Chapter 28. Fourier Series An Eigenvalue Problem.


 Kelley Stone
 3 years ago
 Views:
Transcription
1 Chpter 28 Fourier Series Every time I close my eyes The noise inside me mplifies I cn t escpe I relive every moment of the dy Every misstep I hve mde Finds wy it cn invde My every thought And this is why I find myself wke Filure Tom Sher (Assemblge 23) 28. An Eigenvlue Problem. A self djoint eigenvlue problem. Consider the eigenvlue problem y + λy = 0, y() = y(π), y () = y (π). 330
2 We rewrite the eqution so the eigenvlue is on the right side. L[y] y = λy We demonstrte tht this eigenvlue problem is self djoint. v L[u] L[v] u = v u v u = [ vu ] π + v u [ v u] π v u = v(π)u (π) + v()u () + v (π)u(π) v ()u() = v(π)u (π) + v(π)u (π) + v (π)u(π) v (π)u(π) = 0 Since Green s Identity reduces to v L[u] L[v] u = 0, the problem is self djoint. This mens tht the eigenvlues re rel nd tht eigenfunctions corresponding to distinct eigenvlues re orthogonl. We compute the Ryleigh quotient for n eigenvlue λ with eigenfunction φ. We see tht the eigenvlues re nonnegtive. λ = [ φφ ] π + φ φ φ φ = φ(π)φ (π) + φ()φ () + φ φ φ φ = φ(π)φ (π) + φ(π)φ (π) + φ φ φ φ = φ φ φ φ Computing the eigenvlues nd eigenfunctions. Now we find the eigenvlues nd eigenfunctions. First we consider the cse λ = 0. The generl solution of the differentil eqution is y = c + c 2 x. 33
3 The solution tht stisfies the boundry conditions is y = const. Now consider λ > 0. The generl solution of the differentil eqution is ( ) ( ) y = c cos λx + c 2 sin λx. We pply the first boundry condition. Then we pply the second boundry condition. y() = y(π) ( c cos ) ( λπ + c 2 sin ) ( ) ( ) λπ = c cos λπ + c 2 sin λπ ( ) ( ) ( ) ( ) c cos λπ c 2 sin λπ = c cos λπ + c 2 sin λπ ( ) c 2 sin λπ = 0 y () = y (π) c λ sin ( ) ( λπ + c 2 λ cos ) ( ) ( ) λπ = c λ sin λπ + c 2 λ cos λπ ( ) ( ) ( ) ( ) c sin λπ + c 2 cos λπ = c sin λπ + c 2 cos λπ ( ) c sin λπ = 0 ( λπ ) To stisify the two boundry conditions either c = c 2 = 0 or sin = 0. The former yields the trivil solution. The ltter gives us the eigenvlues λ n = n 2, n Z +. The corresponding solution is y n = c cos(nx) + c 2 sin(nx). There re two eigenfunctions for ech of the positive eigenvlues. We choose the eigenvlues nd eigenfunctions. λ 0 = 0, φ 0 = 2 λ n = n 2, φ 2n = cos(nx), φ 2n = sin(nx), for n =, 2, 3,
4 Orthogonlity of Eigenfunctions. We know tht the eigenfunctions of distinct eigenvlues re orthogonl. In ddition, the two eigenfunctions of ech positive eigenvlue re orthogonl. cos(nx) sin(nx) dx = [ ] π 2n sin2 (nx) = 0 Thus the eigenfunctions {, cos(x), sin(x), cos(2x), sin(2x)} re n orthogonl set Fourier Series. A series of the eigenfunctions φ 0 = 2, φ() n = cos(nx), φ (2) n = sin(nx), for n is ( n cos(nx) + b n sin(nx) ). This is known s Fourier series. (We choose φ 0 = so ll of the eigenfunctions hve the sme norm.) A firly generl 2 clss of functions cn be expnded in Fourier series. Let f(x) be function defined on < x < π. Assume tht f(x) cn be expnded in Fourier series f(x) ( n cos(nx) + b n sin(nx) ). (28.) Here the mens hs the Fourier series. We hve not sid if the series converges yet. For now let s ssume tht the series converges uniformly so we cn replce the with n =. 333
5 We integrte Eqution 28. from to π to determine 0. f(x) dx = 2 0 f(x) dx = π 0 + dx + n cos(nx) + b n sin(nx) dx ( n cos(nx) dx + b n 0 = π f(x) dx = π 0 f(x) dx Multiplying by cos(mx) nd integrting will enble us to solve for m. f(x) cos(mx) dx = cos(mx) dx ( n cos(nx) cos(mx) dx + b n ) sin(nx) dx ) sin(nx) cos(mx) dx All but one of the terms on the right side vnishes due to the orthogonlity of the eigenfunctions. f(x) cos(mx) dx = m cos(mx) cos(mx) dx ( ) f(x) cos(mx) dx = m 2 + cos(2mx) dx m = π f(x) cos(mx) dx = π m f(x) cos(mx) dx. 334
6 Note tht this formul is vlid for m = 0,, 2,... Similrly, we cn multiply by sin(mx) nd integrte to solve for b m. The result is b m = π f(x) sin(mx) dx. n nd b n re clled Fourier coefficients. Although we will not show it, Fourier series converge for firly generl clss of functions. Let f(x ) denote the left limit of f(x) nd f(x + ) denote the right limit. Exmple For the function defined the left nd right limits t x = 0 re f(x) = { 0 for x < 0, x + for x 0, f(0 ) = 0, f(0 + ) =. Result Let f(x) be 2πperiodic function for which Fourier coefficients n = π f(x) cos(nx) dx, b n = π f(x) dx exists. Define the f(x) sin(nx) dx. If x is n interior point of n intervl on which f(x) hs limited totl fluctution, then the Fourier series of f(x) ( n cos(nx) + b n sin(nx) ), converges to 2 (f(x ) + f(x + )). If f is continuous t x, then the series converges to f(x). 335
7 Periodic Extension of Function. Let g(x) be function tht is rbitrrily defined on x < π. The Fourier series of g(x) will represent the periodic extension of g(x). The periodic extension, f(x), is defined by the two conditions: f(x) = g(x) for π x < π, f(x + 2π) = f(x). The periodic extension of g(x) = x 2 is shown in Figure Figure 28.: The Periodic Extension of g(x) = x 2. Limited Fluctution. A function tht hs limited totl fluctution cn be written f(x) = ψ + (x) ψ (x), where ψ + nd ψ re bounded, nondecresing functions. An exmple of function tht does not hve limited totl fluctution 336
8 is sin(/x), whose fluctution is unlimited t the point x = 0. Functions with Jump Discontinuities. Let f(x) be discontinuous function tht hs convergent Fourier series. Note tht the series does not necessrily converge to f(x). Insted it converges to ˆf(x) = 2 (f(x ) + f(x + )). Exmple Consider the function defined by { x for π x < 0 f(x) = π 2x for 0 x < π. The Fourier series converges to the function defined by 0 for x = x for π < x < 0 ˆf(x) = π/2 for x = 0 π 2x for 0 < x < π. The function ˆf(x) is plotted in Figure Lest Squres Fit Approximting function with Fourier series. Suppose we wnt to pproximte 2πperiodic function f(x) with finite Fourier series. f(x) N ( n cos(nx) + b n sin(nx)) Here the coefficients re computed with the fmilir formuls. Is this the best pproximtion to the function? Tht is, is it possible to choose coefficients α n nd β n such tht f(x) α N (α n cos(nx) + β n sin(nx)) 337
