Tag: Functions

Find functions satisfying given conditions

by RoRi

December 20, 2015

Find functions satisfying the given conditions in each of the following cases.

$\displaystyle{ f'(x^2) = \frac{1}{x} }$ for $x> 0$ and $f(1) = 1$ .
$\displaystyle{ f'(\sin^2 x) = \cos^2 x}$ for all $x$ , and $f(1) = 1$ .
$\displaystyle{ f'(\sin x) = \cos^2 x}$ for all $x$ and $f(1) = 1$ .
$\displaystyle{ f'(\log x) = \begin{cases} 1 & \text{for } 0 < x \leq 1, \\ x & \text{for } x > 1, \end{cases}}$ and $f(0) = 0$ .

We make the substitution $t = x^2$ . Since $x > 0$ this gives us $x = \sqrt{t}$ . Therefore,
$\begin{align*} f'(x^2) = \frac{1}{x} && \implies && f'(t) &= \frac{1}{\sqrt{t}} \\[9pt] && \implies && \int f'(t) \, dt &= \int \frac{1}{\sqrt{t}} \, dt \\[9pt] && \implies && f(t) &= 2 \sqrt{t} + C. \end{align*}$

Since we are given $f(1) = 1$ we can solve for $C$ ,

$1 = f(1) = 2 \sqrt{1} + C \quad \implies \quad C = -1.$

Therefore,

$f(t) = 2 \sqrt{t} - 1.$
We make the substitution $t = \sin^2 x$ . Since $\sin^2 x + \cos^2 x = 1$ we then have $\cos^2 x = 1 - t$ . Therefore,
$\begin{align*} f'(\sin^2 x) = \cos^2 x && \implies && f'(t) &= 1-t \\[9pt] && \implies && \int f'(t) \, dt &= \int (1-t) \, dt \\[9pt] && \implies && f(t) &= t - \frac{t^2}{2} + C. \end{align*}$

Since we are given that $f(1) = 1$ we can solve for $C$ ,

$1 = f(1) = 1 - \frac{1}{2} + C \quad \implies \quad C = \frac{1}{2}.$

Therefore,

$f(t) = t - \frac{t^2}{2} + \frac{1}{2}.$

This formula is valid for $0 \leq t \leq 1$ since $t = \sin^2 x$ and $0 \leq \sin^2 x \leq 1$ for all $x$ .
We make the substitution $t = \sin x$ . Since $\sin^2 x + \cos^2 x =1$ we then have $\cos^2 x = 1 - t^2$ . So,
$\begin{align*} f'(\sin x) = \cos^2 x && \implies && f'(t) &= 1-t^2 \\[9pt] && \implies && \int f'(t) \, dt &= \int (1-t^2) \, dt \\[9pt] && \implies && f(t) &= t - \frac{t^3}{3} + C. \end{align*}$

Since we are given that $f(1) = 1$ we can solve for $C$ ,

$1 = f(1) = 1 - \frac{1}{3} + C \quad \implies \quad C = \frac{1}{3}.$

Therefore,

$f(t) = t - \frac{t^3}{3} + \frac{1}{3}.$

This is valid for $|t| \leq 1$ since $t = \sin x$ and $-1 \leq \sin x \leq 1$ for all $x$ .
We make the substitution $t = \log x$ . Then, $x = e^t$ and so we have
$f'(\log x) = \begin{cases} 1 & \text{for } 0 < x \leq 1, \\ x & \text{for } x > 1, \end{cases} \quad \implies \quad f'(t) = \begin{cases} 1 & \text{for } 0 < x \leq 1, \\ e^t & \text{for } x > 1. \end{cases}$

So, we consider the two cases separately. If $0 < x \leq 1$ then we have $t < 0$ and

$f'(t) = 1 \quad \implies \quad \int f'(t) \, dt = \int 1 \, dt \quad \implies \quad f(t) = t + C.$

If $x > 1$ then we have $t \geq 0$ and

$f'(t) = e^t \quad \implies \quad \int f'(t) \,dt = \int e^t \, dt \quad \implies \quad f(t) = e^t + C.$

