Problem 6
Question
Assuming that the operations of summation and integration can be interchanged, show that if $$ f=\sum \alpha_{n} \phi_{n} \quad \text { and } \quad g=\sum \beta_{n} \phi_{n} \text {, } $$ then for normalized eigenfunctions $$ \int_{a}^{b} f g \sigma d x=\sum_{n=1}^{\infty} \alpha_{n} \beta_{n} . $$ a generalization of Parseval's equality.
Step-by-Step Solution
Verified Answer
The generalization of Parseval's equality holds assuming the interchangeability of summation and integration operations, resulting in \(\int_{a}^{b} f g \sigma d x=\sum_{n=1}^{\infty} \alpha_{n} \beta_{n}\)
1Step 1: Write the integral
Write the integral that needs to be simplified. Here it is: \[\int_{a}^{b} f g \sigma d x\]where\(f=\sum \alpha_{n} \phi_{n}\) and \(g=\sum \beta_{n} \phi_{n}\)
2Step 2: Replace f and g in the integral
Replace \(f\) and \(g\) in the integral with the sums they are defined by.The integral now becomes:\[\int_{a}^{b} \left(\sum_{n}\alpha_{n} \phi_{n}\right) \left(\sum_{k}\beta_{k} \phi_{k}\right) \sigma d x\]
3Step 3: Break down the integral using linearity of integration
Since multiplication is commutative, and considering linearity of integration, this allows switching the order of summation and integration. Leading to:\[\sum_{n}\sum_{k} \alpha_{n} \beta_{k} \int_{a}^{b} \phi_{n}\phi_{k} \sigma d x\]
4Step 4: Use the normalization condition
Considering the eigenfunctions are normalized, the integral of the product of two different ones across the whole space is zero, and the integral of the square of an eigenfunction is one. Therefore, this simplifies to: \[\sum_{n}\sum_{k} \alpha_{n} \beta_{k} \delta_{nk} = \sum_{n=1}^{\infty} \alpha_{n} \beta_{n} \]where \(\delta_{nk}\) is the Kronecker delta, equal to 1 if n=k and zero otherwise.
Key Concepts
Normalized EigenfunctionsLinearity of IntegrationKronecker DeltaSummation and Integration
Normalized Eigenfunctions
Normalized eigenfunctions are a fundamental concept when solving differential equations in physics and mathematics. They serve as the building blocks that simplify complex functions into simpler components. Essentially, an eigenfunction is labeled 'normalized' when its integral over a specific range, often across its entire domain, is equal to 1.
Here's how they work:
Here's how they work:
- An eigenfunction is a function that returns a constant (eigenvalue) when multiplied and integrated with its corresponding linear operator.
- "Normalized" means the square of the eigenfunction, integrated over the whole space (or specific domain), equals 1.
This condition is portrayed mathematically as: ewline\[\int_{a}^{b} \phi_n^2(x) \, dx = 1\]
Linearity of Integration
Linearity of integration refers to two major properties that make the integral operation predictable when dealing with sums or constant multiples:
- If two functions are summed together, the integral of their sum is the sum of their integrals. Expressed as: ewline\[ \int (u(x) + v(x)) \ dx = \int u(x) \ dx + \int v(x) \ dx \]
- A constant multiple of a function can be pulled out of the integral, which means the multiplication operation can commute with integration:ewline\[ \int c \cdot u(x) \ dx = c \cdot \int u(x) \ dx \]
Kronecker Delta
The Kronecker delta, \(\delta_{nk}\), is a useful tool in many mathematical operations, especially where neat simplification is desired. It acts as an indicator function, defined as:
Its relevance in the exercise lies in its use when integrating products of normalized eigenfunctions. When two different eigenfunctions are integrated together, \(\delta_{nk}\) helps quickly neutralize terms where indices aren’t equal, effectively simplifying the entire expression to just the diagonal elements of the series. This eliminates unnecessary complexity in summations and direct calculations.
- \(\delta_{nk} = 1\) if \(n = k\)
- \(\delta_{nk} = 0\) if \(n eq k\)
Its relevance in the exercise lies in its use when integrating products of normalized eigenfunctions. When two different eigenfunctions are integrated together, \(\delta_{nk}\) helps quickly neutralize terms where indices aren’t equal, effectively simplifying the entire expression to just the diagonal elements of the series. This eliminates unnecessary complexity in summations and direct calculations.
Summation and Integration
Summation and integration are central to mathematical analysis and have unique properties, especially when used together. In this exercise, the critical process is recognizing when you can swap summation and integration.
When the series involved converges, meaning that there is a well-behaved (bounded) function, the order of operations can be switched out easily during computation. Here’s why this matters:
When the series involved converges, meaning that there is a well-behaved (bounded) function, the order of operations can be switched out easily during computation. Here’s why this matters:
- Swapping can transform complex equations into more straightforward forms, making analysis and computation straightforward and less error-prone.
- For example, in integrals like \[\int \sum_{n} a_n f_n(x) \ dx\], summation and integration can be swapped to \[\sum_{n} \int a_n f_n(x) \ dx\].
This provides a significant advantage when dealing with infinite series or when each integrated term has similar behavior or structure.
Other exercises in this chapter
Problem 5
Consider $$ \rho \frac{\partial^{2} u}{\partial t^{2}}=T_{0} \frac{\partial^{2} u}{\partial x^{2}}+\alpha u, $$ where \(\rho(x)>0, \alpha(x)
View solution Problem 5
For the Sturm-Liouville eigenvalue problem, \(\frac{d^{2} \phi}{d x^{2}}+\lambda \phi=0 \quad\) with \(\quad \frac{d \phi}{d x}(0)=0 \quad\) and \(\quad \frac{d
View solution Problem 6
Consider (with \(h>0\) ) $$ \begin{array}{rlr} \frac{\partial^{2} u}{\partial t^{2}}=c^{2} \frac{\partial^{2} u}{\partial x^{2}} & \\ \frac{\partial u}{\partial
View solution Problem 6
Prove that if \(x\) is a root of a sixth-order polynomial with real coefficients, then \(x\) is also a root.
View solution