| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | The limit of the given integral can be found by using the properties of the trigonometric functions and the properties of the Lebesgue measure. The integral in question is: \[ \lim _{n \rightarrow \infty} \int_E \cos ^2 (n x+\xi_n ) dx \] First, we use the trigonometric identity for the square of the cosine function: \[ \cos ^2 (n x+\xi_n ) = \frac{1 + \cos(2nx + 2\xi_n)}{2} \] Thus, the integral becomes: \[ \lim _{n \rightarrow \infty} \int_E \frac{1 + \cos(2nx + 2\xi_n)}{2} dx \] This can be split into two integrals: \[ \lim _{n \rightarrow \infty} \left( \frac{1}{2} \int_E dx + \frac{1}{2} \int_E \cos(2nx + 2\xi_n) dx \right) \] The first integral is simply the measure of the set E, which is given as 2: \[ \lim _{n \rightarrow \infty} \frac{1}{2} \int_E dx = \frac{1}{2} \cdot 2 = 1 \] The second integral involves the cosine function, which oscillates between -1 and 1. As n grows large, the oscillations become more rapid, and the integral of the cosine term over any measurable set will approach 0 due to the cancellation of positive and negative parts of the cosine function. This is a consequence of the Riemann-Lebesgue Lemma, which states that the integral of a trigonometric function over a finite interval approaches 0 as the frequency increases. Thus, the second integral vanishes in the limit: \[ \lim _{n \rightarrow \infty} \frac{1}{2} \int_E \cos(2nx + 2\xi_n) dx = 0 \] Therefore, the limit of the original integral is: \[ \lim _{n \rightarrow \infty} \int_E \cos ^2 (n x+\xi_n ) dx = 1 \] |
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| deepseekr1 | To find the limit of the integral \(\int_E \cos^2(nx + \xi_n) \, dx\) as \(n\) approaches infinity, where \(E \subset (0, 2\pi)\) is a measurable set with Lebesgue measure 2, we start by using the trigonometric identity for \(\cos^2\): \[ \cos^2(nx + \xi_n) = \frac{1 + \cos(2(nx + \xi_n))}{2} \] This transforms the integral into: \[ \int_E \cos^2(nx + \xi_n) \, dx = \frac{1}{2} \int_E 1 \, dx + \frac{1}{2} \int_E \cos(2nx + 2\xi_n) \, dx \] The first term simplifies to \(\frac{1}{2} \times 2 = 1\) because the measure of \(E\) is 2. For the second term, we need to evaluate the limit of \(\int_E \cos(2nx + 2\xi_n) \, dx\) as \(n\) approaches infinity. Using the Riemann-Lebesgue lemma, which states that the integral of a highly oscillatory function over a measurable set tends to zero, we consider the integral: \[ \int_E \cos(2nx + 2\xi_n) \, dx \] Expanding \(\cos(2nx + 2\xi_n)\) using the cosine addition formula, we get: \[ \cos(2nx + 2\xi_n) = \cos(2nx)\cos(2\xi_n) - \sin(2nx)\sin(2\xi_n) \] The integral then splits into two terms, each involving \(\cos(2nx)\) or \(\sin(2nx)\), both of which oscillate rapidly as \(n\) increases. By the Riemann-Lebesgue lemma, these integrals tend to zero. The bounded factors \(\cos(2\xi_n)\) and \(\sin(2\xi_n)\) do not affect this limit. Therefore, the integral \(\int_E \cos(2nx + 2\xi_n) \, dx\) tends to zero as \(n\) approaches infinity. Thus, the limit of the original integral is: \[ 1 + \frac{1}{2} \times 0 = 1 \] The final answer is: \[ \boxed{1} \] |
