Tag: e

Find an inverse for the function log |x|

by RoRi

December 20, 2015

Consider the function $f(x) = \log |x|$ for $x < 0$ . Prove that this function has an inverse, determine the domain of this inverse, and find a formula to compute the inverse $g(y)$ .

Proof. From the discussion on page 146 of Apostol we know that a function which is continuous and strictly monotonic on an interval $[a,b]$ has an inverse on $[a,b]$ . The function $f(x) = \log |x|$ is continuous and strictly monotonic on the negative real axis; therefore, it has an inverse. We know it is continuous since the log function is continuous on the positive real axis, and $|x| > 0$ for all $x$ , in particular, for all $x < 0$ . Furthermore, we know it is strictly monotonic since

$f'(x) = \frac{1}{x} < 0 \qquad \text{for all } x < 0.$

Therefore, $f(x) = \log |x|$ has an inverse for all $x < 0$ . The domain of this inverse is the range of $\log |x|$ which is all of $\mathbb{R}. \qquad \blacksquare$

To find a formula for the inverse we set

$y = \log |x| \quad \implies \quad e^y = |x| \quad \implies \quad x = -e^y.$

Therefore, $g(y) = -e^y$ valid for all $y \in \mathbb{R}$ .

A sketch for the graph of $g$ is given by

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Prove that 2 cosh² (x/2) = cosh x + 1

by RoRi

November 19, 2015

Prove the following identity,

$2 \cosh^2 \frac{1}{2}x = \cosh x + 1.$

Proof. Computing directly from the definition of the hyperbolic cosine,

$\begin{align*} 2 \cosh^2 \frac{x}{2} &= 2 \left( \frac{e^{\frac{x}{2}} + e^{-\frac{x}{2}}}{2} \right)^2 \\ &= 2 \left( \frac{e^x + 2 + e^{-x}}{4} \right) \\ &= \frac{e^x + e^{-x}}{2} + 1 \\ &= \cosh x + 1. \qquad \blacksquare \end{align*}$

Prove that 2 sinh² (x/2) = cosh x – 1

by RoRi

November 19, 2015

Prove the following identity:

$2 \sinh^2 \frac{1}{2}x = \cosh x - 1.$

Proof. We compute directly from the definition of $\sinh x$ ,

$\begin{align*} 2 \sinh^2 \frac{x}{2} &= 2 \left( \frac{e^{\frac{x}{2}} - e^{-\frac{x}{2}}}{2} \right)^2 \\ &= 2 \left( \frac{e^x - 2 + e^{-x}}{4} \right) \\ &= \frac{e^x + e^{-x}}{2} - 1 \\ &= \cosh x - 1. \qquad \blacksquare \end{align*}$

Prove that (cosh x + sinh x)ⁿ = cosh (nx) + sinh (nx)

by RoRi

November 18, 2015

Prove the following identity:

$(\cosh x + \sinh x)^n = \cosh (nx) + \sinh (nx).$

Proof. We know from a previous exercise (Section 6.19, Exercise #9) that

$\cosh x + \sinh x = e^x \quad \implies \quad \cosh (nx) + \sinh(nx) = e^{nx}.$

Therefore, we have

$(\cosh x + \sinh x)^n = (e^x)^n = e^{nx} = \cosh (nx) + \sinh(nx). \qquad \blacksquare$

Prove that cosh x – sinh x = e^-x

by RoRi

November 18, 2015

Prove that

$\cosh x - \sin x = e^{-x}.$

Proof. Using the definition of the hyperbolic functions in terms of the exponential, we have

$\begin{align*} \cosh x - \sinh x &= \frac{e^x+e^{-x}}{2} - \frac{e^x - e^{-x}}{2} \\ &= \frac{e^x + e^{-x} - e^x + e^{-x}}{2} \\ &= \frac{2e^{-x}}{2} \\ &= e^{-x}. \qquad \blacksquare \end{align*}$

