Fourier Series on a General Interval

Fourier Series on a General Interval

FS04 and FS05 derived the complete real Fourier coefficient set on the special interval [−π,π]:

a0	= ∫ −ππf(x) dx,	(1)
an	= ∫ −ππf(x) cos(nx) dx,	(2)
bn	= ∫ −ππf(x) sin(nx) dx.	(3)

Those formulas are not restricted to functions whose natural period happens to be 2π. The interval [−π,π] is only a convenient dimensionless reference interval.

In this article we stretch or compress that reference interval to

(4)

whose total length is

(5)

The main goals are to derive, rather than memorize,

(6)

(7)

and

(8)

The corresponding series is

(9)

The symbol ∼ is used because the precise conditions under which the infinite series converges to f are postponed to FS14–FS16.

1 Why the old basis cos(nx) and sin(nx) must change

On [−π,π], the full interval has length 2π. The functions

(10)

fit an integer number of cycles into that interval.

If the interval changes to [−L,L], the same functions generally no longer fit an integer number of cycles into one interval of length 2L.

For example, cos x has period 2π. If L = 5, then the interval length is 10, not 2π. There is no reason for cos x to return to the same phase after moving from one copy of the interval to the next.

So the harmonic frequencies must scale with L.

2 Deriving the allowed spatial frequencies from periodicity

Consider a cosine of the form

(11)

where k is a spatial angular frequency, usually called a Wavenumber.

To repeat after the full interval length 2L, require

(12)

for every x.

A cosine returns to the same phase when its argument increases by an integer multiple of 2π. Therefore require

(13)

Divide both sides by 2L:

(14)

The smallest positive wavenumber is obtained for n = 1:

(15)

Here k₀ denotes the fundamental wavenumber. Then

(16)

The corresponding basis functions are

(17)

and

(18)

3 A second derivation by rescaling the coordinate

There is another way to reach the same basis that makes the connection to [−π,π] explicit.

Define the dimensionless variable

(19)

When

(20)

we obtain

(21)

When

(22)

we obtain

(23)

Thus the map

(24)

takes [−L,L] into [−π,π].

Solve Equation (19) for x:

(25)

Differentiate:

(26)

The old basis on the reference interval is

(27)

Substitute

(28)

Then

(29)

and

(30)

So the periodicity derivation and the coordinate-rescaling derivation agree.

4 Orthogonality must also rescale

The coefficient formulas depend on orthogonality, so we must determine the new inner products on [−L,L].

Consider two cosine harmonics:

(31)

Use

(32)

The limits become −π and π, so

Imn(c)

= ∫ −ππ cos(n𝜃) cos(m𝜃) d𝜃.

(33)

FS03 gave

(34)

Therefore

Imn(c)	= (πδmn)	(35)
	= Lδmn.	(36)

Hence

(37)

Exactly the same substitution gives

(38)

and

(39)

5 Norms of the scaled basis functions

Set m = n in Equation (37):

(40)

Therefore

(41)

Similarly,

(42)

The constant function has norm squared

(43)

Hence

(44)

This already predicts the coefficient normalization:

• nonconstant sine and cosine basis functions have norm squared L, so their projection coefficients will contain 1∕L;
• the constant basis has norm squared 2L, so the actual constant projection will contain 1∕(2L).

6 Start with a finite scaled trigonometric expansion

As in FS04 and FS05, begin with a finite expansion so no infinite-series convergence theorem is needed:

(45)

We now extract the coefficients using the same projection operations as before.

7 Derivation of the cosine coefficient on [−L,L]

Choose a positive integer m and multiply Equation (45) by

(46)

Then integrate from −L to L:

	∫ −LLf(x) cos dx	(47)
	= ∫ −LL cos dx	(48)
	+ ∑ n=1Na n ∫ −LL cos cos dx	(49)
	+ ∑ n=1Nb n ∫ −LL sin cos dx.	(50)

The first term vanishes because the nonconstant cosine harmonic has zero mean over one complete period.

The sine-cosine sum vanishes by mixed orthogonality.

Every cosine-cosine term vanishes except the one with n = m.

For that surviving term,

(51)

Thus Equation (50) reduces to

(52)

Divide by L:

(53)

This is Equation (7).

8 Derivation of the sine coefficient on [−L,L]

Now multiply Equation (45) by

(54)

and integrate from −L to L.

