Part CFOURIER ANALYSIS. PARTIALDIFFERENTIALEQUATIONS (PDEs)

215

Part C FOURIER ANALYSIS. PARTIAL

DIFFERENTIAL EQUATIONS (PDEs)

CHAPTER 11 Fourier Analysis

SECTION 11.1. Fourier Series, page 474

Purpose. To derive the Euler formulas (6) for the coefficients of a Fourier series (5) of a given function of period , using as the key property the orthogonality of the trigonometric system.

Main Content, Important Concepts Periodic function

Trigonometric system, its orthogonality (Theorem 1) Fourier series (5) with Fourier coefficients (6) Representation by a Fourier series (Theorem 2) Comment on Notation

If we write instead of in (1), we must do the same in (6.0) and see that (6.0) then becomes (6a) with . This is merely a small notational convenience (but may be a source of confusion to poorer students).

Comment on Fourier Series

Whereas their theory is quite involved, practical applications are simple, once the student has become used to evaluating integrals in (6) that depend on n.

Figure 260 should help students understand why and how a series of continuous terms can have a discontinuous sum.

Comment on the History of Fourier Series

Fourier series were already used in special problems by Daniel Bernoulli (1700–1782) in 1748 (vibrating string, Sec. 12.3) and Euler (Sec. 2.5) in 1754. Fourier’s book of 1822 became the source of many mathematical methods in classical mathematical physics.

Furthermore, the surprising fact that Fourier series, whose terms are continuous functions, may represent discontinuous functions led to a reflection on, and generalization of, the concept of a function in general. Hence the book is a landmark in both pure and applied mathematics. [That surprising fact also led to a controversy between Euler and D. Bernoulli over the question of whether the two types of solution of the vibrating string problem (Secs. 12.3 and 12.4) are identical; for details, see E. T. Bell, The Development of Mathematics, New York: McGraw-Hill, 1940, p. 482.] A mathematical theory of Fourier series was started by Peter Gustav Lejeune Dirichlet (1805–1859) of Berlin in 1829. The concept of the Riemann integral also resulted from work on Fourier series. Later on, these series became the model case in the theory of orthogonal functions (Sec. 5.7). An English translation of Fourier’s book was published by Dover Publications in 1955.

Further Comments on Text and Problems

Figure 260, showing the rectangular periodic wave and partial sums of its Fourier series, is typical and should give the student an intuitive feel for convergence of Fourier series, notably of their behavior near discontinuities. The latter are of great practical importance, as will appear as we proceed.

n⫽ 0 a0

a0>2

2p

The essential property of the trigonometric system is its orthogonality. Other orthogonal systems will follow later in connection with families of solutions of linear ODEs (Legendre polynomials, Bessel functions, etc.).

The integrals needed for Fourier coefficients are those of calculus, but their dependence on n will create a new situation, and the student will need some time and particular attention to become familiar with them.

Problem 25 suggests that students look critically at the speed of convergence, which will depend on the power of in the Fourier coefficients, which in turn depends on continuity of to be developed.

The figures in Probs. 16–21 should help students to get familiar with “piecewise given” functions, as they will occur in practical work on potential problems, heat flow, and so on.

SOLUTION OF PROBLEM SET 11.1, page 482 2.

4. implies

where

Thus has the period . This proves the first statement. The other statement follows by setting

12.

13.

14.

16.

18. is odd.

20.

22. CAS Experiment. Experimental approach to Fourier series. This should help the student obtain a feel for the kind of series to expect in practice, and for the kind and quality of convergence, depending on continuity properties of the sum of the series.

(a) The -periodic function has discontinuities at . The instructor will notice the Gibbs phenomenon (see Sec. 11.7) at the points of discontinuity.

(b) if and if , is continuous

throughout, and the accuracy is much better than in (a).

0⬍ x ⬍p 1⫺ x>p

⫺p_{⬍ x ⬍ 0} f(x)⫽ 1 ⫹ x>p

⫾p f(x)⫽ x(⫺p_{⬍ x ⬍}p)

2p

⫺1

4 sin 4x⫺ 1

6 sin 6x⫺ Á

⫺ 1 2 sin 2x 1

p^{c (2 ⫹}p) sin x⫹ 1

9 (⫺2 ⫹ 3p) sin 3x⫹ 1

25 (2⫹ 5p) sin 5x⫹ Á d f(x)⫺¹2

1 2 ⫹ 2

p ^{asin x ⫹} 1

3 sin 3x⫹1

5 sin 5x⫹ Áb.

2

p ^{sin x⫹} 1

2 sin 2x⫺ 2

9p ^{sin 3x⫺} 1

4 sin 4x⫹ 2

25p ^{sin 5x⫹} 1

6 cos 6x⫹ Á

1

3p²⫺ 4(cos x ⫺¹4 cos 2x⫹¹9 cos 3x⫺ ⫹ Á )

⫹ 2 sin(3x) ⫺ sin(4x) ⫹ 6>5 sin(5x) ⫺ 2>3 sin(6x) ⫹ Á 4 cos(x)

p ^{⫹ 4>9} cos(3x)

p ^⫹ 4 25

cos(5x)

p ⫹ 6 sin(x) ⫹ Á ⫺2 sin(2x) p_{⫺ 8} ^cos(x)p ^⫺

8 9

cos(3x) p ^⫺

8 25

cos(5x) p ^{⫹ Á} a⫽ 1>b.p>a

g(x)

g( x)⫽ f(ax).

f(ax⫹ p) ⫽ f(a[x⫹ (p>a)]) ⫽ f(ax) or g[x⫹ (p>a)] ⫽ g( x), f(x⫹ p) ⫽ f(x)

2p_{>n, 2}p>n, k, k, k>n, k>n f(x)

1>n

(c) has about the same continuity as (b), and the approximation is good.

The coefficients in (a) involve , whereas those in (b) and (c) involve . This is typical. See also CAS Experiment 25.

24. CAS Experiment. The student should recognize the importance of the interval in connection with orthogonality, which is the basic concept in the derivation of the Euler formulas.

For instance, for the integral equals , and the graph suggests orthogonality for a⫽p, as expected.

sin a⫺¹7 sin 7a sin 3x sin 4x

1>n² 1>n

f(x)⫽p²_{⫺ x}²

SECTION 11.2. Arbitrary Period. Even and Odd Functions, Half-Range Expansions, page 483

Purpose. The three topics considered in this section are listed in the title. The three main points are as follows.

