Fourier Series (Even Function)¶

Background¶

Cos Method Fourier Series Even

Educational script demonstrating cos method fourier series even concepts.

Code¶

```python """ Cos Method Fourier Series Even

Educational script demonstrating cos method fourier series even concepts. """

@title Fourier Series On \([-\pi,\pi]\) For Even Function \(g\)¶

import matplotlib.pyplot as plt import numpy as np

======================================================================¶

def main(): f = lambda x : np.sin( (x-1.8)**2 ) g = lambda x: np.array([f(xi) if xi>0 else f(-xi) for xi in x])

n = 10000
theta = np.linspace(-np.pi,np.pi,n)
d_theta = theta[1] - theta[0]
g_theta = g(theta)

deg = 100
g_recovered = np.zeros_like(g_theta)
for k in range(deg):
    A_k = np.sum( g_theta[:-1] * np.cos(k*theta[:-1]) ) * d_theta / np.pi
    B_k = np.sum( g_theta[:-1] * np.sin(k*theta[:-1]) ) * d_theta / np.pi
    if k == 0:
        A_k /= 2
        B_k /= 2
    g_recovered += A_k * np.cos(k*theta) + B_k * np.sin(k*theta)

fig, ax = plt.subplots(1,1,figsize=(12,4))
ax.plot(theta,g_theta,label='Original',lw=10,alpha=0.3)
ax.plot(theta,g_recovered,"--r",label=f'Recovered with {n} Terms')
ax.legend()
plt.show()

if name == "main": main() ```

Exercises¶

Exercise 1. If \(g(x) = f(|x|)\) where \(f(x) = \sin((x-1.8)^2)\), verify that \(g\) is an even function and explain why \(B_k = 0\) for all \(k\).

Solution to Exercise 1

\(g(-x) = f(|-x|) = f(|x|) = g(x)\), confirming \(g\) is even. For even functions, \(B_k = \frac{1}{\pi}\int_{-\pi}^{\pi}g(\theta)\sin(k\theta)d\theta = 0\) because \(g(\theta)\sin(k\theta)\) is an odd function (even times odd = odd), and the integral of an odd function over a symmetric interval is zero.

Exercise 2. For the even function \(g\), the Fourier series reduces to a cosine series. Explain the connection to the COS method for option pricing.

Solution to Exercise 2

The COS method expands the density function on \([a, b]\) using only cosine terms: \(f(x) \approx \sum_{k=0}^{N-1} A_k \cos(k\pi(x-a)/(b-a))\). This works because any function on \([a, b]\) can be even-extended to \([2a-b, b]\), making the cosine series a natural basis. The CF provides the coefficients directly: \(A_k = \frac{2}{b-a}\text{Re}[\varphi(k\pi/(b-a))e^{-ika\pi/(b-a)}]\).

Exercise 3. Compare the convergence of the Fourier series for \(f\) (general) versus \(g\) (even). Does evenness improve convergence?

Solution to Exercise 3

Evenness itself does not improve the convergence rate---it only eliminates the sine terms. However, \(g(x) = f(|x|)\) typically has a kink at \(x = 0\) (unless \(f^\prime(0) = 0\)), which can reduce convergence to \(O(1/N^2)\). For the original \(f\) (smooth), convergence is superalgebraic. So the even extension may actually worsen convergence if it introduces a non-smooth point.

Exercise 4. For 100 expansion terms, both \(f\) and \(g\) show excellent recovery. Estimate the minimum number of terms needed for the error to be below \(10^{-3}\).

Solution to Exercise 4

For smooth \(f\): with exponential convergence \(O(e^{-\alpha N})\), the error reaches \(10^{-3}\) at roughly \(N \approx 7\alpha^{-1} \ln 10 \approx 20\)--30 terms. For \(g\) with a kink: convergence is \(O(1/N^2)\), so \(10^{-3}\) requires \(N \approx 1/\sqrt{10^{-3}} \approx 32\) terms. In practice, about 30--50 terms suffice for both at this tolerance.