If you don't put bounds, I think it's not possible. The graphic of your equation is a cross with four infinite branches and with one paramater, you can describe complicated curves but only with two infinite extensions.
If your variables are bounded, maybe it's possible.
So basically, I wanted to try to make a parametric equation of this equation to infinity and I did some research to transfer an implicit to parametric equation. Still, the case of x^2 + y^2 = 0 is quite complex as you actually said. Because from what I see, when I switch from implicit to parametric, I find some weird things. I actually think you're right, that's impossible.
When you say "If your variables are bounded, maybe it's possible", what do you mean?
1) |cos(t)| / cos(t) = +/- 1 for all values of t ( except when t = multiple of pi/2) so y reduces to +/- sin(t) and x = sin(t) except at t = mult pi/2 when you get a jump discontinuity. But, as t--> pi/2, y --> +/- n.
2) Restricting t to 0<= t <= 2pi only gives you a portion of the 2 lines where x and y are trapped in the square between +/- n. And it doesn't change anything if you let t range over all the real numbers. You are still trapped in the square where x and y are stuck between +/- n.
3) So x = n.sin(t) and y = +/- n.sin(t) except when t = mult of pi/2 where y does not have a value. You might as well use x = n and y = +/-n where you get all of the infinite pairs of solutions and no discontinuities.
x^2 - y^2 = 0 describes 2 straight lines: y = x and y = -x
So let x = t and then y = t or y = -t where t is any real number.
This is sometimes referred to an a "degenerate hyperbola".
For the rectangular hyperbola, x^2 - y^2 = a^2, the asymptotes are the 2 lines y = x and y = -x. As a tends to 0, then the branches of the hyperbola itself actually approach the the 2 lines y = x and and y = -x and when a actually becomes 0 then the hyperbola turns into the 2 lines - hence the term "degenerate"
As a tends to infinity nothing really special happens. The hyperbola still exists has as the same 2 asymptotes, y = x and y = -x . This prevents the hyperbola approaching the 2 vertical lines x = a and x = -a. It just takes larger and larger of values of x to get arbitrarily close to to the asymptotes.
The case where x^2 + y^2 = 0 is the simplest case of all. It describes a single point (0,0)
For the circle x^2 + y^2 = a^2, the radius is a and as a tends to 0 the circle shrinks and approaches the center (0,0) and when a actually takes on the value of zero we get just the point (0,0). Hence a single point is sometimes called a "degenerate circle". As a tends to infinity we just get a larger and larger circle.
The curvature of the circle x^2 + y^2 = a^2 is 1/a so I suppose we can argue the as a tends to infinity the curvature tends to 0 - which is the curvature of a straight line but there is no single equation which can represent such a line.
both of these pairs of equations only pick up the line y=x (i.e. each pair plots the same line) and not the line y=-x and therefore you lose half of the solutions.
The only way to pick up the other half is x = t ; y = -t.
So from above: y=t, x=t or y=-t, x=-t :
You have to change drop one of the - signs to get : y=t, x=t or y=t, x=-t.
If, for example, you set a = 2 then you get x =2t and y = 2t. Let t range over the real numbers you only get one of the 2 straight lines (y =x). You have to include y = +/- at in order to pick up both branches. Same holds true if a is negative. It's still true that x^2 - y^2 = 0 but with x = at and y = at you only pick up half the solutions --- y=x.
x = at ; y = +/-at is the only way to pick up both lines. The "a" is unnecessary here but it doesn't hurt either. It's just simpler to exclude it.
NB: very important - √(t^2), not t because, in effect, it divides by -1 when the arg ≥ π
Derivation:
Make x the subject:
x^2-y^2=0
y^2=-x^2
x=±y
Now we create a method to have 2 sets of points i.e. y=x, y=-x:
t is our parameter.
let r = |t|
let θ = arg(t)
Therefore t = r * cis(θ)
let s = t^2 = r^2 * cis(2θ)
let d = t / √(s)
Therefore d = 1 for 0 ≤ θ < π
-1 for π ≤ θ < 2π
This (using d) gives us a split into 2 sections, which are the same size as the original.
So now, we do the following:
let n = r * cis(2θ) = s / |t|
Now n is the same number at double the angle, which allows us to get any complex number twice from the value of t, which is what we need, it also repeats where the value of d switches from 1 to -1, which is what we need.
let x = n
let y = dn
With this setup, we get every number twice, and then flip the sign for every y value so we get every number paired with the number, and the negated version of this number.
Therefore we have parameterised the relation.
Simplify to only use the pronumeral t:
x = n
= s / |t|
= t^2 / |t|
y = dn
= d * s / |t|
= d * t^2 / |t|
= (t / √(s)) * t^2 / |t|
= t^2 * (t / √(t^2)) / |t|
= t^3 / √(t^2) / |t|
= t^3 / (√(t^2) * |t|)
Therefore, it is possible to parameterise this relation over the complex plane with only one value for x and y for each value of t.
If you don't put bounds, I think it's not possible. The graphic of your equation is a cross with four infinite branches and with one paramater, you can describe complicated curves but only with two infinite extensions.
If your variables are bounded, maybe it's possible.
So basically, I wanted to try to make a parametric equation of this equation to infinity and I did some research to transfer an implicit to parametric equation. Still, the case of x^2 + y^2 = 0 is quite complex as you actually said. Because from what I see, when I switch from implicit to parametric, I find some weird things. I actually think you're right, that's impossible.
