By Kwong-Tin Tang
Pedagogical insights received via 30 years of educating utilized arithmetic led the writer to jot down this set of scholar orientated books. issues corresponding to complicated research, matrix thought, vector and tensor research, Fourier research, indispensable transforms, traditional and partial differential equations are awarded in a discursive kind that's readable and simple to stick to. various truly acknowledged, thoroughly labored out examples including conscientiously chosen challenge units with solutions are used to augment scholars' realizing and manipulative ability. The target is to make scholars cozy and assured in utilizing complex mathematical instruments in junior, senior, and starting graduate courses.
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Extra info for Mathematical Methods for Engineers and Scientists 3: Fourier Analysis, Partial Differential Equations and Variational Methods
Sample text
Find the complex Fourier series of the function f (t) = 0 1 −π < t < 0, 0 < t < π. 1. Since the period is 2π, so p = π, and the complex Fourier series is given by ∞ cn eint f (t) = n=−∞ with c0 = 1 2π 1 cn = 2π π dt = 0 π 1 , 2 e−int dt = 0 1 − e−inπ = 2πni 0 n = even, n = odd. 1 πni Therefore the complex series is f (t) = 1 1 + 2 iπ 1 1 · · · − e−i3t − e−it + eit + ei3t + · · · 3 3 It is clear that c−n = . 1 1 = = c∗n π(−n)i πn(−i) as we expect, sine f (t) is real. Furthermore, since eint − e−int = 2i sin nt, the Fourier series can be written as f (t) = 1 2 + 2 π sin t + 1 1 sin 3t + sin 5t + · · · 3 5 This is also what we expected, since f (t) − 1 2 .
Use the Fourier series for the function whose definition is f (x) = x2 for − 1 < x < 1, and f (x + 2) = f (x), to show that ∞ (a) ∞ (−1)n+1 π2 , = n2 12 n=1 (b) ∞ (c) (−1)n+1 π3 , = (2n − 1)3 32 n=1 1 π2 , = n2 6 n=1 ∞ (d) 1 π4 . 1. 20) with L=1: ∞ (−1)n 4 1 x2 = + 2 cos nπx. 3 π n=1 n2 (a) Set x = 0, so we have x2 = 0, cos nπx = 1. Thus 0= 1 4 + 2 3 π or − It follows that: 1− 4 π2 ∞ (−1)n n2 n=1 ∞ (−1)n 1 = . 2 n 3 n=1 1 1 1 π2 . 7 Properties of Fourier Series (b) With x = 1, the series becomes 1= ∞ (−1)n cos nπ.
6 The Method of Jumps 35 Clearly, In [f (t)] can be evaluated similarly as In [f (t)]. This formula can be used iteratively to find the Fourier coefficient cn for nonzero n, since cn = In [f (t)]/2p. Together with c0 , which is given by a simple integral, these coefficients determine all terms of the Fourier series. For many practical functions, their Fourier series can be simply obtained from the jumps at the points of discontinuity. The following examples will illustrate how quickly this can be done with the sketches of the function and its derivatives.