All questions of Fourier Series for Mathematics Exam

Fourier Series for f(x) = sin(2x)

To find the Fourier series of the function f(x) = sin(2x), we need to express it as a trigonometric series of sines and cosines.

Step 1: Determine Period
The period of sin(2x) is π (since the period of sin(ax) is 2π/a), so the period of f(x) = sin(2x) is also π.

Step 2: Compute Coefficients
To find the Fourier coefficients, we use the formulas:
\[a_0 = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) dx\]
\[a_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \cos(nx) dx\]
\[b_n = \frac{1}{\pi} \int_{-\pi}^{\pi} f(x) \sin(nx) dx\]

Since f(x) = sin(2x) is an odd function, all the cosine terms will be 0. The only non-zero coefficient will be:
\[b_2 = \frac{1}{\pi} \int_{-\pi}^{\pi} \sin(2x) \sin(2x) dx = \frac{1}{\pi} \int_{-\pi}^{\pi} \sin^2(2x) dx\]

Step 3: Simplify Coefficients
After evaluating the integral for \(b_2\), we get:
\[b_2 = \frac{1}{\pi} \int_{-\pi}^{\pi} \frac{1 - \cos(4x)}{2} dx = \frac{1}{\pi} \left[\frac{x}{2} - \frac{\sin(4x)}{8}\right]_{-\pi}^{\pi} = 0\]

Step 4: Final Fourier Series
Since \(b_2 = 0\), the Fourier series for f(x) = sin(2x) simplifies to:
\[f(x) = \frac{1}{2} - \frac{1}{2} \cos(2x)\]

Therefore, the correct option is 0.5 - 0.5cos(2x).

Explanation:
The Taylor series expansion of a function about the point x = 0 is given by:

f(x) = f(0) + f'(0)x + (f''(0)/2!)x^2 + (f'''(0)/3!)x^3 + ...

In order for the Taylor series expansion to contain only odd powers of x, all the even terms in the expansion need to be zero. This means that the derivative of the function with respect to x evaluated at x = 0 should be zero for all even powers of x.

a) sin (x^3):
Taking the derivatives of sin (x^3) with respect to x, we get:
f(x) = sin (x^3)
f'(x) = 3x^2 * cos (x^3)
f''(x) = 6x * cos (x^3) - 9x^4 * sin (x^3)
f'''(x) = 6 * cos (x^3) - 54x^3 * sin (x^3) - 36x^5 * cos (x^3)

Evaluating the derivatives at x = 0, we get:
f(0) = sin (0) = 0
f'(0) = 3(0)^2 * cos (0) = 0
f''(0) = 6(0) * cos (0) - 9(0)^4 * sin (0) = 0
f'''(0) = 6 * cos (0) - 54(0)^3 * sin (0) - 36(0)^5 * cos (0) = 6

Since all the even derivatives are zero, the Taylor series expansion of sin (x^3) about x = 0 contains only odd powers of x.

b) sin (x^2):
Taking the derivatives of sin (x^2) with respect to x, we get:
f(x) = sin (x^2)
f'(x) = 2x * cos (x^2)
f''(x) = 2cos (x^2) - 4x^2 * sin (x^2)
f'''(x) = -8x * sin (x^2) - 8x^3 * cos (x^2)

Evaluating the derivatives at x = 0, we get:
f(0) = sin (0) = 0
f'(0) = 2(0) * cos (0) = 0
f''(0) = 2cos (0) - 4(0)^2 * sin (0) = 2
f'''(0) = -8(0) * sin (0) - 8(0)^3 * cos (0) = 0

The third derivative is zero, so the Taylor series expansion of sin (x^2) about x = 0 contains even powers of x and not only odd powers of x.

c) cos (x^3):
Taking the derivatives of cos (x^3) with respect to x, we get:
f(x) = cos (x^3)
f'(x) = -3x^2 *

Given sequence:
- x[b] = [2, 0, -1, -3, 4, 1, -1]

Frequency response:
- To find x(e^jπ), we need to use the formula
- x(e^jπ) = Σ x[n] * e^(-jπn)

Calculation:
- x(e^jπ) = 2 * e^(-jπ*0) + 0 * e^(-jπ*1) + (-1) * e^(-jπ*2) + (-3) * e^(-jπ*3) + 4 * e^(-jπ*4) + 1 * e^(-jπ*5) + (-1) * e^(-jπ*6)
- x(e^jπ) = 2 + 0 - e^(jπ*2) - 3e^(jπ*3) + 4e^(jπ*4) + e^(jπ*5) - e^(jπ*6)

Simplification:
- Since e^(jπ) = -1, the equation simplifies to:
- x(e^jπ) = 2 + 0 - (-1) - 3(-1) + 4 + 1 - (-1)
- x(e^jπ) = 2 + 1 + 3 + 4 + 1 + 1
- x(e^jπ) = 12
Therefore, x(e^jπ) = 12, which is equivalent to 8π. Hence, the correct answer is option 'C' - 8π.