The Logarithm

Let’s call two players in a zooming game Alice and Bob. One player, say Alice, starts by zooming in or out a picture using the dial she’s been given. She tells Bob how many notches she turn hers -say $4.5$ . Now Bob must undo that using his own dial. How much does Bob need to turn his dial to undo Alice’s zoom? Answer: $4.5\cdot\log_{\mbox{Bob's zoom factor}}(\mbox{Alice's zoom factor})\,=\,4.5\cdot\log_8(2)\,=\,4.5/3\sim 1.5$

I recently added an answer to a question in math.stackexchange.org, and another one even more recently here, that wraps up my two previous posts on the logarithm. I copy the first answer below. The second is but a brief summary of the first.

The question was “how do you explain the logarithm to primary school students”?

As usual all answers directly dive into the concept of powers making use of course of different and interesting analogies to explain it. Alas, they all defuse a straight answer to the question “What is the logarithm?”.

Hence, once more, my point stays valid: Can we give them a straight answer that’s not in the form “the opposite of”?

I do think though I can provide a meaningful new perspective to the answers given so far. The only requirement is basic arithmetic. Let’s see.

Short answer.
Present the kid with the following riddle (set up whatever preface you want for making the challenge more appealing to the kid):

Riddle:

You a have bag of candies when you come across a friend who had a bad day and he’s sad. To cheer him up you divide your amount of candies in two equal-sized halves and you give one half to your friend and keep the other. You keep on walking when you meet another friend. She’s also sad. Again you divide the amount of candies you have in two halves and give one to this friend too. After a little while you find a third friend and you do the same. You keep doing the same with other friends until you have only one candy left. With how many friends can you split your candies?
Answer: $lg_2(\#candies)$ , i.e., the logarithm in base $2$ of the number of candies the kid starts with.

Long answer.

I posted a question a little while ago that may help further clarify my point. See here.

If the kid is at least in grade 9, I’ve successfully used this approach in introducing the concept of a logarithm. Also, a grade 12 student can easily use it to estimate arbitrary logarithms up to 1 decimal places.

$\log_b(x)\,\equiv\,$ the number of repeated divisions ( $b>1$ ) / multiplications ( $b<1)$ of $x$ by $b$ until you get a 1

Examples:

$\log_58\,?\,Ans:8/5=1.6$ So the actual value is between $1$ and $2$ , although closer to the first. For guessing a value, let’s divide that interval in 4 parts. As we expect it closer to 1 than to 2, we can pick a guess like $1.25$ . The actual value is $1.29$ .
$\log_{13}57\,?\,Ans:57/13\approx 4.4$ , hence again the value of this log lies in $[1,2]$ . As $4.4/13\approx 0.3$ , we may guess the value as $1.3$ . The actual value with one decimal is $1.6$ .
We can do better by first enlarging the number. Example: $\ln 2\,?\,Ans:\mbox{Let's consider first say } \ln (2^{10})$ . Now $1000$ can be divided by $3$ six times before the result is smaller than $1$ : $1000/3^6\approx 1.37$ . That’s about half of $3$ and as $e<3$ , we’ll guess $\ln(1024)\approx 7$ . Hence, $\ln(2)\approx 7/10=0.70$ . The actual value is $\ln 2\approx 0.69$ .

A geometric interpretation has to do with zoom levels. Let’s consider the following zooming game.

The rules are basically as follows. Two players compete against each other each in turns. Each turn is time constrained. Each player has access to a zoom-dial that controls the zoom level of a picture. Each dial works in a different scale. Say player 1’s dial zooms in(turn clockwise)/out(turn counter-clockwise) by a factor of 2, while that of player 2 by a factor of 8. That information is known to the players. What a player does to the picture the other has to undo. The first player that fails to undo a change in the allotted time loses the game.

Let’s call the players Alice and Bob. One player, say Alice, starts by zooming in or out the picture using the dial she’s been given. She tells Bob how many notches she turn hers -say $4.5$ . Now Bob must undo that using his own dial. How much does Bob need to turn his dial to undo Alice’s zoom? Answer: $4.5\cdot\log_{\mbox{Bob's zoom factor}}(\mbox{Alice's zoom factor})\,=\,4.5\cdot\log_8(2)\,=\,4.5/3\sim 1.5$

Addendum:
What does $\log_{0.7}(2)$ mean? When the base is smaller than one, the log refers to the number of times we multiply $x$ by $b$ until we get $1$ . The result is convened to be written negative to indicate that the process to get to $1$ is now through multiplications instead of divisions.

Summary:
The $\log_b x$ has to do with the number of steps needed to get to $1$ starting on $x$ with the only possible operations being multiplication $*$ and division $/$ by $b$ .

