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Apply the Intermediate Value Theorem to f on the closed interval [0, 1]. The function is continuous on that interval, and f (0) = −1, while f (1) = 1 − cos(1) > 0. Thus there is some point c ∈ (0, 1) such that f (c) = 0 as required. 32. Exercise. Show there is at least one root of the equation x − e−x = 0 in the interval (0, 1). 33. Corollary. Let f be continuous on the compact interval [a, b], and assume there is some constant h such that f (a) < h and f (b) > h. Then there is a point c with a < c < b such that f (c) = h.
We sometimes refer to f as being continuous on a compact interval. Such an f has some very nice properties. 30. Theorem (Intermediate Value Theorem). Let f be continuous on the compact interval [a, b], and assume that f (a) < 0 and f (b) > 0. Then there is some point c with a < c < b such hat f (c) = 0. Proof. We make no attempt to prove this formally, but sketch a proof with a pair of sequences and a repeated bisection argument. It is also noted that each hypothesis is necessary. 31. Example. Show there is at least one root of the equation x − cos x = 0 in the interval [0, 1].
Let Sn = a + (a + r) + (a + 2r) + · · · + (a + nr), Sn = (a + nr) + (a + (n − 1)r) + · · · + (a + r) + a, so 2Sn = (2a + nr) + (2a + nr) + · · · + (2a + nr), and (a + (a + nr)) Sn = (n + 1). 2 Note that if r > 0 then Sn → ∞ as n → ∞, while if r < 0 then Sn → −∞ as n → ∞. We next consider a geometric progression (or series): Let Sn = a + ar + ar 2 + · · · + ar n , rSn = ar + ar 2 + ar 3 + · · · + ar n+1 , so (1 − r)Sn = a(1 − r n+1 ), and a(1 − r n+1 ) Sn = if r = 1. 1−r Note that if |r| < 1, then Sn → a as n → ∞.
Advanced Calculus And Analysis by I. Craw