Question

Evaluate \(\int_{-1}^{1} P_{0}(x) P_{2}(x) d x\) to show that these functions are orthogonal on (-1,1).

Short answer

The integral evaluates to 0, confirming orthogonality.

Step-by-step solution

Understand the polynomials involved

The polynomials involved are Legendre polynomials. For this problem, we are dealing with the 0th-degree polynomial, \( P_0(x) \), and the 2nd-degree polynomial, \( P_2(x) \). Their definitions are: \[P_0(x) = 1\] and \[P_2(x) = \frac{1}{2}(3x^2 - 1)\]

Set up the integral

The integral we need to evaluate is: \[\int_{-1}^{1} P_{0}(x) P_{2}(x) \, d x\] Substitute the expressions for \( P_0(x) \) and \( P_2(x) \) into the integral: \[\int_{-1}^{1} 1 \cdot \frac{1}{2}(3x^2 - 1) \, d x\]

Simplify the integrand

Simplify the integrand: \[\int_{-1}^{1} \frac{1}{2}(3x^2 - 1) \, d x\] This simplifies to: \[\frac{1}{2}\int_{-1}^{1} (3x^2 - 1) \, d x\]

Split the integral

Split the integral into two parts: \[\frac{1}{2} \left( \int_{-1}^{1} 3x^2 \, d x - \int_{-1}^{1} 1 \, d x \right)\]

Evaluate each part of the integral

Evaluate \(\int_{-1}^{1} 3x^2 \, d x\) and \(\int_{-1}^{1} 1 \, d x\) separately: \[\int_{-1}^{1} 3x^2 \, d x = 3 \left. \frac{x^3}{3} \right|_{-1}^{1} = 3 \left( \frac{1^3}{3} - \frac{(-1)^3}{3} \right) = 3 \left(\frac{1}{3} - \frac{-1}{3}\right) = 3 \left( \frac{1 + 1}{3} \right) = 3 \cdot \frac{2}{3} = 2\] \[\int_{-1}^{1} 1 \, d x = \left. x \right|_{-1}^{1} = 1 - (-1) = 1 + 1 = 2\]

Combine results

Now combine the results: \[\frac{1}{2} (2 - 2) = \frac{1}{2} \cdot 0 = 0\]

Conclusion about orthogonality

Since the integral evaluates to 0, this confirms that \( P_0(x) \) and \( P_2(x) \) are orthogonal over the interval \((-1, 1)\).

Key concepts

Orthogonal Functions

Orthogonal functions are functions that, when multiplied together and integrated over a specific interval, have a result of zero. This property is essential in many areas of mathematics and engineering because it helps in simplifying complex problems. For example:

• Signal processing: Orthogonal functions can represent signals without overlap.
• Data compression: Orthogonality aids in efficient data representation.

To determine if two functions are orthogonal, we evaluate the integral of their product over a given interval. If the result is zero, the functions are orthogonal over that interval.

Legendre Polynomials

Legendre Polynomials are a set of orthogonal polynomials that arise in solving certain types of differential equations. They are especially useful in physics and engineering. The Legendre polynomials, denoted as \( P_n(x) \), follow a specific recurrence relation and are defined on the interval \((-1, 1)\). Some key points about Legendre polynomials:

• \( P_0(x) = 1 \)
• \( P_1(x) = x \)
• \( P_2(x) = \frac{1}{2}(3x^2 - 1) \)

These polynomials are very useful in expansions and approximating functions within the given interval.

Integral Evaluation

Integral evaluation involves finding the value of an integral, which is essentially the area under a curve. In our case, we evaluated: \[ \int_{-1}^{1} P_{0}(x) P_{2}(x) dx = \int_{-1}^{1} 1 \cdot \frac{1}{2}(3x^2 - 1) \, dx \] This requires understanding how to set up the integral, simplify the integrand, and then compute the result. Breaking down the steps simplifies complex integrals, making them manageable:

• Identify the functions involved.
• Substitute the expressions into the integral.
• Simplify the integrand.
• Split into simpler integrals if necessary and then solve.

Polynomial Orthogonality

Polynomial orthogonality means that the polynomials, when multiplied together and integrated over a specific interval, yield zero. This property allows us to work with a set of functions that are mutually exclusive, simplifying mathematical problems.

• For instance, in our exercise, \( P_0(x) \) and \( P_2(x) \) are orthogonal because their product integrated over \(-1, 1\) equals zero.

The integral \[ \int_{-1}^{1} P_{0}(x) P_{2}(x) dx \] was calculated step by step, finally showing that the integral equals zero, proving their orthogonality.

Definite Integrals in Mathematics

Definite integrals are integrals evaluated over a specific interval, providing a numerical result. For example:

• Determine the area under a curve between two points.
• Used in physics for concepts like work and energy.

In our exercise, the definite integral \(\int_{-1}^{1} P_{0}(x) P_{2}(x) \, dx \) was evaluated. Steps like splitting the integral and evaluating each part individually play a crucial role. Definite integrals often depend on fundamental theorem of calculus, which ties together differentiation and integration.