Lecture Notes for Math 251: ODE and PDE. Lecture 33: 10.4 Even and Odd Functions

Transcription

1 Lecture Notes for Math 51: ODE and PDE. Lecture : 1.4 Even and Odd Functions Shawn D. Ryan Spring 1 Last Time: We studied what a given Fourier Series converges to if at a. 1 Even and Odd Functions Before we can appy the discussion from Section 1. to the Fourier Sine and Cosine Series, we need to review some facts about Even and Odd Functions. Reca that an even function is a function satisfying g( x) = g(x). (1) This means that the graph y = g(x) is symmetric with respect to the y-axis. An odd function satisfies g( x) = g(x) () meaning that its graph y = g(x) is symmetric with respect to the origin. Exampe 1. A monomia x n is even if n is even and odd if n is odd. cos(x) is even and sin(x) is odd. Note tan(x) is odd. There are some genera rues for how products and sums behave: (1) If g(x) is odd and h(x) is even, their product g(x)h(x) is odd. () If g(x) and h(x) are either both even or both odd, g(x)h(x) is even. () The sum of two even functions or two odd functions is even or odd, respectivey. To remember the rues consider how many negative signs come out of the arguments. EXERCISE: Verify these rues. (4) The sum of an even and an odd function can be anything. In fact, any function on (, ) can be written as a sum of an even function, caed the even part, and an odd function, caed the odd part. df dx and (5) Differentiation and Integration can change the parity of a function. If f(x) is even, x f(s)ds are both odd, and vice versa. The graph of an odd function g(x) must pass through the origin by definition. This aso tes us that if g(x) is even, as ong as g () exists, then g () =. 1

2 Definite Integras on symmetric intervas of odd and even functions have usefu properties (odd)dx = and (even)dx = (even)dx () Given a function f(x) defined on (, ), there is ony one way to extend it to (, ) to an even or odd function. The even extension of f(x) is f(x) for < x < (4) f( x) for < x <. This is just its refection across the y-axis. Notice that the even extension is not necessariy defined at the origin. The odd extension of f(x) is f(x) for < x < f odd (x) = f( x) for < x <. (5) for x = This is just its refection through the origin. 1.1 Fourier Sine Series Each of terms in the Fourier Sine Series for f(x), sin( nπx ), is odd. As with the fu Fourier Series, each of these terms aso has period. So we can think of the Fourier Sine Series as the expansion of an odd function with period defined on the entire ine which coincides with f(x) on (, ). One can show that the fu Fourier Series of f odd is the same as the Fourier Sine Series of f(x). Let 1 A + A n cos( nπx be the Fourier Series for f odd (x), with coefficients given in Section 1. A n = 1 But f odd is odd and cos is even, so their product is again odd. B n = 1 ) + B n sin( nπx ) (6) f odd (x) cos( nπx )dx = (7) But both f odd and sin are odd, so their product is even. B n = = f odd (x) sin( nπx )dx (8) f odd (x) sin( nπx )dx (9) f(x) sin( nπx )dx, (1)

3 which are just the Fourier Sine coefficients of f(x). Thus, as the Fourier Sine Series of f(x) is the fu Fourier Series of f odd (x), the -periodic odd function that the Fourier Sine Series expands is just the periodic extension f odd. This goes both ways. If we want to compute a Fourier Series for an odd function on (, ) we can just compute the Fourier Sine Series of the function restricted to (, ). It wi amost converge to the origina function on (, ) with the ony issues occurring at any jump discontinuities. The ony works for odd functions. Do not use the formua for the coefficients of the Sine Series, uness you are working with an odd function. Exampe. Write down the odd extension of f(x) = x on (, ) and compute its Fourier Series. To get the odd extension of f(x) we wi need to see how to refect f across the origin. What we end up with is the function x < x < f odd (x) =. (11) x < x < Now. what is the Fourier Series of f odd (x)? By the previous discussion, we know that is wi be identica to the Fourier Sine Series of f(x), as this wi converge on (, ) to f odd. So we have where A n = = Thus the desired Fourier Series is f odd (x) = A n sin( nπx ), (1) ( x) sin( nπx )dx (1) [ ( x) nπ cos( nπx ) n π sin(nπx )] (14) = nπ. (15) f odd (x) = π 1 n sin(nπx ). (16) You might wonder how we were abe a few ectures ago to compute the Fourier Sine Series of a constant function ike f(x) = 1 which is even. It is important to remember that if we are computing the Fourier Sine Series for f(x), it ony needs to converge to f(x) on (, ), where issues of evenness and oddness do not occur. The Fourier Sine Series wi converge to the odd extension of f(x) on (, ). Exampe. Find the Fourier Series for the odd extension of f(x) = < x < x < x <. (17)

