188 METHODS OF MATHEMATICAL PROOF, PART I Chapter 5
let {A,, A,,... , A,,,) be a collection of m + 1 sets. We claim that A, = A, =
...- - A,,,. Now since {A,, A,,... , A,) is a collection of m sets, then
A, = A, =.. = A, by the induction hypothesis. For the same reason, A, =
A,=...- -A,= A,,,. Hence we conclude A, =A, = ... =A,= A,,,, as
desired.
- This exercise contains a list of basic properties of summation notation and is
included primarily as a reference. These properties are particularly relevant to
the theoretical development of the definite integral and to series solutions of dif-
ferential equations, as well as to proving the binomial theorem (Exercise 16). A
feature of these formulas, not emphasized in the text's treatment of "summation,"
is the summing from an arbitrary starting point, rather than necessarily from m = 1.
Let us assume the following as a definition: () z=, x, = z=, x, - Lm:: x,.
Note that the special cases m = 1 of (c) and (e) were proved by induction on n in
the text (Examples 5 and 6). Assume the truth of the case m = 1 in the remainder
of the properties and use the preceding definition () to prove (assuming m, n E N,
m < n, c E R, all x, E R):
16. (a) Use induction to prove the binomial theorem: If x and y are real num-
bers and n is a positive integer, then (x + y)" = s=, e)~''-~y~. [Recall the result
from Exercise 15(f) and the formula (" ') = (;) + 6") from Exercise 7(a), Article
5.1.1
(6) Use specialization to conclude from the result in (a) that 2" = z=, (;) (recall
Exercise 11, Article 1.5).
17. Critique and complete (instructions in Exercise 11, Article 4.1).
(a) THEOREM For all positive integers n and for all real numbers x such that
sin x # 0 (i.e., x is not an integral multiple of n), I;,, cos (2k - l)x =
(sin 2nx)/2 sin x.
"Proof" Define S in the usual manner. To prove the theorem (i.e., to prove S = N),
we must verify (i) and (ii) of Theorem 1:
(i) The case n =^1 is the equation cos x = sin 2x12 sin x, which is true by the
double-angle formula for sine.
(ii) Assume m E S. To prove m + 1 E S, we must show that
Zm.2,' cos (2k - l)x = (sin 2(m + 1)x)/2 sin x
Now Lrn:,' cos (2k - l)x = Lm=, cos (2k - l)x + cos (2m + l)x, which by in-
duction hypothesis equals [sin 2mx/2 sin x] + cos (2m + 1)x. But the latter
quantity equals
[sin 2mx + 2 sin x cos (2m + 1)x]/2 sin x, which equals
[sin 2mx + 2 sin x(cos 2mx cos x - sin 2mx sin x)]/2 sin x
