Demonstration of a theorem about the order observed in the sums of divisors

8 Pages·1760·0.08 MB·English

by  Leonhard Euler

Demonstration of a theorem about the order
5 observed in the sums of divisors∗
0
0
2

ul Leonhard Euler†
J

1
1

Some time ago, I had come upon a theorem by which the natural num-

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bers were seen to be illuminated to no small degree, with which the order
v
contained in the sum of divisors from the series of natural numbers arising
1
0 in procession may be represented from each other. Namely, I revealed that if
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for each of the natural numbers 1,2,3,4,5,6,7,8, etc. all of its divisors are
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0 gathered in a single sum, and in turn these sums of divisors are arranged in
5 a series, which will be
0
/
O 1,3,4,7,6,12,8,15,13,18,12,28,14,24,24,31,18, etc.,
H
that this series will be recurrent, in which each of the following terms may
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h be determined from the preceding by a particular ladder relationship. Yet,
t
a although this order is therefore observed as most worthy, with it hardly
m
possible for it to be doubted for this series to be bound by some specific law,
:
v I had been permitted to arrive at no demonstration firm enough to me of this
i
X order, even though I had approached the matter in many ways. Indeed, I
r had induced the observation of this order, providing that the follow formula
a
for an infinite product which I had considered:
2 3 4 5 6 7
s = (1−x)(1−x )(1−x )(1−x )(1−x )(1−x )(1−x ) etc.,
∗OriginallypublishedasDemonstratio theorematis circa ordinem in summis divisorum
observatum, Novi Commentarii academiae scientiarum Petropolitanae 5 (1760), 75–83,
and republished in Leonhard Euler, Opera Omnia, Series 1: Opera mathematica, Volume
2, Birkh¨auser, 1992. A copy of the original text is available electronically at the Euler
Archive, at http://www.eulerarchive.org. This paper is E244 in the Enestr¨om index.
†Dateoftranslation: July10,2005. TranslatedfromtheLatinbyJordanBell,3rdyear
undergraduate in Honours Mathematics, School of Mathematics and Statistics, Carleton
University,Ottawa,Ontario,Canada. Email: [email protected]. Thistranslation
was written during an NSERC USRA supervised by Dr. B. Stevens.
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whose expansion I had concluded by induction, to be
2 5 7 12 15 22 26 35 40
s = 1−x−x +x +x −x −x +x +x −x −x + etc.,
where the order taken by the differences of the exponents of x is manifest;
namely, the series of differences will be
1,1,3,2,5,3,7,4,9,5,11,6,13,7,15,8, etc.
It is apparent by selecting every other term for this series to be comprised
from the series of odd numbers and from the series of all integral numbers.
To be sure, I have inferred by induction the following condition: s = 1−x−
x2+x5+x7−x12−x15+ etc., if indeed it were s = (1−x)(1−xx)(1−x3)(1−
x4)(1−x5) etc., yet I had not been able to bring forth a solid demonstration
of the equality of these two formulas. For on the previous occasion on which
I had from this source drawn forth this order in the sums of divisors, I
had not prevailed firmly with a demonstration; but I had indicated then
