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22[camera]Algebra
[Special message from Ralph Leighton, with audio from the lecture.]
[f22-00]
[f22-00]
22-1Addition and multiplication
In our study of oscillating systems we shall have occasion to use one
of the most remarkable, almost astounding, formulas in all of
mathematics. From the physicist's point of view we could bring forth
this formula in two minutes or so, and be done with it. But science
is as much for intellectual enjoyment as for practical utility, so
instead of just spending a few minutes on this amazing jewel, we
shall surround the jewel by its proper setting in the grand design of
that branch of mathematics which is called elementary algebra.
Now you may ask, "What is mathematics doing in a physics lecture?" We
have several possible excuses: first, of course, mathematics is an
important tool, but that would only excuse us for giving the formula
in two minutes. On the other hand, in theoretical physics we discover
that all our laws can be written in mathematical form; and that this
has a certain simplicity and beauty about it. So, ultimately, in
order to understand nature it may be necessary to have a deeper
understanding of mathematical relationships. But the real reason is
that the subject is enjoyable, and although we humans cut nature up
in different ways, and we have different courses in different
departments, such compartmentalization is really artificial, and we
should take our intellectual pleasures where we find them.
Another reason for looking more carefully at algebra now, even though
most of us studied algebra in high school, is that that was the first
time we studied it; all the equations were unfamiliar, and it was
hard work, just as physics is now. Every so often it is a great
pleasure to look back to see what territory has been covered, and
what the great map or plan of the whole thing is. Perhaps some day
somebody in the Mathematics Department will present a lecture on
mechanics in such a way as to show what it was we were trying to
learn in the physics course!
The subject of algebra will not be developed from the point of view
of a mathematician, exactly, because the mathematicians are mainly
interested in how various mathematical facts are demonstrated, and
how many assumptions are absolutely required, and what is not
required. They are not so interested in the result of what they
prove. For example, we may find the Pythagorean theorem quite
interesting, that the sum of the squares of the sides of a right
triangle is equal to the square of the hypotenuse; that is an
interesting fact, a curiously simple thing, which may be appreciated
without discussing the question of how to prove it, or what axioms
are required. So, in the same spirit, we shall describe
qualitatively, if we may put it that way, the system of elementary
algebra. We say elementary algebra because there is a branch of
mathematics called modern algebra in which some of the rules such as
$ab = ba$, are abandoned, and it is still called algebra, but we
shall not discuss that.
To discuss this subject we start in the middle. We suppose that we
already know what integers are, what zero is, and what it means to
increase a number by one unit. You may say, "That is not in the
middle!" But it is the middle from a mathematical standpoint, because
we could go even further back and describe the theory of sets in
order to derive some of these properties of integers. But we are not
going in that direction, the direction of mathematical philosophy and
mathematical logic, but rather in the other direction, from the
assumption that we know what integers are and we know how to count.
If we start with a certain number $a$, an integer, and we count
successively one unit $b$ times, the number we arrive at we call
$a+b$, and that defines addition of integers.
Once we have defined addition, then we can consider this: if we start
with nothing and add $a$ to it, $b$ times in succession, we call the
result multiplication of integers; we call it $b$ times $a$.
Now we can also have a succession of multiplications: if we start
with $1$ and multiply by $a$, $b$ times in succession, we call that
raising to a power: $a^b$.
Now as a consequence of these definitions it can be easily shown that
all of the following relationships are true: \begin{equation} \begin
{alignedat}{4} &(\text{a})&\quad &a+b=b+a&\quad\quad &(\text{b})&\
quad &a+(b+c)=(a+b)+c\\ &(\text{c})&\quad &ab=ba&\quad\quad &(\text
{d})&\quad &a(b+c)=ab+ac\\ &(\text{e})&\quad &(ab)c=a(bc)&\quad\quad
&(\text{f})&\quad &(ab)^c=a^cb^c\\ &(\text{g})&\quad &a^ba^c=a^
{(b+c)}&\quad\quad &(\text{h})&\quad &(a^b)^c=a^{(bc)}\\ &(\text{i})&
\quad &a+0=a&\quad\quad &(\text{j})&\quad &a\cdot 1=a\\ &(\text{k})&\
quad &a^1=a \end{alignedat} \label{Eq:I:22:1} \end{equation} \begin
{equation} \begin{alignedat}{4} &(\text{a})&\quad&a+b=b+a\\ &(\text
{b})&\quad&a+(b+c)=(a+b)+c\\ &(\text{c})&\quad&ab=ba\\ &(\text{d})&\
quad&a(b+c)=ab+ac\\ &(\text{e})&\quad&(ab)c=a(bc)\\ &(\text{f})&\quad
&(ab)^c=a^cb^c\\ &(\text{g})&\quad&a^ba^c=a^{(b+c)}\\ &(\text{h})&\
quad&(a^b)^c=a^{(bc)}\\ &(\text{i})&\quad&a+0=a\\ &(\text{j})&\quad&a
\cdot 1=a\\ &(\text{k})&\quad&a^1=a \end{alignedat} \label{Eq:I:22:1}
\end{equation} These results are well known and we shall not belabor
the point, we merely list them. Of course, $1$ and $0$ have special
properties; for example, $a + 0$ is $a$, $a$ times $1= a$, and $a$ to
the first power is $a$.
