Quartic Proof?

Question

Hi- I've been working on this for hours, and I've gotten nowhere. If anyone could help, it would be greatly appreciated. A W-shaped quartic function has two points of inflection (call them W and X). The line that intersects W and X also intersects Y and Z, two other points on the quartic function. I have to prove that the ratio YW:WX:XZ is constant (1:1.62:1, to be specific). How would I go about doing that? Thanks!

Answer 1

Without loss of generality, and for convenience, we can first assume a monic quartic x^4 + ax^3 + bx^2 + cx +d = 0.
Then put into reduced form by x = X - a/4 to get
f(y) = X^4 + pX^2 + qX + r = 0.

Take the second derivative to find the inflection pts

f''(X) = 12X^2 + 2p = 0
X = ±√(-p/6)
Now substitute these back into original equation to find the Y values.

X = ±√(-p/6) ; Y = -5p^2/36 +r ± q√(-p/6)

Now find the equation of the line between the two:

Y = qX + r - 5p^2/36

Set this equal to the original equation to find the other points of intersection with the quartic.

qX + r - 5p^2/36 = X^4 + pX^2 + qX + r
or
X^4 + pX^2 + 5p^2/36 = 0
(X^2 + p/6)(X^2 + 5p/6) = 0
**********************************
I really didnt have to go any farther at this point, as the root difference ratio would be the same as the distance between intersection points ratio:
(1 - (-1)):(√5-1) which gives (√5+1)/2 , which is what you were looking for, but here it is anyway:

**********************************
So we find the other two points are X = ±√(-5p/6)
and Y = -5p^2/36 +r ± q√(-5p/6)

Now we have the coordinates of the four points and can use the distance formula to finish.

Let m = √(-p/6) and n = -5p^2/36 +r
The coordinates are then:
(±m, n ± m) for the inflection pts and,
(±m√5, n ±m√5) for the others

D1 between the inflection pts is 2m√2
D2 between an inflection pt and one of the others is 2m√(3-√5)

and D2/D1 = (√5+1)/2 ≈ 1.62 the golden section!!!

Answer 2

Had me for a moment with the "W-shaped", but I see what you mean now. OK, the points of inflection are going to be of the upside down V in the middle of the W.
Let f(x) be the quartic function:
f(x) = ax^4 + bx^3 + cx^2 + dx + e
Take the first two derivatives: so that you know that
f"(W) = f"(X) = 0
Let g(x) be a straight line connection Y and Z, the two points that live on f(x); you will need this at the end for your ratios.
You know that f(x1) =X and f(x2) =W
You know that the top of the W has a slope of 0, ie f' there = 0
You have now five equations and five unknowns and you can solve the unknows, plug back into the general quartic, and use that to find the distance via the straight line for your ratios.
There are at least two other ways to do this, but this is the easiest I can think of.

Answer 3

Ditto to the above. So far I've managed to do some simplifications, translations, scalings, etc to reduce it to the following:
g is any quartic such that g''(1) = g''(-1) = g(1) = g(-1) = 0. Show its roots are in that proportion. If I get it, I'll let you know.

Alright, here we go.

First of all, we can translate the graph so its point of inflections are at x=-a and x=a without changing anything much. Lets call that f(x).
The equation of the line through those points is:
y - f(a) = (f(a)-f(-a))/(2a) * (x-a)
We want to find where that intersects y=f(x). So we want
f(x) - f(a) = (f(a)-f(-a))/(2a) * (x-a)
Lets put everything on the LHS and call that g(x); so:
g(x) = f(x) - f(a) - (f(a)-f(-a))/(2a) * (x-a).
Now, note that g''(x) = f''(x), and we have reduced the problem to what I said above; namely that given a quartic g(x) with g(a) = g(-a) = g''(a) = g''(-a) = 0, show that the differences between the roots are in that proportion.
Now, we can apply one more simplification - since we want ratios, we can shrink the x axis by a factor of a and assume a = 1.
Now, g''(1) = g''(-1) = 0, so g''(x) = A(x^2-1), or Ax^2 - A.
Integrate that twice:
g(x) = A/12 x^4 - A/2 x^2 + cx + d.
Just to simplify, I'll let B = A/12
g(x) = Bx^4 - 6Bx^2 + cx + d.
Now, sub in 1 and -1. You get two equations equal to 0 which only differ in the cx term, so c = 0.
Also, g(1) = B-6B+D, so D = 5B.
So g(x) = Bx^4 - 6Bx^2 + 5B = B(x^2-5)(x^2-1).
We want that to equal 0, so the roots are plus or minus rt5 and plus or minus 1 (which we knew about).
Thus the required lengths are (-1+rt5), 2, (rt5-1).
Dividing all 3 by (-1+rt5) gives 1:2/(rt5 - 1):1, which is precisely what you want :)
Whew!