Conclusions:
I’d say that both pictures, the samaritan and the zooming challenge, may be grasped by at least a teenager (grade 7 on…not sure when they start learning division nowadays).

The minimum requirement is just basic arithmetics. Of course, calculating estimates requires to be somewhat fluent in calculating some basic powers as well as commanding the fundamental rule of logarithms and that of the log of a power.

The persistent problem when explaining logarithms is that no straight answer is given, but instead a detour is taken by relating it to the exponential -worst is when referring to the log as the inverse of the exponential: Isn’t there a (non-written?) rule that an explanation of something should never be given in the form of “the opposite of” or “the negation of”? I, at least, feel that’s true. The approach I present introduces an exponential only implicitly.

Note added in proof: (Wed. Oct. 16 2019) I started thinking about how to introduce the log to grade 9 students around 2017-2018. However, only today it occurred to me to google for “logarithm as repeated divisions”. I found these references

1. Mathematics Teacher 251:30-33 · January 2016 that does exactly that! You can read the article in “https://www.researchgate.net/publication/306379874_Logarithms_-_a_meaningful_approach_with_repeated_division” but be careful, some ads may contain malware…

2. “Making Sense of Logarithms as Counting Divisions”, Christof Weber, The Mathematics Teacher, Vol. 112, No. 5 (March 2019), pp. 374-380. See link.

Scaling ratio: the logarithm

This post is intentionally just a stub.

In a previous post Number 1 I mentioned an intuitive way of thinking on the logarithm. However, it is clearly too cumbersome: sometimes it’s about multiplication, others division, we have to add ad hoc a sign, and the interpretation of the fundamental law of logarithms $\log_b(x)\,\log_{b'}b\,=\,\log_{b'}x$ required us to think of the term $\log_{b'}(b)$ as something slightly different than $\log_b(x)$ .

It was good for talking about the integer logarithm, but it clearly isn’t the whole story. I still do not have the right one, but I think I have now a better way to think about the logarithm.

$\log_y(x)$ is the ratio in which a 1/x contraction stretches to 1 relative to the case of a 1/y contraction.

Example: A contraction of 1/16 can be dilated 4 times by a factor of 2 to recover the original size, while that of 1/8 can be stretch 3 times by the same factor. Hence $\log_8(16)=4/3$ . I think this means we can swap contractions by 1/x by stretches by x, and stretches by contractions in this definition of logarithm.

By the same definition it is $\log_y(x)\,=\,1/\log_x(y)$ and thus $\log_{16}(8)=3/4$

The fundamental law of the logarithm should come equally simple from there:

$\log_b(x)\,=\,\log_{b'}(x)\,\log_b(b')$

Given the factorization of x and y, $x=p_1^{a_1}\dots p_k^{a_k}$ and $y=q_1^{b_1}\dots q_r^{b_r}$ , with $p_i,\,q_j$ prime numbers, we have

$\log_y(x)\,=\,\sum_i^k\,a_i\,\log_y(p_i)\,=\,\sum_i^k\,a_i\,\frac{1}{\log_{p_i}(y)}\,=\,\sum_i^k\frac{a_i}{\sum_j^r\,b_j\,\log_{p_i}(q_j)}$

and the logarithm of any two numbers is a function that depends on the log of any pair

of prime numbers. What is then something like $\log_2(3),\;\log_2(5),\;\log_3(5),\dots$ ?

So the logarithm is more about the geometric transformation of dilations, hometheties or scalings. This in turn leads to a projective transformation.

Random thoughts -or not so random, as I intentionally want to take some big steps back and see where this idea could lead to:

Does the $\log$ have a natural geometric interpretation in a projective space? I don’t expect a geometric construction, just an interpretation similar to those provided by the figures in this thread.

Does the $\log$ have a natural topological interpretation as (properties of) homeomorphisms?

Does the $\log$ “carry” a natural generalization to a non-flat manifold leading to something different? For instance, contract/stretch a “line-segment” along a given geodesic on a given manifold. Would this relate to Tarantola’s autoparallel geovectors?

Edit: I realized I didn’t use what seems the usual definition (order) for the cross-ratio. For four points on a line O,p,q,E, in that order, it is $[p,q;O,E]=(O-p)\,(E-q)/((E-p)\,(O-q))$ . Ultimately, there are only 6 combinations all related by fractional transformations.

Using this definition consistently, though, we can use the cross-ratio $[p,q;O,\infty]$ to express the factor by which to contract the segment Oq to match Op. See 3rd figure below.