Therefore, we have the function

$f(t) = \begin{cases} t + C & \text{if } t < 0, \\ e^t + C & \text{if } t \geq 0. \end{cases}$

Now, to solve for $C$ we use the condition that $f(0) = 0$ . (Here we’re going to assume we want to make the function continuous at $t = 0$ , i.e., that the two pieces of this piecewise definition take the same value at 0 so that the limits from the left and right would be equal.) Therefore, we have

$0 = f(0) = 0 + C \quad \implies \quad C = 0$

and

$0 = f(0) = e^0 + C \quad \implies \quad C = -1.$

Thus, the function is given by

$f(t) = \begin{cases} t & \text{if } t < 0, \\ e^t - 1 & \text{if } t \geq 0. \end{cases}$

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*}$

Formula for counting lattice points in the ordinate set of a function

by RoRi

July 17, 2015

For a nonnegative function $f$ defined on an interval $[a,b]$ define the set

$S = \{ (x,y) \mid a \leq x \leq b, \ 0 < y \leq f(x) \}.$

(i.e., the region enclosed by the graph of the function and the $x$ -axis between the vertical lines at $x=a$ and $x=b$ ). Then prove

$\text{\# of lattice points in } S = \sum_{n=a}^b [f(n)],$

where $[f(n)]$ is the greatest integer less than or equal to $f(n)$ .

We can help ourselves by drawing a picture to get a good idea of what is going on, then turn that intuition into something more rigorous. The picture is as follows:

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In the picture, we can see the number of lattice points in the ordinate set of $f(x)$ (not including the $x$ -axis since the question stipulates $S$ contains the points $0 < y \leq f(x)$ ). At each integer between $a$ and $b$ , we count the number of lattice points beneath $[f(x)]$ , the greatest integer less than or equal to $f(x)$ . Then we need to turn this intuition from the picture into a proof:

Proof. Let $n \in \mathbb{Z}$ with $a \leq n < b$ . We know such an $n$ exists since $a,b \in \mathbb{Z}$ with $a < b$ . Then, the number of lattice points in $S$ with first element $n$ is the number of integers $y$ such that $0 < y \leq f(n)$ . But, by definition, this is $[f(n)]$ (the greatest integer less than or equal to $f(n)$ ). Summing over all integers $n, \ a \leq n \leq b$ we have,

$\text{\# of lattice points in } S = \sum_{n=a}^b [ f(n) ]. \qquad \blacksquare$

Deduce and prove a formula for [nx]

by RoRi

July 16, 2015

Use the previous exercise (1.11, #4 parts (d) and (e)) to deduce a formula for $[nx]$ and prove this formula is correct.

Claim:

$[nx] = \sum_{k=0}^{n-1} \left[ x + \frac{k}{n} \right].$

Proof. Let $[x] = m$ , then we have

$m \leq x < m+1 \quad \implies \quad nm \leq nx < nm+n.$

Thus, there exists some $j \in \mathbb{Z}$ with $0 \leq j < n$ such that

$nm +j \leq nx < nm+j+1.$

Hence, $[nx] = nm+j$ . Thus,

$m+\frac{j}{n} \leq x < m + \frac{j+1}{n}.$

Then, for each $k \in \mathbb{Z}$ with $0 \leq k < n-j$ we have

$\begin{alignat*}{2} &m+\frac{j+k}{n} \leq x + \frac{k}{n} < m+\frac{j+k+1}{n} & \qquad \qquad & (\text{adding } \frac{k}{n}) \\ \implies \quad & m \leq x + \frac{k}{n} < m+1 & & (\frac{j+k}{n} < 1 \text{ since } k < n-j) \\ \implies \quad &\left[ x + \frac{k}{n} \right] = m = [x] && \text{for } 0 \leq k < n-j. \end{alignat*} \end{align*}$

