看板 Math 關於我們 聯絡資訊
※ 引述《zako1113 (那個人)》之銘言: : 被學生問這道題, 但是完全沒有頭緒 : 背景是只有1科微積分 : f(x) defined on [0,1], f(x)>0, f"(x)<=0, : given a constant 0<M<0.5, show that : for any t such that M < t < 1-M, any s belong to (0,1), : f(t) >= M*f(s). Since f"(x)<=0, f is a concave function. i.e. f( (1-w)*x+w*y ) >= (1-w)*f(x)+w*f(y). for any w belong to (0,1) and 0 <= x,y <= 1. Now fixed M, t and s satisfying above conditions, let w = M, y = s and x = (t-M*y)/(1-M). It is easy to show that 0 <= x <= 1. Hence f(t)=f( (1-M)*x+M*y ) >= (1-M)*f(x)+M*f(y) >= M*f(y) = M*f(s). The last inequality is from f(x)>0. -- ※ 發信站: 批踢踢實業坊(ptt.cc), 來自: 140.112.104.91 ※ 文章網址: https://www.ptt.cc/bbs/Math/M.1450261534.A.45D.html
zako1113 : 明白了 感謝 12/16 19:34