Fourier Series Expansion A Comprehensive Guide With Examples
In mathematical analysis, Fourier series play a crucial role in representing periodic functions as an infinite sum of sines and cosines. This representation is invaluable in various fields, including physics, engineering, and signal processing, for analyzing and synthesizing complex waveforms. The process of expressing a function in terms of its Fourier series is known as Fourier analysis. This article delves into the expansion of functions into Fourier sine and cosine series over the interval (0, π). We will explore the underlying principles, methods, and examples to provide a comprehensive understanding of this essential topic. Understanding Fourier series expansion is not just a theoretical exercise; it's a powerful tool that allows us to decompose complex functions into simpler, more manageable components. By representing functions as a sum of sines and cosines, we can analyze their frequency content, filter out unwanted noise, and even predict their behavior over time. The ability to expand functions into Fourier series opens doors to solving a wide range of problems in science and engineering. The sine and cosine series, in particular, are fundamental in solving boundary value problems related to differential equations, such as those arising in heat transfer, wave propagation, and electrical circuits. Each term in the Fourier series corresponds to a specific frequency component of the function. The coefficients of these terms, known as Fourier coefficients, determine the amplitude and phase of each component. By examining the Fourier coefficients, we can gain insights into the function's spectral characteristics, such as its dominant frequencies and the distribution of energy across the frequency spectrum. This information is invaluable in various applications, from audio processing to image compression.
Fourier Sine Series Expansion
The Fourier sine series is a representation of a function as an infinite sum of sine functions. This expansion is particularly useful for functions defined on an interval (0, L) that satisfy certain boundary conditions. The general form of the Fourier sine series is:
f(x) = Σ[bₙ * sin(nπx/L)] (sum from n=1 to ∞)
where the coefficients bₙ are given by:
bₙ = (2/L) ∫[f(x) * sin(nπx/L) dx] (integral from 0 to L)
These coefficients represent the amplitude of each sine component in the series. The process of determining these coefficients is a key step in obtaining the Fourier sine series representation of a function. Let's explore how to expand specific functions into Fourier sine series over the interval (0, π). This involves calculating the sine coefficients and expressing the function as an infinite sum of sine terms. Consider the function sin⁷(x). To expand this into a Fourier sine series, we first need to calculate the coefficients bₙ. This involves evaluating the integral of sin⁷(x) multiplied by sin(nx) over the interval (0, π). This integral can be solved using trigonometric identities and integration techniques. The resulting coefficients will then be used to construct the Fourier sine series representation of sin⁷(x). Similarly, for the function x(π - x), we need to calculate the sine coefficients by integrating x(π - x) multiplied by sin(nx) over the interval (0, π). This integral can be solved using integration by parts. The resulting coefficients will then be used to construct the Fourier sine series representation of x(π - x). For the piecewise function f(x) defined as x for 0 < x < π/2 and π - x for π/2 < x < π, we need to calculate the sine coefficients by splitting the integral into two parts, one over (0, π/2) and the other over (π/2, π). This is because the function has different definitions over these intervals. The resulting coefficients will then be used to construct the Fourier sine series representation of f(x). Finally, for the function x sin(x), we need to calculate the sine coefficients by integrating x sin(x) multiplied by sin(nx) over the interval (0, π). This integral can be solved using trigonometric identities and integration techniques. The resulting coefficients will then be used to construct the Fourier sine series representation of x sin(x).
(a) sin⁷(x)
To expand sin⁷(x) into a Fourier sine series over (0, π), we need to determine the coefficients bₙ using the formula:
bₙ = (2/π) ∫[sin⁷(x) * sin(nx) dx] (integral from 0 to π)
This integral can be evaluated using trigonometric identities and integration by parts. Specifically, we can use the identity:
sin⁷(x) = (1/64) * (35sin(x) - 21sin(3x) + 7sin(5x) - sin(7x))
Substituting this into the integral, we get:
bₙ = (2/π) ∫[(1/64) * (35sin(x) - 21sin(3x) + 7sin(5x) - sin(7x)) * sin(nx) dx] (integral from 0 to π)
Now, we can use the orthogonality property of sine functions, which states that:
∫[sin(mx) * sin(nx) dx] (integral from 0 to π) = (π/2) if m = n, and 0 otherwise.
