Problem 1 - Advanced Engineering Mathematics

Question

Show that the given functions are orthogonal on the indicated interval. $$ f_{1}(x)=x, f_{2}(x)=x^{2} ; \quad[-2,2] $$

Step-by-Step Solution

Answer

The functions are orthogonal on \([-2, 2]\).

Step 1: Understand Orthogonality

Two functions \( f_1(x) \) and \( f_2(x) \) are considered orthogonal on an interval \( [a, b] \) if the integral of their product over that interval is zero. This means we need to calculate \( \int_{-2}^{2} f_1(x) \, f_2(x) \, dx = 0 \)

Step 2: Set Up the Integral

We need to calculate the integral \( \int_{-2}^{2} x \cdot x^2 \, dx \). This simplifies to \( \int_{-2}^{2} x^3 \, dx \).

Step 3: Evaluate the Integral

To evaluate \( \int_{-2}^{2} x^3 \, dx \), we first find the antiderivative of \( x^3 \), which is \( \frac{x^4}{4} \). Then we calculate \[ \left[ \frac{x^4}{4} \right]_{-2}^{2} = \frac{2^4}{4} - \frac{(-2)^4}{4} \] \[ = \frac{16}{4} - \frac{16}{4} = 4 - 4 = 0 \]

Step 4: Conclude Orthogonality

Since \( \int_{-2}^{2} x^3 \, dx = 0 \), the functions \( f_1(x) = x \) and \( f_2(x) = x^2 \) are orthogonal on the interval \([-2, 2]\).

Key Concepts

Integral Calculus

Integral calculus is like the study of accumulation. It helps us find the total growth or reduction over an interval.
In mathematical terms, it allows us to find the area under a curve between two points on the x-axis.
Think of it as adding up a continuous series of infinitely small pieces.

• The integral of a function provides a way to calculate the accumulation of quantities.
• The notation for integrals uses the integral sign \( \int \), the function to be integrated, and the interval of integration.
• Finding the integral of a function involves finding its antiderivative.

In our example, we are integrating the product of two functions, \( x \) and \( x^2 \), over the interval from -2 to 2.
By integrating \( x \cdot x^2 \), which simplifies to \( x^3 \), we compute the area under the curve \( x^3 \) from \(-2\) to \(2\).
This leads to establishing whether the functions are orthogonal by checking if this area is zero.

Orthogonality

In mathematics, orthogonality is a concept often used to show independence between functions or vectors.
Treat orthogonal functions like mathematical siblings who don't interact. If their interaction measured over an interval results in zero, they are orthogonal.

• In the context of functions, it means their product integrated over a specified interval equals zero.
• This is a fundamental concept in linear algebra and calculus, particularly in function spaces.
• Orthogonal functions can simplify problems and equations, such as in Fourier series and transformations.

For the functions \( f_{1}(x) = x \) and \( f_{2}(x) = x^2 \), to discover if they are orthogonal, calculate \( \int_{-2}^{2} x \cdot x^2 \).
This simplifies to finding if \( \int_{-2}^{2} x^3 \) equals zero. A zero result confirms orthogonality, indicating these functions act independently over the given interval.

Function Evaluation

Function evaluation involves performing calculations or substitutions on functions to get meaningful results.
This is especially important when trying to prove properties like orthogonality.
In our task, evaluating functions means calculating the integral of \( f_{1}(x) \cdot f_{2}(x) \) over the interval from -2 to 2.

• The first step is combining and simplifying the functions before integration so that \( f_1(x) \cdot f_2(x) = x \cdot x^2 = x^3 \).
• Next, find the antiderivative of \( x^3 \), which is \( \frac{x^4}{4} \).
• Finally, use the limits of integration in the antiderivative, calculate \( \left[ \frac{x^4}{4} \right]_{-2}^{2} \), and conclude the integral equals zero.

Function evaluation this way helps in making logical conclusions, such as showing that \( f_1 \) and \( f_2 \) are indeed orthogonal on the specified interval.