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Rational Exponential Expressions and a Conjecture Concerning π and e
Author(s): W. S. Brown
Source: The American Mathematical Monthly, Vol. 76, No. 1 (Jan., 1969), pp. 28-34
Published by: Mathematical Association of America
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28
RATIONAL EXPONENTIAL
EXPRESSIONS
[January
The TheoryofNumbers,4thed.,Oxford,1960.
4. G. H. Hardyand E. M. Wright,
betweenconsecutiveprimes,Quart.J. Math. Oxford,
5. A. E. Ingham,On the difference
Ser.(2) 8 (1937) 255-266.
Indag.Math.,12 (1950)57-58.
functions,
6. L. Kuipers,Prime-representing
Bull.Amer.Math. Soc., 53 (1947) 604.
function,
7. W. H. Mills,A prime-representing
8. I. Niven,Functionswhichrepresentprimenumbers,Proc. Amer.Math. Soc., 2 (1951)
753-755.
Proc.Amer.Math. Soc., 3 (1952) 706-712.
9. 0. Ore,On theselectionofsubsequences,
C. R. Acad. Sci.
Sur une formuledonnanttous les nombrespremiers,
10. W. Sierpinski,
Paris,235 (1952) 1078-1079.
11. B. R. Srinivasan,Formulaeforthe nthprime,J. Indian Math. Soc., (N.S.) 25 (1961)
33-39; 26 (1962) 248.
functionand an associatedformulafor the nth prime,I, II.
An arithmetical
12.
35 (1962) 68-71,72-75.
NorskeVid. Selsk.Forh.Trondheim,
13. C. P. Willans,On formulaeforthe nthprimenumber,Math. Gaz., 48 (1964) 413-415.
58 (1951) 616-618.
thisMONTHLY,
function,
A prime-representing
14. E. M. Wright,
J. LondonMath. Soc., 29 (1954) 63-71.
functions,
A class ofrepresenting
15. -
RATIONAL EXPONENTIAL EXPRESSIONS AND A
CONJECTURE CONCERNING 7rAND e
MurrayHill,New Jersey
W. S. BROWN, Bell TelephoneLaboratories,
Abstract. One of the most controversialand least well defined of mathematical problemsis the problemof simplification.The recentupsurgeof interest
in mechanizedmathematicshas lent new urgencyto this problem,but so farvery
littlehas been accomplished.This paper attemptsto shed lighton the situation
by introducingthe class of rational exponentialexpressions,definingsimplification withinthis class, and showingconstructivelyhow to achieve it. It is shown
that the only simplifiedrational exponential expressionequivalent to 0 is 0 itself,provided that an easily stated conjectureis true. However the conjecture,
if true, will surelybe difficultto prove, since it asserts as a special case that 'r
and e are algebraicallyindependent,and no one has yet been able to prove even
the much weaker conjecture that ir+e is irrational.
1. Introduction.Basically simplificationmeans the application of mathematical identitiesto transforma given expressioninto an equivalent expression
satisfyingsome desired criteria.
If the class of admissible expressionsis too broad, many dilemmas,ofwhich
the followingis typical, arise. A basic identityforsimplificationof expressions
involvingthe logarithmfunctionis
(1)
logZlZ2
=
logz1 + log Z2
whichholds everywhereon the Riemann surfacewhose pointshave the form
(2)
z = rei9,r >
O-
<oo
<0oo.
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1969]
RATIONAL EXPONENTIAL
EXPRESSIONS
29
On this surfaceone must abandon the identity
(3)
e2ir =
1
whichis fundamentalforsimplificationof expressionsinvolvingthe exponential
function.This identitycan be saved if we definethe logarithmfunctionover a
cut plane (for example, by restricting0 in (2) to the interval -7 r<0 <7r), but
then (1) is lost.
Surprisingly,it is not even necessaryto introducecomplexnumbersor multivalued functionsin order to get into insurmountabledifficulty.Consider the
class of expressionsgeneratedfromthe integers,the real constantswrand log 2,
and the real indeterminatex by application of the rational operations,the sine
function,the exponential function,the absolute value function,and substitution. Richardson [1] has shown that there is no algorithmto test whetheror
not a given expressionin this class is identicallyzero. Caviness [21 has shown
that the same result applies to the smaller class of expressionsgenerated as
above, but with log 2 and the exponentialfunctionomitted,provided that one
assumes the unsolvabilityof Hilbert's tenth problem [3]. The two proofsare
essentially the same except that Caviness begins with this assumption, while
Richardsonbeginswitha theoremof Davis, Putnam, and Robinson [4].
