Fourier Sine and Constant Coefficient Derivation Exercises

FS05E1: Fourier Sine- and Constant-Coefficient Derivation Exercises

FS05 completed the real Fourier coefficient set on [−π,π]:

(1)

(2)

and

(3)

This companion concentrates on the two pieces introduced most directly in FS05: the sine coefficients b_n and the constant projection a₀∕2. The goal is to make the coefficient formulas behave like projection tools rather than formulas to memorize.

The interval remains [−π,π] throughout. Rescaling to [−L,L] belongs to FS06.

1 Three reusable checks

Sine-coefficient extraction check

A sine-coefficient extraction check verifies that multiplication by sin(mx) followed by integration over a complete period removes the constant term, all cosine terms, and all sine harmonics except the mth one.

For a finite trigonometric polynomial,

(4)

the check should end with

(5)

Constant-projection mean check

A constant-projection mean check verifies both

(6)

and

(7)

The second expression is the mean value of f over one complete period.

Sine-projection residual check

After projecting f onto sin(mx),

(8)

define

(9)

A correct projection satisfies

(10)

2 Problems

All problems are stated before the hints and complete worked solutions.

Problem 1: isolate one sine coefficient from a known finite sum

Let

(11)

Evaluate

(12)

by expanding every contribution. Do not simply read the answer from the given function.

Problem 2: rebuild the general sine-coefficient formula

Assume

(13)

Multiply by sin(mx), integrate over [−π,π], and derive Equation (3) for 1 ≤ m ≤ N.

Problem 3: recover the constant coefficient and the mean

For

(14)

compute a₀ by integration. Then identify the actual constant term a₀∕2 and the mean value of f over one period.

Problem 4: sine projection from the general projection formula

Use

(15)

with

(16)

on [−π,π] to derive Equation (3) directly.

Problem 5: residual orthogonality for the ramp

For

(17)

the first sine coefficient is

(18)

Define

(19)

Verify directly that

(20)

Problem 6: odd-function shortcut for b_n

Suppose f is odd. Starting from Equation (3), prove

(21)

Problem 7: even functions have no sine coefficients

Suppose f is even. Prove that

(22)

for every positive integer n.

Problem 8: the odd ramp f(x) = x

For

(23)

find a₀, all a_n, and all b_n. Carry out the integration by parts explicitly and show that

(24)

Write the resulting real Fourier series.

Problem 9: odd square wave

Define

(25)

Find a₀, all a_n, and all b_n. Show explicitly why only odd sine harmonics survive.

Problem 10: a DC offset plus an odd ramp

Let

(26)

Determine a₀, a_n, and b_n. Explain why adding the constant 2 changes only the constant projection and leaves all nonzero harmonic coefficients unchanged from the ramp example.

Problem 11: finite-width odd pulse

For 0 < α < π, define

(27)

Find all Fourier coefficients. Show that

(28)

Problem 12: a two-level waveform with nonzero mean

Define

(29)

Find the constant projection and all sine coefficients. Interpret the result by writing f as a constant plus an odd square wave.

Problem 13: normalized sine basis

Define

(30)

Show that ∥ϕ_n∥ = 1. If

(31)

show that

(32)

Then verify that

(33)

Problem 14: energy bookkeeping for a finite orthogonal sum

Let

(34)

Compute

(35)

using orthogonality rather than expanding the entire square. Then verify the result by identifying the squared norms of the constant and sine components.

Problem 15: direct coefficient check for a mixed finite sum

For

(36)

compute a₀, b₃, and b₅ using their integral formulas. Explain why the cosine term does not contaminate either sine coefficient.

Problem 16: derive the complete coefficient bookkeeping for a shifted square wave

Let

(37)

Find a₀, all a_n, and all b_n. Then rewrite the function as

(38)

where q(x) is the odd square wave from Problem 9, and show that this decomposition predicts the same coefficients immediately.

3 Hints

1. Expand the integral into five terms. Only the sin 4x sin 4x term has a nonzero full-period integral.
2. Use the three orthogonality relations from FS03. Keep the n = m sine-sine term separate.
3. Every nonconstant sine or cosine harmonic integrates to zero over [−π,π].
4. Evaluate ⟨sin(mx), sin(mx)⟩ before simplifying the projection formula.
5. Evaluate ∫ x sin xdx by parts, then subtract 2 ∫ sin ²xdx.
6. If f and sin(nx) are both odd, what is the parity of their product?
7. An even function times an odd function is odd.
8. Use odd symmetry first. For the remaining integral choose u = x and dv = sin(nx) dx.
9. Use odd symmetry. Then evaluate ∫ ₀^π sin(nx) dx.
10. Split f = 2 + x before integrating.
11. Use odd symmetry and note that the function vanishes outside [0,α] on the positive half interval.
12. First calculate the mean. Then write the waveform as 2+q(x) where q is an odd square wave of amplitude 1.
13. The norm squared of sin(nx) is π.
14. Orthogonal components contribute additively to the squared norm.
15. Apply the coefficient formulas directly. The mixed sine-cosine inner products vanish.
16. The average of the two levels is 3 and half their difference is 2.

