Prove that for n∈N^∗ Σ_(p=0) ^(n−1) [x+(p/n)]=[nx]


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Question Number 74980 by ~blr237~ last updated on 05/Dec/19

$Prove that for n \in N^{*}$ $\underset{p = 0}{\sum^{n - 1}} [x + \frac{p}{n}] = [nx]$
	Answered by mind is power last updated on 05/Dec/19
	$let f(x)=Σ_(p=0) ^(n−1) [x+(p/n)] f(x+(1/n))=Σ_(p=0) ^(n−1) [x+(1/n)+(p/n)]=Σ_(p=1) ^n [x+(p/n)]=Σ_(p=1) ^(n−1) [x+(p/n)]+[x+1]=1+[x]+Σ_(p=1) ^(n−1) [x+(p/n)] =1+Σ_(p=0) ^(n−1) [x+(p/n)]=1+f(x)....E g(x)=[nx] g(x+(1/n))=[n(x+(1/n))]=1+[nx] ⇒we have juste to show for x∈[0,(1/n)[ x∈[0,(1/n)[⇒x+(p/n)<(1/n)+((n−1)/n)<1,∀p∈[0,n−1] ⇒f(x)=0 x<(1/n)⇒nx<1⇒g(x)=0 g(x)=f(x)=0,∀x∈[0,(1/n)[⇒f(x)=g(x),∀x∈R withe ∀x∈R if x∈[0,(1/n)[ we are donne x=((E(nx))/n)+x−((E(nx))/n) x−((E(nx))/n)∈[0,(1/n)[ E(nx)≤nx⇒x−((E(nx))/n)>0 nx−1<E(nx) ⇒((E(nx))/n)>x−(1/n)⇒x−((E(nx))/n)<(1/n) ⇒f(x)=f(((E(nx))/n)+{x−((E(nx))/n)}) E(nx)+f(x−((E(nx))/n))=k since f=0 in [0,(1/n)[ g(((E(nx))/n)+x−((E(nx))/n))=E(nx)+g(x−((E(nx))/n))=E(nx) since g=0 in [0,(1/n)[ ⇒g(x)=f(x),∀x∈R$
	$let f (x) = \underset{p = 0}{\sum^{n - 1}} [x + \frac{p}{n}]$ $f (x + \frac{1}{n}) = \underset{p = 0}{\sum^{n - 1}} [x + \frac{1}{n} + \frac{p}{n}] = \underset{p = 1}{\sum^{n}} [x + \frac{p}{n}] = \underset{p = 1}{\sum^{n - 1}} [x + \frac{p}{n}] + [x + 1] = 1 + [x] + \underset{p = 1}{\sum^{n - 1}} [x + \frac{p}{n}]$ $= 1 + \underset{p = 0}{\sum^{n - 1}} [x + \frac{p}{n}] = 1 + f (x) . . . . E$ $g (x) = [nx]$ $g (x + \frac{1}{n}) = [n (x + \frac{1}{n})] = 1 + [nx]$ $\Rightarrow we have juste to show for x \in [0, \frac{1}{n} [$ $x \in [0, \frac{1}{n} [\Rightarrow x + \frac{p}{n} < \frac{1}{n} + \frac{n - 1}{n} < 1, \forall p \in [0, n - 1]$ $\Rightarrow f (x) = 0$ $x < \frac{1}{n} \Rightarrow nx < 1 \Rightarrow g (x) = 0$ $g (x) = f (x) = 0, \forall x \in [0, \frac{1}{n} [\Rightarrow f (x) = g (x), \forall x \in R withe$ $\forall x \in R if x \in [0, \frac{1}{n} [we are donne$ $x = \frac{E (nx)}{n} + x - \frac{E (nx)}{n}$ $x - \frac{E (nx)}{n} \in [0, \frac{1}{n} [$ $E (nx) ⩽ nx \Rightarrow x - \frac{E (nx)}{n} > 0$ $nx - 1 < E (nx) \Rightarrow \frac{E (nx)}{n} > x - \frac{1}{n} \Rightarrow x - \frac{E (nx)}{n} < \frac{1}{n}$ $\Rightarrow f (x) = f (\frac{E (nx)}{n} + {x - \frac{E (nx)}{n}})$ $E (nx) + f (x - \frac{E (nx)}{n}) = k since f = 0 in [0, \frac{1}{n} [$ $g (\frac{E (nx)}{n} + x - \frac{E (nx)}{n}) = E (nx) + g (x - \frac{E (nx)}{n}) = E (nx)$ $since g = 0 in [0, \frac{1}{n} [$ $\Rightarrow g (x) = f (x), \forall x \in R$
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