Tag: Means

Find a value which minimizes a given sum

by RoRi

October 13, 2015

Let $a_1, \ldots, a_n \in \mathbb{R}$ . Prove

$\sum_{k=1}^n (x-a_k)^2$

is minimal when $x$ is the arithmetic mean of $a_1, \ldots, a_n$ .

Proof. Recall that the arithmetic mean of $a_1, \ldots, a_n \in \mathbb{R}$ is given by

$x = \frac{a_1 + \cdots + a_n}{n}.$

Now, consider the given sum

$\sum_{k=1}^n (x-a_k)^2.$

To find the minimum we first take the derivative, using the linearity of the derivative over this (finite) sum, (i.e., using that $(f+g)' = f' + g'$ ),

$\begin{align*} \left( \sum_{k=1}^n (x-a_k)^2 \right)' &= \sum_{k=1}^n ((x - a_k)^2)' \\ &= \sum_{k=1}^n 2(x-a_k). \end{align*}$

Setting this equal to 0 and solving,

$\begin{align*} \sum_{k=1}^n 2(x - a_k) = 0 && \implies && 2 \left(\sum_{k=1}^n x - \sum_{k=1}^n a_k \right) &= 0 \\ && \implies && nx - \sum_{k=1}^n a_k &= 0 \\ && \implies && nx &= \sum_{k=1}^n a_k \\ && \implies && x &= \frac{a_1 + \cdots + a_n}{n}. \qquad \blacksquare \end{align*}$

Prove that different functions may have the same average

by RoRi

September 8, 2015

Let $f$ be a continuous, strictly monotonic function on $\mathbb{R}_{>0}$ with inverse $g$ , and let $a_1 < a_2 < \cdots < a_n$ be given positive real numbers. Then define,

$M_f = g \left( \frac{1}{n} \sum_{i=1}^n f(a_i) \right).$

This $M_f$ is called the mean of $a_1, \ldots, a_n$ with respect to $f$ . (When $f(x) = x^p$ for $p \neq 0$ , this coincides with the $p$ th power mean from this exercise).

Show that if $h(x) = af(x) + b$ with $a \neq 0$ , then $M_h = M_f$ .

Proof. Let $h(x) = af(x) + b$ with $a \neq 0$ . Then, $h$ has an inverse since it is strictly monotonic (since it is the composition of $f$ and the linear function $ax+b$ , both of which are strictly monotonic for $a \neq 0$ ). Its inverse is given by

$h^{-1}(x) = f^{-1} \left( \frac{x-b}{a} \right).$

So,

$\begin{align*} M_h &= h^{-1} \left( \frac{1}{n} \sum_{i=1}^n h(a_i) \right) \\ &= h^{-1} \left( \frac{1}{n} \sum_{i=1}^n (af(a_i) + b) \right) \\ &= h^{-1} \left( \frac{a}{n} \left(\sum_{i=1}^n f(a_i)\right) + b \right)\\ &= f^{-1} \left(\frac{\frac{a}{n} \left(\sum_{i=1}^n f(a_i)\right) + b - b}{a} \right)\\ &= g \left( \frac{1}{n} \sum_{i=1}^n f(a_i) \right) \\ &= M_f. \qquad \blacksquare \end{align*}$

Show the mean of a strictly monotonic function lies in an interval

by RoRi

September 7, 2015

Let $f$ be a continuous, strictly monotonic function on $\mathbb{R}_{>0}$ with inverse $g$ , and let $a_1 < a_2 < \cdots < a_n$ be given positive real numbers. Then define,

$M_f = g \left( \frac{1}{n} \sum_{i=1}^n f(a_i) \right).$

This $M_f$ is called the mean of $a_1, \ldots, a_n$ with respect to $f$ . (When $f(x) = x^p$ for $p \neq 0$ , this coincides with the $p$ th power mean from this exercise).

Show that

$a_1 < M_f < a_n.$

Proof. Since $f$ is strictly monotonic on the positive real axis and $a_1 < a_2 < \cdots < a_n$ are $n$ positive reals, we know $f$ is strictly increasing or strictly decreasing, and correspondingly we have,

$f(a_1) < f(a_2) < \cdots < f(a_n) \qquad \text{or} \qquad f(a_1) > f(a_2) > \cdots > f(a_n).$

First, assume $f$ is striclty increasing, then

$f(a_1) < \cdots < f(a_n) \quad \implies \quad \frac{1}{n} \sum_{i=1}^n f(a_1) < \frac{1}{n} \sum_{i=1}^n f(a_i) < \frac{1}{n} \sum_{i=1}^n f(a_n).$

Since $f$ is strictly increasing so is its inverse $g$ (by Apostol’s Theorem 3.10); thus, we have

$\begin{align*} &g \left( \frac{1}{n} \sum_{i=1}^n f(a_1) \right) < g \left( \frac{1}{n} \sum_{i=1}^n f(a_i) \right) < g \left( \frac{1}{n} \sum_{i=1}^n f(a_n) \right) \\[8pt] &\implies \quad g(f(a_1)) < M_f < g(f(a_n)) \\ &\implies \quad a_1 < M_f < a_n. \end{align*}$

