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Monte314

Math: Off on a Tangent

26 posts in this topic

32tangency_zpsthk3jfed.png

(Ugly Text Version)

Spoiler

Is it possible for a polynomial of degree 2 (a parabola) to be tangent at two distinct point to a polynomial of degree 3?
(It is assumed that the leading coefficients of the polynomials are *not* zero)

For two curves y=f(x) and y=g(x) to be tangent at x=x1, the curves must intersect there and have a common tangent line there.  Analytically, this means that both the functions and their derivatives are equal at x1. 

 

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Posted (edited)

I believe this is the kind of question which requires several takes.  But I have got thus far:

Spoiler

Given any cubic polynomial, let's pick any random point on the curve.  On that point, we have a fixed point and the corresponding slope of tangent at that point.  It turns out that if we specify a point plus the slope of tangent which is nonzero at that point, you can only have one single possible quadratic equation which fits such criteria.  (Proof will be given after I have figured out the full solution.)

My hunch says this is not possible if the point has nonzero slope, as you see that is a lack of flexibility to adjust the quadratic equation to our own liking.  We are forced to adjust the cubic equation in this case.  But I think the problem gets very complicated and I will need more time to tidy my thoughts up.

If the point chosen has slope zero, it seems to give the quadratic equation a little bit more flexibility.  But my hunch keeps saying no.

 

Who knows.  I'm a cat.  :meow:

Hope these help.

Edited by K27

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The problem allows you to change both the quadratic and the cubic, as well as the points of tangency.  There are, therefore, nine degrees of freedom: a, b, c, d, r, s, t, x1, and x2; and four non-linear equations.

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Yes, but we have to start somewhere, right?  If I hold all those variables open, I will have too many information to process.

Again, I'm still in the stage of playing with concrete examples.  So no pattern has emerged yet.  :)

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This is what I did, too.  But I wrote a piece of code to generate test cases using a Monte Carlo technique.

Once I was pretty sure I knew what the answer was, I made some simplifying assumptions, and cranked out the proof.

HINT:  If there is a solution, I can rigidly move it around in the plane.  So, WLOG, I can make one of the points of tangency the origin, can't I?  Hmmm?

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Spoiler

It's possible if all the tangent points are the same point, otherwise not.  E.g., Q(x) = x^3, P(x) = x^2

One way of conceptualizing the problem to see all cases is to look at Q(x) - P(x). That'll give you a graph of how Q and P vary together. Q - P is just a cubic, and so it has either 3 real roots, or 1 real root and two imaginary roots.

If Q and P can have two points of being tangent, then Q-P must have two points of being tangent to the x-axis. In other words, we've just done a simple mathematical mapping, where Q becomes Q-P and P becomes P-P, and P-P = 0, or the x-axis.

The tangent points, therefore must be the inflection points of Q-P. But there exist only TWO inflection points (points where the slope of Q-P is zero) because the derivative of Q-P is a quadratic with exactly two roots.

Now, I can do a lot of specific algebraic argumentation at this point, or I can hand-wave the rest of it. Well, I'm just going to hand-wave, because I'm going for conveying conceptual understanding and not proving things absolutely. Each inflection point changes the direction of Q-P, e.g., it's going up, hits inflection point and starts going down, then hits other inflection point and goes back up. Or it could be down->up->down. That means that if the inflection points are different points, then only ONE of them can be tangent to the x-axis at a time. The only time that both inflection points can be on the x-axis at the same time is if they are the same point. In other words, only if there is no down->up->down or up->down->up of the Q-P curve.

Anyway, that's what the behavior looks like. The logic for Q-P implies that when you add P back in and compare Q to P, Q can only be tangent to P at a single point, unless both possible tangent points are the same point.

 

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Physics Dragon jndiii has provided a correct analysis.

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Posted (edited)

Spoiler

I approached this by looking at Q(x) - P(x) and Q'(x) - P'(x).  The conditions imply that tangent points need to lie on the X axis in the graph of the former and need to be an inflection point of the former as well.  Since the former is a cubic, I am leaning toward not being able to get two distinct points out of this, but before I get to any more proof of that, I have been interrupted for some rather unimportant "real" work.

 

Edited by Warrior

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Warrior, your approach is one the several that will work.

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Posted (edited)

After two nights of better sleep, I realised that

Spoiler

You can always find a normal cubic looking curve (which has two turning points), and try to move around a quadratic until they touch at two points.  It's like placing a bowl over a wavy-shaped table, and there will be two contact points.  It is doable.

I simply have no idea how to construct a convincing proof.  I'll continue thinking.

 
 
...... added to this post 21 minutes later:
 

Erm.

Spoiler

I got into result that tells me that's actually impossible.  I'll explain tomorrow after I sleep.

