Exponential Recurrence Relation

[Woxor]

I'm assuming this is the proper forum to post this in, even though I'm asking rather than testing whether anyone knows the answer to this (because I don't).

 

Does anyone know how to solve [math]a_{n+1}=e^{a_{n}}[/math] explicitly for [math]a_{n}[/math] in terms of [math]n[/math] and [math]a_{0}[/math]? I would greatly appreciate knowing both the solution and the method used to obtain it, since my own studies have taken me where a solution would be extremely helpful. I haven't been able to determine whether this is a simple textbook equation or a problem entirely unsolved by the mathematical community.

[Dave]

Hmm. I could be misinterpreting this, but it seems like a pretty trivial difference equation. Here's my reasoning:

 

Say you've got some starting value a₀. Then we have:

 

[math]a_1 = e^{a_0}, a_2 = e^{e^{a_0}}, a_3 = e^{e^{e^{a_0}}}[/math] etc...

[Woxor]

Those values are correct, but I'm looking for something more or less like this: [math]a_{n}=a_{0}*f(n)[/math], except with an actual, explicit function.

 

EDIT: Also, I realize that "[math]a_{n}=\phi^{n}(a_{0})[/math] where [math]\phi(x)=e^{x}[/math]" is the solution, but I need something even more explicit than that. In other words, what is [math]\phi^{n}[/math] in terms of [math]n[/math]?

[bishnu]

There might not even be an answer to your question...an explict formula for iteratitive functions is sometimes impossible to create because often time functions of that nature create tons of complexity, especially when there is a buried variable such as a0...in my opinion espcially since the the equation rapidly diverges for most values if not all values of a0 you would be better off using a programmable calculator to rapidly do the iterations for you.

[Woxor]

Unfortunately, that's no good since I need to work with the function ([math]\phi^{n}[/math]) analytically -- numerical values aren't too hard to obtain.

 

As for the initial value: it seems as though [math]a_{0}=0[/math] would suit my purposes fine, so there's not necessarily a buried variable. As for the divergence, there's nothing vital about the base -- [math]a_{n+1}=e^{\frac{a_{n}}{e}}[/math] would be just as enlightening. (Interestingly, [math]e^{\frac{1}{e}}[/math] seems to be the highest base value for which this converges.)

 

But yeah, I am certainly worried as to whether it's even possible. I haven't taken any classes specifically on recurrence relations, and I haven't worked much with the Reimann zeta function, so I was just wondering if anyone out there actually happens to know one way or the other.

[bishnu]

well this is kindof off the topic but i can show you the limits of infinite powers and whyit could help you..sadly its variables cant be solved algebratically,

 

[math]x_(n+1)=x_n^d[/math]

[math]x_{final}=x_0^{d^{d^{d^{d^{...}}}}}[/math] you interate to infinty and set you x0 to d you get an infinte amount of powersokay here is how to solve for the values

[math]\ln{x_f}=\ln{d}*d^{d^{d^{d^{d^{...}}}}}[/math]

no since the number of d's are infinte you can subsitiute the orginal equation in so you get [math]\ln{x_f}=\ln{d}*x_f[/math]

[math]d=x_f^{\tfrac{1}{x_f}}[/math]

now this cant be solved algebratically but it can be solved using newtons method

now lets pretend that that the equation is the inverse of the one above so you plug x_f in instead of d...now use dervatives to find the max value of the function has a max at e and then decreases so hence forth the max value of d=e^(1/e), since the function we just anaylised is the inverse of the function for the value of the power tree we find that the inverse function is a function up to the point e^(1/e) hence the max valule d can be on converge is e^(1/e) where it converges nicely to e. Thats how to solve the power tree i hope that makes sense and will help you a little.

Edit sorry about the error in one of my equations it should be fixed now