Q6Mathematics I

Question

Find the value of $a_{0}$ for the function $f (x) = ∣ x ∣$ in the interval $(- π, π)$ .

f (x) = ∣ x ∣, x \in (- π, π)

Answer 1

To systematically evaluate the primary Fourier coefficient $a_{0}$ for the absolute value function $f (x) = ∣ x ∣$ defined precisely over the symmetric mathematical interval $[- π, π]$ , we utilize the standard definition for the Fourier series coefficient:

a_{0} = \frac{1}{π} \int_{- π}^{π} f (x) d x = \frac{1}{π} \int_{- π}^{π} ∣ x ∣ d x

Before attempting to integrate, we must strictly analyze the mathematical symmetry of the given function. Because $∣ - x ∣ = ∣ x ∣$ for all possible real values of $x$ , the given function $f (x) = ∣ x ∣$ is classified rigorously as an even function. A fundamental property of definite integrals states that when integrating an even function over a perfectly symmetric interval $[- a, a]$ , the integral can be mathematically simplified to exactly twice the integral evaluated over the positive half $[0, a]$ .

Applying this strict rule of calculus, the expression transforms to:

a_{0} = \frac{2}{π} \int_{0}^{π} ∣ x ∣ d x

Since $x$ is strictly positive in the newly defined integration interval $[0, π]$ , the absolute value function simplifies directly to $x$ . We can now execute the standard power rule integration:

a_{0} = \frac{2}{π} \int_{0}^{π} x d x = \frac{2}{π} [\frac{x ^{2}}{2}]_{0}^{π}

Evaluating the definite integral precisely at the upper limit $π$ and the lower limit $0$ yields:

a_{0} = \frac{2}{π} (\frac{π ^{2}}{2} - \frac{0 ^{2}}{2}) = \frac{2}{π} (\frac{π ^{2}}{2}) = π

Therefore, the mathematically exact value of the required Fourier coefficient $a_{0}$ is $π$ .

Q6Mathematics I

Question

Answer