 

Edited by K27

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Here, I would like to introduce one of the technique which I love so much in maths, and the reason why it's my favourite tool — proof by contradiction.

Spoiler

I mean, you can start off from a totally wrong premise, even if you truly believed so.  Then, you faithfully do all your logical deductions ... and arrive at some unresolvable contradiction that may put the whole universe in whack.  No sweat, you can write a final sentence to save your face:

"Since we arrive at contradiction, here we have proven that the assumption is wrong, and therefore the opposite of the statement is true ... QED."

Neat, eh?  And you can tell people that's what you have always thought to turn out.  No one knows.

Spoiler

Shhhhh.

 

So here comes my proof, with the help of Monte's hint:

Spoiler

I assumed such quadratic and cubic polynomial exist.  So we have two equations.  Since we can translate the curves in the plane WLOG, just move one of the intersecting point to the origin.  Why is this convenient?  Because for the cubic, P(0) = 0.  And for the quadratic, Q(0) = 0.  Which yields:

P(x) = ax^3 + bx^2 + cx, Q(x) = rx^2 + sx

Then, we take derivative.  From the quadratic, we know that Q'(x) = 2rx + s.  At the origin, Q'(0) = s.  Doing the same thing on the cubic, we get P'(0) = 3a(0)^2 + 2b(0) + c = c.  Since they should agree at the origin, we have c = s, or s = c.  Symmetry is a good thing.

Now update what we've got:

P(x) = ax^3 + bx^2 + sx
Q(x) = rx^2 + sx
P'(x) = 3ax^2 + 2bx + s
Q'(x) =2 rx + s

Since they meet at another point again, with the same slope, say at x = x' where x' ≠ 0, so we will have P(x') = Q(x') and P'(x') = Q'(x').

Skipping the algebra, the first pair will leads to x' = (r-b)/a, and the second leads to x' = 2(r-b)/(3a).  Since they should agree, so (r-b)/a = 2(r-b)/(3a).  This gives a(r-b) = 0 after some algebra.  Q(x) is still a cubic polynomial, and it simply means a ≠ 0.  This leads to the inevitable conclusion that r=b.

What's the problem?  Plugging in back into x', we will get x'=0.  So we get to a contradiction.  This leads to the conclusion that a quadratic curve is never going to tangent to a cubic curve at 2 distinct points.

But referring to a remark I made earlier:

Spoiler

What about the "bowl" you place on a cubic wave sort of table that looks very feasible in one's own imagination?  I would now argue, since the impossibility is so clearly shown in the mathematical proof, that there is something to the curvature of a cubic curve that makes your quadratic bowl just avoids contact with the local maximum as you slides the quadratic curve down while touching against the arm of the cubic which extends to the positive infinity.  And if you pass the local maximum without touching it, there is no reason you are going to touch another point in the curve while you try to move your quadratic equation around.  It's the curvature that does the trick.

 

Edited by K27

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17 minutes ago, K27 said:

P'(x) = 3ax^2 + 2bx + c
Q'(x) =2 rx + s

Same tangent line=> Same first derivative yes? 

Rearranging gives 3ax^2+ x( 2b- 2r)+ c- s= 0

As long as the determinant of the above quadratic is greater than zero, OP would be possible yes?

Meaning only if 4b^2- 8br+ 4r^2- 12ac+ 12as > 0 

This is achievable for large values of b and s accompanied by small values for c and r for instance. 

 

Edited by zonsop

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1 minute ago, zonsop said:
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Same tangent line=> Same first derivative yes? 

Rearranging gives 3ax^2+ x( 2b- 2r)+ c- s= 0

As long as the determinant of the above quadratic is greater than zero, OP would be possible yes?

Meaning only if 4b^2- 8br+ 4r^2- 12ac+ 12as > 0 

This is achievable by for large values of b and s accompanied by small values for c and r for instance. 

 

Problem with this is,

Spoiler

I tried it too.  But it only tells you it could be possible, but no conclusive answer.

The proof I provided is a resolute no.

 

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I'm confused now... Is the OP not simply asking whether it's possible? So lowest hanging fruit being to substitute real values that makes it so? 

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Spoiler

But I have produced evidence that it is absolutely impossible.  How can you argue with that?

Have you actually read through my proof, really?

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Spoiler
25 minutes ago, K27 said:

P(x) = ax^3 + bx^2 + cx, Q(x) = rx^2 + sx

Hm... original equations were- 

Q(x)= ax^3+ bx^2+ cx+ d

P(x)= rx^2+ sx+ t

 

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Spoiler
Quote

Since we can translate the curves in the plane WLOG, just move one of the intersecting point to the origin.  Why is this convenient?  Because for the cubic, P(0) = 0.  And for the quadratic, Q(0) = 0.  Which yields:

P(x) = ax^3 + bx^2 + cx, Q(x) = rx^2 + sx

The bolded part implies that d = 0 and t = 0.  If your problem is with me skipping details, I hope next time you will try to exercise your thinking department and fill in these little details yourself.  That would be highly appreciated.  :)

 

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It is likely that what you've proven is that in the event that d= t, the proposition is not possible. To this extent, I could agree with your proof.

