Is it true that
$\sqrt{1+xa}\approx(1+x)\sqrt{\frac{a}{2}}$ for a$\gg$1
I took $x=1$, then the above becomes $\sqrt{1+a}\approx\sqrt{2a}$ which is untrue for very large $a$. But it is asked to prove the above by using Taylor's Series.
Using Taylor series to prove $\sqrt{1+xa}\approx(1+x)\sqrt{\frac{a}{2}}$
2
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real-analysis
taylor-expansion
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0What limit are you considering? Large $a \to \infty$, fixed $x$? In that case the result is false. – 2017-01-25
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0Here $a$ is fixed, and it is a function of $x$. – 2017-01-25
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0So what is the point you want to take the Taylor series at? "$\approx$" only makes sense if some variable is assumed to approach some value. – 2017-01-25
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1No, it doesn't appear to be true, – 2017-01-25
1 Answers
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The expression in the OP is false. Simply let $x=0$; certainly $\sqrt{1+a\cdot 0}=1$ is not asymptotically $(1+0)\sqrt{a/2}=\sqrt{a/2}$.
For $x>0$ and fixed, we have for $a\to \infty$
$$\sqrt{1+ax}=\sqrt{ax}\sqrt{1+\frac1{ax}}\sim \sqrt{ax}\left(1+\frac{1}{2ax}\right)$$