Prove that for every real number x,
[x]+[x+1/n]+[x+2/n]+.........[x+(n-1)/n] = [nx]
[x]+[x+\frac{1}{n}]+[x+\frac{2}{n}]+...+[x+\frac{n-1}{n}]=[nx]
Where [] denotes greatest integer function
Because no one has been able to solve or even take one single good step, a hint... Dividing 1 in n parts, any fraction will lie in one of these n parts
(Edit: Latexified!)
1(Edited)
if we consider the following inequalities (Edited)
x-1<[x]≤x
x+1/n -1 <[x+1/n]≤x+1/n
.
.
.
.
.and so on
.till
x+(n-1)/n -1 <[x+(n-1)/n]≤x+(n-1)/n
adding up all the equalities on the right hand side we
get
l.h.s=(n-1)x+((n-1)(n-2)/2n)
as ((n-1)(n-2)/2n)>1
therefore
we can write it as [nx]
i hope i am correct...pls verify
1is this thinking correct
we observe that the terms being added to x are
tr = r/n where r ranges from 0 to n-1
thus the terms added to x are less than 1 thus every term in the expansion becomes
x+x+x+x+x+... + nterms = nx
62slightly different...
I mean the spirit will be the same but the total sum will be different...
341A typical dumbass solution would be start by assuming that
1 - \frac{k}{n} < \{x\} < 1 - \frac{k-1}{n} for some k lying between 1 and n-1
341But there is a beautiful solution, which starts by looking at the function:
f(x) = [nx] - \left([x] + \left[x + \frac{1}{n} \right ] +...+ \left[x +\frac{n-1}{n} \right] \right)
Check out what is f \left(x + \frac{1}{n} \right )-f(x)
62Dont worry philip.. keep trying :)
After Prophet sir's hint, it has become a piece of cake ... still give it a go..
@ Prophet sir, I now know why you have been such a fan of his... I am reading a couple of books these days that I should have long back.. Let me see if I should read this one too :)
