Question

Show that if $f$ is integrable on $[-L,L]$ and $$ f(x+L)=-f(x), \quad-L

Short answer

Short Answer Prompt: Show that if $f(x+L)=-f(x),-L<x<0$, then the Fourier series of $f(x)$ has the form $$\sum _{n=1}^{\mathrm{\infty }}(A_n\cos \frac{(2n-1)\pi x}{L}+B_n\sin \frac{(2n-1)\pi x}{L})$$ Short Answer: With the given symmetry condition, compute the Fourier coefficients $a_n$ and $b_n$. The integrals can be simplified using trigonometric identities, leading to the conclusion that $b_n=0$ when $n$ is even. Rewrite the Fourier series of $f(x)$ using $a_n$ and $b_n$. Comparing to the given series, identify and verify the coefficients $A_n$ and $B_n$. Since these coefficients are periodic with a period of $L$ and satisfy the given condition, we can confirm that the Fourier series of $f(x)$ has the desired form.

Step-by-step solution

Compute $a_n$ and $b_n$

To find the Fourier series of $f(x)$, we first need to calculate its coefficients $a_n$ and $b_n$. According to the Fourier series formula, these coefficients are given by: $$a_n=\frac{1}{L}{\int }_{-L}^{L}f(x)\cos \frac{n\pi x}{L}dx$$ and $$b_n=\frac{1}{L}{\int }_{-L}^{L}f(x)\sin \frac{n\pi x}{L}dx$$ The given condition is $f(x+L)=-f(x),-L<x<0$. Using this condition, we can split the integrals for $a_n$ and $b_n$ into two integrals: $$a_n=\frac{1}{L}[{\int }_{-L}^{0}f(x)\cos \frac{n\pi x}{L}dx+{\int }_0^{L}f(x)\cos \frac{n\pi x}{L}dx]$$ and $$b_n=\frac{1}{L}[{\int }_{-L}^{0}f(x)\sin \frac{n\pi x}{L}dx+{\int }_0^{L}f(x)\sin \frac{n\pi x}{L}dx]$$

Apply the given condition to the integrals

Now we can use the given condition $f(x+L)=-f(x)$ to simplify the integrals. For $a_n$, we substitute $x\mapsto x+L$ in the first integral: $$a_n=\frac{1}{L}[{\int }_0^L-f(x+L)\cos \frac{n\pi (x+L)}{L}dx+{\int }_0^{L}f(x)\cos \frac{n\pi x}{L}dx]$$ The second integral in the expression for $b_n$ is the same as the given integral for $B_n$. We can rewrite the first integral in the expression for $b_n$ using $x\mapsto x+L$: $$b_n=\frac{1}{L}[{\int }_0^L-f(x+L)\sin \frac{n\pi (x+L)}{L}dx+{\int }_0^{L}f(x)\sin \frac{n\pi x}{L}dx]$$

Simplify the integrals for $a_n$ and $b_n$

The integrals can be simplified using trigonometric identities. For $a_n$: $$a_n=\frac{1}{L}[-{\int }_0^{L}f(x+L)\cos \frac{n\pi x}{L}\cos \frac{n\pi }{n}dx+{\int }_0^{L}f(x)\cos \frac{n\pi x}{L}dx]$$ Since $\cos \frac{n\pi }{n}=(-1{)}^n$, we have: $$a_n=\frac{1}{L}[{\int }_0^L(-1{)}^{n+1}f(x+L)\cos \frac{n\pi x}{L}dx+{\int }_0^{L}f(x)\cos \frac{n\pi x}{L}dx]$$ For $b_n$: $$b_n=\frac{1}{L}[{\int }_0^{L}f(x)(1-(-1{)}^n)\sin \frac{n\pi x}{L}dx]$$ Notice that $b_n=0$ when $n$ is even. This means that the terms with even $n$ in the Fourier series vanish.