4 Complete worked solutions

Solution 1: isolate one sine coefficient from a known finite sum

We must evaluate

∫ −ππf(x) sin 4xdx

= ∫ −ππ sin 4xdx.

(39)

Distribute the integral:

	∫ −ππ sin 4xdx + ∫ −ππ cos x sin 4xdx	(40)
	−∫ −ππ sin 2x sin 4xdx + ∫ −ππ cos 3x sin 4xdx	(41)
	+ ∫ −ππ sin 24xdx.	(42)

The first integral vanishes because a sine harmonic has zero full-period average. The second and fourth vanish by sine-cosine orthogonality. The third vanishes because the two sine harmonics have different indices. The last integral is

(43)

Therefore

(44)

The extraction procedure recovers the coefficient multiplying sin 4x.

Solution 2: rebuild the general sine-coefficient formula

Start with

(45)

Multiply by sin(mx):

f(x) sin(mx)	= sin(mx)	(46)
	+ ∑ n=1Na n cos(nx) sin(mx)	(47)
	+ ∑ n=1Nb n sin(nx) sin(mx).	(48)

Integrate from −π to π:

∫ −ππf(x) sin(mx) dx	= ∫ −ππ sin(mx) dx	(49)
	+ ∑ n=1Na n ∫ −ππ cos(nx) sin(mx) dx	(50)
	+ ∑ n=1Nb n ∫ −ππ sin(nx) sin(mx) dx.	(51)

Now apply orthogonality.

The constant-sine integral is zero:

(52)

Every cosine-sine inner product is zero:

(53)

For the sine-sine terms,

(54)

Therefore every sine term disappears except n = m:

(55)

Divide by π:

(56)

Solution 3: recover the constant coefficient and the mean

Use

(57)

Thus

a0	= ∫ −ππ dx	(58)
	=	(59)
	= 12.	(60)

The Fourier series contains the constant term

(61)

The mean is

∫ −ππf(x) dx	= (12π)	(62)
	= 6.	(63)

So the constant term and the mean agree, as they must.

Solution 4: sine projection from the general projection formula

The projection coefficient of f onto ϕ is

(64)

Set

(65)

The numerator is

(66)

The denominator is

⟨sin(mx), sin(mx)⟩	= ∫ −ππ sin 2(mx) dx	(67)
	= π.	(68)

Therefore

(69)

This is exactly b_m.

Solution 5: residual orthogonality for the ramp

We need

(70)

Separate the terms:

(71)

For the first integral, x sin x is even, so

(72)

Integrate by parts with

(73)

so

(74)

Then

∫ 0πx sin xdx	= 0π + ∫ 0π cos xdx	(75)
	= π.	(76)

Thus

(77)

Also,

(78)

Therefore

∫ −ππr 1(x) sin xdx	= 2π − 2π	(79)
	= 0.	(80)

The residual is orthogonal to the basis function used in the projection.

Solution 6: odd-function shortcut for b_n

If f is odd, then sin(nx) is also odd. The product of two odd functions is even:

(81)

Therefore

bn	= ∫ −ππf(x) sin(nx) dx	(82)
	= ∫ 0πf(x) sin(nx) dx.	(83)

This is the odd-function shortcut.

Solution 7: even functions have no sine coefficients

If f is even and sin(nx) is odd, their product is odd:

(84)

An odd integrand has zero integral over a symmetric interval:

(85)

Hence

(86)

Solution 8: the odd ramp f(x) = x

Because x is odd,

(87)

For the sine coefficient,

bn

= ∫ 0πx sin(nx) dx.

(88)

Use integration by parts:

(89)

so

(90)

Then

∫ 0πx sin(nx) dx	= 0π + ∫ 0π cos(nx) dx	(91)
	= − + 0π	(92)
	= −.	(93)

Therefore

bn	=	(94)
	= .	(95)

Thus

(96)

The convergence symbol reminds us that pointwise convergence details are treated later in the series.

Solution 9: odd square wave

The square wave is odd, so

(97)

Use the odd-function shortcut:

bn	= ∫ 0π sin(nx) dx	(98)
	= 0π	(99)
	=	(100)
	= .	(101)

If n is even,

(102)

so

(103)

If n is odd,

(104)

so

(105)

Hence

(106)

Solution 10: a DC offset plus an odd ramp

Write

(107)

The constant part contributes only to a₀.

First,

a0	= ∫ −ππ(2 + x) dx	(108)
	=	(109)
	= 4.	(110)

Thus

(111)

For n ≥ 1,

an

= ∫ −ππ(2 + x) cos(nx) dx.

(112)

The constant-cosine integral is zero, and x cos(nx) is odd. Therefore

(113)

For the sine coefficient,

bn

= ∫ −ππ(2 + x) sin(nx) dx.