In this discussion we must also assume a few other properties like
continuity and ordering, which are very hard to define; we will let
the rigorous theory do it. Furthermore, it is definitely true that we
have written down too many "rules"; some of them may be deducible
from the others, but we shall not worry about such matters.
22-2The inverse operations
In addition to the direct operations of addition, multiplication, and
raising to a power, we have also the inverse operations, which are
defined as follows. Let us assume that $a$ and $c$ are given, and
that we wish to find what values of $b$ satisfy such equations as $a
+ b = c$, $ab = c$, $b^a = c$. If $a + b= c$, $b$ is defined as $c -
a$, which is called subtraction. The operation called division is
also clear: if $ab = c$, then $b = c/a$ defines division--a solution
of the equation $ab = c$ "backwards." Now if we have a power $b^a =
c$ and we ask ourselves, "What is $b$?," it is called the $a$th root
of $c$: $b = \sqrt[a]{c}$. For instance, if we ask ourselves the
following question, "What integer, raised to the third power,
equals $8$?," then the answer is called the cube root of $8$; it
is $2$. Because $b^a$ and $a^b$ are not equal, there are two inverse
problems associated with powers, and the other inverse problem would
be, "To what power must we raise $2$ to get $8$?" This is called
taking the logarithm. If $a^b = c$, we write $b = \log_ac$. The fact
that it has a cumbersome notation relative to the others does not
mean that it is any less elementary, at least applied to integers,
than the other processes. Although logarithms come late in an algebra
class, in practice they are, of course, just as simple as roots; they
are just a different kind of solution of an algebraic equation. The
direct and inverse operations are summarized as follows: \begin
{equation} \begin{alignedat}{5} &(\text{a})&&\quad \text{addition}&&\
quad &&(\text{a}')&&\quad \text{subtraction}\\ & &&\quad a+b=c&&\quad
&& &&\quad b=c-a\\ &(\text{b})&&\quad \text{multiplication}&&\quad &&
(\text{b}')&&\quad \text{division}\\ & &&\quad ab=c&&\quad && &&\quad
b=c/a\\ &(\text{c})&&\quad \text{power}&&\quad &&(\text{c}')&&\quad \
text{root}\\ & &&\quad b^a=c&&\quad && &&\quad b=\sqrt[a]{c}\\ &(\
text{d})&&\quad \text{power}&&\quad &&(\text{d}')&&\quad \text
{logarithm}\\ & &&\quad a^b=c&&\quad && &&\quad b=\log_ac\\ \end
{alignedat} \label{Eq:I:22:2} \end{equation}
Now here is the idea. These relationships, or rules, are correct for
integers, since they follow from the definitions of addition,
multiplication, and raising to a power. We are going to discuss
whether or not we can broaden the class of objects which $a$, $b$,
and $c$ represent so that they will obey these same rules, although
the processes for $a + b$, and so on, will not be definable in terms
of the direct action of adding $1$, for instance, or successive
multiplications by integers.
22-3Abstraction and generalization
When we try to solve simple algebraic equations using all these
definitions, we soon discover some insoluble problems, such as the
following. Suppose that we try to solve the equation $b = 3 - 5$.
That means, according to our definition of subtraction, that we must
find a number which, when added to $5$, gives $3$. And of course
there is no such number, because we consider only positive integers;
this is an insoluble problem. However, the plan, the great idea, is
this: abstraction and generalization. From the whole structure of
algebra, rules plus integers, we abstract the original definitions of
addition and multiplication, but we leave the rules (22.1) and (22.2
), and assume these to be true in general on a wider class of
numbers, even though they are originally derived on a smaller class.
Thus, rather than using integers symbolically to define the rules, we
use the rules as the definition of the symbols, which then represent
a more general kind of number. As an example, by working with the
rules alone we can show that $3 - 5 = 0 - 2$. In fact we can show
that one can make all subtractions, provided we define a whole set of
new numbers: $0 - 1$, $0 - 2$, $0 - 3$, $0 - 4$, and so on, called
the negative integers. Then we may use all the other rules, like $a(b
+ c) = ab + ac$ and so forth, to find what the rules are for
multiplying negative numbers, and we will discover, in fact, that all
of the rules can be maintained with negative as well as positive
integers.
So we have increased the range of objects over which the rules work,
but the meaning of the symbols is different.
One cannot say, for instance, that $-2$ times $5$ really means to
add $5$ together successively $-2$ times. That means nothing. But
nevertheless everything will work out all right according to the
rules.
An interesting problem comes up in taking powers. Suppose that we
wish to discover what $a^{(3-5)}$ means. We know only that $3 - 5$ is
a solution of the problem, $(3 - 5) + 5 = 3$. Knowing that, we know
that $a^{(3-5)}a^5 = a^3$. Therefore $a^{(3-5)} = a^3/a^5$, by the
definition of division. With a little more work, this can be reduced
to $1/a^2$. So we find that the negative powers are the reciprocals
of the positive powers, but $1/a^2$ is a meaningless symbol, because
if $a$ is a positive or negative integer, the square of it can be
greater than $1$, and we do not yet know what we mean by $1$ divided
by a number greater than $1$!
Onward! The great plan is to continue the process of generalization;
whenever we find another problem that we cannot solve we extend our
realm of numbers. Consider division: we cannot find a number which is
an integer, even a negative integer, which is equal to the result of
dividing $3$ by $5$. But if we suppose that all fractional numbers
also satisfy the rules, then we can talk about multiplying and adding
fractions, and everything works as well as it did before.
