e^{2\pi i k/3}},
\]

so it is one of the three cube roots of \(2\); and

\[
\boxed{f(2^{1/4})=2^{1/4}e^{2\pi i k/4}},
\]

so it is one of the four fourth roots of \(2\).

These equations use the same integer \(k\) arising from \(f(\ln 2)\).

The solution would become valid if there were an additional assumption such as

\[
f(\mathbb R)\subseteq\mathbb R,
\]

because then \(f(\ln 2)\) would be real, forcing \(k=0\). It would also be valid under standard regularity assumptions such as continuity or measurability. Without such an assumption, Step 6—and therefore all later conclusions—is unjustified.

[张旺教授]

Hamel基是什么？能介绍一下吗？

Hamel基只针对f(x+y)=f(x)+f(y), 不适用于指数条件f(e^x)=e^{f(x)}. 所以Hamel基帮不上忙。

[bigbendan]

The solution is not correct for \(f:\mathbb C\to\mathbb C\). The key error occurs in Step 6.

From

\[
e^{f(\ln 2)}=2
\]

the solution concludes

\[
f(\ln 2)=\ln 2.
\]

That implication is valid over \(\mathbb R\), but not over \(\mathbb C\), because the complex exponential is periodic:

\[
e^z=2
\quad\Longleftrightarrow\quad
z=\ln 2+2\pi i k,\qquad k\in\mathbb Z.
\]

Thus all that can be concluded is

\[
\boxed{f(\ln 2)=\ln 2+2\pi i k}
\]

for some integer \(k\).

Consequences

By additivity, \(f(qx)=qf(x)\) for every \(q\in\mathbb Q\). Therefore, for any positive integer \(n\),

\[
f\left(\frac{\ln 2}{n}\right)
=\frac{f(\ln 2)}n
=\frac{\ln 2+2\pi i k}{n}.
\]

Applying the exponential condition gives

\[
\begin{aligned}
f(2^{1/n})
&=f\left(e^{(\ln 2)/n}\right)\\
&=e^{f((\ln 2)/n)}\\
&=e^{(\ln 2+2\pi i k)/n}\\
&=2^{1/n}e^{2\pi i k/n}.
\end{aligned}
\]

Hence the correct conclusions from this argument are:

\[
\boxed{f(\sqrt2)=(-1)^k\sqrt2},
\]

so

\[
f(\sqrt2)\in\{\sqrt2,-\sqrt2\};
\]

\[
\boxed{f(2^{1/3})=2^{1/3}e^{2\pi i k/3}},
\]

so it is one of the three cube roots of \(2\); and

\[
\boxed{f(2^{1/4})=2^{1/4}e^{2\pi i k/4}},
\]

so it is one of the four fourth roots of \(2\).

These equations use the same integer \(k\) arising from \(f(\ln 2)\).

The solution would become valid if there were an additional assumption such as

\[
f(\mathbb R)\subseteq\mathbb R,
\]

because then \(f(\ln 2)\) would be real, forcing \(k=0\). It would also be valid under standard regularity assumptions such as continuity or measurability. Without such an assumption, Step 6—and therefore all later conclusions—is unjustified.

人生轮回。但是一生就限制在一世。