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NOTES
Edited by William Adkins
Nested Square Roots of 2
L. D. Servi
Nested square roots arise in unexpected places. Two such examples are presented and
then validated using a new theorem.
Example 1. The trigonometric table for multiples of π/32 can be represented with
nested square roots:
θ
π
32
π
16
3π
32
π
8
5π
32
3π
16
7π
32
π
4
9π
32
5π
16
11π
32
3π
8
13π
32
7π
16
15π
32
π
2
326
2 cos[θ ]
√
2+ 2+ 2+ 2
√
2+ 2+ 2
√
2+ 2+ 2− 2
2 sin[θ ]
√
2− 2+ 2+ 2
√
2− 2+ 2
2−
√
2+ 2
√
2+ 2− 2− 2
√
2+ 2− 2
√
2+ 2− 2+ 2
√
√
2− 2
√
2− 2− 2− 2
√
2− 2− 2
2−
√
2− 2− 2+ 2
√
2− 2− 2
√
2− 2− 2− 2
√
2− 2+ 2− 2
√
2− 2+ 2
√
2− 2+ 2+ 2
0
2
√
2+ 2− 2+ 2
√
2+ 2− 2
2+
√
2− 2
√
2− 2+ 2
√
2
√
2+ 2− 2
√
2− 2− 2
√
2+ 2
√
2+ 2+ 2− 2
√
2+ 2+ 2
2+
√
2+ 2+ 2
2
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Example 2. The constant π has the following nested square roots representations:

 
 
 
 
 
√
 k


π = lim 2 2 − 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + · · · + 2 + 2 

k→∞ 



 




k square roots

(1)

 
 
 
 
 k

√
2 
2 − 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + · · · + 2 − 2 
= lim 

k→∞ 
3




 




k square roots

(2)









√ 

k−1

2
2
−
2
+
2
+
2
+
= lim 
+
2
+
2
+
2
+
2
+
2
+
·
·
·
+
3
3
·
2

k→∞ 









k square roots
(3)





3
= lim 
k→∞ 
5








√

· 2k−1 2 − 2 + 2 + 2 + 2 + 2 + 2 + 2 + 2 + · · · + 2 − 3 





k square roots
(4)
The relationship between the expressions
b1 π
bk 2 + bk−1 2 + bk−2 2 + · · · + b2 2 + 2 sin
R(bk , . . . , b1 ) =
2
4
April 2003]
NOTES
(5)
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and
f (bk , . . . , b1 ) =
December 11, 2002
10:56 a.m.
bk bk−1 bk bk−1 bk−2
bk bk−1 bk−2 · · · b1
1 bk
−
−
−
− ··· −
2
4
8
16
2k+1
notes.tex
π
(6)
facilitates the derivation of these examples.
Theorem. If b1 lies in the interval [−2, 2] and bi belongs to the set {−1, 0, 1} for
i = 1, then
cos f (bk , . . . , b1 ) = R(bk , . . . , b1 );
(7)
if in addition bk = 0, then
sin f (bk , . . . , b1 ) = R(1, −bk−1 , bk−2 , bk−3 , . . . , b1 ).
(8)
Derivation of Example 1. Example 1 can be confirmed from either a standard trigonometric table or from (7) and
For example, if (b4 , b3 , b2 , b1 ) = (1, −1, 1, −1) then,
(8). √
from (7), cos[13π/32] = 2 − 2 + 2 − 2/2 (so cos[1.28] ≈ .290).
Derivation of Example 2. From (6), f (1, 1, 1, 1, . . . , 1, 1, b1 ) = (2 − b1 )π/2k+1 so
(8) implies that sin[(2 − b1 )π/2k+1 ] = R(1, −1, 1, 1, . . . , 1, 1, b1 ). As k increases
without bound the left-hand side approaches (2 − b1 )π/2k+1 . Hence,


k+1
2
R(1, −1, 1, 1, . . . , 1, 1, b1 )
(9)
π = lim 
k→∞
2 − b1 k terms
if b1 = 2. Equations (1), (2), (3), and (4) follow directly from (9) and (5) by taking b1
to be 1, −1, 4/3, and −4/3, respectively.
Proof of Theorem. From (5) and (6), if ρk = R(bk , . . . , b1 ) and φk = f (bk , . . . , b1 ),
then