Take another example of powers: what is $a^{3/5}$? We know only that
$(3/5)5 = 3$, since that was the definition of $3/5$. So we know also
that $(a^{(3/5)})^5 =$ $a^{(3/5)(5)}=$ $a^3$, because this is one of
the rules. Then by the definition of roots we find that $a^{(3/5)} =
\sqrt[5]{a^3}$.
In this way, then, we can define what we mean by putting fractions in
the various symbols, by using the rules themselves to help us
determine the definition--it is not arbitrary. It is a remarkable fact
that all the rules still work for positive and negative integers, as
well as for fractions!
We go on in the process of generalization. Are there any other
equations we cannot solve? Yes, there are. For example, it is
impossible to solve this equation: $b =$ $2^{1/2} =$ $\sqrt{2}$. It
is impossible to find a number which is rational (a fraction) whose
square is equal to $2$. It is very easy for us in modern days to
answer this question. We know the decimal system, and so we have no
difficulty in appreciating the meaning of an unending decimal as a
type of approximation to the square root of $2$. Historically, this
idea presented great difficulty to the Greeks. To really define
precisely what is meant here requires that we add some substance of
continuity and ordering, and it is, in fact, quite the most difficult
step in the processes of generalization just at this point. It was
made, formally and rigorously, by Dedekind. However, without worrying
about the mathematical rigor of the thing, it is quite easy to
understand that what we mean is that we are going to find a whole
sequence of approximate fractions, perfect fractions (because any
decimal, when stopped somewhere, is of course rational), which just
keeps on going, getting closer and closer to the desired result. That
is good enough for what we wish to discuss, and it permits us to
involve ourselves in irrational numbers, and to calculate things like
the square root of $2$ to any accuracy that we desire, with enough
work.
22-4Approximating irrational numbers
The next problem comes with what happens with the irrational powers.
Suppose that we want to define, for instance, $10^{\sqrt{2}}$. In
principle, the answer is simple enough. If we approximate the square
root of $2$ to a certain number of decimal places, then the power is
rational, and we can take the approximate root, using the above
method, and get an approximation to $10^{\sqrt{2}}$. Then we may run
it up a few more decimal places (it is again rational), take the
appropriate root, this time a much higher root because there is a
much bigger denominator in the fraction, and get a better
approximation. Of course we are going to get some enormously high
roots involved here, and the work is quite difficult. How can we cope
with this problem?
In the computations of square roots, cube roots, and other small
roots, there is an arithmetical process available by which we can get
one decimal place after another. But the amount of labor needed to
calculate irrational powers and the logarithms that go with them (the
inverse problem) is so great that there is no simple arithmetical
process we can use. Therefore tables have been built up which permit
us to calculate these powers, and these are called the tables of
logarithms, or the tables of powers, depending on which way the table
is set up. It is merely a question of saving time; if we must raise
some number to an irrational power, we can look it up rather than
having to compute it. Of course, such a computation is just a
technical problem, but it is an interesting one, and of great
historical value. In the first place, not only do we have the problem
of solving $x=10^{\sqrt{2}}$, but we also have the problem of solving
$10^x = 2$, or $x = \log_{10} 2$. This is not a problem where we have
to define a new kind of number for the result, it is merely a
computational problem. The answer is simply an irrational number, an
unending decimal, not a new kind of a number.
Let us now discuss the problem of calculating solutions of such
equations. The general idea is really very simple. If we could
calculate $10^1$, and $10^{4/10}$, and $10^{1/100}$, and $10^{4/1000}
$ and so on, and multiply them all together, we would get $10^{1.414\
dots}$ or $10^{\sqrt{2}}$, and that is the general idea on which
things work. But instead of calculating $10^{1/10}$ and so on, we
shall calculate $10^{1/2}$, $10^{1/4}$, and so on. Before we start,
we should explain why we make so much work with $10$, instead of some
other number. Of course, we realize that logarithm tables are of
great practical utility, quite aside from the mathematical problem of
taking roots, since with any base at all, \begin{equation} \label
{Eq:I:22:3} \log_b(ac)=\log_ba+\log_bc. \end{equation} We are all
familiar with the fact that one can use this fact in a practical way
to multiply numbers if we have a table of logarithms. The only
question is, with what base $b$ shall we compute? It makes no
difference what base is used; we can use the same principle all the
time, and if we are using logarithms to any particular base, we can
find logarithms to any other base merely by a change in scale, a
multiplying factor. If we multiply Eq. (22.3) by $61$, it is just as
true, and if we had a table of logs with a base $b$, and somebody
else multiplied all of our table by $61$, there would be no essential
difference. Suppose that we know the logarithms of all the numbers to
the base $b$. In other words, we can solve the equation $b^a = c$ for
any $c$ because we have a table. The problem is to find the logarithm
of the same number $c$ to some other base, let us say the base $x$.
We would like to solve $x^{a'} = c$. It is easy to do, because we can
always write $x = b^t$, which defines $t$, knowing $x$ and $b$. As a
matter of fact, $t = \log_b x$. Then if we put that in and solve for
$a'$, we see that $(b^t)^{a'} = b^{a't} = c$. In other words, $ta'$
is the logarithm of $c$ in base $b$. Thus $a' = a/t$. Thus logs to
base $x$ are just $1/t$, which is a constant, times the logs to the
base, $b$. Therefore any log table is equivalent to any other log
table if we multiply by a constant, and the constant is $1/\log_b x$.