The constant term vanishes.

Every cosine-sine term vanishes.

Every sine-sine term with n≠m vanishes.

The surviving equal-harmonic term has norm squared L:

(55)

Therefore

(56)

Divide by L:

(57)

This is Equation (8).

9 Derivation of the constant coefficient

Integrate Equation (45) directly over [−L,L]:

∫ −LLf(x) dx	= ∫ −LL1 dx	(58)
	+ ∑ n=1Na n ∫ −LL cos dx	(59)
	+ ∑ n=1Nb n ∫ −LL sin dx.	(60)

Every nonconstant harmonic integrates to zero over the full period, leaving

(61)

Therefore

(62)

and

(63)

The actual constant term is

(64)

Equation (64) is exactly the mean value of f over an interval of total length 2L.

10 The complete Fourier series on [−L,L]

Collecting the results,

(65)

with

(66)

(67)

and

(68)

The formulas have exactly the same structure as the [−π,π] formulas. Only the scale of the basis and the basis norms have changed.

11 Projection interpretation

Define

(69)

Its squared norm is

(70)

Therefore its projection coefficient is

cn	=	(71)
	= ∫ −LLf(x) cos dx	(72)
	= an.	(73)

The same calculation gives b_n for the sine basis.

For the constant basis,

(74)

Thus all three formulas are one projection rule.

12 Normalized Fourier basis on [−L,L]

The normalized constant basis function is

(75)

For n ≥ 1, define

(76)

and

(77)

These satisfy unit norms and mutual orthogonality.

When normalized basis functions are used, the projection coordinates no longer need a denominator:

(78)

and

(79)

The numerical coordinate changes, but the projected function does not.

13 Parity shortcuts survive the rescaling

Because [−L,L] is symmetric about zero, all parity arguments from FS05 remain valid.

If f is even, then

(80)

for every n, and

(81)

Also,

(82)

If f is odd, then

(83)

and

(84)

14 Example 1: the odd ramp f(x) = x on [−L,L]

Because f(x) = x is odd,

(85)

Only b_n remains:

(86)

Let

(87)

Then

(88)

Use integration by parts:

(89)

Then

(90)

Therefore

∫ 0Lx sin(k nx) dx

= 0L + ∫ 0L cos(k nx) dx.

(91)

The final integral is

(92)

because

(93)

Hence

∫ 0Lx sin(k nx) dx	= −	(94)
	= −	(95)
	= −.	(96)

Therefore

bn	=	(97)
	= .	(98)

Thus

(99)

Setting L = π reproduces the FS05 result

(100)

15 Example 2: f(x) = x² on [−L,L]

The function is even, so

(101)

The constant coefficient is

a0	= ∫ 0Lx2 dx	(102)
	= 0L	(103)
	= .	(104)

Therefore the mean term is

(105)

For n ≥ 1,

(106)

The same two integrations by parts used in FS04 give

(107)

Therefore

(108)

The coefficient scale is proportional to L², exactly as dimensional reasoning predicts because x² has dimensions of length squared.

16 Example 3: a centered pulse on [−L,L]

Let

(109)

This function is even, so

(110)

The constant coefficient is

a0	= ∫ −LLf(x) dx	(111)
	= (2Ad)	(112)
	= .	(113)

Therefore the mean is

(114)

For n ≥ 1,

an	= ∫ 0dA cos dx	(115)
	= 0d	(116)
	= sin .	(117)

The coefficient pattern depends only on the dimensionless width ratio

(118)

17 Recovering the [−π,π] formulas as a special case

Set

(119)

Then

(120)

Also,

(121)

Therefore the general formulas reduce immediately to

(122)

and

(123)

So FS04 and FS05 are exactly the special case L = π of the present derivation.

18 Time-domain notation: period T

For a time-domain problem, the centered interval is naturally written

(124)

where T is the temporal period.

The general-interval formulas are transferred to time by making the mathematical replacements

(125)

This is a replacement of interval notation, not an equation between a physical length and a physical time.

The fundamental angular frequency is therefore

ω0	=	(126)
	= .	(127)

The harmonic phase becomes

	=	(128)
	= nω0t.	(129)

For a T-periodic time signal centered on [−T∕2,T∕2],

(130)

with

(131)

(132)

and

(133)

The actual DC term is

(134)

19 Spatial wavenumber versus temporal frequency

The same mathematics appears in two common notations.

