By Michael Henle

ISBN-10: 0486679667

ISBN-13: 9780486679662

Very good textual content for upper-level undergraduate and graduate scholars exhibits how geometric and algebraic rules met and grew jointly into a big department of arithmetic. Lucid insurance of vector fields, surfaces, homology of complexes, even more. a few wisdom of differential equations and multivariate calculus required. Many difficulties and workouts (some suggestions) built-in into the textual content. 1979 version. Bibliography.

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**Extra resources for A Combinatorial Introduction to Topology (Dover Books on Mathematics)**

**Example text**

Cos ne 2 = 2 sin (~) . Hint: Let z = ei012 and note that the sum in question can be written in terms of z as z - 2n + z-(Zn-Z) + ... + zzn). Summing the geometric progression, we obtain h 1 _(z-zn 2 + z-(Zn-2) + .. , + z2n) 2(1 - zZ) Z2n+ 1 _ Z-(2n+ 1) 10. Evaluate the sum 1 2 Fn+l +--+ ... 2). Fk denotes the kth Fibonacci number (see Exercise 9, 11. Evaluate the sum Hint: By induction, k=l,2, ... ,m, with the understanding that Fo = O. Setting m Fk-lFZk -FkFZk-l = (-llFk, so Fk-l FZk-l Fk FZk ----Thus the sum L~=o(FZk )-1 telescopes.

Solution. Tb express the roots of ZS = 1 in polar form, we just apply the preceeding discussion to obtain the complete set of roots: { cos (2~k) + i sin (2~k) I k= 0, I, 2, 3, 4} . On the other hand, let us find the roots directly by factoring the equation ZS - First of all, ZS - 1 = (z - 1)(Z4 + 1 = O. Z3 + Z2 + + 1). Secondly, Z Hence the complex fifth roots of unity satisfy ( Z2 + 1+-J5 2 z + 1)( Z2 + I--J5 2 z Thus we find the complete set of roots of zS { I, -(I + -J5) ± iJlO 4 - 2-J5 , - + 1) 1 = = O.

In this case, the sequence is constant (xn = Xl for all n) so we must require that Xl be an integer. 0 The sequence a = aI, az, a3, ... is a geometric progression if there is a number r such that for k = 1,2,3, ... , The parameter r is called the common ratio and the nth term in the sequence is given by The sum of a finite geometric progression is easily obtained. If Sn = a + ar + ar 2 +... + arn-l , then r Sn = ar + ar 2 + ar 3 +... + ar n , so (l-r)Sn = a-arn. If r f 1 we can divide by r - 1 and solve for Sn.

### A Combinatorial Introduction to Topology (Dover Books on Mathematics) by Michael Henle

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