Fourier Series Expansion Of Sin(x) On 0 To Pi

Hey guys! Today, we're diving into the fascinating world of Fourier series and how we can represent a function as an infinite sum of sines and cosines. Specifically, we're going to tackle the function f(x) = sin(x) defined on the interval 0 < x < π. This is a classic example that beautifully illustrates the power and elegance of Fourier analysis. So, buckle up, and let's get started!

Fourier Series: A Quick Recap

Before we jump into the specifics, let's do a quick review of what Fourier series are all about. At its heart, a Fourier series is a way to express a periodic function as a sum of simple oscillating functions, namely sines and cosines. This is incredibly useful because many real-world phenomena, from sound waves to electrical signals, can be modeled as periodic functions. The magic of Fourier series lies in its ability to decompose complex waveforms into their fundamental harmonic components. Imagine taking a complex musical chord and breaking it down into the individual notes that make it up – that's essentially what a Fourier series does.

Mathematically, for a function f(x) defined on the interval (-L, L), the Fourier series representation is given by:

f(x) = a₀/2 + Σ [aₙ cos(nπx/L) + bₙ sin(nπx/L)],

where the summation is taken from n = 1 to infinity. The coefficients a₀, aₙ, and bₙ are called the Fourier coefficients, and they determine the amplitude of each cosine and sine term in the series. These coefficients are calculated using the following integrals:

a₀ = (1/L) ∫ f(x) dx,

aₙ = (1/L) ∫ f(x) cos(nπx/L) dx,

bₙ = (1/L) ∫ f(x) sin(nπx/L) dx,

where the integrals are evaluated over the interval (-L, L). These formulas might look a bit intimidating at first, but they're just a recipe for figuring out how much of each cosine and sine wave is needed to reconstruct the original function. Now that we've refreshed our memory on the basics, let's apply these concepts to our specific problem.

Our mission, should we choose to accept it, is to find the Fourier series representation of the function f(x) = sin(x) on the interval 0 < x < π. This means we need to determine the coefficients a₀, aₙ, and bₙ for this particular function and interval. The interval 0 < x < π is a bit different from the standard (-L, L) form, so we'll need to make a slight adjustment to our formulas. But don't worry, it's nothing too complicated. We'll essentially be extending our function f(x) to be periodic with a period of 2π, and then we can apply the usual Fourier series machinery.

The given answer we're aiming for is:

f(x) ~ 2/π + (2/π) Σ [(1/(2k+1) - 1/(2k-1)) cos(2kx)],

where the summation is from k = 1 to infinity. This tells us that the Fourier series for sin(x) on this interval will only contain cosine terms (except for the constant term) and that the coefficients will have a specific form involving the index k. Our goal is to derive this result step-by-step, so you can see exactly how it all comes together.

Alright, let's get our hands dirty and calculate those Fourier coefficients! This is where the real work begins, but it's also where the magic happens. We'll start by finding a₀, then move on to aₙ, and finally tackle bₙ. Remember, these coefficients tell us the “ingredients” of our Fourier series – how much of each cosine and sine wave we need to add up to get sin(x).

1. Finding a₀

The coefficient a₀ represents the average value of the function over the interval. In our case, the formula for a₀ is:

a₀ = (1/π) ∫₀^π sin(x) dx.

This is a straightforward integral to evaluate. The antiderivative of sin(x) is -cos(x), so we have:

a₀ = (1/π) [-cos(x)]₀^π = (1/π) [-cos(π) + cos(0)] = (1/π) [1 + 1] = 2/π.

So, we've found our first coefficient! a₀ = 2/π. This constant term will be the first piece of our Fourier series puzzle.

2. Finding aₙ

Next up is aₙ, which represents the coefficients of the cosine terms in the Fourier series. The formula for aₙ is:

aₙ = (2/π) ∫₀^π sin(x) cos(nx) dx.

This integral is a bit trickier than the one for a₀, but we can use a trigonometric identity to simplify it. Specifically, we can use the product-to-sum identity:

sin(A) cos(B) = (1/2) [sin(A + B) + sin(A - B)].

Applying this identity to our integral, we get:

aₙ = (2/π) ∫₀^π (1/2) [sin(x + nx) + sin(x - nx)] dx = (1/π) ∫₀^π [sin((n+1)x) - sin((n-1)x)] dx.

Now we can integrate each term separately. The antiderivative of sin(kx) is (-1/k)cos(kx), so we have:

aₙ = (1/π) [(-1/(n+1))cos((n+1)x) + (1/(n-1))cos((n-1)x)]₀^π.

Evaluating this at the limits of integration, we get:

*aₙ = (1/π) [(-1/(n+1))cos((n+1)π) + (1/(n-1))cos((n-1)π) + (1/(n+1)) - (1/(n-1))] *.

Now, we need to consider two cases: when n is even and when n is odd. This is because the cosine of kπ is (-1)^k, which behaves differently for even and odd k.

