Fourier Series by Sebastian Vattamattam

FOURIER SERIES PROF. SEBASTIAN VATTAMATTAM

The Fourier series is named after the French scientist and mathematician Joseph Fourier(1768 - 1830), who used them in his work on heat conduction. Areas of its application include electrical engineering, vibration analysis, acoustics, optics, signal and image processing, and data compression. In some of the problems below, you are asked to derive a result : π/4 = 1 − 1/3 + 1/5 − 1/7 + .... This is usually referred to as Gregory’e Series. James Gregory(1638 - 1675), a Scottish Mathematician, proved it in 1671. But, you might be surprised to know that the same series with proof is contained in a Malayalam book called Yuktibhasha witten in AD 1530. In fact, the series was derived by a Malayalee mathematician, Madhava of Sangamagrama, in the 14th century, i.e. three centuries before James Gregory. This result was used by Madhava to calculate approximate values of π. Today it is often referred to, at least by non-Indians, as Madhava-Gregory series. 1. FOURIER SERIES - General Formula The expansion is based on Dirichlet Conditions named after Johann Peter Gustav Dirichlet (1805 - 1859). Derived from the French ‘De Richelet’, Dirichlet is pronounced ‘dirishl´e. Definition 1.1. Dirichlet Conditions Let a, b ∈ R, a ≤ b, f : [a, b] → R. The function f is said to satisfy the Dirichlet Conditions in the interval [a, b] if (1) f is periodic, (2) f has a finite number of discontinuities in each period, and, (3) f has at the most a finite number of maxima and minima. Theorem 1.2. If f satisfies the Dirichlet conditions over the interval [a, b], then for a point x ∈ [a, b] of continuity, P∞ 2πnx + b sin f (x) = a20 + n=1 an cos 2πnx n b−a b−a , Rb 2 2πnx where an = b−a a f (x) cos b−a Rb 2 and bn = b−a f (x) sin 2πnx b−a a This formula being applicable to all the different cases, it is advisable to remember this formula, rather than learning the individual formula for each case. Because of this convenience we don’t treat the individual cases separately. Theorem 1.3. The Fourier series converges to 12 (f (x + 0) + f (x − 0)) if x is a point of discontinuity in (a, b) and to 21 (f (a + 0) + f (b − 0)) if x = a or x = b. 2. Expansion of even or odd functions Definition 2.1. A function f : [−L, L] → R is even if f (−x) = f (x), ∀x ∈ [−L, L] and it is odd if f (−x) = −f (x), ∀x ∈ [−L, L]. 1

PROF. SEBASTIAN VATTAMATTAM

The graph of an even function is symmetric about the y-axis and that of an odd function is symmetric about the origin Theorem 2.2. Suppose f : [−L, L] → R satisfies the dirichlet conditions. Then, RL (1) If f is even on (−L, L), an = L2 0 f (x) cos πnx L and bn = 0. RL (2) If f is odd on (−L, L), then an = 0, and bn = L2 0 f (x) sin πnx L Two important integrals: R ax (1) eax sin(bx)dx = a2e+b2 (a sin(bx) − b cos(bx)) R ax ax (2) e cos(bx)dx = a2e+b2 (a cos(bx) + b sin(bx)) Theorem 2.3. Bernoulli’s Law R R R R R RR uvdx = u vdx − u0 vdx + u00 vdx − ...., where u and v are functions of x. That is, after the first term, write the derivative of the first and the integral of the second factor of the previous term. 3. SOLVED PROBLEMS Problem  3.1. Find the Fourier expansion of the function  0 if −2 < x < −1, k if −1 < x < 1, f (x) =  0 if 1 < x < 2 Deduce the Madhava-Gregory Series,1 − 1/3 + 1/5 − ... = π/4 The function is defined on (−2, 2). f (x) = f (−x) = k for −1 < x < 1, where f (x) 6= 0. Therefore f is even and hence bn = 0. Note that the graph is symmetric about the y-axis. a = −2,Rb = 2, b − a =R 4. R1 2 2 a0 = 24 −2 f (x)dx = 0 f (x)dx = 0 kdx = k R1 2k nπ Similarly, an = 0 k cos πnx 2 dx = πn sin 2 , n 6= 0 (3.1)

f (x)

