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Mathematics · Advanced university

Sine coefficient from orthogonality

This uses the real Fourier convention on [−π,π]. Distinct sine modes are orthogonal.

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Method and conditions

This uses the real Fourier convention on [−π,π]. Distinct sine modes are orthogonal.

bₙ = (1/π)∫₋π^π f(x)sin(nx) dx

Worked examples

For f(x) = (-15)sin(2x) + 5sin(5x) on [−π,π], find the Fourier sine coefficient b_2, using b_j=(1/π) integral of f(x)sin(jx). Round the final result to 4 decimal places if needed.

Cross terms vanish by orthogonality. The integral of sin²(2x) is π, leaving coefficient -15.

-15

For f(x) = (10)sin(9x) + 5sin(13x) on [−π,π], find the Fourier sine coefficient b_9, using b_j=(1/π) integral of f(x)sin(jx). Round the final result to 4 decimal places if needed.

Cross terms vanish by orthogonality. The integral of sin²(9x) is π, leaving coefficient 10.

10

Try it

  1. For f(x) = (18)sin(9x) + 5sin(17x) on [−π,π], find the Fourier sine coefficient b_9, using b_j=(1/π) integral of f(x)sin(jx). Round the final result to 4 decimal places if needed.

  2. For f(x) = (9)sin(9x) + 5sin(14x) on [−π,π], find the Fourier sine coefficient b_9, using b_j=(1/π) integral of f(x)sin(jx). Round the final result to 4 decimal places if needed.

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Answers and marking notes
  1. 18

    Cross terms vanish by orthogonality. The integral of sin²(9x) is π, leaving coefficient 18.

  2. 9

    Cross terms vanish by orthogonality. The integral of sin²(9x) is π, leaving coefficient 9.

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