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Prove that different functions may have the same average

by RoRi

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 pth 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

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 pth 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*}

Calculate the average power in an electrical circuit

by RoRi

Let e(t) and i(t) denote the current and voltage, respectively, in a circuit at time t. Define

    \[ e(t) = 160 \sin t, \qquad i(t) = 2 \sin \left( t - \frac{\pi}{6} \right). \]

Then, we define the average power by the integral equation

    \[ A = \frac{1}{T} \int_0^T e(t) i(t) \, dt, \]

where T is the period of the voltage and current. Find T and calculate the average power.

First, since the voltage is given by e(t) = 160 \sin t we know it is periodic with period 2 \pi, so T = 2 \pi. Then, we compute the average power,

    \begin{align*} A &= \frac{1}{2 \pi} \int_0^{2 \pi}320 \sin t \sin \left(t - \frac{\pi}{6} \right) \, dt \\ &= \frac{160}{\pi} \int_0^{2 \pi} \sin t\left( \sin t \cos \frac{\pi}{6} - \sin \frac{\pi}{6} \cos t \right) \, dt \\ &= \frac{160}{\pi} \left( \cos \frac{\pi}{6} \int_0^{2 \pi} \sin^2 t \, dt - \sin \frac{\pi}{6} \int_0^{2 \pi} \sin t \cos t \, dt \right) \\ &= \frac{160}{\pi} \left( \frac{\sqrt{3}}{2} \int_0^{2 \pi} (1 - \cos 2t) \, dt - \frac{1}{4} \int_0^{2 \pi} \sin 2t \, dt \right) \\ &= \frac{80 \sqrt{3}}{2 \pi} (2\pi - 0) - \frac{40}{\pi}\cdot 0 \\ &= 80 \sqrt{3}. \end{align*}

Compute average voltage and root-mean-square of voltage

by RoRi

Denote the voltage in a circuit at time t by e(t). Let

    \[ e(t) = 3 \sin 2t. \]

Calculate the following:

  1. Average voltage over the interval 0 \leq t \leq \frac{\pi}{2}.
  2. The root-mean-square of the voltage
  1. Denote the average of e(t) by A(e) and compute,

        \begin{align*} A(e) &= \frac{\int_0^{\frac{\pi}{2}} 3 \sin 2t \, dt}{\int_0^{\frac{\pi}{2}} dt} \\ &= \frac{6}{\pi} \int_0^{\frac{\pi}{2}} \sin 2t \, dt \\ &= \frac{3}{\pi} \int_0^{\pi} \sin t \, dt \\ &= \frac{3}{\pi}(1 - \cos \pi) \\ &= \frac{6}{\pi}. \end{align*}

  2. The root-mean-square is given by the square root of the function e^2 over the interval 0 \leq t \leq \frac{\pi}{2}. This gives us

        \begin{align*} \text{root-mean-square } &= \left( \frac{2}{\pi} \int_0^{\frac{\pi}{2}} 9 \sin^2 2t \, dt \right)^{\frac{1}{2}} \\ &= \left( \frac{9}{\pi} \int_0^{\pi} \sin^2 t \, dt \right)^{\frac{1}{2}} \\ &= \sqrt{\frac{9}{2}} \\ &= \frac{3}{\sqrt{2}} = \frac{3 \sqrt{2}}{2}. \end{align*}

    Where we used the formula from the solution of Example 3, p. 101 of Apostol, to compute the integral \int_0^{\pi} sin^2 x \, dx = \frac{\pi}{2}.

Prove some more properties of the averages of functions on an interval

by RoRi

Define A_a^b (f) to be the average of a function f on the interval [a,b],

    \[ A_a^b (f) := \frac{1}{b-a} \int_a^b f(x) \, dx. \]

  1. If c \in \mathbb{R} with a < c < b, prove there exists t \in \mathbb{R} with 0 < t < 1 such that

        \[ A_a^b (f) = t A_a^c (f) + (1-t)A_c^b (f). \]

  2. Prove part (a) holds for weighted averages of functions where for a nonnegative weight function w we define the weighted average of f on [a,b] by

        \[ \frac{\int_a^b w(x) f(x) \, dx}{\int_a^b w(x) \, dx}. \]

  1. Proof. Let

        \[ t = \frac{c-a}{b-a}. \]

    Then, since a < c < b, we have 0 < t < 1. Furthermore,

        \[ 1-t = 1 - \frac{a-c}{a - b} = \frac{a-b-a+c}{a-b} = \frac{c-b}{a-b}. \]