b1 π


sin
for k = 1,

4
ρk =
(10)


 bk √2 + 2ρ
for k = 1,
k−1
2
and

(2 − b1 )π





4



φ

k−1


2
φk =
π





2




φ

 π − k−1
2
for k = 1 and b1 ∈ {−1, 0, 1},
for k = 1 and bk = 1,
(11)
for k = 1 and bk = 0,
for k = 1 and bk = −1.
We demonstrate (7) by proving by induction that cos[φk ] = ρk for k = 1, 2, . . . . For
k = 1, this follows directly from (10) and (11), since cos[(2 − b1 )π/4] = sin[b1 π/4]
for all b1 . Suppose that cos[φκ−1 ] = ρκ−1 for a fixed κ ≥ 2. From the half-angle cosine
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formula (noting that the restrictions on bi imply that 0 ≤ φi ≤ π for all i) we have
φκ−1
bκ (12)
2 + 2 cos[φκ−1 ].
bκ cos
=
2
2
But from (11) we see that for κ ≥ 2 the left-hand side of (12) equals cos[φκ ] for bκ = 1,
0, or −1, while from (10) the right-hand side of (12) equals ρκ . Hence, cos[φκ ] = ρκ ,
so (7) holds for k = κ and, by induction,
for all values of k.
√
Since sin[π/4] = sin[3π/4] = 2/2, (8) follows for k = 1. Using the half-angle
sine formula, (7), and the definition of ρk−1 , we obtain for k > 1
sin[φk−1 /2] = .5 2 − 2 cos[φk−1 ] = .5 2 − 2ρk−1 = R(1, −bk−1 , bk−2 , . . . , b1 ).
From (11), the left-hand side equals sin[φk ] for bk = 1 or −1, whence (8) holds for
k > 1 as well.
Related results. The series,
(2 − b1 )π
(2 − b1 )π 3
(2 − b1 )π
2k+1
2k+1
sin
−
=
3! + · · ·
2 − b1
2k+1
2 − b1
2k+1
2k+1
≈ π + O(2−2k )
suggests that obtaining m digits of accuracy from (1)–(4) requires taking
k≈
m log2 (10)
≈ 1.7m.
2
As a result, (1)–(4) are not as computationally efficient as modern algorithms that
require only log m iterations [1], [2].
Also, equation (1) is somewhat related to the formula