This simplifies the calculation of the coefficients. By applying this property, we can find the non-zero coefficients and construct the Fourier sine series for sin⁷(x). The resulting series will be a sum of sine terms with specific coefficients determined by the integral evaluation. This example demonstrates how trigonometric identities and the orthogonality property can be used to efficiently calculate the Fourier sine series coefficients for a given function. The ability to manipulate trigonometric expressions and apply integration techniques is crucial in obtaining the Fourier series representation. Understanding the properties of sine and cosine functions, such as their orthogonality, is essential for simplifying the calculations and arriving at the final series representation. The Fourier sine series expansion of sin⁷(x) highlights the power of Fourier analysis in decomposing a complex function into its fundamental sine components. This representation can be used to analyze the function's frequency content and to reconstruct the function from its constituent sine waves. The process of finding the Fourier sine series coefficients involves a combination of trigonometric manipulation, integration techniques, and the application of orthogonality properties. The result is a series representation that captures the essential features of the original function in terms of sine waves.
(b) x(π - x)
To expand x(π - x) into a Fourier sine series over (0, π), we need to calculate the coefficients bₙ using the formula:
bₙ = (2/π) ∫[x(π - x) * sin(nx) dx] (integral from 0 to π)
This integral can be evaluated using integration by parts. Let's break down the integration process step by step. First, we choose u = x(π - x) and dv = sin(nx) dx. Then, we find du = (π - 2x) dx and v = -cos(nx)/n. Applying integration by parts, we get:
∫[x(π - x) * sin(nx) dx] = -x(π - x)cos(nx)/n + ∫[(π - 2x)cos(nx)/n dx]
Now, we need to integrate the second term. We again use integration by parts, choosing u = (π - 2x) and dv = cos(nx)/n dx. Then, we find du = -2 dx and v = sin(nx)/n². Applying integration by parts again, we get:
∫[(π - 2x)cos(nx)/n dx] = (π - 2x)sin(nx)/n² + ∫[2sin(nx)/n² dx] = (π - 2x)sin(nx)/n² - 2cos(nx)/n³
Combining these results, we have:
∫[x(π - x) * sin(nx) dx] = -x(π - x)cos(nx)/n + (π - 2x)sin(nx)/n² - 2cos(nx)/n³
Now, we need to evaluate this expression at the limits of integration, 0 and π. After evaluating the limits and simplifying, we can obtain the coefficients bₙ. The resulting coefficients will then be used to construct the Fourier sine series representation of x(π - x). This example illustrates the application of integration by parts in calculating the Fourier sine series coefficients. The process involves multiple applications of integration by parts and careful evaluation of the limits of integration. The resulting series represents the function x(π - x) as a sum of sine terms with specific coefficients determined by the integral evaluation. The Fourier sine series expansion of x(π - x) demonstrates the power of Fourier analysis in representing a polynomial function using trigonometric functions. This representation can be used to analyze the function's behavior and to approximate the function using a finite number of terms in the series.
