Fourier Sine Series Of Sinx

Decomposing sin(x): Unveiling the Fourier Sine Series of sin(x)

The Fourier series, a powerful tool in mathematical analysis, allows us to represent periodic functions as an infinite sum of sine and cosine functions. This decomposition is crucial in various fields, from signal processing and image analysis to solving partial differential equations. While the concept might seem abstract, understanding how to find the Fourier series of even seemingly simple functions like sin(x) is fundamental to mastering this powerful technique. This article will delve into the specific case of the Fourier sine series of sin(x), addressing common challenges and providing a step-by-step approach. We will discover intriguing results that highlight important aspects of Fourier analysis.

1. Understanding the Fourier Sine Series

Before diving into the specifics of sin(x), let's briefly recall the definition of a Fourier sine series. A function f(x) defined on the interval [0, L] can be represented by its Fourier sine series as:

f(x) ≈ Σn=1∞ bn sin(nπx/L)

where the coefficients bn are given by:

bn = (2/L) ∫0L f(x) sin(nπx/L) dx

This series utilizes only sine functions, making it particularly useful for representing odd functions or functions defined on an interval where odd extension is appropriate. The interval [0, L] is crucial; the series is tailored to this specific interval.

2. The Case of f(x) = sin(x) on [0, π]

Let's consider the function f(x) = sin(x) on the interval [0, π]. This choice simplifies the calculation significantly. Notice that we've chosen the interval [0, π], which is a natural choice given the periodicity of sin(x). In this case, L = π. The formula for the coefficients bn becomes:

bn = (2/π) ∫0π sin(x) sin(nx) dx

This integral can be solved using various techniques, including integration by parts or trigonometric identities. A useful trigonometric identity is:

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

Applying this identity, we get:

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

Now, let's evaluate the integral for different values of 'n':

For n = 1: b1 = (1/π) ∫0π [1 - cos(2x)] dx = 1

For n ≠ 1: bn = (1/π) [sin((1-n)x)/(1-n) - sin((1+n)x)/(1+n)] evaluated from 0 to π. This evaluates to 0 because sin(kπ) = 0 for any integer k.

Therefore, the Fourier sine series of sin(x) on the interval [0, π] is simply:

sin(x) = sin(x)

3. Implications and Interpretations

The result that the Fourier sine series of sin(x) on [0, π] is simply sin(x) itself might seem trivial at first. However, it highlights several important points:

Orthogonality of sine functions: The orthogonality of sine functions over the interval [0, π] is crucial. The integral of the product of two different sine functions (with integer multiples of π/L as arguments) over this interval is zero. This property is the foundation of the Fourier series.

Choice of interval: The choice of the interval [0, π] was deliberate. If we had chosen a different interval, the result would have been different, potentially leading to a more complex series.

Function properties: The fact that we obtain a simple representation stems from the fact that sin(x) is already a sine function and is defined over a suitable interval for its fundamental period.

4. Addressing Common Challenges

A common challenge arises when students attempt to find the Fourier sine series of functions that aren't already expressed as sine functions. The key is to carefully evaluate the integral for the coefficients bn using appropriate integration techniques and trigonometric identities. Another challenge lies in choosing the appropriate interval [0, L]. This choice depends on the function's behaviour and the desired representation.

5. Summary

This exploration of the Fourier sine series of sin(x) on the interval [0, π] demonstrates the power and elegance of Fourier analysis. While the result might appear straightforward, it reveals the underlying principles of orthogonality and the importance of choosing the appropriate interval for the series. The simplicity of the result underscores the inherent compatibility between sin(x) and its sine series representation over this specific interval. This understanding provides a solid foundation for tackling more complex Fourier series problems involving other functions and intervals.

Frequently Asked Questions (FAQs)

1. Why use a Fourier sine series instead of a full Fourier series (with cosines)? A Fourier sine series is advantageous when dealing with odd functions or when boundary conditions dictate the use of only sine terms, particularly in solving partial differential equations.

2. What if the interval is different from [0, π]? If the interval is [0, L], the series becomes Σn=1∞ bn sin(nπx/L), and the coefficients bn will change accordingly. The integral limits and the argument within the sine functions will adjust based on the new interval.

3. Can we find the Fourier sine series of any function? Yes, in theory, any function that satisfies certain conditions (e.g., piecewise smoothness) can be represented by a Fourier sine series on a specified interval [0, L].

4. How does the convergence of the Fourier sine series behave? The convergence depends on the function's smoothness. For a continuous and differentiable function, the convergence is generally good. Discontinuities can lead to Gibbs phenomenon near the discontinuities.

5. What are some applications of the Fourier sine series? Applications are abundant: solving heat equations with specific boundary conditions, analyzing periodic signals with odd symmetry, image processing (particularly for odd-symmetric images), and modeling physical phenomena with odd-symmetric properties.

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Fourier Trigonometric Series | GeeksforGeeks 1 Aug 2024 · Fourier Trigonometric Series is a powerful tool for expressing a periodic function f(x) as a sum of sine and cosine functions. This representation is particularly useful because sines and cosines are the fundamental building blocks of periodic functions .

Dr. Z.’s Calc5 Lecture 9 Handout: Fourier Cosine and Sine Series Fourier Sine Series (for Odd Functions) The Fourier series of an odd function f(x) on the interval (−π,π) is the sine series (no cosines show up!) X∞ n=1 b n sinnx , where b n = 2 π Z π 0 f(x)sinnxdx , Half Range Expansion If a function f(x) is only deﬁned on (0,π), then we can extend it to (−π,π) to either get an even

3.2: Fourier Trigonometric Series - Mathematics LibreTexts From our discussion in the last section, we see that The Fourier series is periodic. The periods of $\cos nx$ and $\sin nx$ are $\frac{2\pi}{n}$. Thus, the largest period, $T = 2π$, comes from the $n = 1$ terms and the Fourier series has period $2π$.

