| 模型 | 模型答案 | 正确答案 |
|---|---|---|
| 金声玉亮2.0 | To find the Lagrange error bound for the fifth-degree Taylor polynomial approximation of \(f(x) = \sin(x)\) centered at \(x=0\), we need to first understand the formula for the Lagrange error bound. Given a function \(f(x)\) and its Taylor polynomial \(P_n(x)\) of degree \(n\) centered at \(a\), the Lagrange error bound \(E_n(x)\) for approximating \(f(x)\) by \(P_n(x)\) is given by: \[ E_n(x) = \frac{M}{(n+1)!} |x-a|^{n+1} \] where \(M\) is an upper bound for the absolute value of the \((n+1)\)-th derivative of \(f\) on the interval between \(a\) and \(x\). For the given problem, we have \(f(x) = \sin(x)\), \(P_5(x)\) is the fifth-degree Taylor polynomial centered at \(x=0\), and we are interested in the Lagrange error bound for approximating \(\sin(1)\) using \(P_5(1)\). First, we need to find the \((n+1)\)-th derivative of \(f(x) = \sin(x)\). Since \(n=5\), we are interested in the 6th derivative of \(\sin(x)\), which is: \[ f^{(6)}(x) = \sin(x) \] The maximum absolute value of \(\sin(x)\) on any interval is 1. Hence, \(M = 1\). Now, we can find the Lagrange error bound for approximating \(\sin(1)\) using \(P_5(1)\): \[ E_5(1) = \frac{M}{(5+1)!} |1-0|^{5+1} = \frac{1}{6!} \cdot 1^6 = \frac{1}{720} \] Therefore, the Lagrange error of the polynomial approximation to \(\sin(1)\) is \(\frac{1}{720}\). **Answer: \(\boxed{\frac{1}{720}}\)** |
0.000198 |
| deepseekr1 | To find the Lagrange error of the fifth-degree Taylor polynomial approximation for \( f(x) = \sin(x) \) centered at \( x = 0 \) when approximating \( \sin(1) \), we start by recalling the Taylor polynomial and the Lagrange remainder formula. The fifth-degree Taylor polynomial for \( \sin(x) \) centered at 0 is: \[ P_5(x) = x - \frac{x^3}{6} + \frac{x^5}{120} \] The Lagrange remainder term \( R_n(x) \) for a Taylor polynomial of degree \( n \) is given by: \[ R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!} (x - a)^{n+1} \] where \( c \) is some number between \( a \) and \( x \). For the fifth-degree polynomial (\( n = 5 \)) centered at \( a = 0 \) and evaluated at \( x = 1 \), the remainder term \( R_5(1) \) is: \[ R_5(1) = \frac{f^{(6)}(c)}{6!} \cdot 1^6 \] The sixth derivative of \( \sin(x) \) is \( f^{(6)}(x) = -\sin(x) \). Therefore, substituting into the remainder formula: \[ R_5(1) = \frac{-\sin(c)}{6!} \cdot 1^6 = \frac{-\sin(c)}{720} \] Thus, the Lagrange error of the polynomial approximation to \( \sin(1) \) is: \[ \boxed{-\dfrac{\sin(c)}{720}} \] for some \( c \) between 0 and 1. |
0.000198 |
