## i as an Infinitely Tiered Exponent

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- falseinfinities
**Posts:**3**Joined:**Thu Jan 16, 2014 6:44 am UTC

### i as an Infinitely Tiered Exponent

Starting a title with i is one of the hardest things I've ever done.

Now that that's over, I can explain. (And keep in mind that i is the imaginary square root of −1.)

We (being anybody who knows) know, through a series of proofs that I will not restate here, that e^(i*x) = cos x + i * sin x.

Now, I just say that x is pi/2 (radians), giving me:

e^(i*pi/2) = cos (pi/2) + i * sin (pi/2)

e^(i*pi/2) = 0 + i

e^(i*pi/2) = i

Does anybody see where I am going here?

Yep, you guessed it!

Substitution. We replace all occurrences of i on the left-hand side of our equation with e^(i*pi/2). But, in doing so, we introduce one more instance of i into the equation, and we repeat the substitution once more. And again. And again… Forever.

So we have: [math]i = e^(i*pi/2) = e ^ ( (pi/2) * e ^ ( (pi/2) * e ^ ( (pi/2) * … ) ) )[/math]

I was just wondering if I did anything wrong here… it is kinda strange to see i written as an infinite exponentiation of real numbers.

Now that that's over, I can explain. (And keep in mind that i is the imaginary square root of −1.)

We (being anybody who knows) know, through a series of proofs that I will not restate here, that e^(i*x) = cos x + i * sin x.

Now, I just say that x is pi/2 (radians), giving me:

e^(i*pi/2) = cos (pi/2) + i * sin (pi/2)

e^(i*pi/2) = 0 + i

e^(i*pi/2) = i

Does anybody see where I am going here?

Yep, you guessed it!

Substitution. We replace all occurrences of i on the left-hand side of our equation with e^(i*pi/2). But, in doing so, we introduce one more instance of i into the equation, and we repeat the substitution once more. And again. And again… Forever.

So we have: [math]i = e^(i*pi/2) = e ^ ( (pi/2) * e ^ ( (pi/2) * e ^ ( (pi/2) * … ) ) )[/math]

I was just wondering if I did anything wrong here… it is kinda strange to see i written as an infinite exponentiation of real numbers.

Last edited by falseinfinities on Thu Jan 16, 2014 7:59 pm UTC, edited 1 time in total.

### Re: i as an Infinite Exponent

Well, it is quite clear that i is not the *limit* of the power tower e

On the other hand, we may consider the equation z = e

A somewhat-related concept is the infinite power tower i

^{½πe^…}, because every term in the sequence of partial towers is real.On the other hand, we may consider the equation z = e

^{½πz}, which can be solved with the Lambert W function. And i is indeed one of the solutions to that equation. Furthermore, it is easy enough to show that there are no real solutions, so the power tower cannot converge to any real number.A somewhat-related concept is the infinite power tower i

^{i^(i^…)}, which has been discussed on this forum here and does converge to a complex number.wee free kings

### Re: i as an Infinitely Tiered Exponent

It seems quite problematic to me to take an equation and substitute it into itself. For instance, if you have the equation x = x^2, you can't substitute x = x^2 into the right hand side to get x = x^4, x^8, etc. It seems to me that this is basically what you're trying to do here.

- falseinfinities
**Posts:**3**Joined:**Thu Jan 16, 2014 6:44 am UTC

### Re: i as an Infinitely Tiered Exponent

Wow... I didn't even know there was a Lambert W function... But that is exactly what I was looking for. But what do you mean by every term in the power sequence being real? If you mean every ½πe, then I agree.

If you mean e

We have f(2)=e

=e

=cis(½(-πi))=cos(-½πi)+i*sin(-½πi)

=cos(½πi)-i*sin(½πi)

This seems to be a complex number... But... Oh... This can just be converted to use hyperbolic functions...

=cosh(½π)+sinh(½π)

=e^(½π)

Which is obviously real. Especially so because that's where I started, and it was obviously real from the beginning. But going in circles is fun, sometimes. So in the end, I do believe you...