(c) f(x) = { x if 0 < x < π/2; π - x if π/2 < x < π }
To expand the piecewise function f(x) into a Fourier sine series over (0, π), we need to calculate the coefficients bₙ using the formula:
bₙ = (2/π) ∫[f(x) * sin(nx) dx] (integral from 0 to π)
Since f(x) is defined piecewise, we need to split the integral into two parts:
bₙ = (2/π) { ∫[x * sin(nx) dx] (integral from 0 to π/2) + ∫[(π - x) * sin(nx) dx] (integral from π/2 to π) }
Now, we need to evaluate each integral separately. Both integrals can be solved using integration by parts. For the first integral, ∫[x * sin(nx) dx] (integral from 0 to π/2), we choose u = x and dv = sin(nx) dx. Then, we find du = dx and v = -cos(nx)/n. Applying integration by parts, we get:
∫[x * sin(nx) dx] = -xcos(nx)/n + ∫[cos(nx)/n dx] = -xcos(nx)/n + sin(nx)/n²
Evaluating this from 0 to π/2, we get:
[- (π/2)cos(nπ/2)/n + sin(nπ/2)/n²] - [0]
For the second integral, ∫[(π - x) * sin(nx) dx] (integral from π/2 to π), we choose u = π - x and dv = sin(nx) dx. Then, we find du = -dx and v = -cos(nx)/n. Applying integration by parts, we get:
∫[(π - x) * sin(nx) dx] = -(π - x)cos(nx)/n - ∫[cos(nx)/n dx] = -(π - x)cos(nx)/n - sin(nx)/n²
Evaluating this from π/2 to π, we get:
[0] - [- (π/2)cos(nπ/2)/n - sin(nπ/2)/n²]
Adding the results of the two integrals and multiplying by (2/π), we can obtain the coefficients bₙ. The resulting coefficients will then be used to construct the Fourier sine series representation of f(x). This example demonstrates how to handle piecewise functions when calculating Fourier sine series coefficients. The process involves splitting the integral into parts corresponding to the different definitions of the function and applying integration by parts to each part. The resulting series represents the piecewise function as a sum of sine terms with specific coefficients determined by the integral evaluation. The Fourier sine series expansion of a piecewise function highlights the versatility of Fourier analysis in representing functions with discontinuities or changes in definition over different intervals. This representation can be used to analyze the function's behavior and to approximate the function using a finite number of terms in the series.
(d) x sin(x)
To expand x sin(x) into a Fourier sine series over (0, π), we need to calculate the coefficients bₙ using the formula:
bₙ = (2/π) ∫[x sin(x) * sin(nx) dx] (integral from 0 to π)
This integral can be evaluated using trigonometric identities and integration by parts. We can use the product-to-sum trigonometric identity:
sin(a)sin(b) = (1/2)[cos(a - b) - cos(a + b)]
Applying this identity, we get:
x sin(x)sin(nx) = (x/2)[cos((n - 1)x) - cos((n + 1)x)]
Substituting this into the integral, we have:
bₙ = (1/π) ∫[x[cos((n - 1)x) - cos((n + 1)x)] dx] (integral from 0 to π)
Now, we need to evaluate the integral. We can split it into two integrals:
bₙ = (1/π) { ∫[x cos((n - 1)x) dx] (integral from 0 to π) - ∫[x cos((n + 1)x) dx] (integral from 0 to π) }
Both integrals can be solved using integration by parts. For the first integral, ∫[x cos((n - 1)x) dx] (integral from 0 to π), we choose u = x and dv = cos((n - 1)x) dx. Then, we find du = dx and v = sin((n - 1)x)/(n - 1). Applying integration by parts, we get:
∫[x cos((n - 1)x) dx] = xsin((n - 1)x)/(n - 1) - ∫[sin((n - 1)x)/(n - 1) dx] = xsin((n - 1)x)/(n - 1) + cos((n - 1)x)/(n - 1)²
Evaluating this from 0 to π, we get:
[πsin((n - 1)π)/(n - 1) + cos((n - 1)π)/(n - 1)²] - [0 + 1/(n - 1)²]
For the second integral, ∫[x cos((n + 1)x) dx] (integral from 0 to π), we choose u = x and dv = cos((n + 1)x) dx. Then, we find du = dx and v = sin((n + 1)x)/(n + 1). Applying integration by parts, we get:
∫[x cos((n + 1)x) dx] = xsin((n + 1)x)/(n + 1) - ∫[sin((n + 1)x)/(n + 1) dx] = xsin((n + 1)x)/(n + 1) + cos((n + 1)x)/(n + 1)²
Evaluating this from 0 to π, we get:
[πsin((n + 1)π)/(n + 1) + cos((n + 1)π)/(n + 1)²] - [0 + 1/(n + 1)²]
Subtracting the results of the two integrals and multiplying by (1/π), we can obtain the coefficients bₙ. The resulting coefficients will then be used to construct the Fourier sine series representation of x sin(x). This example demonstrates the use of trigonometric identities and integration by parts in calculating the Fourier sine series coefficients. The process involves transforming the product of trigonometric functions into a sum or difference of trigonometric functions and applying integration by parts to evaluate the resulting integrals. The Fourier sine series expansion of x sin(x) highlights the power of Fourier analysis in representing a function that is a product of a polynomial and a trigonometric function. This representation can be used to analyze the function's behavior and to approximate the function using a finite number of terms in the series.