| gpto1 | To find the Lagrange error of the fifth-degree Taylor polynomial approximation \( P_5(x) \) for \( \sin(x) \) at \( x = 1 \), we'll use the Lagrange form of the remainder for Taylor series. **Step 1: Write down the Taylor polynomial and the remainder term** The fifth-degree Taylor (Maclaurin) polynomial for \( \sin(x) \) centered at \( x = 0 \) is: \[ P_5(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} = x - \frac{x^3}{6} + \frac{x^5}{120} \] The Lagrange remainder (error) \( R_n(x) \) after \( n \) terms is given by: \[ R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x - a)^{n+1} \] where \( c \) is some value between \( a \) and \( x \). Here, \( n = 5 \), \( a = 0 \), and \( x = 1 \). **Step 2: Compute the \((n+1)\)-th derivative** We need the sixth derivative of \( \sin(x) \) because \( n + 1 = 6 \): - \( f(x) = \sin(x) \) - \( f^{(1)}(x) = \cos(x) \) - \( f^{(2)}(x) = -\sin(x) \) - \( f^{(3)}(x) = -\cos(x) \) - \( f^{(4)}(x) = \sin(x) \) - \( f^{(5)}(x) = \cos(x) \) - \( f^{(6)}(x) = -\sin(x) \) So, \( f^{(6)}(c) = -\sin(c) \). **Step 3: Write down the remainder term** Plugging into the remainder formula: \[ R_5(1) = \frac{f^{(6)}(c)}{6!}(1 - 0)^6 = \frac{-\sin(c)}{720} \] **Step 4: Find the maximum value of the remainder** Since \( c \) is between \( 0 \) and \( 1 \), and \( |\sin(c)| \leq \sin(1) \) in this interval (because \( \sin(x) \) increases from \( 0 \) to \( \sin(1) \) for \( x \) from \( 0 \) to \( 1 \)). So the maximum possible absolute value of the remainder is: \[ |R_5(1)| = \left| \frac{-\sin(c)}{720} \right| = \frac{|\sin(c)|}{720} \leq \frac{\sin(1)}{720} \] **Step 5: Compute the numerical value** Calculate \( \sin(1) \): \[ \sin(1) \approx 0.8415 \] Therefore: \[ |R_5(1)| \leq \frac{0.8415}{720} \approx 0.001169 \] **Answer:** An explicit expression: |Error| ≤ sin 1 divided by 720; that is, \[ |R_5(1)| \leq \frac{\sin(1)}{720} \approx 0.001169 \] |
0.000198 |
| GPT-4o | The Taylor series expansion for \( f(x) = \sin(x) \) centered at \( x = 0 \) is given by: \[ \sin(x) = \sum_{n=0}^\infty \frac{(-1)^n x^{2n+1}}{(2n+1)!} \] This means that the Taylor polynomial of degree 5, \( P_5(x) \), is: \[ P_5(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} \] To find the Lagrange error term when approximating \(\sin(1)\) using \( P_5(x) \), we use the Lagrange form of the remainder \( R_n(x) \): \[ R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!} x^{n+1} \] for some \( c \) in the interval between 0 and \( x \). For \( P_5(x) \), \( n = 5 \), so we need \( R_5(x) \): \[ R_5(x) = \frac{f^{(6)}(c)}{6!} x^6 \] Since \( f(x) = \sin(x) \), we have: \[ f^{(6)}(x) = \sin(x) \] Thus, \[ R_5(x) = \frac{\sin(c)}{6!} x^6 \] When \( x = 1 \), this becomes: \[ R_5(1) = \frac{\sin(c)}{6!} \cdot 1^6 = \frac{\sin(c)}{720} \] Now, we need to determine the maximum possible value of \(\sin(c)\) for \( c \in [0, 1] \). Since the sine function is bounded between -1 and 1 for all real numbers: \[ |\sin(c)| \leq 1 \text{ for all } c \] Therefore, \[ |R_5(1)| \leq \frac{1}{720} \] So, the Lagrange error of the polynomial approximation \( P_5(x) \) for \(\sin(1)\) is: \[ \left| R_5(1) \right| \leq \frac{1}{720} \] This is the maximum possible error when using the fifth-degree Taylor polynomial to approximate \(\sin(1)\). |
0.000198 |
