Sunday, May 14, 2006

Exponential Explorations – Part I

During my visit at the home of Mrs. Balabusta and Psychotoddler, I had the pleasure of meeting Gerald Glazer, a retired math professor. It’s not every weekend that I meet someone with as much interest in recreational math as me (or more!) so it was quite a treat to work on the math puzzles he tossed at me while dining on Mrs. B’s yummy cuisine.

One of the problems he gave me had me thinking for quite a while and even after solving it, led to some interesting discoveries. Allow me to give you the puzzle first, and then after one of you solves it, we’ll explore a little further in Part II.

Here’s the problem.

Define f(x) as follows.

Where the dot dot dot at the top means keep going with the ladder of exponents forever.

Find x such that f(x) = 2. Show your work.

Now, let me make sure you don’t make the mistake I made when I first heard this problem. I got the order of the exponentiation backwards. I initially thought that the function was (((x^x)^x)^x)… Meaning, if we were to plug in 1.2 for x we would calculate 1.2 to the 1.2 power, to the 1.2 power, to the 1.2 power, and so on. That’s NOT right. Since (a^b)^c = a^(bc) that would have meant that f(x)=x^(x x x x…) which for any x > 1 is infinite.

The right way to read the function is f(x) = …x^(x^(x^(x^x))). That means you first start with the top exponent and work your way down the ladder. Meaning, if we were to plug in 1.2 for x the way to calculate it is 1.2 to the 1.2 power. Then 1.2 to

*that*power, then 1.2 to

*that*power, and so on and so on forever. Makes sense?

So find x such that plugging it into that function results in 2.

Comments:

Duh. Another mistake.

But I think log x (y) would have been even less useful.

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Oh, to be a fly on the wall during that lunch conversation. A fly on the wall, safely pinned by a flyswatter, dying a merciful death.

Oh, and there was an excellent Kirk/Spock moment there, where Dr. Bean went through his entire reasoning for the solution of a different problem, and at the last second, I blurted out the answer.

Are you people on the MENSA honor roll, or what!?

I plainly want to say "X marks the spot" as my answer. (Pearl and numbers don't mix; Pearl and words do!)

I plainly want to say "X marks the spot" as my answer. (Pearl and numbers don't mix; Pearl and words do!)

Well, it seems the peanut gallery has descended in force. I’ll have you know that the lovely and talented Annabel Lee has already solved the problem and sent me an email. Won’t anyone else take a whack at it? Oven? Stacey? Beuler?

Ralphie: Wait till Part II, that’ll really have you reaching for the arsenic.

Psychotoddler: Hey! I’m not hearing enough gratitude out of you for giving your dad-in-law a much needed vent for his recreational math. Admit it. My visit was the best thing that happened to him in, um, hours.

Torontopearl: Numbers? What numbers? The only number I mentioned was 2. There are hardly any numbers involved, just variables. Give it a try!

Irina: Sorry to harsh your mellow, my friend. Let a smile be your umbrella! Turn that frown upside down!

Ralphie: Wait till Part II, that’ll really have you reaching for the arsenic.

Psychotoddler: Hey! I’m not hearing enough gratitude out of you for giving your dad-in-law a much needed vent for his recreational math. Admit it. My visit was the best thing that happened to him in, um, hours.

Torontopearl: Numbers? What numbers? The only number I mentioned was 2. There are hardly any numbers involved, just variables. Give it a try!

Irina: Sorry to harsh your mellow, my friend. Let a smile be your umbrella! Turn that frown upside down!

THE SOLUTION

Let f(x) = x ^ y, where x is the first x in the

expression and y is the never-ending exponent.

Thanks to the interesting properties of infinite

expressions, we notice that y = x ^ y

x ^ y = 2

x ^ (x ^ y) = 2

x ^ 2 = 2

x = sqrt(2)

Let f(x) = x ^ y, where x is the first x in the

expression and y is the never-ending exponent.

Thanks to the interesting properties of infinite

expressions, we notice that y = x ^ y

x ^ y = 2

x ^ (x ^ y) = 2

x ^ 2 = 2

x = sqrt(2)

Oven: Very nice!

Peanut gallery: Why couldn't all of you be more like Oven?

Look for Part II tomorrow!! We're just scratching the surface!

(Am I killing our readership, or what?)

Peanut gallery: Why couldn't all of you be more like Oven?

Look for Part II tomorrow!! We're just scratching the surface!

(Am I killing our readership, or what?)

Bean, if you forgot to Please EXcuse My Dear Aunt Sally, you need some teenagers to help with math homework.

I'm pondering, without looking at Oven's post. But I'm not very good at your stuff. I think maybe I need some teenagers too.

I'm pondering, without looking at Oven's post. But I'm not very good at your stuff. I think maybe I need some teenagers too.

