Fourier Series

More documents

Recommendations

Info

solutions of singular second order ordinary differential equations that arise when attemptingto solve specific partial differential equations via a particular technique. However, for a setof functions that arise via a quest for analytic equations, they also possess an unexpectedand very useful quality - they can be used to provide very good numerical approximationsto functions, which are generally much better and more reliable than standard polynomialapproximation.ExampleShow that the three functions defined by 0 x 1, 1 x x , 2 x 1 2 3x2 − 1form an orthogonal set on −1,1. Determine the best mean-square approximation to thefunction fx e x on −1,1 in terms of 0 , 1 , 2 .There are two steps to the problem, with the first being to establish that 0 , 1 , 2 forman orthogonal set and the second to obtain the approximation.Step 1. With 0 x 1, 1 x x and 2 x 1 2 3x2 − 1, need to show that the setsatisfies (1.9.1) on −1,1, i.e.verify1−1 m x n x dx gm mnfor all combinations of m,n from 1 to 3. Thus we have to evaluate m n integrals but thereis some symmetry since1−11−11−1 0 x 1 x dx −1 0 x 2 x dx −1 1 x 2 x dx −1111 1 x 0 x dx , 2 x 0 x dx , 2 x 1 x dx ,giving six integrals to be evaluated. Furthermore, 0 and 2 are even functions and 1 isan odd function, so that the integral of products of 0 or 2 with 1 must be zero, i.e.1−11−1 0 x 1 x dx −1 1 x 2 x dx −111 1 x 0 x dx 0 2 x 1 x dx 0.The remaining cross-product integral is easily shown to be zero upon evaluation, so that52
1−1 0 x 2 x dx −11 2 x 0 x dx 0and this confirms (1.9.1) for m ≠ n.The remaining integrals are1 −11 0 x 0 x dx, 1 x 1 x dx and −1 −1enables these integrals to be evaluated as1 2 x 2 x dx . Using the given forms1−11−11−1 0 x 0 x dx 2, 1 x 1 x dx 2 3 , 2 x 2 x dx 2 5 .In terms of (1.9.1),g0 2, g1 2 3 , g2 2 5 ,confirming that the set is orthogonal but not orthonormal.Step 2. We require the coefficients c 0 ,c 1 ,c 2 to be determined so thatis the best fit in the least squares sense.fx e x ~ c 0 0 x c 1 1 x c 2 2 xThis can be obtained by an extension of the results derived in the previous section, togetherwith (1.9.4), to givefx ~ ∑n1f,n gn n x or c n f, ngn.The integrals f, n for n 0,1,2 now need to be evaluated, since the gn ′ s have justbeen calculated. These are1−11−11−1fx 0 x dx e − 1 e ,fx 1 x dx 2 e ,fx 2 x dx e − 7 e ,giving the best fit, when the gns are included, to be53
Page 2 and 3: f 3 x 1 0 ≤ x 1 3 1 x ≤ 2Usi
Page 4 and 5: I 1 0−afx gx dx a0f−u g−u d
Page 6: 1.1.2(c) respectively.10.80.60.40.2
Page 10 and 11: dfdx limx→0fx x − fxx,which i
Page 12 and 13: −fxsinmx dx ∑ a n −n1cosnxsin
Page 14 and 15: than in a formula. Periodicity can
Page 16 and 17: 32.521.510.5000.5 1 1.5 2 2.5 3xFig
Page 18 and 19: |Sx| 12a 0 ∑n1≤ 12a 0 ∑n1
Page 20 and 21: x 1 and fails to satisfy the Diric
Page 22 and 23: 1.4 Half-range Fourier Series1.4.1
Page 24 and 25: f E x 1 fx f−x2 1 2 x − x
Page 26 and 27: defining our new function Fx to beF
Page 28 and 29: Figure 1.4.4. Extension of half-ran
Page 30 and 31: Figure 1.5.1. The periodic function
Page 32 and 33: 1.6 Integration and Differentiation
Page 34 and 35: If fx is continuous in the range
Page 36 and 37: c n 1 2 a n − ib n n 0 1 2 a 0
Page 38 and 39: c n 120e i1−nx dx in − 1 −1n
Page 40 and 41: 1.8 Evaluation and ImplementationMo
Page 42 and 43: The value of the series at these po
Page 44 and 45: ND N x 1 2∑n1Ncosnx ∑ e inx
Page 46 and 47: 1.9 Orthogonal Functions and Fourie
Page 48 and 49: unfortunately it is the same standa
Page 50 and 51: n 1,...N. This is true for all N a
Page 54: e x ~ 1 2 e − 1 e 3x e 5 4 e

Fourier Series

Create successful ePaper yourself

Delete template?

Save as template?