If you mean e

^{½πe^...}for a finite number of stages, (e.g. stage 0 is 1, stage 1 is e, stage 2 is e^{½π}, etc.), on the other hand:We have f(2)=e

^{½π}=e

^{½(-πi)i}=cis(½(-πi))=cos(-½πi)+i*sin(-½πi)

=cos(½πi)-i*sin(½πi)

This seems to be a complex number... But... Oh... This can just be converted to use hyperbolic functions...

=cosh(½π)+sinh(½π)

=e^(½π)

Which is obviously real. Especially so because that's where I started, and it was obviously real from the beginning. But going in circles is fun, sometimes. So in the end, I do believe you...

- Yakk
- Poster with most posts but no title.
**Posts:**11128**Joined:**Sat Jan 27, 2007 7:27 pm UTC**Location:**E pur si muove

### Re: i as an Infinitely Tiered Exponent

LaserGuy wrote:It seems quite problematic to me to take an equation and substitute it into itself. For instance, if you have the equation x = x^2, you can't substitute x = x^2 into the right hand side to get x = x^4, x^8, etc. It seems to me that this is basically what you're trying to do here.

Why not? If x = x^2, then x=x^16 as well.

One of the painful things about our time is that those who feel certainty are stupid, and those with any imagination and understanding are filled with doubt and indecision - BR

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

Last edited by JHVH on Fri Oct 23, 4004 BCE 6:17 pm, edited 6 times in total.

### Re: i as an Infinitely Tiered Exponent

LaserGuy wrote:It seems quite problematic to me to take an equation and substitute it into itself. For instance, if you have the equation x = x^2, you can't substitute x = x^2 into the right hand side to get x = x^4, x^8, etc. It seems to me that this is basically what you're trying to do here.

It is not problematic, it just might introduce spurious solutions. Going with your example, if x=x

^{2}then necessarily x=x

^{4}, but the later has introduce two new solutions not present in the original.

falseinfinities wrote:Wow... I didn't even know there was a Lambert W function... But that is exactly what I was looking for. But what do you mean by every term in the power sequence being real? If you mean every ½πe, then I agree.

If you mean e^{½πe^...}for a finite number of stages, (e.g. stage 0 is 1, stage 1 is e, stage 2 is e^{½π}, etc.), on the other hand:

I mean the sequence given by z

_{0}= 0 and z

_{k+1}= e

^{½πzk}which is: {0, 1, e

^{½π}, e

^{½πe^(½π)}, e

^{½πe^(½πe^(½π))}, …}.

falseinfinities wrote:We have f(2)=e^{½π}

=e^{½(-πi)i}

=cis(½(-πi))=cos(-½πi)+i*sin(-½πi)

=cos(½πi)-i*sin(½πi)

This seems to be a complex number... But... Oh... This can just be converted to use hyperbolic functions...

=cosh(½π)+sinh(½π)

=e^(½π)

Which is obviously real. Especially so because that's where I started, and it was obviously real from the beginning. But going in circles is fun, sometimes. So in the end, I do believe you...

So, you're okay here, but be careful when you try to "factor" complex exponents. In particular, something like (e

^{a})

^{b}is, in general, not equal to e

^{ab}. For example, let a = 1+2πi and b = 1-2πi.

wee free kings

- MartianInvader
**Posts:**808**Joined:**Sat Oct 27, 2007 5:51 pm UTC

### Re: i as an Infinitely Tiered Exponent

You also have to be very careful when turning a repeated operation into a limit, *especially* if you're doing something at the end of the operation that gets "pushed out to infinity" (in your case, multiplying by i). After all, I could just as easily say:

0 = 1*0

0 = 1*1*0

0 = 1*1*1*...*0

0 = 1*1*1*1*1*1...

0 = 1

Aha! I've proved that 0=1, and all of mathematics is flawed!

0 = 1*0

0 = 1*1*0

0 = 1*1*1*...*0

0 = 1*1*1*1*1*1...

0 = 1

Aha! I've proved that 0=1, and all of mathematics is flawed!

Let's have a fervent argument, mostly over semantics, where we all claim the burden of proof is on the other side!

- falseinfinities
**Posts:**3**Joined:**Thu Jan 16, 2014 6:44 am UTC

### Re: i as an Infinitely Tiered Exponent

MartianInvader wrote:You also have to be very careful when turning a repeated operation into a limit, *especially* if you're doing something at the end of the operation that gets "pushed out to infinity" (in your case, multiplying by i). After all, I could just as easily say:

0 = 1*0

0 = 1*1*0

0 = 1*1*1*...*0

0 = 1*1*1*1*1*1...

