Orthogonality of Sines and Cosines

FS03: Orthogonality of Sines and Cosines

FS02 introduced the function inner product

(1)

and defined two functions to be orthogonal when that inner product is zero.

Fourier series become practical because different integer harmonics of sine and cosine are orthogonal over a complete fundamental period.

The purpose of this lecture is to derive that statement from elementary trigonometry and integration. Nothing will be taken on faith.

The central results on [−π,π] are

(2)

(3)

and

(4)

for positive integers m and n.

We will derive each formula separately. We will then extend the result to the interval [−L,L] and explain exactly why this orthogonality allows a Fourier coefficient to be isolated by multiplication and integration.

1 Formal definitions owned by FS03

Trigonometric orthogonality

Two trigonometric functions are orthogonal on an interval when their function inner product on that interval vanishes. For example,

(5)

are orthogonal on [−π,π] for distinct positive integers m and n because

(6)

Harmonic orthogonality

Harmonic orthogonality is the special orthogonality among distinct integer harmonics of a common fundamental frequency when the inner product is taken over a complete fundamental period.

If the fundamental angular frequency is ω₀, the functions

(7)

for integer n form the harmonic family. Distinct members are orthogonal over any complete interval of length

(8)

Full-period cancellation

A periodic sine or cosine with a nonzero integer number of cycles over an integration interval has zero integral over that complete set of cycles. We will call this full-period cancellation.

On [−π,π], for any nonzero integer k,

(9)

and

(10)

Kronecker delta

The Kronecker delta is the discrete symbol

(11)

It lets us write an entire orthogonality table compactly. For example,

(12)

for positive integers m,n.

Trigonometric basis norm

The trigonometric basis norm is the function norm of one basis function under the chosen inner product. On [−π,π],

(13)

for n ≥ 1.

The constant function has a different norm:

(14)

Normalized trigonometric basis

Dividing every nonzero basis function by its norm gives a normalized trigonometric basis. On [−π,π],

(15)

are normalized basis functions. Together they form an orthonormal trigonometric family.

2 Why integer harmonics are the natural periodic family

FS01 established that a periodic function with fundamental period 2π naturally uses integer harmonics

(16)

The function cos(nx) completes exactly n cycles while x runs through any interval of length 2π. The same is true of sin(nx).

This integer-cycle property is the geometric source of the cancellation that follows.

3 The elementary cancellation lemma

Before multiplying two harmonics together, we need one simple integral result.

Let k be a nonzero integer. Consider

(17)

Integrate directly:

Ic(k)	= −ππ	(18)
	= −.	(19)

For integer k,

(20)

and

(21)

Therefore

(22)

Now consider

(23)

One direct method is integration:

Is(k)	= −ππ	(24)
	= − + .	(25)

Because cosine is even,

(26)

so the two terms cancel:

(27)

There is also a symmetry argument. The function sin(kx) is odd, so its integral over the symmetric interval [−π,π] vanishes automatically.

These two equations are the cancellation engine used throughout the rest of the lecture.

4 Deriving the product-to-sum identities

Orthogonality integrals contain products such as

(28)

Products are inconvenient to integrate directly. The product-to-sum identities turn them into sums of single sinusoids, which can then use the cancellation lemma.

We derive the identities rather than quote them.

4.1 Cosine times cosine

Start from the angle-addition formulas

(29)

and

(30)

Add them:

cos(A − B) + cos(A + B)

= 2 cos A cos B.

(31)

Divide by 2:

(32)

4.2 Sine times sine

Subtract the two cosine addition formulas:

cos(A − B) − cos(A + B)

= 2 sin A sin B.

(33)

Therefore

(34)

4.3 Sine times cosine

Use

(35)

and

(36)

Add them:

sin(A + B) + sin(A − B)

= 2 sin A cos B.

(37)

Hence

(38)

5 Cosine-cosine orthogonality when m≠n

Consider

(39)

where m and n are distinct positive integers.

