Proving the Golden Ratio for a W-Shaped Quartic Function

[Paragon]

I formed the following statement: A "W"-shaped quartic function f(x) has two points of inflection B and C. A line through the points B, C passes through f(x) again at A and D. The ratio AB:BC:CD simplifies to 1 : \phi : 1. So, AB = CD and \phi = 1.61803399..., also known as the golden number.

Alright, I want to prove that every "W"-shaped quartic satisfies this golden ratio AB:BC:CD. The proof might then be extended to quartics which are not strictly "W"-shaped.

I'm not quite sure where to start. But, I thought of using the fact that:

\phi = \frac{\phi + 1}{\phi}

which can be rewritten as:

x^{2} - x- 1 = 0

which would give \phi as one of the roots.

Note, that the second derivate of a general quartic function:

F(x) = x^{4} + a_{3}x^{3} + a_{2}x^{2} + a_{1}x + a_{0}

would (I believe) be the following quadratic:

F^{''}(x) = 12x^{2} + 6a_{3}x + 2a_{2}

the roots of the quadratic above would give the points of inflection of F(x).

I have been sitting for several hours, trying to find a proof, but I can't even express the AB:BC:CD ratio for a general quartic. Any suggestions would be great!

 

[Doodle Bob]

It looks like it's not quite true as you've stated. From what I can tell, the ratio that ends up as 1:phi:1, is the ratio of the x-values for the points of intersection of the line passing through the two inflection points with the quadric.
Since you know this will only work with quadrics that have two distinct inflections points, I would suggest that you start with the second derivative -- i.e. a quadratic function with two distinct zeros -- and work backwards towards the quadric. Since your conjecture is not affected by translations or reflections through vertical lines, you can assume that one of the inflection points is the origin and is also a zero of the quadric. This will give you a nice form for F(x) in terms of the constants that you needed for the second derivative. Use this to find the equation of the line connecting the two inflection points of F(x), and use that to find the other two points of intersection.

 

[tookoolfoskool]

hey paragon
wud u happen to be in the IB Diploma?

 

[Paragon]

hey paragon
wud u happen to be in the IB Diploma?

Nope

PS: ... and that is not mah meth =)

 

[too_stoned]

that kid is so from IB Math HL

 

[antevante]

Yes, he is really doing his Ib Hl Maths-portfolio!

I solved it in this way:

1. find the general solutions to the second derivative of the quartic.
2.using the factor theorem, divide the original quartic with (x-root 1 of second derivative) and (x-root 2 of second derivative), you can use synthetic division for example.
3. Point out that the second derivative and the "new" quadratic has the same line of symetry, this proves that AB:CD is 1:1
4. The distance between the line of symetry and the "second derivative root" times sqrt(5) is the distance between the line of symetry and the "new" quadratic root. This gives the relationship AB:BC = 1: 2/(sqrt(5)-1)=fi

This argument ought to hold for "not strictly W-shaped" quartics as well...

Good luck!

 

[diction]

Yes, he is really doing his Ib Hl Maths-portfolio!

I solved it in this way:

1. find the general solutions to the second derivative of the quartic.
2.using the factor theorem, divide the original quartic with (x-root 1 of second derivative) and (x-root 2 of second derivative), you can use synthetic division for example.
3. Point out that the second derivative and the "new" quadratic has the same line of symetry, this proves that AB:CD is 1:1
4. The distance between the line of symetry and the "second derivative root" times sqrt(5) is the distance between the line of symetry and the "new" quadratic root. This gives the relationship AB:BC = 1: 2/(sqrt(5)-1)=fi

This argument ought to hold for "not strictly W-shaped" quartics as well...

Good luck!

Can you explain a bit more of step 3, about how the line of symetry gets the ratio? And how do you know to have the root of f"(x) times sqrt(5) is the distance...especially on why you decided to use sqrt(5)? Thank you very much... =)

 

[Paragon]

1. find the general solutions to the second derivative of the quartic.
2.using the factor theorem, divide the original quartic with (x-root 1 of second derivative) and (x-root 2 of second derivative), you can use synthetic division for example.
3. Point out that the second derivative and the "new" quadratic has the same line of symetry, this proves that AB:CD is 1:1
4. The distance between the line of symetry and the "second derivative root" times sqrt(5) is the distance between the line of symetry and the "new" quadratic root. This gives the relationship AB:BC = 1: 2/(sqrt(5)-1)=fi

This is nice. But...

This argument ought to hold for "not strictly W-shaped" quartics as well...

If that is what I think it is, then you proved the desired golden ratio for a general quartic with any two distinct points of inflection (that's that I did). As the ratio have to involve two distict points of inflexion, there is, I think, nothing more to extent. Hence, you probably have done the last two tasks in one 'calculation', which is not what is asked for.

 

[Paragon]

hey paragon
wud u happen to be in the IB Diploma?

...

that kid is so from IB Math HL

...

Yes, he is really doing his Ib Hl Maths-portfolio!

I'm sorry, but I don't speak English...

 

[too_stoned]

the general formula is giving me some trouble...wanna give me a subtle hint? cause i don't fully understand what u explained to paragon

 

[HBar]

Through a strange string of events, half of my class ended up having to do this, and the other half didn't. I didn't, but this thread interested me. Anyway, I've been messing around with this a little, and I found a general formula to create a quartic function based on two points of inflection (POI). I don't know if this helps at all, but here goes.

From the POIs: (x_1, y_1), (x_2, y_2)

-\frac{1}{12}x_1^4 + \frac{1}{3}x_2x_1^3 + dx_1 + e = y_1
-\frac{1}{12}x_2^4 + \frac{1}{3}x_1x_2^3 + dx_2 + e = y_2

Plug in the POIs and solve the system of equations for "d" and "e".
The quartic is then in the form:

\frac{1}{12}x^4 - \frac{x_2+x_3}{6}x^3 + \frac{x_1x_2}{2}x^2 + dx + e = f(x)

The variables "d" and "e" are the ones found by solving the system of equations. The points x_1 and x_2 are simply from the POIs.

That was my first time using LaTex, so please bear with me. If anyone is interested, I can also post a brief explanation of how I found it.

 

[Paragon]

If anyone is interested, I can also post a brief explanation of how I found it.

Alright, I quess you double-integrated some quadratic, but I don't mind you to explain a bit more.

I think that I got the solution by now, but I'm interested in this, too. In particular, look at this page: http://mathworld.wolfram.com/GoldenRectangle.html

Note that this is only my personal theoretic thoughts (copyright), but... As you can see on that page, it is possible to construct a 'golden rectangle' out of a line intersecting a quartic in four distict places. It is also not impossible to construct a 'golden spiral' out of a 'golden rectangle'. Note that the 'golden rectangles' constructed on the 'golden spiral' interscect. But, there's a 'golden function' for the 'golden spiral', so... i believe that there is a connection. I haven't really thought about this, but someone might...

 

[forgetmenot]

hi there!
ok so I am doing my HL portfolio, and those who have done it will probably know that step 5 and 6 are the hardest (for me at least).
5. form a conjecture and formally prove it using a general quartic functions.
6. extend this investigation to other quartics functions that are not strictly of a 'w' shape.

anyone know what to do for those 2 points? I am terribly lost and any feedback would be absolutely awesome.

thank you in advance =)