Explain Why a Root is Close to...

Sep 2014
3
0
me
Show that1.jpg

Use this to explain why2.jpg is close to, but slightly less than, 10.05.





~ I know why the first part equals the other, but what relevance has that to the explanation?
 

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Feb 2014
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United States
Show thatView attachment 31607

Use this to explain whyView attachment 31609 is close to, but slightly less than, 10.05.





~ I know why the first part equals the other, but what relevance has that to the explanation?
Let's deal with the general proposition first.

$x = \sqrt{N + 1} - \sqrt{N} \implies x\left(\sqrt{N + 1} + \sqrt{N}\right) = \left(\sqrt{N + 1} - \sqrt{N}\right) \left(\sqrt{N + 1} + \sqrt{N}\right) = \left(\sqrt{N + 1}\right)^2 -\left(\sqrt{N}\right)^2 \implies$

$x\left(\sqrt{N + 1} + \sqrt{N}\right) = (N + 1) - N = 1 \implies x = \dfrac{1}{\sqrt{N + 1} + \sqrt{N}} \implies \sqrt{N + 1} - \sqrt{N} = \dfrac{1}{\sqrt{N + 1} + \sqrt{N}} \implies$

$\sqrt{N + 1} = N + \dfrac{1}{\sqrt{N + 1} + \sqrt{N}} < N + \dfrac{1}{2\sqrt{N}}.$

Good now?
 
Sep 2014
3
0
me
lol, I'm even more lost now. I know how to make one equal the other by multiplying top and bottom by one so the appearance is altered,
but all of that, I'm lost in and don't know how to explain the root 101 still. sorry.
 

romsek

MHF Helper
Nov 2013
6,836
3,079
California
lol, I'm even more lost now. I know how to make one equal the other by multiplying top and bottom by one so the appearance is altered,
but all of that, I'm lost in and don't know how to explain the root 101 still. sorry.
let $N=100$

$\sqrt{101}-\sqrt{100}=\dfrac{1}{\sqrt{101}+\sqrt{100}}$

$\sqrt{101}=\sqrt{100}+\dfrac{1}{\sqrt{101}+\sqrt{100}}$

$\sqrt{101}\approx \sqrt{100}+\dfrac{1}{2\sqrt{100}}$

it's a bit less than this since we decreased the denominator slightly in the approximation

$\sqrt{101}\approx 10 + \dfrac{1}{20}=10.05$
 
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Nov 2013
263
59
Australia
Okay, I expect you are already overloaded but I might try anyway. LOL

You already know that the first part equals the second part so I will start from there.

Consider
$\sqrt{101}+\sqrt{100}>\sqrt{100}+\sqrt{100}=10+10=20$

so
$\sqrt{101}+\sqrt{100}>20$ Just by a really little bit

so
$\frac{1}{\sqrt{101}+\sqrt{100}}<\frac{1}{20}$

$\frac{1}{\sqrt{101}+\sqrt{100}}<0.05$ Just by a really little bit

-------------------------------------------------------------

$\begin{array}{rlll}\sqrt{101}-\sqrt{100}&=&\frac{1}{\sqrt{101}+\sqrt{100}}<0.05&\\\\ \sqrt{101}-\sqrt{100}&<&0.05 \qquad &\mbox{Here I just left the middle bit out}\\\\\sqrt{101}-10&<&0.05 &\mbox{Now I can just solve it as an inequality}\\\\\sqrt{101}&<&0.05+10 &\\\\\sqrt{101}&<&10.05\qquad &\mbox{But just by a tiny bit}&\\\\\end{array}$
 
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