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| Title of object: |
asymptotic estimate |
| Canonical Name: |
AsymptoticEstimate |
| Type: |
Definition |
| Created on: |
2006-06-13 11:46:27 |
| Modified on: |
2006-06-13 23:33:56 |
| Classification: |
msc:11N37 |
Revision comment (for changes between this and next version):
| Changes for correction #8258 ('Notation'). |
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Content:
An asymptotic estimate is an estimate that involves the use of $O$, $o$, or $\sim$. Examples of these are:
\begin{center}
\begin{tabular}{rll}
$\displaystyle \sum_{n \le x} \mu^2(n)$ & $= \frac{6}{\pi^2}x+O(\sqrt{x})$ & (see convolution method for more details) \\
$\displaystyle \pi(x)$ & $\sim \frac{x}{\log x}$ & (see prime number theorem for more details) \end{tabular}
\end{center}
Unless otherwise specified, asymptotic estimates are typically valid for $x \to \infty$. An example of an asymptotic estimate that is different from those above in this aspect is:
$$\cos x=1-\frac{x^2}{2}+O(x^4) \text{ for } |x|<1$$
Note that the above estimate would be undesirable for $x \to \infty$, as the error would be larger than the estimate. Such is not the case for $|x|<1$, though.
Tools that are useful for obtaining asymptotic estimates include the Euler-Maclaurin summation formula, Abel's lemma, the convolution method, and the Dirichlet hyperbola method.
If $A \subseteq \mathbb{N}$, then an asymptotic estimate for $\displaystyle \sum_{n \le x} \chi_A(x)$, where $\chi_A$ denotes the characteristic function of $A$, enables one to determine the asymptotic density of $A$ using the formula
$$\lim_{x \to \infty} \frac{1}{x} \sum_{n \le x} \chi_A(x)$$
provided the limit exists. The upper asymptotic density of $A$ and the lower asymptotic density of $A$ can be computed in a similar manner using $\limsup$ and $\liminf$, respectively. (See \PMlinkname{asymptotic density}{AsymptoticDensity} for more details.)
For example, $\mu^2$ is the characteristic function of the squarefree natural numbers. Using the asymptotic estimate above yields the asymptotic density of the squarefree natural numbers:
\begin{center}
$\begin{array}{ll}
\displaystyle \lim_{x \to \infty} \frac{1}{x} \sum_{n \le x} \mu^2(n) & \displaystyle =\lim_{x \to \infty} \frac{1}{x} \left( \frac{6}{\pi^2}+O(\sqrt{x}) \right) \\
\vspace{5mm}
& \displaystyle =\lim_{x \to \infty} \frac{6}{\pi^2}+O \left( \frac{\sqrt{x}}{x} \right) \\
\vspace{5mm}
& \displaystyle =\frac{6}{\pi^2} \end{array}$
\end{center} |
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