Some algebraic manipulation.

Feb 2020
5
0
Germany
I've just got back to study some subjects that require maths, and my skill is pretty rusty. Could anyone help me with putting bounds on the following expression?

For \(\displaystyle k\) a positive integer, and \(\displaystyle \delta \in (0,1)\): \(\displaystyle \frac{\sqrt{(1- \delta)^{2k} + 4\delta} \ - (1-\delta)^k}{2\delta}\)
 

romsek

Math Team
Sep 2015
2,871
1,607
USA
Well it's bounded below by 1. I suspect it increases without bound as $k\to \infty$
 
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Dec 2015
972
128
Earth
$ \lim_{\delta \rightarrow 1 }
\frac{\sqrt{(1- \delta)^{2k} + 4\delta} \ - (1-\delta)^k}{2\delta} =0$.

$
\lim_{\delta \rightarrow 0 }
\frac{\sqrt{(1- \delta)^{2k} + 4\delta} \ - (1-\delta)^k}{2\delta} =\lim_{\delta \rightarrow 0 } \dfrac{4\delta }{4\delta }=1
$.
 
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romsek

Math Team
Sep 2015
2,871
1,607
USA
$\lim_{\delta \rightarrow 1 }
\frac{\sqrt{(1- \delta)^{2k} + 4\delta} \ - (1-\delta)^k}{2\delta} =0$.

$
\lim_{\delta \rightarrow 0 }
\frac{\sqrt{(1- \delta)^{2k} + 4\delta} \ - (1-\delta)^k}{2\delta} =\lim_{\delta \rightarrow 0 } =\lim_{\delta \rightarrow 0 } \dfrac{4\delta }{2\delta }=2
$.
Mathematica returns a limit of 1 as $\delta \to 0$