Tag: e

Prove that a function satisfying a given equation must be the exponential function

by RoRi

November 16, 2015

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

$f'(x) = cf(x) \qquad \text{for all } x$

prove that we must have

$f(x) = Ke^{cx}$

for some constant $K$ .

Proof. Let $g(x) = f(x)e^{-cx}$ . Then we can compute the derivative,

$g'(x) = -cf(x)e^{-cx} + f'(x) e^{-cx}.$

Since we know $f'(X) = cf(x)$ we then have

$g'(x) = -cf(x)e^{-cx} + cf(x)e^{-cx} = 0.$

Therefore, $g(x) = K$ for some constant $K$ and we have

$K = f(x)e^{-cx} \quad \implies \quad f(x) = Ke^{cx}. \qquad \blacksquare$

Deduce a limit relation using the derivative of e^cx

by RoRi

November 15, 2015

Consider the function $f(x) = e^{cx}$ for a constant $c$ . First, show that $f'(0) = c$ and then prove

$\lim_{x \to 0} \frac{e^{cx}-1}{x} = c.$

Proof. If $f(x) = e^{cx}$ then we have

$f'(x) = ce^{cx} \quad \implies \quad f'(0) = ce^0 = c.$

From the definition of the derivative we also know

$\begin{align*} &&f'(0) &= \lim_{h \to 0} \frac{f(0+h) - f(0)}{h} \\ \implies && c&= \lim_{h \to 0} \frac{e^{ch} - 1}{h} \\ \implies && c &= \lim_{x \to 0} \frac{e^{cx}- 1}{x}. \qquad \blacksquare \end{align*}$

Show that a given function satisfies a functional equation

by RoRi

November 15, 2015

Define a function for $a > 0$ by

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

Prove that

$f(x+y) + f(x-y) = 2 f(x) f(y).$

Proof. First, we compute $f(x+y)$ and $f(x-y)$ using the definition of $f$ above, and the properties of the exponential $a^x$ established in the previous exercise (Section 6.17, Exercise #36), we have

$f(x+y) = \frac{a^{2(x+y)+1}}{2a^xa^y}, \qquad f(x-y) = \frac{a^{2x} + a^{2y}}{2a^x a^y}.$

Therefore,

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

Prove basic properties of the exponential function

by RoRi

November 15, 2015

A real number $a$ raised to a real exponent $x$ is defined by

$a^x = e^{x \log a}.$

Prove the following properties:

$\log a^x = x \log a$ .
$(ab)^x = a^x b^x$ .
$a^x a^y = a^{x+y}$ .
$(a^x)^y = (a^y)^x = a^{xy}$ .
For $a \neq 1$ , $y = a^x$ if and only if $x = \log_a y$ .

$a^x = e^{x \log a}$

$e^x$

$E(x)$

$e^x = E(x)$

Proof.
$\log a^x = \log (e^{x \log a}) = x \log a \log e = x \log a. \qquad \blacksquare$
Proof.
$(ab)^x = e^{x \log (ab)} = e^{x \log a + x \log b} = e^{x \log a}e^{x \log b} = a^x b^x. \qquad \blacksquare$
Proof.
$a^x a^y = (e ^{x \log a})(e^{y \log a}) = e^{x \log a + y \log a} = e^{(x+y)\log a} = a^{x+y}. \qquad \blacksquare$
Proof.
$\begin{align*} (a^x)^y &= e^{y \log a^x} = e^{xy \log a} = a^{xy} \\ (a^y)^x &= e^{x \log a^y} = e^{xy \log a} = a^{xy}. \qquad \blacksquare \end{align*}$
Proof. Assume $a \neq 1$ . For the forward direction, assume $y = a^x$ . We have
$y = a^x = e^{x \log a} \quad \implies \quad \log y = \log \left(e^{ x \log a} \right) = x \log a.$

Therefore, since $a \neq 1$ we have $\log a \neq 0$ , and so

$x = \frac{\log y}{\log a} \quad \implies \quad x = \log_a y.$

Conversely, if $x = \log_a y$ then we have

$x \log a = log y \quad \implies \quad e^{x \log a} = e^{\log y} \quad \implies \quad a^x = y. \qquad \blacksquare$