k square roots

√ 


∞

2
+
2
+
·
·
·
+
2
+
2

2

=




π
2
k=1 




(13)
first discovered by François Viète in the sixteenth century [1, p. 686]. Equation (13)
can be rederived by first noting that z = π/2 in (7) gives cos[z2−k ] = R(1
k ), where
−k
1
k is a k-dimensional vector of ones. From [3, p. 44], 1/z = ∞
k=1 cos[z2 ]/ sin[z].
∞
Hence, 2/π = k=1 R(1
k ), which is equivalent to (13).
Clearly, many related expressions can be similarly derived.
ACKNOWLEDGMENTS. The author thanks G. Strang of the Massachusetts Institute of Technology for a
lecture related to the discrete cosine transform at MIT Lincoln Laboratory, which inspired this note, as well
as valuable comments and advice on earlier drafts. He also thanks D. J. Daley (Australian National University), B. Epstein (MIT Lincoln Laboratory), R. S. Pinkham (Stevens Institute of Technology), and E. Wolman
(George Mason University) for their thoughtful comments. This paper is dedicated to my children, Amelia and
Joseph, who I hope will read this paper someday.
April 2003]
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This work is sponsored by the United States Air Force under Air Force Contract #F19628-00-C-0002.
Opinions, interpretations, recommendations and conclusions are those of the author and are not necessarily
endorsed by the United States Government.
REFERENCES
1. L. Berggren, J. Borwein, and P. Borwein, Pi: A Source Book, 2nd ed., Springer-Verlag, New York, 2000.
2. R. P. Brent, Fast multiple-precision evaluation of elementary functions, J. Assoc. Comp. Machine. 23
(1976) 242–251.
3. D. Zwillinger, CRC Standard Mathematical Tables and Formulae, 30th ed., CRC Press, Boca Raton, FL,
1996.
MIT Lincoln Laboratory, Lexington, MA 02420
[email protected]
Strategies for The Weakest Link
Nigel Boston
1. INTRODUCTION. The television program The Weakest Link recently came to
the U.S. from the U.K. It has drawn a lot of interest because of the acerbic style of its
host, Anne Robinson, but it also raises some interesting questions of strategy that we
address in this paper.
First, let us outline how the game is played. In round one, eight players are asked
questions in turn, with each correct answer in sequence raising the pot. Incorrect answers send the pot back to zero. Before his or her question is asked, the next player
can bank the money so far accumulated. At the end of each round, one player is voted
off and eventually one player takes home whatever money has been banked over seven
rounds, the last round counting double.
Let’s consider how well the players did in, for example, the show that aired on
May 7, 2001. The winner took home $60,500. Since, however, 78 out of the 117 questions (doubling up the last round) were answered correctly, he could have taken home
$78,000, if everyone had banked the $1,000 gained from each correct answer. If everyone had waited instead for a sequence of two correct answers (worth $2,500) before
banking and then banked, the winner would have received $70,000. If the strategy had
been to bank only after sequences of three, four, or five correct answers in a row, then
the winner would have gained $60,000, $50,000, or $50,000 respectively. There were
no sequences of six correct answers in a row.
The example suggests that the best strategy might be to bank after every correct
answer. In what follows, we use mathematics (some combinatorics/probability theory)
to substantiate this claim somewhat, although there are some strange twists in the tale.
2. THE PROBLEM. We will model the situation in the following way. Let us assume that the probability that a contestant answers a question correctly is p. For the
time being we assume that this probability is the same for each contestant. Using the
example cited earlier, p might be taken to be 78/117 = 2/3—this seems typical for
this game. In a round of n questions (in the early rounds, n is usually about 18), the
amount banked by various strategies will be computed.
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First, suppose that each player banks after every correct answer. This is expected to
