My Math Forum  

Go Back   My Math Forum > College Math Forum > Number Theory

Number Theory Number Theory Math Forum


Thanks Tree5Thanks
Reply
 
LinkBack Thread Tools Display Modes
May 11th, 2019, 03:19 PM   #1
Senior Member
 
Joined: Dec 2015
From: somewhere

Posts: 596
Thanks: 87

Thumbs up 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.
idontknow is offline  
 
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.
mathman is offline  
May 12th, 2019, 03:09 PM   #3
Senior Member
 
Joined: Aug 2012

Posts: 2,355
Thanks: 737

Quote:
Originally Posted by mathman View Post
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 $\sqrt x$.
Maschke is offline  
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.
idontknow is offline  
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
skipjack is offline  
May 12th, 2019, 09:29 PM   #6
Senior Member
 
Joined: Aug 2012

Posts: 2,355
Thanks: 737

Quote:
Originally Posted by idontknow View Post
$\displaystyle \lfloor y \rfloor > 1+ \int_{1}^{12} \sqrt{x} dx \approx 28 \; \Rightarrow \lfloor y \rfloor =1+28=29$.
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.
Thanks from idontknow

Last edited by Maschke; May 12th, 2019 at 10:10 PM.
Maschke is offline  
May 12th, 2019, 10:01 PM   #7
Senior Member
 
Joined: Dec 2015
From: somewhere

Posts: 596
Thanks: 87

Quote:
Originally Posted by Maschke View Post
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.
idontknow is offline  
May 12th, 2019, 10:05 PM   #8
Senior Member
 
Joined: Dec 2015
From: somewhere

Posts: 596
Thanks: 87

Quote:
Originally Posted by skipjack View Post
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)
Simply add +1 to 28 or add +2 to 27 .

Last edited by idontknow; May 12th, 2019 at 10:11 PM.
idontknow is offline  
May 12th, 2019, 10:08 PM   #9
Senior Member
 
Joined: Aug 2012

Posts: 2,355
Thanks: 737

Quote:
Originally Posted by idontknow View Post
I posted the riemann sum but without proof (to save time).The solution is correct.
Oh I don't doubt it!
Maschke is offline  
May 12th, 2019, 11:49 PM   #10
Global Moderator
 
Joined: Dec 2006

Posts: 20,926
Thanks: 2205

Try it for just 6 terms.
skipjack is offline  
Reply

  My Math Forum > College Math Forum > Number Theory

Tags
integer, part



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Integer part idontknow Elementary Math 3 January 19th, 2019 04:56 AM
Integer part Dacu Algebra 7 April 17th, 2015 05:05 AM
Equation integer part gerva Algebra 18 January 6th, 2015 06:38 AM
integer part of S Albert.Teng Algebra 2 September 23rd, 2012 03:46 AM
More about part integer Fernando Number Theory 1 April 13th, 2012 06:19 AM





Copyright © 2019 My Math Forum. All rights reserved.