This paper attempts to shed light on the simplificationproblem by introducing the class of rational exponential (REX) expressions,definingsimplification withinthis class, and showingconstructivelyhow to achieve it.
The definitionissuch that distinctsimplifiedexpressionsmay be equivalent.
However, it is shown that the only simplifiedREX expressionequivalent to 0
is 0 itself,provided that an easily stated conjectureis true. Unfortunately,the
conjecture,if true, will surely be difficultto prove, since it asserts as a special
case that 7rand e are algebraicallyindependent,and no one has yet been able
to prove even the much weaker conjecture that ir+e is irrational [5].
Section 2 definesthe class of REX expressions,and definessimplificationfor
expressionsin this class. The fundamentalconjectureis discussed in Section 3,
the zero-equivalence theorem for simplifiedexpressionsin Section 4, and the
simplificationalgorithmin Section 5.
2. Definitions.The rationalexponential(RE X) expressionsare those which
are obtained by addition, subtraction,multiplication,division,and substitution
startingwith the (rational) integers,the constants7rand i, some indeterminates
, ZN (denoted collectively by z), and the exponential function. The
.*.
Zl
rational exponentialfunctions are those which can be represented by REX
expressions.
Note that all roots of unity are REX. It is convenient to introduce the
abbreviations
(4)
cm= exp(iir/2n)
m ? 1.
In particularc1 = exp(iir/2)= i.
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30
RATIONAL EXPONENTIAL
EXPRESSIONS
[January
A REX expressionis equivalentto 0 if it is identically0 when viewed as an
analytic functionof z. A REX expressionis equivalentto U (that is, undefined)
if it involves division by a subexpressionequivalent to 0. For example, the expression 2 (z1- z1) is equivalent to 0, while the expression 3+1/2 (z1- z1) is
equivalent to U. The symbol U is consideredto be a REX expression,but the
constructionof more complicated expressions involving it is forbidden.Two
REX expressionsare equivalentif both are equivalent to U or if theirdifference
is equivalent to 0.
process [6]
Any REX expressioncan be transformedby a straightforward
into an equivalent weakly simplifiedREX expression.A REX expressionis said
to be weaklysimplifiedifit is 0 or U or ifit has the form
f(exp pi, ...
g(exp pi,
, exp p., z, r,Coim)
,
exp pn, z, 7r)
where
(a) f and g are relativelyprimenonzeropolynomials;
(If g is the polynomial ? 1, we shall permit,and indeed insist,that it be
omitted,with the sign off being reversedif necessary.Althoughwe require that the polynomialf be nonzero, the numerator of (5) might
neverthelessbe equivalent to 0. For example, it mightbe the expression
exp(zl+z2) -exp(zi)exp(z2). In this case f is the nonzero polynomial
definedbyf(a, b, c) = a-bc.)
(b) the degreeoff in cor is less than the degreeof the minimalpolynomialof
corn;and
(c) Pi, ... , pn(On0_) are distinctweakly simplifiedREX expressionsother
than 0 or U.
A weakly simplifiedREX expressionis said to be simplifiedif it is 0 or U; or if
it has the form(5), and
, pnare simplified,and
(d) pi,
(e) the set {pi, * * * , p, ivr} is linearlyindependentover the rationals.
The significanceof this last conditionwill become clear in Section 4.
. ..
3. The fundamental conjecture. This section discusses the fundamental
conjecture on which our zero-testalgorithmdepends. The simplificationalgorithmalso depends on it, but only in that no use is made of the presentlyunknownidentitieswhose existencewould be impliedby its falsity.
is thattheonlyalgebraic
ROUGH STATEMENT. Roughlyspeaking,theconjecture
relationsinvolving7rand thez's and exponentialsofrationalexponentialexpressions
are thosewhichfollowdirectly
fromthefact thatrootsofunityare algebraicnumbers
and fromtheidentities
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1969]
RATIONAL EXPONENTIAL
31
EXPRESSIONS
exp(0) = 1
exp(i7r) = -1
(6)
exp(z1+ z2) = exp(zi) exp(Z2).
PRECISE STATEMENT. Let pl, . . ., Pnbe nonzerorationalexponentialexpressions such thattheset {p1, * * I, pn, i1r} is linearlyindependentovertherationals.
Then theset {exp pi, * , exp Pn,z, ir} is algebraicallyindependentovertherationals.
Proof oftheConverse.Suppose the set {pi, . . ., pPn,i7 } were linearlydepen, an such that
dent over the rationals. Then there would exist integersao,
n
= O.
aoir + Eaip
(7)
i=l
Using (6),
n
(8)
II
(exp
pi)ai
-
a
}
, exp pn, and a fortiori
the set {exp pi,
so the set {exp pi,
would be algebraicallydependent over the rationals.