occur np times and will yield 1000np dollars.
Second, suppose that each player banks after a sequence of two correct answers
and only then. We compute the expected yield as follows. The outcome of a round can
be given as a sequence of length n of ones and zeros, with one representing a correct
and zero an incorrect answer. So, for instance, if p = 2/3 and n = 3, then we get 000
(all three answers incorrect) with probability 1/27, 001 with probability 2/27, . . . , and
111 with probability 8/27. Our strategy will lead to $0 being banked with probability
11/27 and to $2,500 being banked with probability 16/27 (corresponding to the cases
of 011, 110, 111). Then our expected yield is about $1,480. Compare this with the
expected yield of $2,000 if the first strategy had been pursued instead.
The question then is to compute, for general p, n, and r , the expected yield if the
players pursue a strategy of banking after every sequence of r correct answers and
only then. We will call this the r th strategy.
3. THE COMPUTATION. Let p be the probability of a correct answer and
q = 1 − p the probability of an incorrect answer. Let f (n, r, p) denote the expected
number of occurrences of disjoint runs 11 . . . 1 of length r in a binary sequence of
length n. The amount banked after r correct answers is currently given by v[r ] dollars,
where v is the vector [1000, 2500, 5000, 10000, 25000, 50000, 75000, 125000]. Thus
v[r ] f (n, r, p) dollars will be the expected payoff from the r th strategy in a round of
length n.
For instance, continuing our earlier example, f (3, 2, p) = p 3 + 2 p 2 q = p 3 (1 +
2q/ p), since each of 111, 110, and 011 has one occurrence of 11 and it occurs with
probability p 3 , p 2 q, and p 2 q, respectively. It is clear that f (n, r, p) can be written
in the form p n Fn (r, x), where Fn is a polynomial of degree n − r in q/ p and so in
x := 1 + q/ p = 1/ p.
Lemma. Let Hn (r, x) = nx n−1 − (n + 1)x n − (n − r )x n+r −1 + (n − r + 1)x n+r . Let
n(mod r ) be denoted e, so e is an element of {0, 1, 2, . . . , r − 1}. Then
Fn (r, x) = (Hn (r, x) − He (r, x))/(1 − x r )2
Sketch of proof. The identity
Fn (r, x) + Fn+1 (r, x) + · · · + Fn+r −1 (r, x) = nx n−1
(∗)
was first experimentally observed and then proved. Fixing p and r , we denote
f (n, r, p) by E n . The identity (∗) is then equivalent to
E n+r −1 + p E n+r −2 + · · · + pr −1 E n = npr .
For simplicity, we explain why this holds in the case r = 2; the general case is
similar. Consider a binary sequence of length n + 1. The probability of position i
(1 ≤ i ≤ n) being the start of a 11 is p 2 and so we expect np 2 such occurrences.
This counts all runs 11 but we only want disjoint runs. Every 111 or 1111 gives a
discrepancy of 1 between np 2 and E n+1 . Every 11111 or 111111 gives a discrepancy
of 2, and so on. This discrepancy is accounted for by p E n , whence the formula.
If 1 ≤ n < r , then E n (and so Fn (r, x)) is clearly 0. The identity (∗) therefore determines Fn (r, x) for all n. The formula for Fn given was actually found by summing
derivatives of geometric series. To prove its correctness once found simply takes mathematical induction using (∗).
April 2003]
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We can now answer the main question. For given n and p, for which r is
v[r ] p n Fn (r, p −1 ) (the expected yield from applying the r th strategy) greatest?
The following PARI gp program gives the answer (denoted j (n, p)) for any chosen
n and p.
v=[1000,2500,5000,10000,25000,50000,75000,125000]
h(n,r,x)=n∗x∧(n-1)-(n+1)∗x∧n-(n-r)∗x∧(n+r-1)+(n-r+1)∗x∧(n+r)
f(n,r,p)=p∧n∗(h(n,r,p∧-1)-h(r∗frac(n/r),r,p∧-1))/(1-p∧-r)∧2
j(n,p)=a=0;b=0;for(r=1,8,if(v[r]∗f(n,r,p)>a,a=v[r]∗f(n,r,p);
b=r, ));b
The following commands are useful in understanding the ranges of values of p for
which a given strategy is best. The function k(n, r, s) gives the crossover probability (almost always unique) for changing between the r th and sth strategies, and l(n)
tabulates these values for a given n.
k(n,r,s)=solve(p=.001,.999,v[r]∗f(n,r,p)-v[s]∗f(n,s,p))
