Why does the odd symmetry of a square wave remove constant and cosine terms from its Fourier expansion?

Conditions

• The function is periodic with period $2\pi$.
• The function satisfies $f(-x) = -f(x)$ almost everywhere.

Reasoning, step by step

1. Identify the parity of the square wave: it is an odd function.
2. Recall that cosine terms and the constant term in a Fourier series are even functions.
3. Apply the orthogonality property: the inner product of an odd function and an even function over a symmetric domain is zero.
4. Conclude that the Fourier coefficients for these even components are zero.

Example

For the square wave defined as 1 on $(0,\pi)$ and -1 on $(-\pi,0)$, the script states: 'Odd symmetry removes the constant and cosine terms from its Fourier expansion.'

Common misconceptions

• Believing that only sine terms appear because the function is discontinuous; continuity is not required for parity arguments.
• Confusing odd/even symmetry with half-wave symmetry.

Watch the explanation

Explore next

• What happens to the Fourier series if the square wave has half-wave symmetry instead?
• Do the values at the jump points affect the calculation of these coefficients?