Rearrangement on Conditionally Convergent Integrals in Analogy to Series

Transcription

1 Electronic Proceedings of Undergraduate Mathematics Day, Vol. 3 (8), No. 6 Rearrangement on Conditionally Convergent Integrals in Analogy to Series Edward J Timko University of Dayton Dayton OH timkoedz@notes.udayton.edu Abstract Rearrangements on conditionally convergent series suggests the existence of a similar process for integrals, here also referred to as rearrangement. In this document, a general theorem concerning rearrangement for conditionally convergent integrals is presented, as well as supporting theorems and a corollary to the general theorem. The corollary reads: Let f : + + be a continuous function with an everywhere negative and monotone increasing derivative. If ( )x f(x)dx is conditionally convergent, then z, there exists an arrangement on ( )x f(x)dx such that z = ( )x f(x)dx. ISSN: 54-86

2 Electronic Proceedings of Undergraduate Mathematics Day, Vol. 3 (8), No. 6 Preliminary Theorems and Lemmas Note: Recall that e iθ = cos(θ) + i sin(θ) and ( ) n = e in. Definition. For any function f(x) integrable on [, ) such that f(x)dx = f(j + u)du a rearrangement on f(x)dx is a rearrangement of the terms in the series f(j + u) for each u on [, ]. Theorem. Let {a j } and {b j} be positive, real, decreasing sequences which converge to zero, where a j b j for all j. If a j a j+ b j b j+ j then ( ) j a j ( ) j b j. Proof. From the hypothesis, both sequences are decreasing, and therefore a j a j+ > and b j b j+ >. For n ( ) j a j+ = a l a l+ + a l+ a l a l+n a l+n n j= = b l b l+ + b l+ b l b l+n b l+n ( ) j b j+l. n j= with both sums being positive. Since a n b n, it follows that and so n, which implies < < n j= ( ) j a j+l n j= j= j= ( ) j b j+l n n ( ) j a j+l ( ) j b j+l n+l n+l ( ) j a j ( ) j b j. j=l j=l ISSN: 54-86

3 Electronic Proceedings of Undergraduate Mathematics Day, Vol. 3 (8), No. 6 Since a j, b j as j, it follows that both series are convergent. Knowing this, let l = k + and n, to yield that ( ) j a j ( ) j b j. Theorem. Let f be a positive, decreasing function on +, integrable such that ( )x f(x)dx is conditionally convergent. Then ( )j f(j + u) is conditionally convergent u [, ]. Proof. Note that, for all u [, ] and for all x +, This implies Let N it follows that such that N >. Since x + u < x +. f(x + ) < f( x + u). N N f( x )dx = f(x)dx < Taking the limit that N yields that f(x)dx < N f(j) N f(j + u). f(j + u). Since f is integrable on +, it follows that f(x)dx is finite. Therefore, if the integral f(x)dx then f(x)dx, and therefore f(j +u). Suppose that ( )x f(x)dx L. It then follows ɛ >, M > such that y ( ) x f(x)dx L < ɛ y M. Thus, for M a b it follows by the triangle inequality that b ( ) x f(x)dx a b ( ) x f(x)dx L + ( ) x f(x)dx L < ɛ. a ISSN: 54-86

4 Electronic Proceedings of Undergraduate Mathematics Day, Vol. 3 (8), No. 6 Since f(x) is decreasing, positive, and finite x +, it follows that f must converge to some value, say c, as x. Since c f(x), it follows for x [n, n + /] with n that cos(x)f(x) c cos(x) cos(x)f(x) from which follows that n+/ c cos(x)dx Knowing that it follows that n c n+/ n n+/ n n+/ n cos(x)dx = cos(x)f(x)dx. cos(x)f(x)dx. A similar argument will show that the same is true for the sine function. From this, it follows that c n+/ ( ) x f(x)dx. Let n M. Then it follows that c n+/ ( ) x f(x)dx < ɛ n n n M and therefore c < ɛ. Therefore c =. Thus, f(x) as x, and therefore f(u + j), u [, ] as j. From this, one can conclude that the series ( )j f(j + u) is convergent. Thus, if ( )x f(x)dx is conditionally convergent, then ( )j f(j + u) is conditionally convergent u [, ]. Lemma. Let f be a positive, decreasing, integrable function on every finite interval of +. Then for j. Proof. Note that f(u + j) cos udu = f(u + j) cos(u)du / / = f(/ + j) =. f(u + j) cos udu + f(/ + j) cos udu + cos udu / f(u + j) cos udu / f(/ + j) cos udu 3 ISSN: 54-86

5 Electronic Proceedings of Undergraduate Mathematics Day, Vol. 3 (8), No. 6 Lemma. Let f(x) be positive, decreasing, and integrable on every finite subinterval of + such that u [, ] and j, f(u + j) f(u + j + ) f(j) f(j + ); u [, ] the series j= ( )j f(j + u) is convergent. Then ( ) u j=k ( )j f(j + u) is integrable on [, ] for each positive k, and ( ) u+j f(u + j)du = ( ) u+j f(u + j)du. Proof. First to show integrability. Since j=k ( )j f(j + u) is convergent it follows that f(j + u) as j. Thus, from the proof of Theorem n, k with n k, n ( ) j f(u + j) n ( ) j f(j). j=k It follows from the hypothesis that f is integrable on [j, j + ], and therefore f(u + j) is integrable for u [, ]. Fix k. Define h n (u) = j=k n ( ) j f(u + j). j=k Since the finite sum of integrable functions is integrable, h n (u) is integrable for each n. Therefore u [, ], there exists a finite h(u) such that lim n h n (u) = h(u). Also h n (u) h n (). Choose M = sup{ h k (), h k+ (),...}. Since h n () is convergent and always finite, such a number exists. Therefore, h n (u) M for u [, ] and n k. Therefore, by Lebesgue Dominated Convergence Theorem, h(u) is integrable on [, ]. Note that ( ) u h(u) = cos(u)h(u) + i sin(u)h(u), and that cos(u) and sin(u) are integrable [, ]. Since the product of two integrable functions integrable, it follows that cos(u)h(u) and sin(u)h(u) are integrable [, ], and therefore ( ) u h(u) is integrable on [, ]. Now, in order to conserve space, define g(x) = ( ) x f(x). Thus, for any finite k = k + 4 ISSN: 54-86

