Wednesday, May 17, 2006

Exponential Explorations – Part II

(Part I is here.)

I want to thank everyone who gave Part I a try and encourage everyone who didn’t to at least follow along. There will be bits of this that everyone will be able to do, though other bits will be quite challenging.

To review the solution of Part I, the insight lies in realizing that if the chain of exponents is infinite, then adding another x at the bottom makes no difference. So

f(x) = x ^ f(x)

If we want f(x) to be equal to 2, we just plug 2 in for f(x).

2 = x ^ 2, so x = sqrt(2).

Let’s demonstrate that numerically and play with some different values of x. This is where an Excel spreadsheet comes in handy. Open a new one. (You may be able to do this with a calculator, but it’ll be a pain.) In cell A2 (we’re leaving A1 blank for now) type

**=sqrt(2)**

and hit enter. The cell should display a number around 1.4142. Widen the A column by dragging the thin line between the header of column A and column B so that you can see as many of the digits of the square root of 2 as Excel can calculate. (You can also double click on that line. That automatically sizes a column to the width of its fattest cell.) Now in cell A3 type

**=$a$2^a2**

and hit enter. Quick Excel lesson: the dollar signs make the first cell reference absolute, meaning “no matter where this formula gets copied, always point to cell A2.” The second cell reference without the dollar signs is relative, meaning “wherever this formula gets copied, always point to the cell one above the cell that the formula is in.”

Cell A3 should have a number about 1.63 if you’ve done everything right. Now click back on cell A3 to select it. The right lower corner of the selected cell is a square black drag box that I circled below for you.

That drag box lets you easily copy from one cell to others. Grab the drag box with your pointer and drag it straight down the A column so that you cover about 100 or two hundred cells. What you have now is a whole bunch of cells that each calculate “the value in A2 taken to the power of the cell immediately above me”. So if we put x in A2, the rest of the column calculates progressively close approximations of f(x).

You should be able to scroll down and see that the numbers eventually converge to 2 and remain 2 ever after, which is just the result we were hoping for.

With me so far? Good.

Now what if in Part I asked you to find x such that f(x) = 3? Using the same solution we would come up with x = the cube root of 3, or 3^(1/3). Let’s plug that in. In cell A2 type

**=3^(1/3)**

and press enter. A2 becomes about 1.442 and the entire column recalculates, and we scroll down eagerly anticipating the numbers in the column to settle down to 3. But they don’t. They settle to about 2.478.

What’s going on here?

Let’s see if we can get a higher answer by just plugging in bigger numbers. If we type 2 in to A2, the entire function blows up after just three iterations. That makes sense. f(2) just gets bigger without limit and doesn’t converge. Try 1.5 in A2. Same thing. It blows up, but it takes a little longer. But the square root of two and the cube root of three which are over 1.4 but less than 1.5 both converge.

Here’s the challenge for this part, and I admit, it’s hard. Find the upper limit of convergence of f(x) exactly. That is, find the highest x for which f(x) converges to some finite number. You can play with the spreadsheet to give you some ideas, but ultimately to get an exact (not a numerical) answer, you’ll need to put the spreadsheet away and do some math. By exactly, I mean give the answer as a mathematical expression, not as a number. For example the exact answer of Part I was “the square root of 2” whereas 1.41 (or any number of digits) would not have been exact.

Feel free to use the comments to collaborate with others and share ideas or partial results. If there’s no progress, every day or two I’ll give you a step towards the solution.

Comments:

Actually, I wasn't. I was stumped and waiting for a hint. ;o) Hopefully I'll have time to work on this tomorrow (today).

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Actually, I wasn't as concerned about your challenge as about why 2 worked and 3 didn't. So I tried some values in between. (and realized why you left A1 blank for now) As y increases from 2, the resulting x gets farther from the value it should be converging on, eventually approaching values slightly greater than 1. But then, if x is greater than 1.5, y increases exponentially. It reminds me of a stepwise function. But then I realized that any number taken to the inverse power of itself is between 1 and 1.5 exclusive. So now I'm pondering the function f(x) = x^(1/x). But I need sleep.