(3.2)

f (0)

k 2k 1 πx 1 3πx 1 5πx + cos − cos + cos − ... 2 π 1 2 3 2 5 2 k 2k 1 1 + − + −... 2 π 1 3

From the definition of the function, f (0) = k. Equating these two values of f (0) we get the required series. Note In the Fourier series for f (x), we have given a suitable value to x in order to get the numerical series. This can be done in any of the Fourier expansions. Problem 3.2. Find the Fourier series of −x + 1 if −π ≤ x < 0, f (x) = x+1 if 0 ≤ x ≤ π Deduce: 1/12 + 1/32 + 1/52 + ... = π 2 /8

FOURIER SERIES

The function f is even. Therefore bn = 0 a = −π, b = π, b − a = 2π. (3.3)

Z π 2 f (x)dx 2π −π Z 2 π = f (x)dx π 0 Z 2 π (x + 1)dx = π 0 = π/2 =

Similarly, Z 2 π (x + 1) cos(nx)dx π 0 π sin(nx) − cos(nx) 2 (x + 1) −1× = π n n2 0 2 n = [(−1) − 1] πn2 π 1 1 1 f (x) = π+2 2 − 4 12 + 32 + 52 + ... π 1 1 1 f (0) = π+2 2 − 4 12 cos x + 32 cos 3x + 52 cos 5x + ... ...............(1) But, from the definition, f (0) = 1...................(2) From (1) and (2) the result follows. an

(3.4)

Problem 3.3. Find the Fourier series for f (t) = 1 − t2 , −1 < t < 1 Deduce: 1/12 − 1/22 + 1/32 − ... = π 2 /12 f (−t) = f (t), f is even. bn = 0 a = −1, b = 1, b − a = 2

(3.5)

2 2

f (t)dt Z

−1 1

(1 − t2 )dt

4/3

Similarly, Z (3.6)

(1 − t2 ) cos(πnt)dt

1 sin(πnt) − cos(πnt) 2 (1 − t2 ) − (−2t) × πn π 2 n2 0 −4(−1)n = π 2 n2

n P∞ f (t) = 32 − π42 n=1 (−1) cos(πnt) 2 n P∞ (−1)n 2 4 f (0) = 3 − π2 n=1 n2 ...........(1)

PROF. SEBASTIAN VATTAMATTAM

But, from the definition, f (0) = 1...................(2) Equating the two values of f (0) we get the result. Problem 3.4. Find the Fourier series for 14 (π − x)2 , 0 < x < 2π The interval is (0, 2π), b − a = 2π R 2π 1 2 a0 = 2π (π − x)2 dx = π 2 /6 0 4 Similarly, Z 2π 1 2 (3.7) an = (π − x)2 cos(πnt)dt 2π 0 4 1 − cos(πnt) 2 sin(πnt) = 2 (1 − t ) − (−2t) × πn π 2 n2 0 n −4(−1) = π 2 n2 n P ∞ f (t) = 32 − π42 n=1 (−1) n2 cos(πnt)

4. Fourier Integrals Theorem 4.1. Fourier Integral Theorem The R ∞ function f : (−L, L) → R satisfies the Dirichlet’s conditions and |f (t)|dt < ∞. Then if f is continuous at x ∈ (−L, L), −∞ (4.1)

f (x) =

1 π

∞

f (t) cos λ(t − x)dtdλ 0

−∞

If f is discontinuous at x ∈ (−L, L), 1 π

(4.2)

∞

f (t) cos λ(t − x)dtdλ = 0

−∞

1 [f (x + 0) + f (x − 0)] 2

Definition 4.2. Fourier Sine and Cosine Integrals cos λ(t − x) = cos(λt) cos(λx) + sin(λt) sin(λx) So, equation 4.1 becomes (4.3) f (x) =

