Summer courses almost over, "children's" books in the works.

This summer is one I will remember. I taught both differential equations and calculus II for the first time. Seeing these subjects from the instructors side has really opened my eyes to all kinds of details I never noticed before. One of the most striking new insights is how much these two courses have in common. They both rely deeply on sequences and series. Sequences are like a hallway in mathematics, one that connects many many many rooms.

I am working on two math book projects. The first is a Japanese-styled art book on the topic of sine and cosine. It's inspired by many of the lessons I taught this summer.

Japanese-style book about sine and cosine

I want to bring all of the different ways that sine and cosine are presented in elementary and undergraduate mathematics in to one (long) pictorial document. I start with the differential equation, y''+y=0 then solved it (using the series method from differential equations) producing \sin x = \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n+1)!} x^{2n+1} and \cos x = \sum^{\infty}_{n=0} \frac{(-1)^n}{(2n)!} x^{2n}

Next I wanted a pictorial way to relate these power series to the unit circle. I have found it in this spiral (the first image shows how it is constructed as an involution):

Constructing The Involute Pinwheel

a sequence of involutes: the vertical and horizontal components will form the power series for sine and cosine respectively.

The vertical and horizontal components will form the power series for sine and cosine respectively. Take the series of vertical line segments: \sin x = A_1A_2 - A_3A_4 + A_5A_6 - \cdots and so on, the segments repeatedly over and under-shoot the accutal value of sine. The full paper by Leo S. Gurin, "A problem", can be found here.

I'm going to incorporate Gurin's spiral in to my book. I want to show the power series literally flying out of it, like they have come to life. I wonder if I can make it like the famous drawing of the sine curve projecting out of the unit circle?

Naturally, I already have planned to put that diagram in my booklet.

Work in progress

The Japanese-style book is perfect for series and periodic functions It's one long continuous piece of paper:

Yet very compact:

I'm also working on a very silly book about hypercycloids (that's the "math" name for the shapes drawn by spirographs, did I mention I collect spirographs?):

I'm trying to make it like a children's book, fun, light, a little silly:

I can't wait to share the final product.

Answers to homework.

1. Find an equation of the tangent to the curve y = e^x that is parallel to x−4y = 1.

Y = \frac{1}{4}x + \frac{1-\ln (1/4)}{4}

(It is better to leave the natural log in the equation, don't find a decimal value... Though you may use such value for graphing.)

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2. Find an equation of the tangent to the curve y = e^x that will pass through the origin.

The the tangent to the curve at x_0 will have the form y = e^{x_0}(x+1-x_0) ... Plug in (0,0) and you find you are seeking the x-value where the natural log is zero. So the equation is y= ex.

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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.)

Counting using products.

Today I proved the binomial theorem for my class. I did an informal proof showing them how, if we consider the set containing the alphabet: S = \{ a,b ,c, d, ..., z \} and the following product.

\displaystyle \prod_{s \in S} (1+s_i) = (1+a)(1+b)(1+c)\cdots (1+z)

expanding this produces a sum with every combination of the letter a-z:

\displaystyle \prod_{s \in S} (1+s_i) = 1 + a + b + \cdots + z + ab + ac + ad + \cdots + az + bd + be + \cdots + bz + \cdots + abc + abd + \cdots + abcd + \cdots + abcdefghijklmnopqrstuvwxyz

if we count all of the terms in this sum with, say three elements, we have the number combinations of 26 items taken 3 at a time or 26 \choose 3. Next, we make every letter x:

(1+x)^{26} = 1 + x + x + \cdots + x + xx + xx + xx + \cdots + xx + xx + xx + \cdots + xx + \cdots + xxx + xxx + \cdots + xxxx + \cdots + xxxxxxxxxxxxxxxxxxxxxxxxx

(Pretty silly looking I know!) But, now we can see that the coefficients of the terms in the expansion are given by the same formula used to count combinations in probability.

That was today's lesson. What I've noticed is that this manner of counting is the same basic idea as what I've been looking at with integer partitions. (There we use a different product and coefficents to count.)

I wonder if this type of process has a name, or if it is further formalized in any contexts?

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.

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 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.

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!