Problem 3
Question
Show that the given functions are orthogonal on the indicated interval. $$ f_{1}(x)=e^{x}, f_{2}(x)=x e^{-x}-e^{-x} ; \quad[0,2] $$
Step-by-Step Solution
Verified Answer
The functions are orthogonal on the interval \([0, 2]\) as their inner product is zero.
1Step 1: Definition of Orthogonal Functions
Two functions \(f_1(x)\) and \(f_2(x)\) are orthogonal on the interval \([a, b]\) if their inner product is zero. The inner product is defined as: \(\int_{a}^{b} f_1(x) \cdot f_2(x) \, dx = 0\). In this problem, we need to show that \( \int_{0}^{2} e^{x} \cdot (x e^{-x} - e^{-x}) \, dx = 0 \).
2Step 2: Distribute the Functions for Integration
To simplify the integration process, we distribute \(f_1(x) = e^x\) across \(f_2(x)\) so that: \(f_1(x) \, \cdot \, f_2(x) = e^x \cdot x e^{-x} - e^x \cdot e^{-x}\). This simplifies to two separate integrals: \(x\) and \(1\), as \(e^x \cdot e^{-x} = 1\). Thus, we need to calculate \( \int_{0}^{2} x \, dx - \int_{0}^{2} 1 \, dx \).
3Step 3: Evaluate Each Integral
Evaluate the first integral: \( \int_{0}^{2} x \, dx \). The antiderivative of \(x\) is \( \frac{x^2}{2} \). Evaluating from 0 to 2 gives:\[ \left[ \frac{x^2}{2} \right]_{0}^{2} = \frac{2^2}{2} - \frac{0^2}{2} = 2. \]Now, evaluate the second integral: \( \int_{0}^{2} 1 \, dx = \left[ x \right]_{0}^{2} = 2 - 0 = 2. \).
4Step 4: Subtract the Results
Subtract the result of the second integral from the first:\[ 2 - 2 = 0. \]The result of this subtraction is 0, showing that the inner product of \(f_1(x)\) and \(f_2(x)\) over the interval \([0, 2]\) is zero.
Key Concepts
Inner ProductIntegrationAntiderivatives
Inner Product
The concept of the inner product is crucial in understanding when functions are orthogonal. In simple terms, the inner product for functions is like a dot product for vectors. It's a way of multiplying two functions together over a specific interval to see how they interact.
For functions, the inner product is often expressed as an integral. Specifically, the inner product of two functions \(f_1(x)\) and \(f_2(x)\) over an interval \([a, b]\) is defined by the integral \(\int_{a}^{b} f_1(x) \cdot f_2(x) \, dx\).
The beauty of the inner product is that it provides a measure of the 'angle' between two functions. If the integral results in zero, the functions are considered orthogonal, meaning they are 'perpendicular' in functional terms without overlap, just like perpendicular vectors in geometry.
For functions, the inner product is often expressed as an integral. Specifically, the inner product of two functions \(f_1(x)\) and \(f_2(x)\) over an interval \([a, b]\) is defined by the integral \(\int_{a}^{b} f_1(x) \cdot f_2(x) \, dx\).
The beauty of the inner product is that it provides a measure of the 'angle' between two functions. If the integral results in zero, the functions are considered orthogonal, meaning they are 'perpendicular' in functional terms without overlap, just like perpendicular vectors in geometry.
Integration
Integration is a fundamental concept in mathematics, particularly in calculus. It can be thought of as the reverse process of differentiation, focusing on finding the total area under a curve.
In the context of orthogonal functions, integration is used to evaluate the inner product to determine if it equals zero. This involves taking the integral of the product of two functions over a specified interval.
In the context of orthogonal functions, integration is used to evaluate the inner product to determine if it equals zero. This involves taking the integral of the product of two functions over a specified interval.
- The process involves simplifying the product of the two functions when possible, as seen in the exercise where the product \(e^x \cdot (x e^{-x} - e^{-x})\) was split into the simpler components \(x\) and \(1\).
- This simplification makes the integrals easier to handle.
- Careful calculation and step-by-step evaluation are crucial for accurate results.
Antiderivatives
Antiderivatives are the building blocks of integration. To integrate a function effectively, you need to find its antiderivative first.
An antiderivative is essentially a function that, when differentiated, gives you the original function. For example, the antiderivative of \(x\) is \(\frac{x^2}{2}\).
In the process of determining whether functions are orthogonal, calculating antiderivatives is vital as it allows us to evaluate definite integrals over a given interval. In the step-by-step solution:
An antiderivative is essentially a function that, when differentiated, gives you the original function. For example, the antiderivative of \(x\) is \(\frac{x^2}{2}\).
In the process of determining whether functions are orthogonal, calculating antiderivatives is vital as it allows us to evaluate definite integrals over a given interval. In the step-by-step solution:
- The antiderivative of \(x\) was calculated as \(\frac{x^2}{2}\).
- Evaluating this from 0 to 2 gives the area under the curve for \(x\), leading to the result of 2.
- Similarly, the antiderivative of the constant 1 is simply \(x\), evaluated from 0 to 2 again giving a result of 2.
Other exercises in this chapter
Problem 3
Determine whether the function is even, odd, or neither. $$ f(x)=x^{2}+x $$
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In Problems, find the Fourier series of \(f\) on the given interval. $$ f(x)=\left\\{\begin{array}{lr} 1, & -1
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Consider \(y^{\prime \prime}+\lambda y=0\) subject to the periodic boundary conditions \(y(-L)=y(L), y^{\prime}(-L)=y^{\prime}(L)\). Show that the eigenfunction
View solution Problem 4
Show that the given functions are orthogonal on the indicated interval. $$ f_{1}(x)=\cos x, f_{2}(x)=\sin ^{2} x ; \quad[0, \pi] $$
View solution