ERIC EJ1093398: The Riemann Zeta Zeros from an Asymptotic Perspective

2015·0.49 MB·English

by  ERIC

The Riemann zeta zeros from an
asymptotic perspective
Ken Grant
New South Wales
klplg@hotmail .com
Bernard Riemann (1826–66) was a student of Carl Friedrich Gauss (1777–
1855) at the University of Gottingen in Lower Saxony. Since his youth,
Gauss had conjectured that the distribution of the primes could approximately
be given by
N
,
ln(N)
the approximation becoming better as N became larger. In time, this
conjecture became known as the prime number conjecture and then as the prime
number theorem or PNT. Today we write this as
N
π(N)~
ln(N)
where π(N) represents the number of primes less than a given number N.
A second way to write the PNT is π(x) ~ Li(x) where Li(x) is called the log
integral function and represents the definite integral of 1 as t varies from
ln(t)
2 to x.
In 1859, on the occasion of being elected as a corresponding member
of the Berlin Academy, Bernard Riemann presented a lecture with the title
On the Number of Primes Less Than a Given Magnitude, in which he presented
a mathematics formula, derived from complex integration, which gave a
precise count of the primes on the understanding that one of the terms in the
2 formula, which depended on a knowledge of the non-trivial zeros of the zeta
o.
9 n function, could be evaluated. Riemann had calculated some of the non-trivial
2
ol. zeros and found them all to have a real part equal to 0.5. He conjectured
v
nal that every zero of the zeta function had a real part equal to 0.5. This became
r
u
Jo known as the Riemann conjecture which evolved into the Riemann hypothesis as
s
c
ati more supporting evidence became available.
m
he Riemann’s 1859 paper can be found in Stephen Hawking’s 2005
at
M book God Created the Integers. Riemann’s solution is given using complex
r
o
ni number integration and is assessable to university students studying higher
e
S
an mathematics courses. A more assessable solution for school students who are
ali
r
st studying advanced mathematics (e.g., Specialist Mathematics in the Australian
u
A
10
T
h
Curriculum: Mathematics) is given in John Derbyshire’s very readable 2012 book e
R
ie
Prime Obsession. m
a
n
The aim of this paper is to show that the zeros of the Riemann zeta function n
z
e
ζ(s)=∑∞ 1, where s =a+ib ta z
k=1ks er
o
s
fr
all have the real part equal to one half. o
m
a
This is demonstrated in two ways or methods. Each method shows that n
a
s
there can be only one value for the real part a, and since we know that there ym
p
are zeros with a = 12 , then the proof is complete. totic
p
e
r
s
p
Notation e
c
tiv
(a) The Riemann zeta function e
∑∞ ⎛ 1⎞ = 1 + 1 + 1 + 1 +…=ζ(s)
k=1⎝ks⎠ 1s 2s 3s 4s

where s = a + ib and a and b are real.
(b) The alternating Riemann zeta function
∑∞ ⎛ 1(–1)k−1⎞ = 1 − 1 + 1 − 1 +…=ζ (s)
k=1⎝ks ⎠ 1s 2s 3s 4s a

where s = a + ib and a and b are real.
(c) The symbol ~ between two functions or series means that the two
are asymptotically equal to one another. One series might represent
a convergent series in the variable x and the other might represent a
divergent series in x but the two may be equal to one another to some
degree, say four decimal places, when a finite number of terms of the
divergent series is used. An example of this is given below:
∑∞ 1 ~1+B0 −B1 +B2 −B3 +B4 −B5… (*)
k=1k2 21 22 23 24 25 26

Now this is an example of an asymptotic series. We are approximating a
slowly converging series with a diverging series. The numbers B , B , B , B ,
0 1 2 3
… represent the Bernoulli numbers 1,−1, 1,0,− 1 ,0, 1 ,0,− 1 ,0,…, with
2 6 30 42 30
B = 0 for k ≥ 1.
2k+1
The value of the right-hand side of (*) for varying numbers of non-zero A
u
s
terms is (to 6 decimal places): tr
a
lia
One term: 1 n
S
e
Two terms: 1.5 nio
r
Three terms: 1.625 M
a
th
Four terms: 1.645833 e
m
a
Five terms: 1.644792 tic
s
J
Six terms: 1.64498 o
u
r
n
Seven terms: 1.644913 a
l v
o
Eight terms: 1.64495 l. 2
9
n
o
. 2
11
nt 2
ra Now the exact value of the left-hand side of (*) is π = 1.644934 (to 6
G 6
decimal places).
