By Kwong-Tin Tang
Pedagogical insights received via 30 years of educating utilized arithmetic led the writer to put in writing this set of scholar orientated books. issues reminiscent of complicated research, matrix thought, vector and tensor research, Fourier research, vital transforms, traditional and partial differential equations are provided in a discursive kind that's readable and straightforward to stick to. various basically said, thoroughly labored out examples including rigorously chosen challenge units with solutions are used to reinforce scholars' knowing and manipulative ability. The target is to make scholars cozy and assured in utilizing complicated mathematical instruments in junior, senior, and starting graduate courses.
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Additional info for Mathematical Methods for Engineers and Scientists 3: Fourier Analysis, Partial Differential Equations and Variational Methods
Suppose that f (t), shown in Fig. 10, is a periodic function with a period 2p. It is piecewise continuous. The locations of the discontinuity are at t1 , t2 , . . , tN −1 , counting from left to right. The two end points t0 and tN may or may not be points of discontinuity. Let f (t+ i ) be the right-hand limit of the function as t approaches ti from the right, and f (t− i ), the left-hand limit. At each discontinuity ti , except at two end points t0 and tN = t0 + 2p, we deﬁne a jump Ji as − Ji = f (t+ i ) − f (ti ).
2 n=1 32 1 Fourier Series However, the complex Fourier coeﬃcients are easier to compute, so we ﬁrst express it as a complex Fourier series ∞ cn eint f (t) = n=−∞ with cn = 1 2π π et e−int dt = −π 1 1 e(1−in)t 2π 1 − in π . −π Since e(1−in)π = eπ e−inπ = (−1)n eπ , e−(1−in)π = e−π einπ = (−1)n e−π , eπ − e−π = 2 sinh π, so cn = (−1)n (−1)n 1 + in (eπ − e−π ) = sinh π. 2π(1 − in) π 1 + n2 Now an = cn + c−n = (−1)n 2 sinh π, π 1 + n2 bn = i(cn − c−n ) = − (−1)n 2n sinh π. π 1 + n2 Thus, the Fourier series is given by ex = sinh π 2 sinh π + π π ∞ (−1)n (cos nt − n sin nt).
8 Fourier Series and Diﬀerential Equations 55 Ans. f (t) = 4 2 − π π 1 1 1 cos 2t + cos 4t + cos 6t + · · · 3 15 35 . 7. Use the method of jumps to ﬁnd the half range (a) cosine and (b) sine Fourier expansions of g(t), which is deﬁned only over the range 0 < t < 1 as g(t) = t − t2 , 0 < t < 1. Hint: (a) For the half-range cosine expansion, the function must be symmetrically extended to negative t. That is, we have to expand into a Fourier series the even function f (t) deﬁned as f (t) = g(t) = t − t2 0 < t < 1, g(−t) = −t − t2 −1 < t < 0.