Datasets:
vnpthack
/
vnpt_data

Dataset card Data Studio Files
xet
 Community
1
text
stringlengths
0
6.73k
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 1s set
up. Ít is merely a question of saving time; iŸ 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 # = 10Y2, but we also have the problem of solving 10 = 2, or # = logig 2.
This is not a problem where we have to defne 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 oŸ calculating solutions of such equations.
The general idea is really very simple. If we could caleulate 101, and 10, and
101/10 and 10/1900 and so on, and multiply them all together, we would get
10114: or 10Y2, and that is the general idea on which things work. But instead
of calculating 10119 and so on, we shall caleulate 101/2, 101/4, and so on. Before
we start, we should explain why we make so mụch 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,
logg(ac) = logy ø + logy e. (22.3)
--- Trang 395 ---
W© are all familiar with the fact that one can use this fact in a practical way to
multiply numbers iŸ we have a table of logarithms. The only question is, with
what base ö shall we compute? It makes no diference what base is used; we
can use the same principle all the time, and iŸ 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. IÝ we multiply Eq. (22.3) by 61, ¡it is Just as true,
and ïif we had a table of logs with a base ö, and somebody else multiplied all of
our table by 61, there would be no essential diference. Suppose that we know
the logarithms of all the numbers to the base b. In other words, we can solve
the equation b# = c for any c because we have a table. 'Phe problem is to ñnd
the logarithm of the same number c to some other base, let us say the base #.
We would like to solve #° = e. It is easy to do, because we can always write
z = bÝ, which delnes , knowing z and b. As a matter of fact, £ = log,ø. Then
if we put that in and solve for a”, we see that (bf)%“ = b* = e. In other words,
ta! is the logarithm of ein base b. Thus ø' = ø/£. Thus logs to base # are just
1/f, which is a constant, tìmes the logs to the base, ð. Therefore any log table is
equivalent to any other log table iŸ we multiply by a constant, and the constant
is 1/log,ø. 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 fnd an
answer to that later. At the moment we shall just use the base 10.)
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, is given in the
third column. Thus 10! = 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.* Ủsing 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 1005, 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 bet6er than that; we clearly need more information. 5o we
take the square root again, and find 101/4, which is 1.77828. Now we have the
logarithm of more numbers than we had before, 1.250 is the logarithm of 17.78
* 'TThere is a definite arithmetic procedure, but the easiest way to fnd the square root oŸ any
number X is to choose some ø fairly close, find N/a, average q = sia + (N/a)], and use this
avcrage a” for the next choice for ø. The convergence is very rapid—the number of significant
figures doubles each time.
--- Trang 396 ---
Table 22-1
Successive Square Roots of Ten
1 1024 10.00000 9.00
1/2 512 3.16228 4.32
1/4 256 1.77828 3.113
1/8 128 1.33352 2.668
1/16 64 1.15478 2.476
1/32 32 1.074607 2.3874
1/64 16 1.036633 2.3445
1/128 8 1.018152 2.3234?
1/256 4 1.0090350 2.3130194
1/512 2 1.0045073 2.3077 °3
1/1024 1 1.0022511 2.3051 2°
A/1024 A 1 +0.0022486A 2.3025
(A => 0)
and, incidentally, if it happens that somebody asks for 105, we can get it,
because that is 10(0-5+0:25): ït js therefore the produet of the second and third
numbers. lÝ 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 1000th power of 101/190 to get back to 10, so we had better not sbart
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 cach 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 ¿øke all the square roots, we øwess at the ultimate
limit. When we take a small fraction A/1024 as A approaches zero, what will
the answer be? Of course it will be some number close to 1 + 0.0022511A. Not
--- Trang 397 ---
exactly 1 + 0.0022511A, however—we can get a better value by the following
tríck: we subtract the 1, and then divide by the power s. This ought to correc
all the excesses to the same value. We see that they are very closely equal. Ät
the top of the table they are not equal, but as they come down, they get cÌoser
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,