9 Figure 28.2: Grph of ˆf(x). would give better pproximtion? Lest squred error fit. The most common criterion for finding the best fit to function is the lest squres fit. The best pproximtion to function is defined s the one tht minimizes the integrl of the squre of the devition. Thus if f(x) is to be pproximted on the intervl x b by series N f(x) c n φ n (x), (28.2) 338
10 the best pproximtion is found by choosing vlues of c n tht minimize the error E. N 2 E f(x) c n φ n (x) dx Generlized Fourier coefficients. We consider the cse tht the φ n re orthogonl. For simplicity, we lso ssume tht the φ n re relvlued. Then most of the terms will vnish when we interchnge the order of integrtion nd summtion. ( b N N N ) E = f 2 2f c n φ n + c n φ n c m φ m dx E = E = E = f 2 dx 2 We complete the squre for ech term. E = f 2 dx + N c n fφ n dx + f 2 dx 2 f 2 dx + N N m= m= N c n fφ n dx + N φ 2 n dx ( c 2 n ( c n Ech term involving c n is nonnegtive, nd is minimized for N c n c m φ n φ m dx N c 2 n φ 2 n dx ) φ 2 n dx 2c n fφ n dx fφ n dx φ2 n dx ) 2 ( fφ ) 2 n dx b φ2 n dx c n = fφ n dx φ2 n dx. (28.3) 339
11 We cll these the generlized Fourier coefficients. For such choice of the c n, the error is Since the error is nonnegtive, we hve E = f 2 dx N c 2 n φ 2 n dx. N f 2 dx c 2 n φ 2 n dx. This is known s Bessel s Inequlity. If the series in Eqution 28.2 converges in the men to f(x), lim N E = 0, then we hve equlity s N. f 2 dx = φ 2 n dx. This is Prsevl s equlity. c 2 n Fourier coefficients. Previously we showed tht if the series, f(x) = ( n cos(nx) + b n sin(nx), converges uniformly then the coefficients in the series re the Fourier coefficients, n = f(x) cos(nx) dx, b n = π π 340 f(x) sin(nx) dx.
12 Now we show tht by choosing the coefficients to minimize the squred error, we obtin the sme result. We pply Eqution 28.3 to the Fourier eigenfunctions. n = b n = 0 = f 2 dx dx = π 4 f cos(nx) dx cos2 (nx) dx = π f sin(nx) dx sin2 (nx) dx = π f(x) dx f(x) cos(nx) dx f(x) sin(nx) dx 28.4 Fourier Series for Functions Defined on Arbitrry Rnges If f(x) is defined on c d x < c + d nd f(x + 2d) = f(x), then f(x) hs Fourier series of the form Since f(x) ( ) ( ) nπ(x + c) nπ(x + c) n cos + b n sin. d d c+d c d ( ) nπ(x + c) cos 2 dx = d the Fourier coefficients re given by the formuls n = d b n = d c+d c d c+d c d c+d c d ( ) nπ(x + c) sin 2 dx = d, d ( ) nπ(x + c) f(x) cos dx d ( ) nπ(x + c) f(x) sin dx. d 34
13 Exmple Consider the function defined by x + for x < 0 f(x) = x for 0 x < 3 2x for x < 2. This function is grphed in Figure The Fourier series converges to ˆf(x) = (f(x ) + f(x + ))/2, for x = 2 x + for < x < 0 ˆf(x) = for x = 0 2 x for 0 < x < 3 2x for x < 2. ˆf(x) is lso grphed in Figure The Fourier coefficients re n = 2 ( ) 2nπ(x + /2) f(x) cos dx 3/2 3 = 2 5/2 ( ) 2nπx f(x /2) cos dx 3 /2 3 = 2 /2 ( ) 2nπx (x + /2) cos dx / /2 ( ) 2nπx (4 2x) cos dx 3 3/2 3 = ( 2nπ (nπ) sin 2 3 3/2 /2 ) [ ( nπ 2( ) n nπ + 9 sin 3 ( ) 2nπx (x /2) cos dx 3 )] 342
14 Figure 28.3: A Function Defined on the rnge x < 2 nd the Function to which the Fourier Series Converges. b n = 3/2 = 2 3 = /2 /2 /2 /2 ( ) 2nπ(x + /2) f(x) sin dx 3 ( ) 2nπx f(x /2) sin dx 3 ( ) 2nπx (x + /2) sin dx + 2 3/2 ( ) 2nπx (x /2) sin dx 3 3 / /2 ( ) 2nπx (4 2x) sin dx 3 3/2 3 ) [ ( nπ ) ( nπ )] 2( ) n nπ + 4nπ cos 3 sin 3 3 = 2 (nπ) 2 sin2 ( nπ 3 343
15 28.5 Fourier Cosine Series If f(x) is n even function, (f( x) = f(x)), then there will not be ny sine terms in the Fourier series for f(x). The Fourier sine coefficient is b n = π f(x) sin(nx) dx. Since f(x) is n even function nd sin(nx) is odd, f(x) sin(nx) is odd. b n is the integrl of n odd function from to π nd is thus zero. We cn rewrite the cosine coefficients, n = π = 2 π Exmple Consider the function defined on [0, π) by f(x) = The Fourier cosine coefficients for this function re n = 2 π = { π /2 0 0 f(x) cos(nx) dx f(x) cos(nx) dx. { x for 0 x < π/2 π x for π/2 x < π. x cos(nx) dx + 2 π π/2 for n = 0, 4 8 cos ( ) ( ) nπ πn 2 2 sin 2 nπ for n. 4 (π x) cos(nx) dx In Figure 28.4 the even periodic extension of f(x) is plotted in dshed line nd the sum of the first five nonzero terms in the Fourier cosine series re plotted in solid line. 344
16 Figure 28.4: Fourier Cosine Series Fourier Sine Series If f(x) is n odd function, (f( x) = f(x)), then there will not be ny cosine terms in the Fourier series. Since f(x) cos(nx) is n odd function, the cosine coefficients will be zero. Since f(x) sin(nx) is n even function,we cn rewrite the sine coefficients b n = 2 π 0 f(x) sin(nx) dx. 345
17 Exmple Consider the function defined on [0, π) by { x for 0 x < π/2 f(x) = π x for π/2 x < π. The Fourier sine coefficients for this function re /2 b n = 2 x sin(nx) dx + 2 π 0 π = 6 ( nπ ) ( nπ ) πn cos sin π/2 (π x) sin(nx) dx In Figure 28.5 the odd periodic extension of f(x) is plotted in dshed line nd the sum of the first five nonzero terms in the Fourier sine series re plotted in solid line Complex Fourier Series nd Prsevl s Theorem By writing sin(nx) nd cos(nx) in terms of e ınx nd e ınx we cn obtin the complex form for Fourier series ( n cos(nx) + b n sin(nx) ) = ( ) n 2 (eınx + e ınx ) + b n ı2 (eınx e ınx ) = ( 2 ( n ıb n ) e ınx + ) 2 ( n + ıb n ) e ınx where = n= c n e ınx ( 2 n ıb n ) for n c n = 0 2 for n = 0 ( 2 n + ıb n ) for n. 346
18 Figure 28.5: Fourier Sine Series. The functions {...,e ıx,, e ıx, e ı2x,...}, stisfy the reltion e ınx e ımx dx = e ı(n m)x dx { 2π for n = m = 0 for n m. Strting with the complex form of the Fourier series of function f(x), f(x) c n e ınx, 347
19 we multiply by e ımx nd integrte from to π to obtin If f(x) is relvlued then f(x) e ımx dx = c n e ınx e ımx dx c m = f(x) e ımx dx 2π c m = f(x) e ımx dx = f(x)(e 2π 2π ımx ) dx = c m where z denotes the complex conjugte of z. Assume tht f(x) hs uniformly convergent Fourier series. f 2 (x) dx = = 2π = 2π = 2π ( m= n= ( n= c n c n ( c m e ımx )( n= c n e ınx ) dx [ ] 4 ( n + ıb n )( n ıb n ) ) ( 2 n + b 2 n) [ ] ) 4 ( n ıb n )( n + ıb n ) This yields result known s Prsevl s theorem which holds even when the Fourier series of f(x) is not uniformly convergent. 348
