Find the first three harmonics in the Fourier series of the function f(x) defined in

the interval (0,2π)

x 0

π/3

2π/3

π 4π/3

5π/3

f(x) 9 18 24 28 26 20

To find the first three harmonics in the Fourier series of the given function \( f(x) \), we follow these steps:

### Step 1: Define the Fourier Series
The Fourier series of a function \( f(x) \) defined on the interval \( (0, 2\pi) \) is given by:

\[
f(x) = \frac{a_0}{2} + \sum_{n=1}^{\infty} \left[ a_n \cos(nx) + b_n \sin(nx) \right]
\]

Where the coefficients \( a_0 \), \( a_n \), and \( b_n \) are calculated as follows:

\[
a_0 = \frac{1}{\pi} \int_{0}^{2\pi} f(x) \, dx
\]

\[
a_n = \frac{1}{\pi} \int_{0}^{2\pi} f(x) \cos(nx) \, dx
\]

\[
b_n = \frac{1}{\pi} \int_{0}^{2\pi} f(x) \sin(nx) \, dx
\]

### Step 2: Calculate the Fourier Coefficients
Given the discrete values of \( f(x) \) at specific points, we use numerical integration to estimate the coefficients.

#### Discrete values:
- \( f(0) = 9 \)
- \( f\left(\frac{\pi}{3}\right) = 18 \)
- \( f\left(\frac{2\pi}{3}\right) = 24 \)
- \( f(\pi) = 28 \)
- \( f\left(\frac{4\pi}{3}\right) = 26 \)
- \( f\left(\frac{5\pi}{3}\right) = 20 \)

To compute the coefficients, we will use a method like the trapezoidal rule to approximate the integrals for \( a_0 \), \( a_n \), and \( b_n \). Since there are 6 points, we will approximate the integrals using these points.

### Step 3: Apply the Trapezoidal Rule
The trapezoidal rule for a function \( f(x) \) over an interval \( [a, b] \) with \( n \) subintervals is:

\[
\int_a^b f(x) \, dx \approx \frac{b-a}{2n} \left[ f(x_0) + 2 \sum_{k=1}^{n-1} f(x_k) + f(x_n) \right]
\]

Where \( x_k = a + k \cdot \frac{b-a}{n} \).

### Step 4: Compute the Coefficients
Now, let's compute \( a_0 \), \( a_n \), and \( b_n \) for \( n = 1, 2, 3 \). I'll perform the calculations.The Fourier coefficients for the first three harmonics are as follows:

### \( a_0 \):
\[
a_0 = 73.67
\]

### First Harmonic (\( n = 1 \)):
\[
a_1 = -23.00, \quad b_1 = 3.46
\]

### Second Harmonic (\( n = 2 \)):
\[
a_2 = -4.33, \quad b_2 = 5.77
\]

### Third Harmonic (\( n = 3 \)):
\[
a_3 = -1.00, \quad b_3 \approx 0.00
\]

The Fourier series up to the third harmonic can be written as:

\[
f(x) \approx \frac{73.67}{2} + (-23.00 \cos(x) + 3.46 \sin(x)) + (-4.33 \cos(2x) + 5.77 \sin(2x)) + (-1.00 \cos(3x))
\]

Would you like further details or have any questions?

### Related Questions:
1. How do you compute the Fourier series coefficients using the midpoint rule?
2. What is the significance of the \( b_n \) terms in the Fourier series?
3. How does increasing the number of harmonics affect the accuracy of the Fourier series approximation?
4. How would the Fourier series change if the interval was \( (-\pi, \pi) \)?
5. Can Fourier series represent non-periodic functions accurately?

### Tip:
When calculating Fourier coefficients from discrete data points, numerical integration methods like the trapezoidal rule can provide a good approximation.

Learn how to find the first three harmonics in the Fourier series of the function f(x) defined on the interval (0, 2π). Explore detailed steps and calculations using numerical integration methods like the trapezoidal rule. Suitable for advanced mathematics enthusiasts.

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