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Question Number 74980 by ~blr237~ last updated on 05/Dec/19
Prove that for n∈N^∗ Σ_(p=0) ^(n−1) [x+(p/n)]=[nx]
Provethatforn∈N∗∑n−1p=0[x+pn]=[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
letf(x)=∑n−1p=0[x+pn]f(x+1n)=∑n−1p=0[x+1n+pn]=∑np=1[x+pn]=∑n−1p=1[x+pn]+[x+1]=1+[x]+∑n−1p=1[x+pn]=1+∑n−1p=0[x+pn]=1+f(x)….Eg(x)=[nx]g(x+1n)=[n(x+1n)]=1+[nx]⇒wehavejustetoshowforx∈[0,1n[x∈[0,1n[⇒x+pn<1n+n−1n<1,∀p∈[0,n−1]⇒f(x)=0x<1n⇒nx<1⇒g(x)=0g(x)=f(x)=0,∀x∈[0,1n[⇒f(x)=g(x),∀x∈Rwithe∀x∈Rifx∈[0,1n[wearedonnex=E(nx)n+x−E(nx)nx−E(nx)n∈[0,1n[E(nx)⩽nx⇒x−E(nx)n>0nx−1<E(nx)⇒E(nx)n>x−1n⇒x−E(nx)n<1n⇒f(x)=f(E(nx)n+{x−E(nx)n})E(nx)+f(x−E(nx)n)=ksincef=0in[0,1n[g(E(nx)n+x−E(nx)n)=E(nx)+g(x−E(nx)n)=E(nx)sinceg=0in[0,1n[⇒g(x)=f(x),∀x∈R

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