Introductory lecture notes on Partial Differential Equations - c Anthony Peirce. ot to be copied, used, or revised without explicit written permission from the copyright owner. ecture 6: Bessel s Inequality, Parseval s Theorem, Energy convergence Compiled 4 August 7 In this lecture we consider the counterpart of Pythagoras Theorem for functions whose square is integrable. Square integrable functions are associated with functions describing physical systems having finite energy. For a finite Fourier Series involving terms we derive the so-called Bessel Inequality, in which can be taken to infinity provided the function f is square integrable. The Bessel Inequality is shown to reduce to an equality if and only if the Fourier Series S nx converges to f in the energy norm. The result is known as Parseval s Formula, which has profound consequences for the completeness of the Fourier Basis {, cos, sin }. We see that Parseval s Formula leads to a new class of sums for series of reciprocal powers of n. Key Concepts: Convergence of Fourier Series, Bessel s Inequality, Paresval s Theorem, Plancherel theorem, Pythagoras Theorem, Energy of a function, Convergence in Energy, completeness of the Fourier Basis. 6 Bessel s Inequality and Parseval s Theorem: 6. Bessel s Inequality Definition et fx be a function that is square-integrable on, ] i.e., fx ] dx <, in which case we write f, ]. Consider the Fourier Series associated with fx, namely; et ow fx a S x a + b n sin + b n sin S. fx S x ] f x fxs x + S x
Consider the least-square error defined to be ] E f, S fx S x ] dx f x dx fxs x dx + { f, f f, S + S, S } S x dx ow In addition, f, S S, S a a a a + a n ] a a n + b n fxs x dx fx dx + a n + a n + b n. E f, S ] ] + b n sin dx cos nx dx + b n fx cos sin nx dx dx + b n fx sin dx fx S x ] { } a dx f, f a n + b n ow since E f, S ] fx S x ] dx it follows that a a n + b n where Ef] is known as the energy of the -periodic function f. f x dx f, f Ef]
Fourier Series 3 Theorem Bessel s Inequality: et f, ] then a a n + b n in particular the series a a n + b n is convergent. f x dx 6. Bessel s Inequality, Components of a Vector and Pythagoras Theorem 6.. D Analogue Consider a D vector f, which is decomposed into components in terms of two orthogonal unit vectors ê and ê, i.e. f a ê + a ê ow f f f a ê + a ê a ê + a ê a + a since ê k are orthogonal unit vectors f a + a which is Pythagoras Theorem. 6.. 3D Analogue Suppose we wish to expand a 3-vector f in terms of a set of basis vectors {ê, ê }. Bessel s Inequality assumes the form a + a f Since the subspace span {ê, ê } which represents a plane in R 3 does not include the whole of R 3 the vector a ê + a ê f represents the orthogonal projection of f onto span {ê, ê }. If we include the third basis vector ê 3 in the basis, then the span {ê, ê, ê 3 } R 3. In this case the set {ê, ê, ê 3 } are linearly independent and of full rank and thus span the complete space R 3. {ê, ê, ê 3 } are in this case said to form a complete set. In this case f a ê + a ê + a 3 ê 3 and f a +a +a 3 so that Bessel s Inequality assumes the form of an equality, which in this trivial case reduces to Pythagoras Theorem. For a set of functions, that are complete, the equivalent of Pythagoras Theorem is Parseval s Theorem.
4 6.3 Parseval s Theorem Theorem Parseval s Identity et f, ] then the Fourier coefficients a n and b n satisfy Parseval s Formula If and only if a a n + b n lim f x dx Ef] fx S x ] dx. In this case the The east Square Error assumes the form E f, S ] fx S x ] dx a a n + b n n+ a f a x dx a n + b n a n + b n a n + b n 6. 6.3. Parseval s Theorem for odd functions Theorem 3 Parseval s Identity for odd functions et fx b n sin < x <. Then ] fx dx b n. Proof: fx ] dx m m b m b n sin b m b n δ mn mx sin dx 6. b n. 6.3 Example 6. Recall for x, ], fx x 4 n+ sin. n fx dx x dx 4 4 x 3 3 6 n n n 6.4
ote: n Also note that n 4 For Fourier Sine Components: 6 4. 6 Fourier Series 5 evens n m m Example 6. Consider fx x, < x <. odds m + m 4 + m m+ m+ m + 6 4 8. 6.5 fx dx b n. 6.6 The Fourier Series Expansion is: x 3 + 4 n n cosnx. 6.7 et By Parseval s Formula: n 3 4 cos n x 4 3 + 4 x 4 dx x 5 5 4 k k 3 + 6 4 9 + 6 k+ n 4 n n cos n 6.8 k k k. 6.9 n 4 9 5 45 4 45 8 9 9 6. 4 9 ζ4, 6. n4 where ζ is the Riemann Zeta Function defined by: ζs, s σ + iτ, σ Re{s} > 6. ns