$$(x - 2)^{9} = x^9 - 18\, x^8 + 144 \, x^7 + o(x^7) = x^9(1-18x^{-1}+144 x^{-2} + o(x^{-2}))$$

$$\sqrt{(x - 2)^{9}} = x^{9/2}(1 - 9 x^{-1} + \frac{63}{2} x^{-2} + o(x^{-2}))$$

Then

$$f(x)=x^{9/2} - (x - 2)^{9/2} = 9 x^{7/2} - \frac{63}{2} x^{5/2} +o(x^{5/2})$$

Then for$x$ large enough:

$$9 x^{7/2} - \frac{63}{2} x^{5/2} < f(x) < 9 x^{7/2} $$

You can also bound, if you prefer$ f(x) > (9-\epsilon) x^{7/2} $ for some$\epsilon >0$ and$n > n_0(\epsilon)$

Added: Alternatively, the Taylor expansion of$g(x)=x^{9/2}$ is

$$g(x+a) = x^{9/2} + \frac92 x^{7/2} a + \frac{63}{8} x^{5/2}a^2+ \cdots$$

Then$$x^{9/2} - (x - 2)^{9/2} = g(x)-g(x-2) = 9 x^{7/2} - \frac{63}{2} x^{5/2} +\cdots$$ etc

Use the binomial theorem (expansion valid for$n > 2$):

$\begin{align*} f(n) &= n^{4.5} - (n - 2)^{4.5} \\ &= n^{4.5} - n^{4.5} (1 - 2 / n)^{4.5} \\ &= n^{4.5} \left( 1 - \sum_{k \ge 0} \binom{4.5}{k}\left(\frac{2}{n}\right)^k \right) \\ &= n^{4.5} \left( 1 - \frac{9}{n} - \frac{63}{2 n^2} - O(n^{-3})\right)\end{align*}$

• 2
  $\begingroup$Check your answer. The leading order terms $n^{4.5}$ should cancel.$\endgroup$

  Gary

  – Gary

  2020-08-28 11:44:11 +00:00

  CommentedAug 28, 2020 at 11:44

• asymptotics