For a spatially periodic function with period 2L,

(135)

and

(136)

For a time-periodic function with period T,

(137)

and

(138)

The basis functions are therefore written as either

(139)

or

(140)

20 Dimensional check

If x has dimensions of length, then

(141)

is dimensionless because x∕L is dimensionless.

The trigonometric functions therefore receive dimensionless arguments as required.

The coefficient formula

(142)

also has the same dimensions as f:

(143)

The same is true for b_n and a₀.

21 Why the interval must cover one complete period

The formulas above use [−L,L] because that interval has length 2L, which is one complete period of the chosen basis.

If one integrates over a shorter interval without changing the basis, the cancellation relations can fail.

The important principle is not the location of the endpoints by itself. The important principle is that the integration interval span a complete period over which the chosen harmonics are mutually orthogonal.

Later articles will use intervals such as

(144)

or

(145)

The basis can be shifted accordingly. The present centered interval is especially convenient because even/odd parity shortcuts remain available.

22 A practical coefficient workflow on [−L,L]

For a new periodic function on [−L,L]:

1. Identify the half-period L.
2. Write the fundamental wavenumber

   (146)

3. Write the basis

   (147)

4. Check parity before integrating.
5. Compute the mean term

   (148)

6. Compute surviving cosine coefficients.
7. Compute surviving sine coefficients.
8. Check dimensions.
9. Set L = π as a quick consistency test when the example can be compared with an earlier result.

23 Common mistakes

1. Using cos(nx) and sin(nx) on an arbitrary interval without rescaling the argument.
2. Writing the fundamental wavenumber as 2π∕L instead of π∕L when the full period is 2L.
3. Forgetting that

   (149)

4. Using 1∕π in the coefficient formula after the interval has changed.
5. Using 1∕(2L) for a_n or b_n. The 1∕(2L) factor belongs to the actual constant projection a₀∕2.
6. Forgetting that the equal-harmonic sine and cosine norm squared is L, not 2L.
7. Forgetting to rescale dx when using

   (150)

8. Treating k₀ = π∕L as dimensionless when x carries units. A spatial wavenumber has units of inverse length.
9. Confusing the harmonic number n with the physical wavenumber k_n.
10. Assuming convergence at discontinuities before the convergence theory of FS14–FS16 has been developed.

24 Practice exercises

1. Map [−3, 3] to [−π,π] by defining an appropriate dimensionless variable 𝜃.
2. For a function of period 10 m in space, find L, k₀, k₂, and k₅.
3. Prove by substitution that

   (151)

4. Evaluate

   (152)

5. Derive the normalized constant basis function on [−L,L].
6. Derive the normalized third cosine harmonic on [−L,L].
7. For f(x) = x on [−2, 2], calculate b₁, b₂, and b₃.
8. For f(x) = x² on [−2, 2], calculate a ₀, a₁, and a₂.
9. For the centered pulse in Equation (109), set A = 3, L = 4, and d = 1. Find the mean and the first four cosine coefficients.
10. Show explicitly that the general ramp coefficient reduces to the FS05 result when L = π.
11. A time signal has period T = 8 s. Find ω₀, ω₃, and the coefficient prefactor appearing in the centered-period formulas.
12. Explain the difference between the statements “the interval is [−L,L]” and “the period is 2L.”

25 What FS06 has added

The Fourier formulas from FS04 and FS05 were not special formulas tied to the number π. They were projection formulas written on a convenient reference interval.

The rescaling

(153)

shows how the entire construction moves to any centered interval of length 2L.

The basis changes from

(154)

to

(155)

while the nonconstant basis norm squared changes from π to L.

That one scaling change explains the coefficient prefactor 1∕L.

26 Summary

For a function with period 2L, the fundamental wavenumber is

(156)

The harmonic basis is

(157)

On [−L,L],

(158)

(159)

and mixed sine-cosine inner products vanish.

Therefore

(160)

(161)

and

(162)

The actual mean or DC term is

(163)

For a time signal centered on [−T∕2,T∕2], replace the mathematical half-interval L by T∕2. The same structure is then written using

(164)

FS07 will now apply these formulas to the first complete nonsinusoidal waveform: the square wave.

References