20 Result Prsevl s Theorem. If f(x) hs the Fourier series then f(x) ( n cos(nx) + b n sin(nx)), f 2 (x) dx = π π ( 2 n + b 2 n) Behvior of Fourier Coefficients Before we jump hipdeep into the grunge involved in determining the behvior of the Fourier coefficients, let s tke step bck nd get some perspective on wht we should be looking for. One of the importnt questions is whether the Fourier series converges uniformly. From Result 2.2. we know tht uniformly convergent series represents continuous function. Thus we know tht the Fourier series of discontinuous function cnnot be uniformly convergent. From Section 2.2 we know tht series is uniformly convergent if it cn be bounded by series of positive terms. If the Fourier coefficients, n nd b n, re O(/n α ) where α > then the series cn be bounded by (const) /nα nd will thus be uniformly convergent. Let f(x) be function tht meets the conditions for hving Fourier series nd in ddition is bounded. Let (, p ), (p, p 2 ), (p 2, p 3 ),...,(p m, π) be prtition into finite number of intervls of the domin, (, π) such tht on ech intervl f(x) nd ll it s derivtives re continuous. Let f(p ) denote the left limit of f(p) nd f(p + ) denote the right limit. f(p ) = lim f(p ǫ), f(p+ ǫ 0 + ) = lim f(p + ǫ) ǫ 0 + Exmple The function shown in Figure 28.6 would be prtitioned into the intervls ( 2, ), (, 0), (0, ), (, 2). 349
21 Figure 28.6: A Function tht cn be Prtitioned. Suppose f(x) hs the Fourier series f(x) n cos(nx) + b n sin(nx). We cn use the integrl formul to find the n s. n = π = π ( p f(x) cos(nx) dx f(x) cos(nx) dx + p2 ) f(x) cos(nx) dx + + f(x) cos(nx) dx p p m 350
22 Using integrtion by prts, = ( [ ] p [ ] p2 [ ] ) π f(x) sin(nx) + f(x) sin(nx) + + f(x) sin(nx) nπ p p m ( p p2 ) f (x) sin(nx) dx + f (x) sin(nx) dx + f (x) sin(nx) dx nπ p p m = { [f(p ) f(p + ) ] sin(np ) + + [ f(p nπ m) f(p + m) ] } sin(np m ) π f (x) sin(nx) dx n π = n A n n b n where A n = π m sin(np j ) [ f(p j ) f(p+ j )] j= nd the b n re the sine coefficients of f (x). Since f(x) is bounded, A n = O(). Since f (x) is bounded, b n = π f (x) sin(nx) dx = O(). Thus n = O(/n) s n. (Actully, from the RiemnnLebesgue Lemm, b n = O(/n).) 35
23 Now we repet this nlysis for the sine coefficients. b n = π = π = nπ f(x) sin(nx) dx p2 ) f(x) sin(nx) dx + f(x) sin(nx) dx + + f(x) sin(nx) dx p p m { [f(x) ] p cos(nx) + [ f(x) cos(nx) ] p 2 p + + [ f(x) cos(nx) ] } π p m + ( p p2 f (x) cos(nx) dx + f (x) cos(nx) dx + nπ p ( p = n B n + n n ) f (x) cos(nx) dx p m where B n = ( )n [ ] f() f(π) π π m cos(np j ) [ f(p j ) f(p+ j )] j= nd the n re the cosine coefficients of f (x). Since f(x) nd f (x) re bounded, B n, n = O() nd thus b n = O(/n) s n. With integrtion by prts on the Fourier coefficients of f (x) we could find tht n = n A n n b n where A n = π m j= sin(np j)[f (p j ) f (p + j )] nd the b n re the sine coefficients of f (x), nd b n = n B n + n n where B n = ( )n π [f () f (π)] π m j= cos(np j)[f (p j ) f (p + j )] nd the n re the cosine coefficients of f (x). 352
24 Now we cn rewrite n nd b n s n = n A n + n 2B n n 2 n b n = n B n + n 2A n n 2b n. Continuing this process we could define A (j) n nd B (j) n so tht n = n A n + n 2B n n 3A n n 4B n + b n = n B n + n 2A n + n 3B n n 4A n. For ny bounded function, the Fourier coefficients stisfy n, b n = O(/n) s n. If A n nd B n re zero then the Fourier coefficients will be O(/n 2 ). A sufficient condition for this is tht the periodic extension of f(x) is continuous. We see tht if the periodic extension of f (x) is continuous then A n nd B n will be zero nd the Fourier coefficients will be O(/n 3 ). Result Let f(x) be bounded function for which there is prtition of the rnge (, π) into finite number of intervls such tht f(x) nd ll it s derivtives re continuous on ech of the intervls. If f(x) is not continuous then the Fourier coefficients re O(/n). If f(x), f (x),...,f (k 2) (x) re continuous then the Fourier coefficients re O(/n k ). If the periodic extension of f(x) is continuous, then the Fourier coefficients will be O(/n 2 ). The series n cos(nx) cn be bounded by M /n2 where M = mx ( n + b n ). Thus the Fourier series converges to f(x) uniformly. n Result If the periodic extension of f(x) is continuous then the Fourier series of f(x) will converge uniformly for ll x. If the periodic extension of f(x) is not continuous, we hve the following result. 353
25 Result If f(x) is continuous in the intervl c < x < d, then the Fourier series is uniformly convergent in the intervl c + δ x d δ for ny δ > 0. Exmple Different Rtes of Convergence. A Discontinuous Function. Consider the function defined by f (x) = { for < x < 0, for 0 < x <. This function hs jump discontinuities, so we know tht the Fourier coefficients re O(/n). Since this function is odd, there will only be sine terms in it s Fourier expnsion. Furthermore, since the function is symmetric bout x = /2, there will be only odd sine terms. Computing these terms, b n = 2 sin(nπx) dx 0 [ ] = 2 nπ cos(nπx) 0 = 2 ( ( )n nπ ) nπ = { 4 nπ for odd n 0 for even n. The function nd the sum of the first three terms in the expnsion re plotted, in dshed nd solid lines respectively, in Figure Although the three term sum follows the generl shpe of the function, it is clerly not good pproximtion. 354
26 Figure 28.7: Three Term Approximtion for Function with Jump Discontinuities nd Continuous Function. A Continuous Function. Consider the function defined by x for < x < /2 f 2 (x) = x for /2 < x < /2 x + for /2 < x <. 355
27 Figure 28.8: Three Term Approximtion for Function with Continuous First Derivtive nd Comprison of the Rtes of Convergence. Since this function is continuous, the Fourier coefficients will be O(/n 2 ). Also we see tht there will only be odd sine terms in the expnsion. b n = = 2 /2 /2 0 ( x ) sin(nπx) dx + /2 /2 x sin(nπx) dx + x sin(nπx) dx + 2 ( x) sin(nπx) dx /2 = 4 (nπ) sin(nπ/2) { 2 4 ( ) (n )/2 for odd n (nπ) = 2 0 for even n. 356 /2 ( x + ) sin(nπx) dx
