Tag: Inequalities

Prove some inequalities of sin x

by RoRi

December 20, 2015

For all $x > 0$ prove that

$x - \frac{x^3}{6} < \sin x < x.$

From the first exercise of this section on inequalities, we know $\sin x < x$ for all $0 < x < \frac{\pi}{2}$ . But since $\sin x \leq 1$ for all $x$ and $1 < \frac{\pi}{2}$ we have the inequality on the right immediately,

$\sin x < x \qquad \text{for all } x > 0.$

For the inequality on the left let

$f(x) = \sin x - x + \frac{x^3}{6}.$

Then, we’ll consider the first two derivatives of $f$ to show that it is positive for all $x > 0$ .

$\begin{align*} f'(x) &= \cos x - 1 + \frac{x^2}{2} \\ f''(x) &= -\sin x + x = x - \sin x. \end{align*}$

We know from the inequality on the right that $\sin x < x$ for all $x > 0$ . Hence, $f''(x) > 0$ for all $x > 0$ . Therefore, $f'(x)$ is increasing on the positive real axis. Since

$f'(0) = \cos 0 - 1 + \frac{0^2}{2} = 0$

we then have that $f'(x) > 0$ for all $x> 0$ . Hence, $f(x)$ is increasing on the positive real axis, and since

$f(0) = \sin 0 - 0 + \frac{0^3}{6} = 0$

we have that $f(x) > 0$ for all $x >0$ . Thus, for all $x>0$ ,

$\sin x - x + \frac{x^3}{6} > 0 \quad \implies \quad x - \frac{x^3}{6} < \sin x. \qquad \blacksquare$

Prove inequalities of the log of 1 + 1/x

by RoRi

December 19, 2015

For $x > 0$ prove that

$\frac{1}{x + \frac{1}{2}} < \log \left( 1 + \frac{1}{x} \right) < \frac{1}{x}.$

Before the proof, a picture for this one is probably useful. In the graph below we are going to get the inequality on the right by comparing the area under the graph of $\frac{1}{x}$ from $x$ to $x+1$ (blue curve) and the area of the rectangle under the blue dashed line from $x$ to $x+1$ . We’ll show that the area of the rectangle is $1/x$ and the area under the curve is $\log \left( 1 + \frac{1}{x} \right)$ .

Rendered by QuickLaTeX.com

Proof. First, we note that

$\log \left( 1 + \frac{1}{x} \right) = \log \left( \frac{x+1}{x} \right) = \log (x+1) - \log x.$

Furthermore,

$\int_x^{x+1} \frac{1}{t} \, dt = \log (x+1) - \log x.$

Therefore we have

$\log \left( 1 + \frac{1}{x} \right) = \int_x^{x+1} \frac{1}{t} \, dt.$

(In the picture this is the area under the curve between $x$ and $x+1$ .)
Since the function $\frac{1}{t}$ is strictly decreasing on the positive real axis (since it’s derivative is $-\frac{1}{t^2}$ is negative everywhere) we know that for any fixed $x> 0$ we have $\frac{1}{x} > \frac{1}{t}$ for every $t \in (x, x+1)$ . Therefore, by the monotone property of the integral,

$\int_x^{x+1} \frac{1}{t} \, dt < \int_x^{x+1} \frac{1}{x} \, dt = \frac{1}{x} \int_x^{x+1} dt = \frac{1}{x}.$

(Note the $\frac{1}{x}$ inside the integral is a constant here since we have chosen some fixed $x>0$ . For any fixed $x$ the point $\frac{1}{x}$ is the point on the curve $f(x)$ at the left end of the interval. So, this is saying the integral of the curve $\frac{1}{t}$ from $x$ to $x+1$ is less than the integral of the rectangle of height $\frac{1}{x}$ from $x$ to $x+1$ .) This gives us the inequality on the right that we wanted,