So we see that the right-hand side is equal to the left-hand side correct
to four decimal places if seven terms of the right-hand side series are used.
However, if more than 11 terms of the series on the right-hand side are used,
then the two sides are no longer equal correct to four decimal places. This
is mainly because the first eight Bernoulli numbers lie between zero and 1.1
in magnitude while the next eight Bernoulli numbers lie between 1.1 and
27 298 231.1 in magnitude, which far outpace the denominators of successive
terms which are only increasing by a factor of 2.
Method one
In this method, in s = a + ix, a is considered constant and x is variable and
ζ (a + ix) is found asymptotically in the form (A + A x + A x2 + A x3 + A x4 +
a 0 1 2 3 4
…) + i(C + C x + C x2 + C x3 + C x4 + …) where the A and C are real numbers.
0 1 2 3 4 i i
Justification for assuming such a result is as follows:
ζ (s)=(1−21−s)ζ(s)
a
=(1−21−s)∑∞ 1
k=1ks
=(1−21−s)∑∞ ⎛ 1 ⎞⎡⎣cos(xln(k))−isin(xln(k))⎤⎦
k=1⎝ka⎠

Now, cos(xln(k)) and sin(xln(k)) can be written in terms of power series
in the variable x.
Also, (1 – 21–s) = (1 – 21–(a+ix)) can be written as a series in x with real
coefficients.
Hence we have an infinite sum of terms of the form S (x) + iS (x) multiplied
1 2
by a series in x. This will give an expression in x with real coefficients. This is
the proposal and justification.
However, the method to be used is not that outlined above. Instead, we
equate ζ (s) and (A + A x + A x2 + A x3 + A x4 + …) + i(C + C x + C x2 + C x3
a 0 1 2 3 4 0 1 2 3
2 + C x4 + …) with the understanding that they are asymptotically equal.
o. 4
n
9
2
vol. So, let ζa(s) = (A0 + A1x + A2x2 + A3x3 + A4x4 + …)
rnal + i(C0 + C1x + C2x2 + C3x3 + C4x4 + …) (A)
u
o
J
s
c
mati We wish ultimately to find expressions for the coefficients Ai and Ci. Then we
athe can look for solutions to ζa(s) = 0 in the domain 0 < x < 1.
M So to begin:
r
nio 1 1 1 1
e − + − +…
n S 1a+ix 2a+ix 3a+ix 4a+ix
a
ustrali =(A0+A1x+A2x2+A3x3+A4x4 +…)+i(C0+C1x+C2x2+C3x3+C4x4 +…) (1)
A
12
T
h
Let x = 0 in equation (1) e
R
1 1 1 1 ie
m
− + − +…=A +iC
1a 2a 3a 4a 0 0 an
n
z
e
As the left-hand side is real, the right-hand side is real. So ta
z
e
1 1 1 1 r
o
C =0and A = − + − +… s
0 0 1a 2a 3a 4a fro
m
a
Now use differentiation of equation (1) to find more coefficients. The first n
a
s
differentiation gives: ym
p
⎛(ln(2) ln(3) ln(4) ⎞ to
i⎜ − + −…⎟ tic
⎝ 2a+ix 3a+ix 4a+ix ⎠ p
e
r
s
=(A +2A x+3Ax2+…)+i(C +2C x+C x2+…) (2) pe
1 2 3 1 2 3 ctiv
e
Letting x = 0 in equation (2) we obtain
⎛(ln(2) ln(3) ln(4) ⎞
i⎜ − + −…⎟ =A +iC
⎝ 2a 3a 4a ⎠ 1 1

Equating real and imaginary parts, we find
(ln(2) ln(3) ln(4)
A =0andC = − + −…
1 1 2a 3a 4a

Subsequent differentiation and setting x = 0 gives the following results:
⎛ 1 1 1 1 1 ⎞
A =1−⎜ − + − + +…⎟
0 ⎝2a 3a 4a 5a 6a ⎠
C =0
0
A =0
1
ln2 ln3 ln4 ln5
C = − + − +…
1 2a 3a 4a 5a
1 ⎛ln22 ln23 ln24 ln25 ⎞
A = − + − +…
2 2!⎝⎜ 2a 3a 4a 5a ⎠⎟
C =0
2
A =0
3
1 ⎛ln32 ln33 ln34 ln35 ⎞
C =− − + − +…
3 3!⎝⎜ 2a 3a 4a 5a ⎠⎟
1 ⎛ln42 ln43 ln44 ln45 ⎞
A =− − + − +…
4 4!⎝⎜ 2a 3a 4a 5a ⎠⎟
C =0 Au
4 s
tr
A =0 a
4 lia
n
1 ⎛ln52 ln53 ln54 ln55 ⎞ Se
C5 = 5!⎝⎜ 2a − 3a + 4a − 5a +…⎠⎟ nior M
a
1 ⎛ln62 ln63 ln64 ln65 ⎞ the
A = − + − +… m
6 6!⎝⎜ 2a 3a 4a 5a ⎠⎟ atic
s
J
In general, A = 0 if n is odd; C = 0 if n is even; o
n n ur
A =(−1)2n+1⎛ 1 ⎞∑∞ (−1)k lnnk if n is even and n ≠0 nal v
n ⎝n!⎠ k=2 ka ol. 2
9
C =(−1)n2−1⎛ 1 ⎞∑∞ (−1)k lnnk if n is odd no. 2
n ⎝n!⎠ k=2 ka

13
nt
Gra So the first aim has been achieved, that is, we have found ζa(a + ix)
asymptotically in the form
(A + A x + A x2 + A x + A x4 + …) + i(C + C x + C x2 + C x3 + C x4 + …).
