Tag: Logarithm

Some true/false questions

by RoRi

December 19, 2015

For each statement, prove that it is true or show that it is false.

$2^{\log 5} = 5^{\log 2}$ .
$\displaystyle{\log_2 5 = \frac{\log_3 5}{\log_2 3}}$ .
$\displaystyle{ \sum_{k=1}^n \frac{1}{\sqrt{k}} < 2 \sqrt{n}}$ for every $n \geq 1$ .
$1 + \sinh x \leq \cosh x$ for all $x$ .

True.
Proof. We can compute using the definition of the exponential

$2^{\log 5} = e^{\log \left( 2^{\log 5} \right)} = e^{\log 5 \cdot \log 2} = e^{\log 2 \cdot \log 5} = e^{\log \left(5^{\log 2} \right)} = 5^{\log 2}. \qquad \blacksquare$
False.
On the left we have

$\log_2 5 = \frac{\log 5}{\log 2}.$

While on the right we have,

$\frac{\log_3 5}{\log_2 3} = \frac{\frac{\log 5}{\log 3}}{\frac{\log 3}{\log 2}}} = \frac{\log 5 \log 2}{(\log 3)^2} = \left( \frac{\log 5}{\log 2} \right) \left( \frac{\log 2}{\log 3} \right)^2.$

But since $\frac{\log 2}{\log 3} \neq 1$ , these two quantities cannot be equal.
True.
Proof. The proof is by induction. For the case $n = 1$ on the left we have

$\sum_{k=1}^1 \frac{1}{\sqrt{k}} = 1.$

While on the right we have

$2 \sqrt{1} = 2.$

Therefore, indeed for the case $n = 1$ .

$\sum_{k=1}^n \frac{1}{\sqrt{k}} < 2 \sqrt{n}.$

Assume then that the statement is true for some positive integer $m$ . Then,

$\begin{align*} \sum_{k=1}^{m+1} \frac{1}{\sqrt{k}} &= \frac{1}{\sqrt{m+1}} + \sum_{k=1}^m \frac{1}{\sqrt{k}} \\[9pt] &< \frac{1}{\sqrt{m+1}} + 2 \sqrt{m} &(\text{Ind. hyp.}) \\[9pt] &= \frac{2\sqrt{m}\sqrt{m+1} + 1}{\sqrt{m+1}} \\[9pt] &= 2 \left( \frac{\sqrt{m^2 + m} + \frac{1}{2}}{\sqrt{m+1}} \right) \\[9pt] &< 2 \left( \frac{\sqrt{m^2+m+\frac{1}{4}} + \frac{1}{2}}{\sqrt{m+1}} \right) \\[9pt] &= 2 \left( \frac{\sqrt{(m+\frac{1}{2})^2} + \frac{1}{2}}{\sqrt{m+1}} \right)\\[9pt] &= 2 \left( \frac{m+\frac{1}{2} +\frac{1}{2}}{\sqrt{m+1}} \right) \\[9pt] &= 2 \sqrt{m+1}. \end{align*}$

Thus, the inequality holds for the case $m+1$ ; hence, it holds for all positive integers $n. \qquad \blacksquare$
False.
From the definitions of $\sinh$ and $\cosh$ we have

$\sinh x =\frac{e^x - e^{-x}}{2}, \qquad \cosh x = \frac{e^x + e^{-x}}{2}.$

Using these definitions, the inequality states

$\begin{align*} 1 + \sinh x \leq \cosh x && \implies && 1 + \frac{e^x - e^{-x}}{2} &\leq \frac{e^x + e^{-x}}{2} \\[9pt] && \implies && 2 + e^x - e^{-x} &\leq e^x + e^{-x} \\[9pt] && \implies && 2 \leq 2e^{-x}. \end{align*}$

However, this is false if $x < 0$ since $e^{-x} > 1$ for $x < 0$ .

Find all x satisfying equations given in terms of sinh

by RoRi

December 19, 2015

Let $c$ be the number such that $\sinh c = \frac{3}{4}$ . Find all $x$ that satisfy the given equations.

$\log (e^x + \sqrt{e^{2x} + 1}) = c$ .
$\log (e^x - \sqrt{e^{2x} - 1}) = c$ .