l(n)=for(r=1,7,for(s=r+1,8,print([r,s,k(n,r,s)])))
Looking, for instance, at the case n = 18, we see that for p < 0.563, the first strategy
gives the greatest expected yield. For 0.563 < p < 0.695, the sixth strategy is best.
For p > 0.695, the eighth strategy is best.
This same pattern holds whenever n ≥ 16 or 8 ≤ n ≤ 13. For n = 14 and 15, there
are new wrinkles to the tale. For instance, for n = 15 the seventh strategy actually
beats the eighth strategy for p > 0.951. (Note, however, that this is artificial. It arises
from the possibility of two runs of seven ones but impossibility of two runs of eight
ones. Such runs are unheard of in practice and the model would break down since the
limit of $125,000 winnings per round would be reached. This limit has so negligible an
effect otherwise as not to change the optimal strategy.) The limiting solution as n tends
to infinity, ignoring the $125,000 limit, is to go with the first strategy if p < 0.529, the
eighth if p > 0.647, and the sixth otherwise.
4. SIMULATIONS. The simulations come from the show itself. We tabulate data
from seven shows in May and June 2001:
Date
5/7
5/14
5/21
5/22
5/28
6/4
6/11
Total
“p”
0.667
0.675
0.500
0.574
0.667
0.653
0.587
0.619
Won
60.5
107.5
42
71
81
74.5
83.5
520
1
78
85
55
70
82
81
74
525
2
70
77.5
42.5
65
82.5
70
70
477.5
3
60
80
40
30
80
75
80
445
4
50
100
40
40
90
90
70
480
5
50
150
25
50
100
125
125
625
6
0
100
50
50
100
50
250
600
7
0
75
0
75
75
75
75
375
8
0
125
0
0
0
0
125
250
The table gives the proportion of questions answered correctly, the amount (in thousands of dollars) actually won, and how much the winning contestant would have taken
home if he or she had pursued the r th strategy (1 ≤ r ≤ 8). Compare this with the theoretical expectations of the amount won (to the nearest thousand dollars) under each
strategy based on the empirical values of p:
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“p”
0.619
1
546
2
501
3
479
4
502
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5
686
6
764
7
641
10:56 a.m.
notes.tex
8
599
5. CONCLUSIONS. Our assumptions would seem to support the sixth strategy,
since p is typically in that range. Looking at the data, the fifth and sixth strateges do
seem to do best in the long run. The safest bet is the first strategy (a good choice for the
cautious), which usually does better than the contestants’ chosen strategy. It will pretty
much guarantee at least $70,000 in winnings, but might not satisfy the ambitious. Attempting the fifth or sixth strategy is much more of a risk and will of course incur
repeated derision by Anne Robinson in the rounds that nothing is banked. I thank the
anonymous referee for running a simulation of 100,000 rounds with p = 0.619 (the
average over all the seven shows) and observing that under the sixth strategy nothing
would be banked for the whole show about 8% of the time. Another reason to bank
regularly is that voting ties are resolved by the “strongest link.” If two players have the
same scoring record, then the player who has banked more is regarded as stronger.
Voting strategies bring in many other issues. Recent shows suggest that being the
best player can backfire on you near the end of the show, when players want to vote off
their competition. For instance, on the last two shows from which data was collected
(6/4/01 and 6/11/01), the strongest player was voted off halfway through the program
with the second strongest player ultimately winning. It seems advisable to get some
questions wrong (on purpose if necessary—just be sure to answer quickly). Also, one
should make sure to vote for people certain to be voted off—revenge is commonly cited
as a reason for choosing someone to vote for. Lastly, on at least three occasions, women
have admitted to forming alliances in advance (something the game rules allow) in
order to ensure that the overall winner was a woman. This could lead to a backlash
whereby men vote women off out of fear of becoming targets of female alliances.
Further work will consider mixed strategies (i.e., with a probability pr of banking