for the demonstration of this to rest on the demonstration of the equality
between these two formulas exhibited in an infinite way. With I now having
obtained such a demonstration, the former order in the sums of divisors is
detected as not more true, with it recognized, not only are they able to be
demonstrated, which will follow attendantly just as I settle the conditions,
butmost deservedly, thegroundsofproofbetween these truthswillbeableto
beassignedfirmdemonstrations. Notanydoubtmayremainofthisproperty,
the demonstration of the truth of which rests on each of these propositions,
which I will now set forth and demonstrate:
Proposition I
If the product s = (1 + α)(1 + β)(1 + γ)(1 + δ)(1 + ε)(1 + ζ)(1 + η) etc.,
consisting of infinitely many factors, were converted into a single series, it
will be:
s = (1+α)+β(1+α)+γ(1+α)(1+β)+δ(1+α)(1+β)(1+γ)
+ε(1+α)(1+β)(1+γ)(1+δ)+ζ(1+α)(1+β)(1+γ)(1+δ)(1+ε)+ etc.
Demonstration
2
Sinceforthisseriesthefirsttermis(1+α)andthesecondisequaltoβ(1+α),
the sum of the first and second will be equal to (1 +α)(1+β): if now the
third term γ(1+α)(1+β) were added, it will produce (1+α)(1+β)(1+γ):
moreover, if the fourth term were added, which is δ(1+α)(1+β)(1+γ), the
sum will be equal to (1+α)(1+β)(1+γ)(1+δ). Then proceeding in this
way infinitely, the sum of the whole series, that is of all the terms of it, will
be induced to this product: (1+α)(1+β)(1+γ)(1 +δ)(1+ε)(1+ζ) etc.
From this it is clear that if it were
s = (1+α)(1+β)(1+γ)(1+δ)(1+ε)(1+ζ) etc.
for it to be in turn
s = (1+α)+β(1+α)+γ(1+α)(1+β)+δ(1+α)(1+β)(1+γ)+ etc.
Proposition II
If it were s = (1−x)(1−xx)(1−x3)(1−x4)(1−x5)(1−x6) etc., this product,
consisting of infinitely many factors, will be reduced to this series:
3 2 4 2 3
s = 1−x−xx(1−x)−x (1−x)(1−x )−x (1−x)(1−x )(1−x ) etc.
Demonstration
If this form s = (1−x)(1−xx)(1−x3)(1−x4)(1−x5) etc. is compared with
the preceding form s = (1+α)(1+β)(1+γ)(1+δ)(1+ε) etc., it is manifest
for it to be: α = −x;β = −x2;γ = −x3;δ = −x4;ε = −x5; etc. Therefore
with the values given in the series whose product is made as equal to s duly
substituted, the truth of this proposition stands open, namely for it to be:
3 2 4 2 3
s = 1−x−xx(1−x)−x (1−x)(1−x )−x (1−x)(1−x )(1−x )− etc.
Proposition III
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If it were s = (1 − x)(1 − x2)(1 − x3)(1 − x4)(1 −x5)(1 − x6)(1 − x7) etc.,
by expanding through multiplying this infinite product, disposing the terms
according to the powers of x:
1 2 5 7 12 15 22 26 35 40 51 57
s = 1−x −x +x +x −x −x +x +x −x −x +x +x − etc.
the reckoning of which series is the same as that which was related formerly.
Demonstration
Since it is s = (1−x)(1−x2)(1−x3)(1−x4)(1−x5)(1−x6)(1−x7) etc., it
will be s = 1−x−xx(1−x)−x3(1−x)(1−x2)−x4(1−x)(1−x2)(1−x3)
etc. Putting s = 1−x−Axx, it will be:
2 2 3
A = 1−x+x(1−x)(1−xx)+x (1−x)(1−x )(1−x )+ etc.
Each of the following terms are expanded by the factor 1−x, and arranged
in the following way:
−x −x(1−xx) −x3(1−x2)(1−x3) − etc.
A= +1 +x(1−xx) +x2(1−x2)(1−x3) +x3(1−x2)(1−x3)(1+x4 [sic]) + etc.
and by bringing up the terms from below it will be:
3 5 2 7 2 3 9 2 3 4
A = 1−x −x (1−x )−x (1−x )(1−x )−x (1−x )(1−x )(1−x )− etc.
It is put A = 1−x3 −Bx5, and it will be
2 2 2 3 4 2 3 4
B = 1−x +x (1−x )(1−x )+x (1−x )(1−x )(1−x )+ etc.,
