I-m-0-1-2-m-x-3-m-n-m-1-2-n-x-3-n-dx-then-find-the-value-of-I-m-1-I-m-

Question Number 197919 by universe last updated on 04/Oct/23

I_{m} = \int_{0}^{1} (\frac{⌊ 2^{m} x ⌋}{3^{m}} \underset{n = m + 1}{\sum^{\infty}} \frac{⌊ 2^{n} x ⌋}{3^{n}}) d x

then find the value of

I = \underset{m = 1}{\sum^{\infty}} I_{m} = ?

Answered by witcher3 last updated on 06/Oct/23

2^{m} x = t

\Leftrightarrow I_{m} = \frac{1}{2^{m}} \int_{0}^{2^{m}} \frac{[t]}{3^{2 m}} \sum_{n ⩾ m + 1} \frac{[2^{n - m} t]}{3^{n - m}}

\Rightarrow 18^{m} I_{m} = \int_{0}^{2^{m}} [t] \sum_{n ⩾ 1} \frac{[2^{n} t]}{3^{n}} dt

= \underset{k = 0}{\sum^{2^{m} - 1}} \int_{k}^{k + 1} k . \sum_{n ⩾ 1} \frac{[2^{n} t]}{3^{n}} dt, 2^{n} t = y

= \sum_{k} \sum_{n} \int_{2^{n} k}^{(k + 1) 2^{n}} \frac{k}{6^{n}} [y] dy = 18^{m} I_{m}

\int_{2^{n} k}^{(k + 1) 2^{n}} [y] dy = \underset{j = 0}{\sum^{2^{n} - 1}} \int_{2^{n} k + j}^{2^{n} k + 1 + j} [y] dy

= \underset{j = 0}{\sum^{2^{n} - 1}} 2^{n} k + jdy

= 2^{2 n} k + 2^{n - 1} (2^{n} - 1) =

18^{m} I_{m} = \sum_{k} \sum_{n ⩾ 1} \frac{k (2^{2 n} k + 2^{n - 1} (2^{n} - 1)}{6^{n}}

= \sum_{k} \sum_{n ⩾ 1} k^{2} {(\frac{2}{3})}^{n} + \frac{k}{2} . {(\frac{2}{3})}^{n} - \frac{k}{2} {(\frac{2}{6})}^{n}

= \underset{k = 0}{\sum^{2^{m} - 1}} . ({2 k}^{2} + \frac{3}{4} k) = \frac{(2^{m} - 1) (2^{m}) (2^{m + 1} - 1)}{3} + \frac{3}{4} 2^{m - 1} (2^{m} - 1)

I_{m} = \frac{2^{3 m + 1} - 3 . 2^{2 m} + 2^{m}}{3 . 18^{m}} + \frac{3 . 2^{2 m - 1} - 3 . 2^{m}}{4 . 18^{m}}

I = \sum_{m ⩾ 1} \frac{2}{3} . {(\frac{2}{3})}^{2 m} - {(\frac{2}{9})}^{m} + \frac{1}{3} . \frac{1}{9^{m}} + \frac{3}{8} . {(\frac{2}{9})}^{m} - \frac{3}{4} . {(\frac{1}{9})}^{m}

easy from here

\sum_{n ⩾ 1} a^{n} = \frac{a}{1 - a}, \forall ∣ a ∣< 1

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