0 = 1

Aha! I've proved that 0=1, and all of mathematics is flawed!

Sure, I agree with what you're saying, but your example is slightly flawed.

It's kinda hard to count things after infinity, so 1*1*1*1*1*1*...*0 is like saying (*gulp*) the difference between 1 and 0.999... . The last element cannot exist after an infinite number of elements beforehand.

Therefore, you cannot say that 0 = 1*1*1*1*1... because it is completely different from your previous equations.

The way around this is to treat any infinite sum, product, etc. as a single entity (put it in brackets).

Then, you get the more correct, 0 = 0*(1*1*1*...) = (1*1*1...)*0

### Re: i as an Infinitely Tiered Exponent

falseinfinities wrote:MartianInvader wrote:You also have to be very careful when turning a repeated operation into a limit, *especially* if you're doing something at the end of the operation that gets "pushed out to infinity" (in your case, multiplying by i). After all, I could just as easily say:

0 = 1*0

0 = 1*1*0

0 = 1*1*1*...*0

0 = 1*1*1*1*1*1...

0 = 1

Aha! I've proved that 0=1, and all of mathematics is flawed!

Sure, I agree with what you're saying, but your example is slightly flawed.

It's kinda hard to count things after infinity, so 1*1*1*1*1*1*...*0 is like saying (*gulp*) the difference between 1 and 0.999... . The last element cannot exist after an infinite number of elements beforehand.

Therefore, you cannot say that 0 = 1*1*1*1*1... because it is completely different from your previous equations.

The way around this is to treat any infinite sum, product, etc. as a single entity (put it in brackets).

Then, you get the more correct, 0 = 0*(1*1*1*...) = (1*1*1...)*0

I don't see how it's fundamentally different. I mean, the only difference is that here, you can assume associativity. For illustration, why not create a new operation &. Who knows how its defined - it doesn't matter. We'll write x & y for an application of this operation to two variables. We could write

x = a & x

And correctly infer

x = a & (a & x)

x = a & (a & (a & x))

but then we might write

x = a & (a & (a & ...))

And voila, the x is gone! It's not at the end, because there is no end because it just keeps nesting. It's gone. If you made it so that

a & x = e^(pi/2*x)

then this is exactly what you did. If you made it

a & x = 1 * x

then this is what MartianInvader did. The difference you see is that, for multiplication, it is not necessary to put in as many parenthesis as I did, because (ab)c=a(bc) - the associative law. But why should we decide that an infinite application of something which is associative magically gets an extra term where something which is not does not? Surely the structure of

x = 1 * (1 * (1 * (1 * ...)))

has no room for an x!

Mathematical hangover (n.): The feeling one gets in the morning when they realize that that short, elementary proof of the Riemann hypothesis that they came up with at midnight the night before is, in fact, nonsense.

- phlip
- Restorer of Worlds
**Posts:**7572**Joined:**Sat Sep 23, 2006 3:56 am UTC**Location:**Australia-
**Contact:**

### Re: i as an Infinitely Tiered Exponent

A common way of resolving this problem is by defining a sequence:

a

a

and then asking whether the limit a

For example, for f(x) = 0.9 + x/10, which is is essentially equivalent to a

I don't know what the case is for f(x) = e

a

_{0}= ka

_{n}= f(a_{n-1})and then asking whether the limit a

_{∞}depends on k.For example, for f(x) = 0.9 + x/10, which is is essentially equivalent to a

_{∞}= 0.999..., we find that a_{∞}= 1 regardless of the value of k. However, for f(x) = 1 * x, trivially a_{∞}= k, so just saying f(f(f(f(...)))) isn't well defined without knowing what the value is "infinitely deep" in the nesting.I don't know what the case is for f(x) = e

^{x*pi/2}, but it would be an interesting thing to find out.Code: Select all

`enum ಠ_ಠ {°□°╰=1, °Д°╰, ಠ益ಠ╰};`

void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);}

### Re: i as an Infinitely Tiered Exponent

phlip wrote:I don't know what the case is for f(x) = e^{x*pi/2}, but it would be an interesting thing to find out.