28 The function nd the sum of the first three terms in the expnsion re plotted, in dshed nd solid lines respectively, in Figure We see tht the convergence is much better thn for the function with jump discontinuities. A Function with Continuous First Derivtive. Consider the function defined by f 3 (x) = { x( + x) for < x < 0 x( x) for 0 < x <. Since the periodic extension of this function is continuous nd hs continuous first derivtive, the Fourier coefficients will be O(/n 3 ). We see tht the Fourier expnsion will contin only odd sine terms. b n = = x( + x) sin(nπx) dx + x( x) sin(nπx) dx = 4( ( )n ) (nπ) 3 = { 4 (nπ) 3 for odd n 0 for even n. 0 x( x) sin(nπx) dx The function nd the sum of the first three terms in the expnsion re plotted in Figure We see tht the first three terms give very good pproximtion to the function. The plots of the function, (in dshed line), nd the three term pproximtion, (in solid line), re lmost indistinguishble. In Figure 28.8 the convergence of the of the first three terms to f (x), f 2 (x), nd f 3 (x) re compred. In the lst grph we see closeup of f 3 (x) nd it s Fourier expnsion to show the error. 357
29 28.9 Gibb s Phenomenon The Fourier expnsion of f(x) = { for 0 x < for x < 0 is f(x) 4 π n sin(nπx). For ny fixed x, the series converges to 2 (f(x ) + f(x + )). For ny δ > 0, the convergence is uniform in the intervls + δ x δ nd δ x δ. How will the nonuniform convergence t integrl vlues of x ffect the Fourier series? Finite Fourier series re plotted in Figure 28.9 for 5, 0, 50 nd 00 terms. (The plot for 00 terms is closeup of the behvior ner x = 0.) Note tht t ech discontinuous point there is series of overshoots nd undershoots tht re pushed closer to the discontinuity by incresing the number of terms, but do not seem to decrese in height. In fct, s the number of terms goes to infinity, the height of the overshoots nd undershoots does not vnish. This is known s Gibb s phenomenon Integrting nd Differentiting Fourier Series Integrting Fourier Series. Since integrtion is smoothing opertion, ny convergent Fourier series cn be integrted term by term to yield nother convergent Fourier series. Exmple Consider the step function f(x) = { π for 0 x < π for π x <
30 Figure 28.9: Since this is n odd function, there re no cosine terms in the Fourier series. b n = 2 π sin(nx) dx π 0 [ = 2 ] π n cos(nx) 0 = 2 n ( ( )n ) = { 4 n for odd n 0 for even n. 359
31 Integrting this reltion, f(x) oddn 4 sin nx n Since this series converges uniformly, 4 oddn x n 4 2 oddn f(t) dt F(x) = = = 4 x oddn x oddn oddn oddn 4 n oddn 4 n sin(nt)dt sin(nt) dt [ 4 ] x n n cos(nt) 4 n 2( cos(nx) + ( )n ) n 4 2 oddn cos(nx) n 2 { cos(nx) x π for π x < 0 = F(x) = n 2 x π for 0 x < π. The vlue of the constnt term is 4 oddn n = 2 F(x) dx = 2 π 0 π. 360
32 Thus π 4 oddn { cos(nx) x π for π x < 0 = n 2 x π for 0 x < π. Differentiting Fourier Series. Recll tht in generl, series cn only be differentited if it is uniformly convergent. The necessry nd sufficient condition tht Fourier series be uniformly convergent is tht the periodic extension of the function is continuous. Result The Fourier series of function f(x) cn be differentited only if the periodic extension of f(x) is continuous. Exmple Consider the function defined by { π for 0 x < π f(x) = for π x < 0. f(x) hs the Fourier series f(x) oddn 4 sin nx. n The function hs derivtive except t the points x = nπ. Differentiting the Fourier series yields f (x) 4 cos(nx). oddn For x nπ, this implies 0 = 4 cos(nx), oddn 36
33 which is flse. The series does not converge. This is s we expected since the Fourier series for f(x) is not uniformly convergent. 362
Physics 116C Solution of inhomogeneous ordinary differential equations using Green s functions
Physics 6C Solution of inhomogeneous ordinry differentil equtions using Green s functions Peter Young November 5, 29 Homogeneous Equtions We hve studied, especilly in long HW problem, second order liner
More informationThe Regulated and Riemann Integrals
Chpter 1 The Regulted nd Riemnn Integrls 1.1 Introduction We will consider severl different pproches to defining the definite integrl f(x) dx of function f(x). These definitions will ll ssign the sme vlue
More informationAbstract inner product spaces
WEEK 4 Abstrct inner product spces Definition An inner product spce is vector spce V over the rel field R equipped with rule for multiplying vectors, such tht the product of two vectors is sclr, nd the
More informationc n φ n (x), 0 < x < L, (1) n=1
SECTION : Fourier Series. MATH4. In section 4, we will study method clled Seprtion of Vribles for finding exct solutions to certin clss of prtil differentil equtions (PDEs. To do this, it will be necessry
More informationMATH34032: Green s Functions, Integral Equations and the Calculus of Variations 1
MATH34032: Green s Functions, Integrl Equtions nd the Clculus of Vritions 1 Section 1 Function spces nd opertors Here we gives some brief detils nd definitions, prticulrly relting to opertors. For further
More informationAdvanced Calculus: MATH 410 Notes on Integrals and Integrability Professor David Levermore 17 October 2004
Advnced Clculus: MATH 410 Notes on Integrls nd Integrbility Professor Dvid Levermore 17 October 2004 1. Definite Integrls In this section we revisit the definite integrl tht you were introduced to when
More informationBest Approximation. Chapter The General Case
Chpter 4 Best Approximtion 4.1 The Generl Cse In the previous chpter, we hve seen how n interpolting polynomil cn be used s n pproximtion to given function. We now wnt to find the best pproximtion to given
More informationUNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE
UNIFORM CONVERGENCE MA 403: REAL ANALYSIS, INSTRUCTOR: B. V. LIMAYE 1. Pointwise Convergence of Sequence Let E be set nd Y be metric spce. Consider functions f n : E Y for n = 1, 2,.... We sy tht the sequence
More informationSummary: Method of Separation of Variables
