22. Periodic Functions and Fourier Series

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1 November 29, Periodic Functions and Fourier Series 1 Periodic Functions A real-valued function f(x) of a real variable is called periodic of period T > 0 if f(x + T ) = f(x) for all x R. For instance the functions sin(x), cos(x) are periodic of period 2π. It is also periodic of period 2nπ, for any positive integer n. So, there may be infinitely many periods. If needed we may specify the least period as the number T > 0 such that f(x + T ) = f(x) for all x, but f(x + s) f(x) for 0 < s < T. For later convenience, let us consider piecewise C 1 functions f(x) which are periodic of period 2 > 0 where is a positive real number. Denote this class of functions by P er (R). Note that for each integer n, the functions cos( nπx ), sin(nπx are in examples of such functions. Also, note that if f(x), g(x) P er (R), and α, β are constants, then αf + βg is also in P er (R). In particular, any finite sum a k ( a m cos( mπx m=1 ) + b m sin( mπx ) ) is in P er (R). Here the numbers a 0, a m, b m are constants. )

2 November 29, Fourier Series The next result shows that in many cases the infinite sum f(x) = a m=1 ( a m cos( mπx ) + b m sin( mπx ) ) (1) determines a well-defined function f(x) which again is in P er (R). An infinite sum as in formula (1) is called a Fourier series (after the French engineer Fourier who first considered properties of these series). Fourier Convergence Theorem. et f(x) be a piecewise C 1 function in P er (R). Then, there are constants a 0, a m, b m (uniquely defined by f) such that at each point of continuity of f(x) the expression on the right side of (1) converges to f(x). At the points y of discontinuity of f(x), the series converges to 1 (f(y ) + f(y+)). 2 The values f(y ), f(y+) denote the left and right limits of f as x y, respectively. That is, f(y ) = lim x y,x<y f(x), f(y+) = lim x y,x>y f(x).

3 November 29, It turns out that the constants a 0, a m, b m above are determined by the formulas a 0 = 1 a m = 1 f(x)dx (2) f(x) cos(mπx)dx, and (3) b m = 1 f(x) sin(mπx)dx. (4) We will justify this a bit later, but for now, let us use these formulas to compute some Fourier series. The constants a, a m, b m are called the Fourier coefficients of f. Example 1. et f(x) be defined by x, 2 x < 0 f(x) = x, 0 x < 2 f(x + 4) = f(x) for all x. Determine the Fourier coefficients of f. Note that the graph of this function f(x) looks like a triangular wave. Here = 2, and we compute 0 a 0 = 1 2 ( x)dx = = 2, 2 0 xdx

4 November 29, and, for m > 0, a m = 1 2 b m = ( x) cos(mπx 2 )dx ( x) sin(mπx 2 )dx x cos(mπx 2 )dx 2 x sin(mπx 0 2 )dx. To compute these integrals, we note that, integration by parts gives the formulas x cos(ax)dx = x a sin(ax) sin(ax) dx a = x a x sin(ax) x sin(ax)dx = cos(ax) + a a 2 et us compute a m. First note that the integrand sin(ax) + cos(ax) a 2 (5) h(x) = f(x)cos( mπx 2 ) satisfies h( x) = h(x). Hence, the two integrals are equal and a m = 2(1/2) 2 0 x cos(mπx 2 )dx = 2 0 x cos(mπx 2 )dx Using the formula for (5) above, we get

5 November 29, a m = 2 x cos(mπx 0 2 )dx = 2 = 0 + = x sin( mπx mπ 2 ) + 2 mπ 4 m 2 π 2 (cos(mπ) 1) m odd 0, m even 8, (mπ) 2 2 cos( mπx 2 ) 2 A similar calculation shows that b m = 0 for all m. We will see later that this last fact follows from the fact that f( x) = f(x) for all x. Example 2. et f(x) be defined by f(x) = 0, 3 < x < 1, 1, 1 < x < 1, 0, 1 < x < 3 f(x + 6) = f(x) for all x. Find the Fourier series for f. Here = 3, and we compute 0 a 0 = 1 3 = f(x)dx 1 1 f(x)dx

6 November 29, For m > 0, we have = 2 3 a n = 1 3 = 1 1 cos(nπx 3 )dx 1 nπ sin(nπx = 2 nπ sin(nπ 3 ) 3 ) 1 1 b n = sin(nπx 1 3 )dx 1 = nπ cos(nπx = 1 nπ = 0 3 ) 1 1 cos( nπ 3 ) cos( nπ 3 ) 3 Justification of the Fourier coefficient formulas We need the following basic facts about the integrals of certain products of sines and cosines.

