FourierSeries-1

FourierSeries-1 - Fourier Series When the French...

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Fourier Series When the French mathematician Joseph Fourier (1768–1830) was trying to solve a prob- lem in heat conduction, he needed to express a function as an infinite series of sine and cosine functions: Earlier, Daniel Bernoulli and Leonard Euler had used such series while investigating prob- lems concerning vibrating strings and astronomy. The series in Equation 1 is called a trigonometric series or Fourier series and it turns out that expressing a function as a Fourier series is sometimes more advantageous than expanding it as a power series. In particular, astronomical phenomena are usually periodic, as are heartbeats, tides, and vibrating strings, so it makes sense to express them in terms of periodic functions. We start by assuming that the trigonometric series converges and has a continuous func- tion as its sum on the interval , that is, Our aim is to find formulas for the coefficients and in terms of . Recall that for a power series we found a formula for the coefficients in terms of deriv- atives: . Here we use integrals. If we integrate both sides of Equation 2 and assume that it’s permissible to integrate the series term-by-term, we get But because is an integer. Similarly, . So y ± ² f ± x ² dx ± 2 a 0 x ² sin nx ± 0 n y ² cos ± 1 n sin ³ ² ± 1 n ´ sin n ² sin ± ² n ²µ ± 0 ± 2 a 0 ³ ´ n ± 1 a n y ² cos ³ ´ n ± 1 b n y ² sin y ² f ± x ² ± y ² a 0 ³ y ² ´ n ± 1 ± a n cos ³ b n sin ² c n ± f ± n ² ± a ²· n ! f ± x ² ± ¸ c n ± x ² a ² n f b n a n ² µ x µ f ± x ² ± a 0 ³ ´ n ± 1 ± a n cos ³ b n sin ² 2 ´ ² , µ f ± x ² ³ b 1 sin x ³ b 2 sin 2 x ³ b 3 sin 3 x ³¶¶¶ ± a 0 ³ a 1 cos x ³ a 2 cos 2 x ³ a 3 cos 3 x f ± x ² ± a 0 ³ ´ n ± 1 ± a n cos ³ b n sin ² 1 f 1
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and solving for gives To determine for we multiply both sides of Equation 2 by (where is an integer and ) and integrate term-by-term from to : We’ve seen that the first integral is 0. With the help of Formulas 81, 80, and 64 in the Table of Integrals, it’s not hard to show that for all and So the only nonzero term in (4) is and we get Solving for , and then replacing by , we have Similarly, if we multiply both sides of Equation 2 by and integrate from to , we get We have derived Formulas 3, 5, and 6 assuming is a continuous function such that Equation 2 holds and for which the term-by-term integration is legitimate. But we can still consider the Fourier series of a wider class of functions: A piecewise continuous function on is continuous except perhaps for a finite number of removable or jump disconti- nuities. (In other words, the function has no infinite discontinuities. See Section 2.4 for a discussion of the different types of discontinuities.) ± a , b ² f n ± 1, 2, 3, . . . b n ± 1 ± y ² f ³ x ´ sin nx dx 6 ² sin mx n ± 1, 2, 3, . . . a n ± 1 y ² f ³ x ´ cos 5 n m a m y ² f ³ x ´ cos ± a m a m y ² cos cos ± µ 0 for n ² m for n ± m m n y ² sin cos ± 0 ± a 0 y ² cos ³ ´ n ± 1 a n y ² cos cos ³ ´ n ± 1 b n y ² sin cos 4 y ² f ³ x
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FourierSeries-1 - Fourier Series When the French...

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