I'm just not sure it holds for all other cases. Please could you contradict my proof directly? I would like to be corrected because I'd love for mathematics to always be completely watertight, i.e. I hope we come to the same conclusion about the OP.

I don't think this is one of the unsolved maths mysteries out there. :p And those do exist! :sulk:

 
 
...... added to this post 11 minutes later:
 

I got a really nasty looking value for the points where the graphs intersect-

x= + or - sqrt[(2b- 2r)^2- 4(3)(c-s)] / 2(3a)

Not going to expand that. :dead:

 
 
...... added to this post 23 minutes later:
 

Here, let's give just the one set of possible values so that the OP is possible. We need only one set to prove that the OP is possible. I'm assuming that they are integers for simplicity sake. 

a= 2, b= 10, c= 1, d= 1, r= 1, s= 3, t= 4

P(x)= 2x^3+ 10x^2+ x+ 1
P'(x)= 6x^2+ 20x+ 1

Q(x)= x^2+ 3x+ 4
Q'(x)= 2x+ 3

P'(x)= Q'(x)
6x^2+ 20x+ 1= 2x+ 3
6x^2+ 18x- 2= 0
3x^2+ 9x- 1= 0

x1= +sqrt[81-4(3)(-1)]/ 2(3)= 1.60728
x2= -1.60728

Edited by zonsop

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Hmmm okay.  I'll justify why I can safely assume d=t=0.

Spoiler

I have already mentioned in my proof that the curves can be (using Monte's words) rigidly translated in the whole cartesian plane.  This means as long as we are taking care of the relative position of the cubic and the quadratic, we don't really have to worry about the actual position of both.  (Sounds familiar?  Relativity? ;))

This gives us advantage, because when the absolute position is irrelevant, we can freely choose where to anchor the curves.  So in this case, I chose to posit one of the common tangential point of the two curves at the origin.

Anything I proved for this particular curve can be applied to every other kind of curve in the same class, because they are just being translated to elsewhere.  Stretching is handled by the first coefficient, namely a and r in this question.

So, although you may think I am just proving for a specific case, but the fact that all other cases are just rigid translation of my specific case justifies my extension of this specific case to the other cases.  Or, I should have made it clearer by attaching WLOG at the very beginning of my proof, if that's where the confusion arise.

As for the problem of your proof --

Spoiler

Simply plot those curves you arrived at.  They just don't touch at two distinct points.  They don't work.

Screen%20Shot%202017-04-21%20at%2019.34.

Apart from the fact that it does not work, conceptually, why it went wrong:

Spoiler

By considering a delta > 0, you are only trying to get to the point where P(x) = Q(x) and P'(x) = Q'(x) both has two distinct solution.  The problem is you are not trying to make the two sets of the distinct solutions in agreement in each other.  Your inequalities are "disjoint" to each other, that the existence of two distinct solutions to each system is still not connected between the two systems.

As stated in the OP, you are required to find two points where the two curves touches, and that the tangential slope at those points have to agree with each other as well.  This mandates the solutions of the two sets to agree with each other.

When you only consider the inequalities, you are not making sure that they agree with each other.  So the two points that you find in one system having two distinct intersection points, does not necessarily leads to them having the same slope at that point, as clearly demonstrated by the plot I attached in the previous spoiler.

 

Edited by K27
spelling

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I think the relativity part of maths is where you lost me. :p But I read jndiii's analysis a couple of times and my limited maths is able to handle that!

Does my fundamental error lie in assuming that two such points exist in the first place?

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14 minutes ago, zonsop said:

Does my fundamental error lie in assuming that two such points exist in the first place?

No!  Because even in my own proof, I did exactly that!

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In order for two curves to have the same tangent, they must cross at a common point, and have a tangent with the same slope.  If we assume b, c, d and s and t are 0, then the common point is (0,0), and the slope is 0.

However, from this point, the slope of the lines at any given point proceed at ax^2/y * (-(mod(y)), whereas the slope of the cubed curve proceeds at rx^3/y.  Obviously, these will never meet.

So, the question remains as to whether the factors we've left out wobble the slope of the tangent to cause them to meet again.

That's all I have time for, now.

 

 

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K27's instructive solution, and helpful defense of it, are the kinds of worthwhile discussions you hope will come out of threads like this one.

Thanks to all participants!

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