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hardest topic (1 Viewer)

McLake

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Formula:

n /(1/a1 + 1/a2 + 1/a3 + ..... 1/an)

Explanation:

n is divided by the sum of fractions to n
[In the case where k = -1, for a HM]
 

spice girl

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An example? try this:

Given a1 + a2 + ... + an = k, and a1, a2, ..., an > 0

Prove 1/(a1) + 1/(a2) + ... + 1/(an) >= (n^2)/k
 

spice girl

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You just reverse the process and answers become questions...it's not hard to make these things up...

and beside, these AM-whateverM inequalities often begin like: if a1 + a2 + ... + an = something, then blablabla..
 
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Weisy

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I like all this stuff, it's fascinating

but do we have to know harmonic mean at all for the HSC? It's only that it was never even mentioned to us.

Also in the example that spice girl gave, n is integral?

It would be great to see the proof of that example...I've tried, except I don't think my working's legal. And I've also never done a problem like it before.
 

BlackJack

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Yeah, it's integral...:D n is the number of terms 'a'.

I haven't heard of HM before either. I wonder what it means... I don't think we'd need this in the HSC though, harmonic means aren't mentioned in the syllabus... (however, if they lead us up to a proof...)
 

spice girl

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Knowing the harmonic mean isn't necessary, but in 4umaths, using any method is legal. Personally i've never used the harmonic mean crap, there's usually another way. But anyways:

a1 + a2 + ... + an = k

(a1 + a2 + ... + an)/n = k/n

You look at the expression: 1/(a1) + 1/(a2) + ... + 1/(an) and you recognise it as part of an HM expression. So you use AM >= HM:

(a1 + a2 + ... + an)/n >= n / (1/(a1) + 1/(a2) + ... + 1/(an))

k/n >= n / (1/(a1) + 1/(a2) + ... + 1/(an))
thus 1/(a1) + 1/(a2) + ... + 1/(an) >= (n^2)/k
 

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