(114)

The constant-sine integral is zero. The remaining integral is exactly the ramp coefficient from Problem 8:

(115)

Thus

(116)

Adding a constant shifts the mean but does not change the nonzero harmonic content.

Solution 11: finite-width odd pulse

The pulse is odd, so

(117)

Use the odd-function shortcut. On the positive half interval the function equals A only from 0 to α:

bn	= ∫ 0πf(x) sin(nx) dx	(118)
	= ∫ 0α sin(nx) dx	(119)
	= 0α	(120)
	= .	(121)

This formula contains the square wave as the special case α = π.

Solution 12: a two-level waveform with nonzero mean

The waveform has level 1 for half the period and level 3 for the other half.

Compute the constant coefficient:

a0	=	(122)
	=	(123)
	= 4.	(124)

Therefore

(125)

Now write

(126)

where

(127)

The constant 2 has no sine coefficients, so f has the same sine coefficients as q:

(128)

Thus

(129)

for even n, and

(130)

for odd n.

Solution 13: normalized sine basis

Compute the norm:

∥ϕn∥2	= ∫ −ππ2dx	(131)
	= ∫ −ππ sin 2(nx) dx	(132)
	= (π)	(133)
	= 1.	(134)

Thus ∥ϕ_n∥ = 1.

The coefficient in the normalized basis is

cn	= ⟨f,ϕn⟩	(135)
	= ∫ −ππf(x)dx	(136)
	= ∫ −ππf(x) sin(nx) dx.	(137)

Since

(138)

we obtain

(139)

Then

cnϕn(x)	=	(140)
	= bn sin(nx).	(141)

The coordinate changes when the basis is normalized, but the projected function does not.

Solution 14: energy bookkeeping for a finite orthogonal sum

Write

(142)

The three components

(143)

are mutually orthogonal on [−π,π].

Therefore the squared norm is the sum of the squared component norms:

∫ −ππf(x)2 dx	= ∫ −ππ22 dx + 9 ∫ −ππ sin 2xdx + 16 ∫ −ππ sin 22xdx	(144)
	= 4(2π) + 9π + 16π	(145)
	= 33π.	(146)

This is a finite-dimensional version of the energy bookkeeping that later develops into Parseval’s theorem.

Solution 15: direct coefficient check for a mixed finite sum

The function is

(147)

For the constant coefficient,

a0	= ∫ −ππf(x) dx	(148)
	= (−2π)	(149)
	= −2.	(150)

Thus the actual constant term is

(151)

For b₃,

b3

= ∫ −ππf(x) sin 3xdx.

(152)

The constant-sine term vanishes. The cosine-sine term vanishes. The sin 5x sin 3x term vanishes because the harmonic indices differ. Only

(153)

survives, so

(154)

Similarly,

(155)

The cosine term does not contaminate either sine coefficient because cosine and sine harmonics are orthogonal over the full interval.

Solution 16: complete coefficient bookkeeping for a shifted square wave

The waveform is

(156)

The mean of the two half-period levels is

(157)

Therefore we expect

(158)

Verify directly:

a0	=	(159)
	=	(160)
	= 6.	(161)

Now write

(162)

where q(x) is the odd square wave from Problem 9.

The constant 3 contributes no nonzero harmonic coefficients. Because q is odd, it contributes no cosine coefficients. Hence

(163)

The sine coefficients are twice those of the unit-amplitude odd square wave:

bn	= 2	(164)
	= .	(165)

Thus for even n,

(166)

and for odd n,

(167)

So the waveform has the Fourier representation

(168)

This final problem illustrates the clean separation

(169)

5 Further practice

1. For

   (170)

   compute a₀ and b₂ directly from the integral formulas.

2. For f(x) = x, verify explicitly that the second sine-projection residual

   (171)

   is orthogonal to sin 2x.

3. For

   (172)

   explain from parity which Fourier coefficients must vanish before performing any integration.

4. Compute b₁ and b₂ for f(x) = x³.
5. For the finite-width odd pulse of Problem 11, set α = π∕2 and determine which first eight sine coefficients vanish.
6. For

   (173)

   find the complete coefficient set using results already derived in this companion.

7. For

   (174)

   compute ∫ _−π^πf(x)² dx using orthogonality.

8. A waveform equals −2 on (−π, 0) and 6 on (0,π). Decompose it into a mean plus an odd square wave and infer its Fourier coefficients.
9. Let

   (175)

   For f(x) = 5 sin 3x, compute the normalized-basis coordinate and reconstruct the projection.

10. Explain in words why adding a constant to any periodic function can change a₀ but cannot change any a_n or b_n for n ≥ 1.

6 What these exercises establish

The problems above reinforce four structural ideas.

• A sine coefficient is a projection onto one sine basis function.
• The constant Fourier term is the mean value of the function over one period.
• Parity can eliminate entire coefficient families before any integration is performed.
• Orthogonality separates coefficient extraction into independent one-dimensional projection problems.

The coefficient set on [−π,π] is now complete. The next article changes the interval itself and asks how the basis frequencies and normalization constants must change.

References