Fourier Cosine Series Expansion
The Fourier cosine series is a representation of a function as an infinite sum of cosine functions. This expansion is particularly useful for functions defined on an interval (0, L) that are even or have even extensions. The general form of the Fourier cosine series is:
f(x) = a₀/2 + Σ[aₙ * cos(nπx/L)] (sum from n=1 to ∞)
where the coefficients a₀ and aₙ are given by:
a₀ = (2/L) ∫[f(x) dx] (integral from 0 to L) aₙ = (2/L) ∫[f(x) * cos(nπx/L) dx] (integral from 0 to L) for n ≥ 1
These coefficients represent the amplitude of each cosine component in the series. The coefficient a₀ represents the DC component or the average value of the function over the interval (0, L). Let's explore how to expand specific functions into Fourier cosine series over the interval (0, π). This involves calculating the cosine coefficients and expressing the function as an infinite sum of cosine terms. Consider the function sin⁸(x). To expand this into a Fourier cosine series, we first need to calculate the coefficients a₀ and aₙ. This involves evaluating the integral of sin⁸(x) and sin⁸(x) multiplied by cos(nx) over the interval (0, π). These integrals can be solved using trigonometric identities and integration techniques. The resulting coefficients will then be used to construct the Fourier cosine series representation of sin⁸(x). Similarly, for the function x(π - x), we need to calculate the cosine coefficients by integrating x(π - x) and x(π - x) multiplied by cos(nx) over the interval (0, π). These integrals can be solved using integration by parts. The resulting coefficients will then be used to construct the Fourier cosine series representation of x(π - x). For the piecewise function f(x) (the same as in the sine series example), we need to calculate the cosine coefficients by splitting the integral into two parts, one over (0, π/2) and the other over (π/2, π). This is because the function has different definitions over these intervals. The resulting coefficients will then be used to construct the Fourier cosine series representation of f(x). The Fourier cosine series expansion is a powerful tool for representing even functions or functions with even extensions. This representation is particularly useful in applications where the function is symmetric about the y-axis or where the boundary conditions require a cosine series representation. The process of finding the Fourier cosine series coefficients involves evaluating integrals that may require trigonometric identities, integration by parts, or other integration techniques. The resulting series provides a representation of the function in terms of cosine waves, which can be used to analyze the function's frequency content and to reconstruct the function from its constituent cosine components.
(a) sin⁸(x)
To expand sin⁸(x) into a Fourier cosine series over (0, π), we need to determine the coefficients a₀ and aₙ using the formulas:
a₀ = (2/π) ∫[sin⁸(x) dx] (integral from 0 to π) aₙ = (2/π) ∫[sin⁸(x) * cos(nx) dx] (integral from 0 to π)
These integrals can be evaluated using trigonometric identities and integration techniques. Specifically, we can use the power-reducing formulas and the identity:
sin²(x) = (1 - cos(2x))/2
Applying this repeatedly, we can express sin⁸(x) in terms of cosines of multiple angles. Then, we can use the orthogonality property of cosine functions, which states that:
∫[cos(mx) * cos(nx) dx] (integral from 0 to π) = (π/2) if m = n ≠ 0, π if m = n = 0, and 0 otherwise.