$\log \left( 1 + \frac{1}{x} \right) < \frac{1}{x}.$

Now, to prove the inequality on the left we know that $x > 0$ if and only if $\frac{1}{x} > 0$ . Thus, the given inequality holds for all $x>0$ if and only if it holds for all $\frac{1}{x} > 0$ . Therefore,

$\frac{1}{x+\frac{1}{2}} < \log \left( 1+ \frac{1}{x} \right)$

if and only if

$\frac{1}{\frac{1}{x} + \frac{1}{2}} < \log \left( 1 + \frac{1}{1/x} \right) \implies \frac{x}{1+\frac{1}{2} x} < \log (1+x).$

Now, consider

$\begin{align*} f(x) &= \log(1+x) - \frac{x}{1+\frac{1}{2}x} \\[9pt] f'(x) &= \frac{1}{1+x} - \frac{1+\frac{1}{2}x - \frac{1}{2}x}{(1+\frac{1}{2}x)^2} \\[9pt] &= \frac{1}{1+x} - \frac{1}{1+x+\frac{1}{4}x^2} \\[9pt] &> 0 \end{align*}$

for all $x > 0$ . Therefore, $f(x)$ is increasing on the positive real axis. Since

$f(0) = \log (1+0) - \frac{0}{1+\frac{1}{2}\cdot0} = 0,$

we have that $f(x) > 0$ for all $x > 0$ . Hence, we obtain the inequality on the left

$\frac{1}{x+\frac{1}{2}} < \log \left( 1+\frac{1}{x} \right). \qquad \blacksquare$

Prove the inequality 2x/π < sin x < x for 0 < x < π/2

by RoRi

December 19, 2015

Prove the inequality

$\frac{2}{\pi}{x} < \sin x < x \qquad \text{for} \quad 0 < x < \frac{\pi}{2}.$

Proof. Define a function $f(x) = x - \sin x$ . Then, since $\cos x < 1$ for $0 < x < \frac{\pi}{2}$ we have

$f'(x) = 1 - \cos x > 0 \qquad \text{for } 0 < x < \frac{\pi}{2}.$

Since $f(0) = 0$ and $f$ is increasing on $0 < x < \frac{\pi}{2}$ (since its derivative is positive) we have

$f(x) > 0 \quad \implies \quad x - \sin x > 0 \quad \implies \quad x > \sin x.$

For the inequality on the left (which is much more subtle), we want to show

$\frac{2}{\pi}x < \sin x \qquad \text{which implies} \qquad frac{2}{\pi} < \frac{\sin x}{x}.$

So, we consider the function

$f(x) = \frac{\sin x}{x} \quad \implies \quad f'(x) = \frac{x \cos x - \sin x}{x^2}.$

We know

$\lim_{x \to 0} f(x) = \lim_{x \to 0} \frac{\sin x}{x} = 1$

and

$f \left( \frac{\pi}{2} \right) = \frac{2}{\pi}.$

Now, if we can show that $f(x)$ is decreasing on the whole interval $\left( 0, \frac{\pi}{2} \right)$ then we will be done (since this would mean $f(x) > \frac{2}{\pi}$ on the whole interval since if it were less than $\frac{2}{\pi}$ somewhere then it would have to increase to get back to $\frac{2}{\pi}$ on the right end of the interval).

To show $f(x)$ is decreasing on the whole interval we will show that its derivative is negative. To that end, define a function

$g(x) = x \cos x - \sin x$

(This is the numerator in the expression we got for $f'(x)$ . Since we know the denominator of that expression is always positive, we are going to show this $g(x)$ is always negative to conclude $f'(x)$ is always negative.) Now, $g(0) = 0$ and

$g'(x) = \cos x - x \sin x - \cos x = -x \sin x < 0$

for $0 < x < \frac{\pi}{2}$ . Thus, $g'(x)$ is negative, and so $g(x)$ is decreasing. Since $g(0) = 0$ we then have $g(x)$ is negative on the whole interval. Therefore, $f'(x) \leq 0$ on the interval. Hence, $f(x)$ is decreasing. Hence, we indeed have