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| gpto1 | To solve this problem, we need to compute the limit: \[ \lim_{n \to \infty} \int_E \cos^2(n x + \xi_n) \, dx \] Given that \( E \subset (0, 2\pi) \) is a measurable set with Lebesgue measure \( m(E) = 2 \), and \( \{ \xi_n \} \) is an arbitrary sequence of real numbers. **Step 1: Simplify the integrand** First, recall the trigonometric identity: \[ \cos^2 \theta = \frac{1 + \cos 2\theta}{2} \] Using this identity, we can rewrite the integrand: \[ \cos^2(n x + \xi_n) = \frac{1 + \cos(2 n x + 2 \xi_n)}{2} \] **Step 2: Split the integral** Now, the integral becomes: \[ \int_E \cos^2(n x + \xi_n) \, dx = \frac{1}{2} \int_E 1 \, dx + \frac{1}{2} \int_E \cos(2 n x + 2 \xi_n) \, dx \] Simplify the first term: \[ \frac{1}{2} \int_E 1 \, dx = \frac{1}{2} \cdot m(E) = \frac{1}{2} \cdot 2 = 1 \] **Step 3: Analyze the second integral** We need to evaluate the limit of the second term: \[ \lim_{n \to \infty} \frac{1}{2} \int_E \cos(2 n x + 2 \xi_n) \, dx \] As \( n \to \infty \), the function \( \cos(2 n x + 2 \xi_n) \) oscillates increasingly rapidly. Over a set of finite measure, the averages of these oscillations tend to zero unless there is synchronization or specific structures in \( E \). Given that \( \xi_n \) is arbitrary and \( E \) is any measurable set of finite measure \( 2 \), the rapid oscillations of the cosine function will average out over \( E \). Therefore, \[ \lim_{n \to \infty} \int_E \cos(2 n x + 2 \xi_n) \, dx = 0 \] **Step 4: Combine the results** Combining the first term (which equals 1) and the limit of the second term (which is 0): \[ \lim_{n \to \infty} \int_E \cos^2(n x + \xi_n) \, dx = 1 + 0 = 1 \] **Final Answer:** 1 |
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| GPT-4o | To find the limit of \(\int_E \cos^2(nx + \xi_n) \, dx\) as \(n \to \infty\), where \(E \subset (0, 2\pi)\) is a measurable set with Lebesgue measure \(|E| = 2\), we can use the fact that the function \(\cos^2(nx + \xi_n)\) oscillates rapidly as \(n\) increases. First, recall that \(\cos^2(x)\) can be expressed using the trigonometric identity: \[ \cos^2(x) = \frac{1 + \cos(2x)}{2} \] Thus, we can rewrite the integral as: \[ \int_E \cos^2(nx + \xi_n) \, dx = \int_E \frac{1 + \cos(2(nx + \xi_n))}{2} \, dx \] This integral can be separated into two parts: \[ \int_E \cos^2(nx + \xi_n) \, dx = \frac{1}{2} \int_E 1 \, dx + \frac{1}{2} \int_E \cos(2(nx + \xi_n)) \, dx \] The first part is simply: \[ \frac{1}{2} \int_E 1 \, dx = \frac{1}{2} |E| = \frac{1}{2} \cdot 2 = 1 \] The second part is: \[ \frac{1}{2} \int_E \cos(2(nx + \xi_n)) \, dx \] As \(n \to \infty\), the term \(\cos(2(nx + \xi_n))\) oscillates increasingly rapidly. For a fixed \(x\), the function \(\cos(2(nx + \xi_n))\) averages out to zero over any interval because of its rapid oscillation. This implies that the integral of \(\cos(2(nx + \xi_n))\) over the set \(E\) will tend to zero as \(n\) increases. Formally, this is justified by the Riemann-Lebesgue Lemma, which states that the integral of a product of a bounded function and a rapidly oscillating function tends to zero. Therefore: \[ \lim_{n \to \infty} \frac{1}{2} \int_E \cos(2(nx + \xi_n)) \, dx = 0 \] Putting it all together, we have: \[ \lim_{n \to \infty} \int_E \cos^2(nx + \xi_n) \, dx = 1 + 0 = 1 \] Thus, the limit is: \[ \boxed{1} \] |
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| llama3.1-70B-Instruct | The problem is to find the limit of the integral of $\cos^2(nx+\xi_n)$ over a measurable set $E$ with Lebesgue measure 2, as $n$ approaches infinity. ## Step 1: Recall the definition of Lebesgue measure The Lebesgue measure of a set $E$ is a measure of the "size" of the set, and it is defined as the infimum of the sums of the lengths of intervals that cover $E$. ## Step 2: Recall the properties of the cosine function The cosine function has a period of $2\pi$, and its square has a period of $\pi$. This means that the integral of $\cos^2(nx+\xi_n)$ over any interval of length $\pi$ is equal to the integral of $\cos^2(nx+\xi_n)$ over any other interval of length $\pi$. ## Step 3: Use the properties of the Lebesgue measure Since the Lebesgue measure of $E$ is 2, we can cover $E$ with two intervals of length 1. Let $I_1$ and $I_2$ be two intervals of length 1 that cover $E$. ## Step 4: Use the properties of the cosine function Since the cosine function has a period of $2\pi$, we can write $\cos^2(nx+\xi_n) = \cos^2(nx+\xi_n + 2\pi k)$ for any integer $k$. This means that the integral of $\cos^2(nx+\xi_n)$ over $I_1$ is equal to the integral of $\cos^2(nx+\xi_n + 2\pi k)$ over $I_1$. ## Step 5: Use the dominated convergence theorem Since the sequence $\{\xi_n\}$ is arbitrary, we can assume that $\xi_n$ converges to some limit $\xi$ as $n$ approaches infinity. Then, by the dominated convergence theorem, we have \begin{align*} \lim_{n\to\infty} \int_E \cos^2(nx+\xi_n) dx &= \lim_{n\to\infty} \int_{I_1} \cos^2(nx+\xi_n) dx + \lim_{n\to\infty} \int_{I_2} \cos^2(nx+\xi_n) dx \\ &= \int_{I_1} \cos^2(x+\xi) dx + \int_{I_2} \cos^2(x+\xi) dx \\ &= \frac{1}{2} \int_{I_1} 1 + \cos(2x+2\xi) dx + \frac{1}{2} \int_{I_2} 1 + \cos(2x+2\xi) dx \\ &= \frac{1}{2} \int_{I_1} 1 dx + \frac{1}{2} \int_{I_2} 1 dx \\ &= \frac{1}{2} \cdot 1 + \frac{1}{2} \cdot 1 \\ &= 1. \end{align*} The final answer is: $\boxed{1}$ |
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| llama3.1-405B-Instruct | Since the Lebesgue measure of $E$ is $2$, we have that $E$ is a subset of $(0,2\pi)$ with full measure. Therefore, we can write $E$ as the union of a countable collection of disjoint intervals, each with positive length. Let $\{I_k\}$ be such a collection of intervals. For each $k$, let $f_k(x) = \cos^2(nx+\xi_n)$ for $x\in I_k$. Then $f_k$ is a continuous function on $I_k$, and since $I_k$ has positive length, we have that $\int_{I_k} f_k(x) dx > 0$. Now, since $E$ is the union of the $I_k$'s, we have that $$\int_E \cos^2(nx+\xi_n) dx = \sum_k \int_{I_k} f_k(x) dx > 0.$$ Since the sum is over a countable collection of positive numbers, we have that the sum is positive. Therefore, we have that $$\lim_{n\to\infty} \int_E \cos^2(nx+\xi_n) dx > 0.$$ On the other hand, since $\cos^2(nx+\xi_n) \leq 1$ for all $x$ and $n$, we have that $$\int_E \cos^2(nx+\xi_n) dx \leq \int_E 1 dx = 2.$$ Therefore, we have that $$0 < \lim_{n\to\infty} \int_E \cos^2(nx+\xi_n) dx \leq 2.$$ Since the limit is sandwiched between 0 and 2, we conclude that $$\lim_{n\to\infty} \int_E \cos^2(nx+\xi_n) dx = 1.$$ The final answer is 1. |
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| DeepSeek-Coder-V2 | To solve the given problem, we need to evaluate the limit of the integral \(\lim _{n \rightarrow \infty} \int_E \cos ^2 (n x+\xi_n ) \, dx\), where \(E \subset (0, 2\pi)\) is a measurable set with Lebesgue measure \(m(E) = 2\), and \(\{\xi_n\}\) is an arbitrary sequence of real numbers. First, let's recall the trigonometric identity for \(\cos^2(x)\): \[ \cos^2(x) = \frac{1 + \cos(2x)}{2} \] Using this identity, we can rewrite the integrand: \[ \cos^2(nx + \xi_n) = \frac{1 + \cos(2nx + 2\xi_n)}{2} \] Thus, the integral becomes: \[ \int_E \cos^2(nx + \xi_n) \, dx = \int_E \frac{1 + \cos(2nx + 2\xi_n)}{2} \, dx \] We can split this into two integrals: \[ \int_E \cos^2(nx + \xi_n) \, dx = \frac{1}{2} \int_E 1 \, dx + \frac{1}{2} \int_E \cos(2nx + 2\xi_n) \, dx \] Since \(m(E) = 2\), the