I’m trying to compute a sequence ‘$p$’ such that $m(C(p))=1/10$. where ‘$m$’ is ‘measure’ and ‘$C$’ is ‘Cantor set’
This is my approach: at step n, I remove $2^{n-1}$ times $p^n$, so the total subset I remove at step n has a measure of $\sum 2^{n-1}p^n$
now $\sum 2^{n-1}p^n = \frac{p}{1-2p}$ , which I force to sum up to $9/10$
this equation solves for $p=9/28$
therefore the sequence I’m looking for is: $9/28, 9/56, \dots ,(9/28)^n, \dots$
Is my reasoning correct? thanks! Alessandro
Assume that you start with the interval $[0, 1]$ having length $1$. Let $l_n$ be the length after step $n$ with $l_0=1.$ Let $r_n$ be the ratio of the current length that you keep in every step, i.e. $l_n = r_n l_{n-1}.$ You want $\lim_{n\to\infty} l_n = 1/10.$ But $l_n = r_1 r_2 r_3 \cdots r_n$ so you want to find a sequence $r_1, r_2, r_3, \ldots$ such that $$\prod_{n=1}^{\infty} r_n = \frac{1}{10}.$$
Since we are not very used to products it might be easier to rewrite this as a sum by taking the logarithm: $$\sum_{n=1}^{\infty} \lg r_n = -1.$$ Here, $\lg$ is the logarithm with respect to base 10.