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Math Help - Trouble finding N in epsilon-N proof

  1. #1
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    Trouble finding N in epsilon-N proof

    I'm trying to prove lim(\sqrt{n^2+n}-n)=\frac{1}{2}. I'm driving myself crazy trying to figure out what N should be.

    I have \sqrt{n^2+n}-n = \frac{n}{\sqrt{n^2+n}+n}
    so \epsilon >|\frac{1}{2} - \frac{n}{\sqrt{n^2+n}+n}|. Okay to drop the abs value,
    so \frac{\epsilon-\frac{1}{2}}{n} > -\frac{1}{\sqrt{n^2+n}+n}.
    so -\frac{\epsilon-\frac{1}{2}}{n} < \frac{1}{\sqrt{n^2+n}+n}.
    So -\frac{n}{\epsilon-\frac{1}{2}}-n < \sqrt{n^2+n}.

    But the left side is undefined at \epsilon > \frac{1}{2}, and worse...it could be negative or positive. So I don't think I can proceed, right? I don't think I need to consider cases, so I think I've already done something wrong. I've tried switching the terms in the distance formula, but I end up with a similar problem (left side could be positive or negative, so I can't square it without losing the inequality).

    Thanks,

    Eric.
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  2. #2
    Super Member PaulRS's Avatar
    Joined
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    Quote Originally Posted by Eric Reed View Post
    I'm trying to prove lim(\sqrt{n^2+n}-n)=\frac{1}{2}. I'm driving myself crazy trying to figure out what N should be.

    I have \sqrt{n^2+n}-n = \frac{n}{\sqrt{n^2+n}+n}
    so \epsilon >|\frac{1}{2} - \frac{n}{\sqrt{n^2+n}+n}|. Okay to drop the abs value,
    so \frac{\epsilon-\frac{1}{2}}{n} > -\frac{1}{\sqrt{n^2+n}+n}.
    so -\frac{\epsilon-\frac{1}{2}}{n} < \frac{1}{\sqrt{n^2+n}+n}.
    So -\frac{n}{\epsilon-\frac{1}{2}}-n < \sqrt{n^2+n}.

    But the left side is undefined at \epsilon > \frac{1}{2}, and worse...it could be negative or positive. So I don't think I can proceed, right? I don't think I need to consider cases, so I think I've already done something wrong. I've tried switching the terms in the distance formula, but I end up with a similar problem (left side could be positive or negative, so I can't square it without losing the inequality).

    Thanks,

    Eric.
    Note that: <br />
\tfrac{n}<br />
{{\sqrt {n^2  + n}  + n}} = \tfrac{{\tfrac{n}<br />
{n}}}<br />
{{\tfrac{{\sqrt {n^2  + n}  + n}}<br />
{n}}} = \tfrac{1}<br />
{{\sqrt {1 + \tfrac{1}<br />
{n}}  + 1}}<br />

    You have: <br />
\left| {\tfrac{1}<br />
{{\sqrt {1 + \tfrac{1}<br />
{n}}  + 1}} - \tfrac{1}<br />
{2}} \right| < \varepsilon  \Leftrightarrow  - \varepsilon  < \tfrac{1}<br />
{{\sqrt {1 + \tfrac{1}<br />
{n}}  + 1}} - \tfrac{1}<br />
{2} < \varepsilon <br />

    <br />
1 + \tfrac{1}<br />
{n} > 1\therefore \sqrt {1 + \tfrac{1}<br />
{n}}  + 1 > 2\therefore \tfrac{1}<br />
{2} > \tfrac{1}<br />
{{\sqrt {1 + \tfrac{1}<br />
{n}}  + 1}}<br /> <br />
thus: <br />
\tfrac{1}<br />
{{\sqrt {1 + \tfrac{1}<br />
{n}}  + 1}} - \tfrac{1}<br />
{2} < 0 < \varepsilon <br />

    So you only have to work on the LHS: <br />
 - \varepsilon  < \tfrac{1}<br />
{{\sqrt {1 + \tfrac{1}<br />
{n}}  + 1}} - \tfrac{1}<br />
{2}<br />

    <br />
 - \varepsilon  < \tfrac{1}<br />
{{\sqrt {1 + \tfrac{1}<br />
{n}}  + 1}} - \tfrac{1}<br />
{2} \Leftrightarrow \varepsilon  > \tfrac{1}<br />
{2} - \tfrac{1}<br />
{{\sqrt {1 + \tfrac{1}<br />
{n}}  + 1}} \Leftrightarrow \tfrac{1}<br />
{{\sqrt {1 + \tfrac{1}<br />
{n}}  + 1}} > \tfrac{1}<br />
{2} - \varepsilon <br />
now this ineq. is obvious if <br />
\varepsilon  \geqslant \tfrac{1}<br />
{2}<br />
for the LHS is positive, thus we may assume that: <br />
\varepsilon  < \tfrac{1}<br />
{2}<br />