0 1 2 3 3 4 0 1 2 3 4
In fact, noting that some of the coefficients are zero, we can write:
ζ (a + ix) ~ (A + A x2 + A x4 + A x6 + A x8 + …) + i(C x + C x3 + C x5 + C x7 + …)
a 0 2 4 6 8 1 3 5 7
Notice that the imaginary part of ζ (a + ix) has x as a factor.
a
Now let us go to our second aim which is to consider the equation
ζ (a + ix) = 0 for 0 < x < 1.
a
So we have
(A + A x2 + A x4 + A x6 + A x8 + …) + i(C x + C x3 + C x5 + C x7 + …) = 0 + i.0
0 2 4 6 8 1 3 5 7
Equating real and imaginary parts, we have
(A + A x2 + A x4 + A x6 + A x8 + …) = 0 and (C x + C x3 + C x5 + C x7 + …) = 0.
0 2 4 6 8 1 3 5 7
Let these two equations be rearranged as follows:
The first becomes
⎛ x2A x4A x6A ⎞
A 1+ 2 + 4 + 6 +… =0
0⎝⎜ A0 A0 A0 ⎠⎟
and the second becomes
⎛ x2C x4C x6C ⎞
C x 1+ 3 + 5 + 7 +… =0
1 ⎝⎜ C1 C1 C1 ⎠⎟
Dividing these last two equations by the non-zero numbers A and C
0 1
respectively, we have
⎛ x2A x4A x6A ⎞ ⎛ x2C x4C x6C ⎞
1+ 2 + 4 + 6 +… =0 and x 1+ 3 + 5 + 7 +… =0
⎜ ⎟ ⎜ ⎟
⎝ A0 A0 A0 ⎠ ⎝ C1 C1 C1 ⎠
Now let us consider solutions of ζ (a + ix) = 0. These must also be solutions
a
of the above two equations. Also we know that x = 0 is not a solution of
ζ (a + ix) = 0. So we conclude that the three equations
a
⎛ x2A x4A x6A ⎞ ⎛ x2C x4C x6C ⎞
1+ 2 + 4 + 6 +… =0, 1+ 3 + 5 + 7 +… =0, ζ (a+ix)=0
⎝⎜ A A A ⎠⎟ ⎝⎜ C C C ⎠⎟ a
0 0 0 1 1 1
have the same solutions and that since the first two series are monic, they are
equal.
A C
Equating coefficients of the terms in x, we have 2 = 3 .
A C
A C 0 1
Then it follows that 2 − 3 =0.
