Question

Which of the following statements about Fourier series is/are correct?

• A. The Fourier series of an even function contains only cosine terms
• B. The Fourier series of an odd function contains only cosine terms
• C. The Fourier series of an odd function contains only sine terms
• D. The Fourier series of an even function contains only sine terms

Hint:

Work out whether \(f(x)\cos(n\omega x)\) and \(f(x)\sin(n\omega x)\) are even or odd for an even \(f(x)\) and for an odd \(f(x)\), then recall that an odd function integrates to zero over a symmetric interval.

Answer 1

The Correct Option is A, C

Instead of working through the coefficient integrals directly, we can reason from what a cosine graph and a sine graph look like, and what "even" and "odd" mean for a function.

1. Recall the symmetry definitions: A function is even when \(f(-x) = f(x)\), meaning its graph is a mirror image about the y-axis, like \(\cos x\) itself. A function is odd when \(f(-x) = -f(x)\), meaning its graph has point symmetry about the origin, like \(\sin x\) itself.
2. Match the building blocks to the symmetry: Every cosine term \(\cos(n\omega x)\) is itself an even function, and every sine term \(\sin(n\omega x)\) is itself an odd function. A sum of purely even building blocks can only ever build an even function, and a sum of purely odd building blocks can only ever build an odd function.
3. Apply this to option (A) and (D): Since an even target function can only be built from even building blocks, its Fourier series can only use the constant term and cosine terms; it cannot pick up any sine terms without breaking its own even symmetry. So statement (A) is correct and statement (D), which claims only sine terms for an even function, is false.
4. Apply this to option (B) and (C): An odd target function has no constant part (a constant is even, so it cannot appear) and no cosine part, since cosine terms are even and would break the required odd symmetry. Its series can only be built from sine terms. So statement (C) is correct and statement (B), which claims only cosine terms for an odd function, is false.

This matches exactly what the direct integral calculation gives: even functions expand in cosines only, odd functions expand in sines only.

Let's summarize:

• Cosine terms are even building blocks, sine terms are odd building blocks.
• An even function's series can only use even building blocks, so only cosine terms appear.
• An odd function's series can only use odd building blocks, so only sine terms appear.

So the correct statements are (A) and (C).