When you say "If your variables are bounded, maybe it's possible", what do you mean?
Hello ! Sorry to be so late.
If x (and y) are bounded between -n and n (open interval), you may try :
x=n.sin(t) and y=n.tan(t).∣cos(t)∣ for 0 ≤ t ≤ 2𝜋 but excluding 𝜋/4 and 3𝜋/4
I think it's ok : check, please!
Not trying to be a PITA, but:
1) |cos(t)| / cos(t) = +/- 1 for all values of t ( except when t = multiple of pi/2) so y reduces to +/- sin(t) and x = sin(t) except at t = mult pi/2 when you get a jump discontinuity. But, as t--> pi/2, y --> +/- n.
2) Restricting t to 0<= t <= 2pi only gives you a portion of the 2 lines where x and y are trapped in the square between +/- n. And it doesn't change anything if you let t range over all the real numbers. You are still trapped in the square where x and y are stuck between +/- n.
3) So x = n.sin(t) and y = +/- n.sin(t) except when t = mult of pi/2 where y does not have a value. You might as well use x = n and y = +/-n where you get all of the infinite pairs of solutions and no discontinuities.
x^2 - y^2 = 0 describes 2 straight lines: y = x and y = -x
So let x = t and then y = t or y = -t where t is any real number.
This is sometimes referred to an a "degenerate hyperbola".
For the rectangular hyperbola, x^2 - y^2 = a^2, the asymptotes are the 2 lines y = x and y = -x. As a tends to 0, then the branches of the hyperbola itself actually approach the the 2 lines y = x and and y = -x and when a actually becomes 0 then the hyperbola turns into the 2 lines - hence the term "degenerate"
As a tends to infinity nothing really special happens. The hyperbola still exists has as the same 2 asymptotes, y = x and y = -x . This prevents the hyperbola approaching the 2 vertical lines x = a and x = -a. It just takes larger and larger of values of x to get arbitrarily close to to the asymptotes.
The case where x^2 + y^2 = 0 is the simplest case of all. It describes a single point (0,0)
For the circle x^2 + y^2 = a^2, the radius is a and as a tends to 0 the circle shrinks and approaches the center (0,0) and when a actually takes on the value of zero we get just the point (0,0). Hence a single point is sometimes called a "degenerate circle". As a tends to infinity we just get a larger and larger circle.
The curvature of the circle x^2 + y^2 = a^2 is 1/a so I suppose we can argue the as a tends to infinity the curvature tends to 0 - which is the curvature of a straight line but there is no single equation which can represent such a line.
y=t, x=t or y=-t, x=-t.
both of these pairs of equations only pick up the line y=x (i.e. each pair plots the same line) and not the line y=-x and therefore you lose half of the solutions.
The only way to pick up the other half is x = t ; y = -t.
So from above: y=t, x=t or y=-t, x=-t :
You have to change drop one of the - signs to get : y=t, x=t or y=t, x=-t.
Otherwise you lose the line y = -x.
y=at, x=at, a not equal to zero. It is such a easy question. Why create complexity.
Why create complexity
Here's why.
If, for example, you set a = 2 then you get x =2t and y = 2t. Let t range over the real numbers you only get one of the 2 straight lines (y =x). You have to include y = +/- at in order to pick up both branches. Same holds true if a is negative. It's still true that x^2 - y^2 = 0 but with x = at and y = at you only pick up half the solutions --- y=x.
x = at ; y = +/-at is the only way to pick up both lines. The "a" is unnecessary here but it doesn't hurt either. It's just simpler to exclude it.
Also y=-at, x=at, y=at, x=-at, and y=-at, x=-
*x=-at in above
Sorry, I missed the addendum:
Also y=-at, x=at, y=at, x=-at, and y=-at, x=-at ?
y=-at, x=at or y=at,x=at is all that is needed.
The comma is an implied "and" and the "or" is necessary.
Here's a way to do it over the complex plane:
Parameterised form:
x = t^2 / |t|
y = t^3 / (√(t^2) * |t|)
NB: very important - √(t^2), not t because, in effect, it divides by -1 when the arg ≥ π
Derivation:
Make x the subject:
x^2-y^2=0
y^2=-x^2
x=±y
Now we create a method to have 2 sets of points i.e. y=x, y=-x:
t is our parameter.
let r = |t|
let θ = arg(t)
Therefore t = r * cis(θ)
let s = t^2 = r^2 * cis(2θ)
let d = t / √(s)
Therefore d = 1 for 0 ≤ θ < π
-1 for π ≤ θ < 2π
This (using d) gives us a split into 2 sections, which are the same size as the original.
So now, we do the following:
let n = r * cis(2θ) = s / |t|
Now n is the same number at double the angle, which allows us to get any complex number twice from the value of t, which is what we need, it also repeats where the value of d switches from 1 to -1, which is what we need.
let x = n
let y = dn
With this setup, we get every number twice, and then flip the sign for every y value so we get every number paired with the number, and the negated version of this number.
Therefore we have parameterised the relation.
Simplify to only use the pronumeral t:
x = n
= s / |t|
= t^2 / |t|
y = dn
= d * s / |t|
= d * t^2 / |t|
= (t / √(s)) * t^2 / |t|
= t^2 * (t / √(t^2)) / |t|
= t^3 / √(t^2) / |t|
= t^3 / (√(t^2) * |t|)
Therefore, it is possible to parameterise this relation over the complex plane with only one value for x and y for each value of t.