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## What does the imaginary number i equal. I know that i^2 is -1, but what is i?

George C.
Featured 4 months ago

Here's a more formal way to introduce $i$ than is normally done at school...

#### Explanation:

The imaginary unit $i$ is usually introduced as "the" square root of $- 1$, but note that $- 1$ has two square roots. So how do we tell which is which?

A more formal way to do things is to define the arithmetic of complex numbers in terms of arithmetic on pairs of real numbers, i.e. elements of ${\mathbb{R}}^{2}$ ...

For any real numbers $a , b , c , d$, define addition and multiplication as follows:

$\left(a , b\right) + \left(c , d\right) = \left(a + c , b + d\right)$

$\left(a , b\right) \cdot \left(c , d\right) = \left(a c - b d , a d + b c\right)$

Then we find that:

• There is an identity (i.e. zero) for addition:

$\left(a , b\right) + \left(0 , 0\right) = \left(0 , 0\right) + \left(a , b\right) = \left(a , b\right)$

• Every element has an additive inverse:

$\left(a , b\right) + \left(- a , - b\right) = \left(- a , - b\right) + \left(a , b\right) = \left(0 , 0\right)$

$\left(\left(a , b\right) + \left(c , d\right)\right) + \left(e , f\right) = \left(a , b\right) + \left(\left(c , d\right) + \left(e , f\right)\right)$

$\left(a , b\right) + \left(c , d\right) = \left(c , d\right) + \left(a , b\right)$

• There is an identity (i.e. one) for multiplication:

$\left(a , b\right) \cdot \left(1 , 0\right) = \left(1 , 0\right) \cdot \left(a , b\right) = \left(a , b\right)$

• Every non-zero element has a multiplicative inverse:

$\left(a , b\right) \cdot \left(\frac{a}{{a}^{2} + {b}^{2}} , - \frac{b}{{a}^{2} + {b}^{2}}\right) = \left(\frac{a}{{a}^{2} + {b}^{2}} , - \frac{b}{{a}^{2} + {b}^{2}}\right) \cdot \left(a , b\right) = \left(1 , 0\right)$

• Multiplication is associative:

$\left(\left(a , b\right) \cdot \left(c , d\right)\right) \cdot \left(e , f\right) = \left(a , b\right) \cdot \left(\left(c , d\right) \cdot \left(e , f\right)\right)$

• Multiplication is commutative:

$\left(a , b\right) \cdot \left(c , d\right) = \left(c , d\right) \cdot \left(a , b\right)$

• Multiplication is left and right distributive over addition:

$\left(a , b\right) \cdot \left(\left(c , d\right) + \left(e , f\right)\right) = \left(\left(a , b\right) \cdot \left(c , d\right)\right) + \left(\left(a , b\right) \cdot \left(e , f\right)\right)$

$\left(\left(a , b\right) + \left(c , d\right)\right) \cdot \left(e , f\right) = \left(\left(a , b\right) \cdot \left(e , f\right)\right) + \left(\left(c , d\right) \cdot \left(e , f\right)\right)$

In summary: the elements of ${\mathbb{R}}^{2}$ form a field under the operations of addition and multiplication defined.

• The function from $\mathbb{R}$ to ${\mathbb{R}}^{2}$ defined by:

$f \left(x\right) = \left(x , 0\right)$

preserves addition, multiplication, etc. so is a field homomorphism.

So we can say that this function embeds the real numbers in the complex numbers - for that is the field that we have defined.

Effectively we can say that $\left(a , 0\right)$ is a real number and $\left(a , b\right)$ is a non-real complex number if $b \ne 0$.

We can then change notation and write our complex numbers as:

$a + b i$

where $a , b \in \mathbb{R}$ and $i = \left(0 , 1\right)$ is the imaginary unit.

There's our definition:

$i = \left(0 , 1\right)$

## How do you solve standard cubic equations, and is there a general formula like there is for quadratics?

George C.
Featured 3 months ago

Here's the trigonometric method for the case where the cubic equation has $3$ real roots...