    <br />
\tfrac{1}<br />
{{\sqrt {1 + \tfrac{1}<br />
{n}}  + 1}} > \tfrac{1}<br />
{2} - \varepsilon  \Leftrightarrow \tfrac{1}<br />
{{\tfrac{1}<br />
{2} - \varepsilon }} - 1 > \sqrt {1 + \tfrac{1}<br />
{n}} <br />
this makes sense since <br />
\tfrac{1}<br />
{{\tfrac{1}<br />
{2} - \varepsilon }} - 1 > \tfrac{1}<br />
{{\tfrac{1}<br />
{2} - 0}} - 1 = 1<br />
so: <br />
n > \tfrac{1}<br />
{{\left( {\tfrac{1}<br />
{{\tfrac{1}<br />
{2} - \varepsilon }} - 1} \right)^2  - 1}}<br />
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  3. #3
    Super Member flyingsquirrel's Avatar
    Joined
    Apr 2008
    Posts
    802
    Hello
    Quote Originally Posted by Eric Reed View Post
    I'm trying to prove lim(\sqrt{n^2+n}-n)=\frac{1}{2}. I'm driving myself crazy trying to figure out what N should be.

    I have \sqrt{n^2+n}-n = \frac{n}{\sqrt{n^2+n}+n}
    so \epsilon >|\frac{1}{2} - \frac{n}{\sqrt{n^2+n}+n}|. Okay to drop the abs value,
    so \frac{\epsilon-\frac{1}{2}}{n} > -\frac{1}{\sqrt{n^2+n}+n}.
    so -\frac{\epsilon-\frac{1}{2}}{n} < \frac{1}{\sqrt{n^2+n}+n}.
    So -\frac{n}{\epsilon-\frac{1}{2}}-n < \sqrt{n^2+n}.

    But the left side is undefined at \epsilon > \frac{1}{2}, and worse...it could be negative or positive. So I don't think I can proceed, right? I don't think I need to consider cases, so I think I've already done something wrong. I've tried switching the terms in the distance formula, but I end up with a similar problem (left side could be positive or negative, so I can't square it without losing the inequality).
    One can try working with an equality instead of working with an inequality : the function f:\begin{array}{l} \mathbb{R}^+\mapsto\left[0,\frac12\right]\\x\mapsto \sqrt{x^2+x}-x\end{array} (EDIT : see (*)) is an increasing one since it's derivative, which is f'(x)=\frac{2x+1}{2\sqrt{x^2+x}}-1, is always positive :

    f'(x)>0 \Longleftrightarrow 2x+1-2\sqrt{x^2+x}>0 \Longleftrightarrow 4x^2+4x+1>4x^2+4x \Longleftrightarrow 1>0

    This is interesting because as \forall x\geq0, \,0\leq f(x)\leq \frac12\, , if one solves \varepsilon = \left| \frac{1}{2}-f(x_0)\right| for x_0 one gets \varepsilon > \left| \frac{1}{2}-f(n)\right| for all n>x_0.

    Let's do it : \varepsilon = \left| \frac{1}{2}-f(x_0)\right|\Longleftrightarrow \varepsilon = \frac{1}{2}-\sqrt{x_0^2+x_0}+x_0\Longleftrightarrow \left(\frac{1}{2}-\varepsilon-x_0\right)^2=\sqrt{x_0^2+x_0}^2

    Expanding we get \left(\frac{1}{2}-\varepsilon\right)^2 -2\left(\frac{1}{2}-\varepsilon\right)x_0+x_0^2=x_0^2+x_0 thus x_0=\frac{\left(\frac{1}{2}-\varepsilon\right)^2}{1-2\left(\frac{1}{2}-\varepsilon\right)}=\frac{\left(\frac{1}{2}-\varepsilon\right)^2}{\varepsilon}

    EDIT : (*)  \sqrt{x^2+x}-x\leq \frac12 \Longleftrightarrow x^2+x \leq \frac14+x+x^2 which is always true.
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  4. #4
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    Joined
    Jun 2008
    Posts
    6
    Thanks to both! I gotta get to the point that I can see these simplifications right away. flying squirrel -> unfortunately, we don't have derivatives yet so I don't think I'd get credit for this approach. (Edit...heh... I didn't notice your edit, so I think it's fine...thanks!)
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