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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}} > 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>


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Revision as of 07:12, 20 July 2017

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}} > 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} $

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