consider the following function $f(n) =$ $n^{4.5}-$$(n-2)^{4.5}$ I want to calculate a good tighter upper bound for it. I calculated a lower bound for it like - $n^{4.5} - (n-2)^{4.5} >= n^{4.5} - \sqrt n * (n-2)^3$ Now if we expand $(n-2)^3$ the $n^{4.5}$ term will cancel out and we will have a $n^{3.5}$ term
So $f(n) \geq c * n^{3.5}$ for some constant $c$ we can say...
But getting a good upper bound seems confusing to me... Also not sure if this lower bound is a tight one.
Thanks
$$(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 $n > 2$):
$\begin{align*} f(n) &= n^{4.5} - (n - 2)^{4.5} \\ &= n^{4.5} - n^{4.5} (1 - 2 / n)^{4.5} \\ &= n^{4.5} \left( 1 - \sum_{k \ge 0} \binom{4.5}{k}\left(\frac{2}{n}\right)^k \right) \\ &= n^{4.5} \left( 1 - \frac{9}{n} - \frac{63}{2 n^2} - O(n^{-3})\right) \end{align*}$