Use Equation (32) with

(40)

Then

(41)

Insert this into the integral:

Icc	= ∫ −ππ cos((n − m)x) dx	(42)
	+ ∫ −ππ cos((n + m)x) dx.	(43)

Because n≠m,

(44)

Because both are positive,

(45)

Both n − m and n + m are nonzero integers. Therefore Equation (22) applies to both integrals:

(46)

and

(47)

Hence

(48)

Thus distinct cosine harmonics are orthogonal.

6 Cosine-cosine self-inner product when m = n

The previous derivation cannot simply substitute m = n and say both terms vanish. One of the product-to-sum frequencies becomes zero. That case must be handled separately.

Set m = n:

(49)

From Equation (32), with A = B = nx,

cos 2(nx)	=	(50)
	= .	(51)

Therefore

Icc(n)

= ∫ −ππ1 dx + ∫ −ππ cos(2nx) dx.

(52)

The second integral vanishes by full-period cancellation. The first is just the interval length:

(53)

Hence

Icc(n)	= (2π)	(54)
	= π.	(55)

Thus

(56)

This is not orthogonality to itself. It is the squared norm of the basis function.

Combining the two cases gives Equation (2).

7 Sine-sine orthogonality when m≠n

Now consider

(57)

Use Equation (34):

(58)

Thus

Iss	= ∫ −ππ cos((n − m)x) dx	(59)
	− ∫ −ππ cos((n + m)x) dx.	(60)

If m≠n, both frequencies are nonzero integers. Each integral vanishes. Therefore

(61)

Distinct sine harmonics are orthogonal.

8 Sine-sine self-inner product when m = n

Set m = n:

(62)

From Equation (34), with A = B = nx,

sin 2(nx)	=	(63)
	= .	(64)

Then

Iss(n)	= ∫ −ππ1 dx − ∫ −ππ cos(2nx) dx	(65)
	= (2π) − 0	(66)
	= π.	(67)

Thus

(68)

Combining the cases gives Equation (3).

9 Sine-cosine orthogonality

Now consider

(69)

There are two complementary ways to prove that this integral is zero.

9.1 Method 1: parity

The function

(70)

is odd:

(71)

The function

(72)

is even:

(73)

The product of an odd function and an even function is odd. Therefore

(74)

is odd. The integral of an odd function over a symmetric interval vanishes:

(75)

This proof works even when m = n.

9.2 Method 2: product-to-sum

Use Equation (38):

(76)

Integrate:

Isc	= ∫ −ππ sin((n + m)x) dx	(77)
	+ ∫ −ππ sin((n − m)x) dx.	(78)

If n≠m, both sine terms have integer frequencies and both integrals vanish.

If n = m, then the second sine is

(79)

point by point, while the first integral still vanishes.

Thus in every case

(80)

10 The constant function is also part of the Fourier basis

A real Fourier series contains a constant component in addition to sine and cosine harmonics. Therefore we must also check how the constant function interacts with the harmonics.

For n ≥ 1,

(81)

Likewise,

(82)

Thus the constant function is orthogonal to every nonzero sine and cosine harmonic on the complete interval.

Its norm is

∥1∥2	= ⟨1, 1⟩	(83)
	= ∫ −ππ1 dx	(84)
	= 2π.	(85)

Therefore

(86)

This different norm explains why the constant Fourier coefficient will carry a different normalization factor from the sine and cosine coefficients.

11 The complete orthogonality table on [−π,π]

For positive integers m,n,

(87)

(88)

and

(89)

Also,

(90)

Finally,

(91)

These equations are the algebraic core of real Fourier series.

12 Norms and the normalized trigonometric family

From Equations (56) and (68),

(92)

and

(93)

The constant function has

(94)

Therefore define

(95)

(96)

and

(97)

Each has unit norm. Different members are orthogonal. Thus the family is orthonormal.

13 Worked Example 1: verify one cosine-cosine pair directly

Evaluate

(98)

Use product-to-sum:

cos(2x) cos(5x)	=	(99)
	= ,	(100)

because cosine is even.