AE2 Mathematics Solutions to Example Sheet 2: Fourier Series f(x) | sin x| on (−π, π) with L = π: f(x) is an even function so bn = 0. On [0, π] we have = | sin x| = sin x. where cos nπ = (−1)n. = 1 & the function is neither odd nor even. thus giving the answer. The odd extension of f(x) originally defined on 0. x) sin(nπx) dx.

Differential Equations - Fourier Sine Series - Pauls Online Math … 16 Nov 2022 · In this section we define the Fourier Sine Series, i.e. representing a function with a series in the form Sum( B_n sin(n pi x / L) ) from n=1 to n=infinity. We will also define the odd extension for a function and work several examples finding the Fourier Sine Series for a function.

Fourier Sine Series Examples - MIT Fourier Sine Series Examples 16th November 2007 The Fourier sine series for a function f(x) deﬁned on x ∈ [0,1] writes f(x) as f(x) = X∞ n=1 b n sin(nπx) for some coefﬁcients b n. Because of orthogonality, we can compute the b n very simply: for any given m, we integrate both sides against sin(mπx). In the summation, this gives zero ...

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Fourier Sine Series -- from Wolfram MathWorld 22 May 2025 · If f (x) is an odd function, then a_n=0 and the Fourier series collapses to f (x)=sum_ (n=1)^inftyb_nsin (nx), (1) where b_n = 1/piint_ (-pi)^pif (x)sin (nx)dx (2) = 2/piint_0^pif (x)sin (nx)dx (3) for n=1, 2, 3, .... The last equality is true because f (x)sin (nx) = [-f (-x)] [-sin (-nx)] (4) = f (-x)sin (-nx).

fourier series of $|\\sin x|$ - Mathematics Stack Exchange The function $x\mapsto f(x):=|\sin x|$ is even and $\pi$-periodic; therefore $f$ has a Fourier series of the form $$f(x)={a_0\over2}+\sum_{k=1}^\infty a_k \cos(2kx)$$ with $$a_k={2\over\pi}\int_0^\pi f(x)\cos(2k x)\ dx={2\over\pi}\int_0^\pi \sin x\cos(2k x)\ dx\ .$$ It follows that $$\eqalign{a_k&={1\over\pi}\int_0^\pi\left(\sin\bigl((1+2k)x ...

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Fourier Analysis Guide: Unlocking Sine & Cosine Secrets 3 Nov 2024 · Sine and cosine waves are the building blocks of Fourier analysis. A sine wave is a continuous wave that oscillates smoothly between its maximum and minimum values, with a phase shift of 90 degrees relative to the cosine wave. The cosine wave, on the other hand, starts at its maximum value and decreases to its minimum value over one period.

Fourier Sine Series - Mathphysics.com The Fourier sine series, in contrast, is well adapted to functions which are zero at x=0 and x=L, since all the functions sin(n \pi x/L) have this property. A related but less obvious property that begs for the use of sine functions is an odd symmetry: f(-x) = - f(x).

Fourier Series of The Sine Function - Mathematics Stack Exchange I'm going to give the short answer, here. Use orthogonality. The function sin $\left(\frac{\pi x}{L}\right)$ is orthogonal to the function sin $\left(\frac{n\pi x}{L}\right)$, for all $n\neq 1$, on the interval $(-L,L)$. Where $n=1$, the product of these functions is sin $^2\left(\frac{\pi x}{L}\right)$.

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Fourier Series Calculator - Sage Calculator A Fourier Series is a mathematical way to represent a periodic function as an infinite sum of sine and cosine functions. This powerful tool transforms a function of time (a signal) into a function of frequency, which is especially useful in signal processing, vibration analysis, acoustics, and quantum physics.

9.4: Fourier Sine and Cosine Series - Mathematics LibreTexts 18 Nov 2021 · The Fourier series simplifies if f(x) f (x) is an even function such that f(−x) = f(x) f (− x) = f (x), or an odd function such that f(−x) = −f(x) f (− x) = − f (x). Use will be made of the following facts. The function cos(nπx/L) cos (n π x / L) is an even function and sin(nπx/L) sin (n π x / L) is an odd function.

Fourier sine and cosine series - Wikipedia In mathematics, particularly the field of calculus and Fourier analysis, the Fourier sine and cosine series are two mathematical series named after Joseph Fourier. In this article, f denotes a real -valued function on which is periodic with period 2 L.

Trigonometric Fourier Series - GeeksforGeeks 13 Aug 2024 · Fourier Sine Series: In order to find Fourier sine series for a odd function defined over time interval (0,\text {T}) (0,T) we extend the time period to \text { (-T,T)} (-T,T) and since it is odd function it follows the property f (-t) = -f (t) f (−t) = −f (t), the Fourier transform of f (t) f (t) can be transformed into following equation: 2.

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Fourier Sine And Cosine Series - Prod Logs Hub 12 Oct 2024 · The Fourier sine and cosine series are a fundamental tool in the field of mathematics, particularly in the realm of signal processing and analysis. Developed by French mathematician and physicist Joseph Fourier, these series allow us to represent periodic functions as a sum of sine and cosine terms, which can be incredibly useful for understanding and …

Fourier Sine Series Of Sinx

Decomposing sin(x): Unveiling the Fourier Sine Series of sin(x)

1. Understanding the Fourier Sine Series

2. The Case of f(x) = sin(x) on [0, π]

3. Implications and Interpretations

4. Addressing Common Challenges

5. Summary

Frequently Asked Questions (FAQs)

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