THE NEW THEORY.
When I first realised how the graphs of y = xn behaved, I was so excited and
in such a frenzy to get the Autograph versions in a PowerPoint that I
obviously rushed the necessary detail of the theory too much.
I will endeavour to make a better job of it now!
The problem at first was the change from real y values and complex x values to
real x values and complex y values but on reflection it has helped me
understand more thoroughly how it all works.
Another problem with the new theory was that when we square an equation we
introduce an extra solution which is not a solution of the original equation.
eg
If √(x + 2) = x – 4
then x + 2 = x2 – 8x + 16
so
0 = x2 – 9x + 14
0 = (x – 7)(x – 2)
Hence x = 7 or 2
BUT if we test x = 2 in the original equation we get:
Lhs = √4 = 2 but Rhs = 2 – 4 = – 2
So the “solution” x = 2 is not a solution of the original equation but it is a
solution of the squared equation!
The only valid solution of the original equation is x = 7
Check: Lhs = √9 = 3 Rhs = 7 – 4 = 3
__________________________________________________________
Consider Y = x0.5 or x ½
When x < 0 there are no real Y values (ie only complex Y values) so if we
express the equation as Y = y + iz = x0.5 where z values are the complex part of
the Y values, we can proceed as follows: y + iz = x0.5
Square the equation: (y + iz)2 = x1
y2 – z2 + 2yzi = x1
(N.B. We need x to be the subject of the equation)
The x values are to be real numbers < 0 so 2yzi = 0
which implies that either z = 0 or y = 0
If z = 0, the equation just becomes the original y2 = x or y = x0.5
If y = 0, the equation becomes (-z)2 = x1 or x = z2
In Autograph, the basic curve for x ≥ 0 is entered as:
y = t^0.5, x = t, z = 0
The “phantom y value graph” for x ≤ 0 is entered as:
x = t^(2), z = t, y = 0
The graph in 3D is as shown.
To help me check the validity of what I was
doing I used my graphics calculator to find
the Y value when x = – 1
Y = (– 1)0.5 = 0 + i
P
See point P in the diagram on the left
This graph is remarkably similar to the graph of y = x2 and its phantom in the
“normal” phantom graph theory for real y values.
NORMAL ORIENTATION
The difference of course is that x = y2 is a full parabola but y = x ½ is only the
top half of the parabola since the power ½ does not mean y = ±√x
It is more obvious when rotated.
Q
Q
P
P
If we use y = x ½ and calculate y when x = – 1 we get one solution y = +i as in the
graph on the left. The graph on the right would imply that y = +i or – i
The extra solution appears because we squared the original equation.
Consider Y = x0.25 or x ¼
When x < 0 there are no real Y values (ie only complex Y values) so if we
express the equation as Y = y + iz = x0.25 where z values are the complex part
of the Y values, we can proceed as follows: y + iz = x0.25
Square the equation: (y + iz)2 = x0.5
(y2 – z2) + 2yzi = x0.5
Squaring again to get an equation in the form x = f(y,z)
so (y2 – z2)2 – 4y2z2 + i4yz(y2 – z2) = x
(N.B. We need x to be the subject of the equation for autograph)
If z = 0, the equation just becomes the original y4 = x or y = x0.25
If y = 0, the equation becomes z4 = x so x = z0.25
If y = ± z, the equation becomes -4z4 = x
By squaring the equation TWICE we introduce more non valid solutions!
The situation is very similar to the case of y = x4 in the “normal” theory of
Phantom Graphs.
Y = x4
Rotated version of y = x4
P
The graph of Y = x0.25 is below:
P
Q
By squaring the equation twice we introduced a
Q lot of extra parts to the graph.
Testing Y = (– 1)0.25 = 0.7 + 0.7 i
The graph above has 4 values for x = -1
(see black dots above)
The graph on the left shows the only valid parts
of the graph where P is at the point
x = - 1, Y = 0.7 + 0.7i
Autograph equations for red curve are
y = t^.25, x = t, z = 0
Autograph equations for purple phantom are
y = t, z = t, x = −4t⁴
Consider Y = x0.75 or x ¾
When x < 0 there are no real Y values (ie only complex Y values) so if we
express the equation as Y = y + iz = x0.75 where z values are the complex part
of the Y values, we can proceed as follows: y + iz = x0.75
Square the equation: (y + iz)2 = x1.5
(y2 – z2) + 2yzi = x1.5
Squaring again to get an equation in the form x = f(y,z)
so (y2 – z2)2 – 4y2z2 + i4yz(y2 – z2) = x3
(N.B. We need x to be the subject of the equation for autograph)
If z = 0, the equation just becomes the original y4 = x3 or y = x0.75
If y = 0, the equation becomes z4 = x3 so x = z0.75
If y = ± z, the equation becomes -4z4 = x3 so x = (-4) 1/3 z4/3
Checking the Y value when x = -1 we get Y = (-1).75 = -0.7 + 0.7i
As before this graph will have 1 branch for x > 0 and one phantom branch for
x<0
Autograph equations for red curve are
y = t^.75, x=t, z=0
Autograph equations for purple
phantom are
y = −t, z = t, x = −(4)^(1/3)t^(4/3)
Whenever there is a choice of ± , we
check Y = (-1)^0.75 = -0.7 + 0.7i
This check shows that the phantom we
obtained is the one that goes through
the point x = -1, Y = -0.7 + 0.7i as on
diagram left.