$\frac{\sin x}{x} > \frac{2}{\pi} \quad \implies \quad \sin x > \frac{2}{\pi}x$

for all $x \in \left( 0, \frac{\pi}{2} \right). \qquad \blacksquare$

Prove that x – (1/3)x³ < arctan x if x is positive

by RoRi

November 29, 2015

Consider the function

$f(x) = \arctan x - x + \frac{x^3}{3}.$

Compute the derivative $f'(x)$ and use the sign of the derivative to prove the inequality

$x - \frac{x^3}{3} < \arctan x \qquad \text{for all } x > 0.$

Proof. First, we compute the derivative

$\begin{align*} f'(x) &= \frac{1}{1+x^2} - 1 + x^2 \\ &= \frac{x^4}{1+x^2}. \end{align*}$

Therefore, $f'(x) > 0$ for all $x \neq 0$ . This implies $f(x)$ is strictly increasing if $x \neq 0$ , and in particular, $f(x)$ is increasing for $x > 0$ . This implies

$\begin{align*} f(x) > f(0) && \implies && \arctan x - x + \frac{x^3}{3} &> 0 \\ && \implies && x - \frac{x^3}{3} &< \arctan x. \qquad \blacksquare \end{align*}$

Prove formulas for the partial derivatives of x^y

by RoRi

November 17, 2015

Define a function of two variables, with $x> 0$ ,

$f(x,y) = x^y.$

Prove

$\frac{\partial f}{\partial x} = yx^{y-1} \qquad \text{and} \qquad \frac{\partial f}{\partial y} = x^y \log x.$

Proof. First, we write,

$f(x,y) = x^y = e^{ y \log x}.$

Then we compute the derivatives using the chain rule and formulas for the derivatives of the exponential and logarithm,

$\frac{\partial f}{\partial x} = e^{y \log x} \cdot \frac{y}{x} = yx^{y-1}.$

And,

$\frac{\partial f}{\partial y} = e^{y \log x} \cdot \log x = x^y \log x. \qquad \blacksquare$

Prove some inequalities of the exponential function

by RoRi

November 17, 2015

For $n \in \mathbb{Z}_{>0}$ and for $x \in \mathbb{R}_{>0}$ prove the following inequalities,

$\left( 1 + \frac{x}{n} \right)^n < e^x, \qquad e^x < \left( 1 - \frac{x}{n} \right)^{-n},$

where we assume $x < n$ in the second inequality. Use this to show

$2.5 < e < 2.99.$

Proof. From the previous exercise (Section 6.17, Exercise #41) we know

$e^x > 1 + x$

for $x > 0$ . But by assumption we have $x > 0$ and $n> 0$ ; hence, $\frac{x}{n} > 0$ . Therefore this inequality must hold for $\frac{x}{n}$ :

$e^{\frac{x}{n}} > 1 + \frac{x}{n} \quad \implies \quad e^x > \left( 1 + \frac{x}{n} \right)^n.$

Again, by the previous exercise we have

$e^{-x} > 1 - x$

for $x > 0$ . Since $x > 0$ and $n > 0$ this implies $\frac{x}{n} > 0$ . Therefore,

$\begin{align*} e^{-\frac{x}{n}} > 1 - \frac{x}{n} && \implies && e^{-x} > \left( 1 - \frac{x}{n} \right)^n \\ && \implies && e^x < \left( 1 - \frac{x}{n} \right)^{-n}. \qquad \blacksquare \end{align*}$

Letting $n = 6$ , we have

$\left( 1 + \frac{1}{6} \right)^6 = 2.52 < e < \left(1 - \frac{1}{6} \right)^{-6} = 2.99.$

