futurebird – Page 4 – mathematical sandbox

Looks can be deceiving!

Look at the graph in the previous post. I was right! They don't intersect at a common point, rather the limit of the consecutive intersections is \left(1, \frac{1}{6} \right) .

Close up of the intersections of y^n + x^n -6xy = 0 for the first ten integers.

This is just numeric, from Mathematica ... I'll try to prove it later...

By the way an answer to the question of fining a curve orthogonal to the follium of Descartes is x^3 +y^3 +6xy= the reflection in the line y=-x. Do you know of any other trivial answers?

Goodnight! Tomorrow: Back to the generating functions, Owen has been bugging me!

Orthogonal Curves.

Points that call out to you.

The Follium of Descarts and it's family based on varible exponents.
The Follium of Descarts and it's family based on varible exponents.

x^3+y^3 = 6xy, x^4+y^4 = 6xy, x^5+y^5 = 6xy ... x^n+y^n = 6xy

You get the idea. Naturally I want to know what those intersections are! I need to go for a run before it gets dark, but It seems that if we fix x=1 then as n \rightarrow \infty, y \rightarrow \frac{1}{6}

Maybe the spots that look like common points aren't after all. Maybe it's just a very tight sequence of intersections.


A curve A is orthogonal to B is the tangent lines at their intersections are perpendicular. Thinking of the follium of Descartes again, give the equation for a function orthogonal to the follium.

The Follium of Descartes a=2

I have found one trivial answer that requires no complex calculations. (Do, you see it?) On the other hand, it's quite hard to find an orthogonal curve that passes through (3,3). We'd consider doing this since (3,3) is one of three rational points on this curve in the first quadrant, and the tangent there is -1... so it seems like it might be an easy starting point. For the three rational points here are the slopes in the first quadrant:

(3, 3), Slope: - 1

(36/217, 216/217) , Slope:  about 2.9

(216/217, 36/217) , Slope:  about 0.34

I was hoping the rational points I found earlier would at least yield pretty slopes... but no dice.  Using two of the above we can find a parabola. I did it. It was not pretty.

Counting integer partitions

My close friend Owen introduced me to [[wiki|generating function|generating function]]s last year. I was not impressed at first since they seemed cumbersome... but now I see how powerful they are for counting partitions. So, I will share my understanding of this matter. This post is based on my readings of The Theory of Partitions, by George Andrews.

The generating function for a sequence a0, a1, a2... is the power series whose coefficients are the a_I. That is: f(q)= \sum_{i \geq}^{\infty} a_iq^i.

[[wiki|Partition (number theory)|p(n)]] is used to represent the number partitions on an integer n. Extending this idea, given a set H. (eg. H={1, 3, 7} or H= \mathbb{N}) p( "H" , n) will represent the number of partitions of n, using only elements in H. So, we can see that p(n) = p("\mathbb{N}" , n). If H={1, 3, 7} then p( "H" , 10)= 6 because we have:

1+1+1+1+1+1+1+1+1+1 = 10
3+1+1+1+1+1+1+1 = 10
7+1+1+1 = 10
3+3+1+1+1+1 = 10
3+3+3+1 = 10
3+7 = 10

Clearly, if H is even a moderately large set or if n is bigger than, say, 10 the question becomes very complex very quickly. (I would love to see a 3D bar graph the showed z = p("H_y", x) with H_y = {1,2, 3 ... y } I say bar graph because the partition function works on the integers, is there an analytic continuation of p? I should look in to that!)

Note, H represents the set and "H" represents the set of partitions with elements in H. If
H={1, 3, 7} then "H" is a very large set! Now I'll introduce a few more characters to this scene:

"H_{0}" = \mathcal{O} The set of partitions in to odd parts.

"H"_{\leq d} Partitions where no part in H appears more than d times.

\mathcal{D} Set of partitions with distinct parts.

Next I'll connect all of this to the idea if generating functions. Let

f(q) = \sum_{n \geq} p("H", n)q^n and
f_d(q) = \sum_{n \geq} p("H"_{\leq d}, n)q^n

These are generating functions for the sequence of partitions of n with parts in H and the sequence of partitions of n with distinct parts in H.

Then for |q| <1 or another way we could say The proof is very lovely. The main idea is that the exponents of q will contain a linear combination of elements in H, hence the when we group together terms with the same exponent it is the same thing as counting the linear combinations of elements in H that add up to some integer... That means the coefficients in the power series are the number of partitions! Very neat.

Fixed points.

"Prove that if f,g are continuous functions on [0,1] with range [0,1], then there exists a point c such the f(c)=g(c)"

This is a variation on a popular problem given to encourage students to use the [[wiki|intermediate value theorem|intermediate value theorem]]. The proof involves a function h(x)=f(x)-g(x), this function can't be exclusively positive or negative, so we can find c so that h(c)=0.

The proof is really about getting the given situation in to a form where it will "plug in" to the definition nicely. It has no deeper insight about why this happens. For that we must look at the fixed point theorems.
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Rational points on the folium.

My students are studying implicit differentiation, we look at the folium of Descartes:

x^3 + y^3- 6xy =0

They found the equation of the tangent at (3,3)

For their homework I want to have them look at another point, but I wanted it to be rational, I almost gave them the origin... But that's good. The expression given by implicit differentiation is undefined there. (Though, one can find the slope of the two tangents using limits.)

Next I searched for rational points between 0 and 3. I'm confidant that there are only two x values that give at least one rational y value these are 36/217 and 216/217. (I used the rational root theorem to show this.)

Sadly, these numbers are too messy to use in a homework! Especially when they are mad at me for giving a very tough problem the other day.

Each of these x values has 3 corresponding y values, what are they?
Well, it's time to look for another relation... I want something that will let them find more than one tangent line at a given x value. Oh, and it must be rational... And pretty!