This permits us to choose a particular base, and for convenience we
take the base $10$. (The question may arise as to whether there is
any natural base, any base in which things are somehow simpler, and
we shall try to find an answer to that later. At the moment we shall
just use the base $10$.)
Table 22-1Successive Square Roots of Ten
Power $s$ $1024\,s$ $10^s$ $(10^s-1)/s$
$1\ $9.00\hphantom{00^
phantom{/ $1024$ $10.00000\hphantom{00}$ {000}}$
1024}$
$1/2\ $\phantom $\phantom{1}3.16228\ $4.32\hphantom{00^
phantom {1}512$ hphantom{00}$ {000}}$
{000}$
$1/4\ $\phantom $\phantom{1}1.77828\ $3.113\hphantom{0^
phantom {1}256$ hphantom{00}$ {000}}$
{000}$
$1/8\ $\phantom $\phantom{1}1.33352\ $2.668\hphantom{0^
phantom {1}128$ hphantom{00}$ {000}}$
{000}$
$1/16\ $\phantom $\phantom{1}1.15478\ $2.476\hphantom{0^
phantom {10}64$ hphantom{00}$ {000}}$
{00}$
$1/32\ $\phantom $\phantom{1}1.074607\ $2.3874\hphantom{^
phantom {10}32$ hphantom{0}$ {000}}$
{00}$
$1/64\ $\phantom $\phantom{1}1.036633\ $2.3445\hphantom{^
phantom {10}16$ hphantom{0}$ {000}}$
{00}$
$1/128\ $\phantom $\phantom{1}1.018152\
phantom {100}8$ hphantom{0}$ $2.3234^{211}$
{0}$
$1/256\ $\phantom
phantom {100}4$ $\phantom{1}1.0090350$ $2.3130^{104}$
{0}$
$1/512\ $\phantom $2.3077^{\phantom{1}
phantom {100}2$ $\phantom{1}1.0045073$ 53}$
{0}$
$1/1024$ $\phantom $\phantom{1}1.0022511$ $2.3051^{\phantom{1}
{100}1$ 26}$
$\phantom{00}\Big\
downarrow\hspace 3ex^
{26}$
$\Delta/ $\phantom $1+0.0022486\Delta\ $\raise.5ex\overline{\
1024$ {102}\ overleftarrow{\kern kern 1em}2.3025$
Delta$ 1.5em}$
$(\Delta\
to 0)$
Now let us see how to calculate logarithms. We begin by computing
successive square roots of $10$, by cut and try. The results are
shown in Table 22-1. The powers of $10$ are given in the first
column, and the result, $10^s$, is given in the third column. Thus
$10^1 = 10$. The one-half power of $10$ we can easily work out,
because that is the square root of $10$, and there is a known, simple
process for taking square roots of any number.^1 Using this process,
we find the first square root to be $3.16228$. What good is that? It
already tells us something, it tells us how to take $10^{0.5}$, so we
now know at least one logarithm, if we happen to need the logarithm
of $3.16228$, we know the answer is close to $0.50000$. But we must
do a little bit better than that; we clearly need more information.
So we take the square root again, and find $10^{1/4}$, which
is $1.77828$. Now we have the logarithm of more numbers than we had
before, $1.250$ is the logarithm of $17.78$ and, incidentally, if it
happens that somebody asks for $10^{0.75}$, we can get it, because
that is $10^{(0.5+0.25)}$; it is therefore the product of the second
and third numbers. If we can get enough numbers in column $s$ to be
able to make up almost any number, then by multiplying the proper
things in column 3, we can get $10$ to any power; that is the plan.
So we evaluate ten successive square roots of $10$, and that is the
main work which is involved in the calculations.
Why don't we keep on going for more and more accuracy? Because we
begin to notice something. When we raise $10$ to a very small power,
we get $1$ plus a small amount. The reason for this is clear, because
we are going to have to take the $1000$th power of $10^{1/1000}$ to
get back to $10$, so we had better not start with too big a number;
it has to be close to $1$. What we notice is that the small numbers
that are added to $1$ begin to look as though we are merely dividing
by $2$ each time; we see $1815$ becomes $903$, then $450$, $225$; so
it is clear that, to an excellent approximation, if we take another
root, we shall get $1.00112$ something, and rather than actually take
all the square roots, we guess at the ultimate limit. When we take a
small fraction $\Delta/1024$ as $\Delta$ approaches zero, what will
the answer be? Of course it will be some number close to $1+0.0022511
\,\Delta$. Not exactly $1+0.0022511\,\Delta$, however--we can get a
better value by the following trick: we subtract the $1$, and then
divide by the power $s$. This ought to correct all the excesses to
the same value. We see that they are very closely equal. At the top
of the table they are not equal, but as they come down, they get
closer and closer to a constant value. What is the value? Again we
look to see how the series is going, how it has changed with $s$. It
changed by $211$, by $104$, by $53$, by $26$. These changes are
obviously half of each other, very closely, as we go down. Therefore,
if we kept going, the changes would be $13$, $7$, $3$, $2$ and $1$,
more or less, or a total of $26$. Thus we have only $26$ more to go,
and so we find that the true number is $2.3025$. (Actually, we shall
later see that the exact number should be $2.3026$, but to keep it
realistic, we shall not alter anything in the arithmetic.) From this
table we can now calculate any power of $10$, by compounding the
power out of $1024$ths.