| llama3.1-70B-Instruct | To find the Lagrange error of the polynomial approximation to sin(1), we need to calculate the remainder term of the Taylor series expansion of sin(x) centered at x=0. The Taylor series expansion of sin(x) centered at x=0 is: sin(x) = x - x^3/3! + x^5/5! - x^7/7! + ... The fifth-degree Taylor polynomial approximation to sin(x) is: P_5(x) = x - x^3/3! + x^5/5! To find the Lagrange error, we need to calculate the remainder term R_5(x) = sin(x) - P_5(x). Using the Taylor series expansion, we can write: R_5(x) = -x^7/7! + x^9/9! - ... Now, we need to find the value of R_5(1). Substituting x=1 into the expression for R_5(x), we get: R_5(1) = -1/7! + 1/9! - ... Using a calculator, we can approximate the value of R_5(1) to be: R_5(1) ≈ -0.0001984126984 Therefore, the Lagrange error of the polynomial approximation to sin(1) is approximately -0.0001984126984. |
0.000198 |
| llama3.1-405B-Instruct | The fifth-degree Taylor polynomial approximation for f(x) = sin(x) centered at x = 0 is given by: P5(x) = x - x^3/3! + x^5/5! To find the Lagrange error of the polynomial approximation to sin(1), we need to find the remainder term R5(x) = f(x) - P5(x). Using the Lagrange error bound formula, we have: |R5(x)| ≤ M \* |x-0|^6 / 6! where M is the maximum value of the sixth derivative of f(x) = sin(x) on the interval [0, 1]. The sixth derivative of f(x) = sin(x) is f^(6)(x) = -sin(x), which has a maximum value of 1 on the interval [0, 1]. Therefore, M = 1. Plugging in x = 1, we get: |R5(1)| ≤ 1 \* |1-0|^6 / 6! |R5(1)| ≤ 1/6! |R5(1)| ≤ 1/720 So, the Lagrange error of the polynomial approximation to sin(1) is less than or equal to 1/720. |
0.000198 |
| DeepSeek-Coder-V2 | To solve the problem of finding the Lagrange error of the fifth-degree Taylor polynomial approximation for \( f(x) = \sin(x) \) centered at \( x = 0 \) when approximating \( \sin(1) \), we follow these steps: 1. **Identify the Taylor polynomial and its degree**: The fifth-degree Taylor polynomial \( P_5(x) \) for \( \sin(x) \) centered at \( x = 0 \) is given by: \[ P_5(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} \] 2. **Determine the Lagrange error bound formula**: The Lagrange error bound for the Taylor polynomial approximation is given by: \[ E_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!} (x - a)^{n+1} \] where \( \xi \) is some number between \( a \) and \( x \), and \( f^{(n+1)}(\xi) \) is the \((n+1)\)-th derivative of \( f \) evaluated at \( \xi \). 3. **Identify the function and its derivatives**: For \( f(x) = \sin(x) \), the derivatives are: \[ f'(x) = \cos(x), \quad f''(x) = -\sin(x), \quad f'''(x) = -\cos(x), \quad f^{(4)}(x) = \sin(x), \quad f^{(5)}(x) = \cos(x), \quad f^{(6)}(x) = -\sin(x) \] Since we are considering the fifth-degree Taylor polynomial, we need the sixth derivative: \[ f^{(6)}(x) = -\sin(x) \] 4. **Evaluate the error bound**: The error bound for the fifth-degree Taylor polynomial approximation is: \[ E_5(x) = \frac{f^{(6)}(\xi)}{6!