Answer 4

No answers after 12 minutes? I'll signal you that I'm thinking about it and post progress later. My idea is this: For a general quartic (to make it easier, make it monic), write the quadratic equation whose roots give the inflection points. I'm thinking it may be possible to write the equation of WX without actually solving this (sum & prod of roots?), then do something else with sum & prod of roots you get when solving this line's intersection with the quartic. Already tried it? I will anyway.

Answer 5

let f(x) = ax^4 + bx³ + cx² + dx + k

f'(x) = 4ax³ + 3bx² + 2cx + d
f"(x) = 12ax² + 6bx + 2c
= 0 for points of inflection

Thus 12ax² + 6bx + 2c = 0
ie 6ax² + 3bx + c = 0
So x = (-3b ± √[9b² - 24ac])/(12a) with 9b² - 24ac > 0 for real and distinct roots ie for real and distinct points of inflexion

Let the roots be α and β with α > β
Note α + β = -3b/6a = -b/2a and αβ = c/6a

So y1 = aα^4 + bα³ + cα² + dα + k
and y2 = aβ^4 + βα³ + cβ² + dβ + k

ie W ≡(β, aβ^4 + bβ³ + cβ² + dβ + k) and X ≡(α, aα^4 + bα³ + cα² + dα + k)

Slope WX = Δy/Δx
= [(aα^4 + bα³ + cα² + dα + k) - (aβ^4 + bβ³ + cβ² + dβ + k)/(α - β)
= [a(α^4 - β^4) + b(α³ - β³) + c(α² - β²)+ d(α - β)]/(α - β)
= a(α³ + α²β + αβ² + β³] + b(α² + αβ + β²) + c(α + β) + d
= a(α³ + 3α²β + 3αβ² + β³ - 2α²β - 2αβ²) + b(α² + 2αβ + β² - αβ) + c + d
= a((α + β)³ - 2αβ(α + β)) + b((α + β)² - αβ) + c(α + β) + d
= a(b/2a)³ + 2c/6a * b/2a) + b((b/2a)² - c/6a) + c.c/6a + d
= b³/8a² + bc/6a² + b²/4a² - bc/6a + c²/6a + d
= (3b³ + 4bc + 6b² - 4abc + 4ac² + 24a²d)/(24a²)

So [y - (aα^4 + bα³ + cα² + dα + k)]/(x - α)
= (3b³ + 4bc + 6b² - 4abc + 4ac² + 24a²d)/(24a²)
so y - (aα^4 + bα³ + cα² + dα + k)
= (x - α)(3b³ + 4bc + 6b² - 4abc + 4ac² + 24a²d)/(24a²)
= (3b³ + 4bc + 6b² - 4abc + 4ac² + 24a²d)/(24a²) * x
- α(3b³ + 4bc + 6b² - 4abc + 4ac² + 24a²d)/(24a²)

So y = (3b³ + 4bc + 6b² - 4abc + 4ac² + 24a²d)/(24a²) * x
- α(3b³ + 4bc + 6b² - 4abc + 4ac² + 24a²d)/(24a²)
+ (aα^4 + bα³ + cα² + dα + k)

Now find Y and Z and you are there!! Yuck!!!!!

I like what Scott is doing ... MUCH neater