美文网首页
一、算法-数学基础

一、算法-数学基础

作者: 天色将变 | 来源:发表于2020-02-18 21:30 被阅读0次
指数运算
  • XA * XB = XA+B
  • XA / XB = XA-B
  • (XA ) B = XAB
  • XA + XA = 2XA
  • 2A + 2A = 2 * 2A = 2A+1
对数运算
  • XA = B,则 A = logXB。如:23 = 8,则 3 = log28;43 = 64,则 3 = log464;
  • log28 通常简写为log8,所有以2为底数的对数,2都可以简写。以其他数为底数的对数,底数不能简写。
  • logAB = logA+logB,A,B>0;如:log(8*4) = log8+log4 = 3+2 = 5 = log32,32 = 25
  • logA/B = logA-logB,A,B>0;如:log(8/4) = log8-log4 = 3-2 = 1 = log2,2 = 21
  • logAB = (logCB) / (logCA),A,B,C>0,A不等于1
    证明:
    令:X = logAB,Y = logCB,Z = logCA,有X= Y / Z
    so:B = AX , B = CY, A = CZ
    代入:B = (CZ)X = CY
    得:ZX = Y
  • log(AB)=BlogA。( log(A * A * A.....*A) = logA+logA+....+logA = BlogA )
  • 对于所有的X>0,logX < X
级数运算
  • 20+21+22+....2n = 2n+1-1
  • 当A>1时,A0+A1+A2+....An = (An+1-1) / (A-1)
    证明:
    S = A0+A1+A2+....An
    AS = A1+A2+....An+An+1
    S - AS = A0 - An+1
    S = (An+1-1) / (A-1)
  • 当0<A<1时,A0+A1+A2+....An <= 1/(1-A)
    证明:
    S = A0+A1+A2+....An
    AS = A1+A2+....An+An+1
    S - AS = A0 - An+1
    S = (1 - An+1) / (1 - A)
    此时,由于A<1,所以 An+1当n趋向于无限大时,An+1无限趋向于0,因此S <= 1 / (1 - A)
  • 1+2+3+...+n = n(n+1)/2
    证明:
    S = 1 + 2 + 3 +...+ n-2 + n-1 + n
    S = n + n-1 + n-2 +...+3 + 2 + 1
    相加:
    2S= n+1 + n+1 +....+ n+1 = n(n+1)
    S = n(n+1)/2

相关文章

网友评论

      本文标题:一、算法-数学基础

      本文链接:https://www.haomeiwen.com/subject/bbfnfhtx.html