Shown in the correct sequence:
Y = x0.25
Y = x0.5
Y = x0.75
Y = x1
Consider Y = x1.5
Y = y + iz = x1.5 where z values are the complex part of the Y values, we can
proceed as follows: y + iz = x1.5
Square the equation: (y + iz)2 = x3
y2 – z2 + 2yzi = x3
or (y2 – z2 + 2yzi)1/3= x
The x values are real numbers < 0 so 2yzi = 0
which implies that either z = 0 or y = 0
If z = 0, the equation just becomes the original y2/3= x or y = x1.5
If y = 0, the equation for x becomes –z2 = x3 or x = -z(2/3)
Autograph equations for the red curve are:
y = t^1.5, x = t, z = 0
Autograph equations for the purple phantom
x = −t^(2/3), z = t, y = 0
Whenever there is a choice of ± , we check
Y = (-1)1.5 = 0 – i
This check shows that the phantom we
obtained is the one that goes through the point
x = -1, Y = 0 – i as on diagram left.
Consider Y = x1.25
Y = y + iz = x1.25 where z values are the complex part of the Y values, we can
proceed as follows: y + iz = x1.25
Square the equation: (y + iz)2 = x2.5
(y2 – z2) + 2yzi = x2.5
Squaring again to get an equation in the form x = f(y,z)
so (y2 – z2)2 – 4y2z2 + i4yz(y2 – z2) = x5
(N.B. We need x to be the subject of the equation for autograph)
If z = 0, the equation just becomes the original y4 = x5 or y = x1.25
If y = 0, the equation becomes z4 = x5 so x = z1.25
𝟏
𝟒
If y = ± z, the equation becomes -4z4 = x5 or 𝒙 = −𝟒𝟓 𝒛𝟓
Autograph equations for the red curve are:
y = t^1.25, x = t, z = 0
Autograph equations for the purple phantom
x = −(4)^(1/5)t^(4/5), z = t, y = t
Whenever there is a choice of ± , we check
Y = (-1)1.25 = –0.7 – 0.7 i
This check shows that the phantom we obtained
is the one that goes through the point
x = -1, Y = –0.7 – 0.7 i as on diagram left.
Consider Y = x1.75
Y = y + iz = x1.75 where z values are the complex part of the Y values, we can
proceed as follows: y + iz = x1.75
Square the equation: (y + iz)2 = x3.5
(y2 – z2) + 2yzi = x3.5
Squaring again to get an equation in the form x = f(y,z)
so (y2 – z2)2 – 4y2z2 + i4yz(y2 – z2) = x7
(N.B. We need x to be the subject of the equation for autograph)
If z = 0, the equation just becomes the original y4 = x7 or y = x1.75
If y = 0, the equation becomes z4 = x7 so x = z1.75
𝟏
𝟒
If y = ± z, the equation becomes -4z4 = x7 so x = -𝟒𝟕 𝒛𝟕
Autograph equations for the red curve are:
y = t^1.75, x = t, z = 0
Autograph equations for the purple phantom
x = −(4)^(1/7)t^(4/7), z = t, y = t
Whenever there is a choice of ± , we check
Y = (-1)1.75 = +0.7 – 0.7 i
This check shows that the phantom we
obtained is the one that goes through the point
x = -1, Y = +0.7 – 0.7 i as on diagram left.
Shown in the correct sequence:
Y = x1.25
Y = x1.5
Y = x1.75
Y = x2
Consider Y = x2.5
Y = y + iz = x2.5 where z values are the complex part of the Y values, we can
proceed as follows: y + iz = x2.5
Square the equation: (y + iz)2 = x5
y2 – z2 + 2yzi = x5
or (y2 – z2 + 2yzi)1/5 = x
The x values are real numbers < 0 so 2yzi = 0
which implies that either z = 0 or y = 0
If z = 0, the equation just becomes the original y2/5 = x or y = x2.5
If y = 0, the equation for x becomes –z2 = x5 so that x = -z(2/5)
Autograph equations for the red curve are:
y = t^2. 5, x = t, z = 0
Autograph equations for the purple phantom
x = t^(2/5), z = t, y = t
Whenever there is a choice of ± , we check
Y = (-1)2.5 = 0 + i
This check shows that the phantom we obtained
is the one that goes through the point
x = -1, Y = 0 + i as on diagram left.
