(a) Let F(x) be a periodic function that is odd; that is,...

(a) Let $F(x)$ be a periodic function that is odd; that is, $F(x)=-F(-x) .$ The Fourier expansion of such a function requires only sine functions: $$ F(x)=\sum_{n=0}^{\infty} B_{n} \sin \left(\frac{2 n \pi x}{\lambda}\right) $$ Following the suggestions in Problem $6.35$, prove that $$ B_{m}=\frac{2}{\lambda} \int_{0}^{\lambda} F(x) \sin \left(\frac{2 m \pi x}{\lambda}\right) d x $$ (Note that the sine series has no $n=0$ term, since $\sin 0=0 .$ ) (b) Use this result to show that the Fourier coefficients $B_{n}$ of the "sawtooth" function in Fig. $6.22$ are zero for $n$ even and that $$ B_{n}=(-1)^{(n-1) / 2} \frac{8}{\pi^{2} n^{2}} \quad \text { for } n \text { odd } $$

(a) Let $F(x)$ be a periodic function that is odd; that is, $F(x)=-F(-x) .$ The Fourier expansion of such a function requires only sine functions:
$$
F(x)=\sum_{n=0}^{\infty} B_{n} \sin \left(\frac{2 n \pi x}{\lambda}\right)
$$
Following the suggestions in Problem $6.35$, prove that
$$
B_{m}=\frac{2}{\lambda} \int_{0}^{\lambda} F(x) \sin \left(\frac{2 m \pi x}{\lambda}\right) d x
$$
(Note that the sine series has no $n=0$ term, since $\sin 0=0 .$ ) (b) Use this result to show that the Fourier coefficients $B_{n}$ of the "sawtooth" function in Fig. $6.22$ are zero for $n$ even and that
$$
B_{n}=(-1)^{(n-1) / 2} \frac{8}{\pi^{2} n^{2}} \quad \text { for } n \text { odd }
$$

Key Concepts

Fourier Sine Series

When a function is odd and periodic, its Fourier series includes only sine terms. The Fourier sine series takes advantage of the function's odd symmetry, which effectively nullifies the cosine coefficients due to the antisymmetry of the integrand. This series is particularly useful in solving problems with boundary conditions that enforce odd symmetry.

Fourier Series

Fourier series decompose a periodic function into a sum of sines and cosines (or complex exponentials), each representing a distinct frequency component. This decomposition is fundamental in analyzing and synthesizing periodic signals across mathematics, physics, and engineering, as it provides insight into the behavior of functions in both time and frequency domains.

Odd Functions

An odd function satisfies the symmetry F(x) = -F(-x), which has significant implications for its Fourier representation. Specifically, in Fourier series, odd functions have the property that their expansion in terms of sines and cosines results in a cancellation of cosine (even) terms, leaving only sine terms in the series. This symmetry simplifies the analysis and computation of Fourier coefficients.

Orthogonality of Sine Functions

Sine functions of different frequencies are orthogonal over a complete period, meaning that the integral of the product of two sines with different frequencies is zero. This property is crucial when computing Fourier coefficients, as it allows one to isolate individual sine components by integrating the product of the function and a sine term over its period.

Determination of Fourier Coefficients

The process of finding the Fourier coefficients involves multiplying the function by a sine or cosine function, integrating over a full period, and then normalizing the result. For an odd function, the integration is often performed over a symmetric interval or reduced to half the period due to symmetry, thereby streamlining the calculation of the sine coefficients.

Sawtooth Function in Fourier Analysis

The sawtooth function is a classic example in Fourier analysis that illustrates how discontinuous periodic functions can be represented as an infinite sum of sine (or cosine) functions. In the case of an odd sawtooth wave, the Fourier sine series reveals that only certain harmonics contribute to the series—specifically, the coefficients for even harmonic numbers vanish and the odd harmonics have alternating signs and decreasing amplitudes, which reflects the function's discontinuous and linear behavior.

Transcript

Please give Ace some feedback