Find the derivative of (sin x)^{cos x} + (cos x)^{sin x}

by RoRi

November 13, 2015

Find the derivative of the function

$f(x) = (\sin x)^{\cos x} + (\cos x)^{\sin x}.$

Rewrite each of the exponentials in the expression for $f(x)$ using the definition of the exponential function,

$f(x) = e^{\cos x \log (\sin x)} + e^{\sin x \log (\cos x)}.$

Now, we take the derivative directly using the chain rule and the formula for the derivative of the exponential function

$\begin{align*} f'(x) &= \left( e^{\cos x \log (\sin x)} \right) \left( \cos x \log (\sin x) \right)' \\ & \qquad + \left(e^{\sin x \log (\cos x)} \right) \left( \sin x \log (\cos x) \right)' \\[9pt] &= (\sin x)^{\cos x} \left( - \sin x \log (\sin x) + \frac{(\cos x)^2}{\sin x} \right) \\ & \qquad + (\cos x)^{\sin x} \left( \cos x \log (\cos x) - \frac{(\sin x)^2}{\cos x} \right) \\[9pt] &= (\sin x)^{1 + \cos x} \left( \cot^2 x - \log (\sin x)\right) - (\cos x)^{1+\sin x} \left( \tan^2 x - \log (\cos x) \right). \end{align*}$

Find the derivative of log (e^x + (1+e^2x)^1/2)

by RoRi

November 13, 2015

Find the derivative of the following function:

$f(x) = \log \left( e^x + \sqrt{1+e^{2x}} \right).$

We compute using the chain rule and the formulas for derivatives of logarithms and exponentials,

$\begin{align*} f'(x) &= \left( \frac{1}{e^x + \sqrt{1+e^{2x}}} \right) \left( e^x + \frac{e^{2x}}{\sqrt{1+e^{2x}}} \right) \\[9pt] &= \frac{ e^x \sqrt{1+e^{2x}} + e^{2x}}{\left( e^x + \sqrt{1+e^{2x}} \right) \left( \sqrt{1+e^{2x}} \right)} \\[9pt] &= \frac{e^x (e^x + \sqrt{1+e^{2x}} )}{\left( e^x + \sqrt{1+e^{2x}} \right) \left( \sqrt{1+e^{2x}} \right)} \\[9pt] &= \frac{e^x}{\sqrt{1+e^{2x}}} \\[9pt] &= e^x \left( 1+ e^{2x} \right)^{-\frac{1}{2}}. \end{align*}$

Find the derivative of (e^x – e^-x)/(e^x + e^-x)

by RoRi

November 13, 2015

Find the derivative of the following function:

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

We can compute this derivative directly using the quotient rule,

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

Find the derivative of (1+x)(1+e^x²)

by RoRi

November 13, 2015

Find the derivative of the following function:

$f(x) = (1+x)\left( 1+ e^{x^2} \right).$

We can compute this derivative by multiplying out the expression for $f(x)$ and using the product and chain rules. First, we multiply out the expression,

$\begin{align*} f(x) &= (1+x) \left(1+e^{x^2} \right) \\ & = 1 + x + e^{x^2} + xe^{x^2}. \end{align*}$

Then, we take the derivative,

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

Prove integration formulas for e^axcos (bx) and e^axsin (bx)

by RoRi

November 13, 2015

Let $a$ and $b$ be constants with at least one of them nonzero and define

$A = \int e^{ax} \cos (bx) \, dx, \qquad B = \int e^{ax} \sin (bx) \, dx.$

Using integration by parts, establish the following formulas for constants $C_1, C_2$ ,

$aA - bB = e^{ax} \cos (bx) + C_1, \qquad aB + bA = e^{ax} \sin (bx) + C_2.$

Using these formulas prove the following integration formulas,

$\begin{align*} \int e^{ax} \cos (bx) \, dx &= \frac{e^{ax} (a \cos (bx) + b \sin (bx))}{a^2 + b^2} + C, \\ \int e^{ax} \sin (bx) \, dx &= \frac{e^{ax} (a \sin (bx) - b \cos (bx))}{a^2 + b^2} + C. \end{align*}$

To establish the formula $aA - bB = e^{ax} \cos (bx) + C_1$ we use integration by parts letting

$\begin{align*} u &= \cos (bx), & du &= -b \sin (bx) \, dx \\ dv &= e^{ax} \, dx & v &= \frac{1}{a} e^{ax}. \end{align*}$

