Evaluation

Input	Output
What does the following represent: $\int_{-\infty}^\infty e^{-x^2} \,dx = \sqrt{\pi}$	OpenAI: GPT-4 Turbo PublicPass B The expression $\int_{-\infty}^\infty e^{-x^2} \,dx = \sqrt{\pi}$ represents the integral of the Gaussian function $e^{-x^2}$ over the entire real line. The result of this integral, $\sqrt{\pi}$ , emerges from a deep connection between the exponential function and the geometry of the real number line, specifically related to the normal distribution in statistics. This integral is vitally important in probability theory, statistics, and various fields of physics and engineering. Here's how this result can be understood and derived: Understanding $e^{-x^2}$ The function $e^{-x^2}$ rapidly approaches zero as $x$ moves away from zero (either positively or negatively), more quickly than any power of $1/x$ . It's a bell-shaped curve, symmetrical about the y-axis, known as the Gaussian function or bell curve. The value of $\int_{-\infty}^\infty e^{-x^2} \,dx = \sqrt{\pi}$ This integral, which evaluates the entire area under the Gaussian curve, is famously computed using a clever double integration technique involving polar coordinates: Square the Integral: Consider $I = \int_{-\infty}^\infty e^{-x^2} \,dx$ . If we square this integral, we obtain: [ I^2 = \left(\int_{-\infty}^\infty e^{-x^2} ,dx \right) \left(\int_{-\infty}^\infty e^{-y^2} ,dy \right) = \int_{-\infty}^\infty \int_{-\infty}^\infty e^{-x^2}e^{-y^2} ,dy,dx. ] We can factorize the exponent in the integrand: [ I^2 = \int_{-\infty}^\infty \int_{-\infty}^\infty e^{-(x^2 + y^2)} ,dx,dy. ] Change to Polar Coordinates: By converting the Cartesian coordinates $(x, y)$ to polar coordinates $(r, \theta)$ , where $x = r\cos\theta$ and $y = r\sin\theta$ , the expression $x^2 + y^2 = r^2$ simplifies the double integral, which becomes: [ I^2 = \int_0^{2\pi} \int_0^\infty e^{-r^2} r ,dr,d\theta. ] Through a substitution of $u = r^2$ , $du = 2r \,dr$ , this turns into: [ I^2 = \int_0^{2\pi} \int_0^\infty e^{-u} \frac{1}{2} ,du,d\theta = \frac{1}{2}\int_0^{2\pi} ,d\theta = \pi, ] where the inner integral solves to $1/2$ as integrating $e^{-u}$ from 0 to $\infty$ yields 1. Final Computation: Thus, $I^2 = \pi$ , and taking the square root on both sides gives $I = \sqrt{\pi}$ . This computation not only has theoretical importance but also applications in fields where normal distributions or related notions of smoothing and decay are utilized.

Input

Output

What does the following represent: $\int_{-\infty}^\infty e^{-x^2} \,dx = \sqrt{\pi}$

PublicPass

The expression $\int_{-\infty}^\infty e^{-x^2} \,dx = \sqrt{\pi}$ represents the integral of the Gaussian function $e^{-x^2}$ over the entire real line. The result of this integral, $\sqrt{\pi}$ , emerges from a deep connection between the exponential function and the geometry of the real number line, specifically related to the normal distribution in statistics. This integral is vitally important in probability theory, statistics, and various fields of physics and engineering. Here's how this result can be understood and derived:

Understanding $e^{-x^2}$

The function $e^{-x^2}$ rapidly approaches zero as $x$ moves away from zero (either positively or negatively), more quickly than any power of $1/x$ . It's a bell-shaped curve, symmetrical about the y-axis, known as the Gaussian function or bell curve.

The value of $\int_{-\infty}^\infty e^{-x^2} \,dx = \sqrt{\pi}$

This integral, which evaluates the entire area under the Gaussian curve, is famously computed using a clever double integration technique involving polar coordinates:

Square the Integral: Consider $I = \int_{-\infty}^\infty e^{-x^2} \,dx$ . If we square this integral, we obtain: [ I^2 = \left(\int_{-\infty}^\infty e^{-x^2} ,dx \right) \left(\int_{-\infty}^\infty e^{-y^2} ,dy \right) = \int_{-\infty}^\infty \int_{-\infty}^\infty e^{-x^2}e^{-y^2} ,dy,dx. ] We can factorize the exponent in the integrand: [ I^2 = \int_{-\infty}^\infty \int_{-\infty}^\infty e^{-(x^2 + y^2)} ,dx,dy. ]
Change to Polar Coordinates: By converting the Cartesian coordinates $(x, y)$ to polar coordinates $(r, \theta)$ , where $x = r\cos\theta$ and $y = r\sin\theta$ , the expression $x^2 + y^2 = r^2$ simplifies the double integral, which becomes: [ I^2 = \int_0^{2\pi} \int_0^\infty e^{-r^2} r ,dr,d\theta. ] Through a substitution of $u = r^2$ , $du = 2r \,dr$ , this turns into: [ I^2 = \int_0^{2\pi} \int_0^\infty e^{-u} \frac{1}{2} ,du,d\theta = \frac{1}{2}\int_0^{2\pi} ,d\theta = \pi, ] where the inner integral solves to $1/2$ as integrating $e^{-u}$ from 0 to $\infty$ yields 1.
Final Computation: Thus, $I^2 = \pi$ , and taking the square root on both sides gives $I = \sqrt{\pi}$ .

This computation not only has theoretical importance but also applications in fields where normal distributions or related notions of smoothing and decay are utilized.

