What Functions Satisfy These Specific Recursive Conditions?

[lfdahl]

Define the numbers $e_k$ by $e_0 = 0$, $e_k = \exp(e_{k-1})$ for $k \geq 1$. Determine the functions, $f_k$, for which

\[f_0(x) = x, \;\;\;\;f’_k = \frac{1}{f_{k-1}f_{k-2}\cdot\cdot\cdot f_0}\;\;\;\; for\;\;\; k \geq 1.\]

on the interval $[e_k, \infty)$, and all $f_k(e_k) = 0.$

 

[lfdahl]

Hint:

Spoiler

Prove by induction, that $$f_k(x) = \ln^kx$$

- the $k$-fold composition of $\ln$ with itself.

 

[lfdahl]

Suggested solution:

Spoiler

We show by induction, that $f_k(x) = \ln^kx$, the k-fold composition of $\ln$ with itself. For $k = 0$, we have $f_0(x) = x = \ln^0 x$. Now assume, that $f_k(x) = \ln^kx$ for some $k \geq 0$. Then

\[f'_{k+1}(x) = \frac{1}{f_k(x)f_{k-1}(x)\cdot \cdot \cdot f_0(x)} = \frac{f'_k(x)}{f_k(x)}.\]

So

\[f_{k+1}(x) = \int \frac{f'_k(x)}{f_k(x)}dx = \ln f_k(x) + C = \ln \ln^kx+C = \ln^{k+1}x+C\]

for some constant, $C$. But

\[0 = f_{k+1}(e_{k+1}) = \ln^{k+1}e_{k+1}+C = C.\]

Hence $f_{k+1}(x) = \ln^{k+1}x$ establishing the induction.