after r consecutive correct answers). Calculations so far suggest that the pure strategies
(where each pr is 0 or 1) yield the greatest expected winnings. The assumption of a
constant p can also be relaxed (on 6/11 individual values of p varied between 0.438
and 0.750) and further classes of sequential strategies investigated. On the other hand,
since many of the contestants have trouble with elementary mathematical problems
(e.g., with questions like: What is the square root of 121?), expecting them to implement some sophisticated scheme is perhaps impractical. Thus banking after every
correct answer (particularly for weaker contestants) seems a good idea.
The author recently learned that Paul Coe of Dominican University announced similar results in his January 7, 2002 talk at the Joint Mathematics Meetings in San
Diego. See http://www.ams.org/amsmtgs/2049 abstracts/973-t1-634.pdf for his talk
announcement, dated September 15, 2001. The current paper was received by the
American Mathematical Monthly on June 18, 2001.
ACKNOWLEDGMENTS. The author thanks Andrew Singer and Ralf Koetter for very helpful discussions
and input on this topic. He was partially supported by NSF grant DMS-9970184.
Departments of Mathematics and ECE, University of Wisconsin, Madison, WI 53706
[email protected]
April 2003]
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Proofs of Korovkin’s Theorems via
Inequalities
Mitsuru Uchiyama
In this note we give quite simple proofs of Korovkin’s theorems (see [2] and [1]) by
making use of inequalities related to variance.
Let C[a, b] be the vector space of all real-valued continuous functions on [a, b],
and let T be a linear transformation on C[a, b]. Throughout this note, we assume that
any linear transformation under discussion is bounded. We say that T is positive if
T f ≥ 0 whenever f ≥ 0 on [a, b]. We denote max{| f (x)| : a ≤ x ≤ b} by f (the
norm of f ) and max{T ( f ) : f ≤ 1} by T (the operator norm of T ). A subspace
A of C[a, b] is called a subalgebra if f g belongs to A whenever f and g are members
of A.
The following theorem is called Korovkin’s first theorem.
Theorem A. Let {Tn } be a sequence of positive linear transformations on C[a, b]. If
Tn (h) − h → 0 (n → ∞)
(1)
holds for h = 1, x, and x 2 , then it holds for every h in C[a, b].
Lemma 1. Let A be a norm-closed subalgebra of C[a, b] that contains 1. If T is a
positive linear transformation on A with T (1) ≤ 1, then
V (h) := T (h 2 ) − T (h)2 ≥ 0
for every h in A. Moreover, for f, g, and k in A:
|T ( f g) − T ( f )T (g)|2 ≤ V ( f )V (g),
(2)
T ( f g) − T ( f )T (g) ≤ V ( f )1/2 V (g)1/2 ,
T ( f g) − T ( f )T (g) ≤ V ( f )
1/2
V (g) + V (k)
(3)
1/2
.
(4)
Proof. Since T ((h + t)2 ) ≥ 0 for every h and every real constant function t, we get
T (h 2 ) + 2t T (h) + t 2 T (1) ≥ 0
for all t, which implies that T (h)2 − T (h 2 ) T (1) ≤ 0—hence, that T (h 2 ) − T (h)2 ≥ 0,
because 0 ≤ T (1) ≤ 1. The substitution of f + tg for h in this inequality yields
t 2 (T (g 2 ) − T (g)2 ) + 2t{T ( f g) − T ( f )T (g)} + T ( f 2 ) − T ( f )2 ≥ 0
for all t. This implies (2), from which (3) follows immediately. The estimate
0 ≤ V (g) ≤ V (g) + V (k)
then gives (4).
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Proof of Theorem A. Since Tn (1) − 1 → 0, Tn (1) converges to 1. By considering
Tn /Tn (1) in place of Tn , we may assume that Tn (1) ≤ 1 for all n. This implies that
Tn ≤ 1. By (3),
Tn (x f ) − Tn (x)Tn ( f )2 ≤ Tn (x 2 ) − Tn (x)2 Tn ( f 2 ) − Tn ( f )2 (5)
holds for arbitrary f in C[a, b]. Since
Tn ( f 2 ) − Tn ( f )2 ≤ 2 f 2
and
lim Tn (x)2 = x 2 = lim Tn (x 2 ),
so the right-hand side of (5) tends to 0. Therefore, if Tn ( f ) − f → 0, then
Tn (x f ) − (x f ) → 0. Thus (1) holds for h = x m for m = 0, 1, 2, . . . . Because
Tn is linear, (1) is true for every polynomial p.
Since Tn ≤ 1 for every n, the theorem follows from the Weierstrass theorem asserting the norm-density of polynomials in C[a, b] (see [3, p. 159]).
Let C2π be the space of real-valued continuous functions f on [−π, π] such that
f (−π) = f (π). Then C2π is a closed subalgebra of C[−π, π] and 1 belongs to C2π .