#### Explanation:

Let us deal with the reduced cubic:

${t}^{3} + p t + q = 0$

(see other solution for how to get here if your cubic is more general)

Consider the substitution:

$t = k \cos \theta$

Then our equation becomes:

${k}^{3} {\cos}^{3} \theta + k p \cos \theta + q = 0$

We want to use the trigonometric identity:

$\cos 3 \theta = 4 {\cos}^{3} \theta - 3 \cos \theta$

So dividing our equation by ${k}^{3} / 4$ we find:

$4 {\cos}^{3} \theta + \frac{4 p}{k} ^ 2 \cos \theta + \frac{4 q}{k} ^ 3 = 0$

Then we want:

$\frac{4 p}{k} ^ 2 = - 3$

So:

${k}^{2} = - \frac{4 p}{3}$

and we can choose:

$k = \sqrt{- \frac{4 p}{3}} = \frac{2}{3} \sqrt{- 3 p}$

Note also:

${k}^{4} = {\left(- \frac{4 p}{3}\right)}^{2} = \frac{16 p}{9}$

Then:

$\frac{4 q}{k} ^ 3 = \frac{4 q k}{k} ^ 4 = \frac{4 q k}{\frac{16 p}{9}} = \frac{9 q k}{4} = \frac{3}{2} q \sqrt{- 3 p}$

So our equation becomes:

$4 {\cos}^{3} \theta - 3 \cos \theta + \frac{3}{2} q \sqrt{- 3 p} = 0$

That is:

$\cos 3 \theta = - \frac{3}{2} q \sqrt{- 3 p}$

So:

$3 \theta = \pm {\cos}^{- 1} \left(- \frac{3}{2} q \sqrt{- 3 p}\right) + 2 n \pi \text{ } n \in \mathbb{Z}$

Hence distinct values:

$\cos \theta = \cos \left(\frac{1}{3} {\cos}^{- 1} \left(- \frac{3}{2} q \sqrt{- 3 p}\right) + \frac{2 n \pi}{3}\right) \text{ } n = 0 , 1 , 2$

Then $t = k \cos \theta$, so the distinct roots of our original cubic are:

${t}_{n} = \frac{2}{3} \sqrt{- 3 p} \cos \left(\frac{1}{3} {\cos}^{- 1} \left(- \frac{3}{2} q \sqrt{- 3 p}\right) + \frac{2 n \pi}{3}\right) \text{ } n = 0 , 1 , 2$

Footnote

If not all of the roots are real, then $\left\mid - \frac{3}{2} q \sqrt{- 3 p} \right\mid > 1$. The formula we derived is still true, but we have to deal with complex values of ${\cos}^{- 1}$.

Alternatively, note that $\cosh \theta$ also satisfies:

$\cosh 3 \theta = 4 {\cosh}^{3} \theta - 3 \cosh \theta$

So we can follow a similar derivation to find:

${t}_{n} = \frac{2}{3} \sqrt{- 3 p} \cosh \left(\frac{1}{3} {\cosh}^{- 1} \left(- \frac{3}{2} q \sqrt{- 3 p}\right) + \frac{2 n \pi}{3} i\right) \text{ } n = 0 , 1 , 2$

I have not really used this hyperbolic method, since in the cases where it makes most sense we can use Cardano's method anyway.

## Is #e^x# the unique function of which derivative is itself? Can you prove it?

George C.
Featured 2 months ago

Almost. If $k$ is any constant then $f \left(x\right) = k {e}^{x}$ satisfies $f ' \left(x\right) = f \left(x\right)$.

Uniqueness of ${e}^{x}$ is given by requiring $f \left(x\right)$ to be defined everywhere, and $f \left(0\right) = 1$.

#### Explanation:

Any function of the form $f \left(x\right) = k {e}^{x}$ satisfies $f ' \left(x\right) = f \left(x\right)$

In order for ${e}^{x}$ to be the only solution we need an extra condition such as $f \left(0\right) = 1$

Suppose $f \left(x\right)$ is a well behaved function from $\mathbb{R}$ to $\mathbb{R}$ with $f \left(0\right) = 1$.

