Minimize and maximize
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From: hagman on 4 Oct 2009 11:46 On 4 Okt., 14:31, Kaba <n...(a)here.com> wrote: > Hi, > > Let > > n in N (positive integers) > s in R+ (positive reals) > > f : [1, n] sub N -> R : > f(m) = m^2 (s - 1)^2 / n + 2 m^2 (s - 1) / n + m = (s^2 -1) m^2 / n + m > > Problem: > > 1) Find m_min which minimizes f > > 2) Find m_max which maximizes f > > Any ideas? If n=1, then trivially m_min = m_max = 1. Hence assume n>=2 from now on. We have f(m+1) - f(m) = (s^2-1)*(2m+1)/n + 1. If s>=1, the right hand side is positive, i.e. f is a strictly increasing function, hence m_min = 1, m_max = n. Next assume sqrt(2)/2 <= s < 1. Then for m<n f(m+1) - f(m) = 1 - (1-s^2)*(2m+1)/n >= 1 - (m+ 1/2)/n >0 Hence f is still strictly increasing and m_min = 1, m_max = n. For the rest assume 0 < s < sqrt(2)/2 and let a = 1-s^2 (thus 1/2 < a < 1). Then f(m) = -a m^2 / n + m = -a/n * (m - n/(2a))^2 + n/(4a) This is a downward parabola, hence the maximum is at the integer closest to n/(2a) (observe that n/2 < n/(2a) < n): m_max = round(n/(2-2s^2)) and the minimum occurs at the boundary, i.e. either at 1 or at n. We have to compare f(1) </=/> f(n) <-> (s^2 -1) / n + 1 </=/> (s^2 -1) n + n <-> n - 1 </=/> s^2 (n^2 -1) <-> 1 </=/> s^2 (n+1) Thus m_min = 1 if s^2*(n+1) > 1 and m_min = n if s^2*(n+1) < 1 If s^2*(n+1) = 1, the minimum is occurs at both ends. (similarly, the maximum is taken at two values if n/(2-2s^2) is exactly halfway between two integers).
From: Kaba on 4 Oct 2009 12:31 Kaba wrote: > If s > 1 / sqrt(2), then f is decreasing in [1, n]. Thus > m_min = f(n) > m_max = f(1) Clearly not of my best days. m_min = n m_max = 1 > Otherwise, if s <= 1 / sqrt(2): > m_min = min {f(floor(m)), f(floor(m) + 1)} > m_max = max {f(1), f(n)} m_min = floor(m), if f(floor(m)) <= f(floor(m) + 1) floor(m) + 1, otherwise m_max = 1, if f(1) >= f(n) n, otherwise > Right? > > In the latter case: > > m_min = min {f(floor(m)), f(floor(m) + 1)} = > f(floor(m)), if m - floor(m) <= 1 / 2 > f(floor(m) + 1), otherwise m_min = floor(m), if m - floor(m) <= 1 / 2 floor(m) + 1, otherwise > m_max = max {f(1), f(n)} = > f(1), if s >= 1 / sqrt(n + 1) > f(n), otherwise m_max = 1, if s >= 1 / sqrt(n + 1) n, otherwise -- http://kaba.hilvi.org
From: Kaba on 4 Oct 2009 14:03 Thanks hagman, that seems correct. -- http://kaba.hilvi.org
From: Kaba on 4 Oct 2009 18:09 Thanks for help, Tonio. -- http://kaba.hilvi.org
From: Stephen J. Herschkorn on 5 Oct 2009 01:19
Tonico wrote: >On Oct 4, 2:31 pm, Kaba <n...(a)here.com> wrote: > > >>Let >> >>n in N (positive integers) >>s in R+ (positive reals) >> >>f : [1, n] sub N -> R : >>f(m) = m^2 (s - 1)^2 / n + 2 m^2 (s - 1) / n + m >> >>Problem: >> >>1) Find m_min which minimizes f >> >>2) Find m_max which maximizes f >> >>Any ideas? >> >> > > > >Yes: what about adding some parentheses to understand clearly what is >numerator and what ia denominator? This is the least one could expect >from a graduate student... > > It seemed clear of me. {[m^2 (s - 1)^2] / n} + {[2 m^2 (s - 1)] / n} + m (Or would you prefer even more parentheses, brackets, and braces?) Don't you know standard precedence of arithmetic operations? This is the least once could expect from someone who chastises graduate students... -- Stephen J. Herschkorn sjherschko(a)netscape.net Math Tutor on the Internet and in Central New Jersey |