} (x - 0)^6 = \frac{-\sin(\xi)}{720} x^6 \] Since we are approximating \( \sin(1) \), we set \( x = 1 \): \[ E_5(1) = \frac{-\sin(\xi)}{720} \cdot 1^6 = \frac{-\sin(\xi)}{720} \] 5. **Determine the maximum error**: The maximum value of \( \sin(\xi) \) for \( \xi \) in the interval \([0, 1]\) is 1 (since \( \sin(x) \) reaches its maximum value of 1 at \( x = \frac{\pi}{2} \), but we are only considering up to \( x = 1 \)). Therefore, the maximum error is: \[ E_5(1) = \frac{-1}{720} = -\frac{1}{720} \] Since we are interested in the absolute value of the error: \[ |E_5(1)| = \frac{1}{720} \] Thus, the Lagrange error of the polynomial approximation to \( \sin(1) \) is: \[ \boxed{\frac{1}{720}} \] |
0.000198 |
| Qwen2-72B-Instruct | To find the Lagrange error bound for the fifth-degree Taylor polynomial approximation of \(f(x) = \sin(x)\) centered at \(x=0\), we need to first understand the formula for the Lagrange error bound. Given a function \(f(x)\) and its Taylor polynomial \(P_n(x)\) of degree \(n\) centered at \(a\), the Lagrange error bound \(E_n(x)\) for approximating \(f(x)\) by \(P_n(x)\) is given by: \[ E_n(x) = \frac{M}{(n+1)!} |x-a|^{n+1} \] where \(M\) is an upper bound for the absolute value of the \((n+1)\)-th derivative of \(f\) on the interval between \(a\) and \(x\). For the given problem, we have \(f(x) = \sin(x)\), \(P_5(x)\) is the fifth-degree Taylor polynomial centered at \(x=0\), and we are interested in the Lagrange error bound for approximating \(\sin(1)\) using \(P_5(1)\). First, we need to find the \((n+1)\)-th derivative of \(f(x) = \sin(x)\). Since \(n=5\), we are interested in the 6th derivative of \(\sin(x)\), which is: \[ f^{(6)}(x) = \sin(x) \] The maximum absolute value of \(\sin(x)\) on any interval is 1. Hence, \(M = 1\). Now, we can find the Lagrange error bound for approximating \(\sin(1)\) using \(P_5(1)\): \[ E_5(1) = \frac{M}{(5+1)!} |1-0|^{5+1} = \frac{1}{6!} \cdot 1^6 = \frac{1}{720} \] Therefore, the Lagrange error of the polynomial approximation to \(\sin(1)\) is \(\frac{1}{720}\). **Answer: \(\boxed{\frac{1}{720}}\)** |
0.000198 |
| baichuan4 | To find the Lagrange error of the fifth-degree Taylor polynomial approximation for $f(x) = \sin(x)$, centered at $x=0$, we first need to determine the Taylor polynomial $P_5(x)$. The Taylor series for $f(x) = \sin(x)$ around $x=0$ is given by: $$ \sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + \cdots $$ Therefore, the fifth-degree Taylor polynomial $P_5(x)$ is: $$ P_5(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} $$ The Lagrange error formula for a Taylor polynomial of degree $n$ approximating a function $f(x)$ around a point $a$ is: $$ R_n(x) = \frac{f^{(n+1)}(z)}{(n+1)!}(x-a)^{n+1} $$ where $z$ is some number between $a$ and $x$. For our problem, $n=5$, $f(x) = \sin(x)$, $a=0$, and we want to approximate $f(1) = \sin(1)$. The $(n+1)$th derivative of $\sin(x)$ is $f^{(6)}(x) = -\sin(x)$. Substituting these values into the Lagrange error formula, we get: $$ R_5(1) = \frac{-\sin(z)}{6!}(1-0)^6 = \frac{-\sin(z)}{720} $$ Since we don't know the exact value of $z$, we can bound the error by using the maximum value of $|\sin(z)|$ on the interval $[0, 1]$, which is $1$. Therefore, the maximum possible error is: $$ |R_5(1)| \leq \frac{1}{720} $$ So, the Lagrange error of the polynomial approximation to $\sin(1)$ is at most $\frac{1}{720}$. |
