Graphs of the form y = x n for n ≥ 1

Graphs of the form y = x n all go through the point (1, 1) The even powers, y = x 2, x 4, x 6 … also go through ( – 1, 1) and the odd powers, y = x 1, x 3, x 5 … go through ( – 1, – 1) So I will use – 1 ≤ x ≤ 1 as the domain for each graph

y = x 1

y = x 2

y = x 3

y = x 4

y = x 5

However, if we increase the power n in steps of 0.1 a strange thing happens. When n = 1, 2, 3, 4, 5… we get the familiar curves as we just saw but when n takes the values 1.1, 1.2, 1.3, …..up to 1.9 the left hand side of the curve vanishes! The same thing happens for n = 2.1 to 2.9 and again for n = 3.1 to 3.9 etc The left hand side of the curve appears to vanish for non whole number values of n because it does not exist in 2 dimensional space!

y = x 1

y = x 1.5 ?

y = x 2

y = x 2.5 ?

y = x 3

y = x 3.5 ?

y = x 4

y = x 4.5 ?

y = x 5

This is different from the normal “phantom graph” theory where we have REAL y values and a COMPLEX x plane

In this situation we have REAL x values and a COMPLEX y plane.

y = x 1

y = x 1.25

y = x 1.5

y = x 1.75

y = x 2

y = x 2.25

y = x 2.5

y = x 2.75

y = x 3

y = x 3.25

y = x 3.5

y = x 3.75

y = x 4

y = x 4.25

y = x 4.5

y = x 4.75

y = x 5

THEORY. Consider y = x 2.5 When x < 0 there are no real y values (ie only complex y values) so if we express the equation as y + iz = x 2.5 where z values are in the complex y plane, we can proceed as follows: y + iz = x 2.5 Square the equation: (y + iz) 2 = x 5 y 2 – z 2 + 2yzi = x 5 or (y 2 – z 2 + 2yzi) 1/5 = x The x values are real numbers < 0 so 2 yz i = 0 which implies that either z = 0 or y = 0 If z = 0, the equation just becomes the original y = x 2.5 If y = 0, the equation for x becomes –z 2 = x 5 So that x = -z (2/5) In Autograph, the basic curve for x ≥ 0 is entered as: y = t^2.5, x = t, z = 0 The “phantom y value graph” for x ≤ 0 is entered as: x = −t^(2/5), z = t, y = 0

Similarly for y = x 2.25 we raise the equation to the power 4 (y + iz) 4 = x 9 (y 2 – z 2 + 2yzi) 2 = x 9 so (y 2 – z 2 ) 2 – 4y 2 z 2 + i4yz(y 2 – z 2 ) = x 9 The x values are real numbers < 0 so i4yz(y 2 – z 2 ) = 0 Which implies that either z = 0 or y = 0 or y = ± z When squaring (or quadrupling) equations we often introduce unwanted extra solutions so we have to check which are appropriate. I checked using x = – 1 so (– 1) 2.25 = i If z = 0, the equation just becomes the original y = x 2.25 If y = 0, the equation becomes -4z 4 = x 9 so x = -4 (1/9) z (4/9) In Autograph, the basic curve for x ≥ 0 is entered as: y = t^2.25, x = t, z = 0 The “phantom y value graph” for x ≤ 0 is entered as: x = −(4)^(1/9)t^(4/9), z = t, y = t N.B. In this case I chose y = + z and y = + t because (-1) 2.25 = i and squaring the equation introduces extra solutions.

Similarly for y = x 2.75 we raise the equation to the power 4 (y + iz) 4 = x 11 (y 2 – z 2 + 2yzi) 2 = x 11 so (y 2 – z 2 ) 2 – 4y 2 z 2 + i4yz(y 2 – z 2 ) = x 11 The x values are real numbers < 0 so i4yz(y 2 – z 2 ) = 0 Which implies that either z = 0 or y = 0 or y = ± z When squaring or quadrupling equations we often introduce unwanted extra solutions so we have to check which are appropriate. I checked using x = – 1 so (– 1) 2.25 = 0.7 – 0.7i If z = 0, the equation just becomes the original y = x 2.25 If y = 0, the equation becomes -4z 4 = x 11 so x = -4 (1/11) z (4/11) In Autograph, the basic curve for x ≥ 0 is entered as: y = t^2.75, x = t, z = 0 The “phantom y value graph” for x ≤ 0 is entered as: x = −(4)^(1/11)t^(4/11), z = t, y = -t (N.B. In this case I chose y = – z and y = – t because (-1) 2.25 = 0.7 – 0.7i and squaring the equation introduces extra solutions.)