The transition from period to period 2L amounts to a linear transformation in x.

From (5), (6) in Sec. 11.1 it produces (5), (6) in this section.

For even functions the Fourier series reduces to a cosine series (hence a series without sine terms) with coefficients .

For odd functions the Fourier series reduces to a sine series with coefficients For period the corresponding simpler formulas are separately listed in the Summary.

Typical illustrations of all this are shown in Examples 1–5.

The third and last topic is half-range expansions, typically illustrated in Example 6 and Fig. 272. This will be applied to physical problems of vibrations and heat conduction in the next chapter.

In the problem set we start with some general questions on even and odd functions (Probs. 1–7), followed by Fourier series developments for various periods (Probs. 8–17) and by some general problems (Probs. 18–22).

Finally, half-range expansions are needed in Probs. 23–30, cosine series as well as sine series in each case, that is, functions given on an interval are to be represented on as a cosine series as well as a sine series, of a function of period 2L, obtained by extending the given function from to

as an even or odd function, respectively.

⫺L ⬉ x ⬉ L

0⬉ x ⬉ L f苲(x)

⫺L ⬉ x ⬉ L f(x) 0⬉ x ⬉ L

2p

(6**).

(5**) (6*)

(5*)

2p

a

–1 – 0.5

0.5

2␲

–2␲ –␲ ␲

1

Section 11.1. Integral in Problem 24 as a function of a

SOLUTIONS TO PROBLEM SET 11.2, page 490 1. Neither, even, odd, odd, neither

2. Even, even, neither, odd, even

4. Odd for sums and for products of an odd number of factors,

Even for products of an even number of factors.

6. Odd. This is important in connection with the integrand in the Euler formulas for the Fourier coefficients. It implies the simplification of the Fourier series of an odd function to a Fourier sine series and of the Fourier series of an even function to a Fourier cosine series.

8. Even, . Cf. Prob. 12 in Sec. 11.1. The Fourier series is

9. Odd,

10. Odd, . Cf. also Prob. 21 in Sec. 11.1. The series is

11. Even,

12. Even. The series is

14. Even, , full-wave rectification of a cosine current. The series is

15. Odd,

16. Odd, if if , series

17. Even,

18.

⫽ 50V0冮_ⴚ1>200^{1>200cos 100(n⫹ 1)pt dt⫹ 50V0}冮_ⴚ1>200^{1>200cos 100(n⫺ 1)pt dt ;}

a_n⫽ 100V0冮_ⴚ1>200^{1>200cos 100pt cos 100npt dt}

bn⫽ 0, a0⫽ V0

p^,

L⫽ 1. 1>2 ⫺ 4 cos(px)

p^{2 ⫺ 4>9}

cos(3px) p^{2 ⫺}

4 25

cos(5px) p^{2 ⫹ Á}

⫺ 1

p^{asin 2}px⫹ 1

2 sin 4px⫹1

3 sin 6px⫹ Áb

sin 5px⫹ Áb 2

p^{3 a(}p²_{⫺ 4) sin} px⫹ 1

27 (9p²_{⫺ 4) sin 3}px⫹ 1

125(25p²_{⫺ 4)} 0⬍ x ⬍ 1

⫺1 ⬍ x ⬍ 0, f(x)⫽ x² L⫽ 1, f(x)⫽ ⫺x²

L⫽p. ⫺p⁴ (sin x⫺¹9 sin 3x⫹25¹ sin 5x⫹ Á ) 2

p ^⫹ 4 p ^a

1

1ⴢ 3 cos 2px⫺ 1

3ⴢ 5 cos 4px⫹ 1

5ⴢ 7 cos 6px⫺ ⫹ Áb . L⫽¹2

2 3⫹ 4

p^2acos px

2 ⫺1

4 cos px⫹ 1

9 cos 3px

2 ⫺ 1

16 cos 2px⫹ 1

25 cos 5px

2 ⫹ Áb L⫽ 1, ⫺p⁴² (cos p x⫺4¹ cos 2p x⫹¹9 cos 3px⫺ Á )

8 p ^asin

px 4 ⫹ 1

2sin px 2 ⫹ 1

3 sin 3px

4 ⫹ Áb L⫽ 4

L⫽ 2. ⫺p⁴ (sin (1>2px)⫹ 1>3 sin (3>2px)⫹ 1>5 sin (5>2px)⫹ Á ) 1>2 ⫹ 4 cos(px)

p^{2 ⫹ 4>9}

cos(3px) p^{2 ⫹}

4 25

cos(5px) p^{2 ⫹ Á} L⫽ 1

f(⫺x) ⫽ f1(⫺x) Á f2k⫹1(⫺x) ⫽ (⫺1)^2k⫹1 f₁(x) Á f_2k⫹1(x)⫽ ⫺f(x).

2k⫹ 1

hence the series is

20. Set . Then

Hence, the results.

22. Note that the functions in Prob. 8 and Prob. 17 are related as follows.

The function in Prob. 8 is

and the function in Prob. 17 is

It can readily be verified that and hence the result.

24. (a)

(b)

25. (a)

(b) 1>2 p⫹ 2 sin (x) ⫹ 2 sin (2x) ⫹ 2>3 sin (3x) ⫹ 2>5 sin (5x) ⫹ 2>3 sin (6x) ⫹ Á 1>2 p⫹ 2 cos (x) ⫺ 2>3 cos (3x) ⫹ 2>5 cos (5x) ⫹ Á

⫹ 1

7 sin 7px 4 ⫹ 1

9 sin 9px 4 ⫺ 1

5 sin 5px

2 ⫹ 1

11 sin 11px

4 ⫹ Áb

2 p^asin

px

4 ⫺ sin px 2 ⫹ 1

3 sin 3px 4 ⫹ 1

5 sin 5px 4 ⫺ 1

3 sin 3px 2 1

2⫺ 2 p^acos

px 4 ⫺ 1

3 cos 3px 4 ⫹ 1

5 cos 5px

4 ⫺ ⫹ Áb L⫽ 4,

f17⫽ 1 ⫺ f8

f17⫽ e⫺x ⫺1 ⬍ x and x ⬍ 0 x 0⬍ x and x ⬍ 1 f₈⫽ e⫺x ⫺1 ⬍ x and x ⬍ 0

x 0⬍ x and x ⬍ 1

(1⫹¹4⫹¹9⫹16¹ ⫹25¹ Á).

