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Toeplitz Sequence Class Notes
Heidi Tucker and Matt Marks
October 26, 2007
1 Toeplitz Sequence
A Toeplitz Sequence begins by giving an increasing sequence of rational num- bers s 1 ≤ s 2 ≤ s 3 ≤ .... This matches up with another sequence of rationals tn that moves in a descending order: ... ≤ t 3 ≤ t 2 ≤ t 1. To this we add a requirement that (∀i, j)(si ≤ tj )
. But a gap is unwanted and can be avoided by “squeezing the numbers”. We add a final requirement that
lim x→∞
tn − sn = 0
. Basically, every real number is defined by a Toeplitz sequence, namely its decimal expansion. For example,
2 = 1. 41459 .... defines a T-sequence as follows. With this decimal one can take
1 < 1. 4 < 1. 41 < 1. 414 < .... < 1. 415 < 1. 42 < 1. 5 < 2
And every time it is the next digit in the sequence that is considered. From the above statement, it is known that t 1 − s 1 ≤ 1 and that t 2 − s 2 ≤ 101. From this it is possible to conclude that tn − sn ≤ (^10) n^1 − 1. It is also easy to see here that every si ≤ tj.
2 The Story of e
From there the class went to the story of e and compound interest. In order to compute interest there must be an interest rate x, and a principal a 0. Assuming somebody’s interest is compounded yearly, the principal will advance to a 1 = a 0 + xa 0 = a 0 (1 + x) at the end of the year. The next time around it would look like a 2 = a 1 + xa 1 = a 1 (1 + x) = a 0 (1 + x)^2. Interest can be compounded quarterly as well. To compound quarterly throughout the year take
a 1 = a 0 (1 +
x 4
), a 2 = a 0 (1 +
x 4
)^2 , a 4 = a 0 (1 +
x 4
)^4.
a 4 represents the end of the year. In order to compound monthly take:
a 1 = a 0 (1 +
x 12
And at the end of the 12 month cycle the final product will look like
a 12 = a 0 (1 +
x 12
)^12
The more something is compounded the larger the return on interest is. If a person had a contract where interest were to be compounded every day of the year, he or she would earn more money than a person who has had his or her interest compounded every month. But by looking at the three above examples of compounding interest it is easy to see that they all have something in common. That is the expression of the form
(1 +
x n
)n.
Simplifiying for now with x = 1 we next demonstrate Toeplitz sequence s 1 ≤ s 2 ≤ s 3 ≤ ... ≤ t 3 ≤ t 2 ≤ t 1 with sn = (1 + (^) n^1 )n. This gives a definition of e as limn→∞ (1 + (^) n^1 )n^ = e. So if sn = (1 + (^) n^1 )n, then s 1 = (1 + 11 )^1 = 2 and s 2 = (1 + 12 )^2 = 94 = 2.25 and s 3 = (1 + 13 )^3 = 2.37. Now there is a lemma
(1 +
n
)n^ < (1 +
n + 1
)n+1.
This is worked out by realizing that (^1) n > (^) n+1^1. Then when 1 is added to
each side the sum is (1 + (^1) n ) > (1 + (^) n+1^1 ). But once these are taken to their respected powers there comes a problem.
(1 +
n
)n^ > (1 +
n + 1
)n^ < (1 +
n + 1
)n+1.
We take sn and tn and rewrite them simplify doing arithmetic with them:
sn = (1 +
n
)n^ = (
n + 1 n
)n
tn = (1 −
n
)−n^ = (
n − 1 n
)−n^ = (
n n − 1
)n
Some examples for sn:
s 2 = (
)^2
s 3 = (
)^3
s 4 = (
)^4
Some examples for tn:
t 3 = (
)^3
t 4 = (
)^4
t 5 = (
)^5
Look at the Geometric Inequality above, with b = n+1 n > a = n n+2+.
(1 + (^1) n )n+1^ − (1 + (^) n+1^1 )n+ (1 + (^) n^1 ) − (1 + (^) n+1^1 )
≤ (n + 1)(1 +
n
)n
Transform this lemma so that sn is integrated into it using (n + 1)(1 + (^) n^1 )n^ = (n + 1)sn by following the steps:
(1) sn(1+^
1 1 n^ )−sn+ n −^ n^1 +
(2) sn(1+ (^1) n )−sn+ nn+1(n+1)−n
sn(1+ (^1) n )−sn+ 1 n(n+1)
(4) (sn(1 + (^1) n ) − sn+1)(n(n + 1))
(5) sn(1 + (^) n^1 ) − sn+1 ≤ (^) n(nn+1+1) sn
Watch closely, the underline portions will cancel out...
(6) sn + (^1) n sn − sn+1 ≤ (^) n^1 sn
(7) sn ≤ sn+
Using this reasoning, how can we determine that tn+1 ≤ tn? WELL.... Dur- ing class we first determined what a and b were.
an+1^ = tn(1 −
n
)−^1 = (1 −
n
)−n(1 −
n
)−^1 = (
n n − 1
)n(
n n − 1
n n − 1
)n+
Thus, because an+1^ = ( (^) n−n 1 )n+1, a = (^) nn− 1. To find b:
bn+1^ = tn+1 = (1 −
n + 1
)−(n+1)^ = (
n + 1 n
)n+
Thus, because bn+1^ = (n+1 n )n+1, b = n+1 n.
Simply as a checker, we can set these in a relation and cross multiply by positive numbers so as not to change the relation. We find which of the two, a = (^) nn− 1 or b = n+1 n is larger.
n n − 1
n + 1 n Which gives us n^2 ∼ n^2 − 1.
This, quite obviously shows that a is larger than b. Using the earlier lemma again,
(n + 1)an^ ≤
(1 + (^1) n )n+1^ − (1 + (^) n+1^1 )n+ (1 + (^) n^1 ) − (1 + (^) n+1^1 )
≤ (n + 1)(1 +
n
)n
While using
tn = (1 −
n
)−n^ = (
n − 1 n
)−n^ = (
n n − 1
)n
We can craftily incorporate tn in like such:
n(
n + 1 n
n + 1 n
)n^ ≤
tn(1 − (^1) n )−^1 − tn+ ( (^) nn− 1 ) − (n+1 n )
≤ (n + 1)(
n n − 1
)n
We can eventually determine tn+1 ≤ tn. Exercise: Write the conclusion, pulling all the parts of this proof together.