Let us now actually calculate a logarithm, because the process we
shall use is where logarithm tables actually come from. The procedure
is shown in Table 22-2, and the numerical values are shown in Table
22-1 (columns 2 and 3).
Table 22-2Calculation of a logarithm: $\boldsymbol{\log_{10} 2}$
$2 \div 1.77828 = 1.124682$
$1.124682 \div 1.074607 = 1.046598$, etc.
$\
therefore $2=(1.77828)(1.074607)(1.036633)(1.0090350)(1.000573)$
\,$
$\phantom{2}=10^{\biggl[\dfrac{1}{1024}\mbox
{(256+32+16+4+0.254)}\biggr]}=10^{\biggl[\dfrac{308.254}
{1024}\biggr]}$
$\phantom{2}=10^{0.30103}\phantom{(256+32+16+4}\biggl(\
dfrac{573}{2249}=0.254\biggr)$
$\
therefore $\log_{10}2=0.30103$
\,$
Suppose we want the logarithm of $2$. That is, we want to know to
what power we must raise $10$ to get $2$. Can we raise $10$ to the $1
/2$ power? No; that is too big. In other words, we can see that the
answer is going to be bigger than $1/4$, and less than $1/2$. Let us
take the factor $10^{1/4}$ out; we divide $2$ by $1.778\dots$, and
get $1.124\dots$, and so on, and now we know that we have taken away
$0.250000$ from the logarithm. The number $1.124\dots$, is now the
number whose logarithm we need. When we are finished we shall add
back the $1/4$, or $256/1024$. Now we look in the table for the next
number just below $1.124\dots$, and that is $1.074607$. We therefore
divide by $1.074607$ and get $1.046598$. From that we discover that
$2$ can be made up of a product of numbers that are in Table 22-1, as
follows: \begin{equation*} 2 = (1.77828)(1.074607)(1.036633)
(1.0090350)(1.000573). \end{equation*} \begin{gather*} 2 = (1.77828)
(1.074607)(1.036633)\;\times\\ (1.0090350)(1.000573). \end{gather*}
There was one factor $(1.000573)$ left over, naturally, which is
beyond the range of our table. To get the logarithm of this factor,
we use our result that $10^{\Delta/1024} \approx 1+ 2.3025 \Delta/
1024$. We find $\Delta= 0.254$. Therefore our answer is $10$ to the
following power: $(256 + 32 + 16 + 4 + 0.254)/1024$. Adding those
together, we get $308.254/1024$. Dividing, we get $0.30103$, so we
know that the $\log_{10} 2 = 0.30103$, which happens to be right
to $5$ figures!
This is how logarithms were originally computed by Mr. Briggs of
Halifax, in 1620. He said, "I computed successively $54$ square roots
of $10$." We know he really computed only the first $27$, because the
rest of them can be obtained by this trick with $\Delta$. His work
involved calculating the square root of $10$ twenty-seven times,
which is not much more than the ten times we did; however, it was
more work because he calculated to sixteen decimal places, and then
reduced his answer to fourteen when he published it, so that there
were no rounding errors. He made tables of logarithms to fourteen
decimal places by this method, which is quite tedious. But all
logarithm tables for three hundred years were borrowed from
Mr. Briggs' tables by reducing the number of decimal places. Only in
modern times, with the WPA and computing machines, have new tables
been independently computed. There are much more efficient methods of
computing logarithms today, using certain series expansions.
In the above process, we discovered something rather interesting, and
that is that for very small powers $\epsilon$ we can calculate $10^\
epsilon$ easily; we have discovered that $10^\epsilon = 1+ 2.3025\
epsilon$, by sheer numerical analysis. Of course this also means that
$10^{n/2.3025} = 1+ n$ if $n$ is very small. Now logarithms to any
other base are merely multiples of logarithms to the base $10$. The
base $10$ was used only because we have $10$ fingers, and the
arithmetic of it is easy, but if we ask for a mathematically natural
base, one that has nothing to do with the number of fingers on human
beings, we might try to change our scale of logarithms in some
convenient and natural manner, and the method which people have
chosen is to redefine the logarithms by multiplying all the
logarithms to the base $10$ by $2.3025\dots$ This then corresponds to
using some other base, and this is called the natural base, or
base $e$. Note that $\log_e (1 + n) \approx n$, or $e^n \approx 1+ n$
as $n\to0$.
It is easy enough to find out what $e$ is: $e = 10^{1/2.3025\dots}$
or $10^{0.434310\dots}$, an irrational power. Our table of the
successive square roots of $10$ can be used to compute, not just
logarithms, but also $10$ to any power, so let us use it to calculate
this natural base $e$. For convenience we transform $0.434310\dots$
into $444.73/1024$. Now, $444.73$ is $256 + 128 + 32 + 16 + 8 + 4 +
0.73$. Therefore $e$, since it is an exponent of a sum, will be a
product of the numbers \begin{equation*} (1.77828)\!(1.33352)\!
(1.074607)\!(1.036633)\!(1.018152)\! (1.009035)\!(1.001643) = 2.7184.
