Math Problem Statement
Choose any function f (x) that is continuous in the interval [−1, 1], and express it as f (x) = N ′P i=0 α∗ i Ti(x), where N ′P i=0 ui = 1 2u0 + u1 + · · ·+uN . Find the coefficients α∗ i using the orthogonality condition proved in (1).
Solution
To solve this problem, we will express the continuous function defined on the interval in terms of Chebyshev polynomials of the first kind, , and then find the coefficients using the orthogonality condition of Chebyshev polynomials.
Step 1: Chebyshev Polynomials and Series Expansion
The Chebyshev polynomials of the first kind, , are a set of orthogonal polynomials on the interval with the weight function .
Any continuous function on the interval can be expanded as a series of Chebyshev polynomials: where are the expansion coefficients to be determined.
Step 2: Orthogonality of Chebyshev Polynomials
The Chebyshev polynomials satisfy the following orthogonality condition over the interval :
\begin{cases} 0 & \text{if } i \neq j \\ \frac{\pi}{2} & \text{if } i = j \neq 0 \\ \pi & \text{if } i = j = 0 \end{cases}$$ This orthogonality condition allows us to extract the coefficients $$\alpha_i^*$$ by multiplying both sides of the expansion by $$T_j(x)$$ and integrating. ### Step 3: Finding the Coefficients $$\alpha_i^*$$ To find the coefficients $$\alpha_i^*$$, multiply both sides of the expansion by $$\frac{T_j(x)}{\sqrt{1 - x^2}}$$ and integrate over $$[-1, 1]$$: $$\int_{-1}^{1} \frac{f(x) T_j(x)}{\sqrt{1 - x^2}} dx = \sum_{i=0}^{N} \alpha_i^* \int_{-1}^{1} \frac{T_i(x) T_j(x)}{\sqrt{1 - x^2}} dx$$ By the orthogonality condition, the right-hand side simplifies to: $$\int_{-1}^{1} \frac{f(x) T_j(x)}{\sqrt{1 - x^2}} dx = \alpha_j^* \begin{cases} \pi & \text{if } j = 0 \\ \frac{\pi}{2} & \text{if } j \neq 0 \end{cases}$$ Thus, the coefficients $$\alpha_j^*$$ are given by: $$\alpha_0^* = \frac{1}{\pi} \int_{-1}^{1} \frac{f(x) T_0(x)}{\sqrt{1 - x^2}} dx$$ $$\alpha_j^* = \frac{2}{\pi} \int_{-1}^{1} \frac{f(x) T_j(x)}{\sqrt{1 - x^2}} dx \quad \text{for} \, j \geq 1$$ ### Step 4: Conclusion The function $$f(x)$$ can be expanded as: $$f(x) = \sum_{i=0}^{N} \alpha_i^* T_i(x)$$ where the coefficients $$\alpha_i^*$$ are determined by the integrals: $$\alpha_0^* = \frac{1}{\pi} \int_{-1}^{1} \frac{f(x)}{\sqrt{1 - x^2}} dx$$ $$\alpha_j^* = \frac{2}{\pi} \int_{-1}^{1} \frac{f(x) T_j(x)}{\sqrt{1 - x^2}} dx \quad \text{for} \, j \geq 1$$ Let me know if you'd like any details or further clarification! Here are some related questions to explore: 1. What are the properties of Chebyshev polynomials of the first kind? 2. How can we use Chebyshev polynomials in numerical interpolation? 3. What is the difference between Chebyshev polynomials of the first kind and second kind? 4. How does the orthogonality condition simplify finding the coefficients in other polynomial expansions? 5. What is the role of the weight function $$\frac{1}{\sqrt{1 - x^2}}$$ in this expansion? **Tip**: Orthogonal polynomials like Chebyshev polynomials are very useful in approximation theory and numerical analysis because they minimize the error in polynomial interpolation.Ask a new question for Free
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Math Problem Analysis
Mathematical Concepts
Orthogonal Polynomials
Chebyshev Polynomials
Fourier-like Series Expansion
Formulas
f(x) = Σα∗i Ti(x)
Orthogonality condition: ∫_{-1}^{1} [Ti(x)Tj(x)] / √(1 - x^2) dx
α0∗ = (1/π) ∫_{-1}^{1} [f(x)] / √(1 - x^2) dx
αj∗ = (2/π) ∫_{-1}^{1} [f(x)Tj(x)] / √(1 - x^2) dx for j ≥ 1
Theorems
Orthogonality of Chebyshev Polynomials
Chebyshev Series Expansion
Suitable Grade Level
Undergraduate (Advanced Calculus or Numerical Methods)
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