The following theorem is called Korovkin’s second theorem.
Theorem B. Let {Tn } be a sequence of positive linear transformations of C2π . If (1)
holds for h = 1, sin x, and cos x, then it holds for every h in C2π .
Proof. As in the proof of Theorem A, there is no loss of generality in assuming that
Tn (1) ≤ 1. By (4), we have for every f in C2π :
Tn ( f sin x) − Tn ( f )Tn (sin x)2
≤ Tn ( f 2 ) − Tn ( f )2 Tn (sin2 x) − Tn (sin x)2 + Tn (cos2 x) − Tn (cos x)2 = Tn ( f 2 ) − Tn ( f )2 Tn (1) − Tn (sin x)2 − Tn (cos x)2 ≤ 2 f 2 Tn (1) − Tn (sin x)2 − Tn (cos x)2 → 2 f 2 1 − sin2 x − cos2 x = 0
as n → ∞. This implies that Tn ( f sin x) → f sin x whenever Tn f → f . We see
similarly that Tn ( f cos x) → f cos x in this situation. Thus (1) is true for h =
sinm x cosk x for all nonnegative integers m and n, which ensures that it is valid
for h = sin mx cos kx for all such m and n. Thus (1) holds for every trigonometric
polynomial p, and since the latter functions are dense in C2π (see [3, p. 190]), for
every h in C2π .
We next consider the space C(D) of complex-valued continuous functions f on the
closed unit disk D = {z : z ≤ 1} in the complex plane. A linear transformation T
on C(D) is called positive if T ( f ) ≥ 0 whenever f ≥ 0. It is easy to see that if T is
positive then T ( f ) = T ( f ). With a proof similar to that of Lemma 1, we have:
Lemma 2. If T is a positive linear transformation on C(D) with T (1) ≤ 1, then
V (h) := T (h2 ) − T (h)2 ≥ 0
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for every h in C(D). Moreover, for f and g in C(D) it is the case that
|T ( f g) − T ( f )T (g)| ≤ V ( f )1/2 V (g)1/2 ,
T ( f g) − T ( f )T (g) ≤ V ( f )1/2 V (g)1/2 .
(6)
We now state the complex Korovkin theorem:
Theorem C. Let {Tn } be a sequence of positive linear transformations on C(D). If (1)
holds for h = 1, z, and z2 , then it holds for every h in C(D).
Proof. We may assume that Tn (1) ≤ 1. It is evident that (1) holds for h whenever it
holds for h. The estimate (6) garantees that (1) holds for h = z m z n for all nonnegative
integers m and n, hence for every polynomial in z and z. By Stone’s theorem (see [3,
p. 165]) the set of all such polynomials is dense in C(D), so (1) holds for every h in
C(D).
Remark. If T is a positive linear functional on the real space C[0, 1] with T (1) = 1,
!1
then there is a Borel probability measure µ on [0, 1] such that T ( f ) = 0 f dµ for ev!
!
1
1
ery f in C[0, 1]. Then T ( f 2 ) − T ( f )2 = 0 f 2 dµ − ( 0 f dµ)2 ≥ 0 is the variance
of f . We have shown in [4] that (6) remains valid if C[a, b] is replaced by an arbitrary
noncommutative C ∗ -algebra .
REFERENCES
1. F. Altomare and M. Campiti, Korovkin-Type Approximation Theory and Its Applications, Walter de
Gruyter, Berlin, 1994.
2. P. P. Korovkin, On convergence of linear positive operators in the space of continuous functions,Dokl.
Akad.Nauk SSSR (N.S.) 90 (1953) 961–964 (Russian).
3. W. Rudin, Principles of Mathematical Analysis, 3rd ed., McGraw-Hill, New York, 1976.
4. M. Uchiyama, Korovkin-type theorems for Schwarz maps and operator monotone functions in C ∗ algebras, Math. Z. 230 (1999) 785–797.
Department of Mathematics, Fukuoka University of Education, Munakata, Fukuoka, 811-4192, Japan
[email protected]
The Zeros of the Partial Sums of the
Exponential Series
Peter Walker
"
Denote by sn (z) the partial sum of the exponential series: sn (z) = nk=0 z k /k! .
Since the zeros of sn grow asymptotically in proportion to n, it is convenient to
rescale, and to consider pn (z) = sn (nz). The asymptotic distribution of the zeros of sn
was given by G. Szegö [7] and J. Dieudonné [2], who showed that for large n the zeros
of pn cluster along the curve K ,
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$
# $
%
K = z : $ze1−z $ = 1, |z| ≤ 1 ,
with argument uniformly distributed with respect to arg(ze1−z ).
√
It was shown by Bucholtz in [1] that all zeros of pn lie within distance 2e/ n of
K and that no zero lies either on K or in the bounded component of C − K where
|ze1−z | < 1. We strengthen his elementary argument to show that there are no zeros
