Chapter 11: Problem 3
Find the Fourier cosine series. $$ f(x)=x^{2}-2 L x ; \quad[0, L] $$
Short Answer
Expert verified
Answer: The Fourier cosine series of the function f(x) = x^2 - 2Lx on the interval [0, L] is f(x) = (L^2)/3.
Step by step solution
01
Define the function and the interval
The function is given as:
$$
f(x) = x^2 - 2Lx
$$
And the interval for the function is: [0, L].
02
Write down the general form of the Fourier cosine series
The Fourier cosine series has the following general form:
$$
f(x) = \frac{A_0}{2} + \sum_{n=1}^{\infty}A_n\cos(\frac{n\pi x}{L})
$$
where A0 and An are the Fourier coefficients.
03
Calculate the Fourier coefficients (A0, An)
We need to find the coefficients A0 and An by integrating over the [0, L] interval.
The A0 coefficient is given by:
$$
A_0 = \frac{2}{L}\int_{0}^{L} f(x)dx
$$
So, let's calculate A0 for our function:
$$
A_0 = \frac{2}{L}\int_{0}^{L} (x^2 - 2Lx)dx
$$
$$
A_0 = \frac{2}{L}\left[\frac{x^3}{3} - Lx^2\right]_{0}^{L}
$$
$$
A_0 = \frac{2}{L}\left[\frac{L^3}{3} - L^3\right] = \frac{2}{3}L^2
$$
Next, let's calculate the An coefficients. An is given by:
$$
A_n = \frac{2}{L} \int_{0}^{L} f(x) \cos(\frac{n\pi x}{L}) dx
$$
So, we have:
$$
A_n = \frac{2}{L} \int_{0}^{L} (x^2 - 2Lx) \cos(\frac{n \pi x}{L}) dx
$$
Now, we need to integrate this expression with respect to x:
To integrate x^2*cos(nπx/L) dx, use integration by parts:
Let u = x^2, dv = cos(nπx/L) dx
Then, du = 2x dx, v = L/nπ * sin(nπx/L)
Integrating the expression:
$$
A_n = \frac{2}{L}\left[\frac{L}{n\pi}(x^2\sin(\frac{n\pi x}{L}))\right]_{0}^{L} - \frac{2}{L}\int_{0}^{L} \frac{2L}{n\pi}x\sin(\frac{n\pi x}{L})dx
$$
First integral is 0. Use integration by parts again for the second integral:
Let u = x, dv = sin(nπx/L) dx
Then, du = dx, v = -L/nπ * cos(nπx/L)
Now, the expression becomes:
$$
A_n = -\frac{4L}{n^2\pi^2}\left[\frac{-L}{n\pi}(x\cos(\frac{n\pi x}{L}))\right]_{0}^{L} + \frac{4L^2}{n^3\pi^3}\int_{0}^{L} \cos(\frac{n\pi x}{L})dx
$$
$$
A_n = -\frac{4L^3}{n^3\pi^3}\left[\sin(\frac{n\pi x}{L})\right]_{0}^{L}
$$
As sin(nπ) = 0 for all integer values of n, An = 0
04
Substitute the coefficients in the Fourier cosine series formula.
Since we have calculated A0 and An, let's substitute them into the Fourier cosine series formula:
$$
f(x) = \frac{A_0}{2} + \sum_{n=1}^{\infty}A_n\cos(\frac{n\pi x}{L})
$$
$$
f(x) = \frac{L^2}{3} + 0
$$
05
Simplify the Fourier cosine series.
As all the An coefficients are 0, the Fourier cosine series is simplified to:
$$
f(x) = \frac{L^2}{3}
$$
That's the Fourier cosine series for the given function f(x) = x^2 - 2Lx on the interval [0, L].
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Fourier coefficients
Fourier coefficients are the building blocks for representing a periodic function as a sum of simpler trigonometric functions. They serve as the weights or scale factors before each trigonometric component in the series. In the Fourier cosine series, these coefficients are crucial for capturing the cosines to form an approximation of the original function. The Fourier cosine series uses two main types of coefficients: \( A_0 \) and \( A_n \). For
- \( A_0 \): This coefficient determines the mean value of the function over the interval. It is given by the integral \( A_0 = \frac{2}{L}\int_{0}^{L} f(x)dx \), which you solve by integrating the function over its interval.
- \( A_n \): These coefficients contribute to the series terms of \( \cos(\frac{n\pi x}{L}) \) where \( n \) is a positive integer. This is obtained through \( A_n = \frac{2}{L} \int_{0}^{L} f(x) \cos(\frac{n\pi x}{L})dx \), needing integration by parts for complex functions.
integration by parts
Integration by parts is a technique that simplifies the integration of products of functions. It's inspired by the product rule for differentiation and is widely used when one part of the integrand is easily integrable.The method is mathematically expressed as:\[ \int u \, dv = uv - \int v \, du \]Here, \( u \) and \( dv \) are parts of the original function being integrated. You choose them based on their ability to simplify the integral process when substituted.
Practical Use
To effectively use integration by parts:- Identify parts: Choose \( u \) as the portion that becomes simpler when differentiated (like \( x^2 \)), and \( dv \) to be easily integrated part (such as \( \cos(\frac{n\pi x}{L})dx \)).
- Differentiate and Integrate: Compute \( du \) and \( v \) based on your choice, then integrate and differentiate respectively.
cosine functions
Cosine functions are key components of the Fourier cosine series. These functions, represented as \( \cos(x) \), oscillate between -1 and 1 and are crucial for describing certain periodic behaviors. In engineering and physics, they model phenomena that repeat regularly over time or space.
Role in Fourier Series
Cosine functions help reconstruct the given periodic function segmentally in Fourier series:- The term \( \cos(\frac{n\pi x}{L}) \) represents these functions in the series, where \( n \) indicates harmonic frequencies explainable by integer multiples.'
- Within integrals, they work to find coefficients that ensure the approximated function stays true to its periodic path on a stipulated interval.
mathematical series
A mathematical series is essentially the sum of sequential elements, like numbers or functions, often represented in an ordered manner. They are central in approaches where approximation of functions can be improved by aggregating terms.
Fourier Series
Fourier cosines series are special cases of mathematical series focusing on enclosing a function in terms of cosine terms. Their series take the form:\[ f(x) = \frac{A_0}{2} + \sum_{n=1}^{\infty} A_n \cos \left(\frac{n \pi x}{L}\right) \]This breaks a complex waveform into simpler oscillating components.- Each term in the series represents a layer of approximation refined by adding or subtracting harmonics of the fundamental frequency.
- The summation aspect allows for continuous refinement, improving the approximation to closely follow the original function.