f(⫺1) ⫽ 1 ⫺ 1>3 ⫽ 2>3 ⫽p⁴²

x⫽ ⫺1

⫹ 2V0

p ^a 1

1ⴢ 3 cos 200pt⫺ 1

3ⴢ 5 cos 400pt⫹ 1

5ⴢ 7 cos 600pt⫺ ⫹ Áb V0

p^⫹ V0

2 cos 100pt

26. (a)

(b)

⫹a1 5 ⫹ 2

25p^{b sin 5x ⫺} 1

6 sin 6x⫹a1 7 ⫺ 2

49pb sin 7x ⫹ Á a1 ⫹ 2

p^{b sin x ⫺} 1

2 sin 2x⫹a1 3 ⫺ 2

9p^{b sin 3x ⫺} 1 4 sin 4x

⫹ 1

49 cos 7x⫹ 1

81 cos 9x⫹ 1

50 cos 10x⫹ 1

121 cos 11x⫹ Áb 3p

8 ⫺ 2

p ^{acos x ⫹} 1

2 cos 2x⫹ 1

9 cos 3x⫹ 1

25 cos 5x⫹ 1 18 cos 6x L⫽p,

0 0

x ␲

␲

The student should be invited to find the two functions that the sum of the series represents. This can be done by graphing

and then , where

The first of these functions is discountinuous, the coefficients being proportional to , whereas is continuous, its Fourier coefficients being proportional to , so that they go to zero much faster than the others.

28. (a)

(b)

30. Shift by . This cosine series is obtained. The sine series needs to be multiplied by . The result is that the coefficients of odd multiples of remain the same, whereas the coefficients of even multiples of are multiplied by .

SECTION 11.3. Forced Oscillations, page 492

Purpose. To show that mechanical or electrical systems with periodic but nonsinusoidal input may respond predominantly to one of the infinitely many terms in the Fourier series of the input, giving an unexpected output; see Fig. 277, where the output frequency is essentially five times that of the input.

Further Comments on Text and Problems

Example 1 in the text is sufficient to show the general idea and is typical of the problems.

Problems 2 and 20 are particularly important. In fact, students should decrease the damping term in (4), letting it approach zero. This will give an additional better understanding of the present situation.

SOLUTIONS TO PROBLEM SET 11.3, page 494 2. For (and as before) the amplitudes are

An increase of the damping constant increases for all n, hence it decreases all amplitudes .

4. is given by the sine series in Example 1 with . The new is n times the old. Hence is now so large that the output is practically a cosine vibration having five times the input frequency.

6.

7.

⫽ ⫺1.19, ⫺2.78, ⫺5.26, 4.76, 2.27, 0.0417

y⫽ C1sin vt⫹ C2 cos vt⫹ a(v) cos t, a(v) ⫽ 1>(v²⫺ 1)

y⫽ C2 sin (vt)⫹ C1 cos (vt)⫹ (⫺b²⫹ v²) cos(at)⫹ (v²⫺ a²) cos(bt) v⁴⫹ (⫺b²⫺ a²)v²⫹ a²b² C5

Cn

k⫽ ⫺1 r

r

C_n

D_n c(⬎ 0)

C₇⫽ 0.0742, C9⫽ 0.0005, C11⫽ 0.0001.

C₅⫽ 0.0021,

C₁⫽ 0.0265, C3⫽ 0.0035, c⫽ 0.05

k⫽ 49

⫺1 p

⫺1 p p 2L

p ^asin px

L ⫺ 1

2 sin 2px L ⫹1

3 sin 3px L ⫺ 1

4 sin 4px

L ⫹ ⫺ Áb L

2 ⫺ 4L p^{2 acos}

px L ⫹ 1

9 cos 3px L ⫹ 1

25 cos 5px

L ⫹ Áb

1>n² f2

1>n

f2(x)⫽ bx>2 if ⫺p>2 ⬍ x ⬍p_>2 p>2 ⫺ x>2 if p>2 ⬍ x ⬍ 3p_>2.

(2>p)(sin x⫺ 3^ⴚ2 sin 3x⫹ ⫺ Á ) ⫽ f2

sin x⫺¹2 sin 2x⫹¹3 sin 3x⫹ ⫺ Á ⫽ x>2

8. The given is

and

The corresponding Fourier series, a Fourier cosine series, is

Substituting this into the above ODE and solving it gives the answer

10.

11.

14. The Fourier series is a Fourier sine series, as given and derived in Example 1 of Sec. 11.1 with coefficients

(n odd).

Hence the ODE must be solved with the right side

(n odd).

The steady-state solution of this ODE is

where

with as in Prob. 13.

So we have , etc. and no cosine terms in the Fourier series on the right side of the ODE. In the solution the damping constant appears with the cosine terms (and in ), causing a phase shift, which is zero if . Also, increasing c increases , hence it decreases the amplitudes; this is physically understandable.

16. For the right side we have the Fourier sine series

with the coefficients if and if

. Substitution of this series into the ODE gives y⫽ A1 cos t⫹ B1 sin t⫹ A3 cos 3t⫹ B3 sin 3t⫹ Á 11, Á

n⫽ 3, 7, bn⫽ 4>(n²p) n⫽ 1, 5, 9, Á , bn⫽ ⫺4>(n²p) 4

p ^{asin t ⫺} 1

9 sin 3t⫹ 1

25 sin 5t⫺ ⫹ Áb Dn

c⫽ 0 Dn

b1⫽ 4>p, b2⫽ 0, b3⫽ 4>(3p) Dn⫽ (1 ⫺ n²)²⫹ n²c²

An⫽ ⫺4nc>(npDn), Bn⫽ (4 ⫺ 4n²)>(npDn) y⫽ a^ⴥ

n⫽1 n odd

(A_n cos nt⫹ Bn sin nt) r_n(t)⫽ (4>np) sin nt

b_n⫽ 4>(np) (k⫽ 1)

y⫽ C1 cos vt⫹ C2 sin vt⫺ 4 p^a

sin t v²⫺ 1 ⫹ 1

3 sin 3t v²⫺ 9⫹ 1

5 sin 5t

v²⫺ 25⫹ Áb

⫺ 1

3# _5(v2⫺ 16) cos 4t⫺ 1

5 # _7(v2⫺ 36) cos 6t⫺ Á y⫽ C1 cos vt⫹ C2 sin vt⫹ 1

2v² ⫺ 1

1# _{3 (v}2⫺ 4) cos 2t

⫺ 1

3# _5(v2⫺ 16) cos 4t⫹ ⫺ Á .