Consider Y = x2.25
Y = x2.25 we raise the equation to the power 4
(y + iz)4 = x9
(y2 – z2 + 2yzi)2 = x9
so (y2 – z2)2 – 4y2z2 + i4yz(y2 – z2) = x9
The x values are real numbers < 0 so i4yz(y2 – z2) = 0
Which implies that either z = 0 or y = 0 or y = ± z
If z = 0, the equation just becomes the original y4 = x9 or y = x2.25
If y = 0, the equation becomes z4 = x9 or z = x2.25
If y = ± z, the equation becomes -4z4 = x9 so x = -4(1/9) z(4/9)
Autograph equations for the red curve are:
y = t^2. 25, x = t, z = 0
Autograph equations for the purple phantom
x = −(4)^(1/9)t^(4/9), z = t, y = t
Whenever there is a choice of ± , we check
Y = (– 1)2.25 = 0.7 + 0.7i
This check shows that the phantom we obtained
is the one that goes through the point
x = -1, Y = +0.7 + 0.7 i as on diagram left.
Consider Y = x2.75
Similarly for Y = x2.75 we raise the equation to the power 4
(y + iz)4 = x11
(y2 – z2 + 2yzi)2 = x11
so (y2 – z2)2 – 4y2z2 + i4yz(y2 – z2) = x11
The x values are real numbers < 0 so i4yz(y2 – z2) = 0
Which implies that either z = 0 or y = 0 or y = ± z
(N.B. In this case I chose y = – z and so y = – t because (-1)2.25 = 0.7 – 0 .7i
and squaring the equation introduces extra solutions.)
If z = 0, the equation just becomes the original y4 = x11 or y = x2.25
If y = 0, the equation becomes z4 = x11 so x = z(4/11)
If y = ± z, the equation becomes -4z4 = x11so x = -4(1/11) z(4/11)
Autograph equations for the red curve are:
y = t^2. 75, x = t, z = 0
Autograph equations for the purple phantom
x = −(4)^(1/11)t^(4/11), z = t, y = - t
Whenever there is a choice of ± , we check:
Y = (– 1)2.25 = +0.7 – 0.7i
This check shows that the phantom we chose is
the one that goes through the point
x = -1, Y = +0.7 – 0.7 i as on diagram left.
Shown in the correct sequence:
Y = x2.25
Y = x2.5
Y = x2.75
Y = x3
SIMILARLY:
The Autograph equations for:
Y = x3.25 are:
y = t^3.25, x = t, z = 0
x = −(4)^(1/13)t^(4/13), z = −t, y = −t
Whenever there is a choice of ± , we check
Y = (– 1)3.25 = -0.7 – 0.7i
This check shows that the phantom we
chose is the one that goes through the point
x = -1, Y = -0.7 – 0.7 i as on diagram right.
________________________________________________________________
The Autograph equations for:
Y = x3.5 are:
y = t^3.5, x = t, z = 0
x = −t^(2/7), z = −t, y = 0
Whenever there is a choice of ± , we check
Y = (– 1)3.5 = 0 – i
This check shows that the phantom we
chose is the one that goes through the point
x = -1, Y = 0 – i as on diagram right.
________________________________________________________________
The Autograph equations for:
Y = x3.75 are:
y = t^3.75, x = t, z = 0
x = −(4)^(1/13)t^(4/13), z = −t, y = +t
Whenever there is a choice of ± , we check
Y = (– 1)3.75 = +0.7 – 0.7i
This check shows that the phantom we
chose is the one that goes through the point
x = -1, Y = +0.7 – 0.7 i as on diagram right.
The Autograph equations for:
Y = x4.25 are:
y = t^4.25, x = t, z = 0
x = −(4)^(1/17)t^(4/17), z = +t, y = +t
Whenever there is a choice of ± , we check
Y = (– 1)4.25 = +0.7 + 0.7i
This check shows that the phantom we
chose is the one that goes through the point
x = -1, Y = +0.7 + 0.7 i as on diagram right.
________________________________________________________________
The Autograph equations for:
Y = x4.5 are:
y = t^4.5, x = t, z = 0
x = −t^(2/9), z = +t, y = 0
Whenever there is a choice of ± , we check
Y = (– 1)4.5 = 0 + i
This check shows that the phantom we
chose is the one that goes through the point
x = -1, Y = 0 + i as on diagram right.
________________________________________________________________
The Autograph equations for:
Y = x4.75 are:
y = t^4.75, x = t, z = 0
x = −(4)^(1/17)t^(4/17), z = +t, y = -t
Whenever there is a choice of ± , we check
Y = (– 1)4.75 = -0.7 + 0.7i
This check shows that the phantom we
chose is the one that goes through the point
x = -1, Y = -0.7 + 0.7 i as on diagram right.