Then, for $n-j \leq k < n$ , we have

$\begin{alignat*}{3} &m+\frac{j+k}{n} \leq x + \frac{k}{n} < m + \frac{j+k+1}{n} \\ \implies \quad & m+1 \leq x + \frac{k}{n} < m+2 & \qquad \qquad & (\frac{j+k}{n} \geq 1 \text{ since } n-j \leq k) \\ \implies \quad & \left[ x + \frac{k}{n} \right] = m+1 = [x] + 1 && \text{for } n-j \leq k < n. \end{alignat*}\end{align*}$

So,

$\begin{align*} \sum_{k=0}^{n-1} \left[ x+ \frac{k}{n} \right] &= \sum_{k=0}^{n-j-1} \left[ x + \frac{k}{n} \right] + \sum_{k=n-j}^{n-1} \left[ x + \frac{k}{n} \right] \\ &= (n-j)[x] + j([x] + 1) \\ &= n[x] + j \\ &= nm+j \\ &= [nx]. \qquad \blacksquare \end{align*}$

Prove some properties of the greatest integer function

by RoRi

July 16, 2015

For any $x \in \mathbb{R}$ we denote the greatest integer less than or equal to $x$ by $[x]$ . Prove the following properties of the function $[x]$ :

$[x+n] = [x] + n$ for any integer $n$ .
$[-x] = \displaystyle{\begin{cases} -[x] & \text{if } x \text{ is an integer,} \\ -[x]-1 & \text{otherwise.} \end{cases}}$
$[x+y] = [x]+[y]$ or $[x]+[y]+1$ .
$[2x] = [x] + \left[ x + \frac{1}{2} \right]$ .
$[3x] = [x] + \left[ x + \frac{1}{3} \right] + \left[ x + \frac{2}{3} \right]$ .

Proof. Let $[x+n] = m$ for some integer $m$ . Then,
$\begin{align*} m \leq x+n < m+1 &&\implies && m-n &\leq x < m-n+1 \\ &&\implies && [x] &= m-n \\ &&\implies && [x] + n &= m. \end{align*}$

But, we defined $m=[x+n]$ . Thus, $[x+n] = [x] + n$ for any $n \in \mathbb{Z}. \qquad \blacksquare$
Proof. If $x \in \mathbb{Z}$ , then $x = n$ for some $n \in \mathbb{Z}$ . Hence, $[x]=n$ and
$-x=-n \quad \implies \quad [-x]=-n \quad \implies \quad [-x]=-[x].$

On the other hand, if $x \notin \mathbb{Z}$ , then let $[x] = n$ . This gives us,

$\begin{align*} n \leq x < n+1 &&\implies && -n-1 &< -x < n &(n \neq x \text{ since } x \notin \mathbb{Z}) \\ &&\implies && [-x] &= -n-1 = -[x]-1. \qquad \blacksquare \end{align*}$
Proof. Let $[x] = m$ and $[y] = n$ , then we have
$m \leq x < m+1 \qquad \text{and} \qquad n \leq y < n+1.$

So, adding, we obtain,

$m+n \leq x+y < m+n+2.$

Thus,

$[x+y] = m+n = [x] + [y] \qquad \text{or} \qquad [x+y] = m+n+1 = [x] + [y] + 1. \qquad \blacksquare$
Proof. By part (c) we have,
$[2x] = [x+x] = [x] + [x] \qquad \text{or} \qquad [x] + [x] + 1.$

If $[2x] = [x] + [x]$ , then let $[x] = n$ ,

$\begin{align*} && [2x] &= 2n \\ \implies && 2n &\leq 2x < 2n+1 \\ \implies && n &\leq x < n + \frac{1}{2} \\ \implies && n &\leq x + \frac{1}{2} < n+1 \\ \implies && \left[ x + \frac{1}{2} \right] &=n . \end{align*}$

Thus, $[2x] = 2n = n+n = [x] + \left[ x + \frac{1}{2} \right]$ .
On the other hand, if $[2x] = [x] + [x] + 1$ , then let $[x] = n$ , and