2 A C
o. 0 1
n Note here that it should be remembered that the equal signs being used
9
2
ol. above are not actually equal signs but rather they represent approximations
v
nal to a certain number of decimal places, or asymptotically equal signs.
r
u
Jo A C
s Now let y= 2 − 3
c
mati A0 C1
he and using the values of the coefficients found previously, we have
at
M ⎛ln22 ln23 ln24 ln25 ⎞ ⎛ln32 ln33 ln34 ln35 ⎞
r − + − +… − + − +…
Senio y=⎛⎜ 1⎞⎟⎝⎜ 2a 3a 4a 5a ⎠⎟ +⎛ 1⎞⎝⎜ 2a 3a 4a 5a ⎠⎟
n ⎝2!⎠ ⎛ 1 1 1 1 1 ⎞ ⎝3!⎠ ⎛ln2 ln3 ln4 ln5 ⎞
ralia ⎝1−2a +3a −4a +5a −6a +…⎠ ⎝ 2a − 3a + 4a − 5a +…⎠
st
u
A
14
T
h
Notice that y has been expressed in terms of one unknown, a. This can e
R
ie
be shown on a graph using the software Graphmatica, but first the unknown m
a
n
must be replaced by the variable x. This graph is shown below for the domain n
z
e
0 ≤ x ≤ 1, that is, 0 ≤ a ≤ 1. Each of the four series involved has been truncated ta
z
e
r
to ten terms. o
s
fr
The graph shown in Figure 1 is as it would appear on a hard copy using the o
m
a
Graphmatica software (see Appendix 1 for a larger section of this graph). n
a
s
y
m
p
to
tic
p
e
r
s
p
e
c
tiv
e
Figure 1.
Equations on screen:
y=.5*(f2(x))/(f1(x))+1/6*(h2(x))/(h1(x))
Functions used by these equations:
f=1
f1(x)=(2^–x)*(ln(2))^(2*f–2)–(3^–x)*(ln(3))^(2*f–2)
+(4^–x)*(ln(4))^(2*f–2)–(5^–x)*(ln(5))^(2*f–2)+(6^–x)*(ln(6))^(2*f–2)
–(7^–x)*(ln(7))^(2*f–2)+(8^–x)*(ln(8))^(2*f–2)–(9^–x)*(ln(9))^(2*f–2)
+(10^–x)*(ln(10))^(2*f–2)–(11^–x)*(ln(11))^(2*f–2)
f2(x)=(2^–x)*(ln(2))^(2*f)–(3^–x)*(ln(3))^(2*f)+(4^–x)*(ln(4))^(2*f)
–(5^–x)*(ln(5))^(2*f)+(6^–x)*(ln(6))^(2*f)
–(7^–x)*(ln(7))^(2*f)+(8^–x)*(ln(8))^(2*f) A
u
s
–(9^–x)*(ln(9))^(2*f)+(10^–x)*(ln(10))^(2*f)–(11^–x)*(ln(11))^(2*f) tr
a
lia
h1(x)=(2^–x)*(ln(2))^(2*f–1)–(3^–x)*(ln(3))^(2*f–1) n
S
e
+(4^–x)*(ln(4))^(2*f–1)–(5^–x)*(ln(5))^(2*f–1)+(6^–x)*(ln(6))^(2*f–1) nio
r
–(7^–x)*(ln(7))^(2*f–1)+(8^–x)*(ln(8))^(2*f–1) M
a
th
–(9^–x)*(ln(9))^(2*f–1)+(10^–x)*(ln(10))^(2*f–1)–(11^–x)*(ln(11))^(2*f–1) e
m
a
h2(x)=(2^–x)*(ln(2))^(2*f+1)–(3^–x)*(ln(3))^(2*f+1) tic
s
J
+(4^–x)*(ln(4))^(2*f+1)–(5^–x)*(ln(5))^(2*f+1) o
u
r
n
+(6^–x)*(ln(6))^(2*f+1)–(7^–x)*(ln(7))^(2*f+1) a
l v
o
+(8^–x)*(ln(8))^(2*f+1)–(9^–x)*(ln(9))^(2*f+1) l. 2
9
+(10^–x)*(ln(10))^(2*f+1)–(11^–x)*(ln(11))^(2*f+1) no
. 2
15
nt
Gra In the following conclusion, use is made of the fact that ζ(s) and ζa(s)
have the same zeros, that is, ζ(s) = 0 and ζ (s) = 0 have the same roots, e.g.,
a
s = 0.5 + i14.134725 approximately.
