Fourier Cosine-Coefficient Derivation and Projection Exercises

FS04E1: Fourier Cosine-Coefficient Derivation and Projection Exercises

FS04 derived the cosine coefficient formula

(1)

and the constant coefficient

(2)

This companion is calculation practice for those results. The emphasis is not on memorizing Equation (1). The emphasis is on repeatedly seeing why it is a projection formula, when symmetry can shorten the integral, how the constant term differs from the nonzero harmonics, and how integration by parts produces useful coefficient formulas.

The sine coefficient formula is deliberately not derived here. That belongs to FS05.

1 Three reusable checks

Cosine-coefficient extraction check

A cosine-coefficient extraction check verifies that multiplying a finite trigonometric sum by cos(mx) and integrating over [−π,π] leaves only the mth cosine term multiplied by its norm squared π.

DC-coefficient check

A DC-coefficient check verifies the distinction

(3)

versus the actual constant term in the Fourier series,

(4)

Thus a₀∕2 is the mean value of f over the interval.

Cosine-projection residual check

After projecting f onto a cosine harmonic,

(5)

define the residual

(6)

A correct projection satisfies

(7)

2 Problems

All problems are stated before the hints and complete solutions.

Problem 1: isolate one known cosine coefficient

Let

(8)

Evaluate

(9)

by expanding every term. Do not simply read the answer from f(x).

Problem 2: rebuild the general extraction formula

Assume

(10)

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

Problem 3: extract the constant coefficient from a finite sum

For

(11)

compute a₀ by integration and identify the actual DC component a₀∕2.

Problem 4: projection formula versus Fourier formula

Using the FS02 projection formula

(12)

set

(13)

on [−π,π] and derive the Fourier cosine coefficient formula directly.

Problem 5: residual orthogonality

Let

(14)

and suppose that

(15)

Define

(16)

Verify directly that

(17)

Problem 6: even-function shortcut

Suppose f is even. Show that

(18)

Explain why the shortcut works.

Problem 7: odd functions have zero cosine coefficients

Suppose f is odd. Show, without using the sine coefficient formula, that

(19)

for every n ≥ 0.

Problem 8: the DC coefficient of x²

For

(20)

compute a₀ and the mean value a₀∕2.

Problem 9: cosine coefficients of x²

Starting from

(21)

perform integration by parts twice and derive

(22)

Show every boundary term.

Problem 10: first cosine projection of x²

Using Problems 8 and 9, write the constant-plus-first-cosine approximation

(23)

Evaluate it at x = 0 and x = π and compare with x².

Problem 11: cosine coefficients of |x|

For

(24)

compute a₀ and derive a_n by integration by parts. Simplify the result separately for even and odd n.

Problem 12: centered rectangular pulse

Let

(25)

Compute a₀ and a_n. Evaluate a₁,a₂,a₃,a₄,a₅.

Problem 13: centered pulse of arbitrary half-width

Generalize Problem 12 to

(26)

Derive a₀ and a_n in terms of A and α.

Problem 14: triangular even function

Let

(27)

Compute a₀ and derive the cosine coefficients. Identify which harmonics vanish.

Problem 15: general interval [−L,L]

For the basis

(28)

use the projection formula to derive

(29)

Also derive the corresponding a₀ formula.

Problem 16: normalized-basis coefficient

Define

(30)

If

(31)

show that

(32)

Explain why c_n and a_n are different numbers even though they describe the same projected function.

3 Hints

1. Expand the four terms and use the FS03 orthogonality relations one at a time.
2. Keep the sum finite. The cosine-cosine integral is π only when n = m.
3. Every nonconstant integer sine or cosine integrates to zero over [−π,π].
4. Compute ⟨cos(mx), cos(mx)⟩ explicitly.
5. Use ∫ x² cos xdx from the n = 1 case of Problem 9, or derive it directly twice by parts.
6. The product of two even functions is even.
7. Odd times even is odd.
8. Use evenness before integrating.
9. First choose u = x² and dv = cos(nx)dx; then integrate the remaining x sin(nx) term by parts.
10. From Problem 9, a₁ = −4.
11. |x| is even, so work only on [0,π].
12. Only the interval [−π∕2,π∕2] contributes.
13. Replace π∕2 by α and include amplitude A.
14. Write π −|x| as π − x on [0,π].
15. The norm squared of each nonzero cosine basis function on [−L,L] is L.
16. The normalized basis has norm one, so its coordinate is the inner product itself.