\end{equation*} \begin{align*} (1.&77828)\!(1.33352)\!(1.074607)\!
(1.036633)\;\times\\ &(1.018152)\!(1.009035)\!(1.001643)= 2.7184. \
end{align*} (The only problem is the last one, which is $0.73$, and
which is not in the table, but we know that if $\Delta$ is small
enough, the answer is $1 + 0.0022486\,\Delta$.) When we multiply all
these together, we get $2.7184$ (it should be $2.7183$, but it is
good enough). The use of such tables, then, is the way in which
irrational powers and the logarithms of irrational numbers are all
calculated. That takes care of the irrationals.
22-5Complex numbers
Now it turns out that after all that work we still cannot solve every
equation! For instance, what is the square root of $-1$? Suppose we
have to find $x^2 =-1$. The square of no rational, of no irrational,
of nothing that we have discovered so far, is equal to $-1$. So we
again have to generalize our numbers to a still wider class. Let us
suppose that a specific solution of $x^2 =-1$ is called something, we
shall call it $i$; $i$ has the property, by definition, that its
square is $-1$. That is about all we are going to say about it; of
course, there is more than one root of the equation $x^2 =-1$.
Someone could write $i$, but another could say, "No, I prefer $-i$.
My $i$ is minus your $i$." It is just as good a solution, and since
the only definition that $i$ has is that $i^2=-1$, it must be true
that any equation we can write is equally true if the sign of $i$ is
changed everywhere. This is called taking the complex conjugate. Now
we are going to make up numbers by adding successive $i$'s, and
multiplying $i$'s by numbers, and adding other numbers, and so on,
according to all of our rules. In this way we find that numbers can
all be written in the form $p + iq$, where $p$ and $q$ are what we
call real numbers, i.e., the numbers we have been defining up until
now. The number $i$ is called the unit imaginary number. Any real
multiple of $i$ is called pure imaginary. The most general number,
$a$, is of the form $p+iq$ and is called a complex number. Things do
not get any worse if, for instance, we multiply two such numbers, let
us say $(r + is)(p + iq)$. Then, using the rules, we get \begin
{align} (r + is)(p + iq) &= rp + r(iq) + (is)p + (is)(iq)\notag\\
[1ex] &= rp + i(rq) + i(sp) + (ii)(sq)\notag\\[1ex] \label{Eq:I:22:4}
&= (rp - sq) + i(rq + sp), \end{align} since $ii =$ $i^2 =$ $-1$.
Therefore all the numbers that now belong in the rules (22.1) have
this mathematical form.
Now you say, "This can go on forever! We have defined powers of
imaginaries and all the rest, and when we are all finished, somebody
else will come along with another equation which cannot be solved,
like $x^6 + 3x^2 =-2$. Then we have to generalize all over again!"
But it turns out that with this one more invention, just the square
root of $-1$, every algebraic equation can be solved! This is a
fantastic fact, which we must leave to the Mathematics Department to
prove. The proofs are very beautiful and very interesting, but
certainly not self-evident. In fact, the most obvious supposition is
that we are going to have to invent again and again and again. But
the greatest miracle of all is that we do not. This is the last
invention. After this invention of complex numbers, we find that the
rules still work with complex numbers, and we are finished inventing
new things. We can find the complex power of any complex number, we
can solve any equation that is written algebraically, in terms of a
finite number of those symbols. We do not find any new numbers. The
square root of $i$, for instance, has a definite result, it is not
something new; and $i^i$ is something. We will discuss that now.
We have already discussed multiplication, and addition is also easy;
if we add two complex numbers, $(p + iq) + (r + is)$, the answer is $
(p + r) + i(q + s)$. Now we can add and multiply complex numbers. But
the real problem, of course, is to compute complex powers of complex
numbers. It turns out that the problem is actually no more difficult
than computing complex powers of real numbers. So let us concentrate
now on the problem of calculating $10$ to a complex power, not just
an irrational power, but $10^{(r+is)}$. Of course, we must at all
times use our rules (22.1) and (22.2). Thus \begin{equation} \label
{Eq:I:22:5} 10^{(r+is)}=10^r10^{is}. \end{equation} But $10^r$ we
already know how to compute, and we can always multiply anything by
anything else; therefore the problem is to compute only $10^{is}$.
Let us call it some complex number, $x + iy$. Problem: given $s$,
find $x$, find $y$. Now if \begin{equation*} 10^{is}=x+iy, \end
{equation*} then the complex conjugate of this equation must also be
true, so that \begin{equation*} 10^{-is}=x-iy. \end{equation*} (Thus
we see that we can deduce a number of things without actually
computing anything, by using our rules.) We deduce another
interesting thing by multiplying these together: \begin{equation} \
label{Eq:I:22:6} 10^{is}10^{-is}=10^0=1=(x+iy)(x-iy)=x^2+y^2. \end
{equation} \begin{equation} \begin{gathered} \label{Eq:I:22:6} 10^
{is}10^{-is}=10^0=1\\ =(x+iy)(x-iy)=x^2+y^2. \end{gathered} \end
{equation} Thus if we find $x$, we have $y$ also.