,exp pn,z,7r
ofthefundamental
COROLLARY.
Settingn= 1 and pi =1 in theprecisestatement
we obtainthecorollarythatir and e are algebraicallyindependent.
conjecture,
4. Zero-eqtuivalencetheorem for simplifiedREX expressions. Let p be a
to 0 or U.
simplifiedREX expressionotherthan 0 or U. Then p is notequtivalent
Proof. Since p is simplified,we can writeit in the form(5) where (5a)-(5e)
hold. The proofis by inductionon the depthof p, definedas follows.If n =0 in
(5), then the depth of p is 0. Otherwisethe depth of p is
max(di, * * *, dn) + 1
d, are the depths of Pi, * - *, pnrespectively.
where d1,
Now let d be the depth of p. If d =0, then the proofis straightforward.In
the case d > 0, we assume inductivelythat the theoremis true forexpressionsof
depth less than d. It follows,using (5c) and (5d), that none of the pl, * * *, Pnis
equivalent to U. Thereforeneitherthe numeratornor the denominatorof (5) is
equivalent to U.
Viewed as a polynomialin the elementsof the set
z, }
{exp pi, * , exp pn71
(9)
which is algebraically independent by (5e) and the fundamental conjecture,
This coefficient
the numeratorof (5) has, by (5a), at least one nonzerocoefficient.
is a polynomialin co, (over the integers),whose degree,by (5b), is less than the
degree of the minimal polynomialof o..mIt followsthat this polynomialis not
equivalent to 0, and thereforethe numeratorof (5) is not equivalent to 0. A
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32
RATIONAL EXPONENTIAL
[January
EXPRESSIONS
similarargumentshows that the denominatorof (5) is not equivalent to 0, and
it followsimmediatelythat p is not equivalent to 0 or U.
5. Simplificationalgorithm.The purpose of this algorithmis to transforma
weakly simplifiedREX expression,other than 0 or U, into a simplifiedREX
expression.The given expressionhas the form(5), and conditions(5a) through
(5c) are satisfied.We now definethe innermost
exponentialsin a REX expression
as those whose argumentsdo not involve the exponential function.The algorithmproceeds fromthe innermostexponentials outward.
Step 1 (Initialize). Set k= 0, so that {ql,
, qk} and {ri,
, rk} denote
the emptyset. Now rewrite(5) in the form
, exp, P y ... r,, z c m)
g(exp pi, * , exp pn, rl, * * , r, Z, 7r)
..f(exppi,..
(10)
Step 2 (Begin loop withtestfor completion).At this point the expression(10),
with rl,**,
rk viewed as additional indeterminates,is weakly simplified.
Furthermore,r1, * *, rkare abbreviations forexp ql,
, exp qk respectively,
and ql, * * * qk are simplified expressions of the form
qi=
(1)
fi(ri, *..
Finally, the set {ql, *
theset {ri,
,
rk, z,
gi(r1***
, r_1, z, 7'r,CO.k)
_
rr1s_z, 7rT)
*
k.
, qk, i7r} is linearlyindependentover the rationals,and
r } is algebraically
independent
overtherationals.Note
that the pi, ..., p n may depend on the r1,
rk.
If n is zero in (10), then replace r1,
, rkby exp ql,
, exp qk respecqk playingthe role ofPi,
tively.Now (10) has the form(5) withql, *
pn
and (5a) through(5e) are satisfied.Thereforewe are finished.
Otherwise(that is, if n>0), proceed to Step 3.
Step 3 (Introduceqk1 and rk+l). Let qk+l be the argumentof any innermost
exponential in (10), and replace exp qt+l by the abbreviation r+l. In general,
. pn.Clearly it has the form
qk+l will be a subexpressionof one of the pi,
Pn*
,
(12)
qk+l =
fk+,(ri)
.. *
gkL+l(rly
,
...
rk, 2, 7r CO.)
, ak, z, 7r)
If we replace r1,
, rkby exp ql, . . , exp qk, then (12) has the form(5) with
. . .*, qk playing the role of Pi,
* , pn,and (Sa) through(5e) are satisfied.
Thereforeqk?l is simplified.
Step 4 (Test for linear dependence).Recall that the set {Iq, * * , qk, ir} iS
linearlyindependentover the rationals. If qk+?can be expressedas a linear com* , qk, then we shall rewrite(10) accordingly(see Step
bination of ilr and qi,
Sa). Otherwise,we shall adjoin qk+l to the set (see Step 5b).