4 Worked solutions

Solution 1: isolate one known cosine coefficient

Begin with

I	= ∫ −ππ cos 3xdx	(33)
	=	(34)
	.	(35)

The first three integrals are zero by orthogonality. The final integral is

(36)

Therefore

(37)

The operation has isolated the coefficient of cos 3x.

Solution 2: rebuild the general extraction formula

Multiply the finite expansion by cos(mx):

f(x) cos(mx)	= cos(mx)	(38)
	+ ∑ n=1Na n cos(nx) cos(mx)	(39)
	+ ∑ n=1Nb n sin(nx) cos(mx).	(40)

Integrate over [−π,π]:

∫ −ππf(x) cos(mx) dx	= ∫ −ππ cos(mx) dx	(41)
	+ ∑ n=1Na n ∫ −ππ cos(nx) cos(mx) dx	(42)
	+ ∑ n=1Nb n ∫ −ππ sin(nx) cos(mx) dx.	(43)

The constant-cosine integral vanishes. Every sine-cosine integral vanishes. Every cosine-cosine integral with n≠m vanishes. Only the n = m term remains:

(44)

Divide by π:

(45)

Solution 3: extract the constant coefficient

Integrate the function:

∫ −ππf(x) dx	= ∫ −ππ dx	(46)
	= 12π.	(47)

Therefore

(48)

The actual constant term is

(49)

That agrees with the constant visible in the original finite sum.

Solution 4: projection formula versus Fourier formula

Set

(50)

The projection coefficient is

(51)

The numerator is

(52)

The denominator is

(53)

Therefore

(54)

Thus c = a_m.

Solution 5: residual orthogonality

We must evaluate

(55)

Split the integral:

(56)

The second integral is

(57)

For the first integral, evenness gives

(58)

Integrate by parts twice:

∫ 0πx2 cos xdx	= 0π − 2 ∫ 0πx sin xdx	(59)
	= −2	(60)
	= −2π.	(61)

Hence

(62)

Therefore the residual inner product is

(63)

Solution 6: even-function shortcut

If f is even, then cos(nx) is also even. Their product is even:

(64)

Therefore

(65)

Substitute into the coefficient formula:

(66)

Solution 7: odd functions have zero cosine coefficients

If f is odd and cos(nx) is even, then

(67)

is odd. Hence

(68)

for n ≥ 1, so

(69)

For n = 0 the integrand is simply f(x), which is odd, so

(70)

Solution 8: the DC coefficient of x²

Because x² is even,

a0	= ∫ 0πx2 dx	(71)
	= 0π	(72)
	= .	(73)

Therefore the mean value is

(74)

Solution 9: cosine coefficients of x²

Start with

(75)

Let

(76)

Then

(77)

Thus

∫ 0πx2 cos(nx) dx	= 0π −∫ 0πx sin(nx) dx	(78)
	= −∫ 0πx sin(nx) dx,	(79)

because sin(nπ) = 0.

For the remaining integral, choose

(80)

so

(81)

Then

∫ 0πx sin(nx) dx	= 0π + ∫ 0π cos(nx) dx	(82)
	= −	(83)
	= −.	(84)

Therefore

(85)

Finally,

(86)

Solution 10: first cosine projection of x²

From Problems 8 and 9,

(87)

Therefore

(88)

At x = 0,

(89)

whereas

(90)

At x = π,

(91)

while

(92)

One harmonic is only a low-order projection, so pointwise agreement is not expected to be exact.