$\begin{align*} && [2x] &= 2n+1 \\ \implies && 2n+1 &\leq 2x < 2n+2 \\ \implies && n+\frac{1}{2} &\leq x < n+1 \\ \implies && n +1 &\leq x + \frac{1}{2} < n+2 \\ \implies && \left[ x + \frac{1}{2} \right] &= n+1. \end{align*}$

Thus, $[2x] = n+n+1 = [x] + \left[ x + \frac{1}{2} \right]. \qquad \blacksquare$
Proof. By part (c) we have
$[3x] = [x+x+x] = [x+x] + [x] \qquad \text{or} \qquad [x+x]+[x] + 1.$

And,

$[x+x] = [x] + [x] \qquad \text{or} \qquad [x]+[x]+1.$

So, putting these together we have,

$[3x] = [x]+[x]+[x], \qquad \text{or} \qquad [x]+[x]+[x]+1, \qquad \text{or} \qquad [x]+[x]+[x]+2.$

If $[3x] = [x] + [x] + [x]$ , then, let $[x] = n$ , so

$\begin{align*} && 3n &\leq 3x < 3n+1 \\ \implies && n &\leq x < n + \frac{1}{3} \\ \implies && n &\leq x + \frac{1}{3} < n+1 & \text{and}&& n &\leq x+\frac{2}{3} < n+1 \\ \implies && [x] &= \left[ x+ \frac{1}{3} \right] = \left[ x+ \frac{2}{3} \right] = n. \end{align*}$

Thus, $[3x] = 3n = [x] + \left[ x + \frac{1}{3} \right] + \left[ x + \frac{2}{3} \right]$ .
Next, if $[3x] = [x] + [x] + [x] + 1$ , then let $[x] = n$ , giving us,

$\begin{align*} && 3n+1 &\leq 3x < 3n+1 \\ \implies && n + \frac{1}{3} &\leq x < n + \frac{2}{3} \\ \implies && n + \frac{1}{3} & \leq x + \frac{1}{3} < n+1 &\implies \quad \left[ x + \frac{1}{3} \right] &=n \\ \text{and}, \ \implies && n + 1 &\leq x + \frac{2}{3} < n+2 &\implies \quad \left[ x + \frac{2}{3} \right] &= n+1. \end{align*}$

Thus,

$[3x] = 3n+1 \quad \implies \quad [3x] = [x] + \left[ x + \frac{1}{3} \right] + \left[ x + \frac{2}{3} \right].$

Finally, if $[3x] = [x] + [x] + [x] + 2$ , then let $[x] = n$ , and we have

$\begin{align*} && 3n+2 &\leq 3x < 3n+3 \\ \implies && n+\frac{2}{3} &\leq x < n+1 \\ \implies && n+1 &\leq x+\frac{1}{3} < n+2 & \text{and} && n+1 \leq x+\frac{2}{3} < n+2. \end{align*}$

So,

$[3x] = 3n+2 = [x] + \left[ x + \frac{1}{3} \right] + \left[ x + \frac{2}{3} \right]. \qquad \blacksquare$

Draw the graphs of some functions

by RoRi

July 16, 2015

Draw the graphs of the functions defined below on the interval $[-2,2]$ , and if it is a step function find a partition $P$ such that the function is constant on the open subintervals of $P$ .

$f(x) = x + [x]$ .
$f(x) = x - [x]$ .
$f(x) = [-x]$ .
$f(x) = 2[x]$ .
$f(x) = \left[x+\frac{1}{2} \right]$ .
$f(x) = [x] + \left[ x+ \frac{1}{2} \right]$ .

This is not a step function. The graph is below.
This is not a step function. The graph is below.
This is a step function and it is constant on the open subintervals of the partition, $P = \{ -2,-1,0,1,2 \}$ . The graph is below.
This is a step function and it is constant on the open subintervals of the partition, $P = \{ -2,-1,0,1,2 \}$ . The graph is below.
This is a step function and it is constant on the open subintervals of the partition, $P = \left\{ -2,-\frac{3}{2},-\frac{1}{2},\frac{1}{2},\frac{3}{2},2 \right\}$ . The graph is below.
This is a step function and it is constant on the open subintervals of the partition, $P = \left\{ -2,-\frac{3}{2},-1,-\frac{1}{2},0,\frac{1}{2},1,\frac{3}{2},2 \right\}$ . The graph is below.

