### Saturday, April 16, 2005

## Infinite tower

Here is a cool puzzle I invented recently: Let x = Exp[pi/2], and y=x^x^x^x^... - i.e. y is an infinite tower of exponents of x (and to make the evaluation order explicit we can write it as: y=x^{x^{x^{... ). Find a finite solution for y!

Comments:

<< Home

Solution: y = ±i

First we raise x to both sides, as such: x^(y) = x^(x^x^x^x...) Thus, x^y = y.

Substituting e^(π/2) for x, we now have:

y = (e^(π/2))^y

y = e^(yπ/2)

Take the natural log:

ln(y) = yπ/2

And thus

ln(y)/y = π/2

It now seems we are close to a solution. However, if we examine the left hand part of this equality, we find it cannot equal π/2. Let f(c) = ln(c)/c, then:

f'(c) = (1-ln(c))/(c^2) and f(c) has an extrema when f'(c) = 0. Letting 1-ln(c) equal 0, we find the extremum occurs when c = e. f(e) = 1/e. Inspection shows this is a maximum. Thus max(f(c)) = 1/e < π/2.

As such, ln(y)/y = π/2 has no real solution. However, thus far we have ignored the possibility that y is a complex number. Recall that if z = r*e^(i*t), ln(z) = ln(r) + i*t = ln|z| + iArg(z). Thus [ln|y| + iArg(y)]/y = π/2. By inspection* and the fact that the imaginary part of this expression must be 0, one can find that i is a solution, as seen below:

[ln|i| + iArg(i)]/i

= [ln(1) + i*π/2]/i

= (i*π/2)/i

= π/2

To confirm this result, we go back to y = x^y. Raising both sides to the power (1/y) yield y^(1/y) = x. Thus i^(1/i) = e^(π/2). Since 1/i = -i:

i^(-i) = e^(π/2)

From Euler's equation, i can be written as e^(iπ/2), thus:

(e^(iπ/2))^(-i) = e^(π/2)

e^(-i*iπ/2) = e^(π/2)

e^(π/2) = e^(π/2)

*Actually, it turns out this result can be deduced more rigorously than just from intuition, as follows, and we can find all solutions.

Let y = a + bi, then:

[ln|y| + iArg(y)]/(a + bi) = π/2

[ln|y| + iArg(y)]*(a - bi)/[(a + bi)*(a - bi)] = π/2

[a*ln|y| + a*iArg(y) - bi*ln|y| + b*Arg(y)]/(a^2 + b^2) = π/2

The term a*iArg(y) must equal 0. We know Arg(y) ≠ 0 as this would make y purely real, and there is no real solution. Thus a = 0, and we now have:

[b*Arg(y) - bi*ln|y|]/(b^2) = π/2

[Arg(y) - i*ln|y|]/b = π/2

The term i*ln|y| must also equal 0. Thus ln|y| = 0 implies |y| = 1 implies √(a^2 + b^2) = 1. However, since a = 0:

√(b^2) = 1 implies b = ±1

If we put this result back in we see Arg(i)/1 = π/2, and Arg(-i)/(-1) = -Arg(-i) = π/2 both of which are correct.

It's a pretty cool result that a real number raised by a real number an infinite number of times yields a purely imaginary number. Potentially more interesting is the fact that you have y = (e^a)^(e^a)^(e^a)^... and it turns out that y = e^(ia). However, it appears that this solution does not work in general for any a.

Little brother Ben G.

First we raise x to both sides, as such: x^(y) = x^(x^x^x^x...) Thus, x^y = y.

Substituting e^(π/2) for x, we now have:

y = (e^(π/2))^y

y = e^(yπ/2)

Take the natural log:

ln(y) = yπ/2

And thus

ln(y)/y = π/2

It now seems we are close to a solution. However, if we examine the left hand part of this equality, we find it cannot equal π/2. Let f(c) = ln(c)/c, then:

f'(c) = (1-ln(c))/(c^2) and f(c) has an extrema when f'(c) = 0. Letting 1-ln(c) equal 0, we find the extremum occurs when c = e. f(e) = 1/e. Inspection shows this is a maximum. Thus max(f(c)) = 1/e < π/2.

As such, ln(y)/y = π/2 has no real solution. However, thus far we have ignored the possibility that y is a complex number. Recall that if z = r*e^(i*t), ln(z) = ln(r) + i*t = ln|z| + iArg(z). Thus [ln|y| + iArg(y)]/y = π/2. By inspection* and the fact that the imaginary part of this expression must be 0, one can find that i is a solution, as seen below:

[ln|i| + iArg(i)]/i

= [ln(1) + i*π/2]/i

= (i*π/2)/i

= π/2

To confirm this result, we go back to y = x^y. Raising both sides to the power (1/y) yield y^(1/y) = x. Thus i^(1/i) = e^(π/2). Since 1/i = -i:

i^(-i) = e^(π/2)

From Euler's equation, i can be written as e^(iπ/2), thus:

(e^(iπ/2))^(-i) = e^(π/2)

e^(-i*iπ/2) = e^(π/2)

e^(π/2) = e^(π/2)

*Actually, it turns out this result can be deduced more rigorously than just from intuition, as follows, and we can find all solutions.

Let y = a + bi, then:

[ln|y| + iArg(y)]/(a + bi) = π/2

[ln|y| + iArg(y)]*(a - bi)/[(a + bi)*(a - bi)] = π/2

[a*ln|y| + a*iArg(y) - bi*ln|y| + b*Arg(y)]/(a^2 + b^2) = π/2

The term a*iArg(y) must equal 0. We know Arg(y) ≠ 0 as this would make y purely real, and there is no real solution. Thus a = 0, and we now have:

[b*Arg(y) - bi*ln|y|]/(b^2) = π/2

[Arg(y) - i*ln|y|]/b = π/2

The term i*ln|y| must also equal 0. Thus ln|y| = 0 implies |y| = 1 implies √(a^2 + b^2) = 1. However, since a = 0:

√(b^2) = 1 implies b = ±1

If we put this result back in we see Arg(i)/1 = π/2, and Arg(-i)/(-1) = -Arg(-i) = π/2 both of which are correct.

It's a pretty cool result that a real number raised by a real number an infinite number of times yields a purely imaginary number. Potentially more interesting is the fact that you have y = (e^a)^(e^a)^(e^a)^... and it turns out that y = e^(ia). However, it appears that this solution does not work in general for any a.

Little brother Ben G.

Reminiscent of a high-school math meet problem (College of Charleston 1995ish): x = 1/(-1+(1/(-1+(1/(-1+...

What is x?

Also, google knows that exp(i*pi/2) is i. Of course it does.

David Chandler

What is x?

Also, google knows that exp(i*pi/2) is i. Of course it does.

David Chandler

Hi David! Let's see here ... a ha, the Golden Ratio! - which is then related to the Fibonacci sequence and Logarithmic spirals, so you could probably further expand this out into quite a pleasing puzzle. Hmmm...

Post a Comment
<< Home