Physics 246 Electricity nd Mgnetism I, Fll 26, Lecture 22 1 Summry: Method of Seprtion of Vribles 1. Seprtion of Vribles in Crtesin Coordintes 2. Fourier Series Suggested Reding: Griffiths: Chpter 3, Section
More informationBest Approximation in the 2norm
Jim Lmbers MAT 77 Fll Semester 111 Lecture 1 Notes These notes correspond to Sections 9. nd 9.3 in the text. Best Approximtion in the norm Suppose tht we wish to obtin function f n (x) tht is liner combintion
More informationg i fφdx dx = x i i=1 is a Hilbert space. We shall, henceforth, abuse notation and write g i f(x) = f
1. Appliction of functionl nlysis to PEs 1.1. Introduction. In this section we give little introduction to prtil differentil equtions. In prticulr we consider the problem u(x) = f(x) x, u(x) = x (1) where
More informationDefinite integral. Mathematics FRDIS MENDELU
Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová Brno 1 Motivtion  re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function defined on [, b]. Wht is the re of the
More information1 The Riemann Integral
The Riemnn Integrl. An exmple leding to the notion of integrl (res) We know how to find (i.e. define) the re of rectngle (bse height), tringle ( (sum of res of tringles). But how do we find/define n re
More informationDefinite integral. Mathematics FRDIS MENDELU. Simona Fišnarová (Mendel University) Definite integral MENDELU 1 / 30
Definite integrl Mthemtics FRDIS MENDELU Simon Fišnrová (Mendel University) Definite integrl MENDELU / Motivtion  re under curve Suppose, for simplicity, tht y = f(x) is nonnegtive nd continuous function
More informationMath 554 Integration
Mth 554 Integrtion Hndout #9 4/12/96 Defn. A collection of n + 1 distinct points of the intervl [, b] P := {x 0 = < x 1 < < x i 1 < x i < < b =: x n } is clled prtition of the intervl. In this cse, we
More informationFUNDAMENTALS OF REAL ANALYSIS by. III.1. Measurable functions. f 1 (
FUNDAMNTALS OF RAL ANALYSIS by Doğn Çömez III. MASURABL FUNCTIONS AND LBSGU INTGRAL III.. Mesurble functions Hving the Lebesgue mesure define, in this chpter, we will identify the collection of functions
More informationReview SOLUTIONS: Exam 2
Review SOUTIONS: Exm. True or Flse? (And give short nswer ( If f(x is piecewise smooth on [, ], we cn find series representtion using either sine or cosine series. SOUTION: TRUE. If we use sine series,
More informationOverview of Calculus I
Overview of Clculus I Prof. Jim Swift Northern Arizon University There re three key concepts in clculus: The limit, the derivtive, nd the integrl. You need to understnd the definitions of these three things,
More information63. Representation of functions as power series Consider a power series. ( 1) n x 2n for all 1 < x < 1
3 9. SEQUENCES AND SERIES 63. Representtion of functions s power series Consider power series x 2 + x 4 x 6 + x 8 + = ( ) n x 2n It is geometric series with q = x 2 nd therefore it converges for ll q =
More informationCalculus III Review Sheet
Clculus III Review Sheet 1 Definitions 1.1 Functions A function is f is incresing on n intervl if x y implies f(x) f(y), nd decresing if x y implies f(x) f(y). It is clled monotonic if it is either incresing
More information1 The fundamental theorems of calculus.
The fundmentl theorems of clculus. The fundmentl theorems of clculus. Evluting definite integrls. The indefinite integrl new nme for ntiderivtive. Differentiting integrls. Tody we provide the connection
More informationA REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H. Thomas Shores Department of Mathematics University of Nebraska Spring 2007
A REVIEW OF CALCULUS CONCEPTS FOR JDEP 384H Thoms Shores Deprtment of Mthemtics University of Nebrsk Spring 2007 Contents Rtes of Chnge nd Derivtives 1 Dierentils 4 Are nd Integrls 5 Multivrite Clculus
More informationMath Fall 2006 Sample problems for the final exam: Solutions
Mth 425 Fll 26 Smple problems for the finl exm: Solutions Any problem my be ltered or replced by different one! Some possibly useful informtion Prsevl s equlity for the complex form of the Fourier series
More informationProperties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives
Block #6: Properties of Integrls, Indefinite Integrls Gols: Definition of the Definite Integrl Integrl Clcultions using Antiderivtives Properties of Integrls The Indefinite Integrl 1 Riemnn Sums  1 Riemnn
More informationConvergence of Fourier Series and Fejer s Theorem. Lee Ricketson
Convergence of Fourier Series nd Fejer s Theorem Lee Ricketson My, 006 Abstrct This pper will ddress the Fourier Series of functions with rbitrry period. We will derive forms of the Dirichlet nd Fejer
More informationGreen function and Eigenfunctions
Green function nd Eigenfunctions Let L e regulr SturmLiouville opertor on n intervl (, ) together with regulr oundry conditions. We denote y, φ ( n, x ) the eigenvlues nd corresponding normlized eigenfunctions
More informationMain topics for the First Midterm
Min topics for the First Midterm The Midterm will cover Section 1.8, Chpters 23, Sections 4.14.8, nd Sections 5.15.3 (essentilly ll of the mteril covered in clss). Be sure to know the results of the
More informationReview of Riemann Integral
1 Review of Riemnn Integrl In this chpter we review the definition of Riemnn integrl of bounded function f : [, b] R, nd point out its limittions so s to be convinced of the necessity of more generl integrl.
More informationMath Theory of Partial Differential Equations Lecture 29: SturmLiouville eigenvalue problems (continued).
Mth 412501 Theory of Prtil Differentil Equtions Lecture 29: SturmLiouville eigenvlue problems (continued). Regulr SturmLiouville eigenvlue problem: d ( p dφ ) + qφ + λσφ = 0 ( < x < b), dx dx β 1 φ()
More informationReview of basic calculus
Review of bsic clculus This brief review reclls some of the most importnt concepts, definitions, nd theorems from bsic clculus. It is not intended to tech bsic clculus from scrtch. If ny of the items below
More informationLecture 1. Functional series. Pointwise and uniform convergence.