⫹ 1

2v² ⫹ 1

1 # _{3 (v}2⫺ 4) cos 2t y⫽ C1 cos vt⫹ C2 sin vt

r(t)⫽ 1 2⫹ 1

1 # ₃ ^{cos 2t}⫺ 1

3 # ₅ ^{cos 4t}⫹ 1

5 # ₇ ^{cos 6t} ⫺ ⫹ Á .

⫺p

4 cos t if p

2 ⬍ t ⬍ 3p 2 . r(t)⫽ p

4 cos t if⫺p

2 ⬍ t ⬍ p 2 r (t)

with coefficients

. The damping constant c appears in the cosine terms, causing a phase shift, which is zero if . Also, c increases , hence it decreases the amplitudes, which is physically understandable.

18. The ODE in Probs. 17–19 is the same, except for the changing right sides, whose Fourier series we use term-by-term, as in the text. The solution of the ODE is of the general form

with coefficients obtained by substitution

, in particular, , the ODE being

For the present problem we have the Fourier series

Hence and all the other and with n even are zero. The formula for the is where n is odd. Numerically evaluating the terms, we obtain the solution (the current in the RLC-circuit)

20. with and obtained

as solutions of

SECTION 11.4. Approximation by Trigonometric Polynomials, page 495 Purpose. We show how to find “best” approximations of a given function by trigono- metric polynomials of a given degree N.

Important Concepts Trigonometric polynomial Square error, its minimum (6)

Bessel’s inequality, Parseval’s identity Short Courses. This section can be omitted.

Comment on Quality of Approximation

This quality can be measured in many ways. Particularly important are (i) the absolute value of the maximum deviation over a given interval, and (ii) the mean square error considered here. See Ref. [GenRef7] in App. 1.

⫺ncAn⫹(k⫺ n²)B_n⫽ 0.

(k⫺ n²)A_n⫹ ncB_n⫽ 4>n²p

B_n A_n C_n⫽ 2An2 ⫹ Bn2 ⫽ 4>(n²p1D_n), D_n⫽ (n²⫺ k)²⫹ n²c²

⫺ 0.942 sin 3t ⫹ 0.056 cos 5t ⫺ 0.187 sin 5t ⫹ Á . I⫽ 41.416 ⫺ 12.662 cos t ⫺ 14.069 sin t ⫺ 0.031 cos 3t

⫺800>(pn²) a_n

B_n A_n A₀⫽ 10 ⫹ 10p

100⫹ 100p_⫺ ⁸⁰⁰

p ^{acos t ⫹} 1

9 cos 3t⫹ 1

25 cos 5t⫹ Áb.

I

s

r

A0⫽ a0>10 A_n⫽ ⫺n²⫺ 10

D_n a_n, Bn⫽ 10n

D_na_n, Dn⫽ (n²⫺ 102²⫹ 100n² I⫽ A0⫹ a^ⴥ

n⫽1

1An cos nt⫹ Bn sin nt) D_n

c⫽ 0

An⫽ ⫺ncbn>Dⁿ, Bn⫽ (1 ⫺ n²)bn>Dⁿ, Dn⫽ (1 ⫺ n²)²⫹ n²c²

Further Comments

The ideas in this section play a basic role in more advanced applied and abstract courses.

CAS Experiment 10 will give the student a feel for the size of the error of the present approximation and its size as a function of the number of terms considered, that is, for the rapidity of its decrease.

For other types of interpolation and approximation and for numeric work, see Secs. 19.3, 19.4, and 20.5.

SOLUTIONS OF PROBLEM SET 11.4, page 498

2. since is odd. Calculation of the gives the approximating trigono- metric polynomial

From this, the minimum square error is obtained as shown in Example 1 of the text; note that the present function and that in Example 1 differ just by an additive constant . 4.

5.

6. The function in Prob. 3 is continuous, the function in Prob. 5 is not; indeed, it is the derivative of the function in Prob. 3, and differentiation produces a factor n in each term.

8. The approximating trigonometric polynomial of minimum square error is

From this we obtain , etc.

These values are small; indeed, a graph shows that the four terms given are such that this partial sum of F*approximates f(x)rather accurately.

E*⫽ 0.5951, 0.0292, 0.0292, 0.0066, 0.0066 F⫽ 2

p^⫺ 4 p^a

1 1 #

3 cos 2x⫹ 1 3#

5 cos 4x⫹ 1 5 #

7 cos 6x⫹ Áb , E*⫽ 62.012, 60.739, 60.739, 60.420, 60.420

b_n⫽n²((⫺1)^1ⴙn⫹ 1) F⫽ 4 sin(x) ⫹ 4>3 sin(3x) ⫹ 4>5 sin(5x) ⫹ Á ,

E*⫹⫽ 4.13, 0.99, 0.39, 0.17, 0.1125⁴ cos (5x)⫺ 1>9 cos (6x) ⫹ Á

F⫽ 2>3 p²⫹ 4 cos(x) ⫺ cos(2x) ⫹ 4>9 cos(3x) ⫺ 1>4 cos(4x)

p F⫽ 2 asin x ⫺ 1

2 sin 2x⫹ Á ⫹(⫺1)^N⫹1

N sin Nxb . b_n

f(x)⫽ x a_n⫽ 0

0 x

–␲ ␲

1

Section 11.4. Problem 8

10. CAS Experiment. Factors are the continuity or discontinuity and the speed with which the coefficients go to zero,

For given on some data are as follows ( f, decrease of the coefficients, continuity or not, smallest N such that f(x) ⫺p_{⬍ x ⬍}p E*⬍ 0.1).

1>n, 1>n².

, discontinuous, , continuous,

, discontinuous, for , continuous,

, continuous. For we still have . For f in Prob. 9 we have , continuity,

The functions

with have coefficients proportional to and when .

These data indicate that the whole situation is more complex than one would at first assume. So the student may need your help and guidance.

14. The Fourier series is

and Parseval’s identity (8) gives

Section 11.5. Sturm—Liouville Problems. Orthogonal Functions, page 498

Purpose. Discussion of eigenvalue problems for ordinary second-order ODEs (1) under boundary conditions (2).

Main Content, Important Concepts

Sturm–Liouville equations, Sturm–Liouville problem Reality of eigenvalues

Orthogonality of eigenfunctions

Orthogonality of Legendre polynomials and Bessel functions Short Courses. Omit this section.

