HW1-510-soln

# HW1-510-soln - AMS 510 HW1 Solution 1.49 From x y 20 x 2y 22xy in the same way we can get x 2 z22xz and y 2 z2 2yz then sum the three inequality

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AMS 510 HW1 Solution 1.49 From x y 2 0 x 2 y 2 2xy in the same way, we can get x 2 z 2 2xz and y 2 z 2 2yz then sum the three inequality together, we can get 2x 2 2y 2 2z 2 2xy 2xz 2yz x 2 y 2 z 2 xy xz yz 1.51 To prove x n 1 1 x n 1 x n 1 x n , we need to prove x n 1 x n 1 x x n 1 0 ⇒  x 1 x 2n 1 1 x n 1 0 , when 0 x 1 x – 1 0 and x 2n 1 1 0 , then we can get conclusion; when x 1 x – 1 0 and x 2n 1 1 0 , then we can get conclusion too; when x = 1 , x n 1 1 x n 1 = x n 1 x n 1.52 For a 0 , we can get  a 1 a 2 0 a 1 a 2 a 1 a 2 , For a 0 , we can get  a 1 a 2 0 ⇒ − a 1 a 2 a 1 a 2 , 1.54 if a = b 3 , then 1 3 a 1 a 2 a 3  3 a 1 a 2 a 3 b 1 3 b 2 3 b 3 3 3 b 1 b 2 b 3 b 1 3 b 2 3 b 3 3 3 b 1 b 2 b 3 = b 1 b 2 b 3  b 1 2 b 2 2 b 3 2 b 1 b 2 b 2 b 3 b 1 b 3 = 1 2 b 1 b 2 b 3 [ b 1 b 2 2  b 2 b 3 2  b 1 b 3 2 ] a is positive, so b is positive, then Right Hand Side is not less than zero, so we have b 1 3 b 2 3 b 3 3 3 b 1 b 2 b 3 0 b 1 3 b 2 3 b 3 3 3 b 1 b 2 b 3 1 3 a 1 a 2 a 3  3 a 1 a 2 a 3 1.84 Equation holds for n=1, assuming equation holds for n=k, then for n=k+1 1 3 5  2 k – 1  2 k 1 = k 2 2k 1 = k 1 2 equation holds.

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1.85 Equation holds for n=1, assuming equation holds for n=k, then for n=k+1 1 1 3 1 3 5 1 5 7  1 2 k – 1  2 k 1 1 2 k 1  2 k 3 = k 2 k 1 1 2 k 1  2 k 3 = k 1 2 k 3 equation holds. 1.89 Equation holds for n=1, assuming equation holds for n=k, then for n=k+1 1 3 2 3 3 3 ... k 3  k 1 3 = 1 4 k 2 k 1 2  k 1 3 = k 1 2 1 4 k 2 k 1 = 1 4 k 1 2 k 2 2 equation holds.
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## This note was uploaded on 01/31/2011 for the course AMS 510 taught by Professor Feinberg,e during the Fall '08 term at SUNY Stony Brook.

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HW1-510-soln - AMS 510 HW1 Solution 1.49 From x y 20 x 2y 22xy in the same way we can get x 2 z22xz and y 2 z2 2yz then sum the three inequality

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