This simplifies the calculation of the coefficients. By applying these techniques, we can find the non-zero coefficients and construct the Fourier cosine series for sin⁸(x). The resulting series will be a sum of cosine terms with specific coefficients determined by the integral evaluation. This example demonstrates how trigonometric identities and the orthogonality property can be used to efficiently calculate the Fourier cosine series coefficients for a given function. The ability to manipulate trigonometric expressions and apply integration techniques is crucial in obtaining the Fourier series representation. Understanding the properties of cosine functions, such as their orthogonality, is essential for simplifying the calculations and arriving at the final series representation. The Fourier cosine series expansion of sin⁸(x) highlights the power of Fourier analysis in decomposing a complex function into its fundamental cosine components. This representation can be used to analyze the function's frequency content and to reconstruct the function from its constituent cosine waves. The process of finding the Fourier cosine series coefficients involves a combination of trigonometric manipulation, integration techniques, and the application of orthogonality properties. The result is a series representation that captures the essential features of the original function in terms of cosine waves.
(b) x(π - x)
To expand x(π - x) into a Fourier cosine series over (0, π), we need to calculate the coefficients a₀ and aₙ using the formulas:
a₀ = (2/π) ∫[x(π - x) dx] (integral from 0 to π) aₙ = (2/π) ∫[x(π - x) * cos(nx) dx] (integral from 0 to π)
First, let's calculate a₀:
a₀ = (2/π) ∫[x(π - x) dx] = (2/π) ∫[πx - x² dx] (integral from 0 to π)
Integrating and evaluating at the limits, we get:
a₀ = (2/π) [πx²/2 - x³/3] (evaluated from 0 to π) = (2/π) [π³/2 - π³/3] = (2/π) [π³/6] = π²/3
Now, let's calculate aₙ. This integral can be evaluated using integration by parts. Let's break down the integration process step by step. First, we choose u = x(π - x) and dv = cos(nx) dx. Then, we find du = (π - 2x) dx and v = sin(nx)/n. Applying integration by parts, we get:
∫[x(π - x) * cos(nx) dx] = x(π - x)sin(nx)/n - ∫[(π - 2x)sin(nx)/n dx]
Now, we need to integrate the second term. We again use integration by parts, choosing u = (π - 2x) and dv = sin(nx)/n dx. Then, we find du = -2 dx and v = -cos(nx)/n². Applying integration by parts again, we get:
∫[(π - 2x)sin(nx)/n dx] = (π - 2x)(-cos(nx)/n²) - ∫[2cos(nx)/n² dx] = -(π - 2x)cos(nx)/n² - 2sin(nx)/n³
Combining these results, we have:
∫[x(π - x) * cos(nx) dx] = x(π - x)sin(nx)/n + (π - 2x)cos(nx)/n² + 2sin(nx)/n³
Now, we need to evaluate this expression at the limits of integration, 0 and π. After evaluating the limits and simplifying, we can obtain the coefficients aₙ. The resulting coefficients will then be used to construct the Fourier cosine series representation of x(π - x). This example illustrates the application of integration by parts in calculating the Fourier cosine series coefficients. The process involves multiple applications of integration by parts and careful evaluation of the limits of integration. The resulting series represents the function x(π - x) as a sum of cosine terms with specific coefficients determined by the integral evaluation. The Fourier cosine series expansion of x(π - x) demonstrates the power of Fourier analysis in representing a polynomial function using trigonometric functions. This representation can be used to analyze the function's behavior and to approximate the function using a finite number of terms in the series.