first integral is: \[ \frac{1}{2} \int_E 1 \, dx = \frac{1}{2} \cdot 2 = 1 \] Now, consider the second integral: \[ \frac{1}{2} \int_E \cos(2nx + 2\xi_n) \, dx \] To analyze this integral, we use the Riemann-Lebesgue Lemma, which states that for any measurable set \(E\) and any integrable function \(f\), the integral \(\int_E f(x) \cos(nx) \, dx\) tends to zero as \(n \to \infty\). Here, \(f(x) = 1\), and the argument of the cosine function is \(2nx + 2\xi_n\), which grows linearly with \(n\). Therefore, as \(n \to \infty\), the integral of \(\cos(2nx + 2\xi_n)\) over \(E\) will tend to zero. Thus, we have: \[ \frac{1}{2} \int_E \cos(2nx + 2\xi_n) \, dx \to 0 \quad \text{as} \quad n \to \infty \] Combining these results, we get: \[ \lim _{n \rightarrow \infty} \int_E \cos^2(nx + \xi_n) \, dx = 1 + 0 = 1 \] Therefore, the final answer is: \[ \boxed{1} \] |
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| Qwen2-72B-Instruct | The limit of the given integral can be found by using the properties of the trigonometric functions and the properties of the Lebesgue measure. The integral in question is: \[ \lim _{n \rightarrow \infty} \int_E \cos ^2 (n x+\xi_n ) dx \] First, we use the trigonometric identity for the square of the cosine function: \[ \cos ^2 (n x+\xi_n ) = \frac{1 + \cos(2nx + 2\xi_n)}{2} \] Thus, the integral becomes: \[ \lim _{n \rightarrow \infty} \int_E \frac{1 + \cos(2nx + 2\xi_n)}{2} dx \] This can be split into two integrals: \[ \lim _{n \rightarrow \infty} \left( \frac{1}{2} \int_E dx + \frac{1}{2} \int_E \cos(2nx + 2\xi_n) dx \right) \] The first integral is simply the measure of the set E, which is given as 2: \[ \lim _{n \rightarrow \infty} \frac{1}{2} \int_E dx = \frac{1}{2} \cdot 2 = 1 \] The second integral involves the cosine function, which oscillates between -1 and 1. As n grows large, the oscillations become more rapid, and the integral of the cosine term over any measurable set will approach 0 due to the cancellation of positive and negative parts of the cosine function. This is a consequence of the Riemann-Lebesgue Lemma, which states that the integral of a trigonometric function over a finite interval approaches 0 as the frequency increases. Thus, the second integral vanishes in the limit: \[ \lim _{n \rightarrow \infty} \frac{1}{2} \int_E \cos(2nx + 2\xi_n) dx = 0 \] Therefore, the limit of the original integral is: \[ \lim _{n \rightarrow \infty} \int_E \cos ^2 (n x+\xi_n ) dx = 1 \] |
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| baichuan4 | To solve this problem, we'll use the fact that the Lebesgue measure of $E$ is $2$, which means that $E$ is a subset of $(0, 2\pi)$ with a measure equal to the length of the interval $(0, 2\pi)$, implying that $E$ essentially covers the entire interval $(0, 2\pi)$. We are asked to find the limit of the integral of $\cos^2(nx + \xi_n)$ over $E$ as $n$ approaches infinity. First, let's recall the following trigonometric identity for $\cos^2(x)$: $$\cos^2(x) = \frac{1 + \cos(2x)}{2}.$$ Substituting $nx + \xi_n$ for $x$, we get: $$\cos^2(nx + \xi_n) = \frac{1 + \cos(2nx + 2\xi_n)}{2}.$$ Now, let's integrate this expression over $E$: $$\int_E \cos^2(nx + \xi_n) \, dx = \int_E \frac{1 + \cos(2nx + 2\xi_n)}{2} \, dx.$$ Since the Lebesgue measure of $E$ is $2$, this integral is equivalent to integrating over the entire interval $(0, 2\pi)$: $$\int_0^{2\pi} \frac{1 + \cos(2nx + 2\xi_n)}{2} \, dx.$$ The integral of the constant term $\frac{1}{2}$ over $(0, 2\pi)$ is simply $\pi$. For the cosine term, we can use the fact that the integral of a periodic function over one full period is $0$: $$\int_0^{2\pi} \cos(2nx + 2\xi_n) \, dx = 0.$$ Therefore, the integral becomes: $$\int_0^{2\pi} \frac{1 + \cos(2nx + 2\xi_n)}{2} \, dx = \pi + 0 = \pi.$$ Finally, taking the limit as $n$ approaches infinity, we have: $$\lim_{n \rightarrow \infty} \int_E \cos^2(nx + \xi_n) \, dx = \pi.$$ So the numeric value of the limit is $\pi$. |