Prove some inequalities using the function e^x-1-x

by RoRi

November 17, 2015

Define a function:
$f(x) = e^x - 1 - x \qquad \text{for all } x \in \mathbb{R}.$

Prove that $f'(x) \geq 0$ for all $x \geq 0$ and $f'(x) \leq 0$ for all $x \leq 0$ . Prove the following inequalities are valid for all $x > 0$ ,

$e^x > 1 + x, \qquad e^{-x} > 1 -x.$
Using integration and part (a) prove
$e^x > 1 + x + \frac{x^2}{2!}, \qquad e^{-x} < 1 - x + \frac{x^2}{2!}.$
Using integration and part (a) prove
$e^x > 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}, \qquad e^{-x} > 1 - x + \frac{x^2}{2!} - \frac{x^3}{3!}.$
State and prove a generalization of the inequalities above.

First, we take the derivative of $f(x)$ ,
$f'(x) = e^x - 1 \quad \implies \quad f'(0) = 0.$

Since $e^x - 1$ is strictly increasing everywhere (since $f''(x) = e^x > 0$ for all $x$ ) we have $f'(x) \geq 0$ for $x \geq 0$ and $f'(x) \leq 0$ for $x \leq 0$ . Thus, $f(x)$ has a minimum at $x = 0$ and $f(0) = 0$ so $f(x) > 0$ for all $x \neq 0$ . Therefore, $f(x) > 0$ when $x > 0$ ; hence,

$e^x -1 - x > 0 \quad \implies \quad e^x > 1 + x.$

Furthermore, when $x < 0$ we have

$e^x - 1 - x > 0 \quad \implies \quad e^x > 1 + x.$

Therefore, if $x > 0$ then $-x < 0$ and so we have

$e^{-x} > 1 - x.$
Using the inequalities in part (a) we integrate over the interval $0$ to $x$ ,
$\begin{align*} e^x > 1 + x && \implies && \int_0^x e^t \, dt > \int_0^x (1+t) \, dt \\ && \implies && e^x - 1> \left(t + \frac{t^2}{2}\right)\Bigr \rvert_0^x + C \\ && \implies && e^x - 1> x + \frac{x^2}{2!} \\ && \implies && e^x > 1 + x + \frac{x^2}{2!} \end{align*}$

For the other inequality we proceed similarly,

$\begin{align*} e^{-x} > 1 - x && \implies && \int_0^x e^{-t} \, dt &> \int_0^x (1-t) \, dt \\ && \implies && -e^{-x} + 1 &> x - \frac{x^2}{2}\\ && \implies && e^{-x} &< 1 - x + \frac{x^2}{2!}. \qquad \blacksquare \end{align*}$
We use the same strategy as before, starting with the inequalities we established in part (b),
$\begin{align*} e^x > 1 + x + \frac{x^2}{2!} && \implies && \int_0^x e^t \, dt > \int_0^x \left (1 + t + \frac{t^2}{2!} \right) \, dt \\ && \implies && e^x - 1 > \left( t + \frac{t^2}{2!} + \frac{t^3}{3!}\right)\Bigr \rvert_0^x \\ && \implies && e^x > 1 + x + \frac{x^2}{2!} + \frac{x^3}{3!}. \end{align*}$

Similarly, for the other inequality,

$\begin{align*} e^{-x} < 1 - x + \frac{x^2}{2!} && \implies && \int_0^x e^{-t} \, dt &< \int_0^x \left( 1 - t + \frac{t^2}{2!} \right) \, dt \\ && \implies && -e^{-x} + 1& < x - \frac{x^2}{2!} + \frac{x^3}{3!} \\ && \implies && e^{-x} &> 1 - x +\frac{x^2}{2!} - \frac{x^3}{3!}. \qquad \blacksquare \end{align*}$
Claim: The following general inequalities hold for all $x > 0$ :
$\begin{align*} e^x &> \sum_{k=0}^n \frac{x^k}{k!} & \text{for all } n \in \mathbb{Z}_{>0} \\ e^{-x} &> \sum_{k=0}^n (-1)^k \frac{x^k}{k!} & \text{for all odd } n \in \mathbb{Z}_{>0} \\ e^{-x} &< \sum_{k=0}^n (-1)^k \frac{x^k}{k!} & \text{for all even } n \in \mathbb{Z}_{>0}. \end{align*}$