We can write the Maclaurin series for $f \left(x\right)$ as:

$f \left(x\right) = {\sum}_{n = 0}^{\infty} {a}_{n} {x}^{n} \text{ }$ for some constants ${a}_{0} , {a}_{1} , \ldots$

Then:

$1 = f \left(0\right) = {a}_{0}$

and:

$f ' \left(x\right) = {\sum}_{n = 0}^{\infty} \left(\frac{d}{\mathrm{dx}} {a}_{n} {x}^{n}\right)$

$\textcolor{w h i t e}{f ' \left(x\right)} = {\sum}_{n = 0}^{\infty} \left(n {a}_{n} {x}^{n - 1}\right)$

$\textcolor{w h i t e}{f ' \left(x\right)} = {\sum}_{n = 0}^{\infty} \left(n + 1\right) {a}_{n + 1} {x}^{n}$

Since we want $f ' \left(x\right) = f \left(x\right)$, we have:

${\sum}_{n = 0}^{\infty} \left(n + 1\right) {a}_{n + 1} {x}^{n} = {\sum}_{n = 0}^{\infty} {a}_{n} {x}^{n}$

So equating the coefficients, we find:

$\left\{\begin{matrix}{a}_{0} = 1 \\ {a}_{n + 1} = {a}_{n} / \left(n + 1\right) \text{ for "n >= 0}\end{matrix}\right.$

So:

#1/a_0 = 1/1 = 0!#

#1/a_1 = 1/a_0 = 1!#

#1/a_2 = 2/a_1 = 2!#

#1/a_3 = 3/a_2 = 3!#

and:

#1/a_(n+1) = (n+1) (1/a_n) = (n+1)!#

Hence:

#f(x) = sum_(n=0)^oo x^n/(n!)#

which is one of the definitions of ${e}^{x}$

Then if $k$ is any constant, we find:

$\frac{d}{\mathrm{dx}} k {e}^{x} = k \frac{d}{\mathrm{dx}} {e}^{x} = k {e}^{x}$

So:

$f \left(x\right) = k {e}^{x}$

is also a solution of $f \left(x\right) = f ' \left(x\right)$

Footnote

There are some non well behaved functions that also satisfy $f ' \left(x\right) = f \left(x\right)$

For example:

$f \left(x\right) = \left\{\begin{matrix}\text{undefined" " if " x=0 \\ abs(e^x)" if } x \ne 0\end{matrix}\right.$

## Find the equation of parabola whose focus is(3,0) and directrix is 3x+4y=1?

Douglas K.
Featured 2 months ago

$16 {x}^{2} - 24 x y + 9 {y}^{2} - 144 x + 8 y + 224 = 0$

#### Explanation:

The formula for the Distance from a Point to a Line is:

$d = | a x + b y + c \frac{|}{\sqrt{{a}^{2} + {b}^{2}}} \text{ [1]}$

where $\left(x , y\right)$ is the point and the line is of the form $a x + b y + c = 0$.

The distance from any point $\left(x , y\right)$ on the parabola to the line $3 x + 4 y - 1 = 0$ is:

$d = | 3 x + 4 y - 1 \frac{|}{\sqrt{{3}^{2} + {4}^{2}}}$

$d = | 3 x + 4 y - 1 \frac{|}{5} \text{ [2]}$

The distance from the focus $\left(3 , 0\right)$ to any point, $\left(x , y\right)$ on the parabola is:

$d = \sqrt{{\left(x - 3\right)}^{2} + {\left(y - 0\right)}^{2}} \text{ [3]}$

The definition of a parabola is the locus of points equidistant from its directrix and its focus, therefore, we can derive the equation of the parabola by setting the right side of equation [2] equal to the right side of equation [3]:

$| 3 x + 4 y - 1 \frac{|}{5} = \sqrt{{\left(x - 3\right)}^{2} + {\left(y - 0\right)}^{2}}$

Multiply both sides by 5:

$| 3 x + 4 y - 1 | = 5 \sqrt{{\left(x - 3\right)}^{2} + {\left(y - 0\right)}^{2}}$

Square both sides:

${\left(3 x + 4 y - 1\right)}^{2} = 25 \left({\left(x - 3\right)}^{2} + {\left(y - 0\right)}^{2}\right)$

Expand the squares:

$9 {x}^{2} + 24 x y + 16 {y}^{2} - 6 x - 8 y + 1 = 25 {x}^{2} + 25 {y}^{2} - 150 x + 225$

Simplify:

$- 16 {x}^{2} + 24 x y - 9 {y}^{2} + 144 x - 8 y - 224 = 0$

Multiply both sides by -1:

$16 {x}^{2} - 24 x y + 9 {y}^{2} - 144 x + 8 y + 224 = 0$

The above is the standard Cartesian form for a conic section.