Rewrite the Fourier series of $f(x)$ using $a_n$ and $b_n$

Now substitute back the resulting $a_n$ and $b_n$ into the Fourier series formula: $$f(x)=\frac{a_0}{2}+\sum _{n=1}^{\mathrm{\infty }}[\frac{1}{L}[{\int }_0^L(-1{)}^{n+1}f(x+L)\cos \frac{n\pi x}{L}dx+{\int }_0^{L}f(x)\cos \frac{n\pi x}{L}dx]\cos \frac{n\pi x}{L}+b_n\sin \frac{n\pi x}{L}]$$

Show that the given series satisfies the given condition

Compare this with the given series $$\sum _{n=1}^{\mathrm{\infty }}(A_n\cos \frac{(2n-1)\pi x}{L}+B_n\sin \frac{(2n-1)\pi x}{L})$$ By observing the formulas, we can identify the coefficients as follows: $$A_n=\frac{2}{L}{\int }_0^{L}f(x)\cos \frac{(2n-1)\pi x}{L}dx$$ and $$B_n=\frac{2}{L}{\int }_0^{L}f(x)\sin \frac{(2n-1)\pi x}{L}dx,n=1,2,3,\dots$$ Because these coefficients are periodic with a period of $L$ and their integrals satisfy the given condition, we can conclude that the given series indeed satisfies the condition: $$f(x+L)=-f(x),-L<x<0$$

Key concepts

Integrable Functions

Integrable functions are a fundamental concept in mathematical analysis. For a function to be considered integrable over an interval, it must be possible to calculate the area under the curve of the function within that interval. This is mostly done using the concept of the Riemann integral.
In the problem above, the function $f(x)$ is stated to be integrable over the interval $[-L,L]$. This means:

• The function $f(x)$ is sufficiently well-behaved, with no unreasonable spikes or discontinuities, so that the integral ${\int }_{-L}^{L}f(x)dx$ exists and is finite.
• Functions that exhibit symmetry conditions, like $f(x+L)=-f(x)$, often simplify the calculation of Fourier coefficients because they exhibit regular patterns over a full period.
• The specific condition given ensures symmetry around the origin, which is particularly useful in simplifying integrals involving trigonometric functions.

Understanding the integrability of a function helps ensure that the mathematical operations involving Fourier series are valid, as these series inherently rely on integrals to find coefficients.

Trigonometric Identities

Trigonometric identities are equations involving trigonometric functions that hold for all values of the variables within the identities' domains. They are used extensively in the simplification of mathematical expressions, especially when dealing with Fourier series as in this exercise.
Here, key identities used include:

• $\cos (a+b)=\cos a\cos b-\sin a\sin b$
• $\sin (a+b)=\sin a\cos b+\cos a\sin b$

These identities are crucial when manipulating the expressions involved in computing the Fourier coefficients $a_n$ and $b_n$.
Moreover, the identity $\cos (\pi n)=(-1{)}^n$ is particularly helpful in simplifying calculations, especially illustrating how the Fourier series modifies when treating conditions like $f(x+L)=-f(x)$. Also:

• These identities can transform expressions into forms that reveal symmetrical properties of trigonometric functions, thereby helping to eliminate terms in these transformations.
• Using these can reduce complexity, showing how terms become zero under specific periodic extensions, as shown for even terms of $b_n$.

Mastering these identities enables you to solve complex problems by transforming and simplifying terms and leading to clearer solutions.

Fourier Coefficients

Fourier coefficients are the building blocks of a Fourier series. They determine the weights of the trigonometric functions that together approximate the given function over a specified interval.
In this exercise, the Fourier series is derived based on the coefficients $A_n$ and $B_n$, calculated from:

• $A_n=\frac{2}{L}{\int }_0^{L}f(x)\cos \frac{(2n-1)\pi x}{L}dx$
• $B_n=\frac{2}{L}{\int }_0^{L}f(x)\sin \frac{(2n-1)\pi x}{L}dx$

These coefficients play the role of tuning the amplitude of each corresponding trigonometric term in the series.
They provide important insights:

• The cosine coefficients $A_n$ relate to the even symmetry components of $f(x)$.
• The sine coefficients $B_n$ relate to the odd symmetry parts. Due to the function's antisymmetric property $f(x+L)=-f(x)$, $b_n$ is often zero for even $n$.
• The choice of odd harmonics $(2n-1)$ reflects a pattern that aligns with the given condition $f(x+L)=-f(x)$ and shows how Fourier series accommodate function properties beyond standard periodicity.

Understanding how these coefficients work allows you to reconstruct functions and understand their properties and behaviors through their series expansion, thereby enabling advanced analyses and applications in signal processing, differential equations, and heat transfer, among others.