Solution 11: cosine coefficients of |x|

Because |x| is even,

(93)

Thus

(94)

For n ≥ 1,

(95)

Integrate by parts:

∫ 0πx cos(nx) dx	= 0π −∫ 0π sin(nx) dx	(96)
	= .	(97)

Therefore

(98)

If n is even,

(99)

If n is odd,

(100)

Solution 12: centered rectangular pulse

The pulse is even. Its constant coefficient is

a0	= ∫ 0π∕21 dx	(101)
	= 1.	(102)

Thus its mean value is 1∕2.

For n ≥ 1,

an	= ∫ 0π∕2 cos(nx) dx	(103)
	= sin .	(104)

Hence

a1	= ,	(105)
a2	= 0,	(106)
a3	= −,	(107)
a4	= 0,	(108)
a5	= .	(109)

Solution 13: centered pulse of arbitrary half-width

Again use evenness. The constant coefficient is

a0	= ∫ 0αAdx	(110)
	= .	(111)

Therefore the mean value is

(112)

For n ≥ 1,

an	= ∫ 0α cos(nx) dx	(113)
	= sin(nα).	(114)

Problem 12 follows by setting

(115)

Solution 14: triangular even function

On [0,π],

(116)

The constant coefficient is

a0	= ∫ 0π(π − x)dx	(117)
	= π.	(118)

Hence the mean is π∕2.

For n ≥ 1,

an

= ∫ 0π(π − x) cos(nx)dx.

(119)

The π cos(nx) term integrates to zero, so

(120)

From Problem 11,

(121)

Thus

(122)

For even n,

(123)

For odd n,

(124)

Solution 15: general interval [−L,L]

Take

(125)

The projection formula gives

(126)

FS03 gives

(127)

for n ≥ 1. Therefore

(128)

For the constant basis function,

(129)

The actual constant projection is therefore

(130)

Since the Fourier convention writes this constant as a₀∕2,

(131)

Solution 16: normalized-basis coefficient

The normalized cosine basis function is

(132)

Its coordinate is

cn	=	(133)
	= ∫ −ππf(x) cos(nx)dx.	(134)

But

(135)

Therefore

(136)

The two numerical coefficients differ because they multiply differently scaled basis functions:

an cos(nx)	=	(137)
	= cnϕn(x).	(138)

The projected function is the same.

5 Additional practice

1. For f(x) = 2−3 cos 2x+8 sin 5x+6 cos 6x, isolate the coefficient of cos 6x by integration.
2. Compute a₀ for f(x) = 5 + cos 7x − 4 sin 2x.
3. Show directly that all cosine coefficients of f(x) = x³ vanish on [−π,π].
4. Compute a₀ and a₁ for f(x) = 1 + x².
5. Compute a₂ and a₃ for f(x) = x² using the closed formula from Problem 9.
6. For f(x) = |x|, evaluate a₁,a₂,a₃,a₄.
7. For a centered pulse with A = 3 and α = π∕3, compute a₀,a₁,a₂,a₃.
8. For f(x) = π −|x|, write the constant plus first three nonzero cosine terms.
9. On [−2, 2], derive the coefficient multiplying cos(nπx∕2).
10. Explain why replacing cos(nx) by 5 cos(nx) changes the projection coefficient but not the projected function.

6 What these exercises establish

The main lessons are:

• A Fourier cosine coefficient is an ordinary function-space projection coefficient.
• Orthogonality performs the harmonic isolation.
• The factor 1∕π is the reciprocal norm squared of cos(nx) on [−π,π].
• The quantity a₀∕2, not a₀, is the mean or DC component.
• Evenness can reduce symmetric-interval integrals to twice a half-interval integral.
• Odd functions have no constant or cosine content on a symmetric interval.
• Integration by parts converts polynomial-times-cosine integrals into boundary terms that reveal coefficient decay and sign patterns.
• Piecewise even functions often produce transparent spectral selection rules such as vanishing even harmonics.

References