Chapter 2: Problem 11
Let \(f\) be a \(2 \pi\)-periodic piecewise continuous function and $$ f(x) \sim \sum_{n=1}^{\infty}\left[a_{n} \cos n x+b_{n} \sin n x\right] $$ its Fourier series on \([-\pi, \pi]\). Set $$ g(x)=\int_{-\pi}^{x}[f(t)+f(\pi-t)] d t $$ and let $$ g(x) \sim \frac{A_{0}}{2}+\sum_{n=1}^{\infty}\left[A_{n} \cos n x+B_{n} \sin n x\right] $$ be the Fourier series of \(g\) on \([-\pi, \pi]\). Express the \(A_{n}\) and \(B_{n}\) in terms of \(a_{n}\) and \(b_{n}\).
Short Answer
Step by step solution
Understand the Fourier Series of f(x)
Define the Function g(x)
Analyze the Symmetry of f(t) + f(-t)
Integrate to Find g(x)
Simplify the Integral Expression
Derive Fourier Coefficients A_n and B_n for g(x)
Express A_n and B_n
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with Vaia!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Piecewise Continuous Function
Fourier Coefficients
- The \(a_n\) coefficients correspond to the even, or cosine, parts of the function.
- The \(b_n\) coefficients correspond to the odd, or sine, parts of the function.
Symmetry
- **Even symmetry** where \(f(-x) = f(x)\). Functions with even symmetry have only cosine terms in their series since cosine is even.
- **Odd symmetry** where \(f(-x) = -f(x)\). These functions contain only sine terms due to the odd nature of sine.
- **Half-wave symmetry**, which affects coefficients based on periodicity adjustments.
Integral Transforms
- The integral of a sum of functions can be split into the sum of integrals, allowing for handling individual components one at a time.
- Often, transformations take place over a specific range \( [-\pi, \pi] \), which ties into the periodicity of the functions.