0.000198 |
| doubao | The Taylor series for $\sin x$ is $x - \frac{x^3}{3!} + \frac{x^5}{5!} - \cdots$. So the fifth-degree Taylor polynomial $P_5(x)$ is $x - \frac{x^3}{3!} + \frac{x^5}{5!}$. The Lagrange error bound is given by $R_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!}(x-a)^{n+1}$, where $\xi$ is some point between $a$ and $x$. Here $n=5$, $a=0$, and $x=1$. The sixth derivative of $\sin x$ is $\pm \sin x$. So the absolute value of the sixth derivative is bounded by 1. The Lagrange error is then bounded by $\frac{1}{(6)!}(1-0)^{6} = \frac{1}{720}$. |
0.000198 |
| minimax | To find the Lagrange error bound for the Taylor polynomial approximation of \( f(x) = \sin(x) \) at \( x = 1 \), we need to use the remainder term of the Taylor series, which is given by the Lagrange form of the remainder: \[ R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x - a)^{(n+1)} \] where \( f^{(n+1)}(c) \) is the \( (n+1) \)-th derivative of \( f \) evaluated at some \( c \) in the interval between \( a \) and \( x \), \( (x - a)^{(n+1)} \) is the \( (n+1) \)-th power of \( (x - a) \), and \( n \) is the degree of the polynomial. For the fifth-degree Taylor polynomial centered at \( x = 0 \), we have \( n = 5 \), and \( a = 0 \). The function \( f(x) = \sin(x) \) has derivatives that cycle through \( \sin(x) \), \( \cos(x) \), \( -\sin(x) \), and \( -\cos(x) \). Since \( \sin(x) \) and \( \cos(x) \) are bounded by \( -1 \) and \( 1 \), the maximum value of the \( (n+1) \)-th derivative of \( \sin(x) \) is \( 1 \) for all \( n \). The remainder term for the fifth-degree Taylor polynomial is then: \[ R_5(1) = \frac{f^{(6)}(c)}{6!}(1 - 0)^6 \] Since \( f^{(6)}(x) = \sin(x) \) or \( f^{(6)}(x) = \cos(x) \), the maximum value of \( f^{(6)}(x) \) is \( 1 \) (since the maximum value of \( \sin(x) \) and \( \cos(x) \) is \( 1 \)). Thus, the maximum value of the remainder term is: \[ R_5(1) \leq \frac{1}{6!} \] \[ R_5(1) \leq \frac{1}{720} \] This is the Lagrange error bound for the fifth-degree Taylor polynomial approximation of \( \sin(x) \) at \( x = 1 \). It tells us that the actual error \( |f(1) - P_5(1)| \) is less than or equal to \( \frac{1}{720} \). |
0.000198 |
| kimi | The Taylor polynomial \( P_5(x) \) for \( f(x) = \sin(x) \) centered at \( x = 0 \) is given by: \[ P_5(x) = \sin(0) + \cos(0)x + \frac{1}{2!}\sin(0)x^2 - \frac{1}{3!}\cos(0)x^3 + \frac{1}{4!}\sin(0)x^4 - \frac{1}{5!}\cos(0)x^5 \] Since \( \sin(0) = 0 \) and \( \cos(0) = 1 \), the polynomial simplifies to: \[ P_5(x) = x - \frac{x^3}{6} + \frac{x^5}{120} \] The Lagrange error term for a Taylor polynomial is given by: \[ R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}x^{n+1} \] where \( c \) is some number between \( x \) and the center of the polynomial (in this case, 0), and \( n \) is the degree of the polynomial minus one (so for \( P_5(x) \), \( n = 4 \)). For \( f(x) = \sin(x) \), the sixth derivative is \( \sin(x) \) again, and its value at \( c \) is \( \sin(c) \). So the Lagrange error term for \( P_5(x) \) at \( x = 1 \) is: \[ R_4(1) = \frac{\sin(c)}{5!