7 November 29, cos(mπx )cos(nπx )dx = 0, m n,, m = n 0,, m = n = 0 (6) cos(mπx ) sin(nπx )dx = 0 for all m, n; (7) sin(mπx ) sin(nπx )dx = 0, m n,, m = n 0, 0, m = n = 0 (8) We justify formula (8), leaving the other similar calculations to the reader. First recall some formulas related to the sine and cosine functions. The sum and difference formulas are: cos(α + β) = cos(α)(cos(β) sin(α) sin(β) (9) cos(α β) = cos(α)(cos(β) + sin(α) sin(β). (10) Applying the first formula with α = β gives cos(2α) = cos(α) 2 sin(α) 2

8 November 29, This implies that 1 + cos(2α) = cos(α) 2 + sin(α) 2 + cos(α) 2 sin(α) 2 = 2 cos(α) 2 or the so-called cosine half-angle formula cos(α) 2 = 1 (1 + cos(2α)). 2 Similarly, the sine half-angle formula is sin(α) 2 = 1 (1 cos(2α)). 2 Formulas (9) and (10) imply that cos(α β) cos(α + β) = 2 sin(α) sin(β). Using α = nx, β = mx then gives cos((n m)x) cos((n + m)x) = 2 sin(nx) sin(mx). (11) This implies, for m n and both positive, sin(mπx ) sin(nπx )dx = 1 n)πx cos((m )dx n)πx cos((m )dx 2

9 November 29, If m = n = 0, then = 2π = 0. sin(mπx ) sin(mπx )dx = while if m = n 0, we have sin( (m n)πx ) m n 0 dx = 0, sin(mπx ) sin(mπx )dx = ( sin( mπx )2 ) dx Now, suppose that f(x) = a m 1 = cos( 2mπx = 1 x sin(2mπx 2 =. 2mπ ) sin((m+n)πx ) m + n ) dx a m cos( mπx ) + b m sin( mπx ). (12) Since the integrals of cosine and sine functions over intervals of lengths equal to their periods vanish, we have f(x)dx = (a m 1 a m cos( mπx ) + b m sin( mπx ))dx

10 November 29, = a 0 2 (2) + = a 0 + m 1 m 1 a m cos( mπx )dx b m sin( mπx )dx Analogously, using the orthogonality relations above, we have that, for n 1, f(x) cos(nπx )dx = cos(nπx )(a a m cos( mπx m 1 ))dx + cos(nπx )( b m sin( mπx m 1 ))dx = a m which gives (3). Formula (4) is justified in a similar way. 4 Even and Odd functions A function f(x) is called even if f( x) = f(x) for all x. Analogously, a function f(x) is called odd if f( x) = f(x) for all x. For example, cos(x) is even, and sin(x) is odd. Also, one sees easily that linear combinations of even (odd) functions are again even (odd). The following facts are useful.

11 November 29, The product of two odd functions is even. 2. The product of two even functions is even. 3. The product of and even function and an odd function is odd. The following comments are useful. 4. The Fourier series of a periodic odd function only involves sine terms: i.e.; a m = 0 for all m 1. This is because a m = 1 f(x) cos(mπx )dx and the function f(x) cos( mπx ) is odd. 5. Similarly, the Fourier series of a periodic even function only involves cosine terms: i.e.; b m = 0 for all m Any function f(x) defined on [0, ] (with > 0 has an even extension f even (x) to [, ] and also has an odd extension f odd (x) to [, ]. The definitions are f even (x) = f odd (x) = f(x) 0 x f( x) x 0 f(x) 0 x f( x) x 0

12 November 29, Hence, any piecewise continuous function f(x) on [0, ] can be represented both as Fourier cosine series and a Fourier sine series.