1 Introduction. Lecture 1. Functionl series. Pointwise nd uniform convergence. In this course we study mongst other things Fourier series. The Fourier series for periodic function f(x) with period 2π is
More informationINTRODUCTION TO INTEGRATION
INTRODUCTION TO INTEGRATION 5.1 Ares nd Distnces Assume f(x) 0 on the intervl [, b]. Let A be the re under the grph of f(x). b We will obtin n pproximtion of A in the following three steps. STEP 1: Divide
More informationf(x)dx . Show that there 1, 0 < x 1 does not exist a differentiable function g : [ 1, 1] R such that g (x) = f(x) for all
3 Definite Integrl 3.1 Introduction In school one comes cross the definition of the integrl of rel vlued function defined on closed nd bounded intervl [, b] between the limits nd b, i.e., f(x)dx s the
More informationUNIFORM CONVERGENCE. Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3
UNIFORM CONVERGENCE Contents 1. Uniform Convergence 1 2. Properties of uniform convergence 3 Suppose f n : Ω R or f n : Ω C is sequence of rel or complex functions, nd f n f s n in some sense. Furthermore,
More informationMA 124 January 18, Derivatives are. Integrals are.
MA 124 Jnury 18, 2018 Prof PB s oneminute introduction to clculus Derivtives re. Integrls re. In Clculus 1, we lern limits, derivtives, some pplictions of derivtives, indefinite integrls, definite integrls,
More informationRiemann is the Mann! (But Lebesgue may besgue to differ.)
Riemnn is the Mnn! (But Lebesgue my besgue to differ.) Leo Livshits My 2, 2008 1 For finite intervls in R We hve seen in clss tht every continuous function f : [, b] R hs the property tht for every ɛ >
More information8 Laplace s Method and Local Limit Theorems
8 Lplce s Method nd Locl Limit Theorems 8. Fourier Anlysis in Higher DImensions Most of the theorems of Fourier nlysis tht we hve proved hve nturl generliztions to higher dimensions, nd these cn be proved
More informationAM1 Mathematical Analysis 1 Oct Feb Exercises Lecture 3. sin(x + h) sin x h cos(x + h) cos x h
AM Mthemticl Anlysis Oct. Feb. Dte: October Exercises Lecture Exercise.. If h, prove the following identities hold for ll x: sin(x + h) sin x h cos(x + h) cos x h = sin γ γ = sin γ γ cos(x + γ) (.) sin(x
More information7.2 Riemann Integrable Functions
7.2 Riemnn Integrble Functions Theorem 1. If f : [, b] R is step function, then f R[, b]. Theorem 2. If f : [, b] R is continuous on [, b], then f R[, b]. Theorem 3. If f : [, b] R is bounded nd continuous
More informationMapping the delta function and other Radon measures
Mpping the delt function nd other Rdon mesures Notes for Mth583A, Fll 2008 November 25, 2008 Rdon mesures Consider continuous function f on the rel line with sclr vlues. It is sid to hve bounded support
More informationSection 6.1 Definite Integral
Section 6.1 Definite Integrl Suppose we wnt to find the re of region tht is not so nicely shped. For exmple, consider the function shown elow. The re elow the curve nd ove the x xis cnnot e determined
More informationMAA 4212 Improper Integrals
Notes by Dvid Groisser, Copyright c 1995; revised 2002, 2009, 2014 MAA 4212 Improper Integrls The Riemnn integrl, while perfectly welldefined, is too restrictive for mny purposes; there re functions which
More informationLecture 3 ( ) (translated and slightly adapted from lecture notes by Martin Klazar)
Lecture 3 (5.3.2018) (trnslted nd slightly dpted from lecture notes by Mrtin Klzr) Riemnn integrl Now we define precisely the concept of the re, in prticulr, the re of figure U(, b, f) under the grph of
More informationSTURMLIOUVILLE BOUNDARY VALUE PROBLEMS
STURMLIOUVILLE BOUNDARY VALUE PROBLEMS Throughout, we let [, b] be bounded intervl in R. C 2 ([, b]) denotes the spce of functions with derivtives of second order continuous up to the endpoints. Cc 2
More informationTheoretical foundations of Gaussian quadrature
Theoreticl foundtions of Gussin qudrture 1 Inner product vector spce Definition 1. A vector spce (or liner spce) is set V = {u, v, w,...} in which the following two opertions re defined: (A) Addition of
More informationMATH 174A: PROBLEM SET 5. Suggested Solution
MATH 174A: PROBLEM SET 5 Suggested Solution Problem 1. Suppose tht I [, b] is n intervl. Let f 1 b f() d for f C(I; R) (i.e. f is continuous relvlued function on I), nd let L 1 (I) denote the completion
More informationProf. Girardi, Math 703, Fall 2012 Homework Solutions: 1 8. Homework 1. in R, prove that. c k. sup. k n. sup. c k R = inf
Knpp, Chpter, Section, # 4, p. 78 Homework For ny two sequences { n } nd {b n} in R, prove tht lim sup ( n + b n ) lim sup n + lim sup b n, () provided the two terms on the right side re not + nd in some
More informationImproper Integrals, and Differential Equations
Improper Integrls, nd Differentil Equtions October 22, 204 5.3 Improper Integrls Previously, we discussed how integrls correspond to res. More specificlly, we sid tht for function f(x), the region creted
More informationUnit #9 : Definite Integral Properties; Fundamental Theorem of Calculus
Unit #9 : Definite Integrl Properties; Fundmentl Theorem of Clculus Gols: Identify properties of definite integrls Define odd nd even functions, nd reltionship to integrl vlues Introduce the Fundmentl
More informationPHYSICS 116C Homework 4 Solutions
PHYSICS 116C Homework 4 Solutions 1. ( Simple hrmonic oscilltor. Clerly the eqution is of the SturmLiouville (SL form with λ = n 2, A(x = 1, B(x =, w(x = 1. Legendre s eqution. Clerly the eqution is of
More informationDEFINITION The inner product of two functions f 1 and f 2 on an interval [a, b] is the number. ( f 1, f 2 ) b DEFINITION 11.1.
398 CHAPTER 11 ORTHOGONAL FUNCTIONS AND FOURIER SERIES 11.1 ORTHOGONAL FUNCTIONS REVIEW MATERIAL The notions of generlized vectors nd vector spces cn e found in ny liner lger text. INTRODUCTION The concepts
More informationENGI 9420 Lecture Notes 7  Fourier Series Page 7.01
ENGI 940 ecture Notes 7  Fourier Series Pge 7.0 7. Fourier Series nd Fourier Trnsforms Fourier series hve multiple purposes, including the provision of series solutions to some liner prtil differentil
More informationSuppose we want to find the area under the parabola and above the x axis, between the lines x = 2 and x = 2.