4 on (, ). The Fourier Series for f odd (x) on (, ) wi just be the Fourier Sine Series for f(x) on (, ). The Fourier Sine coefficients for f(x) are A n = f(x) sin( nπx )dx (18) = ( sin(nπx )dx + (x ) ) sin(nπx ) (19) = ( 9 nπ cos(nπx ) + (x ) cos( nπx nπ ) + 9 ) n π sin(nπx ) () = ( 9 nπ (cos(nπ ) 1) 9 nπ cos(nπ) 9 ) n π sin(nπx ) (1) = ( 9 nπ (1 cos(nπ ) + ( 1)n+1 ) 9 ) n π sin(nπ ) () = ( 1 cos( nπ nπ ) + ( 1)n+1 ) nπ sin(nπ ) () and the Fourier Series is f odd (x) = π 1 n [1 cos(nπ ) + ( 1)n+1 nπ sin(nπ )] sin(nπx). (4) EXERCISE: Sketch the Odd Extension of f(x) given in the previous Exampe and write down the formua for it. 1. Fourier Cosine Series Now consider what happens for the Fourier Cosine Series of f(x) on (, ). This is anaogous to the Sine Series case. Every term in the Cosine Series has the form A n cos( nπx ) (5) and hence is even, so the entire Cosine Series is even. So the Cosine Series must converge on (, ) to an even function which coincides on (, ) with f(x). this must be the even extension f(x) < x <. (6) f( x) < x < Notice that this definition does not specify the vaue of the function at zero, the ony restriction on an even function at zero is that, if it exists, the derivative shoud be zero. 4

5 It is straight forward enough to show that the Fourier coefficients of f even (x) coincide with the Fourier Cosine coefficients of f(x). The Euer-Fourier formuas give A n = 1 = = f even (x) cos( nπx )dx (7) f even (x) cos( nπx )dx since f even (x) cos( nπx ) is even (8) f even (x) cos( nπx )dx (9) which are the Fourier Cosine coefficients of f(x) on (, ) B n = 1 f even (x) sin( nπx )dx = () since f even (x) sin( nπx ) is odd. Thus the Fourier Cosine Series of f(x) on (, ) can be considered as the Fourier expansion of f even (x) on (, ), and therefore aso as expansion of the periodic extension of f even (x). It wi converge as in the Fourier Convergence Theorem to this periodic extension. This aso means that if we want to compute the Fourier Series of an even function, we can just compute the Fourier Cosine Series of its restriction to (, ). It is very important that this ony be attempted if the function we are starting with is even. Exampe 4. Write down the even extension of f(x) = x on (, ) and compute its Fourier Series. The even extension wi be x < x <. (1) + x < x < Its Fourier Series is the same as the Fourier Cosine Series of f(x), by the previous discussion. So we can just compute the coefficients. Thus we have 1 A + A n cos( nπx ), () 5

6 where So we have A A n = = = = x) [( nπ = ( f(x)dx = ( x)dx = () f(x) cos( nπx )dx (4) ( x) cos( nπx )dx (5) sin( nπx ) n π( cos(nπ) + cos()) n π cos(nπx )] (6) ) (7) = n π (( 1)n+1 + 1). (8) + Exampe 5. Write down the even extension of f(x) = n π (( 1)n+1 + 1). (9) x < x x (4) and compute its Fourier Series. Using Equation (??) we see that the even extension is x < x < x < < x < x x. (41) 6

7 We just need to compute the Fourier Cosine coefficients of the origina f(x) on (, ). A = = ( / f(x)dx (4) dx + x ) / dx = ( ) = 9 (44) 4 A n = f(x) cos( nπx )dx (45) = ( / cos(nπx )dx + (x ) / ) cos(nπx )dx (46) = ( 9 nπ sin(nπx ) / + (x ) sin( nπx nπ ) / + 9 ) n π cos(nπx ) / (47) = ( 9 nπ sin(nπ ) + 9 (cos(nπ) cos( nπ )) n π ) (48) = 6 ( 1 nπ sin(nπ ) + 1 (( 1) n cos( nπ )) nπ ) (49) = 6 ( 1 nπ nπ (( 1)n cos( nπ )) + 1 ) sin(nπ ). (5) (4) So the Fourier Series is f even = π ( 1 1 ( ( 1) n cos( nπ n nπ )) + 1 ) sin(nπ ) cos( nπx ). (51) HW 1.4 # 1,,8,11,1,16,17 7