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Here's a more formal way to introduce
The imaginary unit
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
For any real numbers
#(a, b) + (c, d) = (a+c, b+d)#
#(a, b) * (c, d) = (ac-bd, ad+bc)#
Then we find that:
There is an identity (i.e. zero) for addition:
#(a, b) + (0, 0) = (0, 0) + (a, b) = (a, b)#
Every element has an additive inverse:
#(a, b) + (-a, -b) = (-a, -b) + (a, b) = (0, 0)#
Addition is associative:
#((a, b) + (c, d)) + (e, f) = (a, b) + ((c, d) + (e, f))#
Addition is commutative:
#(a, b) + (c, d) = (c, d) + (a, b)#
There is an identity (i.e. one) for multiplication:
#(a, b) * (1, 0) = (1, 0) * (a, b) = (a, b)#
Every non-zero element has a multiplicative inverse:
#(a, b) * (a/(a^2+b^2), -b/(a^2+b^2)) = (a/(a^2+b^2), -b/(a^2+b^2)) * (a, b) = (1, 0)#
Multiplication is associative:
#((a, b) * (c, d)) * (e, f) = (a, b) * ((c, d) * (e, f))#
Multiplication is commutative:
#(a, b) * (c, d) = (c, d) * (a, b)#
Multiplication is left and right distributive over addition:
#(a, b) * ((c, d) + (e, f)) = ((a, b) * (c, d)) + ((a, b) * (e, f))#
#((a, b) + (c, d)) * (e, f) = ((a, b) * (e, f)) + ((c, d) * (e, f))#
In summary: the elements of
#f(x) = (x, 0)#
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
We can then change notation and write our complex numbers as:
#a+bi#
where
There's our definition:
#i = (0, 1)#
Here's the trigonometric method for the case where the cubic equation has
Let us deal with the reduced cubic:
#t^3+pt+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 + kp 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
#4 cos^3 theta + (4p)/k^2 cos theta + (4q)/k^3 = 0#
Then we want:
#(4p)/k^2 = -3#
So:
#k^2 = -(4p)/3#
and we can choose:
#k = sqrt(-(4p)/3) = 2/3sqrt(-3p)#
Note also:
#k^4 = (-(4p)/3)^2 = (16p)/9#
Then:
#(4q)/k^3 = (4qk)/k^4 = (4qk)/((16p)/9) = (9qk)/4 = 3/2qsqrt(-3p)#
So our equation becomes:
#4 cos^3 theta-3 cos theta + 3/2qsqrt(-3p) = 0#
That is:
#cos 3 theta = -3/2qsqrt(-3p)#
So:
#3 theta = +-cos^(-1)(-3/2qsqrt(-3p)) + 2npi" "n in ZZ#
Hence distinct values:
#cos theta = cos(1/3 cos^(-1)(-3/2qsqrt(-3p))+(2npi)/3)" "n = 0, 1, 2#
Then
#t_n = 2/3sqrt(-3p) cos(1/3 cos^(-1)(-3/2qsqrt(-3p))+(2npi)/3)" "n = 0, 1, 2#
Footnote
If not all of the roots are real, then
Alternatively, note that
#cosh 3 theta = 4 cosh^3 theta - 3 cosh theta#
So we can follow a similar derivation to find:
#t_n = 2/3sqrt(-3p) cosh(1/3 cosh^(-1)(-3/2qsqrt(-3p))+(2npi)/3i)" "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.
Almost. If
Uniqueness of
Any function of the form
In order for
Suppose
We can write the Maclaurin series for
#f(x) = sum_(n=0)^oo a_n x^n" "# for some constants#a_0, a_1,...#
Then:
#1 = f(0) = a_0#
and:
#f'(x) = sum_(n=0)^oo (d/(dx) a_n x^n)#
#color(white)(f'(x)) = sum_(n=0)^oo (n a_n x^(n-1))#
#color(white)(f'(x)) = sum_(n=0)^oo (n+1) a_(n+1) x^n#
Since we want
#sum_(n=0)^oo (n+1) a_(n+1) x^n = sum_(n=0)^oo a_n x^n#
So equating the coefficients, we find:
#{ (a_0 = 1), (a_(n+1) = a_n/(n+1)" for "n >= 0") :}#
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
Then if
#d/(dx) k e^x = k d/(dx) e^x = k e^x#
So:
#f(x) = k e^x#
is also a solution of
Footnote
There are some non well behaved functions that also satisfy
For example:
#f(x) = { ("undefined" " if " x=0), (abs(e^x)" if "x != 0) :}#
The formula for the Distance from a Point to a Line is:
where
The distance from any point
The distance from the focus
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]:
Multiply both sides by 5:
Square both sides:
Expand the squares:
Simplify:
Multiply both sides by -1:
The above is the standard Cartesian form for a conic section.
Here is a graph of the parabola, the focus and the directrix:
Recall that, for the Hyperbola
asymptotes are given by,
In our Case, the asymptotes are,
Next,
Multiplying by
By
With
Solution set
Alternate notation:
Solve:
1) Examine the related equation to find the boundaries.
or
The solutions to the related equation define boundary points that divide the number line into 5 parts, 3 regions and the 2 boundary points.
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,
In region,
In region,
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
Alternate notation:
2-D Graph of the inequality:
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