in which those terms written before 1−x2 are expanded throughout, and it
will be
−x2 −x4(1−x3) −x6(1−x3)(1−x4)
B =
1 +x2(1−x3) +x4(1−x3)(1−x4) +x6(1−x3)(1−x4)(1−x5)
and once again by bringing up the terms from below it will be obtained:
5 8 3 11 3 4 14 3 4 5
B = 1−x −x (1−x )−x (1−x )(1−x )−x (1−x )(1−x )(1−x )− etc.
It is put B = 1−x5 −Cx3, and it will be
3 3 3 4 6 3 4 5
C = 1−x +x (1−x )(1−x )+x (1−x )(1−x )(1−x )+ etc.,
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where each term is expanded by the factor 1 − x3, so that it is made by
writing as before:
−x3 −x6(1−x4) −x9(1−x4)(1−x5) − etc.
C = 1 +x3(+x4 [sic]) +x6(1−x4)(1−x5) +x9(1−x4)(1−x5)(1−x6) + etc.
from which is assembled:
7 11 4 15 4 5 19 4 5 6
C = 1−x −x (1−x )−x (1−x )(1−x )−x (1−x )(1−x )(1−x )− etc.
It is put C = 1−x7 −Dx11, and it will be
4 4 4 5 8 4 5 6
D = 1−x +x (1−x )(1−x )+x (1−x )(1−x )(1−x )+ etc.,
which goes to this form:
−x4 −x8(1−x5) −x12(1−x5)(1−x6) − etc.
D = 1 +x4(1+x5 [sic]) +x8(1−x5)(1−x6) +x12(1−x5)(1−x6)(1−x7)+ etc.
and it will thus be
9 14 5 19 5 6 24 5 6 7
D = 1−x −x (1−x )−x (1−x )(1−x )−x (1−x )(1−x )(1−x )− etc.;
and if in turn it is put D = 1−x9 −Ex14, it will be discovered in a similar
way:
11 17
E = 1−x −Fx ; and then further:
17 20 15 23 17 26
F = 1−x −Gx ;G = 1−x −Hx ;H = 1−x −Ix ;
etc.
We now reconstitute these successive values, and it will be:
s = 1−x−Axx
2 2 5 7
Ax = x (1−x )−Bx
7 7 5 15
Bx = x (1−x )−Cx
15 15 7 26
Cx = x (1−x )−Dx
26 26 9 40
Dx = x (1−x )−Ex
etc.
For this reason we will have:
2 3 7 5 15 7 26 9 40 11
s = 1−x−x (1−x )+x (1−x )−x (1−x )+x (1−x )−x (1−x )+ etc.,
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or, as it ought to be seen:
2 5 7 12 15 22 26 35 40 51
s = 1−x−x +x +x −x −x +x +x −x −x +x + etc.,
from which at once the rule indicated earlier for the differences of exponents
shines forth most brilliantly.
Proposition IV
or
The principal theorem which is to be
demonstrated
If the formula which is written n denotes the sum of all the divisors of the
number n, and in a similar manRner of less numbers, just as n−α would be
designated by (n−α), the sum of the divisors of the number n, or n, then
hinges on the sRums of the divisors of lesser numbers, so that it willRbe
n= (n−1)+ (n−2)− (n−5)− (n−7)+ (n−12)+ (n−15)
Z Z Z Z Z Z Z
− (n−22)− (n−26)+ (n−35)+ (n−40)− (n−51)− (n−57) etc.
Z Z Z Z Z Z
From this, the following is to be noted:
1◦. The signs + and − of this progression to move back and forth by
pairs of terms.
2◦. The rule for the differences of the numbers 1,2,5,7,12,15,22,26,etc.,
which are 1,3,2,5,3,7,4, etc. to become manifest; from which it is gathered
3zz±z
for all these numbers to be contained in the general formula .
2
3◦. In each case for those terms which are to be taken to be from the
beginning of this progression, which leave after the sign positive numbers;
indeed all the remaining, which set in front of the sign R negative numbers
to be omitted; thus if it were n = 10, it will be 10 = 9R+ 8− 5− 3 =
13+15−6−3 = 18. R R R R R
4◦. In those cases in which the term (n−n) occurs, which will happen
if n is a number of the series 1,2,5,7,1R2,15, etc., for it to be required for
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the given number n itself to be taken in these cases for the value of the
term (n − n) or 0; in such a way, if it were n = 7, then it will be
