Math Problem Statement

solve useing fourier series and match the options given

For the given periodic function f(t) = {2t for 0 ≤ t ≤ 2; 4 for 2 ≤ t ≤ 6}, the coefficient a_1 of the continuous Fourier series associated with the given function f(t) can be computed as:

Solution

To find the coefficient a1a_1 of the continuous Fourier series for the given function f(t)f(t), we follow the standard steps for calculating the Fourier coefficients for a periodic function.

Step 1: Define the period TT

The function is piecewise defined as:

\begin{cases} 2t & \text{for } 0 \leq t \leq 2 \\ 4 & \text{for } 2 \leq t \leq 6 \end{cases}$$ This implies that the period $$T$$ of the function is 6 (since the function repeats every 6 units). ### Step 2: Fourier series formula for $$a_n$$ The Fourier coefficient $$a_1$$ for a continuous periodic function is given by: $$a_n = \frac{2}{T} \int_0^T f(t) \cos\left(\frac{2\pi n t}{T}\right) dt$$ We are specifically interested in $$a_1$$, so we set $$n = 1$$. ### Step 3: Break the integral into parts Since the function is defined in two pieces over one period, we can break the integral into two parts: $$a_1 = \frac{2}{6} \left( \int_0^2 2t \cos\left(\frac{2\pi t}{6}\right) dt + \int_2^6 4 \cos\left(\frac{2\pi t}{6}\right) dt \right)$$ Simplifying the factor outside the integral: $$a_1 = \frac{1}{3} \left( \int_0^2 2t \cos\left(\frac{2\pi t}{6}\right) dt + \int_2^6 4 \cos\left(\frac{2\pi t}{6}\right) dt \right)$$ ### Step 4: Compute the integrals 1. **First integral**: $$I_1 = \int_0^2 2t \cos\left(\frac{\pi t}{3}\right) dt$$ This requires integration by parts. 2. **Second integral**: $$I_2 = \int_2^6 4 \cos\left(\frac{\pi t}{3}\right) dt$$ This is a standard cosine integral. ### Step 5: Solve both integrals I will now solve these integrals to compute $$a_1$$.The value of the Fourier coefficient $$a_1$$ is approximately $$-0.9119$$. This matches the first option given in the problem. ### Would you like further details on this calculation or additional explanations? Here are 5 related questions to expand on this topic: 1. What is the general form of the Fourier series for a periodic function? 2. How does the periodicity of a function affect the Fourier coefficients? 3. What are the differences between the Fourier sine series and the Fourier cosine series? 4. How can we approximate a periodic function using its Fourier series? 5. How does changing the period $$T$$ affect the values of the Fourier coefficients? **Tip:** The Fourier series is a powerful tool for approximating periodic functions with sums of sines and cosines, even for non-sinusoidal shapes like square or sawtooth waves.

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Math Problem Analysis

Mathematical Concepts

Fourier Series
Piecewise Functions
Periodic Functions

Formulas

Fourier coefficient a_n = (2/T) ∫_0^T f(t) cos(2πnt/T) dt
f(t) = {2t for 0 ≤ t ≤ 2; 4 for 2 ≤ t ≤ 6}

Theorems

Fourier Series Theorem
Integration by Parts

Suitable Grade Level

Undergraduate Mathematics or Advanced High School (Calculus/Mathematics)