Find The Half-range Cosine And Sine Series Expansions Of The Function F(t) = 2t + 1 For 0 ≤ T ≤ 1.

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Introduction

In the realm of mathematical analysis, Fourier series stand as a cornerstone for representing periodic functions as an infinite sum of sines and cosines. These series are not merely theoretical constructs; they have profound applications in diverse fields such as signal processing, image analysis, acoustics, and quantum mechanics. This article delves into the fascinating world of half-range Fourier series expansions, focusing on a specific function, f(t) = 2t + 1, defined over the interval 0 ≤ t ≤ 1. We will explore how to derive both the half-range cosine series and the half-range sine series representations of this function. Understanding these expansions is crucial for solving various problems in engineering and physics, where functions are often defined over a limited interval and need to be extended periodically for further analysis. The beauty of Fourier series lies in their ability to decompose complex functions into simpler sinusoidal components, thereby simplifying analysis and enabling efficient computation. Furthermore, we will unravel the theoretical underpinnings of half-range expansions, emphasizing their practical significance in approximating functions and solving boundary value problems. Join us on this journey as we dissect the intricacies of Fourier analysis and illuminate the path to mastering half-range series expansions.

(a) Half-Range Cosine Series Expansion of f(t)

The primary goal here is to express the function f(t) = 2t + 1 as a sum of cosine terms. To achieve this, we first need to extend the function f(t) to an even function over the interval -1 ≤ t ≤ 1. This even extension, denoted as f_e(t), is defined as f_e(t) = f(t) for 0 ≤ t ≤ 1 and f_e(t) = f(-t) for -1 ≤ t < 0. Graphically, this even extension mirrors the original function across the y-axis, creating a symmetrical shape. The half-range cosine series expansion is then given by the formula:

f(t) = A_0/2 + Σ[A_n * cos(nπt/L)] (from n=1 to ∞)

where L is the length of the interval, which in our case is 1. The coefficients A_0 and A_n are calculated using the following integrals:

A_0 = (2/L) ∫[f(t) dt] (from 0 to L)
A_n = (2/L) ∫[f(t) * cos(nπt/L) dt] (from 0 to L)

For our specific function f(t) = 2t + 1 and L = 1, we first compute A_0:

A_0 = 2 ∫[(2t + 1) dt] (from 0 to 1)
    = 2 [t^2 + t]_0^1
    = 2 [(1^2 + 1) - (0^2 + 0)]
    = 2 * 2
    = 4

Next, we calculate A_n:

A_n = 2 ∫[(2t + 1) * cos(nπt) dt] (from 0 to 1)

This integral requires integration by parts. Let's break it down:

Let u = 2t + 1 and dv = cos(nπt) dt. Then, du = 2 dt and v = (1/(nπ))sin(nπt). Applying integration by parts:

∫[(2t + 1) * cos(nπt) dt] = (2t + 1) * (1/(nπ))sin(nπt) - ∫[(1/(nπ))sin(nπt) * 2 dt]
                             = (2t + 1) * (1/(nπ))sin(nπt) + (2/(n^2π^2))cos(nπt)

Now, evaluate the definite integral from 0 to 1:

∫[(2t + 1) * cos(nπt) dt] (from 0 to 1) = [(2(1) + 1) * (1/(nπ))sin(nπ(1)) + (2/(n^2π^2))cos(nπ(1))] - [(2(0) + 1) * (1/(nπ))sin(nπ(0)) + (2/(n^2π^2))cos(nπ(0))]
                                       = [3 * (1/(nπ))sin(nπ) + (2/(n^2π^2))cos(nπ)] - [1 * (1/(nπ))sin(0) + (2/(n^2π^2))cos(0)]
                                       = [0 + (2/(n^2π^2))cos(nπ)] - [0 + (2/(n^2π^2))]
                                       = (2/(n^2π^2))(cos(nπ) - 1)

Thus,

A_n = 2 * (2/(n^2π^2))(cos(nπ) - 1)
    = (4/(n^2π^2))(cos(nπ) - 1)

Since cos(nπ) = (-1)^n, we have:

A_n = (4/(n^2π^2))((-1)^n - 1)

Notice that when n is even, A_n = 0, and when n is odd, A_n = (-8/(n2π2)). Therefore, the half-range cosine series is:

f(t) = 4/2 + Σ[(-8/((2k-1)^2π^2)) * cos((2k-1)πt)] (from k=1 to ∞)
     = 2 - (8/π^2) Σ[cos((2k-1)πt) / (2k-1)^2] (from k=1 to ∞)

This series represents f(t) = 2t + 1 over the interval 0 ≤ t ≤ 1 using only cosine terms.