Mth 43 Section 6. Section 6.: Definite Integrl Suppose we wnt to find the re of region tht is not so nicely shped. For exmple, consider the function shown elow. The re elow the curve nd ove the x xis cnnot
More informationExam 2, Mathematics 4701, Section ETY6 6:05 pm 7:40 pm, March 31, 2016, IH1105 Instructor: Attila Máté 1
Exm, Mthemtics 471, Section ETY6 6:5 pm 7:4 pm, Mrch 1, 16, IH115 Instructor: Attil Máté 1 17 copies 1. ) Stte the usul sufficient condition for the fixedpoint itertion to converge when solving the eqution
More informationReview of Calculus, cont d
Jim Lmbers MAT 460 Fll Semester 200910 Lecture 3 Notes These notes correspond to Section 1.1 in the text. Review of Clculus, cont d Riemnn Sums nd the Definite Integrl There re mny cses in which some
More informationDefinition of Continuity: The function f(x) is continuous at x = a if f(a) exists and lim
Mth 9 Course Summry/Study Guide Fll, 2005 [1] Limits Definition of Limit: We sy tht L is the limit of f(x) s x pproches if f(x) gets closer nd closer to L s x gets closer nd closer to. We write lim f(x)
More informationIntegration Techniques
Integrtion Techniques. Integrtion of Trigonometric Functions Exmple. Evlute cos x. Recll tht cos x = cos x. Hence, cos x Exmple. Evlute = ( + cos x) = (x + sin x) + C = x + 4 sin x + C. cos 3 x. Let u
More informationP 3 (x) = f(0) + f (0)x + f (0) 2. x 2 + f (0) . In the problem set, you are asked to show, in general, the n th order term is a n = f (n) (0)
1 Tylor polynomils In Section 3.5, we discussed how to pproximte function f(x) round point in terms of its first derivtive f (x) evluted t, tht is using the liner pproximtion f() + f ()(x ). We clled this
More informationAdvanced Calculus: MATH 410 Uniform Convergence of Functions Professor David Levermore 11 December 2015
Advnced Clculus: MATH 410 Uniform Convergence of Functions Professor Dvid Levermore 11 December 2015 12. Sequences of Functions We now explore two notions of wht it mens for sequence of functions {f n
More informationAntiderivatives/Indefinite Integrals of Basic Functions
Antiderivtives/Indefinite Integrls of Bsic Functions Power Rule: In prticulr, this mens tht x n+ x n n + + C, dx = ln x + C, if n if n = x 0 dx = dx = dx = x + C nd x (lthough you won t use the second
More informationW. We shall do so one by one, starting with I 1, and we shall do it greedily, trying
Vitli covers 1 Definition. A Vitli cover of set E R is set V of closed intervls with positive length so tht, for every δ > 0 nd every x E, there is some I V with λ(i ) < δ nd x I. 2 Lemm (Vitli covering)
More informationDefinite Integrals. The area under a curve can be approximated by adding up the areas of rectangles = 1 1 +
Definite Integrls 5 The re under curve cn e pproximted y dding up the res of rectngles. Exmple. Approximte the re under y = from x = to x = using equl suintervls nd + x evluting the function t the lefthnd
More informationMath 1431 Section M TH 4:00 PM 6:00 PM Susan Wheeler Office Hours: Wed 6:00 7:00 PM Online ***NOTE LABS ARE MON AND WED
Mth 43 Section 4839 M TH 4: PM 6: PM Susn Wheeler swheeler@mth.uh.edu Office Hours: Wed 6: 7: PM Online ***NOTE LABS ARE MON AND WED t :3 PM to 3: pm ONLINE Approimting the re under curve given the type
More informationCzechoslovak Mathematical Journal, 55 (130) (2005), , Abbotsford. 1. Introduction
Czechoslovk Mthemticl Journl, 55 (130) (2005), 933 940 ESTIMATES OF THE REMAINDER IN TAYLOR S THEOREM USING THE HENSTOCKKURZWEIL INTEGRAL, Abbotsford (Received Jnury 22, 2003) Abstrct. When relvlued
More informationChapter 6 Notes, Larson/Hostetler 3e
Contents 6. Antiderivtives nd the Rules of Integrtion.......................... 6. Are nd the Definite Integrl.................................. 6.. Are............................................ 6. Reimnn
More informationTHE EXISTENCEUNIQUENESS THEOREM FOR FIRSTORDER DIFFERENTIAL EQUATIONS.
THE EXISTENCEUNIQUENESS THEOREM FOR FIRSTORDER DIFFERENTIAL EQUATIONS RADON ROSBOROUGH https://intuitiveexplntionscom/picrdlindeloftheorem/ This document is proof of the existenceuniqueness theorem
More informationThe area under the graph of f and above the xaxis between a and b is denoted by. f(x) dx. π O
1 Section 5. The Definite Integrl Suppose tht function f is continuous nd positive over n intervl [, ]. y = f(x) x The re under the grph of f nd ove the xxis etween nd is denoted y f(x) dx nd clled the
More information1 The fundamental theorems of calculus.
The fundmentl theorems of clculus. The fundmentl theorems of clculus. Evluting definite integrls. The indefinite integrl new nme for ntiderivtive. Differentiting integrls. Theorem Suppose f is continuous
More informationThe Riemann Integral
Deprtment of Mthemtics King Sud University 20172018 Tble of contents 1 Antiderivtive Function nd Indefinite Integrls 2 3 4 5 Indefinite Integrls & Antiderivtive Function Definition Let f : I R be function
More information(4.1) D r v(t) ω(t, v(t))
1.4. Differentil inequlities. Let D r denote the right hnd derivtive of function. If ω(t, u) is sclr function of the sclrs t, u in some open connected set Ω, we sy tht function v(t), t < b, is solution
More informationMath 124B January 24, 2012
Mth 24B Jnury 24, 22 Viktor Grigoryn 5 Convergence of Fourier series Strting from the method of seprtion of vribes for the homogeneous Dirichet nd Neumnn boundry vue probems, we studied the eigenvue probem
More informationMath& 152 Section Integration by Parts
Mth& 5 Section 7.  Integrtion by Prts Integrtion by prts is rule tht trnsforms the integrl of the product of two functions into other (idelly simpler) integrls. Recll from Clculus I tht given two differentible
More information11 An introduction to Riemann Integration
11 An introduction to Riemnn Integrtion The PROOFS of the stndrd lemms nd theorems concerning the Riemnn Integrl re NEB, nd you will not be sked to reproduce proofs of these in full in the exmintion in
More information221B Lecture Notes WKB Method
Clssicl Limit B Lecture Notes WKB Method Hmilton Jcobi Eqution We strt from the Schrödinger eqution for single prticle in potentil i h t ψ x, t = [ ] h m + V x ψ x, t. We cn rewrite this eqution by using
More informationn f(x i ) x. i=1 In section 4.2, we defined the definite integral of f from x = a to x = b as n f(x i ) x; f(x) dx = lim i=1
The Fundmentl Theorem of Clculus As we continue to study the re problem, let s think bck to wht we know bout computing res of regions enclosed by curves. If we wnt to find the re of the region below the
More information0.1 Properties of regulated functions and their Integrals.
MA244 Anlysis III Solutions. Sheet 2. NB. THESE ARE SKELETON SOLUTIONS, USE WISELY!. Properties of regulted functions nd their Integrls.. (Q.) Pick ny ɛ >. As f, g re regulted, there exist φ, ψ S[, b]:
More informationThe final exam will take place on Friday May 11th from 8am 11am in Evans room 60.