Consider the linear dependence equation
ql
,
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1969]
RATIONALEXPONENTIAL EXPRESSIONS
(13)
aoir +
33
k+1
i31
aisq =: ?
whereao,
, ak+1 are undeterminedrationals. Substituting(11) and (12) into
(13), replacing all of the co,,by powers of ci for some sufficiently
large 1, and
rearrangingterms,we obtain a polynomial in rl, * * *, rk, z, wr,and wi whose
degree in wxis less than the degree of the minimalpolynomialof WI and whose
coefficientsare integral linear combinations of a0, * * *, aA+,. Because of the
algebraic independenceof the set {ri, * rk, z, 7r},equation (13) implies that
all of these coefficients
vanish. Thus we obtain a set of homogeneous integral
linear equations in the unknownsao, * * *, aA+4.Since the set {q1, . * q,9t iw}
is linearlyindependent,ak+1 must be nonzero in any nontrivialsolution vector,
and the solution space, if any, must be one dimensional,Using standard linear
methods we can determinewhetheror not a solution exists, and if so, we can
find it.
If a solutionexists,go on to Step 5a. Otherwise,proceed to Step 5b.
Step 5a (Linear dependence-replaceqk+1and rk+1). In this case we can write
,
qk+
(14)
boiir
2co
+
b'q.
Ic
i=1 ci
where bi and ci are relativelyprimeintegersfori 0,
(15)
q
1qi qii
r;
exp(q,)
C'
,a
epq ix q,,
1,
k. Letting
,
* ,k
we have
(16)
exp(ciqi ) = (r')ci,
i
k
1,
and
(17)
rk+1= exp(qk+l)
c*
Ii
(r, ) t
*=1
Substitute (16) and (17) into (10), set k' k, and proceed to Step 6.
Step Sb (Linear independence--adjoin qk?l). In this case the set
I
q... ,
, ir }
is linearlyindependent.Now, in (10), replace r, by d! forall i 1, * , k
and proceed to Step 6.
Step 6 (Simplify(10) weakly).Simplify(10) weakly, treatingrI,
indeterminates.The resultis an expressionof the form
f(exp pt, ' *
exp pnrli
,
Xr,
ZX
)
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k+ 1,
,
as
34
MATHEMATICAL NOTES
[January
, rt,.Note that the set
wherethe p * , p' may dependon the r',
conjecture.
by thefundamental
independent,
{rl,* r** ,r, z, ir} is algebraically
Step 7 (Iterate).Drop all primes,so that (10') replaces(10), and returnto
Step 2.
The authorwishesto thankA. J. Goldsteinformanyvaluable discussions
Acknowledgment.
thecourseofthiswork.
throughout
References
of a real variable,
1. Daniel Richardson,Some unsolvableproblemsinvolvingfunctions
NOTICES, Amer.Math. Soc., 13 (1966) 135.
CarnegieInstituteof Technology
2. B. F. Caviness,On canonicalformsand simplification,
Ph.D. Thesis,1967.
Probleme,Bull. Amer.Math. Soc., 8 (1901-02)437-479.
3. David Hilbert,Mathematische
4. MartinDavis, HilaryPutnam,and JuliaRobinson,The decisionproblemforexponential
equations,Ann.ofMath.,74 (1961)425.
diophantine
1949,p. 84.
Press,Princeton,
University
Numbers,Princeton
5. C. L. Siegel,Transcendental
6. Harry Pollard,The Theoryof AlgebraicNumbers,MAA, Carus MonographNo. 9,
Wiley,New York,1950,p. 38.
MATHEMATICAL
NOTES
EDITED BY DAVID DRASIN
shouldbesenttoDavid Drasin,DivisionofMathematical
forthisDepartment
Manuscripts
IN 47907.
Lafayette,
Sciences,PurdueUniversity,
AN AREA-WIDTH INEQUALITY FOR CONVEX CURVES
of California,
PAULR. CHERNOFF,University
Berkeley
a regionofarea A.
planecurve,bounding
THEOREM.Let C be a closedconvex
Then
Letw(0)denotethewidthofC in the0-direction.
(1)
A <-
+L w(o)w
(o+ -)dO
withequalityif and onlyif C is a circle.
Proof.We shall employa methoddue to A. Hurwitz (see Courant [1] p. 213)
to prove the isoperimetricinequality. It clearlysufficesto establish the theorem
forthe case of C2curvessuch that foreach 0, O< 0 < 2ir,thereis exactlyone point,
say (x(0), y(O)), at which the normal to C makes an angle 0 with the X-axis.
Then, as Courant indicates,thereis a C2 periodicfunctionp(0) such that
(2)
x(O) = p(o) cos 0
-
p'(0) sin 0
y(0) = p(o) sin 0 + p'(0) cos 0.
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