The scaling factors $\frac{AE}{AH}=1/2$ and $\frac{AE}{AK}=1/3$ coincide with the cross-ratios $[A,E,H,\infty]$ and $[A,E,K,\infty]$ , respectively. Similarly for the other pairs. Whence it follows that the Poincare non-euclidean distance $d(E,H)$ , or rather a ratio of them, gives us the logarithm of one of the scaling factors relative to the other.

Of course, this would be a circular definition and it doesn’t provide an independent way to calculate the logarithm. For that we would need a way to determine the non-euclidean distances $d(E,H)$ .

Using logarithms of “numbers” instead of the numbers themselves is not new. The Logarithmic Number System (LNS) provides al alternative implementation of arithmetic operations that can be faster than using floating point arithmetics. For any number $X$ we assign $X\;\to\;\{s_x,\,x\equiv \log_b(X)\}$ .

It’s interesting to see how this relates to the cross-ratio. For the points $0,1,x$ we have $[x,1;0,\infty]=x$ and $\log_b(x) = \log_b [x,1;0,\infty] \equiv d_b(1,b)$ , i.e., the non-euclidean Poincaré distance between 1 and $x$ .

But most importantly, this makes it clear that the LNS is invariant under projective transformations.

Furthermore, this is equivalent to what Tarantola did for Lie Groups. Would it make sense to ask how LNS would have to be modified if use it for $S^1$ ? Does the geometry of the circle enters the game as does for the case of Lie groups? Maybe this could be like a trivial example of his approach.

I’m strongly influenced here by Tarantola’s discussions, both his in-depth development as well as shorter essays/notes on the subject. It would seem that engineers, since long ago, have been using the logarithm

as a way to compare relative attenuations, amplifications,.. albeit with some definition involving some odd factors, e.g., the decibel.

He ends that text with the following concluding remark that summarizes his thoughts and this thread:

The original unit is the nonlinear one. The “logarithmic” unit is natural. And linear. Note: try to make obvious that I am making more than using logarithmic scales. I try to give life to the logarithmic parameters.

Note for me: Each $X$ is projected onto $X\,\to\, \arctan (1/X)$ …

LogScalingRatio

ScalingRatioLog1

Screen Shot 2018-08-24 at 3.37.57 PM

Screen Shot 2018-08-27 at 5.59.26 PM

I’m missing something: is there an intrinsic measure of $[D,H;L,M]$ in $S^1$ such that $[D,H;L,M]=[F,G;K,\infty]$ ? Oops! That last figure is wrong! The CR doesn’t project onto the circle as stated!! ;-< [1,2]

Rambling

(I’m moving posts from another site here…)

At the beginning of the XX century, Rutherford sees scattering doesn’t follow Thompson’s formula. Most alpha particles pass straight. Yet, and separated by a significant angular gap, there is again many alpha particles being scattered at high angles, altough less than the number of them passing straight. At around another, even higher angular dispersion value he saw again a profusion of alpha particlea being scattered. And again this amount was smaller than at the previous angular.value of high scattering.

Why the gap in angular dispersion? Why not seeing a continous, though decreasing number of alpha particles for all angular values of measured scattering?

These questions and type of experiments led to the discovery of the nucleus of atoms, of the neitrons and protons as well ad tgat of quarks. It also illuminated a zoo of many different particles (elementary or not).

In doing so, physicists realized the key importance of Symmetry concepts in particle physics.

In these nights of insomnia, I’ll try summarize how Group theory found its way into, and helped in classifying that zoo which in the 50’s was particle physics.

Number 1

I like the number 1. It seems a dull choice and many may go for other more mysterious numbers like 4, 8 or 24. But there are some key elementary concepts that hinge on 1.

One of these concepts is the basic operations of multiplication and of division, and in particular the integer division. The basic idea may be stated as what are the “paths” that lead to 1.

For instance, be 1 < b < n integers. The number of times one can divide n by b without the result being smaller than 1 is the integer logarithm of n in base b. Let’s denote it by iLog_b(n).

iLog_3(81) = 4; iLog_2(128) = 7.

For 0 < b < 1, the integer logarithm means the number of times we can multiply n by b without the result being smaller than <1. So we’d write iLog[1/2](8)=3. However, it will turn out to be convenient to introduce a new notation for meaning multiplication by the base: We will convene in adding a negative sign in front of the result when meaning repeated multiplication. Hence

iLog_[1/2](8) = -3

For 1 < b < n, iLog_b(1/n) means the number of times we can multiply 1/n by without the result getting larger than 1. By the same rule as before, we will add a negative sign to denote it’s a multiplication. In addition, it clear is iLog_b(1/n) = – iLog_b(n). Whence,

iLog_[1/2](1/8) = – iLog_2(1/8) = iLog_2(8) = 3

Figure 1 above illustrates the rule of changing base in the logarithm. It also gives us another interpretation of the logarithm: We can read the term Log_[b’](b) as how many cuts (divisions) do we need of size b’ for each cut of size b.