If $f$ is strictly decreasing then

$\begin{align*} & f(a_1) > \cdots > f(a_n) \\[8pt] \implies & \frac{1}{n} \sum_{i=1}^n f(a_1) > \frac{1}{n} \sum_{i=1}^n f(a_i) > \frac{1}{n} \sum_{i=1}^n \\[8pt] \implies & g\left( \frac{1}{n} \sum_{i=1}^n f(a_1) \right) < g \left( \frac{1}{n} \sum_{i=1}^n f(a_i) \right) < g \left( \sum_{i=1}^n f(a_n) \right) \\[8pt] \intertext{(where the inequalities reverse since $f$ decreasing implies its inverse, $g$, is also decreasing)} \implies & a_1 < M_f < a_n. \qquad \blacksquare \end{align*}$

Prove connection between power mean and arithmetic mean of a function

by RoRi

September 6, 2015

Let $f$ be a continuous, strictly monotonic function on $\mathbb{R}_{>0}$ with inverse $g$ , and let $a_1 < a_2 < \cdots < a_n$ be given positive real numbers. Then define,

$M_f = g \left( \frac{1}{n} \sum_{i=1}^n f(a_i) \right).$

This $M_f$ is called the mean of $a_1, \ldots, a_n$ with respect to $f$ . (When $f(x) = x^p$ for $p \neq 0$ , this coincides with the $p$ th power mean from this exercise).

Show that

$f(M_f) = \frac{1}{n} \sum_{i=1}^n f(a_i).$

Proof. Since $g$ is the inverse of $f$ we know $f(g(x)) = x$ for all $x$ in the range of $f$ , i.e., for all $x$ such that there is some $c \in \mathbb{R}_{>0}$ such that $f(c) = x$ .

By the definition of $M_f$ then, we have that

$f(M_f) = f \left( g \left( \frac{1}{n} \sum_{i=1}^n f(a_i) \right) \right).$

So, if $\frac{1}{n} \sum_{i=1}^n f(a_i)$ is in the domain of $g$ then we are done. Since $g$ is the inverse of $f$ it’s domain is equal to the range of $f$ . We show that this value is in the range of $f$ using the intermediate value theorem.

Without loss of generality, assume $f$ is strictly increasing (the alternative assumption, that $f$ is strictly decreasing will produce an almost identical argument). Then, since $a_1 < a_2 < \cdots < a_n$ are all positive real numbers we have $f(a_1) < f(a_2) < \cdots < f(a_n)$ . (Here if we’d assumed that $f$ was strictly decreasing the roles inequalities would be reversed.) Then we have,

$f(a_1) = \frac{1}{n} \sum_{i=1}^n f(a_1) < \frac{1}{n} \sum_{i=1}^n f(a_i) < \frac{1}{n} \sum_{i=1}^n f(a_n) = f(a_n).$

Hence, by the intermediate value theorem, since

$\frac{1}{n} \sum_{i=1}^n f(a_i) \in [f(a_1), f(a_n)]$

there must be some $c \in \mathbb{R}_{>0}$ such that

$f(c) = \frac{1}{n} \sum_{i=1}^n f(a_i).$

Thus, $\frac{1}{n} \sum_{i=1}^n f(a_i)$ is in the domain of $g$ , so

$f(M_f) = f \left( g \left( \frac{1}{n} \sum_{i=1}^n f(a_i) \right) \right) = \frac{1}{n} \sum_{i=1}^n f(a_i). \qquad \blacksquare$

Application of the arithmetic mean – geometric mean inequality

by RoRi

July 10, 2015

Use the Arithmetic mean – Geometric Mean inequality to prove that for $a,b,c \in \mathbb{R}_{>0}$ , with $abc =8$ , then

$a+b+c \geq 6 \qquad \text{and} \qquad ab + ac + bc \geq 12.$

Proof. Since $a,b,c$ are positive real numbers from the AM-GM inequality we know,

$\begin{align*} &&G(a,b,c) &\leq M_1 (a,b,c) \\ \implies && (abc)^{1/3} &\leq \left( \frac{a+b+c}{3} \right) \\ \implies && 8 &\leq \frac{(a+b+c)^3}{27} \\ \implies && 27\cdot 8 &\leq (a+b+c)^3 \\ \implies && a+b+c &\geq 6. \end{align*}$

Then, for the second inequality we consider $M_{-1}$ and use the inequality from part (b) of the AM-GM exercise,