    So,

        \begin{align*} tA_a^c (f) + (1-t)A_c^b (f) &= \frac{c-a}{b-a} \cdot \frac{1}{c-a} \int_a^c f(x) \, dx + \frac{c-b}{a-b} \cdot \frac{1}{b-c} \int_c^b f(x) \, dx \\ &= \frac{1}{b-a} \int_a^c f(x) \, dx - \frac{1}{a-b} \int_c^b f(x) \, dx \\ &= \frac{1}{b-a} \left( \int_a^c f(x) \, dx + \int_c^b f(x) \, dx \right)\\ &= \frac{1}{b-a} \int_a^b f(x) \, dx \\ &= A_a^b (f). \qquad \blacksquare \end{align*}

  2. Proof. Let

        \[ t = \frac{\int_a^c w(x) \, dx }{\int_a^b w(x) \, dx}. \]

    Then,

        \[ 1-t = \frac{\int_a^b w(x) \, dx - \int_a^c w(x) \, dx}{\int_a^b w(x) \, dx} = \frac{\int_c^b w(x) \, dx}{\int_a^b w(x) \, dx}. \]

    Thus, 0 < t < 1 (since \int_c^b w(x) \, dx < \int_a^b w(x) \, dx since a < c < b and since w is nonnegative both of these are nonnegative).
    Then,

        \begin{align*} t \cdot A_a^c (f) + (1-t) \cdot A_c^b (f) &= \frac{\int_a^c w(x) \, dx}{\int_a^b w(x) \, dx} \cdot \frac{\int_a^c w(x) f(x) \, dx}{\int_a^c w(x) \, dx} + \frac{\int_c^b w(x) \, dx}{\int_a^b w(x) \, dx} \cdot \frac{\int_c^b w(x) f(x) \, dx}{\int_c^b w(x) \, dx} \\ &= \frac{\int_a^c w(x) f(x) \, dx + \int_c^b w(x) f(x) \, dx}{\int_a^b w(x) \, dx} \\ &= \frac{\int_a^b w(x) f(x) \, dx}{\int_a^b w(x) \, dx} \\ &= A_a^b (f). \qquad \blacksquare \end{align*}

Prove some properties for weighted averages of functions

by RoRi

With reference to the previous exercise which of the following properties are valid for weighted averages of a function on an interval [a,b]. Denote the weighted average of f with a weight function w on [a,b] by A(f).

  1. Additive property: A(f+g) = A(f) + A(g).
  2. Homogeneous property: A(cf) = cA(f) for all c \in \mathbb{R}.
  3. Monotone property: A (f) \leq A(g) if f \leq g on [a,b].

All of these properties are valid for weighted averages.

  1. Proof. We compute,

        \begin{align*} A(f+g) &= \frac{\int_a^b w(x)(f(x) + g(x))\, dx}{\int_a^b w(x) \, dx} \\ &= \frac{\int_a^b w(x)f(x) \, dx + \int_a^b w(x) g(x) \, dx}{\int_a^b w(x) \, dx} & (\text{Add. property of integral})\\ &= \frac{\int_a^b w(x) f(x) \, dx}{\int_a^b w(x) \, dx} + \frac{\int_a^b w(x)g(x) \, dx }{ \int_a^b w(x) \, dx} \\ &= A(f) + A(g). \qquad \blacksquare \end{align*}

  2. Proof. We compute,

        \begin{align*} A(cf) &= \frac{\int_a^b c f(x) w(x) \, dx}{\int_a^b w(x) \, dx} \\ &= c \cdot \frac{\int_a^b w(x) f(x) \, dx}{\int_a^b w(x) \, dx} & (\text{Homogeneous prop. of integrals})\\ &= c A(f). \qquad \blacksquare \end{align*}

  3. Proof. Assume f \leq g on [a,b], then since w is nonnegative (definition of a weight function) we have w(x) f(x) \leq w(x) g(x) for all x \in [a,b]. Next, by the monotone property of the integral we have

        \[ \int_a^b w(x) f(x) \, dx \leq \int_a^b w(x) g(x) \, dx.\]

    Then, since w is nonnegative, \int_a^b w(x) \, dx is also nonnegative and we have

        \begin{align*} \frac{\int_a^b w(x) f(x) \, dx}{\int_a^b w(x) \, dx} &\leq \frac{\int_a^b w(x) g(x) \, dx}{\int_a^b w(x) \, dx} \\ \implies A(f) &\leq A(g). \qquad \blacksquare \end{align*}