*I hope women of the world can forgive my usurpation of the phrase from the Women's Movement, but the idea applies as I return again to the topic of infinite radicals.*

It is said that Ramanujan posed the above problem to the *Journal of Indian Mathematical Society*:in 1911. I use it on my "On This Day in Math" blog for a number fact on January third since it is the third day of the year. Because the problems of analysis from infinite series often dances at the edge (or outside) my understanding of pure mathematics, I always question my assumptions about them, and so for several years I have asked about a seeming extension (or perhaps contraction) of this infinite sequence. What happens when we chop off one layer from the front. My thinking went like this:

If we take the expression and square both sides we get \( 9= 1+2 \sqrt{1+3 \sqrt{1+...}} \)

And doing the obvious arithmetic to clear the preamble before the first radical we arrive at \(4= \sqrt{1+3 \sqrt{1+...}} \)

Now if we repeat the process of squaring and simplifying the result a second time we get \(5= \sqrt{1+4 \sqrt{1+...}} \) and thus, as they say, "to Infinity".

Just to make it easier, I have included in the remainder of this post some earlier thoughts about different infinite nested radicals exploring them on my on...

___________________ Reposted material from Dec, 2009 ________________________________

Recently (2009) someone on the Calculus EDG asked about the value of. I sent a link to some work I had done a while ago exploring the same idea, and extending to finding the value of

. I have picked out some parts below, but you can see the rest at this link. (apparently this link has been lost in the internet . I tried the Wayback machine but it seems to be an incomplete copy. Ir you are way more savvy than me, and who isn't really, then maybe you'll do better and share what you find.) This is a very old Word Document so give it some time to load. Hopefully it is worth while. Dave Renfro then sent me a copy of some papers about the topic, including this one from a 1935 American Mathematical Monthly.

When you take the iterated square root of a number, such as \(x = \sqrt{n+ \sqrt{n+ ...}} \) and then square both sides, you get \(x^2 = x + n\). This means that we can find solutions using basic quadratic solution approaches, and then find solutions that produce integer values of x. The positive solution becomes \( \frac{\sqrt{4n+1}+1}{2} \)

One of the nice things I discovered was that the iterated square roots of 2 was not the only number that gave an integer answer. In fact, 2, 6, 12, 20, 30.... all were equal to integer values... This sequence is the pronic or oblong numbers, which are twice the triangular numbers. These numbers can be expressed as (n)(n+1) . It took me a moment to realize why they are the ones that would work. These are numbers that, when multiplied by four and increased by 1, become perfect squares, \( 4 (n^2+n)+ 1 = 4n^2 + 4n+1 = (2n+1)^2 \). And the square root, being an odd (2n+1) number so that when 1 is added, we get a number divisible by 2.

It seems, according to the Herschfeld article, that the problem was a common topic in the Columbia classes of Dr. Edward Kasner. Kasner, of course, is known for his part in the creation of the term "googol" for 10^100. If your interested in any of these topics, check either or both the links above .

I had not yet tried to consider the roots of the cube root of (a+cube root(a+ .... etc)) and so I wanted to take a shot.. By the same process I had used before, the value would be the solution to x^3-x-n=0 .

If the iterated value was 1, the value approaches about x=~1.32472. For n=2 the value is x=~1.52138. By the time we get to n=6, we get x=2. The actual solution for any n is

OK... that really isn't very much fun to play with, but after some experimenting, I came up with the fact that the following sequence of numbers produced integer values when iterated; 6, 24, 120, 210... ; or perhaps it is more revealing to write them a different way (1*2*3) , (2*3*4), (3*4*5)... so they were sort of the three dimensional pronic numbers, the products of three consecutive integers. (I have never seen a name for these, so I'm introducing hexonic, because they are all divisible by six. Sphenic is also appropriate since it is the Greek root for wedge shaped, but it seems overused for any number with three distinct factors.)

I could not manipulate the above equation to make it clear that these were the only values as I had with the quadratic, but it got me thinking, what if I did fourth roots ? (This is the point where a more clever mathematician would have said hmmmm, squares are solved by x^{2} - x -n=0; and and cubes by x^{3} - x -n=0, maybe there is a pattern)

Extending the solutions for square and cube roots, I tried 1*2*3*4 = 24.... but the solution of \(n^4 - n- 24=0\) was NOT 2; in fact, it was about 2.1617??? (Yep, guess who picked the wrong pattern to pursue?

Exploring I found that n=2 was a solution to \(n^4 - n- 14=0\). And n=3 was a solution when the constant was 78. The sequence is 14, 78, 252, 620, 1290,... These values follow the form n*(n-1)*(n^2+n+1)

I realized, somewhat belatedly, that you could generate these sequences by simply using n^{k} - n ( \(2^4-2=14, 3^4-3 = 78, etc \)for integer values of k, and factoring the same would give you the simplified form of the expression. And it seemed true for all the others. The pronic numbers 2, 6, 12 are \(2^2-2, 3^2-3, 4^2-4\) and I'll let you convince yourself that the cubes root iterations works the same.

After struggling with solving n^3 - n -k=0 I realized that I could just start with values of n, and find out what k came out to be. A very late "aha" moment. So for n=2, 2

^{3}- 2 = ??? and six pops out like Alg I.

There is a familiar quotation about forests and trees that seems to apply here, but it came to me somewhat late.

But sometimes, that's how my mind works... do it the hard way first.

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