Kiwi: You're on the right track. That's why I left A1 blank, so you could put

in A2 and then put various numbers in A1. That helps you explore and think about what's going on. But then working on the limit of convergence is a seperate piece.

Tzipster91: Welcome! It's nice to see you here. This

**=a1^(1/a1)**in A2 and then put various numbers in A1. That helps you explore and think about what's going on. But then working on the limit of convergence is a seperate piece.

Tzipster91: Welcome! It's nice to see you here. This

*is*the test.
Psychotoddler: Thanks for letting me know, but what about you? Don't you wanna play?

Tzipster91: Bring it on!

Tzipster91: Bring it on!

Well, I know *what* the answer is, just haven't had a chance to look up the Taylor series and whatnot that I used to know to explain *why* it's the answer. It's e-asy to see, though...I looked at "x^(1/x)" for x= 2, 2.01, ..., 2.99, 3 to see where the max was.

Annabel: You're a brainy chick! Strong work.

Kiwi: I was going to just spill the entire beans today and give the solution, but I'm delighted you're still working on it. OK. Here's the first few steps to get you started.

I found a graphing calculator very helpful to visualize what was happening. Let’s take the case of x=sqrt(2) and see how f(x) converges on the value two. What f(x) is doing is iterating the function sqrt(2)^x over and over again. You can get a visual sense of that by graphing the following two functions. (You can just copy each one into the graphing calculator and hit enter.) Before you do, I suggest you limit the x and y plotting ranges (in the bottom left text-entry boxes) to 0 to 5.

and

The second function is obviously just the identity line y=x that runs diagonally through the plane.

So after you’ve graphed both functions you should have a display that looks like this. (I’ve added the red markings myself.) The curve sqrt(2)^x crosses the identity line in two places. (You can put the pointer over the plane and see the coordinates of the pointer displayed.) The crossovers are at (2,2) and (4,4) which makes sense, since sqrt(2)^2=2 and sqrt(2)^4=4.

Here’s how I graphically imagine the convergence process that the column in the Excel spreadsheet is doing. Take a look at this picture. I’ve zoomed in by setting the limits of both axes to 1 to 2.5, and again, all the red markings were added by me. Look at the red line that makes a staircase. We’ll start at the lower left of the line and make an analogy between every step on the staircase and every cell going down the Excel column. The first cell in the column is sqrt(2), so the first line I drew is a vertical one at y = sqrt(2). That intersects the curve at (sqrt(2), sqrt(2)^sqrt(2)) which is about (1.41, 1.63). That value is the next cell down the Excel column. Then extending the line horizontally to the identity diagonal gets you to (1.63, 1.63), then up from there until you hit the curve gets you the next cell on the spreadsheet. And so on, and so on. Does that make sense.

So you can see on the picture how successive iterations will get closer and closer to 2.

Try now graphing the analogues for starting with different x. For example, graph (3^(1/3)^x and that’ll show you why it converges at a number other than 3. Then graph 1.5^x and you’ll see why starting the column with 1.5 doesn’t converge at all. That should give you a clear idea about how to figure out the highest x that will allow f(x) to converge.

You can do it!

Kiwi: I was going to just spill the entire beans today and give the solution, but I'm delighted you're still working on it. OK. Here's the first few steps to get you started.

I found a graphing calculator very helpful to visualize what was happening. Let’s take the case of x=sqrt(2) and see how f(x) converges on the value two. What f(x) is doing is iterating the function sqrt(2)^x over and over again. You can get a visual sense of that by graphing the following two functions. (You can just copy each one into the graphing calculator and hit enter.) Before you do, I suggest you limit the x and y plotting ranges (in the bottom left text-entry boxes) to 0 to 5.

**sqrt(2)^x**and

**x**The second function is obviously just the identity line y=x that runs diagonally through the plane.

So after you’ve graphed both functions you should have a display that looks like this. (I’ve added the red markings myself.) The curve sqrt(2)^x crosses the identity line in two places. (You can put the pointer over the plane and see the coordinates of the pointer displayed.) The crossovers are at (2,2) and (4,4) which makes sense, since sqrt(2)^2=2 and sqrt(2)^4=4.