Then we can evaluate $A$ using the formula for integration by parts,

$\begin{align*} &&A &= \int e^{ax} \cos (bx) \, dx \\[9pt] \implies && A&= \frac{1}{a}e^{ax} \cos (bx) + \frac{b}{a} \int e^{ax} \sin (bx) \, dx + C_1 \\[9pt] \implies && aA &= e^{ax} \cos (bx) + bB + C_1 \\[9pt] \implies && aA - bB &= e^{ax} \cos (bx) + C_1. \end{align*}$

To establish the second formula, $aB + bA = e^{ax} \sin (bx) + C_2$ , we use integration by parts again. Let

$\begin{align*} u &= \sin (bx) & du &= b \cos (bx) \, dx \\ dv &= e^{ax} \, dx & v &= \frac{1}{a} e^{ax}. \end{align*}$

Then we have

$\begin{align*} && B&= \int e^{ax} \sin (bx) \, dx \\[9pt] \implies && B &= \frac{1}{a} e^{ax} \sin (bx) - \frac{b}{a} \int e^{ax} \cos (bx) \, dx + C_2 \\[9pt] \implies && aB &= e^{ax} \sin (bx) - bA + C_2 \\[9pt] \implies && aB + bA &= e^{ax} \sin (bx) + C_2. \end{align*}$

This establishes the two requested equations, now we prove the two integral identities.

Proof. Solving for $B$ in the second equation above we have

$aB + bA = e^{ax} \sin (bx) + C_2 \quad \implies \quad B = \frac{1}{a} \left( e^{ax} \sin (bx) - bA) + C_2.$

Plugging this into the first equation we have

$\begin{align*} && aA - bB &= e^{ax} \cos (bx) + C_1 \\[9pt] \implies && aA - \frac{b}{a} \left( e^{ax} \sin (bx) - bA + C_2 \right) &= e^{ax} \cos (bx) + C_1 \\[9pt] \implies && aA - \frac{b}{a} e^{ax} \sin (bx) + \frac{b^2}{a} A &= e^{ax} \cos (bx) + C \\[9pt] \implies && A \left( a + \frac{b^2}{a}\right) &= e^{ax} \cos (bx) + \frac{b}{a} e^{ax} \sin (bx) + C \\[9pt] \implies && A (a^2 + b^2) &= a e^{ax} \cos (bx) + b e^{ax} \sin (bx) + C \\[9pt] \implies && A &= \frac{a e^{ax} \cos (bx) + b e^{ax} \sin (bx)}{a^2+b^2} + C \\[9pt] \implies && \int e^{ax} \cos (bx) \, dx &= \frac{a e^{ax} \cos (bx) + b e^{ax} \sin (bx)}{a^2+b^2} + C. \end{align*}$

Next, for the second integral equation we are asked to prove, we use the formula we obtained for $B$ above,

$B = \frac{1}{a} \left( e^{ax} \sin (bx) - bA \right) + C.$

Then, we use the expression we obtained for $A$ into this,

$B = \frac{1}{a} \left( e^{ax} \sin (bx) - b \left( \frac{a e^{ax} \cos (bx) + b e^{ax} \sin (bx)}{a^2+b^2} \right) \right) + C.$

This implies,

$\begin{align*} && B &= \frac{1}{a} e^{ax} \sin (bx) - \frac{be^{ax} \cos (bx) + \frac{b^2}{a} e^{ax} \sin (bx)}{a^2+b^2} + C \\[9pt] \implies && B &= \frac{(a^2+b^2)e^{ax} \sin (bx) - (ab)e^{ax} \cos (bx) - b^2 e^{ax} \sin (bx)}{a(a^2+b^2)} + C \\[9pt] \implies && B &= \frac{a^2 e^{ax} \sin (bx) - (ab) e^{ax} \cos (bx)}{a(a^2+b^2)} + C \\[9pt] \implies && B &= \frac{a e^{ax} \sin (bx) - b e^{ax} \cos (bx)}{a^2+b^2} + C \\[9pt] \implies && \int e^{ax} \sin (bx) \, dx &= \frac{a e^{ax} \sin (bx) - b e^{ax} \cos (bx)}{a^2+b^2} + C. \qquad \blacksquare \end{align*}$