Almost Convergence.

The title of my last post "Almost Cauchy Sequence" made me wonder if "Almost Cauchy" was already defined in some formal sense. I could not find much on "Almost Cauchy" ... but I did find out about "Almost Convergence." (I think I have ran in to this before... but it is very neat... observe:)

A sequence S = S_n is almost convergent to L if for any \epsilon > 0 we can find an integer n such that the average of n or more consecutive terms in the sequence is within \epsilon of L. Formally, \forall \epsilon > 0 \quad \exists N \ni \quad \displaystyle \left| \frac{1}{n} \sum_{i=0}^{n-1} S_{k-i} -L    \right| < \epsilon \quad \forall n > N, \; k \in \mathbb{N}. (I took this definition from here, and I wonder why they didn't mention that we need k \geq n or else we are looking at negative terms in the sequence... But, maybe there is an interpretation of something like S_{-3}? Or maybe the author thinks that would be obvious? I'm going to ask at the mathematics stack exchange!)

Almost Cauchy Sequence, unfinished business.

I just remembered a question from beginning analysis that my professor mentioned in class:

"Find a sequence A that is not Cauchy, yet any consecutive terms A_n, A_{n+1} will become arbitrarily close."

It was one of those questions that he mentioned in class for us to do on our own, but I never quite got to it. I was thinking about convergence and complete metric spaces today (and about unfinished business, as I've made a lot of my own since I started teaching.) And the question came back to me.

We seek a sequence, A, where \forall \; \epsilon > 0 \quad \exists N \in \mathbb{N} \;  \ni \; \forall n > N \quad |A_n - A_{n+1}| < \epsilon and \exists \epsilon > 0 \; \ni  \quad \forall  N \in \mathbb{N} \quad \exists m,n > N \quad |A_m - A_n| > \epsilon. (Writing the negation of what it means for a sequence to be Cauchy was a little more tricky than I thought...)

The natural log function is the most obvious answer. For large vales of n \ln (n) - \ln (n+1) is small. That is, \displaystyle \lim_{n \rightarrow \infty} \ln \frac{n}{n+1} = 0. This is true becuase after taking EXP of both sides we have: \displaystyle \lim_{n \rightarrow \infty}  \frac{n}{n+1} = 1

Next, it seems obvious that the natural log can't be Cauchy, but just to be certian if we let \epsilon = \frac{1}{2} then, \forall N \in \mathbb{N} if we choose some n > N and let m= ne (so, n,m > N) then

| \ln m - \ln n | = |\ln \frac{m}{n}| = |\ln \frac{ne}{n}| = 1 > \frac{1}{2} = \epsilon.

NOT Cauchy at all!

Ok, I imagine my professor was thinking of the natural log... but, what other functions also have this property? \sqrt[n]{x}? So, I guess this is really asking when \displaystyle \lim_{x \rightarrow \infty} f(x) - f(x+1) = 0 while \displaystyle \lim_{x \rightarrow \infty} f(x) = \pm \infty. It is necesairy but not sufficent for f'(x) > 0 \; \forall x and f''(x) < 0 \; \forall x , both will have the x-axis as a horizontal asymptote.

Nice to have finally gotten around to doing this problem! I think it will be good practice to find more unfinished business in my notes.

Extending L' Hopital. (For my students. )

How can one deal with \lim_{x \to 0^+} x^{x^2}? It appears to be the indeterminate form 0^0. The graph shows that it has a nice limit:

(Notice, how this function produces imaginary vales when  it is between -1 and 0. )

But, this is the wrong kind of indeterminate form for L'Hopital.  If we could somehow show that it was, in fact, the same as \frac{0}{0} we could use the derivative to help us... Here is an extension of L'Hopital that will do just that:

\lim_{x \to c} f(x)^{g(x)} = \exp \lim_{x \to c} \frac{\ln f(x)}{1/g(x)}

See if you can apply it. We will talk more about this in class. If you simply said "we can't use L'Hopital" that is correct for this problem based on what you knew up until now. If you applied it anyway and tried to say that:  \lim_{x \to c} f(x)^{g(x)} =\lim_{x \to c} f'(x)^{g'(x)} ... then that is incorrect! (And I believe it leads to the wrong answer here.)

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.

Quickie.

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.

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!