(b) Half-Range Sine Series Expansion of f(t)

Now, let's explore the half-range sine series representation of f(t) = 2t + 1. This time, we need to extend f(t) as an odd function over the interval -1 ≤ t ≤ 1. This odd extension, denoted as f_o(t), is defined as f_o(t) = f(t) for 0 < t ≤ 1, f_o(t) = -f(-t) for -1 ≤ t < 0, and f_o(0) = 0. The odd extension reflects the original function across both the x-axis and the y-axis, resulting in a function that is symmetric about the origin. The half-range sine series expansion is given by:

f(t) = Σ[B_n * sin(nπt/L)] (from n=1 to ∞)

where L = 1 in our case. The coefficients B_n are calculated using the formula:

B_n = (2/L) ∫[f(t) * sin(nπt/L) dt] (from 0 to L)

Substituting f(t) = 2t + 1 and L = 1, we get:

B_n = 2 ∫[(2t + 1) * sin(nπt) dt] (from 0 to 1)

Again, we employ integration by parts. Let u = 2t + 1 and dv = sin(nπt) dt. Then, du = 2 dt and v = (-1/(nπ))cos(nπt). Applying integration by parts:

∫[(2t + 1) * sin(nπt) dt] = (2t + 1) * (-1/(nπ))cos(nπt) - ∫[(-1/(nπ))cos(nπt) * 2 dt]
                             = -(2t + 1) * (1/(nπ))cos(nπt) + (2/(nπ))∫[cos(nπt) dt]
                             = -(2t + 1) * (1/(nπ))cos(nπt) + (2/(nπ)) * (1/(nπ))sin(nπt)
                             = -(2t + 1) * (1/(nπ))cos(nπt) + (2/(n^2π^2))sin(nπt)

Evaluate the definite integral from 0 to 1:

∫[(2t + 1) * sin(nπt) dt] (from 0 to 1) = [-(2(1) + 1) * (1/(nπ))cos(nπ(1)) + (2/(n^2π^2))sin(nπ(1))] - [-(2(0) + 1) * (1/(nπ))cos(nπ(0)) + (2/(n^2π^2))sin(nπ(0))]
                                       = [-3 * (1/(nπ))cos(nπ) + 0] - [-1 * (1/(nπ))cos(0) + 0]
                                       = (-3/(nπ))cos(nπ) + (1/(nπ))

Thus,

B_n = 2 [(-3/(nπ))cos(nπ) + (1/(nπ))]
    = (2/(nπ))(1 - 3cos(nπ))
    = (2/(nπ))(1 - 3(-1)^n)

Therefore, the half-range sine series expansion is:

f(t) = Σ[(2/(nπ))(1 - 3(-1)^n) * sin(nπt)] (from n=1 to ∞)

This series represents f(t) = 2t + 1 over the interval 0 < t < 1 using only sine terms. This representation is particularly useful in problems where boundary conditions require the function to be zero at the endpoints of the interval.

Conclusion

In conclusion, we have successfully derived both the half-range cosine and sine series expansions for the function f(t) = 2t + 1 over the interval 0 ≤ t ≤ 1. The cosine series provides an even extension of the function, while the sine series provides an odd extension. These expansions are crucial tools in various fields, allowing us to represent functions in terms of trigonometric series and solve complex problems in engineering, physics, and applied mathematics. The process involves extending the function appropriately, calculating the Fourier coefficients, and constructing the series. Understanding these concepts provides a solid foundation for further exploration in Fourier analysis and its diverse applications.