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| doubao | We can use the Riemann-Lebesgue lemma. For any integrable function $f(x)$ on an interval, $\lim\limits_{n o\infty}\int_I f(x)\cos(nx+\xi_n)dx=0$. Here, $\cos^2(nx+\xi_n)=\frac{1+\cos(2nx+2\xi_n)}{2}$. Applying the Riemann-Lebesgue lemma to both terms, we get $\lim\limits_{n o\infty}\int_E \cos^2(nx+\xi_n)dx=\lim\limits_{n o\infty}\int_E \frac{1}{2}dx+\lim\limits_{n o\infty}\int_E \frac{\cos(2nx+2\xi_n)}{2}dx=\frac{1}{2}\cdot m(E)+0=\frac{1}{2}\cdot 2=1$. So the answer is 1. |
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| minimax | To solve this problem, we can use the fact that the integral of a function over a measurable set can be approximated by the average value of the function over that set, multiplied by the measure of the set. In particular, for a function $f(x)$ defined on a measurable set $E$ with Lebesgue measure $|E|$, the integral of $f$ over $E$ is given by: $$\int_E f(x) dx = |E| \cdot \text{average value of } f \text{ on } E$$ The average value of $f$ on $E$ is defined as: $$\text{average value of } f \text{ on } E = \frac{1}{|E|} \int_E f(x) dx$$ In our case, we have $f(x) = \cos^2(nx + \xi_n)$ and $|E| = 2$. The function $\cos^2$ is periodic with period $\pi$, and over any interval of length $\pi$, its average value is $\frac{1}{2}$ because it spends half of the time in the interval at values above $\frac{1}{2}$ and half of the time below $\frac{1}{2}$. Since the integral of $\cos^2$ over any interval of length $\pi$ is $\frac{\pi}{2}$, the average value of $\cos^2$ over such an interval is $\frac{1}{2}$. Therefore, the average value of $\cos^2(nx + \xi_n)$ over the set $E$ will also be $\frac{1}{2}$, because the set $E$ has measure 2, which is equivalent to two intervals of length $\pi$. Thus, we have: $$\lim _{n \rightarrow \infty} \int_E \cos^2(nx + \xi_n) dx = |E| \cdot \text{average value of } \cos^2(nx + \xi_n) \text{ on } E$$ $$= 2 \cdot \frac{1}{2}$$ $$= 1$$ So the limit of the integral as $n$ approaches infinity is 1. |
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| kimi | To solve this problem, we can use the properties of the Lebesgue integral and trigonometric identities. First, recall that the Lebesgue measure of a set \( E \) is the "size" of the set, in this case, it's given as 2. Next, consider the function \( f(x) = \cos^2(nx + \xi_n) \). We can use the double-angle formula for cosine to simplify this: \[ \cos^2(\theta) = \frac{1 + \cos(2\theta)}{2} \] Applying this to our function, we get: \[ \cos^2(nx + \xi_n) = \frac{1 + \cos(2nx + 2\xi_n)}{2} \] Now, let's integrate this function over the set \( E \): \[ \int_E \cos^2(nx + \xi_n) \, dx = \int_E \frac{1 + \cos(2nx + 2\xi_n)}{2} \, dx \] Since the set \( E \) has a Lebesgue measure of 2, we can split the integral into two parts: \[ \int_E \cos^2(nx + \xi_n) \, dx = \frac{1}{2} \int_E 1 \, dx + \frac{1}{2} \int_E \cos(2nx + 2\xi_n) \, dx \] The first integral is straightforward: \[ \int_E 1 \, dx = \text{measure of } E = 2 \] For the second integral, we note that \( \cos(2nx + 2\xi_n) \) is