7 =R 6+ 5− R2− 0 = 12+ 6−3−7 = 8, and if it were n = 12, it
Rwill beR = R11+R 10−R 7− 5+ 0 = 12+18−8−6+12 = 28.
R R R R R R
Demonstration
Theseries z = x 1+x2 2+x3 3+x4 4+x5 5+etc. isformed, inwhich
each power of x wRill be mRultiplieRd by theRsum ofRthe divisors of the exponent
of its power. If now all the sums of divisors are resolved, it is manifest for
this series to be transformed into this form
2 3 4 5 2 4 6 8 10
z = 1(x+x +x +x +x + etc.)+2(x +x +x +x +x + etc.)
3 6 9 12 15 4 8 12 16 20
+3(x +x +x +x +x + etc.)+4(x +x +x +x +x + etc.)
5 10 15 20 25 6 12 18 24 30
+5(x +x +x +x +x + etc.)+6(x +x +x +x +x + etc.)
etc.
which may be summed as a geometric series:
1x 2xx 3x3 4x4 5x5 6x6
z = + + + + + + etc.
1−x 1−xx 1−x3 1−x4 1−x5 1−x6
dx
This form is multiplied by − , and the integral of the product will be
x
zdx
3 4 5
− = l(1−x)+l(1−xx)+l(1−x )+l(1−x )+l(1−x )+ etc.,
Z x
zdx
3 4 5 6
or − = l(1−x)(1−xx)(1−x )(1−x )(1−x )(1−x ) etc.,
Z x
in which the expression after the logarithm sign is the very same which was
zdx
calleduponasequaltos inthepreceding proposition. Itwill be− = ls,
x
and thus by taking the alternate value for s, it will be thus be: R
zdx
2 5 7 12 15 22 26
− = l(1−x−x +x +x −x −x +x +x − etc.),
Z x
dx
whose differential divided by − gives another value for z, namely
x
1x+2x2 −5x5 −7x7 +12x12 +15x15 −22x22 − etc.
z = .
1−x−x2 +x5 +x7 −x12 −x15 +x22 + etc.
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If this value is put as equal to what had been taken, and both sides are
multiplied by the denominator 1 − x − x2 + x5 + x7 − x12 etc., it will be
discovered by disposing all the powers of x after the terms, and by all of
them being collected on the same side:
0= x 1 +x2 2 +x3 3 +x4 4 +x5 5 +x6 6 +x7 7 +x8 8 +x9 9 +x10 10 etc.
R . R R R R R R R R R
.
. − 1 − 2 − 3 − 4 − 5 − 6 − 7 − 8 − 9
. R . R R R R R R R R
. .
. . − 1 − 2 − 3 − 4 − 5 − 6 − 7 − 8
. . R . R . R . R R R R R
. . . . .
. . . . . + 1 + 2 + 3 4 + 5
. . . . . R . R . R R R
. . . . . . .
. . . . . . . + 1 2 + 3
. . . . . . . R . R . R .
. . . . . . . . . .
. . . . . . . . . .
-1 -2 * * +5 * +7 * * *
from which by equating the coefficients of each of the powers of x to
nothing, it will follow to be
1 = 1 6 = 5+ 4− 1
R 2 = 1+2 R 7 = R 6+R 5−R 2−7
R 3 = R 2+ 1 R 8 = R 7+R 6−R 3− 1
R 4 = R 3+R 2 R 9 = R 8+R 7−R 4−R 2
R 5 = R 4+R 3−5 R 10 =R 9+R 8−R 5−R 3
R R R R R R R R
and by attending to the innate character of this equation even slightly it will
stand open for it to be in general:
n = (n−1)+ (n−2)− (n−5)− (n−7)+ (n−12)+ (n−15)− etc.,
Z Z Z Z Z Z Z
with this progression continuing in each case until the sums of negative num-
bers arereached. Hence by this it is clear for the developed numbers 1,2,5,7,
etc. which appear in this formula, to hold until the term 0; from this it is
concluded, in the case in which in the progression obtaineRd for n the term
(n − n) occurs, or 0, for the value of this always to be taRken as equal
Rto the given numberRn: thus the full and perfected demonstration of the
proposed theorem is obtained, which, besides the treatment of infinite series
proceeds with logarithms and differentials; indeed it is not quite natural, but
on account of this is to be estimated that much more remarkable.
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