|
WHAT EVERYONE
SHOULD KNOW ABOUT e AND LOGARITHMS
by M.A. Strong (University
of Pennsylvania)
Where
does “e” come from?
Say
that $1 is put into a bank that gives 100% interest. If interest is
compounded once per year, the value (V) of our investment becomes:
|
V(1) = initial principal x (1 + interest rate)
= $1 x (1 + 1) = $2 |
(where
the number in parentheses following V indicates the frequency of
compounding and “x” means multiplied by).
If
interest is compounded twice a year, an interest rate of 50% (i.e.,
100/2) or .50
per $1 will be used for each 6-month period. Thus at the end of the
first 6 months we would have:
During the next 6 months is
$1.50 would grow to:
|
V(2) =
$1.50 x (1+ 0.50) = $2.25
= $1 x (1 + 0.50) (1 + 0.50)
= $1 x (1+0.50) 2 |
Similarly, if interest is
compounded quarterly, the quarterly interest rate would be 25% (or .25
per dollar).
|
End of 1st quarter: |
$1 x
(1 + 0.25) = $1.25 |
|
2nd
quarter: |
$1.25 x (1+0.25) = [$1 x (1+0.25)] (1+0.25) =$1.5 |
|
3rd
quarter: |
$1.56 x (1+0.25) = {[$1 x (1+0.25)] (1+0.25)} = $1.95 |
|
4th
quarter: |
V(4)
= $1.95 x (1+0.25) = $1 x (1+0.25) (1+0.25) (1+0.25)(1+0.25) |
| |
= $1 x (1+0.25) 4 = $2.44 |
|
Thus: |
V(1)
= $1 x (1 + (1.0 /1) 1 =
$2.00 |
| |
V(2)
= $1 x (1 + (1.0 /2) 2 = $2.25 |
| |
V(4)
= $1 x (1 + (1.0 /4) 4 =
$2.44 |
| In
general: |
V(m)
= $1 x (1 + (1.0 /m)m |
| |
where m represents the frequency of compounding in 1
year. |
The limiting case is where
interest is compounded continuously, i.e., where m becomes
infinite.
This gives us our definition
of e:
|
limit V (m) = limit [$1 X (1 + (1/m)m] =
e (dollars) |
| m +
¥ m + ¥ |
As you can see in the
following diagram, the value of our $1, as it is compounded more and
more frequently (i.e., as m gets larger) does not get infinitely large,
but approaches a limit.
/images/e_loga2.gif)
Generalizations:
1) More years: if
$1 becomes $e after 1 year of continuous compounding, and we leave this
$e in the bank a second year, each dollar in $e will grow to e and the
value of our savings account will be:
|
$e
x e = $e2
which would become
in three years: $e3
or, more generally,
after t years, $et |
2) Different principal
(i.e., initial investment amount): since each $1 grows to $et
in t years, if we invested $A to start with, our final value will be A
times as large or $Aet.
3) Different interest
rate: in the initial formula, replace 100% (or 1.00 per dollar)
with r:
| V(m)
= $1 x (1 + (r/m))m = [(1 + (r/m) (m/r)]
r |
|
|
= [(1 + (1/z)z]r |
if
we let m/r = z and r/m = 1/z |
| |
|
| V(m)
= [(1 + (1/z))z]r |
Since m
®
¥ made (1 + (1/m))m
= e, |
| |
and
if m
®
¥ , z = (m/r)
®
¥
too, |
|
= [e]r |
\
(1 + (1/z))z = e. |
Applying all three generalizations:
| V(m)
= A (1 + (r/m))mt |
$A
rather than $1, is the principal. |
| |
r/m
means that in each of m periods only |
| |
1/mth
of the interest rate is applied. |
| |
mt =
compoundings per year x the number |
| |
of
years, i.e., the total no. of compounding |
| |
|
|
= A [(1 + (r/m))(m/r)] rt |
if
we let m/r = w, then r/m = 1/w as above |
|
= A [(1 + (1/w))w]rt |
|
| |
|
|
= Aert |
thus
m
®
¥ implies m/r=w
®
¥, |
| |
and
(1 + (1/w))w = e |
This
analysis can be applied to any growth process where growth takes place
continuously at a constant rate r, including population growth. Note
that r, previously the interest rate, can no be reinterpreted as the
instantaneous rate of growth of the function Aert.
LOGARITHMS
A
logarithm is the power to which a base must be raised to attain a
particular number:
|
f the base is 4,
then since 42 = 16
log4 16 = 2
base no. base to this
power = no. |
“common” logs use 10 for the base
|
log10
1,000 = 3 (since 103 = 1000)
log10
100 = 2 (since 102 = 100)
log10
10 = 1 (since 101 = 10)
log10
1 = 0 (since 100 = 0)
log10
0.1 = -1 (since 10-1 = (1/10)1 =
(1/10) = 0.1
log10
0.01 = -2 (since 10-2 = (1/10)2 =
(1/100) = 0.01
etc. |
In
general, y = bt
«
t = log b y
In
analytic work, natural logs, using e as the base, are more
convenient to use. Note that loge is frequently written
ln.
|
Since y = bt
+ t = log b y
y = e3
+ 3 + ln y = ln e3
y = en
+ n + ln y = ln en |
Example: What will the rate of growth have to be for a population P to
double in 20 years (i.e., what growth rate is required if population 20
years from now is two time the current population –2*P)?
|
2P = P ert
2 = e 20r
ln
2 = ln e20r
ln
2 = 20r
0.6931 = 20r |
r
= (0.6931/20) = 0.035 or a growth rate of (0.035 * 100) = 3.5
percent
To calculate doubling time, if you know the growth rate:
| |
t=
69.3/r where t is the doubling time and r is the growth rate
express as a % |
|
OR |
t=
0.693/r where r is the growth rate expressed per person (NOT
per cent) |
|
/clear.gif){kind=spacer}
/graypixel.gif){kind=tracking}