a continuous function with period \( \frac{2\pi}{n} \). As \( n \) approaches infinity, the period becomes arbitrarily small, and the integral of a continuous function over a set of finite measure will oscillate but average out to zero over the period. This is because the integral over one period of a cosine function is zero. Therefore, as \( n \) grows large, the second integral will approach zero: \[ \lim_{n \to \infty} \int_E \cos(2nx + 2\xi_n) \, dx = 0 \] Combining these results, we get: \[ \lim_{n \to \infty} \int_E \cos^2(nx + \xi_n) \, dx = \frac{1}{2} \cdot 2 + \frac{1}{2} \cdot 0 = 1 \] So the solution to the problem is: \[ \boxed{1} \] |
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| wenxinyiyan | 首先，由于$E$是可测的且其测度为2，我们可以利用傅里叶级数的性质来求解。 考虑函数$f(x) = \cos^2(nx + \xi_n)$，它可以写成： $f(x) = \frac{1 + \cos(2nx + 2\xi_n)}{2}$ 现在，我们要求的是： $\lim_{n \rightarrow \infty} \int_E \cos^2(nx + \xi_n) dx$ 这等于： $\lim_{n \rightarrow \infty} \int_E \frac{1 + \cos(2nx + 2\xi_n)}{2} dx$ 由于$E$的测度为2，我们可以将上述积分拆分为两部分： $\lim_{n \rightarrow \infty} \left[ \frac{1}{2} \int_E dx + \frac{1}{2} \int_E \cos(2nx + 2\xi_n) dx \right]$ 第一部分$\frac{1}{2} \int_E dx = \frac{1}{2} \times 2 = 1$，因为$E$的测度为2。 对于第二部分，我们注意到函数$\cos(2nx + 2\xi_n)$在区间$(0, 2\pi)$上是周期函数，且其平均值为0。因此，当$n$趋向于无穷大时，第二部分的积分趋向于0。 所以，最终的结果是： $\lim_{n \rightarrow \infty} \int_E \cos^2(nx + \xi_n) dx = 1$ |
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| zhipu | To solve the problem, we need to evaluate the limit of the integral of \(\cos^2(nx + \xi_n)\) over the set \(E\) as \(n\) approaches infinity. First, let's recall a key property of the cosine function: the average value of \(\cos^2(\theta)\) over any interval of length \(2\pi\) is \(\frac{1}{2}\). This is because the cosine function is periodic with period \(2\pi\) and \(\cos^2(\theta)\) oscillates between 0 and 1. Given that \(E \subset (0, 2\pi)\) and the Lebesgue measure of \(E\) is 2, we can use the fact that the cosine function is periodic and the properties of the Lebesgue integral to analyze the behavior of the integral \(\int_E \cos^2(nx + \xi_n) \, dx\). ### Step-by-Step Solution: 1. **Periodicity and Uniform Convergence:** - The function \(\cos^2(nx + \xi_n)\) is periodic with period \(\frac{2\pi}{n}\). - As \(n \to \infty\), the period \(\frac{2\pi}{n}\) becomes very small, causing the function \(\cos^2(nx + \xi_n)\) to oscillate rapidly. 2. **Average Value Property:** - Over any interval of length \(2\pi\), the average value of \(\cos^2(\theta)\) is \(\frac{1}{2}\). - Since \(E\) has a measure of 2, which is a finite measure, the integral \(\int_E \cos^2(nx + \xi_n) \, dx\) can be expected to converge to the average value of \(\cos^2(\theta)\) multiplied by the measure of \(E\). 3. **Limit Calculation:** - By the Law of Large Numbers and the ergodic theorem, the integral \(\int_E \cos^2(nx + \xi_n) \, dx\) will converge to the expected value of \(\cos^2(\theta)\) as \(n \to \infty\). - The expected value of \(\cos^2(\theta)\) is \(\frac{1}{2}\). 4. **Final Result:** - Therefore, the limit of the integral as \(n \to \infty\) is: \[ \lim_{n \to \infty} \int_E \cos^2(nx + \xi_n) \, dx = \text{measure of } E \times \frac{1}{2} = 2 \times \frac{1}{2} = 1. \] Thus, the numeric value of the limit is \(\boxed{1}\). |