Now the problem is how to compute $10$ to an imaginary power. What
guide is there? We may work over our rules until we can go no
further, but here is a reasonable guide: if we can compute it for any
particular $s$, we can get it for all the rest. If we know $10^{is}$
for any one $s$ and then we want it for twice that $s$, we can square
the number, and so on. But how can we find $10^{is}$ for even one
special value of $s$? To do so we shall make one additional
assumption, which is not quite in the category of all the other
rules, but which leads to reasonable results and permits us to make
progress: when the power is small, we shall suppose that the "law"
$10^\epsilon = 1+ 2.3025\epsilon$ is right, as $\epsilon$ gets very
small, not only for real $\epsilon$, but for complex $\epsilon$ as
well. Therefore, we begin with the supposition that this law is true
in general, and that tells us that $10^{is} = 1+ 2.3025\cdot is$, for
$s\to0$. So we assume that if $s$ is very small, say one part
in $1024$, we have a rather good approximation to $10^{is}$.
Now we make a table by which we can compute all the imaginary powers
of $10$, that is, compute $x$ and $y$. It is done as follows. The
first power we start with is the $1/1024$ power, which we presume is
very nearly $1+ 2.3025i/1024$. Thus we start with \begin{equation} \
label{Eq:I:22:7} 10^{i/1024}=1.00000+0.0022486i, \end{equation} and
if we keep multiplying the number by itself, we can get to a higher
imaginary power. In fact, we may just reverse the procedure we used
in making our logarithm table, and calculate the square, $4$th power,
$8$th power, etc., of (22.7), and thus build up the values shown in
Table 22-3. We notice an interesting thing, that the $x$ numbers are
positive at first, but then swing negative. We shall look into that a
little bit more in a moment. But first we may be curious to find for
what number $s$ the real part of $10^{is}$ is zero. The $y$-value
would be $1$, and so we would have $10^{is} = 1i$, or $is = \log_{10}
i$. As an example of how to use this table, just as we calculated $\
log_{10} 2$ before, let us now use Table 22-3 to find $\log_{10} i$.
Table 22-3Successive Squares of $\boldsymbol{10^{i/1024} = 1 +
0.0022486i}$
Power $is$ $1024s$ $10^{is}$
$i/1024$ $\phantom{000}1$ $\phantom{-}1.00000 + 0.00225i$*
$i/512\phantom{0}$ $\phantom{000}2$ $\phantom{-}1.00000 + 0.00450i$
$i/256\phantom{0}$ $\phantom{000}4$ $\phantom{-}0.99996 + 0.00900i$
$i/128\phantom{0}$ $\phantom{000}8$ $\phantom{-}0.99984 + 0.01800i$
$i/64\phantom{00}$ $\phantom{00}16$ $\phantom{-}0.99936 + 0.03599i$
$i/32\phantom{00}$ $\phantom{00}32$ $\phantom{-}0.99742 + 0.07193i$
$i/16\phantom{00}$ $\phantom{00}64$ $\phantom{-}0.98967 + 0.14349i$
$i/8\phantom{000}$ $\phantom{0}128$ $\phantom{-}0.95885 + 0.28402i$
$i/4\phantom{000}$ $\phantom{0}256$ $\phantom{-}0.83872 + 0.54467i$
$i/2\phantom{000}$ $\phantom{0}512$ $\phantom{-}0.40679 + 0.91365i$
$i/1\phantom{000}$ $1024$ $-0.66928 + 0.74332i$
* Should be $0.0022486i$
Which of the numbers in Table 22-3 do we have to multiply together to
get a pure imaginary result? After a little trial and error, we
discover that to reduce $x$ the most, it is best to multiply "$512$"
by "$128$." This gives $0.13056 + 0.99159i$. Then we discover that we
should multiply this by a number whose imaginary part is about equal
to the size of the real part we are trying to remove. Thus we choose
"$64$" whose $y$-value is $0.14349$, since that is closest
to $0.13056$. This then gives $-0.01308 + 1.00008i$. Now we have
overshot, and must divide by $0.99996 + 0.00900i$. How do we do that?
By changing the sign of $i$ and multiplying by $0.99996 - 0.00900i$
(which works if $x^2 + y^2 = 1$). Continuing in this way, we find
that the entire power to which $10$ must be raised to give $i$ is $i
(512 + 128 + 64 - 4 - 2 + 0.20)/1024$, or $698.20i/1024$. If we raise
$10$ to that power, we can get $i$. Therefore $\log_{10} i =
0.68184i$.
22-6Imaginary exponents
Table 22-4Successive Powers of $\boldsymbol{10^{i/8}}$
$p=$ $\text{power}\cdot8/i$ $10^{ip/8}$
$\phantom{0}0$ $\phantom{-}1.00000+0.00000i$
$\phantom{0}1$ $\phantom{-}0.95882+0.28402i$
$\phantom{0}2$ $\phantom{-}0.83867+0.54465i$
$\phantom{0}3$ $\phantom{-}0.64944+0.76042i$
$\phantom{0}4$ $\phantom{-}0.40672+0.91356i$
$\phantom{0}5$ $\phantom{-}0.13050+0.99146i$
$\phantom{0}6$ $-0.15647+0.98770i$
$\phantom{0}7$ $-0.43055+0.90260i$
$\phantom{0}8$ $-0.66917+0.74315i$
$\phantom{0}9$ $-0.85268+0.52249i$
$10$ $-0.96596+0.25880i$
$11$ $-0.99969-0.02620i$
$12$ $-0.95104-0.30905i$
$14$ $-0.62928-0.77717i$
$16$ $-0.10447-0.99453i$
$18$ $+0.45454-0.89098i$
$20$ $+0.86648-0.49967i$
$22$ $+0.99884+0.05287i$
$24$ $+0.80890+0.58836i$
To further investigate the subject of taking complex imaginary
powers, let us look at the powers of $10$ taking successive powers,
not doubling the power each time, in order to follow Table 22-3
further and to see what happens to those minus signs. This is shown
in Table 22-4, in which we take $10^{i/8}$, and just keep multiplying
it. We see that $x$ decreases, passes through zero, swings almost
to $-1$ (if we could get in between $p = 10$ and $p = 11$ it would
obviously swing to $-1$), and swings back. The $y$-value is going
back and forth too.
