Linear Mean Function

I've been looking at functions \(f:\mathbb{R}^n \to \mathbb{R}\) which necessarily satisfy the following 3 properties. Given \( a_1, a_2, \dots a_n \in \mathbb{R}^+ \)

1.f(x1+c,x2+c,,xn+c)=f(x1,x2,,xn)+c2.f(cx1,cx2,,cxn)=cf(x1,x2,,xn)3.i=1naixi=0f(x1,x2,,x3)=0\begin{array} { l l } 1. & f(x_1 + c, x_2 + c, \dots , x_n + c) = f(x_1, x_2, \dots , x_n) + c \\ 2. & f(cx_1, cx_2, \dots , cx_n) = cf(x_1, x_2, \dots, x_n) \\ 3. & \sum_{i=1}^n a_i x_i = 0 \Leftrightarrow f(x_1, x_2, \dots, x_3) = 0 \end{array}

I believe that the the only function that satisfies this is f(x1,x2,,xn)=i=1naixii=1naif(x_1,x_2, \dots, x_n) = \frac{\sum_{i=1}^n a_i x_i}{\sum_{i=1}^n a_i}

I have the proof too while I'll write up later. I'm interested in all of your interpretations of this.

What happens if we only have 2 out of these 3 conditions?

Note by Josh Banister
4 years, 8 months ago

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Let x1x_1, x2x_2, \dots, xnx_n be given real numbers. We seek a value of cc so that a1(x1+c)+a2(x2+c)++an(xn+c)=0.a_1 (x_1 + c) + a_2 (x_2 + c) + \dots + a_n (x_n + c) = 0. Solving for cc, we get c=a1x1+a2x2++xnxna1+a2++an.c = -\frac{a_1 x_1 + a_2 x_2 + \dots + x_n x_n}{a_1 + a_2 + \dots + a_n}.

From the third property, f(x1+c,x2+c,,xn+c)=0f(x_1 + c, x_2 + c, \dots, x_n + c) = 0, so by the first property, f(x1,x2,,xn)=c=a1x1+a2x2++xnxna1+a2++an.f(x_1, x_2, \dots, x_n) = -c = \frac{a_1 x_1 + a_2 x_2 + \dots + x_n x_n}{a_1 + a_2 + \dots + a_n}. Note that the second property is never used.

Jon Haussmann - 4 years, 8 months ago

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The idea behind the second property is what happens if the third one was never there? (What happens now if we only assume the first 2 ideas. Hence the linear part of the title)

Josh Banister - 4 years, 8 months ago

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That's very interesting! I believe there is a geometric interpretation of the result, as the (normalized) distance of the vector x projected onto a.

Calvin Lin Staff - 4 years, 8 months ago

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