Logarithmic Growth

by
K. D.  Stroyan
University of Iowa

copyright 1997 by Academic Press, Inc. - All rights reserved.

Notebook Overview

Logarithm tends to infinity, but it does this VERY SLOWLY.  For example, the harmonic series,  1 + (1/2) + (1/3) + (1/4) + (1/5) + ⋯ + (1/n)  is approximately  Log[ n ].  A mythical super-dooper computer adding ten billion terms per second still would not get to  100  in the age of our solar system, because  making  Log [ n ] = 100 requires  n  to be so large .  This NoteBook helps you estimate how long the super duper computer must run to get to 100.

<Enter> the next computation to see how slow the Log function grows.

Plot[Log[n], {n, 0, 100}]; Plot[Log[n], {n, 0, (100^100)}];

References to the Text

This NoteBook goes with Chapter 28 of the text on High School Review.

Magnitudes of the sum 1 + (1/2) + (1/3) + ⋯ + (1/n) + ⋯

The basic mathematical fact is that in exact arithmetic,

(Underscript[Lim, n -> ∞])[
          1 + (1/2) + (1/3) + ⋯ + (1/n)] =  
        +∞

Numerical machines can't show this very well for two reasons.  First, accuracy problems in floating point arithmetic eventually make the summation useless.  Second, this sum only goes to Infinity logarithmically,

1 + (1/2) + (1/3) + ⋯ + (1/n) = Log[n] + Gn
where

Gn  tends to Euler's second constant EulerGamma ≈ 0.57721566...

N[EulerGamma, 55]

Suppose we had a mythical super duper computer that could add ten billion terms of  1 + 1/2 + 1/3 + 1/4 + ...  each second.  How long would it take the computer to make the sum more than  100?  (If it tends to Infinity, it surely gets to 100 eventually.)

We want
	1 + (1/2) + (1/3) + ⋯ + (1/n)> 100

so

Log[n] > 100 - 0.57722 > 99

Ask Mathematica for the  n  so that  Log[n] = 99, or  n = E^99

\(Clear[n]; 
Solve[Log[n]\  == \ 99, n]\)N[%]

Our mythical SuperDuper computer adds ten billion (1010)terms per second.  Multiply that by sixty seconds per minute, and you have the total number of terms per minute.  Multiply that by sixty minutes per hour times twenty-four hours per day and you have the number of terms the computer can add in one day.  Multiply that by 365 days per years (or 365.2425 if you want to be obnoxiously precise) and you have the number of calculations per year.  Our solar system's age is approximately five billion years, so if you multiply your number by five billion (5 x 10 9) you will have the number of terms that the mythical SuperDuper computer can sum if you let it run for as long as the solar system has existed so far.  In the next cell, set m equal to the number of terms that the computer can sum in one lifetime of the solar system.  Don't forget to use a space each time you multiply!

m  =  ?? ????

Now compute how many ages of the solar system it will take to make

1 + (1/2) + (1/3) + ⋯ + (1/n) > 100

(<Enter> the computation of the next cell.)

N[(n/m)]

Practice Exercises

That's a very long time!  How many centuries is it?

Exercise 28.6.2

How many centuries does it take the Super-Duper computer to add enough terms of the harmonic series to make the sum at least 100?

Symbolic Mathematica summation:

Mathematica knows the limit approximation for the sum of the harmonic series.  Try the following and notice that no computer can sum to 10^100 in recorded history.  (Mathematica knows analytical tricks to approximate the large sum.)

(∑_(k = 1)^10) (1/k)(∑_(k = 1)^100) (1/k)(∑_(k = 1)^1000) (1/k)NSum[(1/k), {k, 1, (10^100)}]NSum[(1/k), {k, 1, (10^100)}] - Log[(10.0^100)]N[EulerGamma]