Mazdak Farrokhzad edited Version2.tex  over 10 years ago

Commit id: c51f3d36c8863cfaf0838dfb7c01d39a30f9d76e

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Substitute in $f_2$ with $f_1$:  $$  f_2(n) = b\sum_{j=i}^{n-1} 1 + c\sum_{j=i}^{n-1} (n-i) = bn + cn cn\sum_{j=i}^{n-1} 1 + c\sum_{j=i}^{n-1} i\\   = bn + cn^2 + \frac{c}{2}n(n-1) = (c + \frac{c}{2})n^2 + (b - \frac{c}{2})n  $$  $$f_1(j,i) = \sum_{k=i}^j d = d \sum_{k=i}^j 1 = d(1 + j - i) = d + dj - di$$  $$