We are given $\sinh c = \frac{3}{4}$ . From the formula for $\sinh$ this means
$\frac{e^c - e^{-c}}{2} = \frac{3}{4}.$

Then, from the given equation we have

$\log (e^x + \sqrt{e^{2x} + 1}) = c \quad \implies \quad e^x + \sqrt{e^{2x} + 1} = e^c.$

Thus,

$e^{-c} = \frac{1}{e^x + \sqrt{e^{2x} + 1}}.$

So, then we have

$\begin{align*} \frac{3}{4} = \frac{e^c - e^{-c}}{2} &= \frac{1}{2} \left( e^x + \sqrt{e^{2x} + 1} - \frac{1}{e^x + \sqrt{e^{2x}+1}} \right) \\[9pt] &= \frac{1}{2} \left( \frac{(e^x + \sqrt{e^{2x}+1})^2 - 1}{e^x + \sqrt{e^{2x}+1}} \right) \\[9pt] &= \frac{e^{2x} + 2e^x \sqrt{e^{2x}+1} + e^{2x}+1 - 1}{2(e^x + \sqrt{e^{2x}+1})} \\[9pt] &= \frac{e^{2x} + e^x \sqrt{e^{2x}+1}}{e^x + \sqrt{e^{2x}+1}} \\[9pt] &= e^x \end{align*}$

Therefore we have

$e^x = \frac{3}{4} \quad \implies \quad x = \log 3 - \log 4 = \log 3 - 2 \log 2.$
There can be no $x$ which satisfy the given equation. As in part (a), we use the definition of $\sinh x$ to obtain the equation,
$\sinh c = \frac{3}{4} \quad \implies \quad e^c - e^{-c} = \frac{3}{2}.$

Next, we use the equation given in the problem to write,

$\begin{align*} &&\log (e^x - \sqrt{e^{2x} - 1}) &= c \\[9pt] \implies && e^x - \sqrt{e^{2x} -1} &= e^c \\[9pt] \implies && \frac{(e^x - \sqrt{e^{2x} - 1})(e^x + \sqrt{e^{2x}-1})}{e^x + \sqrt{e^{2x}-1}} &= e^c \\[9pt] \implies && \frac{e^{2x} - e^{2x} + 1}{e^x + \sqrt{e^{2x}-1}} &= e^c \\[9pt] \implies && \frac{1}{e^x + \sqrt{e^{2x}-1}} &= e^c. \end{align*}$

Furthermore, we can obtain an expression for $e^{-c}$ by considering

$e^x - \sqrt{e^{2x} - 1} &= e^c \quad \implies \quad \frac{1}{e^x - \sqrt{e^{2x} - 1}} &= e^{-c}.$

Putting these expressions for $e^c$ and $e^{-c}$ into our original equation we have

$\begin{align*} \frac{3}{2} &= e^c - e^{-c} \\[9pt] &= \left( \frac{1}{e^x + \sqrt{e^{2x}-1}} \right) - \left( \frac{1}{e^x - \sqrt{e^{2x}-1}} \right) \\[9pt] &= \frac{e^x - \sqrt{e^{2x}-1} - e^x - \sqrt{e^{2x}-1}}{(e^x + \sqrt{e^{2x}-1})(e^x - \sqrt{e^{2x}-1})} \\[9pt] &= \frac{ -2 \sqrt{e^{2x}-1}}{e^{2x} - e^{2x} + 1} \\[9pt] &= -2\sqrt{e^{2x}-1} \end{align*}$

But this implies

$\sqrt{e^{2x}-1} = -\frac{3}{4}$

which is impossible. Hence, there can be no real $x$ satisfying this equation.

Prove the formulas for derivatives of products and quotients

by RoRi

December 19, 2015

Derive the formulas for the derivative of a product and the derivative of a quotient from the corresponding formulas for the derivative of a sum and the derivative of a difference.

We know the derivative rules for sums and differences are:

$\left( f(x) + g(x) \right)' = f'(x) + g'(x) \quad \text{and} \quad \left( f(x) - g(x) \right)' = f'(x) - g'(x).$

To derive the derivative rule for products using logarithmic differentiation we let $h(x) = f(x)g(x)$ and compute

$\begin{align*} h(x) = f(x)g(x) && \implies && \log |h(x)| &= \log |f(x)g(x)| \\[9pt] && \implies && (\log |h(x)|)' &= (\log |f(x)g(x)|)' \\[9pt] && \implies && \frac{h'(x)}{h(x)} &= (\log |f(x)| + \log |g(x)|)' \\[9pt] && \implies && \frac{h'(x)}{h(x)} &= \frac{f'(x)}{f(x)} + \frac{g'(x)}{g(x)} \\[9pt] && \implies && h'(x) &= \left( \frac{f'(x)}{f(x)} + \frac{g'(x)}{g(x)} \right) h(x) \\[9pt] && \implies && h'(x) &= \left( \frac{f'(x)}{f(x)} + \frac{g'(x)}{g(x)} \right) f(x)g(x) \\[9pt] && \implies && (f(x)g(x))' &= f'(x) g(x) + f(x)g'(x). \end{align*}$

This is the usual rule for derivative of a product.