$\begin{align*} && M_{-1} (a,b,c) &\leq G(a,b,c) \\ \implies && \left( \frac{a^{-1} + b^{-1} + c^{-1}}{3} \right)^{-1} &\leq (abc)^{1/3} \\ \implies && \left( \frac{3}{\frac{1}{a} + \frac{1}{b} + \frac{1}{c}} \right) &\leq 8^{1/3} \\ \implies && \frac{27}{\left( \frac{1}{a} + \frac{1}{b} + \frac{1}{c} \right)^3} &\leq 8 \\ \implies && \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right)^3 &\geq \frac{27}{8} \\ \implies && \frac{1}{a} + \frac{1}{b} + \frac{1}{c} &\geq \frac{3}{2} \\ \implies && \frac{bc + ac + ab}{abc} &\geq \frac{3}{2} \\ \implies && ab + ac + bc &\geq \frac{3}{2} \cdot 8 = 12. \qquad \blacksquare \end{align*}$

Prove the arithmetic mean – geometric mean inequality

by RoRi

July 10, 2015

We define the geometric mean $G$ of real numbers $x_1, \ldots, x_n$ by

$G := (x_1 \cdot \cdots \cdot x_n)^{1/n}.$

We also recall the definition of the $p$ th power mean, $M_p$ :

$M_p = \left( \frac{x_1^p + \cdots + x_n^p}{n} \right)^{1/p}.$

Prove the arithmetic mean – geometric mean inequality, i.e., prove $G \leq M_1$ where $M_1$ is the $p$ th power mean with $p=1$ (also known as the arithmetic mean).
For integers $p,q$ with $q < 0 < p$ , prove that $M_q < G < M_p$ if $x_1, \ldots, x_n$ are not all equal.

Proof. First, if $x_1 = \cdots = x_n$ , then

$G = (x_1 \cdot \cdots \cdot x_n)^{1/n} = (x_1^n)^{1/n} = x_1 \qquad \text{and} \qquad M_1 = \left( \frac{x_1 + \cdots + x_n}{n} \right) = \left( \frac{nx_1}{n} \right) = x_1.$

Hence, $G = M_1$ for the case $x_1 = \cdots = x_n$ .
Now, if $x_1, \ldots, x_n$ are not all equal then first we write,

$G^n = \left( (x_1 \cdot \cdots \cdot x_n)^{1/n}\right)^n = x_1 \cdot \cdots \cdot x_n.$

Then using the previous exercise, we proceed,

$\begin{align*} &&\left( \frac{1}{G^n} \right) (x_1 \cdot \cdots \cdot x_n) &= 1 \\ \implies && \left( \frac{x_1}{G} \right) \left( \frac{x_2}{G} \right) \cdot \cdots \cdot \left( \frac{x_n}{G} \right) &= 1 \\ \implies && \frac{x_1}{G} + \frac{x_2}{G} + \cdots + \frac{x_n}{G} &> n & (\text{previous exercise})\\ \implies && x_1 + x_2 + \cdots + x_n &> n \cdot G \\ \implies && M_1 &> G. \end{align*}$

Therefore, if $x_1, \ldots, x_n$ are not all equal then we have the strict inequality $G < M_1$ , as requested $. \qquad \blacksquare$
Proof. We’ll start with the inequality on the right first. So, we want to show $G < M_p$ for any $n$ positive real numbers $x_1, \ldots, x_n$ not all equal, where $p$ is a positive integer. First, we’ll want to observe that if $x_1, \ldots, x_n$ are positive real numbers, not all equal, then $x_1^p, \ldots, x_n^p$ are also positive real numbers, not all equal. So from the definition of $M_p$ and letting $M_p (x_1^p, \ldots, x_n^p)$ denote the $p$ th power mean of the numbers $x_1^p, \ldots, x_n^p$ , we have

$\begin{align*} M_1 (x_1^p, \ldots, x_n^p) &= \left( \frac{x_1^p + \cdots + x_n^p}{n} \right) \\ &= \left( \left( \frac{x_1^p + \cdots + x_n^p}{n} \right)^{1/p} \right)^p \\ &= (M_p(x_1, \ldots, x_n))^p. \end{align*}$

So, the observation is that $(M_p (x_1, \ldots, x_n))^p = M_1(x_1^p, \ldots, x_n^p)$ . Now, using part (a), we have,

$\begin{align*} G(x_1, \ldots, x_n)^p &= (x_1 \cdots x_n)^{p/n}\\ &= (x_1^p \cdots x_n^p)^{1/n} \\ &= G(x_1^p, \ldots, x_n^p) \\ &< M_1 (x_1^p, \ldots, x_n^p) & (\text{part (a)})\\ &= (M_p (x_1, \ldots, x_n))^p. \end{align*}$

Hence, for any positive real numbers $x_1, \ldots, x_n$ , not all equal we have $(G(x_1, \ldots, x_n))^p < (M_p (x_1, \ldots, x_n))^p$ which implies (see this exercise) $G < M_p$ . This gives us the inequality on the right.
Now, for $q$ a negative integer, we must show $M_q < G$ . Since $q < 0$ we know that $-q>0$ and so $G < M_{-q}$ from the inequality we just proved. So,

$G^{-q} < (M_{-q})^{-q} \quad \implies \quad G^q > M_q^q \quad \implies \quad G > M_q.$

Where we have used the same exercise again, and the fact that $G^{-q} = \frac{1}{G^q}. \qquad \blacksquare$