Then

∫ −ππ cos(2x) cos(5x) dx	= ∫ −ππ cos(3x) dx	(101)
	+ ∫ −ππ cos(7x) dx	(102)
	= 0 + 0	(103)
	= 0.	(104)

The two harmonics are orthogonal.

14 Worked Example 2: same cosine harmonic

Evaluate

(105)

Use

(106)

Then

∫ −ππ cos 2(4x) dx	= (2π) + (0)	(107)
	= π.	(108)

This is the norm squared of cos(4x).

15 Worked Example 3: sine-cosine orthogonality by parity

Evaluate

(109)

The sine factor is odd and the cosine factor is even. Their product is odd. Therefore

(110)

No product-to-sum identity is needed.

16 Worked Example 4: orthogonality depends on the interval

Consider

(111)

On [−π,π] they are orthogonal:

(112)

Now integrate on [0,π∕2]:

(113)

Use product-to-sum:

(114)

Then

I	= 0π∕2	(115)
	=	(116)
	=	(117)
	= .	(118)

Thus the same two functions are not orthogonal on this shorter interval.

This is why the integration interval is part of the inner product definition.

17 Generalization from [−π,π] to [−L,L]

Fourier series are often written on an interval of total length 2L rather than 2π. The corresponding harmonic functions are

(119)

and

(120)

We can derive their orthogonality by changing variables.

Let

(121)

Then

(122)

and

(123)

When

(124)

we have

(125)

When

(126)

we have

(127)

Consider the cosine-cosine inner product:

I	= ∫ −LL cos cos dx	(128)
	= ∫ −ππ cos(ny) cos(my) dy.	(129)

Use Equation (87):

I	= (πδmn)	(130)
	= Lδmn.	(131)

Therefore

(132)

Similarly,

(133)

and

(134)

The constant function has

(135)

Therefore on [−L,L] the normalized basis becomes

(136)

(137)

and

(138)

18 Generalization to a time interval of one period

Suppose a periodic signal has period T₀ and fundamental angular frequency

(139)

Its harmonics are

(140)

and

(141)

Over any complete interval of length T₀,

(142)

for positive m,n.

Likewise,

(143)

and

(144)

The starting time t₀ does not matter as long as the integration covers one complete fundamental period.

19 Why distinct integer harmonics are special

The product-to-sum derivation shows exactly what is required.

For different cosine harmonics,

(145)

produces frequencies

(146)

When m and n are integers, both are integers. If m≠n, neither is zero. Each resulting sinusoid therefore completes an integer number of cycles on the chosen full-period interval and integrates to zero.

This explains the connection

(147)

If arbitrary noncommensurate frequencies are chosen, this cancellation is not guaranteed over the same finite interval.

20 The geometric meaning of the Kronecker delta

FS02 showed that orthogonal vectors have zero cross inner product and nonzero self inner product. The same pattern occurs here.

For normalized cosine functions,

(148)

we have

(149)

Thus the Kronecker delta is the infinite-dimensional analogue of the familiar Cartesian relation

(150)

The Fourier sine and cosine functions behave like perpendicular coordinate directions in function space.

21 How orthogonality isolates one harmonic

We now reach the key bridge to FS04 and FS05.

Suppose, for the moment, that a function can be represented by a finite trigonometric sum

(151)

Suppose we want to isolate the coefficient A_m. Multiply both sides by cos(mx):

f(x) cos(mx)	= A0 cos(mx)	(152)
	+ ∑ n=1NA n cos(nx) cos(mx)	(153)
	+ ∑ n=1NB n sin(nx) cos(mx).	(154)

Now integrate over [−π,π]:

∫ −ππf(x) cos(mx) dx	= A 0 ∫ −ππ cos(mx) dx	(155)
	+ ∑ n=1NA n ∫ −ππ cos(nx) cos(mx) dx	(156)
	+ ∑ n=1NB n ∫ −ππ sin(nx) cos(mx) dx.	(157)

Now apply orthogonality.