Similarly, for the derivative of a quotient, let $h(x) = \frac{f(x)}{g(x)}$ and then compute,

$\begin{align*} h(x) = \frac{f(x)}{g(x)} && \implies && \log | h(x)| &= \log \left| \frac{f(x)}{g(x)} \right| \\[9pt] && \implies && (\log|h(x)|)' &= \left( \log \left| \frac{f(x)}{g(x)} \right| \right)' \\[9pt] && \implies && \frac{h'(x)}{h(x)} &= \left( \log |f(x)| - \log |g(x)| \right) ' \\[9pt] && \implies && \frac{h'(x)}{h(x)} &= \frac{f'(x)}{f(x)} - \frac{g'(x)}{g(x)} \\[9pt] && \implies && h'(x) &= \left( \frac{f'(x)}{f(x)} - \frac{g'(x)}{g(x)} \right) h(x) \\[9pt] && \implies && h'(x) &= \left( \frac{f'(x)}{f(x)} - \frac{g'(x)}{g(x)} \right) \frac{f(x)}{g(x)} \\[9pt] && \implies && \left( \frac{f(x)}{g(x)} \right)' &= \frac{f'(x)g(x) - g'(x)f(x)}{f(x)g(x)} \cdot \frac{f(x)}{g(x)} \\[9pt] && \implies && \left( \frac{f(x)}{g(x)} \right)' &= \frac{f'(x)g(x) - g'(x)f(x)}{(g(x))^2}. \end{align*}$

Which is the usual rule for derivative of a quotient.

Compute f(x) + f(1/x) where f satisfies a given integral equation

by RoRi

December 17, 2015

Define a function by an integral equation as follows:

$f(x) = \int_1^x \frac{\log t}{t+1} \, dx \qquad \text{for} \quad x > 0.$

Using this formula compute $f(x) + f(1/x)$ .

Substituting $x$ and $\frac{1}{x}$ into the formula for $f(x)$ we have

$f(x) + f \left( \frac{1}{x} \right) = \int_1^x \frac{\log t}{t+1} \, dt + \int_1^{\frac{1}{x}} \frac{\log t}{t+1} \, dt.$

For the second integral we make the substitution $t = \frac{1}{u}$ , $dt = -\frac{1}{u^2} \, du$ . This gives us

$\begin{align*} f(x) + f \left( \frac{1}{x} \right) &= \int_1^x \frac{\log t}{t+1} \, dt + \int_1^x \frac{\log \left( \frac{1}{u} \right)}{\frac{1}{u} + 1} \left(-\frac{1}{u^2} \right) \, du \\[9pt] &= \int_1^x \frac{\log t}{t+1} \, dt + \int_1^x \frac{\log u}{u+u^2} \, du \\[9pt] &= \int_1^x \left( \frac{\log t}{t+1} + \frac{\log t}{t+t^2} \right) \, dt \\[9pt] &= \int_1^x \left( \frac{t \log t + \log t}{t(t+1)} \right) \, dt \\[9pt] &= \int_1^x \frac{\log t}{t} \, dt \\[9pt] &= \left. \frac{(\log t)^2}{t} \right|_1^x \\[9pt] &= \frac{(\log x)^2}{2}. \end{align*}$

As a check, we evaluate at $x = 2$ ,

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

Prove a property of the derivative if arctangent and the logarithm obey a given relation

by RoRi

November 29, 2015

$\arctan \frac{y}{x} = \log \sqrt{x^2 + y^2}$

prove that

$\frac{dy}{dx} = \frac{x+y}{x-y}.$

Proof. First, we consider the derivatives of the left and right side of the given equation. (Treating $y$ as a function of $x$ and remembering to use the chain rule.) So, for the derivative on the left, we have

$\begin{align*} D \left( \arctan \frac{y}{x} \right) &= \left( \frac{-y}{x^2} + \frac{1}{x} \frac{dy}{dx} \right) \left( \frac{1}{1+\left( \frac{y}{x} \right)^2} \right) \\ &= \frac{x \frac{dy}{dx} - y}{x^2+y^2}. \end{align*}$

On the right we have,

$\begin{align*} D \left( \log \sqrt{x^2+y^2} \right) &= \left( x + y \frac{dy}{dx} \right) \left( \frac{1}{x^2+y^2} \right)\\ &= \frac{x + y \frac{dy}{dx}}{x^2+y^2}. \end{align*}$

