From charlesreid1

(Created page with "=Volume 1= ==Chapter 1: Basic Concepts: Harmonic numbers== Harmonic numbers become important in analyses of algorithms. Define <math> H_n = 1 + \dfrac{1}{2} + \dfrac{1}{3}...")
 
(Fix math syntax error: change > to \geq in the inequality (non-strict, since 1/(2^m+2^m) = 1/2^{m+1}) (via update-page on MediaWiki MCP Server))
 
(2 intermediate revisions by the same user not shown)
Line 10: Line 10:


While it does not occur often in classical mathematics, it crops up more often in algorithm analysis.
While it does not occur often in classical mathematics, it crops up more often in algorithm analysis.
We can make <math>H_n</math> as large as we please from observing that
<math>
H_{2^m} \geq 1 + \dfrac{m}{2}
</math>
This results from the fact that
<math>
H_{2^{m+1}} = H_{2^m} + \dfrac{1}{2^m+1} + \dfrac{1}{2^m + 2} + \dots + \dfrac{1}{2^{m+1}}
</math>
Now we have,
<math>
H_{2^m} + \dfrac{1}{2^m+1} + \dfrac{1}{2^m + 2} + \dots + \dfrac{1}{2^{m+1}} \geq H_{2^m} + \dfrac{1}{2^{m+1}} + \dfrac{1}{2^{m + 1}} + \dots + \dfrac{1}{2^{m+1}}
</math>
and the right side is
<math>
H_{2^m} + \frac{1}{2}
</math>
therefore
<math>
H_{2^m} \geq 1 + \dfrac{m}{2}
</math>
=Flags=
{{AOCPFlag}}
[[Category:Harmonic Numbers]]

Latest revision as of 01:22, 6 June 2026

Volume 1

Chapter 1: Basic Concepts: Harmonic numbers

Harmonic numbers become important in analyses of algorithms. Define

$ H_n = 1 + \dfrac{1}{2} + \dfrac{1}{3} + \dfrac{1}{4} + \dots + \dfrac{1}{n} = \sum_{1 \leq k \leq n} \dfrac{1}{k} \qquad n \geq 0 $

While it does not occur often in classical mathematics, it crops up more often in algorithm analysis.

We can make $ H_n $ as large as we please from observing that

$ H_{2^m} \geq 1 + \dfrac{m}{2} $

This results from the fact that

$ H_{2^{m+1}} = H_{2^m} + \dfrac{1}{2^m+1} + \dfrac{1}{2^m + 2} + \dots + \dfrac{1}{2^{m+1}} $

Now we have,

$ H_{2^m} + \dfrac{1}{2^m+1} + \dfrac{1}{2^m + 2} + \dots + \dfrac{1}{2^{m+1}} \geq H_{2^m} + \dfrac{1}{2^{m+1}} + \dfrac{1}{2^{m + 1}} + \dots + \dfrac{1}{2^{m+1}} $

and the right side is

$ H_{2^m} + \frac{1}{2} $

therefore

$ H_{2^m} \geq 1 + \dfrac{m}{2} $

Flags