Proof. The proof is by induction. We have already established the inequalities for the $n = 1$ case for all three inequalities. Now, to prove the first inequality assume

$e^x > \sum_{k=0}^n \frac{x^k}{k!} \qquad \text{for some } n \in \mathbb{Z}_{>0}.$

Then we have

$\begin{align*} && \int_0^x e^t \, dt &> \int_0^x \left( \sum_{k=0}^n \frac{t^k}{k!} \right) \, dt \\[9pt] \implies && e^x - 1 &> \sum_{k=0}^n \int_0^x \frac{t^k}{k!} \, dt &(\text{linearity of integral}) \\[9pt] \implies && e^x &> 1 + \sum_{k=0}^n \left( \frac{t^{k+1}}{(k+1)!} \right)\Bigr \rvert_0^x \\[9pt] \implies && e^x &> 1 + \sum_{k=0}^n \frac{x^{k+1}}{(k+1)!} \\[9pt] \implies && e^x &> 1 + \sum_{k=1}^{n+1} \frac{x^k}{k!} &(\text{Reindexing})\\ \implies && e^x &> \sum_{k=0}^{n+1} \frac{x^k}{k!}. \end{align*}$

This establishes the first inequality. For the second inequality we have proved the case $n = 1$ already. Assume

$e^{-x} > \sum_{k=0}^n (-1)^k \frac{x^k}{k!} \qquad \text{for some odd } n \in \mathbb{Z}_{>0}.$

Since $n$ is odd we may write $n = 2m+1$ for some nonnegative integer $m$ . We want to show that the inequality must hold for the next odd integer, $n+2$ . We have,

$\begin{align*} &&\int_0^x e^{-t} \, dt&> \int_0^x \left( \sum_{k=0}^n (-1)^k \frac{t^k}{k!} \right) \, dt \\ \implies && -e^{-x} + 1 &> \sum_{k=0}^n (-1)^k \int_0^x \frac{t^k}{k!} \, dt \\ \implies && -e^{-x} &> -1 + \sum_{k=0}^n (-1)^k \frac{x^{k+1}}{(k+1)!} \\ \implies && e^{-x} &< 1 - \sum_{k=0}^n (-1)^k \frac{x^{k+1}}{(k+1)!} \\ \implies && e^{-x} &< 1 - \sum_{k=1}^{n+1} (-1)^{k-1} \frac{x^k}{k!} \\ \implies && e^{-x} &< 1 + \sum_{k=1}^{n+1} (-1)^k \frac{x^k}{k!} \\ \implies && e^{-x} &< \sum_{k=0}^{n+1} (-1)^k \frac{x^k}{k!} \\ \end{align*}$

We want to show that the inequality holds for $n+2$ so we integrate both sides again,

$\begin{align*} && e^{-x} &< \sum_{k=0}^{n+1} (-1)^k \frac{x^k}{k!} \\ \implies && \int_0^x e^{-t} \, dt &< \int_0^x \left( \sum_{k=0}^{n+1} (-1)^k \frac{x^k}{k!} \right) \, dx \\ \implies && -e^{-x} + 1 &< \sum_{k=0}^{n+1} (-1)^k \int_0^x \frac{t^k}{k!} \, dt \\ \implies && -e^{-x} &< -1 + \sum_{k=0}^{n+1} (-1)^k \frac{x^{k+1}}{(k+1)!} + C \\ \implies && -e^{-x} &< -1 + \sum_{k=1}^{n+2} (-1)^{k-1} \frac{x^k}{k!} \\ \implies && e^{-x} &> 1 - \sum_{k=1}^{n+2} (-1)^{k-1} \frac{x^k}{k!} \\ \implies && e^{-x} &> 1 + \sum_{k=1}^{n+2} (-1)^k \frac{x^k}{k!} \\ \implies && e^{-x} &> \sum_{k=0}^{n+2} (-1)^k \frac{x^k}{k!}. \end{align*}$

