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Essay2Solutions - Mathematics 421 Extra Fourier Series...

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Unformatted text preview: Mathematics 421 Extra Fourier Series Solutions Spring 2008 Introduction Few solutions caught the intent of the exercises in the supplement entitled, An operational view of Fourier coefficients, so solutions will be given here. The instructions specified that you were to use the results of the examples along with the shifting theorem and relation developed in Section 5 of the supplement between the series of a function and the series for the derivative of the function. The intent was to show that these methods, along with the use of linearity to combine special series is much more robust (i.e., more likely to lead to correct results) than techniques based on general methods of Calculus applied to the given expression for the given function. The word operational refers to methods that work directly with the Fourier coefficients of a function, depending on the relation of the function to other functions rather than using the definition of the Fourier coefficients as integrals. In practice, the computations are now done almost entirely by computer, but this means that it is most important to assure that the required computation has been correctly described, and the consequences of mistakes are easily detected, possibly more easily detected than a typographical error that led to the unintended computation. These skills are far more important than those used to survive a second semester Calculus course. A. Examples The first example was the function f.x/ D 1 if a < x < a otherwise for some a with 0 < a < p . We found f.x/ D a p C 1 X n D 1 2 n sin n a p cos n x p : Note that formally substituting n for n in the expression for the coefficient a n gives the same value, so c n D c n D a n =2 when converting this cosine series to a complex Fourier series. For g.x/ D x on the interval OE p;p , we found g.x/ D 1 X n D 1 D . 1/ n C 1 2p n sin n x p : Note that formally substituting n for n in the expression for the coefficient b n changes the sign, so c n D ib n =2 and c n D ib n =2 D ib n =2 when converting this sine series to a complex Fourier series....
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Essay2Solutions - Mathematics 421 Extra Fourier Series...

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