Exercise-1.14(new book) (BOMC-2011)

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sm.joty
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Exercise-1.14(new book) (BOMC-2011)

Unread post by sm.joty » Thu Oct 27, 2011 11:51 am

Can anyone explain this problem. I can't understand the phrase " the fractional part".


Prove that for any positive integer n, the fractional part of \[\sqrt{4n^2+n}\]
is smaller than\[\frac{1}{4}\]
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Re: Exercise-1.14(new book) (BOMC-2011)

Unread post by sourav das » Thu Oct 27, 2011 12:13 pm

Fractional part of ($\sqrt 2 = 1.414...$) is ($.414...$)
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Re: Exercise-1.14(new book) (BOMC-2011)

Unread post by sm.joty » Thu Oct 27, 2011 12:16 pm

many many thanks Sourav da. :D
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বাধা-বিঘ্ন না পেরিয়ে
বড় হয়েছে কে কবে.........

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Re: Exercise-1.14(new book) (BOMC-2011)

Unread post by rakeen » Mon Oct 31, 2011 3:22 pm

need hints(the more the better;coz it seems a bit difficult!)
r@k€€/|/

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Re: Exercise-1.14(new book) (BOMC-2011)

Unread post by *Mahi* » Mon Oct 31, 2011 7:43 pm

Try finding a perfect square just greater than $4n^2+n$.
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Re: Exercise-1.14(new book) (BOMC-2011)

Unread post by sourav das » Mon Oct 31, 2011 8:06 pm

Or just less than $4n^2 + n$
You spin my head right round right round,
When you go down, when you go down down......
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Re: Exercise-1.14(new book) (BOMC-2011)

Unread post by *Mahi* » Mon Oct 31, 2011 8:07 pm

sourav das wrote:Or just less than $4n^2 + n$
Nope, the problem says the fractional part smaller than $\frac 1 4$
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Re: Exercise-1.14(new book) (BOMC-2011)

Unread post by nafistiham » Mon Oct 31, 2011 8:10 pm

*Mahi* wrote:Try finding a square integer just greater than $4n^2+n$.
like $9n^{2}$ ?
as $\sqrt{9n^{2}}=3n $ ???
\[\sum_{k=0}^{n-1}e^{\frac{2 \pi i k}{n}}=0\]
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Re: Exercise-1.14(new book) (BOMC-2011)

Unread post by *Mahi* » Mon Oct 31, 2011 8:14 pm

No, something greater than $4n^2+n$ which you can express as perfect square.
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Re: Exercise-1.14(new book) (BOMC-2011)

Unread post by sourav das » Mon Oct 31, 2011 8:16 pm

*Mahi* wrote:
sourav das wrote:Or just less than $4n^2 + n$
Nope, the problem says the fractional part smaller than $\frac 1 4$
Any of them will work. See:
$4n^2$ is just smaller than $4n^2 + n$
So let $\sqrt{4n^2 + n}=2n + x $ here x is the fractional part. then do it in your way.
You spin my head right round right round,
When you go down, when you go down down......
(-$from$ "$THE$ $UGLY$ $TRUTH$" )

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