The partial terms diverge.

- phlip
- Restorer of Worlds
**Posts:**7572**Joined:**Sat Sep 23, 2006 3:56 am UTC**Location:**Australia-
**Contact:**

### Re: i as an Infinitely Tiered Exponent

Derek wrote:The partial terms diverge.

Well sure, for real numbers, e

^{x*pi/2}>> x for all x... but what about complex numbers? Are there any complex numbers that it converges for, other than fixed points? What do the basins look like for each one?

Code: Select all

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void ┻━┻︵╰(ಠ_ಠ ⚠) {exit((int)⚠);}

- Voekoevaka
**Posts:**42**Joined:**Wed Apr 10, 2013 10:29 am UTC**Location:**Over nine thousand.

### Re: i as an Infinitely Tiered Exponent

Ok.

You prooved that i is a root of the equation : x=e

But iterating u

When you try to differenciate the function f(x)=e

You prooved that i is a root of the equation : x=e

^{xπ/2}.But iterating u

_{n+1}==e^{unπ/2}, starting with any real number will make a divergent sequence, going to infinity. Starting with i will make you stay in i.When you try to differenciate the function f(x)=e

^{xπ/2}, you get f'(x)=(π/2)e^{xπ/2}, and by taking its walue on i, you get iπ/2. Its absolute value is π/2, which is bigger than 1, so the point is repulsive. By starting your sequence near i, but not exactly on i, you will be pushed out.I'm a dozenalist and a believer in Tau !

### Re: i as an Infinitely Tiered Exponent

But the magnitude of the derivative is π/2 everywhere. So if I'm interpreting it right, it's repulsive everywhere.

I've made some graphs with my complex function grapher. I'm not sure if some of them are correct, but at least they look pretty. Each picture is a link to a fullscreen-sized version.

The lower left corner of each graph is at -(2 + 3.56i) and the upper right corner is at 2 + 3.56i.

First, here's what f(z) = z looks like:

f(z) = e

f(z) - z

f

f

f

f

f

f

f

Past 24 iterations I'm having trouble telling the zeros apart from specks of dust on the screen, also resolution. But from the looks of it, there might be other points for which the sequence converges.

Disclaimer: not actually a mathematician but an enthusiastic high-schooler.

I've made some graphs with my complex function grapher. I'm not sure if some of them are correct, but at least they look pretty. Each picture is a link to a fullscreen-sized version.

The lower left corner of each graph is at -(2 + 3.56i) and the upper right corner is at 2 + 3.56i.

First, here's what f(z) = z looks like:

**Spoiler:**

^{zπ/2}**Spoiler:**

**Spoiler:**

^{2}(z) - z = f(f(z)) - z**Spoiler:**

^{3}(z) - z**Spoiler:**

^{4}(z) - z**Spoiler:**

^{5}(z) - z**Spoiler:**

^{6}(z) - z**Spoiler:**

^{7}(z) - z**Spoiler:**

^{21}(z) - z**Spoiler:**

Disclaimer: not actually a mathematician but an enthusiastic high-schooler.

### Re: i as an Infinitely Tiered Exponent

Voekoevaka wrote:Ok.

You prooved that i is a root of the equation : x=e^{xπ/2}.

But iterating u_{n+1}==e^{unπ/2}, starting with any real number will make a divergent sequence, going to infinity. Starting with i will make you stay in i.

When you try to differenciate the function f(x)=e^{xπ/2}, you get f'(x)=(π/2)e^{xπ/2}, and by taking its walue on i, you get iπ/2. Its absolute value is π/2, which is bigger than 1, so the point is repulsive. By starting your sequence near i, but not exactly on i, you will be pushed out.

Worse, actually, is that f'(x) happens to equal π/2f(x) in this case, meaning for fixed points x=f(x) it holds that f'(x)=π/2x. So an attractive fixed point would have to be in the ball of radius 2/π around 0. No such fixed points exist. Attractive cycles could exist, but I doubt it.

Mathematical hangover (n.): The feeling one gets in the morning when they realize that that short, elementary proof of the Riemann hypothesis that they came up with at midnight the night before is, in fact, nonsense.

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