# Counterexamples in Analysis (Dover Books on Mathematics) by Bernard R. Gelbaum, John M. H. Olmsted

By Bernard R. Gelbaum, John M. H. Olmsted

Those counterexamples deal typically with the a part of research often called "real variables." the first half the book discusses the true quantity approach, capabilities and boundaries, differentiation, Riemann integration, sequences, endless sequence, extra. The second part examines services of two variables, aircraft units, sector, metric and topological areas, and serve as areas. 1962 version. contains 12 figures.

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Additional resources for Counterexamples in Analysis (Dover Books on Mathematics)

Sample text

Their difference, a number of the form 33 . . 300 . . 0, is a multiple of k and contains a three. Benford’s Law 17 For example For k = 1, N = 3 has 1 · N containing a three. For k = 2, N = 15 has 2N containing a three. For k = 3, N = 10 has 3N containing a three. For k = 4, N = 75 has 4N containing a three. We can construct a number that works simultaneously for the four values of k by stringing together the values of N we have so far, separating them by zeros. For example, for N = 75 00000 10 00000 15 00000 3 we have that each of N , 2N , 3N and 4N contains the digit three.

Call the sequence [ pn ]. i) The frequency sequence of the frequency sequence of [ pn ] is [ pn ]. ii) Adding position numbers to the entries of [ pn ] and to the entries of its frequency sequence produces complementary sequences. ): [ pn ] : 1, 2, 2, 2, 3, 3, 6, 7, 7, 7, 7, 9, 11, 11, 14, 14, 14, 14, 14, 14, 14, 14, 15, . . Its frequency sequence is: [qn ] : 0, 1, 4, 6, 6, 6, 7, 11, 11, 12, 12, 14, 14, 14, 22, . . ) Now compute the frequency sequence for [qn ] to see that is [ pn ]. ) Add the position numbers 1, 2, 3, 4, 5, .

Divide them by k and look at their remainders. As there are only a ﬁnite number of possible remainders, two numbers in the list leave the same remainder. Their difference, a number of the form 33 . . 300 . . 0, is a multiple of k and contains a three. Benford’s Law 17 For example For k = 1, N = 3 has 1 · N containing a three. For k = 2, N = 15 has 2N containing a three. For k = 3, N = 10 has 3N containing a three. For k = 4, N = 75 has 4N containing a three. We can construct a number that works simultaneously for the four values of k by stringing together the values of N we have so far, separating them by zeros.