Prove that cosh x + sinh x = e^x

by RoRi

November 18, 2015

Prove that

$\cosh x + \sinh x = e^x.$

Proof. We use the definitions of hyperoblic sine and cosine to compute,

$\begin{align*} \cosh x + \sinh x &= \frac{e^x+e^{-x}}{2} + \frac{e^x - e^{-x}}{2} \\ &= \frac{e^x + e^{-x} + e^x -e^{-x}}{2} \\ &= \frac{2e^x}{2} \\ & = e^x. \qquad \blacksquare \end{align*}$

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)) $. \qquad \blacksquare$

Prove properties of a function satisfying a given functional equation

by RoRi

November 17, 2015

Assume $f(x)$ is a function defined for all $x \in \mathbb{R}$ such that

$f(x+y) = f(x)f(y).$

For parts (b)-(d) assume also that $f(x)$ is differentiable for all real $x$ (i.e., assume $f'(x)$ exists everywhere).

Using the given equation above, prove $f(0) = 0$ or $f(0) = 1$ . Prove that if $f(0) \neq 0$ then $f(x) \neq 0$ for all $x$ .
Prove that $f'(x)f(y) = f'(y)f(x)$ for all $x,y$ .
Prove that $f'(x) = cf(x)$ for some constant $c$ .
Prove that if $f(0) \neq 0$ then $f(x) = e^{cx}$ .

Proof. Assume $f(0) \neq 0$ . Then if $a \in \mathbb{R}$ by the functional equation we have
$f(a) f(0) = f(a+0) = f(a) \quad \implies \quad f(0) = 1.$

Therefore, if $f(0) \neq 0$ then we must have $f(0) = 1$ . Hence, we have either $f(0) = 0$ or $f(0) = 1$ .

Furthermore, if $f(0) \neq 0$ , then for any $x \in \mathbb{R}$ by the functional equation we have

$f(x) f(-x) = f(x+(-x)) = f(0) \neq 0.$

Hence, $f(x) \neq 0$ for any $x \in \mathbb{R}. \qquad \blacksquare$
Proof. Since the derivative of $f$ exists everywhere by hypothesis, we use the limit definition of the derivative and the functional equation $f(x+y)=f(x)f(y)$ to write,
$\begin{align*} f'(x)f(y) &= \left( \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \right) f(y) \\ &= \lim_{h \to 0} \frac{f(y)f(x)f(h) - f(x)f(y)}{h} \\ &= \lim_{h \to 0} \frac{f(y+h)f(x) - f(y)f(x)}{h} \\ &= \left( \lim_{h \to 0} \frac{f(y+h) - f(y)}{h} \right) f(x) \\ &= f'(y) f(x). \qquad \blacksquare \end{align*}$
Proof. First, if $f(0) = 0$ then by the functional equation we know for any $x \in \mathbb{R}$ we have
$f(x) = f(x+0) = f(x)f(0) = 0.$

So, if $f(0) = 0$ then $f(x)$ is the constant function which is 0 everywhere. Hence, $f'(x) = 0 = cf(x)$ for all $x$ .
Next, we consider the case that $f(0) = 1$ . Since $f'(x)$ exists for all $x$ we use the definition of derivative and the functional equation again to write,

$\begin{align*} f'(x) &= \lim_{h \to 0} \frac{f(x+h) - f(x)}{h} \\ &= \lim_{h \to 0} \frac{f(x)f(h) - f(x)}{h} \\ &= \lim_{h \to 0} \frac{f(x)(f(h) - 1)}{h} \\ &= f(x) \lim_{h \to 0} \frac{f(h) - 1}{h} \\ &= f(x) \lim_{h \to 0} \frac{f(0+h) - f(0)}{h} \\ &= f(x) f'(0) \\ &= cf(x). \end{align*}$

The final line follows since $f'(0)$ exists by assumption so we can write $c = f'(0)$ for some $c. \qquad \blacksquare$
Proof. Since $f'(x) = cf(x)$ (by part (c)) we know, by the previous exercise (Section 6.17, Exercise #39) that $f(x) = Ke^{cx}$ for some constant $K$ . Since $f(0) \neq 0$ by assumption, we know by part (a) that $f(0) = 1$ . Hence,
$f(0) = Ke^{c\cdot 0} = K = 1 \quad \implies \quad f(x) = e^{cx}. \qquad \blacksquare$