Since, the general function passes through points M(0,0) and N(a,0), we could substitute them into the function to find out the values of the two constants C and D. By substituting M(0,0) into the general function, we get:
(0) 4 a(0) 3 (0) C ( x) D 12 6 D=0
By substituting N(a, 0 ) into the general function, we get:
(a) 4 a(a) 3 (0) C (a) 0 12 6
a3 C= 12 By substituting the values of the constants into the general function, we get
x 4 ax 3 a 3 x 12 6 12 4 3 y x 2ax a 3 x y
The line that passes through the two points of inflection as well as cuts the function at two points is y = 0. In order to find the coordinates of the two intersection points L and O of line MN and
y x 4 2ax a 3 x the function, we have to solve the equations y0
By substituting the two equations in each other, we get: 0 = x4 – 2ax + a3x Since we know that two of the roots of the equation has to be 0 and a as M and N are intersections. Hence, (x-a) and (x-0) are factors of the equation. We have: 0 = (x – 0)(x – a)(x2 – ax – a2) And to find the x-coordinate of L and O, we have to solve the equation x2 – ax – a2=0 x2 – ax – a2=0
x
a a 2 4(a 2) 2 aa 5 x 2
x a (1 5 ) 2
L
y
N (a,0) M (0,0) O
x
Dia. 1 A sketch of the general quartic function.
Since points L, M, N and O lies on y = 0, the y-coordinate of all four points must be 0.
a(1 5 ) a(1 a 5 ) ,0) and O is ( ,0) 2 2 Now, we would have to calculate the distance of the three line segments LM, MN and
Hence, L is (
NO. Since distance Using the same method used above, the distance of LM =| (
a(1 5 ) 0) 2 (0 0) 2 | 2
a(1 5 ) | 2 Similarly, the distance of MN
=|
= | (a 0) 2 (0 0) 2 | = |a| Lastly, the distance of NO =| (
a(1 5 ) a) 2 (0 0) 2 | 2 a(1 5 ) 2 ) | 2
= | (
=|
a(1 5 ) | 2
a(1 5 ) a(1 5 ) |: |a| : | | 2 2 2 The ratio can be simplified to 1: | |:1 1 5
The ratio LM:MN:NO is therefore |
1 5 |: 1 2 Thus, we proved that our conjecture that a line passing through the two points of
By rationalization, we get 1:
inflection in a ‘W’ shaped quartic function which meets the function again at two points, the distance between the three line segments would be a fixed ratio namely the Golden Ratio . After know that our conjecture applies to ‘W’ shaped quartic functions, we will test it against other deformed quartic equations.
Quartic function could exist in a few common forms, ‘W’ shaped, ‘M’ shaped or ‘U’ shaped. Now let’s try another ‘M’ shaped quartic function h(x) = -8x4 - 20x3 - 12x2 + x + 15 to see if this ratio occurs again. This time, we define the two inflection points of the function as U and V and the two points that is projected along UV that meets the function again called as T and W. Following the same steps as above, we first take the derivative of the function, we get：
dy = -32x3 – 60x2 - 24x + 1 dx Then its second derivative, we get:
d2y = -96x2 - 120x - 24 dx Let the second derivative be zero: d2y =0 dx -96x2 - 120x - 24 = 0
x = -1 or x = -
1 4 1 . To get the coordinates of the 4
The inflection point occurs when x equals to -1 or -
two inflection points, we have to find the corresponding values of y in each point. This could be achieved by substituting corresponding values of x into h(x):
1 x4 x 1 or ,values corrected to 6 decimal places 457 y 14 y 32
With the coordinates of the two inflection points, we could use the two-point form to find the equation of line TUVW.
457 14 y 14 32 1 x (1) (1) 4
3 21 9 9 y= x+ 4 2 32 32
y=
1 (3x + 115) 8
We therefore get the equation of the line TUVW that passes through the two inflection points (U and V) and cuts the function at two other points (T and W) to be y=
1 (3x + 115). 8
Similarly, we would have to find out the coordinated of P’ and S’ as well as the distance of P’Q’, Q’R’ and R’S’. The coordinates of intersection P’ and S’ given by the software are (,) and (,) respectively and corrected to 6 decimal places. Using the same method mentioned above, the distance of P’Q’ = (0.236068- 1) 2 (27.944272- 23) 2 = 5.096439, cor. to 6 d.p. Similarly, Q’R’ =
(1 3) 2 (23 - 15)2
= 8.246211, cor to 6 d.p. Lastly, R’S’ =
2 (3 4.236068) 2 (15 - 10.055728)
= 5.096439, cor to 6 d.p. Thus, the ratio of P’Q’:Q’R’:R’S’ is 5.096439: 8.346211: 5.096439. By dividing the ratio by 5.096439, it becomes 1:1.618034:1 Besides ‘M’ shaped, quartic functions could also be a deforms U shape. For example, j(x) = x4 + 2x3 - 8x + 3 is a deformed U shape which looks like this: We could test to see if our conjecture works for this case or not. This time, we define the two inflection points of the function as U’ and V’ and the two points that is projected along U’V’ that meets the function again called as T’ and W’. Following the same steps as above, we first take the derivative of the function, we get：
dy = 4x3 + 6x2 - 8 dx
Then its second derivative, we get:
d2y = 4x2 + 6x dx Let the second derivative be zero: d2y =0 dx 4x2 + 6x = 0
x = 0 or x = -
3 2 3 . To get the coordinates of the 2
The inflection point occurs when x equals to 0 or -
two inflection points, we have to find the corresponding values of y in each point. This could be achieved by substituting corresponding values of x into j(x):
3 x2 x 0 or ,values corrected to 6 decimal places 213 y 3 y 16
With the coordinates of the two inflection points, we could use the two-point form to find the equation of line T’U’V’W’.
213 3 y 3 16 3 x0 0 2
-
3 9 165 y+ = x 2 2 16
y=
1 (55x + 24) 8
We therefore get the equation of the line T’U’V’W’ that passes through the two inflection points (U’ and V’) and cuts the function at two other points (T’ and W’) to be y =
1 (55x + 24) 8
Similarly, we would have to find out the coordinated of T’ and W’ as well as the distance of T’U’, U’V’ and V’W’. The coordinates of intersection T’ and W’ given by the software are (,) and (,) respectively and corrected to 6 decimal places.
Using the same method mentioned above, the distance of P’Q’ = (0.236068- 1) 2 (27.944272- 23) 2 = 5.096439, cor. to 6 d.p. Similarly, Q’R’ =
(1 3) 2 (23 - 15)2
= 8.246211, cor to 6 d.p. Lastly, R’S’ =
2 (3 4.236068) 2 (15 - 10.055728)
= 5.096439, cor to 6 d.p. Thus, the ratio of P’Q’:Q’R’:R’S’ is 5.096439: 8.346211: 5.096439. By dividing the ratio by 5.096439, it becomes 1:1.618034:1 From the above, we could conclude that …