}(1)^5 \] \[ R_4(1) = \frac{\sin(c)}{120} \] Since \( c \) is between 0 and 1, we can't determine its exact value, but we know that \( \sin(c) \) will be less than or equal to 1. Therefore, the maximum error will be when \( \sin(c) = 1 \), which gives us: \[ R_4(1) \leq \frac{1}{120} \] So the Lagrange error of the polynomial approximation to \( \sin(1) \) is at most \( \frac{1}{120} \). |
0.000198 |
| wenxinyiyan | 我们要找的是sin(x)在x=1处的拉格朗日误差，使用五阶泰勒多项式近似。首先，我们知道sin(x)的泰勒展开式是： $sin(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} - \frac{x^7}{7!} + ...$ 五阶泰勒多项式$P_5(x)$是： $P_5(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!}$ 拉格朗日误差公式为： $E_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!} \cdot x^{n+1}$ 其中，$f^{(n+1)}(c)$是函数f在区间(0, x)内某一点c的(n+1)阶导数。在我们的情况下，n=5，x=1。 sin(x)的六阶导数是$-sin(x)$，所以$f^{(6)}(c) = -sin(c)$。 拉格朗日误差公式变为： $E_5(1) = \frac{-sin(c)}{6!} \cdot 1^6$ 由于c在(0, 1)之间，sin(c)的值在(0, 1)之间。因此，最大的可能误差发生在sin(c)=1时，此时误差为： $E_5(1) = \frac{-1}{6!} = -\frac{1}{720} \approx -0.0013888888888888888$ 所以，$P_5(x)$近似sin(1)的拉格朗日误差不超过$\frac{1}{720}$。 |
0.000198 |
| zhipu | To find the Lagrange error of the polynomial approximation to \(\sin(1)\) using the fifth-degree Taylor polynomial \(P_5(x)\) for \(f(x) = \sin(x)\), centered at \(x = 0\), we need to follow these steps: 1. **Determine the Taylor polynomial \(P_5(x)\):** The general form of a Taylor polynomial of degree \(n\) for a function \(f(x)\) centered at \(x = a\) is: \[ P_n(x) = f(a) + f'(a)(x - a) + \frac{f''(a)}{2!}(x - a)^2 + \cdots + \frac{f^{(n)}(a)}{n!}(x - a)^n \] For \(f(x) = \sin(x)\) and \(a = 0\), the derivatives of \(\sin(x)\) at \(x = 0\) are: \[ \begin{align*} f(x) &= \sin(x) \quad &f(0) &= 0 \\ f'(x) &= \cos(x) \quad &f'(0) &= 1 \\ f''(x) &= -\sin(x) \quad &f''(0) &= 0 \\ f'''(x) &= -\cos(x) \quad &f'''(0) &= -1 \\ f^{(4)}(x) &= \sin(x) \quad &f^{(4)}(0) &= 0 \\ f^{(5)}(x) &= \cos(x) \quad &f^{(5)}(0) &= 1 \\ \end{align*} \] Therefore, the fifth-degree Taylor polynomial \(P_5(x)\) for \(\sin(x)\) centered at \(x = 0\) is: \[ P_5(x) = 0 + 1 \cdot x + \frac{0}{2!} \cdot x^2 + \frac{-1}{3!} \cdot x^3 + \frac{0}{4!} \cdot x^4 + \frac{1}{5!} \cdot x^5 \] Simplifying, we get: \[ P_5(x) = x - \frac{x^3}{6} + \frac{x^5}{120} \] 2. **Determine the Lagrange error \(R_5(x)\):** The Lagrange error \(R_n(x)\) for the Taylor polynomial approximation is given by: \[ R_n(x) = \frac{f^{(n+1)}(\xi)}{(n+1)!} (x - a)^{n+1} \] where \(\xi\) is some value in the interval \([a, x]\). For \(n = 5\), \(f(x) = \sin(x)\), and \(a = 0\), we need \(f^{(6)}(x)\): \[ f^{(6)}(x) = -\sin(x) \] Thus, the Lagrange error \(R_5(x)\) is: \[ R_5(x) = \frac{f^{(6)}(\xi)}{6!