Conclusion for Method 1
From the graph we see that there is only one intercept in the domain 0 < x < 1;
this occurs at about x = 0.45. This means that there is only one possible value
of a in the domain 0 < a < 1 at about a = 0.45. Now it is known that all non–
trivial Riemann zeta zeros have real part a lying in the interval 0 < a < 1. But
our graph shows that there is only one possible value for a in this interval. And
since it is also known that there are Riemann zeta zeros with real part a = 0.5,
then we conclude that the value of a = 0.45, shown on the above graph, is
an approximation for a = 0.5. Then we finally conclude that if s = a + ib is a
solution to the equation ζ(s) = 0 or ζ (s) = 0 then a = 0.5.
a
Method 2
The second method begins with the asymptotic formula
∑∞ ⎛ 1⎞ ~1+ B0 −B1 + B2s +B4s(s+1)(s+2)
k=1⎝ks⎠ 2s−1(s−1) 2s 2!2s+1 4!2s+3
B s(s+1)(s+2)(s+3)(s+4)
+ 6 +… (B)
6!2s+5
where B are the Bernoulli numbers and s = a + ib.
i
In this method, the intention is firstly to truncate the right-hand side of
the above equation to two terms, let s = a + ix as in Method 1, express the
truncated series in the form S (x) + iS (x), where
1 2
S (x) = (D + D x + D x2 + D x3 + D x4 + …),
1 0 1 2 3 4
S (x) = (E + E x + E x2 + E x3 + E x4 + …),
2 0 1 2 3 4
compare the coefficients of x2 as in Method 1 and produce a graph. Secondly,
truncate the right-hand side of (B) to three terms and do the same as above
to produce a graph; then truncate the right-hand side to four terms, and then
five terms, producing four graphs in total. Finally, a conclusion is arrived at in
2 a similar vein to that in Method 1.
o.
n
9
2
ol. Step 1
v
nal Truncate the right-hand side of (B) to two terms and let s = a + ix.
r
s Jou ∑∞ ⎛ 1⎞ ~1+ B0 =1+B021−s =1+ B021−s
atic k=1⎝ks⎠ 2s−1(s−1) s−1 a+ix−1
m
Mathe =1+B0(eln2)1−s(a−1(+ai−x)1()a−−ix1−ix)
r
enio =1+B (eln2)(1−a)−ix (a−1)−ix
n S 0 (a−1+ix)(a−1−ix)
a
ali
r
st
u
A
16
T
Now multiply both sides by [(a – 1)2 + x2] to obtain he
R
⎡(a−1)2+x2⎤∑∞ ⎛ 1⎞ ~⎡(a−1)2+x2⎤+B 21−a[(a−1)−ix][cos(xln2)−isin(xln2)] iem
⎣ ⎦ k=1⎝ks⎠ ⎣ ⎦ 0 an
n
z
e
x2ln22 x4ln42 ta
Writing cos(xln2)=1− + +… z
e
2! 4! ro
s
x3ln32 x5ln52 fro
and sin(xln2)=xln2− + −… m
3! 5! an
a
s
we arrive at the following: ym
p
⎡(a−1)2+x2⎤ ζ (a+ix) to
⎣ ⎦ a tic
p
~⎧⎨⎩⎡⎣(a−1)2+B021−a(a−1)⎤⎦+⎡⎣⎢1−B021−a(a−1)ln22!2−B021−aln2⎤⎦⎥x2+ terms in x4, x6, …⎫⎬⎭ erspectiv
e
⎧ ⎡ln22 (a−1)ln32⎤ ⎫
+i⎨⎩−B021−a[1+(a−1)ln2]x+B021−a⎣⎢ 2! + 3! ⎦⎥x3+ terms in x5, x7, …⎬⎭
We note that the imaginary part of the above has a factor of x. When this is
factorised, the result can be written in the form:
⎡(a–1)2+x2⎤ζ (a+ix)~(D +D x2+D x4 +…)+ix(E +E x2+E x4 +…)
⎣ ⎦ a 0 2 4 0 2 4
and then as
⎛ D x2 D x4 ⎞ ⎛ E x2 E x4 ⎞
⎡(a–1)2+x2⎤ζ (a+ix)~D 1+ 2 + 4 +… +ixE 1+ 2 + 4 +…
⎣ ⎦ a 0⎝⎜ D D ⎠⎟ 0⎝⎜ E E ⎠⎟
0 0 0 0
Using similar reasoning to that used in Method 1, we can say the following:
The equation
⎡(a–1)2+x2⎤ζ (a+ix)=0
⎣ ⎦ a
implies that
⎛ D x2 D x4 ⎞
1+ 2 + 4 +… =0
⎜ ⎟
⎝ D D ⎠
0 0
and
⎛ E x2 E x4 ⎞
1+ 2 + 4 +… =0
⎜ ⎟
⎝ E E ⎠
0 0
That
⎛ D x2 D x4 ⎞ ⎛ E x2 E x4 ⎞
⎡(a–1)2+x2⎤ζ (a+ix), 1+ 2 + 4 +… and 1+ 2 + 4 +…
⎣ ⎦ a ⎝⎜ D D ⎠⎟ ⎝⎜ E E ⎠⎟
0 0 0 0 A
u
s
have the same zeros; tra
lia
n
That S
e
n
⎛ D D ⎞ ⎛ E E ⎞ io
⎝⎜1+ D x22 + D x44 +…⎠⎟=⎝⎜1+ E x22 + E x44 +…⎠⎟ r Ma
0 0 0 0 th
e
m
Equating the coefficients of x2 in the two series above gives a
D E tics
2 = 2 J
o
D E u
0 0 rn
a
and consequently l vo
D2 − E2 =0 l. 29 n
D0 E0 o. 2
17
nt
ra As in Method 1, please note here that it should be remembered that
G
the equal signs being used above are not actually equal signs but rather
they represent approximations to a certain number of decimal places, or
asymptotically equal signs.