Mth 104: finl informtion The finl exm will tke plce on Fridy My 11th from 8m 11m in Evns room 60. The exm will cover ll prts of the course with equl weighting. It will cover Chpters 1 5, 7 15, 17 21, 23
More informationChapter 7 Notes, Stewart 8e. 7.1 Integration by Parts Trigonometric Integrals Evaluating sin m x cos n (x) dx...
Contents 7.1 Integrtion by Prts................................... 2 7.2 Trigonometric Integrls.................................. 8 7.2.1 Evluting sin m x cos n (x)......................... 8 7.2.2 Evluting
More informationChapter 0. What is the Lebesgue integral about?
Chpter 0. Wht is the Lebesgue integrl bout? The pln is to hve tutoril sheet ech week, most often on Fridy, (to be done during the clss) where you will try to get used to the ides introduced in the previous
More informationSturmLiouville Eigenvalue problem: Let p(x) > 0, q(x) 0, r(x) 0 in I = (a, b). Here we assume b > a. Let X C 2 1
Ch.4. INTEGRAL EQUATIONS AND GREEN S FUNCTIONS Ronld B Guenther nd John W Lee, Prtil Differentil Equtions of Mthemticl Physics nd Integrl Equtions. Hildebrnd, Methods of Applied Mthemtics, second edition
More informationMA Handout 2: Notation and Background Concepts from Analysis
MA350059 Hndout 2: Nottion nd Bckground Concepts from Anlysis This hndout summrises some nottion we will use nd lso gives recp of some concepts from other units (MA20023: PDEs nd CM, MA20218: Anlysis 2A,
More informationCalculus in R. Chapter Di erentiation
Chpter 3 Clculus in R 3.1 Di erentition Definition 3.1. Suppose U R is open. A function f : U! R is di erentible t x 2 U if there exists number m such tht lim y!0 pple f(x + y) f(x) my y =0. If f is di
More informationSection 4.8. D v(t j 1 ) t. (4.8.1) j=1
Difference Equtions to Differentil Equtions Section.8 Distnce, Position, nd the Length of Curves Although we motivted the definition of the definite integrl with the notion of re, there re mny pplictions
More informationMath 61CM  Solutions to homework 9
Mth 61CM  Solutions to homework 9 Cédric De Groote November 30 th, 2018 Problem 1: Recll tht the left limit of function f t point c is defined s follows: lim f(x) = l x c if for ny > 0 there exists δ
More informationk and v = v 1 j + u 3 i + v 2
ORTHOGONAL FUNCTIONS AND FOURIER SERIES Orthogonl functions A function cn e considered to e generliztion of vector. Thus the vector concets like the inner roduct nd orthogonlity of vectors cn e extended
More informationMath 360: A primitive integral and elementary functions
Mth 360: A primitive integrl nd elementry functions D. DeTurck University of Pennsylvni October 16, 2017 D. DeTurck Mth 360 001 2017C: Integrl/functions 1 / 32 Setup for the integrl prtitions Definition:
More information1 E3102: a study guide and review, Version 1.0
1 E3102: study guide nd review, Version 1.0 Here is list of subjects tht I think we ve covered in clss (your milege my vry). If you understnd nd cn do the bsic problems in this guide you should be in very
More informationThe HenstockKurzweil integral
fculteit Wiskunde en Ntuurwetenschppen The HenstockKurzweil integrl Bchelorthesis Mthemtics June 2014 Student: E. vn Dijk First supervisor: Dr. A.E. Sterk Second supervisor: Prof. dr. A. vn der Schft
More informationEuler, Ioachimescu and the trapezium rule. G.J.O. Jameson (Math. Gazette 96 (2012), )
Euler, Iochimescu nd the trpezium rule G.J.O. Jmeson (Mth. Gzette 96 (0), 36 4) The following results were estblished in recent Gzette rticle [, Theorems, 3, 4]. Given > 0 nd 0 < s
More informationSections 5.2: The Definite Integral
Sections 5.2: The Definite Integrl In this section we shll formlize the ides from the lst section to functions in generl. We strt with forml definition.. The Definite Integrl Definition.. Suppose f(x)
More information1. GaussJacobi quadrature and Legendre polynomials. p(t)w(t)dt, p {p(x 0 ),...p(x n )} p(t)w(t)dt = w k p(x k ),
1. GussJcobi qudrture nd Legendre polynomils Simpson s rule for evluting n integrl f(t)dt gives the correct nswer with error of bout O(n 4 ) (with constnt tht depends on f, in prticulr, it depends on
More informationMath 324 Course Notes: Brief description
Brief description These re notes for Mth 324, n introductory course in Mesure nd Integrtion. Students re dvised to go through ll sections in detil nd ttempt ll problems. These notes will be modified nd
More informationFourier Series and Their Applications
Fourier Series nd Their Applictions Rui iu My, 006 Abstrct Fourier series re of gret importnce in both theoreticl nd pplied mthemtics. For orthonorml fmilies of complex vlued functions {φ n}, Fourier Series
More informationu t = k 2 u x 2 (1) a n sin nπx sin 2 L e k(nπ/l) t f(x) = sin nπx f(x) sin nπx dx (6) 2 L f(x 0 ) sin nπx 0 2 L sin nπx 0 nπx
Chpter 9: Green s functions for timeindependent problems Introductory emples Onedimensionl het eqution Consider the onedimensionl het eqution with boundry conditions nd initil condition We lredy know
More informationMath 1B, lecture 4: Error bounds for numerical methods
Mth B, lecture 4: Error bounds for numericl methods Nthn Pflueger 4 September 0 Introduction The five numericl methods descried in the previous lecture ll operte by the sme principle: they pproximte the
More informationProperties of the Riemann Integral
Properties of the Riemnn Integrl Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University Februry 15, 2018 Outline 1 Some Infimum nd Supremum Properties 2
More informationImproper Integrals. Type I Improper Integrals How do we evaluate an integral such as
Improper Integrls Two different types of integrls cn qulify s improper. The first type of improper integrl (which we will refer to s Type I) involves evluting n integrl over n infinite region. In the grph
More informationChapter 5. , r = r 1 r 2 (1) µ = m 1 m 2. r, r 2 = R µ m 2. R(m 1 + m 2 ) + m 2 r = r 1. m 2. r = r 1. R + µ m 1
Tor Kjellsson Stockholm University Chpter 5 5. Strting with the following informtion: R = m r + m r m + m, r = r r we wnt to derive: µ = m m m + m r = R + µ m r, r = R µ m r 3 = µ m R + r, = µ m R r. 4
More informationPartial Derivatives. Limits. For a single variable function f (x), the limit lim
Limits Prtil Derivtives For single vrible function f (x), the limit lim x f (x) exists only if the righthnd side limit equls to the lefthnd side limit, i.e., lim f (x) = lim f (x). x x + For two vribles
More informationWe divide the interval [a, b] into subintervals of equal length x = b a n
Arc Length Given curve C defined by function f(x), we wnt to find the length of this curve between nd b. We do this by using process similr to wht we did in defining the Riemnn Sum of definite integrl:
More information