Now, using the given equation we have

$\begin{align*} \arctan \frac{y}{x} = \log \sqrt{x^2+y^2} && \implies && D \left( \arctan \frac{y}{x} \right) &= D \left( \log \sqrt{x^2+y^2} \right) \\[9pt] && \implies && \frac{x \frac{dy}{dx} - y}{x^2+y^2} &= \frac{x+y\frac{dy}{dx}}{x^2+y^2} \\[9pt] && \implies && x \frac{dy}{dx} - y &= x + y\frac{dy}{dx} \\ && \implies && (x-y) \frac{dy}{dx} &= x + y \\ && \implies && \frac{dy}{dx} &= \frac{x+y}{x-y}. \qquad \blacksquare \end{align*}$

Find the derivative of log (arccos (x^-1/2))

by RoRi

November 29, 2015

Find the derivative of the function

$f(x) = \log \left( \arccos \frac{1}{\sqrt{x}} \right).$

Using the chain rule and the formulas for the derivative of the logarithm and the arccosine we have,

$\begin{align*} f'(x) &= \left( \frac{1}{\arccos \left( \frac{1}{\sqrt{x}} \right)} \right) \left( \frac{-1}{\sqrt{1 - \left( \frac{1}{\sqrt{x}} \right)^2}} \right) \left(\frac{-1}{2x^{\frac{3}{2}}} \right)\\ &= \frac{\sqrt{x}}{2x^{\frac{3}{2}} \sqrt{x-1} \arccos \frac{1}{\sqrt{x}}} \\ &= \frac{1}{2x \sqrt{x-1} \arccos \frac{1}{\sqrt{x}}}. \end{align*}$

The formula is valid for $x > 1$ .

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,

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

Find the derivative of a product of terms (x-a_i)^b_i

by RoRi

November 15, 2015

Find the derivative of the following function:

$f(x) = \prod_{i=1}^n (x-a_i)^{b_i}.$

To take this derivative we want to use logarithmic differentiation. To that end we take the derivative of both sides,

$\begin{align*} \log f(x) &= \log \left( \prod_{i=1}^n (x-a_i)^{b_i} \right) \\ &= \sum_{i=1}^n b_i \log(x-a_i). \end{align*}$

Therefore, taking the derivative of both sides, we have

$\begin{align*} &&\frac{f'(x)}{f(x)} &= \sum_{i=1}^n \frac{b_i}{x-a_i} \\ \implies && f'(x) &= f(x) \left( \sum_{i=1}^n \frac{b_i}{x-a_i} \right) \\ \implies && f'(x) &= \left( \prod_{i=1}^n (x-a_i)^{b_i} \right) \left( \sum_{i=1}^n \frac{b_i}{x-a_i} \right) \end{align*}$

Find the derivative of (x²(3-x)^1/3)/((1-x)(3+x)^2/3)

by RoRi

November 13, 2015

Find the derivative of the function

$f(x) = \frac{x^2(3-x)^{\frac{1}{3}}}{(1-x)(3+x)^{\frac{2}{3}}}.$

First, we take the logarithm of both sides so we can use logarithmic differentiation,

$\log f(x) = 2 \log x + \frac{1}{3} \log (3-x) - \log (1-x) - \frac{2}{3} \log (3+x).$

Then we differentiate both sides,

$\begin{align*} \frac{f'(x)}{f(x)} &= \frac{2}{x} + \frac{1}{3(3-x)} + \frac{1}{1-x} - \frac{2}{3(3+x)} \\[9pt] &= \frac{18-7x}{3x(3-x)} + \frac{7+5x}{3(1-x)(3+x)} \\[9pt] &= \frac{(54 - 57x - 4x^2 + 7x^3) + (21x + 8x^2 - 5x^3)}{3x(3-x)(1-x)(3+x)} \\[9pt] &= \frac{2x^3 + 4x^2 - 36x + 54}{3x(1-x)(3-x)(3+x)}. \end{align*}$

Therefore,

$\begin{align*} f'(x) &= \left( \frac{x^2(3-x)^{\frac{1}{3}}}{(1-x)(3+x)^{\frac{2}{3}}} \right) \left( \frac{2x^3 + 4x^2 - 36x + 54}{3x(1-x)(3-x)(3+x)} \right)\\[9pt] &= \frac{2x^4 + 4x^3 - 36x^2 + 54x}{3(1-x)^2(3-x)^{\frac{2}{3}} (3+x)^{\frac{5}{3}}}. \end{align*}$

Find the derivative of x^1/x

by RoRi

November 13, 2015

Find the derivative of the function

$f(x) = x^{\frac{1}{x}}.$

We take the derivative using logarithmic differentiation. First, taking the logarithm of both sides,

$\log f(x) = \frac{1}{x} \log x = \frac{\log x}{x}.$

Then take the derivative of both sides,

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