Here is a graph of the parabola, the focus and the directrix:

## How do you write an equation for the hyperbola with asymptote #y=+-(4x)/3#, contains #(3sqrt(2,)4)#?

Ratnaker Mehta
Featured 3 weeks ago

${x}^{2} / 9 - {y}^{2} / 16 = 1$.

#### Explanation:

Recall that, for the Hyperbola $S : {x}^{2} / {a}^{2} - {y}^{2} / {b}^{2} = 1$, the

asymptotes are given by, #y=+-b/a*x;" where, "a,b gt 0#.

In our Case, the asymptotes are, $y = \pm \frac{4}{3} \cdot x$.

$\therefore \frac{b}{a} = \frac{4}{3.} \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots \ldots . . \left(1\right)$.

Next, $\left(3 \sqrt{2} , 4\right) \in S \Rightarrow {\left(3 \sqrt{2}\right)}^{2} / {a}^{2} - {4}^{2} / {b}^{2} = 1$.

$\therefore \frac{18}{a} ^ 2 - \frac{16}{b} ^ 2 = 1$.

Multiplying by ${b}^{2}$, we get, $18 \cdot {b}^{2} / {a}^{2} - 16 = {b}^{2}$.

By $\left(1\right)$, then, $18 \cdot {\left(\frac{4}{3}\right)}^{2} - 16 = {b}^{2} , i . e . , {b}^{2} = 16$.

$\therefore b = 4 \left(\because , b > 0\right) , \mathmr{and} , \left(1\right) \Rightarrow a = 3 , \mathmr{and} , {a}^{2} = 9$.

With ${a}^{2} = 9 , \mathmr{and} {b}^{2} = 16$, the reqd. eqn. of the Hyperbola is,

${x}^{2} / 9 - {y}^{2} / 16 = 1$.

## How do you solve #x^2-4x>=5#?

Nimo N.
Featured 2 weeks ago

Solution set
# color(red) (-oo <= x <= -1"," or 5 <= x <= oo #
Alternate notation:
# color(red) ( x in (-oo, -1] uu [5, oo) #

#### Explanation:

Solve:
# x^2 âˆ’ 4x â‰¥ 5 #

1) Examine the related equation to find the boundaries.
# x^2 âˆ’ 4x â‰¥ 5 #
# x^2 âˆ’ 4x - 5 = 0 #
$\left(x - 5\right) \left(x + 1\right) = 0$
$\left(x - 5\right) = 0$
$x = 5$
or
$\left(x + 1\right) = 0$
$x = - 1$

The solutions to the related equation define boundary points that divide the number line into 5 parts, 3 regions and the 2 boundary points.

• Since the inequality is inclusive ("non-strict") the boundary points are in the solution set.

A graph of the number line, with solid dots on (-1), and (5) will help in thinking about the problem.

To check for the solution, use test points, one from each region, in the inequality to see if they qualify:

Try (-2), (0), and (6). It is best not to pick numbers that make the arithmetic difficult.

In region, $- \infty < x < - 1$
# ( -2 )^2 âˆ’ 4( -2) â‰¥ 5 " ?"#
# 4 + 8 â‰¥ 5 " ?"#
# 12 â‰¥ 5 " YES"#

In region, $- 1 < x < 5$
# ( 0 )^2 âˆ’ 4( 0 ) â‰¥ 5 " ?" #
# 0 â‰¥ 5 " NO" #

In region, $5 < x < \infty$
# ( 6 )^2 âˆ’ 4( 6 ) â‰¥ 5 " ?" #
# 36 âˆ’ 24 â‰¥ 5 " ?" #
# 12 â‰¥ 5 " YES" #

We must be careful in stating the solution inequalities. x can not be in both regions at once, since they are disconnected. There is a full region between the two regions in the solution, so we can't use "and", we must use "or".

Solution set
# color(red) (-oo <= x <= -1"," or 5 <= x <= oo #
Alternate notation:
# color(red) ( x in (-oo, -1] uu [5, oo) #

2-D Graph of the inequality:

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