(c) f(x) =
To fully address the Fourier cosine series expansion for the piecewise function f(x), we need the complete definition of the function. Assuming f(x) is the same piecewise function defined earlier for the sine series:
f(x) = { x if 0 < x < π/2; π - x if π/2 < x < π }
We proceed to expand this piecewise function into a Fourier cosine series over (0, π). We need to calculate the coefficients a₀ and aₙ using the formulas:
a₀ = (2/π) ∫[f(x) dx] (integral from 0 to π) aₙ = (2/π) ∫[f(x) * cos(nx) dx] (integral from 0 to π)
Since f(x) is defined piecewise, we need to split the integrals into two parts. First, let's calculate a₀:
a₀ = (2/π) { ∫[x dx] (integral from 0 to π/2) + ∫[(π - x) dx] (integral from π/2 to π) }
Integrating and evaluating at the limits, we get:
a₀ = (2/π) { [x²/2] (evaluated from 0 to π/2) + [πx - x²/2] (evaluated from π/2 to π) } a₀ = (2/π) { (π²/8) + (π² - π²/2) - (π²/2 - π²/8) } = (2/π) { π²/8 + π²/2 - π²/2 + π²/8 } = (2/π) { π²/4 } = π/2
Now, let's calculate aₙ. We need to split the integral into two parts:
aₙ = (2/π) { ∫[x * cos(nx) dx] (integral from 0 to π/2) + ∫[(π - x) * cos(nx) dx] (integral from π/2 to π) }
Now, we need to evaluate each integral separately. Both integrals can be solved using integration by parts. For the first integral, ∫[x * cos(nx) dx] (integral from 0 to π/2), we choose u = x and dv = cos(nx) dx. Then, we find du = dx and v = sin(nx)/n. Applying integration by parts, we get:
∫[x * cos(nx) dx] = xsin(nx)/n - ∫[sin(nx)/n dx] = xsin(nx)/n + cos(nx)/n²
Evaluating this from 0 to π/2, we get:
[(π/2)sin(nπ/2)/n + cos(nπ/2)/n²] - [0 + 1/n²]
For the second integral, ∫[(π - x) * cos(nx) dx] (integral from π/2 to π), we choose u = π - x and dv = cos(nx) dx. Then, we find du = -dx and v = sin(nx)/n. Applying integration by parts, we get:
∫[(π - x) * cos(nx) dx] = (π - x)sin(nx)/n + ∫[sin(nx)/n dx] = (π - x)sin(nx)/n - cos(nx)/n²
Evaluating this from π/2 to π, we get:
[0 - cos(nπ)/n²] - [(π/2)sin(nπ/2)/n - cos(nπ/2)/n²]
Adding the results of the two integrals and multiplying by (2/π), we can obtain the coefficients aₙ. The resulting coefficients will then be used to construct the Fourier cosine series representation of f(x). This example demonstrates how to handle piecewise functions when calculating Fourier cosine series coefficients. The process involves splitting the integral into parts corresponding to the different definitions of the function and applying integration by parts to each part. The resulting series represents the piecewise function as a sum of cosine terms with specific coefficients determined by the integral evaluation. The Fourier cosine series expansion of a piecewise function highlights the versatility of Fourier analysis in representing functions with discontinuities or changes in definition over different intervals. This representation can be used to analyze the function's behavior and to approximate the function using a finite number of terms in the series.
In conclusion, the expansion of functions into Fourier sine and cosine series is a fundamental technique in mathematical analysis with wide-ranging applications. We've explored the methods for calculating Fourier coefficients and representing various functions, including trigonometric functions, polynomials, and piecewise functions, as infinite sums of sines and cosines. These series provide a powerful tool for analyzing the frequency content of functions and for approximating functions using a finite number of terms. Understanding the properties of Fourier series, such as the orthogonality of sine and cosine functions, and mastering integration techniques are crucial for successful Fourier analysis. The examples discussed in this article illustrate the application of these concepts to specific functions, providing a solid foundation for further exploration of Fourier series and their applications in various fields.