Comment on Importance

This theory owes its significance to two factors. On the one hand, boundary value problems involving practically important ODEs (Legendre’s, Bessel’s, etc.) can be cast into Sturm–

Liouville form, so that here we have a general theory with several important particular cases. On the other hand, the theory gives important general results on the spectral theory of those problems.

Comment on Existence of Eigenvalues

This theory is difficult. Quite generally, in problems where we can have infinitely many eigenvalues, the existence problem becomes nontrivial, in contrast with matrix eigenvalue

1 2 ⫹ 1

4 ⫽ 1

p冮_ⴚ^p_p^{cos4 x dx.}

cos² x⫽¹2⫹¹2 cos 2x (k⫽ 1), 5(k⫽ 2),11(k⫽ 3), 21(k⫽ 4)

N⭌ 1 E*⬍ 0.1

1>n² k⫽ 1, 2, 3, 4

f(x)⫽ c (x⫹¹2p)^2k⫺ (¹2p)^2k if ⫺p_{⬍ x ⬍ 0}

⫺(⫺x ⫺¹2p)^2k⫹ (2¹p)^2k if 0⬍ x ⬍p N⫽ 2.

1>n²

E*⫽ 0.1769 N⫽ 200

f⫽ x⁶,1>n

N⫽ 40 f⫽ x⁴,1>n

N⫽ 200 E*⫽ 6.105

f⫽ x³,1>n²

N⫽ 5 f⫽ x²,1>n²

N⫽ 126 f⫽ x,1>n

problems (Chap. 8), where existence is trivial, a consequence of the fact that a polynomial equation ( f not constant) has at least one solution and at most n numerically different ones (where n is the degree of the polynomial).

Move of This Theory from Chap. 5 to Chap. 11

Sturm–Liouville theory is motivated to a large extent by the use of orthogonality in connection with Fourier series and by generalizations from (in the separation of the vibrating string PDE, the wave PDE) to more general linear ODEs (Legendre and Bessel above all).

The Sturm–Liouville material is now close to one of its main applications in PDEs (Chap. 12). More importantly, orthogonality seems more complicated to grasp than other theories and needs digestion time and more maturity than the average student probably has in Part A on ODEs. At least, this is my observation resulting from teaching these matters many times to engineers, physicists, and mathematicians, representing groups of various interests and maturity in applied mathematics.

SOLUTIONS TO PROBLEM SET 11.5, page 503

2. If is a solution of (1), so is because (1) is linear and homogeneous; here, is the eigenvalue corresponding to . Also, multiplying (2) with by c, we see that also satisfies the boundary conditions. This proves the assertion.

4. . Problems 2 to 6 are most useful in applications; they concern situations that appear rather frequently.

6. Perform the differentiations in (1), divide by p, and compare; that is,

Hence . A reason for

performing this transformation may be the discovery of the weight function needed for determining the orthogonality. We see that

This problem shows that a Sturm–Liouville equation is rather general; more precisely, it is equivalent to a general second-order homogeneous linear ODE whose coefficient of the unknown function contains a parameter , the coefficient being of the form

7.

8.

10. We need

From the boundary conditions we obtain

y

r

r

y(0)⫽ A ⫽ y(1)⫽ A cos k ⫹ B sin k y

r

y⫽ A cos kx ⫹ B sin kx l⫽ (mp_>L)², m⫽ 1, 2, Á ; ym⫽ sin (mpx>L) l_m⫽ (mp_>5)², m⫽ 1, 2, Á ; ym⫽ sin (mpx>5)

Q⫹ lR.

y(x) l

r(x)⫽ h(x)p(x)⫽ h(x) exp a冮^{f(x) dxb .}

f⫽ p

r

s

r

r

s

r

p y

r

p ⫹ l r

pb y ⫽ 0.

a⫽ ⫺p, b⫽p, c⫽p, k⫽ 0 z_m

y⫽ ym

y_m l⫽ lm

z_m y_m

y

s

f(x)⫽ 0

Ordering gives

By eliminating A and then requiring (to have , an eigenfunction) or simply by noting that for this homogeneous system to have a nontrivial solution A, B, the determinant of its coefficients must be zero; that is,

;

hence , so that the eigenvalues and eigenfunctions are

;

12. A general solution , of this ODE with constant

coefficients is obtained as usual. The Sturm–Liouville form of the ODE is obtained by using the formulas in Prob. 6,

From this and the boundary conditions we expect the eigenfunctions to be orthogonal on with respect to the weight function . Now, from that general solution

and , we see that we are left with From the second

boundary condition we now obtain

Hence the eigenvalues and eigenfunctions are

13.

14. Team Project. (a) We integrate over x from to 1, hence over defined by

from to 0. Using , we thus obtain

which is zero for integer

(b) Following the hint, we calculate for :

⫽ Á ⫽ (⫺1)^k k

n! 冮₀^ⴥdx^dnⴚknⴚk (xⁿe^ⴚx) dx⫽ 0.

冮₀^{⬁eⴚx xkLn(x) dx⫽}n!¹ 冮₀^⬁xkdx^dnn (xⁿe^ⴚx) dx⫽ ⫺ k

n! 冮₀^⬁xkⴚ1dx^dnnⴚ1ⴚ1 (xⁿe^ⴚx) dx k⬍ n

兰e^ⴚxx^kLn dx⫽ 0 m⫽ n.

⫽ 冮₀^pcos mu cos nu du⫽ 1

2冮₀^{p(cos (m}⫹ n)u ⫹ cos (m⫺ n)u) du,

冮_ⴚ1¹cos (m arc cos x) cos (n arc cos x)(1⫺ x²)^ⴚ1>2dx (1⫺ x²)^ⴚ1>2 dx⫽ ⫺du

x⫽ cos u p ⫺1 u

p⫽ e^6x, q⫽ 9e^6x, r⫽ e^6x, lm⫽ m², ym⫽ e^ⴚ3x sin mx, m⫽ 1, 2, Á l_m⫽ (mp)², y_m⫽ e^2x sin mpx.

y(1)⫽ e² sin k⫽ 0, k ⫽ mp, m⫽ 1, 2, Á . y(1)⫽ 0

y⫽ e^2x sin kx.

y(0)⫽ A ⫽ 0

e^ⴚ4x 0ⱕ x ⱕ 1

(e^ⴚ4xy

r

r

y⫽ e^2x(A cos kx⫹ B sin kx), k ⫽ 1l

y0⫽ 1, ym(x)⫽ cos (2mpx), sin (2mpx), m ⫽ 1, 2, Á . l_m⫽ (2mp)², m ⫽ 0, 1, Á

cos k⫽ 1, k ⫽ 2mp

k(1⫺ cos k)²⫹ k sin² k⫽ k(2⫺ 2 cos k) ⫽ 0 y⫽ 0

B⫽ 0

(k sin k)A⫹ k(1⫺ cos k)B⫽ 0.

