I discovered a new summation formula regarding the consecutive powers of a given base. I noticed, to start, that any given power of 2 could be written as a sum involving the previous powers of two- for example, 4-1 = 2+1, or 32-1 = 16+8+4+2+1, generalized in the formula Sum(i=0, n-1) {2ⁱ } = 2ⁿ - 1. My idea was to expand this formula to include the powers of any base.

So, I first asked- is it even possible to write every power of 3 as a sum of the terms preceding it? Instinct said yes, and with a little guesswork I noticed a pattern: 3-1 = 2(1), 9-1 = 2(1+3), 27-1 = 2(1+3+9), and so on. This led to the general formula, Sum(i=0, n-1) {3ⁱ} = (3ⁿ - 1)/2.]

Before I state the general formula for xⁿ, which you might be able to guess, I would like to note that there is a somewhat non-traditional proof of this kind of formula that I used to verify my results. I'll give an example using 3.

Our goal is to prove that 1+2(1+3+9+...+3^n-1) = 3ⁿ

Multiplying out and adding the 1 to the first term of the sum,

3+2(3)+2(9)+...+2(3^n-1) = 3ⁿ

Adding the first two terms of the LHS,

9+2(9)+2(27)+...+2(3^n-1) = 3ⁿ

Adding the first two terms of the LHS,

27+2(27)+...+2(3^n-1) = 3ⁿ

And so on. This domino process by which each preceeding sum adds with the next term to create a power of three can be quantified by this relationship: 3^k-1 + 2(3^k-1) = 3^k. This series of summations eventually ends with 3ⁿ.

A similar proof can be can be used to show that the formula for 2^n works, or for any consequent base. Here's the general formula: 1+(x-1)*Sum(i=0, n-1) {xⁱ} = xⁿ. With this formula, we can now see what the main hinge of the given proof really is- it is that the first "domino" in the series comes from 1+(x-1)(1) = x, which is exactly what is needed.

After creating this formula and remembering one of my older finds on the same topic, I had the desire to somehow combine the two finds and generate some cool new relationship. (The other formula came up with the same kind of result, a formula for xⁿ, though in that case the formula was a sum of polynomial values instead of exponential values.) I'll have to go back and see if there's any way to reconcile the two.

As fun as Richey Numbers are, it's time to move on.

I've come up with an unusual puzzle to get me back on the random-math track. It's a problem that stumped me through all of 10th grade English class, the year I formally learned of Polar Graphing. I was curious about the graph of r = Θ, which takes the form of a spiral. On the interval [0, π/2] the graph hits a maximum x value. We have the points (x,y) = (0,0) and (0, π/2), but in between there the graph is an unusual curve in the first quadrant. I asked, what was the maximum x value the graph hits in that interval?

I had no idea how to solve it at first. I asked numerous math teachers, and those who didn't completely ignore me weren't sure what to do. This year I'm in Calculus AB, and in light of a recent unit regarding Optimization I was struck with an idea.

We have the relationships between Polar and Cartesian coordinates that x^2+y^2 = r^2, and tan Θ = y/x. The function says that r = Θ, so

x^2 + y^2 = Θ^2

Also,

xtan Θ = y => x^2 tan^2(Θ) = y^2, so

x^2 + x^2 tan^2 (Θ) = Θ^2

x^2(1+tan^2 (Θ)) = Θ^2

Since 1+tan^2 (Θ) = sec^2 (Θ),

x^2 = Θ^2 cos^2 (Θ)

x = Θ cosΘ

Now we have a simple relationship relating x and theta. All we need to do is differentiate and find where the maximums and minimums occur on Θ ε [0, π/2].

dx/dΘ = cos Θ - Θ sin Θ

Θ = cot Θ, a problem best left to the calculator. Approximately, Θ = .86033, and x = Θ cos Θ = .56110.

The maximums in x and y in any interval can be found using the same method; they are the solutions to the equations Θ = cot Θ and Θ = tan Θ, respectively. I have yet to try this out on other polar functions.

Richey Numbers- New Beginnings

I have finalized my findings on Richey Numbers, numbers that can be written in a certain number of ways as the sum of two squares. As it turns out, I was right before, though the specifics were a little off.

A formula I had read about some time ago in Paul J. Nahin's An Imaginary Tale says that any number (x^2+y^2)(u^2+v^2) can always be written as the sum of two squares in two different ways. I overlooked it, thinking it was just a formula for generating 2nd order Richey Numbers- now I realize that it is the formula that generates all Richey Numbers, and is very closely related to the one I came up with, which generates some Richey Numbers.

Since N = (x^2+y^2)(u^2+v^2), we can algebraically re-write the product on the RHS as the sum of two squares, in two different ways, as such:

(x^2+y^2)(u^2+v^2) = (xu+yv)^2 + (xv-yu)^2 = (xv+yu)^2+(xu-yv)^2

(By the way, some mathematician didn't sit thinking about the product on the left for hours on end until he came up with a way to re-write it; there is a method involving complex numbers that Nahin presents in his book. Unfortunately I don't have it right now- the method is interesting though, and I will post it when I get the book back.)

You can also check the algebra out by yourself if you don't believe me.

As for the general nth case Richey Numbers- numbers that can be written in n ways as the sum of two squares- I came to some new conclusions. 1st order Richey Numbers consist of three sets- first, all the primes that aren't of the form 4k+3, i.e. those that are of the form 4k+0, 4k+1 or 4k+2. These are 2 and all those of 4k+1, since there are no primes of the form 4k+0 and 2 is the only one of the form 4k+2. Some examples of these are 5, 13, 17, 29, and 37. You can check that these can be written in exactly one way as the sum of two squares.

Second, the set of integers 2^(2a+1) for some a. This expression can algebraically be re-written as 2^(2n) + 2^(2n), which is the sum of two squares. These numbers include 2, 8, 32, 128...

The last set is the some of the numbers within set of positive integers whose prime factorizations contain only primes from the first set, and only one prime of the form 4k+1. For example, 10 = (2)(5); since both 2 and 5 are primes that aren't of the form 4k+3 10 can be written as the sum of two squares, as 10 = 9+1. 15 = (3)(5) cannot be written as the sum of two squares because 3 is a prime of the form 4k+3.

I have conjectured that the numbers that can be written as the sum of two squares in more than one way are those that contain two or more different factors that are primes of the form 4k+1. For example, 26 = (2)(13) = 25+1 can only be written in exactly one way as the sum of two squares because it only has one factor of the form 4k+1, namely 13. 130 = 2(5)(13) = 121+9 = 81+49 can be written in two ways as the sum of two squares because it has two factors of the form 4k+1, in this case both 5 and 13.

I have yet to look into numbers that can be written in three or four or five ways as the sum of two squares, though in writing my C++ program for this problem some time ago I did notice (after preforming some trial runs) that the prime factorizations of the higher-order Richey Numbers seem to be closely related to earlier order Richey Numbers, just as 2nd order Richey Numbers are closely related to 1st order ones. (For those who know C++ programming, my code checks how many ways a given integer can be written as the sum of two squares, a variable I denoted n. I was making use of it to check for numbers where n=2, but the code can easily be transformed to look for other values of n.)

Having just looked a little further into the matter now, as an early hypothesis it appears that an nth order Richey Number is always the product of n prime factors of the form 4k+1, though I have yet to algebraically show this.