Draw the graphs of the functions and points of intersection for x^2 -2 and 2x^2 + 4x + 1

by RoRi

July 13, 2015

Let $f(x) = x^2 - 2$ and $g(x) = 2x^2 + 4x + 1$ . Find the points of intersection and draw the graph.

The points of intersection are at,

$x^2 - 2 = 2x^2 + 4x + 1 \ \implies \ x^2 + 4x + 3 = 0 \ \implies \ (x+3)(x+1) = 0.$

Thus, they intersect at $x=-3$ and $x=-1$ . These correspond to the points $(-3, 7)$ and $(-1,-1)$ . The graph is below.

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Draw the graphs and find the points of intersection of x and x^3

by RoRi

July 13, 2015

Let $f(x) = x$ and $g(x) = x^3$ . Draw the graphs of $f$ and $g$ and show the points at which they intersect.

First, we can determine algebraically the points of intersection. They are the points such that $f(x) = g(x)$ , i.e.,

$x = x^3 \ \implies \ x = 0 \quad \text{or} \quad x^2 = 1 \ \implies \ x = 0,\pm 1.$

The function values at these points are,

$f(0) = g(0) = 0, \quad f(-1) = g(-1) = -1, \quad f(1) = g(1) = 1.$

Thus, the points of intersection are $(0,0), (-1,-1), (1,1)$ . The graph is below.

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Verify some formulas for the function g(x) = (4-x^2)^(1/2)

by RoRi

July 13, 2015

Let $g(x) \sqrt{4-x^2}$ for $|x| \leq 2$ .

$g(-x) = g(x)$ .
$g(2y) = 2 \sqrt{1-y^2}$ .
$g(1/t) = \frac{\sqrt{4t^2-1}}{|t|}$ .
$g(a-2) = \sqrt{4a-a^2}$ .
$g(s/2) = \frac{1}{2} \sqrt{16-s^2}$ .
$\frac{1}{2+g(x)} = \frac{2-g(x)}{x^2}$ .

Proof.
$g(-x) = \sqrt{4-(-x^2)} = \sqrt{4-x^2} = g(x)$ . Valid for $|x| \leq 2. \qquad \blacksquare$
Proof.
$g(2y) = \sqrt{4-(2y)^2} = \sqrt{4-4y^2} = 2 \sqrt{1-y^2}$ . Valid for $|y| \leq 1. \qquad \blacksquare$ .
Proof.
$\displaystyle{ g(1/t) = \sqrt{4 - \left( \frac{1}{t} \right)^2} = \sqrt{\frac{1}{t^2}(4t^2 - 1)} = \frac{\sqrt{4t^2-1}}{\sqrt{t^2}} = \frac{\sqrt{4t^2-1}}{|t|}}$ . Valid for $|t| \geq \frac{1}{2}. \qquad \blacksquare$
Proof.
$g(a-2) = \sqrt{4-(a-2)^2} = \sqrt{4-a^2 + 4a - 4} = \sqrt{4a - a^2}$ . Valid for $0 \leq a \leq 4. \qquad \blacksquare$
Proof.
$\displaystyle{g \left( \frac{s}{2} \right) = \sqrt{4-\frac{s^2}{4}} = \frac{1}{2} \sqrt{16-s^2}}$ . Valid for $|s| \leq 4. \qquad \blacksquare$
Proof.
$\displaystyle{\frac{1}{2+g(x)} = \frac{1}{2+\sqrt{4-x^2}} = \frac{2-\sqrt{4-x^2}}{4-4+x^2} = \frac{2-g(x)}{x^2}}$ . Valid for $0 < |x| \leq 2. \qquad \blacksquare$