Hence, if the inequality holds for odd $n$ , then it also holds for the next odd number, $n+2$ . Hence, it holds for all positive, odd integers.
The exact same induction argument works for all of the even integers (starting with the $n = 2$ case we proved in part (c)) $. \qquad \blacksquare$

Calculate an approximation of log 7 in terms of log 5

by RoRi

November 8, 2015

With reference to the previous three exercises (here, here, and here) use $x = \frac{1}{6}$ to calculate $\log 7$ in terms of $\log 5$ . Obtain the qpproximation

$1.945907 < \log 7 < 1.945911.$

Letting $x = \frac{1}{6}$ we have

$\log \frac{1+\frac{1}{6}}{1 - \frac{1}{6}} = \log \frac{7}{5} = \log 7 - \log 5.$

Therefore, we apply Theorem 6.5,

$\begin{align*} &&\log \frac{7}{5} &= 2 \left( \frac{1}{6} + \frac{(1/6)^3}{3} + \frac{(1/6)^5}{5} + \frac{(1/6)^7}{7} \right) + R_4 \left( \frac{1}{6} \right) \\ \implies && \log 7 &= \log 5 + 2 \left( \frac{1}{6} + \frac{(1/6)^3}{3} + \frac{(1/6)^5}{5} + \frac{(1/6)^7}{7} \right) + R_4 \left( \frac{1}{6} \right) \\ \implies && 1.945907 &< \log 7 < 1.945911. \end{align*}$

Calculate an approximation of log 5 in terms of log 2

by RoRi

November 8, 2015

With reference to the previous two exercises (here and here) obtain the following inequalities for $\log 5$ ,

$1.609435 < \log 5 < 1.609438.$

If we take $x = \frac{1}{9}$ then we have

$\log \left( \frac{1+x}{1-x} \right) = \log \left( \frac{1+\frac{1}{9}}{1-\frac{1}{9}} \right) = \log 5 - 2 \log 2.$

Therefore, we apply Theorem 6.5 to obtain

$\begin{align*} && &\log \frac{5}{4} = 2 \left( \frac{1}{9} + \frac{(1/9)^3}{3} + \frac{(1/9)^5}{5} + \frac{(1/9)^7}{7} \right) + R_4 \left( \frac{1}{9} \right) \\ \implies && &1.609435 < \log 5 < 1.609438. \end{align*}$

Calculate an approximation of log 3 in terms of log 2

by RoRi

November 8, 2015

Use Theorem 6.5 (pp. 240-241 of Apostol) and the previous exercise to compute $\log 3$ in terms of $\log 2$ . To do this, observe that if $x = \frac{1}{5}$ then

$\frac{1+x}{1-x} = \frac{3}{2}.$

Using this value of $x$ and $m = 5$ establish the inequalities:

$1.098611 < \log 3 < 1.098617.$

Taking $x = \frac{1}{5}$ and $m = 5$ in Theorem 6.5 we have

$\begin{align*} \log \left( \frac{3}{2} \right) &= 2 \left( \frac{1}{5} + \frac{(1/5)^3}{3} + \frac{(1/5)^5}{5} + \frac{(1/5)^7}{7} + \frac{(1/5)^9}{9} \right) + R_5 \left( \frac{1}{5} \right) \\ &= 0.405465 + R_5 \left( \frac{1}{5} \right). \end{align*}$

Then, since $\log \left(\frac{3}{2} \right) = \log 3 - \log 2$ , we combine with the previous exercise (Section 6.11, Exercise #1) to obtain,

$1.098611 < \log 3 < 1.098617.$