(1⫺ cos k)A ⫺ (sin k)B ⫽ 0

SECTION 11.6. Orthogonal Series. Generalized Fourier Series, page 504 Purpose. To show how families (sequences) of orthogonal functions, as they arise in eigenvalue problems and elsewhere, are used in series for representing other functions, and to show how orthogonality becomes crucial in simplifying the determination of the coefficients of such a series by integration.

Main Content, Important Concepts Standard notation

Orthogonal expansion (1), eigenfunction expansion Fourier constants (2)

Fourier–Legendre series (Example 1) Fourier–Bessel series (Example 2)

Orthogonality of Bessel functions (Theorem 1) Mean square convergence

Completeness, also called totality (Theorem 2) Comments on Text

Formula (2) for the Fourier constants (the coefficients of an orthogonal series) generalizes (6) in Sec. 11.2 for the Fourier coefficients of a Fourier series.

Note that for Fourier–Bessel series (9) with coefficients (10) you obtain infinitely many orthogonal families, each consisting of infinitely many Bessel functions.

In many applications of these series (and other orthogonal series), it turns out that one needs only a relatively small number of terms for obtaining a reasonable accuracy.

Completeness of orthogonal families of functions guarantees that the set of given functions to be developed is sufficiently large to be of practical (and theoretical!) interest.

In practical work, numeric methods may be needed for obtaining values of Fourier constants.

SOLUTIONS TO PROBLEM SET 11.6, page 509 1.

2. . This is probably most simply obtained by the method of undetermined coefficients, beginning with the highest power, and . The point of these problems is to make the student aware that these developments look totally different from the usual expansions in terms of powers of x.

3.

4.

6. The series may contain all even powers, not just powers

8. . The size of , that is, the rapidity of convergence seems to depend on the variability of . A discontinuous derivative (e.g., as for occurring in connection with rectifiers) makes it virtually impossible to reach the goal. Let alone when itself is discontinuous. In the present case the series is

Rounding seems to have considerable influence in Prob. 8–13.

10. f(x)⫽ 0.7468P0(x)⫺ 0.4460P2(x)⫹ 0.0739P4(x)⫺ Á , m0⫽ 4 f(x)⫽ 0.95493P1(x)⫺ 1.15824P3(x)⫹ 0.21929P5 (x)⫺ ⫹ Á .

f(x) ƒsin x ƒ

f(x) m0

m0⫽ 5

x^4m.

1

5P0(x)⫹⁴7P2(x)⫹35⁸ P4(x) P0(x), P1(x), ¹₃P0(x)⫹²3P2(x), ³₅P1(x)⫹²5P3(x),

6

5 P₀(x)⫹⁴7 P₂(x)⫹35⁸ P4(x)

P₂(x) x²

2

3P₂(x)⫹ 2P1(x)⫹⁴3P₀(x) 8(P₁(x)⫺ P2(x)⫹ P4(x))

(ym, yn)

12. . Compare with Prob. 13!

14. Team Project. (b) A Maclaurin series has the coefficients We thus obtain

If we set , this becomes

(c) , etc.

(d) We write , etc., and use (21). By integrations by parts, for ,

(e) from (22) with instead of n. In this equation,

the first term on the right equals by (21). The last terms equals , as follows by differentiation of (21).

SECTION 11.7. Fourier Integral, page 510

Purpose. Beginning in this section, we show how ideas from Fourier series can be extended to nonperiodic functions defined on the real line, leading to integrals instead of series.

Main Content, Important Concepts Fourier integral (5)

Existence Theorem 1

Fourier cosine integral, Fourier sine integral, (10)–(13) Application to integration

Short Courses. This section can be omitted.

Comments on Text and Problems

The simplest example on Fourier series can also serve here as an introduction (Example 1 and Fig. 280) to motivate the present extension to Fourier integrals as well as the result in Theorem 1, which we must leave without proof (a reference is given in the text); so the situation is somewhat similar to that on Fourier series near the beginning of the chapter.

⫺Hen

s

xHen

r

n ⫺ 1 nHen⫽ nxHenⴚ1⫺ nHe_nⴚ1

r

⫽ (⫺1)^nⴚmm! 冮_ⴚ⬁^{⬁He0v(nⴚm) dx⫽ 0.}

⫽ (⫺1)^nⴚ1m 冮_ⴚ⬁^{⬁Hemⴚ1v(nⴚ1) dx⫽ Á}

冮_ⴚ⬁^{⬁vHemHen dx⫽ (⫺1)n}冮_ⴚ⬁^{⬁Hemv(n) dx⫽ (⫺1)nⴚ1}冮_ⴚ⬁^⬁Hem

^r

n⬎ m e^ⴚx2>2⫽ v, v⁽ⁿ⁾⫽ dⁿv>dxⁿ

Gx⫽ a a

r

r

f⁽ⁿ⁾(0)⫽ e^x2>2(⫺1)ⁿ dⁿ

dzⁿ(e^ⴚz2>2)`

z⫽x⫽ (⫺1)ⁿ e^x2>2 dⁿ

dxⁿ(e^ⴚx2>2) ⫽ Hen(x).

x⫺ t ⫽ z f⁽ⁿ⁾(0)⫽ dⁿ

dtⁿ (e^txⴚt2>2)`

t⫽0⫽ e^x2>2 dⁿ

dtⁿ (e^{ⴚ(xⴚt)2>2})`

t⫽0. (0)>n!

f⁽ⁿ⁾

a_n⫽ f(t)⫽ a^ⴥ

n⫽0

a_ntⁿ

f(x)⫽ 0.6116P0(x)⫺ 0.7032P2(x)⫹ 0.0999P4(x)⫹ Á , m0⫽ 4

It is interesting that we shall now be able to prove and understand the Gibbs phenomenon in terms of the sine integral and Dirichlet’s discontinuous factor in Example 2. See, in particular, Fig. 283.