Evaluation ID: c3b885bf-1ee0-454b-9bb3-c39dcc245fdb
Created: April 23, 2024

Evaluation Rating

Binary Pass/Fail: Pass

Score: 100

Additional Feedback: The submission provides a comprehensive and accurate explanation of the Gaussian integral and its significance. It goes into detail about the exponential function, the geometry behind the integral, the double integration technique involving polar coordinates, and the final computation. The connection to probability theory, statistics, and various fields of physics and engineering is highlighted effectively.

Factual Consistency: B: the submitted answer is a superset of the expert answer and is fully consistent with it

Raw Output

{
  "id": "gen-kZje5Kog2DC6wACwssgRdpTrV1Yu",
  "model": "openai/gpt-4-turbo",
  "object": "chat.completion",
  "created": 1713841692,
  "choices": [
    {
      "index": 0,
      "message": {
        "role": "assistant",
        "content": "The expression $\\int_{-\\infty}^\\infty e^{-x^2} \\,dx = \\sqrt{\\pi}$ represents the integral of the Gaussian function $e^{-x^2}$ over the entire real line. The result of this integral, $\\sqrt{\\pi}$, emerges from a deep connection between the exponential function and the geometry of the real number line, specifically related to the normal distribution in statistics. This integral is vitally important in probability theory, statistics, and various fields of physics and engineering. Here's how this result can be understood and derived:\n\n### Understanding $e^{-x^2}$\nThe function $e^{-x^2}$ rapidly approaches zero as $x$ moves away from zero (either positively or negatively), more quickly than any power of $1/x$. It's a bell-shaped curve, symmetrical about the y-axis, known as the Gaussian function or bell curve.\n\n### The value of $\\int_{-\\infty}^\\infty e^{-x^2} \\,dx = \\sqrt{\\pi}$\nThis integral, which evaluates the entire area under the Gaussian curve, is famously computed using a clever double integration technique involving polar coordinates:\n\n1. **Square the Integral:**\n   Consider $I = \\int_{-\\infty}^\\infty e^{-x^2} \\,dx$. If we square this integral, we obtain:\n   \\[\n   I^2 = \\left(\\int_{-\\infty}^\\infty e^{-x^2} \\,dx \\right) \\left(\\int_{-\\infty}^\\infty e^{-y^2} \\,dy \\right) = \\int_{-\\infty}^\\infty \\int_{-\\infty}^\\infty e^{-x^2}e^{-y^2} \\,dy\\,dx.\n   \\]\n   We can factorize the exponent in the integrand:\n   \\[\n   I^2 = \\int_{-\\infty}^\\infty \\int_{-\\infty}^\\infty e^{-(x^2 + y^2)} \\,dx\\,dy.\n   \\]\n\n2. **Change to Polar Coordinates:**\n   By converting the Cartesian coordinates $(x, y)$ to polar coordinates $(r, \\theta)$, where $x = r\\cos\\theta$ and $y = r\\sin\\theta$, the expression $x^2 + y^2 = r^2$ simplifies the double integral, which becomes:\n   \\[\n   I^2 = \\int_0^{2\\pi} \\int_0^\\infty e^{-r^2} r \\,dr\\,d\\theta.\n   \\]\n   Through a substitution of $u = r^2$, $du = 2r \\,dr$, this turns into:\n   \\[\n   I^2 = \\int_0^{2\\pi} \\int_0^\\infty e^{-u} \\frac{1}{2} \\,du\\,d\\theta = \\frac{1}{2}\\int_0^{2\\pi} \\,d\\theta = \\pi,\n   \\]\n   where the inner integral solves to $1/2$ as integrating $e^{-u}$ from 0 to $\\infty$ yields 1.\n\n3. **Final Computation:**\n   Thus, $I^2 = \\pi$, and taking the square root on both sides gives $I = \\sqrt{\\pi}$.\n\nThis computation not only has theoretical importance but also applications in fields where normal distributions or related notions of smoothing and decay are utilized."
      },
      "finish_reason": "stop"
    }
  ],
  "usage": {
    "prompt_tokens": 38,
    "completion_tokens": 751,
    "total_tokens": 789,
    "total_cost": 0.02291
  }
}

Raw Rating Output

{
  "id": "gen-mqLaI37AV9FQSaikvyfcn1T8MRoW",
  "model": "openai/gpt-3.5-turbo-0125",
  "object": "chat.completion",
  "created": 1713841696,
  "choices": [
    {
      "index": 0,
      "message": {
        "role": "assistant",
        "content": null,
        "tool_calls": [
          {
            "id": "call_yM0A2ChvbOfjiwxcjLyuOXIR",
            "type": "function",
            "function": {
              "name": "get_eval_rating",
              "arguments": "{\"binary_pass_fail\":true,\"score_out_of_100\":100,\"additional_feedback\":\"The submission provides a comprehensive and accurate explanation of the Gaussian integral and its significance. It goes into detail about the exponential function, the geometry behind the integral, the double integration technique involving polar coordinates, and the final computation. The connection to probability theory, statistics, and various fields of physics and engineering is highlighted effectively.\",\"factual_consistency\":\"B\"}"
            }
          }
        ]
      },
      "logprobs": null,
      "finish_reason": "tool_calls"
    }
  ],
  "usage": {
    "prompt_tokens": 1136,
    "completion_tokens": 98,
    "total_tokens": 1234,
    "total_cost": 0.000715
  }
}

Understanding e−x2e^{-x^2}e−x2

The value of ∫−∞∞e−x2 dx=π\int_{-\infty}^\infty e^{-x^2} \,dx = \sqrt{\pi}∫−∞∞​e−x2dx=π​

Evaluation

Evaluation Rating

Understanding $e^{-x^2}$

The value of $\int_{-\infty}^\infty e^{-x^2} \,dx = \sqrt{\pi}$