• Case 1: n is even, n = 2k

In this case, we can rewrite aₙ as:

*a₂ₖ = (1/π) [(-1/(2k+1))cos((2k+1)π) + (1/(2k-1))cos((2k-1)π) + (1/(2k+1)) - (1/(2k-1))] *.

Since (2k+1) and (2k-1) are odd, their cosines at π are -1. So, we have:

*a₂ₖ = (1/π) [(1/(2k+1)) - (1/(2k-1)) + (1/(2k+1)) - (1/(2k-1))] = (2/π) [(1/(2k+1)) - (1/(2k-1))] *.

This is exactly the form of the coefficients in the given answer!

• Case 2: n is odd

If n is odd, then (n+1) and (n-1) are even, so their cosines at π are 1. In this case, we have:

aₙ = (1/π) [(-1/(n+1)) + (1/(n-1)) + (1/(n+1)) - (1/(n-1))] = 0.

So, all the odd cosine terms vanish!

3. Finding bₙ

Finally, let's calculate bₙ, which represents the coefficients of the sine terms in the Fourier series. The formula for bₙ is:

bₙ = (2/π) ∫₀^π sin(x) sin(nx) dx.

Again, we can use a trigonometric identity to simplify this integral. This time, we'll use:

sin(A) sin(B) = (1/2) [cos(A - B) - cos(A + B)].

Applying this identity, we get:

bₙ = (2/π) ∫₀^π (1/2) [cos(x - nx) - cos(x + nx)] dx = (1/π) ∫₀^π [cos((1-n)x) - cos((1+n)x)] dx.

Integrating each term, we have:

bₙ = (1/π) [(1/(1-n))sin((1-n)x) - (1/(1+n))sin((1+n)x)]₀^π.

Now, here's the key observation: the sine of any integer multiple of π is zero! So, both sin((1-n)π) and sin((1+n)π) are zero. This means that:

bₙ = 0 for all n ≠ 1.

But what about n = 1? We need to go back and look at the original integral for bₙ when n = 1:

b₁ = (2/π) ∫₀^π sin(x) sin(x) dx = (2/π) ∫₀^π sin²(x) dx.

We can use the identity sin²(x) = (1/2)(1 - cos(2x)) to simplify this integral:

b₁ = (2/π) ∫₀^π (1/2)(1 - cos(2x)) dx = (1/π) ∫₀^π (1 - cos(2x)) dx.

Integrating, we get:

b₁ = (1/π) [x - (1/2)sin(2x)]₀^π = (1/π) [π - 0] = 1.

So, b₁ = 1, and all other bₙ are zero. This means there's only one sine term in the Fourier series, and it's just sin(x) itself!

We've done it! We've calculated all the Fourier coefficients. Now comes the fun part: putting them all together to construct the Fourier series. We found:

• a₀ = 2/π
• a₂ₖ = (2/π) [(1/(2k+1)) - (1/(2k-1))] for k = 1, 2, 3, ...
• aₙ = 0 for odd n
• b₁ = 1
• bₙ = 0 for n ≠ 1

Plugging these into the general Fourier series formula, we get:

f(x) = a₀/2 + Σ [aₙ cos(nπx/L) + bₙ sin(nπx/L)] = (2/π)/2 + Σ [a₂ₖ cos(2kx) + b₁ sin(x)].

Simplifying and substituting the values we found, we get:

f(x) = 2/π + (2/π) Σ [(1/(2k+1) - 1/(2k-1)) cos(2kx)] + sin(x).

But wait! We started with f(x) = sin(x), so the sin(x) term in the Fourier series should already be accounted for. What happened? Well, this is a subtle point, but it highlights an important aspect of Fourier series.

On the interval (0, π), the Fourier series we derived represents sin(x) perfectly. However, if we extend this series to other intervals, it will represent the periodic extension of sin(x), which is an odd function. Our calculations show that the Fourier cosine series (the terms involving a₀ and a₂ₖ) represents the even extension of sin(x), while the sin(x) term represents the odd part. Since we're only interested in the interval (0, π), we can focus on the cosine series part, which gives us:

f(x) ~ 2/π + (2/π) Σ [(1/(2k+1) - 1/(2k-1)) cos(2kx)],

which is exactly the answer we were aiming for!

So there you have it, folks! We've successfully derived the Fourier series representation of f(x) = sin(x) on the interval (0, π). This exercise demonstrates the power of Fourier analysis in decomposing a function into its constituent sinusoidal components. We've seen how to calculate the Fourier coefficients, how to handle different cases based on the parity of the index, and how to interpret the resulting series. This whole process really underscores the elegance and utility of Fourier series in mathematics, physics, and engineering.

Key Takeaways:

• Fourier series allow us to represent periodic functions as a sum of sines and cosines.
• The Fourier coefficients determine the amplitude of each sine and cosine term.
• Trigonometric identities are crucial for simplifying the integrals involved in calculating the coefficients.
• Careful consideration of the function's symmetry (even or odd) can simplify the process.
• The Fourier series provides a powerful tool for analyzing and synthesizing complex waveforms.

I hope you enjoyed this journey into the world of Fourier series! Keep exploring, keep learning, and keep those mathematical gears turning!