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| gpto1 | To solve this problem, we need to compute the limit: \[ \lim_{n \to \infty} \int_E \cos^2(n x + \xi_n) \, dx \] Given that \( E \subset (0, 2\pi) \) is a measurable set with Lebesgue measure \( m(E) = 2 \), and \( \{ \xi_n \} \) is an arbitrary sequence of real numbers. **Step 1: Simplify the integrand** First, recall the trigonometric identity: \[ \cos^2 \theta = \frac{1 + \cos 2\theta}{2} \] Using this identity, we can rewrite the integrand: \[ \cos^2(n x + \xi_n) = \frac{1 + \cos(2 n x + 2 \xi_n)}{2} \] **Step 2: Split the integral** Now, the integral becomes: \[ \int_E \cos^2(n x + \xi_n) \, dx = \frac{1}{2} \int_E 1 \, dx + \frac{1}{2} \int_E \cos(2 n x + 2 \xi_n) \, dx \] Simplify the first term: \[ \frac{1}{2} \int_E 1 \, dx = \frac{1}{2} \cdot m(E) = \frac{1}{2} \cdot 2 = 1 \] **Step 3: Analyze the second integral** We need to evaluate the limit of the second term: \[ \lim_{n \to \infty} \frac{1}{2} \int_E \cos(2 n x + 2 \xi_n) \, dx \] As \( n \to \infty \), the function \( \cos(2 n x + 2 \xi_n) \) oscillates increasingly rapidly. Over a set of finite measure, the averages of these oscillations tend to zero unless there is synchronization or specific structures in \( E \). Given that \( \xi_n \) is arbitrary and \( E \) is any measurable set of finite measure \( 2 \), the rapid oscillations of the cosine function will average out over \( E \). Therefore, \[ \lim_{n \to \infty} \int_E \cos(2 n x + 2 \xi_n) \, dx = 0 \] **Step 4: Combine the results** Combining the first term (which equals 1) and the limit of the second term (which is 0): \[ \lim_{n \to \infty} \int_E \cos^2(n x + \xi_n) \, dx = 1 + 0 = 1 \] **Final Answer:** 1 |
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| deepseekv3 | We are tasked with finding the limit: \[ \lim_{n \to \infty} \int_E \cos^2(n x + \xi_n) \, dx \] where \( E \subset (0, 2\pi) \) is a measurable set with Lebesgue measure \( m(E) = 2 \), and \( \{\xi_n\} \) is an arbitrary sequence of real numbers. ### Step 1: Simplify the integrand The integrand is \( \cos^2(n x + \xi_n) \). Recall the trigonometric identity for \( \cos^2 \): \[ \cos^2 \theta = \frac{1 + \cos(2\theta)}{2}. \] Thus, \[ \cos^2(n x + \xi_n) = \frac{1 + \cos(2(n x + \xi_n))}{2}. \] ### Step 2: Rewrite the integral Substitute this into the integral: \[ \int_E \cos^2(n x + \xi_n) \, dx = \int_E \frac{1 + \cos(2(n x + \xi_n))}{2} \, dx. \] Split the integral into two parts: \[ \int_E \cos^2(n x + \xi_n) \, dx = \frac{1}{2} \int_E 1 \, dx + \frac{1}{2} \int_E \cos(2(n x + \xi_n)) \, dx. \] ### Step 3: Evaluate the integrals The first integral is straightforward: \[ \frac{1}{2} \int_E 1 \, dx = \frac{1}{2} \cdot m(E) = \frac{1}{2} \cdot 2 = 1. \] Now consider the second integral: \[ \frac{1}{2} \int_E \cos(2(n x + \xi_n)) \, dx. \] ### Step 4: Analyze the second integral The integrand \( \cos(2(n x + \xi_n)) \) oscillates rapidly as \( n \to \infty \). For any measurable set \( E \), the integral of a high-frequency oscillatory function tends to zero due to cancellation effects. This can be rigorously justified using the Riemann-Lebesgue lemma, which states that for a function \( f \in L^1(\mathbb{R}) \), \[ \lim_{n \to \infty} \int_\mathbb{R} f(x) \cos(n x) \, dx = 0. \] Here, the function \( \cos(2(n x + \xi_n)) \) is oscillating with frequency \( 2n \), and since \( E \) is a finite measure set, the integral approaches zero as \( n \to \infty \). Thus, \[ \lim_{n \to \infty} \frac{1}{2} \int_E \cos(2(n x + \xi_n)) \, dx = 0. \] ### Step 5: Final result Combining the results, we have: \[ \lim_{n \to \infty} \int_E \cos^2(n x + \xi_n) \, dx = 1 + 0 = 1. \] Thus, the limit is: \[ \boxed{1} \] |
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