Fourier cosine and Fourier sine integral in (10) and (11) are analogs of Fourier cosine and Fourier sine series.

Example 3 shows a basic application.

The evaluation of integrals by the present method is shown in Probs. 1–6 for the Fourier integral itself and in Probs. 7–12 and 16–20 for Fourier cosine and sine integrals, respectively.

CAS Experiments 13 and 15 should help the student in gaining additional insight beyond the present formalism.

Project 14 is a first step into transform theory similar to that of the Laplace transform in Chap. 6, the latter being of much greater importance to the engineer.

SOLUTIONS TO PROBLEM SET 11.7, page 517

1. ,

(see Example 3) etc.

2. Use (11) and to get, with the help of (11) in App. 3.1,

3. Use (11);

4. Use (10). Also use (11) in App. A3.1. Take Then

5.

6. Take . Then (11) in this section and (11) in App. A3.1 give

Integrate by parts, obtaining

Now use the first formula in (11) to obtain the result.

7. 2 p 冮^⬁

0

1⫺ cos w

w² cos xw dw B(w)⫽ w⫹ 1

1⫹ (w ⫹ 1)² ⫹ w⫺ 1

1⫹ (w ⫺ 1)² ⫽ 2w³ w⁴⫹ 4.

⫽ 冮₀^{⬁eⴚvsin (w⫹ 1)v dv ⫹} 冮₀^{⬁eⴚv sin (w⫺ 1)v dv.}

B(w)⫽ 2 p 冮0^⬁

pe^ⴚv cos v sin wv dv f(x)⫽pe^ⴚx cos x (x ⬎ 0)

B(w)⫽ 2

p 冮₀^{1⫺ p2x sin wv dv⫽ w cos w}_w^{⫺ sin w2}

A(w)⫽ 1

p冮_ⴚ^p_p^>2_>2^p2cos v cos wv dv⫽ 2 cos ¹₂pw 1⫺ w² . f(x)⫽¹2pcos x.

B⫽p² 兰^0⬁p4 sin wvdv⫽^{1⫺ cos w}2w ^p

B(w)⫽ 冮₀^psin v sin wv dv⫽ sin pw 1⫺ w². f(x)⫽ (p>2) sin x (0 ⬉ x ⬉p)

f(x)⫽pe^ⴚx(x⬎ 0) gives A ⫽ 兰^0⬁ e^ⴚx>2 cos wv dv⫽2(1>4 ⫹ w^{1 2}), B⫽(1>4 ⫹ w^{w 2})

8. , so that the answer is

Although many students will do the actual integration by their CAS, problems of the present type have the merit of illustrating the ideas of integral representations and transforms, a rather deep and versatile creation, and the techniques involved, such as the proper choice of integration variables and integration limits. Moreover, graphics will help in understanding the transformation process and its properties, for instance, with the help of Prob. 18 or similar experiments.

10.

12. , so that the integral

representation is

14. Project. (a) Formula (a1): Setting , we have from (10)

If we again write w instead of p, we obtain (a1).

Formula (a2): From (11) with replaced by we have

where the last equality follows from (10).

Formula (a3) follows by differentiating (10) twice with respect to w,

(b) In Prob. 7 we have

Hence by differentiating twice we obtain A

s

p^{(2wⴚ3 sin w⫺ 2 wⴚ2 cos w⫺ wⴚ1 sin w).}

A⫽ 2

p^{wⴚ1sin w.}

d²A dw² ⫽ ⫺2

p冮₀^{⬁f *}(v) cos wv dv, f*(v)⫽ v²f(v).

B*(w)⫽ 2

p冮₀^ⴥvf(v) sin wv dv⫽ ⫺dA dw vf (v) f(v)

f(ax)⫽ 冮₀^⬁A(w) cos axw dw⫽ 冮₀^{⬁Aapab cos xp dpa .}

wa⫽ p 2

p 冮₀^{ⴥ1⫺ eⴚa (cos wa}⫺ w sin wa)

1⫹ w² cos xw dw.

A⫽ 2

p 冮₀^{aeⴚv cos wv dv⫽ p2 a1⫺ eⴚa (cos wa}1⫹ w^{⫺ w2 sin wa)}b 4

p 冮₀^{ⴥsin aw}⫺ aw cos aw

w³ cos xw dw

2

p 冮0^⬁c a1 ⫺ 2

w²b sin w ⫹ 2

w cos wd cos xw w dw.

A⫽ 2

p 冮₀^{1v2 cos wv dv⫽p2w asin w ⫺} w²² sin w⫹ 2 w cos wb

By (a3) we now get the result, as before,

(c) see Prob. 7. By differentiation,

This agrees with the result obtained by using (11). Note well that here we are dealing with a relation between the two Fourier transforms under consideration.

(d) The derivation of the following formulas is similar to that of (a1)–(a3).

(d1)

(d2) B as in (11)

(d3)

16. From (11) we obtain

so that the answer is

17.

18. gives the integral representation

.

Note that the Fourier sine series of the odd periodic extension of of period has

the Fourier coefficients . Compare this with .

20. 2

p冮₀^⬁w⫺ (w cos w ⫹ sin w)>e

1⫹ w² sin xw dw

B(w) b_n⫽ (2>p)n(1⫹ cos np)>(n²⫺ 1) f(x) 2p

2

p冮₀^{⬁w (1}w^{⫹ cos 2}⫺ 1^pw) sin xw dw B(w)⫽ 2

p冮₀^pcos v sin wv dv⫽ 2 p

w(1⫹ cos pw) w²⫺ 1 2

p 冮^⬁

0

cos w⫺ 1

w sin xw dw f(x)⫽ 2

p冮₀^{ⴥsin aw⫺ aw}w^{2 cos aw} sin xw dw.

B(w)⫽ 2

p冮₀^{avsin wv dv⫽ p2}w² (sin aw⫺ awcos aw) x²f(x)⫽ 冮₀^⬁D*(w) sin xw dw, D*(w)⫽ ⫺d²B

dw². xf(x)⫽ 冮₀^⬁C*(w) cos xw dw, C*(w)⫽ dB

dw, f(bx)⫽ 1

b冮₀^⬁Bawbb sin xw dw (b ⬎ 0) B*(w)⫽ ⫺dA

dw ⫽ ⫺2 p^a

cos w

w ⫺sin w w² b . A(w)⫽ (2sin w)>(pw);

x²f (x)⫽ 2

p冮₀^{⬁c a⫺}w²³ ⫹ 1

wb sin w ⫹ 2

w² cos wd cos xw dw.

