# Solving $f_n=\exp(f_{n-1})$ : Where is my mistake?

I was trying to solve the recurrence $f_n=\exp(f_{n-1})$. My logic was this : $f_n -f_{n-1}=\exp(f_{n-1})-f_{n-1}$. The associated differential equation would then be $\dfrac{dg}{dn}=e^g-g$. if $f(m)=g(m)>0$ for some real $m$ then for $n>m$ we would have $g(n)>f(n)$. Solving the differential equation with seperating the variables gives the solution $g(n)= \mathrm{inv}(\int_c^{n} \dfrac{dt}{e^t-t}+c_1)+c_2$ for some $c,c_1,c_2$. That solutions seems correct since $e^t-t=0$ has no real solution for $t$ so there are no issues with singularities near the real line.

However $\mathrm{inv}(\int_c^{n} \dfrac{dt}{e^t-t}+c_1)+c_2$ Does NOT seem near $f_n$ let alone larger than it !! So where did I make a big mistake in my logic ? And can that mistake be fixed ?

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I assume there is some $f_0$? –  Asaf Karagila Jun 1 '13 at 21:56
Yes. But that value is not so important for the problem im having right ? –  mick Jun 1 '13 at 21:57
Why inv(...) "Does NOT seem near fn let alone larger than it"? –  Did Jun 1 '13 at 21:59
@Did the solution to $f_n$ grows superexponentially whereas the inv... does not , right ?! –  mick Jun 1 '13 at 22:01
Superexponential growth means faster than $e^n , e^{e^n} , ...$ For those unaware of the meaning of that word. –  mick Jun 1 '13 at 22:05
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The problem is that the primitive $\displaystyle\int_\cdot^x\frac{\mathrm dt}{\mathrm e^t-t}$ does not converge to infinity when $x\to+\infty$.
The comparison between $(f_n)$ and $g$ reads $$\int_{f_1}^{f_n}\frac{\mathrm dt}{\mathrm e^t-t}\leqslant n-1,$$ for every $n\geqslant1$. When $n\to\infty$, the LHS converges to a finite limit hence one can be sure that the LHS and the RHS are quite different when $n\to\infty$ and that this upper bound becomes trivial for every $n$ large enough.
Take-home message: to compare the sequence $(f_n)$ solving a recursion $f_{n+1}=f_n+h(f_n)$ and the function $g$ solving the differential equation $g'(t)=h(g(t))$ can be fruitful only when the integral $\displaystyle\int_\cdot^{+\infty}\frac{\mathrm dt}{h(t)}$ diverges.
???    –  Did Jun 2 '13 at 14:07