within distance d of K , where
d=
'
1 & 1/n
2 −1 .
2
2e
More recently, Varga and Carpenter [8] (see also Saff and Varga [6]) have shown by
much deeper methods that all zeros of pn satisfy
|z| ≤ 1, | arg z| ≥ cos
−1
$)
$
(
√
$ 1 − z $ 1/n
n − 2 $$ 1−z $$
$
$
,
≥ τn 2πn $
, ze
n
z $
(1)
where
τn =
1
n!
=1+
+O
√
n
−n
12n
n e
2πn
1
.
n2
This determines a larger zero-free region, due to the presence of the factor n 1/n on the
right-hand side of the last inequality in (1).
Our result is
Theorem 1. The following statements concerning pn hold:
(i) For all n, pn (1) = sn (n) > en /2.
(ii) There are no zeros z of pn with |z| ≤ 1 and |ze1−z | ≤ 21/n .
(iii) There are no zeros z of pn for which dist(z, K ) ≤ (21/n − 1)/(2e2 ).
Proof. (i) Doing n integrations by parts leads to
*
*
1 z n −t
1 ∞ n −t
−z
e sn (z) = 1 −
t e dt =
t e dt.
n! 0
n! z
!n
!∞
Hence, to show that sn (n) > en /2 it is enough to prove 0 t n e−t dt < n t n e−t dt, or
after a change of variable,
*
1
& −t 'n
dt <
te
0
*
∞
&
te−t
'n
dt.
(2)
1
To show (2) we first observe by comparing Taylor coefficients, that eu < 1/(1 − u)
if 0 < u < 1. Now square and integrate over [0, u] to get e2u < (1 + u)/(1 − u) when
0 < u < 1. This can be rewritten as eu (1 − u) < e−u (1 + u), or equivalently, with
g(t) = te−t , as g(1 − u) < g(1 + u) for 0 < u < 1. Hence
*
0
1
& −t 'n
te
dt <
*
2
& −t 'n
te
dt,
1
from which (2) follows.
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"
(ii) (Cf. [1, Theorem 2].) From the definition of pn (z) = nk=0 (nz)k /k! we see that,
if |ze1−z | ≤ 21/n and |z| ≤ 1, then
$
$
∞
k$
$ $$ −nz +
$
(nz)
$
−nz
$1 − e pn (z)$ = $e
$
$
$
k!
k=n+1
$
$
∞
$&
k n−k $
$ 1−z 'n −n + n z $
e
= $ ze
$
$
k! $
k=n+1
≤ 2e−n
∞
+
nk
<1
k!
k=n+1
by (i). Thus pn (z) = 0 for such z, as required.
(iii) Let f (z) = ze1−z , so that f (z) = (1 − z)e1−z and | f (z)| ≤ 2e2 when |z| ≤ 1.
Then if dist(z, K ) ≤ d = (21/n − 1)/(2e2 ) and u is a point of K with |z − u| ≤ d, we
have
| f (z)| ≤ | f (u)| + 2e2 d = 21/n ,
and the asserted result follows from (ii).
Remarks. Part (i) of the theorem is contained in the much harder result of Karamata
[3] and [4, 3.6.16], where it is shown that n −n n! (sn (n) − en /2) increases from 1/2 to
2/3 as n increases from 0 to ∞. It is curious to observe that deleting a single term from
sn reverses the inequality:
sn−1 (n) < en /2,
which follows by comparing the (n − j)th and (n + j − 1)th terms in
n−1 k
+
n
k=0
k!
<
∞
+
nk
k=n
k!
.
Combining this with part (i) of the theorem gives sn − en /2 < n n /n! and hence the
asymptotic relation
''
&
&
sn (n) = en /2 1 + O n −1/2 .
In particular e−n sn (n) → 1/2 as n → ∞. A much sharper result was given by Ramanujan [5], namely, that
& '
nn 2
4
en
+
−
+ O n −2 .
sn (n) =
2
n! 3 135n
ACKNOWLEDGMENTS. It is a pleasure to thank Professor Iossif Ostrovski for bringing the references [6]
and [8] to my attention and for helpful discussion.
REFERENCES
1. J. D. Bucholtz, A characterisation of the exponential series, this M ONTHLY 73 (1966) 121–123.
2. J. Dieudonné, Sur les zeros des polynomes-sections de e z , Bull. des Sciences Math., 2e série, 59
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3. J. Karamata, Sur quelques problèmes posés par Ramanujan, J. Indian Math. Soc. (N.S.) 24 (1960) 343–
365.
4. D. S. Mitrinović, Analytic Inequalities, Springer-Verlag, Berlin, 1970.
5. S. Ramanujan, Collected Papers, p. 26, Cambridge University Press, Cambridge, 1927.
6. E. B. Saff and R. S. Varga, On the zeros and poles of Padé approximants of e z , Numer. Math. 25 (1975)
1–14.
7. G. Szegö, Über eine Eigenschaft der Exponentialreihe, Sitzungsberichte der Berliner Math. Gesellschaft
21 (1922) 50–64; also in Collected Papers, vol. 1, Birkhauser, Boston, 1982, pp. 645–662.
8. R. S. Varga and A. J. Carpenter, Asymptotics for the zeros of the partial sums of e z , in Lecture Notes in
Math. 1435, Springer-Verlag, Berlin, 1990, pp. 201–7.
College of Arts and Science, American University of Sharjah, P.O. 26666, SHARJAH, United Arab Emirates
[email protected]
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