The constant term vanishes:

(158)

Every sine-cosine term vanishes:

(159)

Every cosine-cosine term with n≠m vanishes. Only one term remains:

(160)

But

(161)

Therefore

(162)

So

(163)

FS04 will develop this argument carefully as the derivation of the Fourier cosine coefficients. The purpose here is only to expose the mechanism:

• multiply by the desired basis function,
• integrate,
• orthogonality removes every unwanted basis direction,
• divide by the surviving basis norm squared.

22 A direct connection to FS02 projection

FS02 derived the projection coefficient

(164)

Choose

(165)

Then

(166)

Therefore

(167)

The Fourier cosine coefficient is therefore not a new type of mathematical object. It is an ordinary function-space projection coefficient onto an orthogonal trigonometric basis.

The same will hold for sine coefficients.

23 Common mistakes

1. Assuming two functions are orthogonal because their graphs look different. Orthogonality is an integral condition.
2. Forgetting that the interval is part of the inner product. Functions can be orthogonal on one interval and nonorthogonal on another.
3. Using the m≠n product-to-sum argument when m = n. The difference frequency becomes zero and produces the nonzero norm-squared term.
4. Assuming a function is orthogonal to itself. A nonzero function has positive self inner product.
5. Forgetting that the constant basis function has norm squared 2π, not π.
6. Writing

   (168)

   without specifying the limits. Orthogonality is not an indefinite-integral statement.

7. Forgetting that sine-cosine orthogonality on a symmetric interval follows immediately from parity.
8. Using noninteger frequencies and expecting the same cancellation on [−π,π] without checking the interval.
9. Confusing orthogonal with orthonormal. The unscaled sine and cosine functions are orthogonal but have norm , not 1.
10. Treating the Kronecker delta as a function of x. It is a discrete symbol depending on the indices m and n.

24 Self-check questions

1. Why does cos((n − m)x) cause a special case when n = m?
2. Why does the integral of cos(kx) vanish on [−π,π] for nonzero integer k?
3. Why can sine-cosine orthogonality be proved without product-to-sum identities?
4. What is the norm of cos(7x) on [−π,π]?
5. What is the norm of the constant function 1 on [−π,π]?
6. What scaling makes sin(5x) normalized?
7. What does δ_mn mean when m = 3 and n = 5?
8. What does δ_mn mean when m = n = 4?
9. Why are sin x and sin 2x not necessarily orthogonal on an arbitrary subinterval?
10. In one sentence, explain why multiplying a Fourier sum by cos(mx) and integrating isolates the matching cosine coefficient.

25 Practice exercises

1. Derive Equation (32) from the cosine angle-addition identities.
2. Derive Equation (34) from the same identities.
3. Derive Equation (38) from the sine angle-addition identities.
4. Evaluate

   (169)

5. Evaluate

   (170)

6. Evaluate

   (171)

   using both parity and product-to-sum.

7. Evaluate

   (172)

8. Evaluate

   (173)

9. Find the norms of 1, cos(5x), and sin(5x) on [−π,π].
10. Normalize the three functions in the preceding problem.
11. Show that

    (174)

12. Show that

    (175)

13. Determine whether cos x and cos 2x are orthogonal on [0,π∕2].
14. Determine whether sin x and cos 2x are orthogonal on [−π∕2,π∕2].
15. For

    (176)

    compute

    (177)

    using orthogonality rather than expanding every integral independently.

16. For the same f(x), compute

    (178)

26 What FS03 has established

We now know that the real trigonometric family

(179)

is orthogonal on [−π,π].

The nonzero harmonics have squared norm

(180)

while the constant has squared norm

(181)

The general interval [−L,L] gives squared harmonic norm

(182)

and constant norm squared

(183)

Most importantly, orthogonality gives a mechanism for extracting one component from a trigonometric sum.

That mechanism is exactly what the next lectures formalize:

• FS04 derives the Fourier cosine coefficients,
• FS05 derives the sine coefficients and the constant coefficient.

References

References