Guides / Series

Fourier series: integrate against each sine and cosine, halve the constant term, and let symmetry do half the work

A Fourier series writes a periodic function as a constant plus sines and cosines of whole-number multiples of one base frequency. Each coefficient comes from one integral, because on a full period the sines and cosines are orthogonal: integrating f against cos nx keeps exactly one term of the series and kills the rest. Even functions need only cosines and odd functions only sines, at a jump the series lands halfway, and Parseval's identity turns the coefficients into sums such as 1 + 1/9 + 1/25 + ⋯ = π²/8. Most wrong answers come from forgetting to halve a₀, treating cos nπ as 0, or dividing by π when the period isn't 2π.

Why it works

On [−π, π], the integral of cos mx cos nx is 0 unless m = n, when it is π, and the same holds for the sines; a sine times a cosine always integrates to 0. So multiply the series by cos nx and integrate: every term but one vanishes, and what is left is π a_n. That is the coefficient formula. It works the way a dot product with a unit vector picks out one coordinate.

The same formula with n = 0 gives a₀ as twice the average of f, because cos 0 = 1 integrates to 2π, not π. Writing the constant term as a₀/2 lets one formula serve every n. On an even function, f(x) sin nx is odd and integrates to 0, so only cosines survive; on an odd function only sines do.

Example 1: a step and its value at the jump

Find the series of f(x) = 0 on (−π, 0) and f(x) = 1 on (0, π), repeated with period 2π.

1. a₀ = (1/π)∫₀^π 1 dx = 1f is 0 on the left half, so only (0, π) counts.
2. a_n = (1/π)∫₀^π cos nx dx = 0sin nx/n is 0 at both ends.
3. b_n = (1 − cos nπ)/(nπ)From ∫₀^π sin nx dx = (1 − cos nπ)/n.
4. b_n = 2/(nπ), n oddcos nπ = −1 for odd n and +1 for even n, so the even terms are 0.
5. f ~ 1/2 + (2/π)(sin xThe constant is a₀/2 = 1/2, the average of f …
6. + sin 3x/3 + sin 5x/5 + ⋯)… plus the odd sines.
7. At x = 0 the series is 1/2Every sine is 0 there, so it lands halfway between the left value 0 and the right value 1. At x = π/2, where f = 1, it gives 1 − 1/3 + 1/5 − ⋯ = π/4.

Example 2: an even function, |sin x|

Find the series of f(x) = |sin x|, the output of a full-wave rectifier.

1. b_n = 0|sin x| is even, so only cosines appear.
2. a_n = (2/π)∫₀^π sin x cos nx dxFor an even function, twice the integral over (0, π), where |sin x| = sin x.
3. a₀ = 4/π∫₀^π sin x dx = 2.
4. a_n = −4/(π(n² − 1)), n evenFrom the product-to-sum identity. For odd n the integral is 0.
5. a₂ = −4/(3π)Then a₄ = −4/(15π) and a₆ = −4/(35π).
6. f ~ 2/π − (4/π)Σ c_k cos 2kxThe sum runs over k = 1, 2, 3, …, with n = 2k.
7. c_k = 1/(4k² − 1)The constant is a₀/2 = 2/π ≈ 0.637, the average of |sin x|. Check at x = 0: Σ 1/(4k² − 1) = 1/2, so the series gives 0.

Example 3: a period other than 2π

Find the series of f(x) = 1 − x² on [−1, 1], repeated with period 2.

1. L = 1The period is 2L = 2. Use cos(nπx) and 1/L = 1 in front.
2. b_n = 0f is even.
3. a₀ = ∫₋₁¹ (1 − x²) dx = 4/3So the constant term is 2/3.
4. a_n = 4(−1)ⁿ⁺¹/(n²π²)Integrate (1 − x²)cos(nπx) by parts twice. a₁ = 4/π².
5. Check at x = 02/3 + (4/π²)(1 − 1/4 + 1/9 − ⋯). The alternating sum is π²/12, so the series gives 2/3 + 1/3 = 1 = f(0).

Example 4: Parseval's identity

Use the series of Example 1 to find 1 + 1/3² + 1/5² + ⋯.

1. (1/π)∫ f² dxParseval: this equals a₀²/2 + Σ (a_n² + b_n²).
2. (1/π)∫₀^π 1 dx = 1f² = f, which is 1 on half the period.
3. a₀²/2 = 1/2With a₀ = 1. Leaving this term out is a common slip.
4. Σ b_n² = (4/π²)Σ 1/n²Over odd n, from b_n = 2/(nπ).
5. 1 = 1/2 + (4/π²)Σ 1/n²Put the pieces together.
6. Σ 1/n² = π²/8Over odd n. Check: the even terms add to (1/4)(π²/6), and π²/8 + π²/24 = π²/6.

Common mistakes, and the exact line they happen on

1. Writing a₀ instead of a₀/2

The constant term of Example 2.

1. a₀ = 4/πRight so far.
2. f ~ 4/π − ⋯This is the mistake. The constant term is a₀/2 = 2/π. With 4/π ≈ 1.27, the series says the average of |sin x| is larger than its maximum, 1. The constant term is always the average of f over a period: check it first.

2. Treating cos nπ as 0

The sine coefficients of Example 1.

1. b_n = (1 − cos nπ)/(nπ)Right so far.
2. b_n = 1/(nπ)This is the mistake. cos nπ is (−1)ⁿ, never 0; sin nπ is the one that is 0. The right coefficients are 2/(nπ) for odd n and 0 for even n. f − 1/2 is odd and also symmetric about x = π/2, which rules out the even sines.

3. Dividing by π when the period is 2L

The constant term of Example 3.

1. ∫₋₁¹ (1 − x²) dx = 4/3Right so far.
2. a₀ = (1/π)(4/3)This is the mistake. The factor in front is 1/L, and L = 1 here. Dividing by π makes the constant term 2/(3π) ≈ 0.212, but f runs from 0 to 1 and averages 2/3. The 1/π only belongs when the period is 2π.

Practice

1. Which coefficients of f(x) = x³ on [−π, π] are zero?

   Show answer

   All the a_n, including a₀: x³ is odd.

2. What is the constant term of the series of f(x) = x² + 1 on [−π, π]?

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   The average of f: π²/3 + 1.

3. Read off the coefficients of f(x) = 2 cos x + sin 3x.

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   a₁ = 2, b₃ = 1, and every other coefficient is 0. A sum of sines and cosines is already its own Fourier series.

4. f is −1 on (−π, 0) and 2 on (0, π). What does its series give at x = 0?

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   The average of the two sides, (−1 + 2)/2 = 1/2.

5. For a function of period 6, which cosines appear, and what factor goes in front of the coefficient integrals?

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   L = 3, so cos(nπx/3), with 1/3 in front of integrals over [−3, 3].