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Question Number 200004 by Mingma last updated on 12/Nov/23

	Answered by aleks041103 last updated on 12/Nov/23
	$−1<x<1 f(x)=((x+2^k )/2^(k+1) )=(x/2^(k+1) )+(1/2)⇒(1/2)−(1/2^(k+1) )<f(x)<(1/2)+(1/2^(k+1) ) ⇒⌊(1/2)−(1/2^(k+1) )⌋≤⌊f(x)⌋<⌊(1/2)+(1/2^(k+1) )⌋ k=0: 0≤⌊f(x)⌋<1, ⇒⌊f(x)⌋=0 k≥1: ⌊f(x)⌋=0 ⇒Σ_(k=0) ^∞ ⌊((x+2^k )/2^(k+1) )⌋= { ((1, x=1)),((0, x∈[−1,1))) :} ⇒∫_(−1) ^( 1) Σ_(k=0) ^∞ ⌊((x+2^k )/2^(k+1) )⌋dx=0≠−1$
	$- 1 < x < 1$ $f (x) = \frac{x + 2^{k}}{2^{k + 1}} = \frac{x}{2^{k + 1}} + \frac{1}{2} \Rightarrow \frac{1}{2} - \frac{1}{2^{k + 1}} < f (x) < \frac{1}{2} + \frac{1}{2^{k + 1}}$ $\Rightarrow ⌊ \frac{1}{2} - \frac{1}{2^{k + 1}} ⌋ ⩽ ⌊ f (x) ⌋ < ⌊ \frac{1}{2} + \frac{1}{2^{k + 1}} ⌋$ $k = 0 : 0 ⩽ ⌊ f (x) ⌋ < 1, \Rightarrow ⌊ f (x) ⌋ = 0$ $k ⩾ 1 : ⌊ f (x) ⌋ = 0$ $\Rightarrow \underset{k = 0}{\sum^{\infty}} ⌊ \frac{x + 2^{k}}{2^{k + 1}} ⌋ = {\begin{cases} 1, x = 1 \\ 0, x \in [- 1, 1) \end{cases}$ $\Rightarrow \int_{- 1}^{1} \underset{k = 0}{\sum^{\infty}} ⌊ \frac{x + 2^{k}}{2^{k + 1}} ⌋ d x = 0 \neq - 1$
	Commented by Mingma last updated on 12/Nov/23
	Nice solution
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