D E
Let y= 2 − 2
D E
0 0
Writing this out fully, we have:
⎡ ln22 ⎤ ⎡ln22 (a−1)ln32⎤
1−B 21−a(a−1) −B 21−aln2 +
y= ⎣⎢ 0 2! 0 ⎦⎥−B 21−a ⎣⎢ 2! 3! ⎦⎥
⎡(a−1)2+B 21−a(a−1)⎤ 0 −B 21−a[1+(a−1)ln2]
⎣ 0 ⎦ 0
This represents an equation in one unknown, a.
As mentioned previously, Graphmatica requires us to write equations in terms
of x.
The graph of this equation using Graphmatica is shown in Figure 2 as the
graph y = h1(x) – h2(x).
Figure 2
Equations shown in Figure 2:
2 y = h1(x) – h2(x)
o.
n
9
2
ol. Functions used by these equations:
v
nal f1(x)=(x–1)^2+b0*(x–1)*2^(1–x)
r
u
Jo f2(x)=1–b0*(2^(1–x)*(x–1)*1/2*(ln(2))^2–b0*ln(2)*2^(1–x)
s
c
ati h1(x)=–f2(x)/f1(x)
m
he f3(x)=1*b0*(2^(1–x))*(1+(x–1)*ln(2))
at
M f4(x)=b0*(2^(1–x))*(1/2*(ln(2))^2+(x–1)*1/6*(ln(2))^3)
r
o
ni h2(x)=f4(x)/f3(x)
e
S
n
a
ali
r
st
u
A
18
T
h
From the graph we see that there is only one intercept in the domain e
R
ie
0 < x < 1; this occurs at about x = 0.44. That is, there is only one value of a in m
a
n
the domain 0 < a < 1. As concluded in Method 1, this value of a = 0.44 is an n
z
e
approximation for a = 0.5. ta
z
e
r
o
s
fr
Step 2 o
m
a
Truncate the right-hand side of (B) to three terms. n
a
s
The method used in Step 1 above now results in the following equation and ym
p
graph: totic
f2(a)+ f6(a) f4(a)+ f8(a) p
y= − er
s
f1(a)+ f 5(a) f3(a)+ f7(a) p
e
c
tiv
where e
f1(a)=(a–1)2+b 21–a(a–1)
0
⎛1 ⎞
f2(a)=1–b 21–a⎜ (a–1)ln22+ln2⎟
0 ⎝2 ⎠
f3(a)=–b 21–a(1+(a–1)ln2)
0
⎛ 1 1 ⎞
f4(a)=b 21–a⎜ ln22+ (a–1)ln32⎟
0 ⎝2! 3! ⎠
f 5(a)=–b 2–a(a–1)2
1
⎛ 1 ⎞
f6(a)=–b 2–a⎜1– (a–1)2ln22⎟
1 ⎝ 2! ⎠
f7(a)=–b 2–a(a–1)2ln2
1
f8(a)=–b 2–a⎛ 1(a–1)2ln32–ln2⎞
1 ⎝3! ⎠
From Figure 3, whose equation is y = h7(x) – h8(x), we see that there are
two values of a as intercepts. One at a = –0.1 approximately and the other at
about a = –1.3. The important point to notice for this paper is that there is no
value of a in the domain 0 < a < 1.
A
u
s
tr
a
lia
n
S
e
n
io
r
M
a
th
e
m
a
tic
s
J
o
u
r
n
a
l v
o
Figure 3 l. 2
9
n
o
. 2
19