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 May 11th, 2019, 03:19 PM #1 Senior Member   Joined: Dec 2015 From: somewhere Posts: 596 Thanks: 87 Integer part Compute the integer part of $\displaystyle y=\sqrt{1} +\sqrt{2} +... +\sqrt{12}$.(without calculator) $\displaystyle \lfloor y \rfloor =$? Last edited by idontknow; May 11th, 2019 at 03:25 PM.
 May 12th, 2019, 01:54 PM #2 Global Moderator   Joined: May 2007 Posts: 6,806 Thanks: 716 You need to get the square roots to enough places (2?) and then sum. If you are not allowed to use a calculator, there are at least 2 methods to get the square roots to a few decimal places.
May 12th, 2019, 03:09 PM   #3
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 Originally Posted by mathman You need to get the square roots to enough places (2?) and then sum. If you are not allowed to use a calculator, there are at least 2 methods to get the square roots to a few decimal places.
There are better ways arising from various formulas for the sum of the square roots of the first n positive integers. One such can easily be derived by integrating $\sqrt x$.

 May 12th, 2019, 04:04 PM #4 Senior Member   Joined: Dec 2015 From: somewhere Posts: 596 Thanks: 87 $\displaystyle \lfloor y \rfloor > 1+ \int_{1}^{12} \sqrt{x} dx \approx 28 \; \Rightarrow \lfloor y \rfloor =1+28=29$. Also by AM-GM : $\displaystyle \lfloor y \rfloor =2+ \lfloor 12\cdot (12!)^{1/24} \rfloor$. Last edited by idontknow; May 12th, 2019 at 04:12 PM.
 May 12th, 2019, 07:07 PM #5 Global Moderator   Joined: Dec 2006 Posts: 20,926 Thanks: 2205 Why the "1 +" and "2 +"? Thanks from idontknow
May 12th, 2019, 09:29 PM   #6
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 Originally Posted by idontknow $\displaystyle \lfloor y \rfloor > 1+ \int_{1}^{12} \sqrt{x} dx \approx 28 \; \Rightarrow \lfloor y \rfloor =1+28=29$.
I think the tricky part is that you have to bound the error term. The integral's not exactly the sum of the rectangles and you have to show that the difference isn't big enough to change the floor.

ps -- Couple more thoughts. One, I am not sure where you get your 1 plus the integral. Two, I think it should be the integral from 1 to 13. When I worked it out by hand I got 31. No guarantees. Three, using the fact that $\sqrt x$ is monotone, there's a formula to bound the error.

https://en.wikipedia.org/wiki/Riemann_sum

pps -- Ah you are taking the right Riemann sum. So you are correct, the integral is from 1 to 12. And to answer a question that was asked, the 1 is the leftmost Riemann rectangle with base (0,1) and height $\sqrt 1 = 1$.

ppps -- Right Riemann sum as in rightmost point in the interval. Also right as in correct but that's not the meaning I meant.

Last edited by Maschke; May 12th, 2019 at 10:10 PM.

May 12th, 2019, 10:01 PM   #7
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 Originally Posted by Maschke I think the tricky part is that you have to bound the error term. The integral's not exactly the sum of the rectangles and you have to show that the difference isn't big enough to change the floor. ps -- Couple more thoughts. One, I am not sure where you get your 1 plus the integral. Two, I think it should be the integral from 1 to 13. When I worked it out by hand I got 31. No guarantees. Three, using the fact that $\sqrt x$ is monotone, there's a formula to bound the error. https://en.wikipedia.org/wiki/Riemann_sum
I posted the riemann sum but without proof (to save time).The solution is correct.

May 12th, 2019, 10:05 PM   #8
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 Originally Posted by skipjack Why the "1 +" and "2 +"?
Example : $\displaystyle x\in \mathbb{N}$ and $\displaystyle x>27,33$.
$\displaystyle x>27,33$ means $\displaystyle x>28$ or $\displaystyle x\geq 29$.(we need the upper bound of 27,33 so avoid 28 in interval of x)

Last edited by idontknow; May 12th, 2019 at 10:11 PM.

May 12th, 2019, 10:08 PM   #9
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 Originally Posted by idontknow I posted the riemann sum but without proof (to save time).The solution is correct.
Oh I don't doubt it!

 May 12th, 2019, 11:49 PM #10 Global Moderator   Joined: Dec 2006 Posts: 20,926 Thanks: 2205 Try it for just 6 terms.

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