SECTION 11.8. Fourier Cosine and Sine Transforms, page 518

Purpose. Fourier cosine and sine transforms are obtained immediately from Fourier cosine and sine integrals, respectively, and we investigate some of their properties.

Content

Fourier cosine and sine transforms Transforms of derivatives (8), (9) Comment on Purpose of Transforms

Just as the Laplace transform (Chap. 6), these transforms are designed for solving differential equations. We show this for PDEs in Sec. 12.7.

Short Courses. This section can be omitted.

SOLUTIONS TO PROBLEM SET 11.8, page 522 1.

2.

If you got your answer by a CAS (e.g., by Maple) in a somewhat unusual form, plot it to see that it is correct.

3.

4. This is standard integral of calculus. If you want to do it by your CAS, you may have trouble for general ; however, you should still be able to see what the limit is and your CAS should be able to do evaluation for any fixed number a. For instance, this is the situation for Maple.

5.

6. We have if

By integration,

Hence (5a) with would give

just one of the three terms shown in the answer to Prob. 5. This should show the student the importance of the continuity assumptions in the present and similar cases.

8. The defining integral (1a) has no limit,

(w fixed!).

Similarly for (2a).

冮₀^⬁ k cos wx dx⫽ k lim_x:⬁ sin wx w

⫺2 B 2 p

sin w w³ , f

r

gc(w)⫽ 2 B 2 p

sin w w .

0⬍ x ⬍ 1, f

s

f

s

fˆc (w)⫽ 212(sin(w) ⫺ cos(w)w) 1pw³

a (⬎0) fˆc (w)⫽ 12(1 ⫺ cos(2w))

1pw² f(x)⫽B

2 p冮^⬁

0

fc(w) cos wx dw⫽ e 1 if 0 ⬍ x ⬍ 1

⫺1 if 1⬍ x ⬍ 2.

fˆ_c (w)⫽ 12(⫺1 ⫹ 2 sin(w)w ⫹ cos(2w)) 2pw²

9.

10. Use to obtain from (5b)

hence by collecting terms

and so on.

11.

12. from formula 4 in Table I

(see Sec. 11.10 of text).

14. Formula 4 in Table II (see Sec. 11.10 of text) with gives

On the other hand, by formula 2 in Table II,

Comparison proves

SECTION 11.9 Fourier Transform. Discrete and Fast Fourier Transforms, page 522

Purpose. Derivation of the Fourier transform from the complex form of the Fourier integral; explanation of its physical meaning and its basic propertites.

Main Content, Important Concepts Complex Fourier integral (4) Fourier transform (6), its inverse (7) Spectral representation, spectral density Transforms of derivatives (9), (10) Convolution

Comments on Content

The complex Fourier integral is relatively easily obtained from the real Fourier integral in Sec. 11.7, and the definition of the Fourier transform is then immediate.

Note that convolution differs from that in Chapter 6, and so does the formula (12) in the convolution theorem (we now have a factor ).

Short Courses. This section can be omitted.

12p f * g

f * g

⌫(¹2)⫽ 1p.

f_s(x^ⴚ1>2)⫽ w^ⴚ1>2. f_s(x^ⴚ1>2)⫽B

2 p ^w

ⴚ1>2 ⌫ a1 2b sin p

4 ⫽ 1

1p ^wⴚ1>2⌫a 1 2b . a⫽¹2

f_s(xe^ⴚx2>2)⫽ ⫺fs((e^ⴚx2>2)

r

c(e^ⴚx2>2)⫽ we^ⴚw2>2 12 (w²⫺ 2w sin(w) ⫺ 2 cos(w) ⫹ 2)

1pw³

(a²⫹ w²)f_s( f(x))⫽B 2 p ^w f_s(f

s

2 p ^w, f

s

e^ⴚa12w 1p(w²⫹ 1)

SOLUTIONS TO PROBLEM SET 11.9, page 533

2. This involves a transformation of exponential functions into a sine, as mentioned in Prob. 1. Integration of the defining integral gives

4. By integration of the defining integral we obtain

6. , as obtained by integration and simplification, namely

7.

8. By integration by parts we obtain

⫽ 1

12p(⫺w ⫹ i)² (1⫹ iwe^1⫹iw).

⫽ 1

12p(⫺w ⫹ i)² (1⫹ e^1⫹iw(⫺1 ⫺ i(⫺w ⫹ i)))

⫽ 1

12p^a⫺

e^1⫹iw

1⫹ iw ⫺ 1⫺ e^1⫹iw (1⫹ iw)²b

⫽ 1

12p^a⫺(⫺1)

e^1⫹iw

⫺(1 ⫹ iw) ⫹ e^⫺(1⫹iw)x

⫺(1 ⫹ iw)(1 ⫹ iw) `⁰

ⴚ1b 1

12p 冮_ⴚ1^{0xeⴚxⴚiwx dx⫽} 12¹p^a

xe^ⴚ(1⫹iw)x

⫺(1 ⫹ iw) `⁰

ⴚ1⫹ 1

1⫹ iw冮_ⴚ1^{0 eⴚ(1⫹iw)x dxb}

⫺1>212(⫺1 ⫺ aw²⫹ e^ⴚiwa⫹ ie^ⴚiwawa) 1pw²

⫽ 1

12p 2

1⫹ w² ⫽B 2 p

1 1⫹ w².

⫽ 1

12p^a 1

1⫺ iw ⫹ 1 1⫹ iwb 1

12p ^c冮_ⴚ⬁^{0 ex(1ⴚiw) dx⫹}冮₀^{⬁ex(ⴚ1ⴚiw) dxd}

12>p(1⫹ w²)

⫽ 1

12p

# ¹

k⫺ iw. 1

12p冮_ⴚ⬁^{0 e(kⴚiw)x dx⫽} ₁₂¹p

# ^e

(kⴚiw)x

k⫺ iw `

0

ⴚ⬁

⫽B 2 p

sin (w⫺ 1) w⫺ 1 .

⫽ ⫺i

12p(1⫺ w) 2i sin (1⫺ w) 1

12p冮_ⴚ1^{1e(1ⴚw)ix dx⫽} 12p^⫺i(2⫺ w) (e^(1ⴚw)i⫺ e^ⴚ(1ⴚw)i)