1 π

∞

cos(λt) 0

f (t) cos(λt)dtdλ + −∞

1 π

∞

sin(λt) 0

f (t) sin(λt)dtdλ −∞

R∞ R∞ Let A(λ) := −∞ f (t) cos(λt)dt, and B(λ) := −∞ f (t) sin(λt)dt Then equation 4.3 becomes Z 1 ∞ (4.4) f (x) = [A(λ) cos(λt) + B(λ) sin(λt)]dλ π 0 Problem Express in the fourier integral form the function 4.3. x2 if 0 < x < 1, f (x) = 0 if x > 1

FOURIER SERIES

Z (4.5)

A(λ)

∞

f (x) cos(λx)dx −∞ Z 1

x2 cos(λx)dx

= 0

1 − sin(λx) − cos(λx) 2 sin(λx) = x +2 − 2x λ λ2 λ3 0 sin λ sin λ cos λ = +2 2 −2 3 λ λ λ Z 1 Z ∞ x2 sin(λx)dx f (x) sin(λx)dx = B(λ) =

(4.6)

−∞

1 − cos(λx) − sin(λx) cos(λx) = x2 − 2x + 2 λ λ2 λ3 0 − cos λ sin λ cos λ 2 sin λ = +2 2 +2 3 − λ λ λ λ3 f (x)

∞

cos λ sin λ sin λ + 2 2 − 2 3 ] cos(λx)dλ λ λ λ 0 Z ∞ 1 − cos λ sin λ cos λ 2 sin λ + [ +2 2 +2 3 − ] sin(λx)dλ π 0 λ λ λ λ3 1 π

[

Definition 4.4. Consider 4.3 If f is odd, Z Z ∞ 2 ∞ (4.7) f (x) = sin(λt)dλ f (t) sin(λt)dt π 0 o This is called the Fourier Sine Integral. If f is even, Z Z ∞ 2 ∞ cos(λt)dλ f (t) cos(λt)dt (4.8) f (x) = π 0 o This is called the Fourier Cosine Integral. Problem 4.5. Express as a Fourier sine integral the function 1 if 0 ≤ x ≤ π, f (x) = 0 if x > π R∞ Hence evaluate 0 1−cos(πλ) sin λπdλ λ ExtendR the definitionR to (−∞, 0) so that it is odd. ∞ ∞ f (x) = π2 0 sin(λt)dλ o f (t) sin(λt)dt Z (4.9)

∞

Z f (t) sin(λt)dt

sin(λt)dt o

= (4.10)

f (x)

1 − cos πλ Z λ 2 ∞ 1 − cos πλ sin(λx)dλ π 0 λ

PROF. SEBASTIAN VATTAMATTAM

R ∞ 1−cos λπ Therefore, sin(λx)dλ = π2 f (x) λ 0 π if 0 < x < π, 2 = 0 if x > π x = π is a point of discontinuity, and hence

Z (4.11) 0

∞

1 − cos πλ sin(λπ)dλ λ 1 [f (π − 0] + f (π + 0) 2 = π/4 =

4.1. Complex form of Fourier Integrals. In 4.1, Z Z 1 ∞ ∞ (4.12) f (t) cos λ(t − x)dtdλ f (x) = π 0 −∞ Z ∞ Z ∞ 1 = cos λ(t − x)dλ f (t)dt π 0 −∞ Z ∞ Z ∞ 1 cos λ(t − x)dλ f (t)dt..............(a) = 2π −∞ −∞ R∞ R∞ 1 Also, 0 = 2π sin λ(t − x)dλ −∞ f (t)dt..............(b) −∞ (a) + i(b) gives, Z ∞ Z ∞ 1 [cos λ(t − x) + i sin λ(t − x)]dλ (4.13) f (x) = f (t)dt 2π −∞ −∞ Z ∞Z ∞ 1 = f (t)eiλ(t−x) dtdλ 2π −∞ −∞