By Daniel J. Velleman

ISBN-10: 0883853450

ISBN-13: 9780883853450

**Read Online or Download American Mathematical Monthly, volume 117, number 4, April 2010 PDF**

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**Additional resources for American Mathematical Monthly, volume 117, number 4, April 2010**

**Example text**

Closed-form expressions, of course, are no longer available, but, conceding this fact, why can’t we agree to work with inequalities in place of identities? The systems 1p 1p + 2p 1p + 2p + 3p ≤ ≤ ≤ ··· 1 · 2p 2 · 3p 3 · 4p (50) 13 p 13 p + 23 p 13 p + 23 p + 33 p ≤ ≤ ≤ ··· , 1 p+1 22 p 2 p+1 32 p 3 p+1 42 p (51) and for example, are every bit as attractive as identities (47) and (48). Indeed, since the inequalities are known to be valid whenever p ≥ 1 or p ≤ 0, and to switch direction whenever 0 ≤ p ≤ 1, identities (47) and (48) may be viewed as degenerate versions of (50) and (51)!

Suppose that I is any interval. Then a real-valued function is uniformly continuous on I if and only if it is defined on I and preserves quasi-Cauchy sequences from I . 4169/000298910X480793 328 c THE MATHEMATICAL ASSOCIATION OF AMERICA [Monthly 117 Proof. It is an easy exercise to check that uniformly continuous functions preserve quasi-Cauchy sequences. Conversely, suppose that f defined on I is not uniformly continuous. Then there exists an > 0 such that for any δ > 0 there exist a, b ∈ I with |a − b| < δ but | f (a) − f (b)| ≥ .

8) we may determine C, or eliminate it. Theorem 4. If a1 , . . 9) j =k F(1, i) ei(a1 +···+an ) + F(1, −i) e−i(a1 +···+an ) 2 and F(1, i) ei(a1 +···+an ) − F(1, −i) e−i(a1 +···+an ) = 2i n k=1 F(sin ak , cos ak ) . 9) with Theorems 2 and 3. 11) there and use Lemma 3. Theorem 5. If a1 , . . , an are complex numbers, no two of which differ by an integer multiple of π, and if b1 , . . , bn are complex numbers, then sin(z − b1 ) · · · sin(z − bn ) = cos (a1 + · · · + an − b1 − · · · − bn ) sin(z − a1 ) · · · sin(z − an ) n n j =1 k=1 1≤ j ≤n j =k + sin(ak − b j ) sin(ak − a j ) (A) cot(z − ak ) and cos(z − b1 ) · · · cos(z − bn ) = cos(a1 + · · · + an − b1 − · · · − bn ) cos(z − a1 ) · · · cos(z − an ) n n j =1 k=1 1≤ j ≤n j =k − 322 c sin(ak − b j ) sin(ak − a j ) (B) tan(z − ak ).

### American Mathematical Monthly, volume 117, number 4, April 2010 by Daniel J. Velleman

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