BMAD 62 Report post Posted July 30, 2013 This post will possibly help with "averages part II" given two positive real numbers a and b, there are four ways of computing the "mean": am= arithmetic mean = (a+b)/2 gm = geometric mean = sqrt[a b] hm = harmonic mean = 2 / (1/a + 1/b) rms = root mean square = sqrt[(a^2+b^2)/2] Rank these from smallest to largest proving your ranks. If one type is not always less than another, explain when it is equal or when it is more. Share this post Link to post Share on other sites

0 Rainman 14 Report post Posted July 30, 2013 (edited) A^{2} - G^{2} = (a^{2}+2ab+b^{2})/4 - ab = (a^{2}-2ab+b^{2})/4 = (a-b)^{2}/4, which is at least 0 with equality only for a=b. Hence A is greater than or equal to G, with equality only for a=b. Lemma: (a+b)^{2} is greater than or equal to 4ab, with equality only for a=b. Proof: (a+b)^{2}-4ab = (a-b)^{2}, which is greater than or equal to 0, with equality only for a=b. H = 2/((a+b)/ab) = 2ab/(a+b) (H/G)^{2} = 4ab/(a+b)^{2}, which is less than or equal to 1 by Lemma. Hence H is less than or equal to G, with equality only for a=b. R^{2} - A^{2} = (a^{2}+b^{2})/2 - (a^{2}+2ab+b^{2})/4 = (2a^{2}+2b^{2})/4 - (a^{2}+2ab+b^{2})/4 = (a^{2}-2ab+b^{2})/4 = ((a-b)/2)^{2}, which is at least 0 with equality only for a=b. Conclusion: R>A>G>H, unless a=b in which case they are all equal. Edit: ranking them from smallest to largest, that would be H, G, A, R Edited July 30, 2013 by Rainman Share this post Link to post Share on other sites

This post will possibly help with "averages part II"

given two positive real numbers a and b, there are four ways of computing the "mean":

am= arithmetic mean = (a+b)/2

gm = geometric mean = sqrt[a b]

hm = harmonic mean = 2 / (1/a + 1/b)

rms = root mean square = sqrt[(a^2+b^2)/2]

Rank these from smallest to largest proving your ranks. If one type is not always less than another, explain when it is equal or when it is more.

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