Disclaimer: My hands hurt, so this is the compact version of my figuring, with no fancy math formatting. My name is not LaTex.

f(x)=y, because y is 3 characters shorter. So using the rules of logs:

x=log y (x^x^x^x...)

=x log y (x^x^x^x...)

=x^2 log y (x^x^x^x...)

=x^3 log y (x^x^x^x...)

...

x=x^inf log y (x)

x^(-inf)=log y (x)

x=y^x^-inf

There's probably a way to simplify this, but I don't remember it. I had to search the internet to remind myself how to use logs.

Now, x^-inf is an infinitely tiny number, and so I tried calculating 2^.1, .01, .001, etc. I get x->1. However, the first thing I did before all this figuring was to make a spreadsheet calculating y based on the value of x. (A column of cells =$A$1^A1, where A1=x.) I plug in these increasingly small powers of two, and I get results increasingly close to 1, not 2. So something's wrong.

Originally I was experimenting with values of x between 1 and 2, and it seemed that for x<~1.5, y approached 2, but for x>~1.5, y approached infinity. This doesn't give me a warm fuzzy, but it seems closer to the solution.

I stink at limits. There are no tidy rules, and without rules I'm lost. Now I'm gonna read what Oven said.

f(x)=y, because y is 3 characters shorter. So using the rules of logs:

x=log y (x^x^x^x...)

=x log y (x^x^x^x...)

=x^2 log y (x^x^x^x...)

=x^3 log y (x^x^x^x...)

...

x=x^inf log y (x)

x^(-inf)=log y (x)

x=y^x^-inf

There's probably a way to simplify this, but I don't remember it. I had to search the internet to remind myself how to use logs.

Now, x^-inf is an infinitely tiny number, and so I tried calculating 2^.1, .01, .001, etc. I get x->1. However, the first thing I did before all this figuring was to make a spreadsheet calculating y based on the value of x. (A column of cells =$A$1^A1, where A1=x.) I plug in these increasingly small powers of two, and I get results increasingly close to 1, not 2. So something's wrong.

Originally I was experimenting with values of x between 1 and 2, and it seemed that for x<~1.5, y approached 2, but for x>~1.5, y approached infinity. This doesn't give me a warm fuzzy, but it seems closer to the solution.

I stink at limits. There are no tidy rules, and without rules I'm lost. Now I'm gonna read what Oven said.

See what I mean? Without tidy rules, I make simple things so very difficult! I should stick to logic puzzles. I'm good at them. ;o)

Kiwi: Great work, especially with the excel spreadsheet which will come in handy in Part II. It helps you figure out the neighborhood that the answer is in (like about 1.4) but it doesn't get you the answer exactly: the square root of 2.

Thanks for giving it a whirl.

Thanks for giving it a whirl.

Kiwi - For the record, I went barking up the logarithm tree myself, to no avail. Ultimately, it came down to a flash of insight.

Kiwi: I'm actually not sure what you did wrong. I don't understand how you got to your first line:

x=log y (x^x^x^x...)

if you start with y=x^x^x^x^... and take the log of both sides, you get

log y = (x^x^x^x...) log x

but I don't see how that helps you. If you show me where you got the first line from, I'll try to figure out where things went wrong.

x=log y (x^x^x^x...)

if you start with y=x^x^x^x^... and take the log of both sides, you get

log y = (x^x^x^x...) log x

but I don't see how that helps you. If you show me where you got the first line from, I'll try to figure out where things went wrong.

I tried to use the definition of a log, but sometimes I have a hard time using that right. With my lack of formatting, log y (x) means log base y of x.

x = b^y -> y = log b (x)

so I figured, move the top exponent,

y = x^x^x... -> x = log (x^x^x...)(y)

...duh...never mind.

x = b^y -> y = log b (x)

so I figured, move the top exponent,

y = x^x^x... -> x = log (x^x^x...)(y)

...duh...never mind.

Kiwi: Now I get it. You made the same mistake I initially did in reversing the order that the exponents are taken. (That's why I showed you the way to write the parentheses.) The last operation that's done is the bottom one, not the top, so if you wanna take the log of the infinite chain of exponents, you gotta peel the bottom one, not the top one off.

So you should have

x^x^x^x... = log base x of y

does that make sense?

So you should have

x^x^x^x... = log base x of y

does that make sense?

*you gotta peel the bottom one, not the top one off.*

Duh. Another mistake.

But I think log x (y) would have been even less useful.

Kiwi: I'm not sure it would have been less usefull. It would at least have been true.

Let's see:

x^x^x^x... = log base x of y

that means

y=log base x of y

which means

x^y=y

which puts you on the right track.

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Let's see:

x^x^x^x... = log base x of y

that means

y=log base x of y

which means

x^y=y

which puts you on the right track.

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