Here’s how I graphically imagine the convergence process that the column in the Excel spreadsheet is doing. Take a look at this picture. I’ve zoomed in by setting the limits of both axes to 1 to 2.5, and again, all the red markings were added by me. Look at the red line that makes a staircase. We’ll start at the lower left of the line and make an analogy between every step on the staircase and every cell going down the Excel column. The first cell in the column is sqrt(2), so the first line I drew is a vertical one at y = sqrt(2). That intersects the curve at (sqrt(2), sqrt(2)^sqrt(2)) which is about (1.41, 1.63). That value is the next cell down the Excel column. Then extending the line horizontally to the identity diagonal gets you to (1.63, 1.63), then up from there until you hit the curve gets you the next cell on the spreadsheet. And so on, and so on. Does that make sense.

So you can see on the picture how successive iterations will get closer and closer to 2.

Try now graphing the analogues for starting with different x. For example, graph (3^(1/3)^x and that’ll show you why it converges at a number other than 3. Then graph 1.5^x and you’ll see why starting the column with 1.5 doesn’t converge at all. That should give you a clear idea about how to figure out the highest x that will allow f(x) to converge.

You can do it!

*I'm delighted you're still working on it.*

Actually, I wasn't. I was stumped and waiting for a hint. ;o) Hopefully I'll have time to work on this tomorrow (today).

I’m showing the rest of the solution, so skip this comment if you don’t want to read it.

The graphs we were making on the graphing calculator were all in the form y=n^x where n is the value that we would plug into f(x). The upper limit of convergence happens when n^x is just tangent to the y=x identity line. We are looking for the n that makes this happen.

So, find n such that

y=n^x and

y=x are tangent

That means that the slope of y=n^x is 1 at their point of intersection. To figure out the slope of n^x we take the derivative. We remember that

y = n^x = e^(x log n) [that log is a natural log, or base e]

we also remember that if g is a function of x, d/dx e^g(x)=(e^g(x)) dg/dx

so

dy/dx = (log n)(e^(x log n))

which we can set equal to 1

so we have three equations to work with

[1] (log n)(e^(x log n))=1

[2] y=x

[3] y=e^(x log n)

substituting 3 into 1:

y log n = 1

y = 1/log n

using 2:

x = y = 1/log n

substituting that back into 3:

1/log n = e^((1/log n) log n)

1/log n = e^1 = e

log n = 1/e

so

n = e ^ (1/e) which is about 1.44466…

I love it when e pops up in a result uninvited, and what could be cooler than e to the power of its reciprocal?

Tangentially, for low values of x (like below 0.1) f(x) oscillates between two values rather than converging. The lower limit of convergence when this starts to happen is x=e^(-e). Groovy, no?

I hope you all learned something, kids. But the important thing is that we all had fun.

The graphs we were making on the graphing calculator were all in the form y=n^x where n is the value that we would plug into f(x). The upper limit of convergence happens when n^x is just tangent to the y=x identity line. We are looking for the n that makes this happen.

So, find n such that

y=n^x and

y=x are tangent

That means that the slope of y=n^x is 1 at their point of intersection. To figure out the slope of n^x we take the derivative. We remember that

y = n^x = e^(x log n) [that log is a natural log, or base e]

we also remember that if g is a function of x, d/dx e^g(x)=(e^g(x)) dg/dx

so

dy/dx = (log n)(e^(x log n))

which we can set equal to 1

so we have three equations to work with

[1] (log n)(e^(x log n))=1

[2] y=x

[3] y=e^(x log n)

substituting 3 into 1:

y log n = 1

y = 1/log n

using 2:

x = y = 1/log n

substituting that back into 3:

1/log n = e^((1/log n) log n)

1/log n = e^1 = e

log n = 1/e

so

n = e ^ (1/e) which is about 1.44466…

I love it when e pops up in a result uninvited, and what could be cooler than e to the power of its reciprocal?

Tangentially, for low values of x (like below 0.1) f(x) oscillates between two values rather than converging. The lower limit of convergence when this starts to happen is x=e^(-e). Groovy, no?

I hope you all learned something, kids. But the important thing is that we all had fun.

Sorry, I'm really busy. But I'll definitely work on this and read your solution when I get time. Thanks for keeping me sharp!

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