s of the Fourier Theorem (Sect. 1.3. The Fourier Theorem: Continuous case. : Using the Fourier Theorem. The Fourier Theorem: Piecewise continuous case. : Using the Fourier Theorem. The Fourier Theorem: Continuous case. Theorem (Fourier Series If the function f : [, R R is continuous, then f can be expressed as an infinite series f (x = a + [ ( x a n cos with the constants a n and b n given by a n = 1 b n = 1 ( x f (x cos ( x f (x sin ( x + b n sin dx, n, dx, n 1. Furthermore, the Fourier series in Eq. (?? provides a -periodic extension of f from the domain [, R to R. (1
The Fourier Theorem: Continuous case. Sketch of the Proof: Define the partial sum functions f N (x = a N [ ( x + a n cos with a n and b n given by a n = 1 ( x f (x cos b n = 1 ( x f (x sin ( x + b n sin dx, n, dx, n 1. Express f N as a convolution of Sine, Cosine, functions and the original function f. Use the convolution properties to show that lim f N(x = f (x, N x [,. s of the Fourier Theorem (Sect. 1.3. The Fourier Theorem: Continuous case. : Using the Fourier Theorem. The Fourier Theorem: Piecewise continuous case. : Using the Fourier Theorem.
: Using the Fourier Theorem. f (x = 1 + x x [,, 1 x x [, 1. Solution: In this case = 1. The Fourier series expansion is f (x = a + [ an cos(x + b n sin(x, where the a n, b n are given in the Theorem. We start with a, a = f (x dx = (1 + x dx + (1 x dx. a = (x + x ( + x x 1 (1 = 1 + ( 1 1 We obtain: a = 1. : Using the Fourier Theorem. f (x = 1 + x x [,, 1 x x [, 1. Solution: Recall: a = 1. Similarly, the rest of the a n are given by, a n = Recall the integrals a n = (1 + x cos(x dx + f (x cos(x dx cos(x dx = 1 (1 x cos(x dx. sin(x, and x cos(x dx = x sin(x + 1 n π cos(x.
: Using the Fourier Theorem. f (x = 1 + x x [,, 1 x x [, 1. Solution: It is not difficult to see that a n = 1 sin(x + [ x sin(x + 1 n π cos(x + 1 sin(x 1 [ x sin(x + 1 n π cos(x 1 a n = [ 1 n π 1 [ 1 n π cos( n π cos( 1 n π. We then conclude that a n = n π [ 1 cos(. : Using the Fourier Theorem. f (x = 1 + x x [,, 1 x x [, 1. Solution: Recall: a = 1, and a n = n π [ 1 cos(. Finally, we must find the coefficients b n. A similar calculation shows that b n =. Then, the Fourier series of f is given by f (x = 1 + n π [ 1 cos( cos(x.
: Using the Fourier Theorem. f (x = 1 + x x [,, 1 x x [, 1. Solution: Recall: f (x = 1 + n π [ 1 cos( cos(x. We can obtain a simpler expression for the Fourier coefficients a n. Recall the relations cos( = ( n, then f (x = 1 + f (x = 1 + n π [ 1 ( n cos(x. n π [ 1 + ( n+1 cos(x. : Using the Fourier Theorem. f (x = 1 + x x [,, 1 x x [, 1. Solution: Recall: f (x = 1 + n π [ 1 + ( n+1 cos(x. If n = k, so n is even, so n + 1 = k + 1 is odd, then a k = (k π (1 1 a k =. If n = k 1, so n is odd, so n + 1 = k is even, then a k = (k 1 π (1 + 1 a 4 k = (k 1 π.
: Using the Fourier Theorem. f (x = 1 + x x [,, 1 x x [, 1. Solution: Recall: f (x = 1 + n π [ 1 + ( n+1 cos(x, and a k =, a k = 4 (k 1 π. We conclude: f (x = 1 + k=1 4 (k 1 cos((k 1πx. π s of the Fourier Theorem (Sect. 1.3. The Fourier Theorem: Continuous case. : Using the Fourier Theorem. The Fourier Theorem: Piecewise continuous case. : Using the Fourier Theorem.
The Fourier Theorem: Piecewise continuous case. Recall: Definition A function f : [a, b R is called piecewise continuous iff holds, (a [a, b can be partitioned in a finite number of sub-intervals such that f is continuous on the interior of these sub-intervals. (b f has finite limits at the endpoints of all sub-intervals. The Fourier Theorem: Piecewise continuous case. Theorem (Fourier Series If f : [, R R is piecewise continuous, then the function f F (x = a + [ ( x ( x a n cos + b n sin where a n and b n given by a n = 1 ( x f (x cos b n = 1 ( x f (x sin satisfies that: dx, n, dx, n 1. (a f F (x = f (x for all x where f is continuous; (b f F (x = 1 lim [ x x + discontinuous. f (x + lim x x f (x for all x where f is
s of the Fourier Theorem (Sect. 1.3. The Fourier Theorem: Continuous case. : Using the Fourier Theorem. The Fourier Theorem: Piecewise continuous case. : Using the Fourier Theorem. : Using the Fourier Theorem. Find the Fourier series of f (x = and periodic with period T =. 1 x [,, 1 x [, 1. Solution: We start computing the Fourier coefficients b n ; b n = 1 ( x f (x sin dx, = 1, b n = b n = ( ( sin ( x dx + [ cos(x + 1 (1 sin ( x dx, [ cos(x b n = ( [ 1 [ + cos( + cos( + 1. 1,
: Using the Fourier Theorem. Find the Fourier series of f (x = and periodic with period T =. 1 x [,, 1 x [, 1. Solution: b n = ( [ 1 [ + cos( + cos( + 1. b n = 1 [ [ 1 cos( cos( + 1 = 1 cos(, We obtain: b n = [ 1 ( n. If n = k, then b k = kπ If n = k 1, then b k = hence b k = 4 (k 1π. [ 1 ( k, hence b k =. [ 1 ( k, (k 1π : Using the Fourier Theorem. Find the Fourier series of f (x = and periodic with period T =. Solution: Recall: b k =, and b k = a n = a n = 1 a n = ( ( x f (x cos ( cos ( x dx + [ sin(x 1 x [,, 1 x [, 1. 4 (k 1π. dx, = 1, + 1 (1 cos ( x dx, [ sin(x a n = ( [ 1 [ sin( + sin( 1, a n =.
: Using the Fourier Theorem. Find the Fourier series of f (x = and periodic with period T =. Solution: Recall: b k =, b k = Therefore, we conclude that 1 x [,, 1 x [, 1. 4 (k 1π, and a n =. f F (x = 4 π k=1 1 (k 1 sin( (k 1π x.