Integrals - Series

Log[x]^n

x-kSin[x]

Nested Radical Recursions

Logs of Negative Numbers

Explorations

While admiring my own blog yesterday, I took a trip to one of the links- the "Calculus Problems" one. Upon finding some of them quite interesting, I solved two of the simpler ones; below are my solutions.

Calculus Problems

The attached document below speaks for itself. Though you will have to zoom in a lot- button's on the top bar on the right once you go full screen.

x^y=y^x

With the power of Scribd I can now upload the cool Mathematica work I've done to supplement some of my old posts. Below there are two very long tables with data relating to Richey Numbers, and a couple of extensions.

The first table consists of all the numbers that can be written as the sum of two squares, their prime factorizations, their generators (what squares Mathematica added to get the number) and how many distinct ways they can be written as the sum of two squares. The second table has the same information, only it looks at the sum of two cubes, not two squares. The right column is the one that tells us how many ways that number can be written as the sum of two squares or cubes. That column tells me what numbers to look at- for the first table, those that have a value of 3 or greater are interesting because I've already generalized those that can be written in two or one ways as the sum of two squares. The second table has red in the right column when the value is greater than one, but I still haven't generalized the set of numbers that can be written in one way as the sum of two cubes, let alone two ways. Finally, the green in the left hand column occurs whenever the number in question is a prime or a power of a prime- I noticed some interesting trends among those types of numbers. See what trends you can find, and if you do see something interesting, write a comment!

Sum of Two Powers Table

I recently submitted the solution (proof) to a problem posed in the April 2009 edition of Math Magazine. It is embedded below as a PDF. I hope you enjoy it, though it is a bit lengthy. (There's a button on the top right of the document window to make it full screen, and then a zoom button so the font is big enough to read.)
Game Show Problem

On the Power Tower

I recently looked into the properties of infinite power towers, or an infinite tower of powers. I first asked if the nested power tower with the square root of 2,

n^(n^(n^(n^(..., n=2^(1/2),

converges or not.

So, I simply asked Mathematica, and it turns out that the power tower of square roots of 2 converges to 2.

To prove this, we can set x equal to the tower and solve for it by some nice manipulation. (We can only do this when the number actually converges, but in this case I know it does so we're alright.)

x=n^n^n^n..., n=2^(1/2)

x=n^x

This equation, we can now see, has two solutions: x=2 and x=4. (Because Sqrt[2]^2=2 and Sqrt[2]^4=4.) However, we know that the power tower converges to 2- so why do we get the extraneous solution x=4 when solving analytically?

As it turns out, the power tower converges for any n in the interval (0,2^1/2]. As we've seen, for n=2^1/2, the equation has two solutions. However, it is interesting to note that for n on (1,2^1/2], the equation has two solutions, whereas for n on (0,1], it has exactly one solution. We can see this by looking at the function f(x)=n^x-x defined for n on (0,2^1/2].

In terms of the stated problem, f(x)'s zeroes tell us what the power tower of n converges to. I will now show that on the intervals stated, (0,1] and (1, 2^1/2], the equation has one and two solutions, respectively.

f(x)=n^x-x

f'(x)=ln[n]n^x-1

f'(x)=0 when

1=ln[n]n^x

n^x=1/ln[n]

x ln[n]=-ln[ln[n]]

x=-ln[ln[n]]/ln[n]

f(x) decreases on (-Infinity, -ln[ln[n]]/ln[n])
f(x) increases on (-ln[ln[n]]/ln[n],Infinity)

This is the single critical point of our function, and we can now see why we get two solutions for 1
Why the second zero is trivial, I have no idea. But I do like this problem, as well as some related ones. In writing this post I noticed that a similar "nested" infinite square root runs into a similar, but more obvious problem in the algebra.

Take x=Sqrt[n+Sqrt[n+Sqrt[n+..., out to infinity.

This transforms, just as above, into

x^2 = n+x

x^2-x-n=0

Using the quadratic formula, we obtain

x=(1+-Sqrt[1+4n])/2

Taking n=2 as an example,

x=2 or -1.

Mathematica agrees that the more logical solution is the correct one- naturally, 2 is the answer. Still, it is strange that the -1 shows up in our analytical approach.

Paralleling my other analysis, this function has two solutions most of the time, namely for n>0, and is useless when n<0 because we have no business assuming the nested square root's convergence in that case. (Clearly Sqrt[-.5+Sqrt[-.5+... doesn't converge to any real number, so setting the expression equal to x is against the law.)

There is one more of these that runs into the same problem: the continued fraction [n,n,n,n...] However, I see no need to share the specifics of this one as it comes out to a nearly identical quadratic problem. I leave it to the reader to investigate the case further.

Finally, I'd like to add a quick fun fact about the nested square root and the continued fraction. If we take the value of n=1, in BOTH cases we get the golden ratio, (i.e. Sqrt[1+Sqrt[1+...=1+1/(1+1/(1+1/...=(1+Sqrt[5])/2). You can check it yourself.

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.