Solve a simple exponential recursive equation?

Question

I want to fill a range between two values, not linearly but exponentially. For this I use a starting value L take exponent R, and repeat for N number of steps.

L = 1e-7
R = 0.9997   # needs solving
N = 1000
for _ in range(N):
    L = L ** R
>>> L
6.523-06

Need to solve for R such that end value of L after N repeats is 1e-05.

Edit: This is not the same question as suggested. In the linked question the formula linked is $ f(x) = a \times r^s.$ For my question, the formula is different: $L[t] = L[t-1]^R$ where t is a recursion step with maximum steps N.

Does this answer your question? Convert 60 Linear steps to Exponential from (0 - 5) — Ethan Bolker, Mar 20 '24 at 22:04
$L^{R^N}$ iis the final value, So you need: $R^N\geq \log_{L} (1e-05)$ or $N=\log_R\log_{L}(1e-05).$ Of course, you'll need to pick an integer, and it depends if you want the first integer such that the value is $\geq 1e-05$ or the last integer it is not. — Thomas Andrews, Mar 20 '24 at 22:16
It would help if you stated the problem in math form rather than program form - not everyone here knows Python, nor that ** is exponentiation. This is not a programmer forum. — Thomas Andrews, Mar 20 '24 at 22:23

Matija · Accepted Answer · 2024-03-20T22:37:47.037

Let $c=10^{-7}$, $N=1000$ and $C=10^{-5}$ be the constants from the questions. Recall that $(c^R)^R=c^{R\cdot R}=c^{R^2}$, further $$((c^R)^R)^R=(c^{R^2})^R=c^{R^2\cdot R}=c^{R^3}$$ and so on. Thus, the program computes $c^{R^N}$ and the task is to solve $$c^{R^N}=C$$ for $R$. Taking the logarithm on both sides yields $$R^N\log c=\log C.$$ With basic transformations we get $$R=\left(\frac{\log(C)}{\log(c)}\right)^{1/N}.$$ This is the general solution. For the given numbers this yields $R=(5/7)^{0.001}$.

Solve a simple exponential recursive equation?

1 Answers1