} x^6 = \frac{-\sin(\xi)}{720} x^6 \] 3. **Evaluate the Lagrange error at \(x = 1\):** To find the Lagrange error at \(x = 1\), we substitute \(x = 1\) into the error formula: \[ R_5(1) = \frac{-\sin(\xi)}{720} \cdot 1^6 = \frac{-\sin(\xi)}{720} \] where \(\xi\) is some value in the interval \([0, 1]\). The maximum value of \(|\sin(\xi)|\) on the interval \([0, 1]\) is 1 (since \(\sin(\xi)\) ranges from 0 to \(\sin(1)\), and \(\sin(1) < 1\)). Therefore, the maximum possible Lagrange error is: \[ |R_5(1)| \leq \frac{1}{720} \] So, the Lagrange error of the polynomial approximation to \(\sin(1)\) using the fifth-degree Taylor polynomial \(P_5(x)\) is bounded by \(\frac{1}{720}\). The exact error depends on the specific value of \(\xi\) in the interval \([0, 1]\), but it will not exceed this bound. |
0.000198 |
| gpto1 | To find the Lagrange error of the fifth-degree Taylor polynomial approximation \( P_5(x) \) for \( \sin(x) \) at \( x = 1 \), we'll use the Lagrange form of the remainder for Taylor series. **Step 1: Write down the Taylor polynomial and the remainder term** The fifth-degree Taylor (Maclaurin) polynomial for \( \sin(x) \) centered at \( x = 0 \) is: \[ P_5(x) = x - \frac{x^3}{3!} + \frac{x^5}{5!} = x - \frac{x^3}{6} + \frac{x^5}{120} \] The Lagrange remainder (error) \( R_n(x) \) after \( n \) terms is given by: \[ R_n(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x - a)^{n+1} \] where \( c \) is some value between \( a \) and \( x \). Here, \( n = 5 \), \( a = 0 \), and \( x = 1 \). **Step 2: Compute the \((n+1)\)-th derivative** We need the sixth derivative of \( \sin(x) \) because \( n + 1 = 6 \): - \( f(x) = \sin(x) \) - \( f^{(1)}(x) = \cos(x) \) - \( f^{(2)}(x) = -\sin(x) \) - \( f^{(3)}(x) = -\cos(x) \) - \( f^{(4)}(x) = \sin(x) \) - \( f^{(5)}(x) = \cos(x) \) - \( f^{(6)}(x) = -\sin(x) \) So, \( f^{(6)}(c) = -\sin(c) \). **Step 3: Write down the remainder term** Plugging into the remainder formula: \[ R_5(1) = \frac{f^{(6)}(c)}{6!}(1 - 0)^6 = \frac{-\sin(c)}{720} \] **Step 4: Find the maximum value of the remainder** Since \( c \) is between \( 0 \) and \( 1 \), and \( |\sin(c)| \leq \sin(1) \) in this interval (because \( \sin(x) \) increases from \( 0 \) to \( \sin(1) \) for \( x \) from \( 0 \) to \( 1 \)). So the maximum possible absolute value of the remainder is: \[ |R_5(1)| = \left| \frac{-\sin(c)}{720} \right| = \frac{|\sin(c)|}{720} \leq \frac{\sin(1)}{720} \] **Step 5: Compute the numerical value** Calculate \( \sin(1) \): \[ \sin(1) \approx 0.8415 \] Therefore: \[ |R_5(1)| \leq \frac{0.8415}{720} \approx 0.001169 \] **Answer:** An explicit expression: |Error| ≤ sin 1 divided by 720; that is, \[ |R_5(1)| \leq \frac{\sin(1)}{720} \approx 0.001169 \] |
0.000198 |
| deepseekv3 | server error: Error code: 400 - {'error': {'code': 'RequestTimeOut', 'param': None, 'message': 'Request timed out, please try again later.', 'type': 'RequestTimeOut'}, 'id': 'chatcmpl-fb0d596d-7222-9ffd-92a2-0ac32f54e4aa', 'request_id': 'fb0d596d-7222-9ffd-92a2-0ac32f54e4aa'} | 0.000198 |