341ah bipin's here. Hi hsbhatt here :D
Let a and b be 2 coprime +ve integers and f(x) =[x] (greatest integer function). show that
\sum_{n=1}^{b-1}{f\left( \frac{na}{b}\right)}=\frac{(a-1)(b-1)}{2}
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62Prophet sir, the folklore goes like this..
bipin and nishant know each other from Kharagpur...
Bipin joined goiit.. I started TargetIIT ;)
:)
341and bipn came to targetiit and then they lived happily ever after :D
1HAPPYSSSSSSSSSSSSSSSS
ENDINGSSSSSSSSSSSSSSSSSSSSS!!!!!!!!!!
SRY FOR POOR ENGLISH[3][3][3][3]
62arrey amid all this, you all forgot to find the solution to the brilliant question
341A corollary of this result forms part of proof one of Gauss's jewels in Number Theory known as Quadratic Reciprocity Law. This result too was discovered by the Prince of Mathematicians.
1thanks nishant bhaiyya, i got to know abt u two. and yeah, as Jaiho says, its the perfect happy ending. LOL,
Hint: relationship between [x] and [-x] when x isnt an integer
62arrey it was never bad between us :D
Lol I guess bipin must also be laughing already ;)
1hahaaa....hi hari sir and nishant sir :D and other members of targetiit
bad between us :O .... arey KD (the nick we used for him) came to my room when this site was in trial run and asked me to give my critical suggestions....
he is a high voltage guy and one of the best mathematician, programmer bla bla bla i have ever met face to face and always carrying a refreshing smile on his face....and he is proving himself as he is directly addressing people not as emperor (although he is) but as a member......
chal bhai zada ho gaya :D .... if i wud have an orkut account i wud have surely send the above as testi :)
between solve the problem guys..
1[a/b]+[2a/b]+[3a/b]+...[ra/b]....+[(b-r)a/b]....+[(b-2)a/b]+[(b-1)a/b]
= [a/b]+[2a/b]+[3a/b]+...
+ a +[-ra/b]+....a+[-3a/b] + a+[-2a/b]+a+[-a/b]
if x is not an integer, then [-x]=-1-[x]
here ra/b where r={1,2..b-1} isnt an integer as a,b are coprime.
so we have the above expression as
[a/b]+[2a/b]+[3a/b]+...
+ a -1-[ra/b] +...+a-1-[3a/b]+a-1-[2a/b]+a-1-[a/b]
so we ultimately have a(b-1)/2 -(b-1)/2
= (a-1)(b-1)/2
cheers!!