12. Let xo < x < < x, and let f be continuously differentiable. Show that a f[xo, x₁, x₂] = f[xo, X₁, X₁, X₁, Xi+l..., Xn] axi
12. Let xo < x < < x, and let f be continuously differentiable. Show that a f[xo, x₁, x₂] = f[xo, X₁, X₁, X₁, Xi+l..., Xn] axi
Advanced Engineering Mathematics
10th Edition
ISBN:9780470458365
Author:Erwin Kreyszig
Publisher:Erwin Kreyszig
Chapter2: Second-order Linear Odes
Section: Chapter Questions
Problem 1RQ
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![**Problem 12:**
Given the set of ordered points \( x_0 < x_1 < \cdots < x_n \), and a function \( f \) that is continuously differentiable, demonstrate that:
\[
\frac{\partial}{\partial x_i} f[x_0, x_1, \ldots, x_n] = f[x_0, x_1, \ldots, x_i, x_i, x_{i+1}, \ldots, x_n]
\]
Explanation:
- \( x_0, x_1, \ldots, x_n \) represent ordered input values.
- The notation \( f[x_0, x_1, \ldots, x_n] \) is typically used for divided differences in numerical analysis.
- The derivative with respect to \( x_i \) can be expressed by repeating \( x_i \) in the divided difference set.
This problem involves concepts from calculus and numerical analysis, specifically focusing on divided differences and differentiability.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fad0d55fe-d83b-4711-86a1-cee8ecea510f%2F029aee4e-f627-4455-8e77-68da777572fa%2Fcj6r35s_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem 12:**
Given the set of ordered points \( x_0 < x_1 < \cdots < x_n \), and a function \( f \) that is continuously differentiable, demonstrate that:
\[
\frac{\partial}{\partial x_i} f[x_0, x_1, \ldots, x_n] = f[x_0, x_1, \ldots, x_i, x_i, x_{i+1}, \ldots, x_n]
\]
Explanation:
- \( x_0, x_1, \ldots, x_n \) represent ordered input values.
- The notation \( f[x_0, x_1, \ldots, x_n] \) is typically used for divided differences in numerical analysis.
- The derivative with respect to \( x_i \) can be expressed by repeating \( x_i \) in the divided difference set.
This problem involves concepts from calculus and numerical analysis, specifically focusing on divided differences and differentiability.
![Theorem 1 provides the following formulas:
\[ f[x_0, x_1] = \frac{f[x_1] - f[x_0]}{x_1 - x_0} \]
\[ f[x_0, x_1, x_2] = \frac{f[x_1, x_2] - f[x_0, x_1]}{x_2 - x_0} \]
[...]
In these formulas, \( x_0, x_1, x_2, \ldots \) can be interpreted as independent variables. This leads to equations such as:
\[ f[x_i, x_{i+1}, \ldots, x_{i+j}] = \frac{f[x_{i+1}, x_{i+2}, \ldots, x_{i+j}] - f[x_i, x_{i+1}, \ldots, x_{i+j-1}]}{x_{i+j} - x_i} \]
Equation (13) reflects the concept of divided differences, which is useful in polynomial interpolation. It expresses a recursive relationship between the divided differences of different orders.](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fad0d55fe-d83b-4711-86a1-cee8ecea510f%2F029aee4e-f627-4455-8e77-68da777572fa%2Fw0o2oe6_processed.png&w=3840&q=75)
Transcribed Image Text:Theorem 1 provides the following formulas:
\[ f[x_0, x_1] = \frac{f[x_1] - f[x_0]}{x_1 - x_0} \]
\[ f[x_0, x_1, x_2] = \frac{f[x_1, x_2] - f[x_0, x_1]}{x_2 - x_0} \]
[...]
In these formulas, \( x_0, x_1, x_2, \ldots \) can be interpreted as independent variables. This leads to equations such as:
\[ f[x_i, x_{i+1}, \ldots, x_{i+j}] = \frac{f[x_{i+1}, x_{i+2}, \ldots, x_{i+j}] - f[x_i, x_{i+1}, \ldots, x_{i+j-1}]}{x_{i+j} - x_i} \]
Equation (13) reflects the concept of divided differences, which is useful in polynomial interpolation. It expresses a recursive relationship between the divided differences of different orders.
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