Minimize and maximize
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From: Kaba on 4 Oct 2009 08:31 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 Problem: 1) Find m_min which minimizes f 2) Find m_max which maximizes f Any ideas? -- http://kaba.hilvi.org
From: Tonico on 4 Oct 2009 09:16 On Oct 4, 2:31 pm, 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 > > 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... Tonio
From: Kaba on 4 Oct 2009 10:05 Tonico wrote: > On Oct 4, 2:31 pm, 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 > > > > 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... Sorry:) I have been writing a lot of stuff using the AsciiMath notation lately, under which there is no ambiguity, and I then wrote that expression automatically. The function is: f(m) = (m^2 (s - 1)^2 / n) + 2 (m^2 (s - 1) / n) + m Meanwhile, I figured out that f(m) = (m^2 (s^2 + 1) / n) + m Let us expand the domain of f to reals, then: f'(m) = (2 m (s^2 + 1) / n) + 1 f'(m) = 0 => m = -n / (2 (s^2 + 1)) < 0 < 1 Since f is also a quadratic with a positive leading coefficient, f is increasing in [1, n]. => m_min = f(1), m_max = f(n) -- http://kaba.hilvi.org
From: Tonico on 4 Oct 2009 10:24 On Oct 4, 4:05 pm, Kaba <n...(a)here.com> wrote: > Tonico wrote: > > On Oct 4, 2:31 pm, 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 > > > > 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... > > Sorry:) I have been writing a lot of stuff using the AsciiMath notation > lately, under which there is no ambiguity, and I then wrote that > expression automatically. > > The function is: > > f(m) = (m^2 (s - 1)^2 / n) + 2 (m^2 (s - 1) / n) + m > > Meanwhile, I figured out that > > f(m) = (m^2 (s^2 + 1) / n) + m > How did you figure that? You wrote: f(m) = (m^2 (s - 1)^2 / n) + 2 (m^2 (s - 1) / n) + m = (m^2/n)[(s-1)^2 + 2(s-1)] + m = (m^2)/n)(s-1)(s+1) + m = (m^2/n)(s^2 - 1) + m , and from here: f'(m) = (2m/n)(s^2 - 1) + 1 = 0 <==> m = n/[2(1-s^2)] ==> there's only one critic point at m = n/[2(1-s^2)]. The function f(m) is a parabola (and thus its extreme point is its vertex), but we can't even say whether it is convez upwards or downwards since this depends on its quadratic coefficient (s^2 - 1)/ n, which can be zero (s = 1) or negative or positive, depending on s < 1 or s > 1. Tonio > Let us expand the domain of f to reals, then: > > f'(m) = (2 m (s^2 + 1) / n) + 1 > > f'(m) = 0 > => > m = -n / (2 (s^2 + 1)) < 0 < 1 > > Since f is also a quadratic with a positive leading coefficient, f is > increasing in [1, n]. > > => m_min = f(1), m_max = f(n) > > --http://kaba.hilvi.org- Hide quoted text - > > - Show quoted text -
From: Kaba on 4 Oct 2009 11:28
Tonico wrote: > > The function is: > > > > f(m) = (m^2 (s - 1)^2 / n) + 2 (m^2 (s - 1) / n) + m > > > > Meanwhile, I figured out that > > > > f(m) = (m^2 (s^2 + 1) / n) + m > > > > How did you figure that? You wrote: I made an error in derivation. > f(m) = (m^2 (s - 1)^2 / n) + 2 (m^2 (s - 1) / n) + m = > > (m^2/n)[(s-1)^2 + 2(s-1)] + m = (m^2)/n)(s-1)(s+1) + m = > > (m^2/n)(s^2 - 1) + m , and from here: > > f'(m) = (2m/n)(s^2 - 1) + 1 = 0 <==> m = n/[2(1-s^2)] ==> there's > > only one critic point at m = n/[2(1-s^2)]. > > The function f(m) is a parabola (and thus its extreme point is its > vertex), but we can't even say whether it is convez upwards or > downwards since this depends on its quadratic coefficient (s^2 - 1)/ > n, which can be zero (s = 1) or negative or positive, depending on s < > 1 or s > 1. That's right. I'll restrict the problem such that: s in [0, 1[ sub R. Further, n in N, n >= 2. Let m = n / [2(1 - s^2)] Then: m >= 1 <=> n / [2(1 - s^2)] >= 1 <=> n / 2 >= 1 - s^2 <=> s^2 >= 1 - n / 2 <=> true And: m <= n <=> n / [2(1 - s^2)] <= n <=> 1 / 2 <= 1 - s^2 <=> s^2 <= 1 / 2 <=> s <= 1 / sqrt(2) If s > 1 / sqrt(2), then f is decreasing in [1, n]. Thus m_min = f(n) m_max = f(1) Otherwise, if s <= 1 / sqrt(2): m_min = min {f(floor(m)), f(floor(m) + 1)} m_max = max {f(1), f(n)} 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_max = max {f(1), f(n)} = f(1), if s >= 1 / sqrt(n + 1) f(n), otherwise -- http://kaba.hilvi.org |