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January 4

Product distribution of piecewise defined probability density functions

Can anyone give me a complete treatment / algorithm (the formula in the article is completely inadequate) of the product distribution of any two functions? I am interested in cases where the functions may be piecewise and range across positive and negative values. 68.0.144.214 (talk) 17:32, 1 January 2014 (UTC)[reply]

Having looked at the article, I don't understand the formula for the case of identical distributions each with density function f, surely a single-parameter function. So how can f appear with two parameters in the integral? Why is it not just the case above with the product being f(x)f(z/x)?31.54.113.130 (talk) 20:36, 2 January 2014 (UTC)[reply]

It should be f(x)f(z/x); I fixed the formula in the article so it should be correct now. --RDBury (talk) 02:18, 3 January 2014 (UTC)[reply]

Area of triangle from altitude components

A slightly-edited extract from the triangle article is thus:

Denoting the altitudes from sides a, b, and c respectively as

h_{a}

,

h_{b}

, and

h_{c}

, and denoting the semi-sum of the reciprocals of the altitudes as

H=(h_{a}^{-1}+h_{b}^{-1}+h_{c}^{-1})/2

we have

\mathrm {Area} ^{-1}=4{\sqrt {H(H-h_{a}^{-1})(H-h_{b}^{-1})(H-h_{c}^{-1})}}.

This follows straightforwardly from Heron's formula, substituting (twice area)/altitude for each side length. I've been trying to derive a formula for the triangle's area in terms only of the distances from each vertex to the orthocentre, without success. Can anyone help?31.54.113.130 (talk) 20:22, 4 January 2014 (UTC)[reply]

Let a, b, c be the distances from the vertices A, B, C to the orthocenter O, let D, E, F be the feet of the perpendiculars on sides BC, AC, and AB and let d, e, f be the distances from O to these points. Let α, β, γ be the angles BOF=COE, AOF=COD, BOD=AOE. Then d = c cos β = b cos γ, so c/cos γ = b/cos β and similarly these are equal to a/cos α. Let r be the common value. From α + β + γ = π it's not hard to derive the identity cos²α+cos²β+cos²γ+2cosαcosβcosγ=1. From this a cubic equation r³=r(a²+b²+c²)+2abc follows. Solve this for r, use the cubic equation if you want a formula, and from a, b, c and r, the values d, e, f and the area of the triangle easily follow. For example for a=b=c=1 the equation for r is r³=3r+2, the only positive solution is r=2 from which d=e=f=1/2, s=√3, h=3/2, area = 3√3/4. --RDBury (talk) 05:42, 5 January 2014 (UTC)[reply]

Thanks. The need to solve a cubic equation means that only some values of a, b and c will give an exact value of area; (1,2,3) for example leads to a nasty-looking expression for r via Wolfram, it then being impracticable to continue your outlined steps ending in the area of the triangle. Anyway, my intuition that the three lengths are enough to determine the triangle is justified.31.54.113.130 (talk) 09:31, 5 January 2014 (UTC)[reply]