Figure 22-1
In Fig. 22-1 the dots represent the numbers that appear in Table 22-4
, and the lines are just drawn to help you visually. So we see that
the numbers $x$ and $y$ oscillate; $10^{is}$ repeats itself, it is a
periodic thing, and as such, it is easy enough to explain, because if
a certain power is $i$, then the fourth power of that would be $i^2$
squared. It would be $+1$ again, and therefore, since $10^{0.68i}$ is
equal to $i$, by taking the fourth power we discover that $10^{2.72i}
$ is equal to $+1$. Therefore, if we wanted $10^{3.00i}$, for
instance, we could write it as $10^{2.72i}$ times $10^{0.28i}$. In
other words, it has a period, it repeats. Of course, we recognize
what the curves look like! They look like the sine and cosine, and we
shall call them, for a while, the algebraic sine and algebraic
cosine. However, instead of using the base $10$, we shall put them
into our natural base, which only changes the horizontal scale; so we
denote $2.3025s$ by $t$, and write $10^{is} = e^{it}$, where $t$ is a
real number. Now $e^{it} = x + iy$, and we shall write this as the
algebraic cosine of $t$ plus $i$ times the algebraic sine of $t$.
Thus \begin{equation} \label{Eq:I:22:8} e^{it}=\operatorname{\
underline{\cos}}t+ i\operatorname{\underline{\sin}}t. \end{equation}
What are the properties of $\operatorname{\underline{\cos}} t$ and $\
operatorname{\underline{\sin}} t$? First, we know, for instance, that
$x^2 + y^2$ must be $1$; we have proved that before, and it is just
as true for base $e$ as for base $10$. Therefore $\operatorname{\
underline{\cos}}^2 t+ \operatorname{\underline{\sin}}^2 t= 1$. We
also know that, for small $t$, $e^{it} = 1+it$, and therefore $\
operatorname{\underline{\cos}} t$ is nearly $1$, and $\operatorname{\
underline{\sin}} t$ is nearly $t$, and so it goes, that all of the
various properties of these remarkable functions, which come from
taking imaginary powers, are the same as the sine and cosine of
trigonometry.
Is the period the same? Let us find out. $e$ to what power is equal
to $i$? What is the logarithm of $i$ to the base $e$? We worked it
out before, in the base $10$ it was $0.68184i$, but when we change
our logarithmic scale to $e$, we have to multiply by $2.3025$, and if
we do that it comes out $1.570$. So this will be called "algebraic $\
pi/2$." But, we see, it differs from the regular $\pi/2$ by only one
place in the last point, and that, of course, is the result of errors
in our arithmetic! So we have created two new functions in a purely
algebraic manner, the cosine and the sine, which belong to algebra,
and only to algebra. We wake up at the end to discover the very
functions that are natural to geometry. So there is a connection,
ultimately, between algebra and geometry.
We summarize with this, the most remarkable formula in mathematics: \
begin{equation} \label{Eq:I:22:9} e^{i\theta}=\cos\theta+i\sin\theta.
\end{equation} This is our jewel.
We may relate the geometry to the algebra by representing complex
numbers in a plane; the horizontal position of a point is $x$, the
vertical position of a point is $y$ (Fig. 22-2). We represent every
complex number, $x+iy$. Then if the radial distance to this point is
called $r$ and the angle is called $\theta$, the algebraic law is
that $x+iy$ is written in the form $re^{i\theta}$, where the
geometrical relationships between $x$, $y$, $r$, and $\theta$ are as
shown. This, then, is the unification of algebra and geometry.
Fig. 22-2.$x + iy = re^{i\theta}$.
When we began this chapter, armed only with the basic notions of
integers and counting, we had little idea of the power of the
processes of abstraction and generalization. Using the set of
algebraic "laws," or properties of numbers, Eq. (22.1), and the
definitions of inverse operations (22.2), we have been able here,
ourselves, to manufacture not only numbers but useful things like
tables of logarithms, powers, and trigonometric functions (for these
are what the imaginary powers of real numbers are), all merely by
extracting ten successive square roots of ten!
1. There is a definite arithmetic procedure, but the easiest way to
find the square root of any number $N$ is to choose some $a$
fairly close, find $N/a$, average $a' = \tfrac{1}{2}[a + (N/a)]$,
and use this average $a'$ for the next choice for $a$. The
convergence is very rapid--the number of significant figures
doubles each time. -
Copyright (c) 1963, 2006, 2013 by the California Institute of
Technology, Michael A. Gottlieb and Rudolf Pfeiffer