6 Electronic Proceedings of Undergraduate Mathematics Day, Vol. 3 (8), No. 6 and = = k k + +. Therefore = for any finite positive k. Let u [, ]. Since f is decreasing and positive, it follows that < f(u + j) f(u + j + ), and thus, < f(u + j) f(u + j + ) f(j) f(j + ). Therefore, by Theorem ( ) j f(u + j) Thus, from which follows cos(u) g(j) ( ) j f(j + u) ( ) j f(u + j)du g(j). cos(u) g(j) g(j)du with the integrability of the left-hand side being given by Lemma. Evaluating the right-hand integral, cos(u) ( ) j f(u + j)du g(j)du. The same follows identically for the sine function in place of the cosine function. Therefore, by the Triangle Inequality, g(j)du. 5 ISSN: 54-86

7 Electronic Proceedings of Undergraduate Mathematics Day, Vol. 3 (8), No. 6 Now, since cos u, it follows that f(j) f(u + j) f(u + j) cos(u) f(u + j) f(j). Integrating u over [, ] yields f(j) f(u + j) cos(u)du f(j). Similarly, f(u + j) sin(u)du f(j). By a similar argument, [f(j) f(j + )] [f(u + j) f(u + j + )] cos(u) f(j) f(j + ) and therefore [f(j) f(j + )] [f(u + j) f(u + j + )] cos(u)du f(j) f(j + ) with similar following for the sine function. Note, for u [, ], sin(u), and therefore [f(u + j) f(u + j + )] sin(u). Thus, By Lemma, for j, [f(u + j) f(u + j + )] sin(u)du. For every u [, ] and y ω f(u + j) cos(u)du. [f(u + j) f(u + j + )] cos u f(u + j) f(u + j + ) f(j) f(j + ). Therefore or [f(u + j) f(u + j + )] cos udu f(u + j) cos udu [f(j) f(j + )]du = f(j) f(j + ) f(u + j + ) cos udu f(j) f(j + ). Letting a j = f(u + j) cos udu and b j = f(j), it follows from Theorem that ( ) j f(u + j) cos udu ( ) j f(j) for all k. Similarly, ( ) j f(u + j) sin udu ( ) j f(j) 6 ISSN: 54-86

8 Electronic Proceedings of Undergraduate Mathematics Day, Vol. 3 (8), No. 6. By the Triangle Inequality g(j) and, again by the Triangle Inequality, ( + ) g(j). Since g(j) is convergent, it follows that it is Cauchy, and therefore ɛ >, N such that g(j) < ɛ ( ) k N. + Thus which implies < ɛ < ɛ k N and therefore =. Main Theorem Theorem 3 (Main Theorem). Let f be a positive, decreasing function integrable on any finite subinterval of +. If u [, ] and j it holds that f(u+j) f(u+j +) f(j) f(j +) ( )x f(x)dx is conditionally convergent then z there exists a rearrangement on ( )x f(x)dx such that, z = ( )x f(x)dx. 7 ISSN: 54-86

9 Electronic Proceedings of Undergraduate Mathematics Day, Vol. 3 (8), No. 6 Proof. Note ( ) x f(x)dx = j+ j ( ) x f(x)dx. Making the substitution that u = j + x, it follows that ( ) x f(x)dx = ( ) u+j f(u + j)du. By Theorem, u [, ] it holds that ( )j f(j + u) is conditionally convergent. Therefore, by Lemma ( ) x f(x)dx = ( ) u+j f(u + j)du. Since f(u + j)( )j is conditionally convergent u [, ] it follows that h(u), there exists a rearrangement of the terms in the series such that h(u) = f(u+j)( )j. This constitutes a rearrangement on ( )x f(x)dx. Choose z such that z = z e iθ. Now choose a rearrangement of the terms in f(u + j)( )j for u [, ] such that f(u + j)( ) j = z cos(u θ). Then ( ) x f(x)dx = z. Corollary. Let f be a positive function, integrable on any finite subinterval of +, with an everywhere negative and increasing derivative. If ( )x f(x)dx is conditionally convergent, then z, there exists a rearrangement on f(x)dx such that z = ( )x f(x)dx. Proof. It suffices to show that f satisfies the requirements of the Main Theorem. The requirement of conditional convergence is obviously met. It is also clear that f is decreasing. Since f is continuous, it follows by the Mean Value Theorem that there exists c j (j, j + u) such that f(u + j) f(j) = f (c j )u. Since (j, j + u) (j +, j + u + ) =, it follows that c j < c j+, and f (c j ) < f (c j+ ). Thus, f(u + j) f(j) < f(u + j + ) f(j + ) 8 ISSN: 54-86

10 Electronic Proceedings of Undergraduate Mathematics Day, Vol. 3 (8), No. 6 and f(u + j) f(u + j + ) < f(j) f(j + ). Thus, f satisfies the requirements of the Main Theorem. 9 ISSN: 54-86