Proof. Proceed by contradiction. Suppose that f has more than one root in (0,∞). Let r1 and r2, with r1 < r2, be two consecutive such roots (note that we can always find two consecutive roots r1 and r2 because f (r) > 0 for any root r). From the hypothesis, we have f (r1) > 0 and f (r2) > 0. Since f ′ is continuous at r1, there exists an 0 < 2 < (r2− r1) such that f (x) > 0 in the interval (r1,...

(1) Let X be a real linear space and R1, R2 be finite sequences of elements of X. If lenR1 = lenR2, then ∑ (R1 +R2) = ∑ R1 + ∑ R2. (2) Let X be a real linear space and R1, R2, R3 be finite sequences of elements of X. If lenR1 = lenR2 and R3 = R1−R2, then ∑ R3 = ∑ R1− ∑ R2. (3) Let X be a real linear space, R1, R2 be finite sequences of elements of X, and a be an element of R. If R2 = aR1, then ...

OBJECTIVES The aim of this study was to assess the impact of oral water and intravenous furosemide challenges on blood oxygenation level-dependent magnetic resonance imaging measurements in the kidney and to examine the contribution of R2 (=1/T2) to changes in R2* (=1/T2*). MATERIALS AND METHODS This Health Insurance Portability and Accountability Act-compliant study had institutional review ...