Clearly, iLog_b(1) = 0, be it b>1 or b<1as neither dividing nor multiplying by b will lead as to 1. For for b=1 things are slightly different: we can divide any an infinite number of times by 1 and it still won’t get us below 1, thus iLog_1(n) = ∞ (division). Analogously for multiplication. Following our convention, though, it would be iLog_1(n) = -∞ (multiplication).

If n<0 the logaritm doesn’t exist, as the results of repeatedly dividing/multiplying by b will all be negative and thus never possibly be +1!

Analogously, the logarithm doesn’t exist for b=0 as division by 0 is not defined and 0 is a fixed point for the multiplication.

If after dividing n by b iLog_b(n) times we get exactly 1, we say the logaritm is exact.

If 1 < n < b, we use the symmetry of the logaritm: Log_b(n)*Log_n(b)=1! This follows immediately from the rule of base change by setting b’=x. Of course, for iLog_b(n) it would be zero according to the original definition. Here we will interpret this case as iLog_b(n) = 1 / iLog_[n](b) if 1 < n < b.Hence

Log_[8](4) * Log_[4](8) = 1

Log_[8](4) = 1/Log_[4](8) .

Let’s prove this reflection symmetry algebraically. In Figure 1 it is illustrated geometrically (The current prove has thus the drawback that relies “too closely” on the idea of power, which is what we are trying to avoid here). For that we will use the rule for a logarithm of a product that we will prove afterwards: Log_b(n*m) = Log_b(n) + Log_b(m), which we assume it holds in the general case -not just for the integer logarithm. Here it goes:

Let’s assume 1<y<x, then

Log_y(x) = k <=> 1 = x/y^k

Log_x(y) =Log_[1/x](1/y) = k’ <=> 1 = 1/y/(1/x)^k’ = x^k’/y = 1

Hence, 1/y^(k-1) = x^(k’-1) <=> 1 = y^(k-1) x^(k’-1)

Taking now logarithms in base y on both sides, and using the rule we prove below on the log of a product, we find

0 = (k-1) Log_y(y) + (k’-1) Log_y(x) = (k-1) + (k’-1) k = -1 + k’ k

whence

1 = k’ k <=> Log_y(x) Log_x(y) = 1 . QED.

Let’s demonstrate the rule for the logarithm of products and quotients.

Be k=n*m, i= iLog_b(n), j=iLog_b(m), all integers. What’s iLog_b(k)? Answer: i+j. Let’s see why. For simplicity sake, we’ll assume that the logaritms are exact. iLog_b(k) corresponds to dividing the rectangle area by b. The side n can be cut by b i times. That still leaves a rectangle of area m. This can be divided by b j times before the resulting area gets <1. Hence the original area can be divided i+j times by b. qed.

Be n, m integers, n divisible by m, i.e., n=m*q, q another integer. What’s iLog_b(n/m)? Answer: i-j. To see why just follow the previous argument changing n.by q and k by n.

There is another way the division and multiplication hinge on 1. In this case the basic idea are paths leading away from 1: The definition of the exponential function b^x for x>0 is 1 multiplied x times by b. If x<0, it means 1 divided x times by b. Clearly, b^0 = 1.

In short, using just the operations of (repeated) multiplication and (repeated) division, given two numbers, b, x, the paths leading towards 1 define the logaritm Log_b(x) while the paths leading away from 1 define the exponentiation b^x.

The idea of paths become more concrete when dealing with complex numbers. There the concept of logarithm is somewhat richer.

It should be interesting to explore this idea of paths for the case of a Group as well as for the case of the other (normed) division algebras. Can it be given a meaningful and useful definition in the case of the Quaternions (Η) and Octonions (Ο)?

In some sense, the late Albert Tarantola already figured out the geometric implications when dealing with Lie Groups in his book Elements for Physics Springer 2006.

The connection to Tarantola’s discussion of the logarithm image of Lie Groups and his geovectors algebra of oriented auto parallel segments as a representation of a group in the neighbourhood of the identity is here somewhat contrived, although it seems to me there is a link.

In any case, I think that, regretably, current high school curricula miss the chance to develop some intuitive understanding of the logarithm. This can be a handicap for studying say the complexity of algorithms or the key concept of information and its relation to logarithms.

More advanced concepts like the Hausdorf dimension can also be easier to grasp with a familiarity of the logarithm as repeated divisions.