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Differential Calculus

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20 Jul 2007 21:06:41 IST

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limits problem...plz solve

None

lim [x]+[2x]+[3x]+................[nx]

n²

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Titun

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Joined: 23 Dec 2006

Posts: 374

20 Jul 2007 21:32:40 IST

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[ y ] denotes the greatest integer less than or equal to x

[ y ] = y - { y } where, {y} represents the fractional part of x.

Now, _{[n ]}

_{[infinity ]} ( [x] + [2x] + [3x] + ........+[nx] ) / n²

= _{[n ]}

_{[infinity ]} (x + 2x + 3x + ..........+ nx ) / n² - _{[n ]}

_{[infinity ]} ( {x} + {2x} + ............ + {nx} ) / n²

= _{[n ]}

_{[infinity ]} x . (1 + 2 + .........n) / n² - P

where, P = _{[n ]}

_{[infinity ]} ( {x} + {2x} + ............ + {nx} ) / n²

= _{[n ]}

_{[infinity ]} x . n (n + 1) / 2 . n² - P

= _{[n ]}

_{[infinity ]} x . (1 + 1/n) / 2 - P

= x / 2 - P

We know, that { y } for any real number y represents the fractional part of y
and 0

{ y } < 1

Therefore,
P = _{[n ]}

_{[infinity ]} ( {x} + {2x} + ............ + {nx} ) / n²

_{[n ]}

_{[infinity ]}( 1 + 1 + 1 +1 + 1.................. n times ) / n²

[ since, each term {x}, {2x}, {3x} ...........{nx} < 1 ]

i.e P

_{[n ]}

_{[infinity ]} n / n² = _{[n ]}

_{[infinity ]} 1 / n = 0

i.e P

0

But P can't be negative as each term of the limit is positive. {x}/n², {2x}/n²,.........{nx}/n²each of the terms are positive. So, the required limit has to be non-negative ( i.e either positive or zero )

Hence P must be zero as P can't be negative.

Hence,
_{[n ]}

_{[infinity ]} ( [x] + [2x] + [3x] + ........+[nx] ) / n²

= x / 2 - P = x / 2 - 0 = x / 2

Ans: x / 2

Cheers !!!!!!

Khinar Thukral

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Joined: 8 Feb 2007

Posts: 309

21 Jul 2007 10:24:20 IST

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There can be an easier method also .See we know that the minimum value of the greatest integer function [x] is x-1 and the maximum value is x. . By remembering all this we can rewrite the question as

x-1 <[x]

2(x-1)<[2x]<2x---------------2

3(x-1)<[3x]<3x----------------3
|
|
so on

n(x-1)<[nx]--------------n

adding the equations 1 to n we get

x-1+2x-2+3x-3-------nx-n<[1x]+[2x]+[3x]------[nx]

on separating the terms x+2x+3x---nx from terms ?(1+2+3-----n) and diving both the sides by n^2

(x+2x+3x-----nx)-n/n^2 < the question given<(x+2x+3x-----nx)/n^2

applying the formulae if the summation of n terms

nx(n+1)/2n^2-1/n < the question given < nx(n+1)/2n^2

now taking n common from both the sides of inequality and applying the limits we get

x/2

So the limit is x/2

This method is sandwich theorem.

Khinar Thukral

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Joined: 8 Feb 2007

Posts: 309

21 Jul 2007 10:25:07 IST

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Plzz rate if found easy.

joy francis

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Joined: 19 Feb 2007

Posts: 1802

22 Jul 2007 08:31:31 IST

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Excellent answer given by khinart.

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