1 Defining f(x)
Let:
∞
1
𝑓(𝑥) = − ln(1 − 𝑥) = ∑ 𝑥 𝑖
𝑖
𝑖=1
1.1 Theorem
Let f(n)(x) be the nth derivative of f(x). For example, f(0)(x) = f(x), f(1)(x) =
f’(x), f(2)(x) = f’’(x), and so on. Then:
∞
𝑓 (𝑛) (𝑥) = ∑
𝑖=0
(𝑛 + 𝑖 − 1)! 𝑖
𝑥
𝑖!
1.1.1 Proof
Consider n = 1.
1
1
1
𝑓(𝑥) = 𝑥 + 𝑥 2 + 𝑥 3 + 𝑥 4 + ⋯
2
3
4
𝑓 (1) (𝑥) = 𝑓 ′ (𝑥) = 1 + 𝑥 + 𝑥 2 + 𝑥 3 + ⋯
∞
∞
∞
𝑖=0
𝑖=0
𝑖=0
(1 + 𝑖 − 1)! 𝑖
𝑖!
= ∑ 𝑥𝑖 = ∑ 𝑥𝑖 = ∑
𝑥
𝑖!
𝑖!
The formula holds for n = 1. If we assume that the formula is true for a given n,
and show that implies that the formula holds for m = n+1, then the proof is
complete by induction.
∞
𝑓 (𝑚) (𝑥) = 𝑓 (𝑛+1) (𝑥) = ∑
𝑖=1
(𝑛 + 𝑖 − 1)!
∙ 𝑖 ∙ 𝑥 𝑖−1
𝑖!
Note that the lowest exponent in this series occurs when i – 1 = 0, or when i =
1, so that’s where the summation starts. Now, let m = n+1 and j = i – 1:
𝑓
(𝑚) (𝑥)
∞
∞
𝑗=0
𝑗=0
(𝑚 + 𝑗 − 1)!
(𝑚 + 𝑗 − 1)! 𝑗
=∑
∙ (𝑗 + 1) ∙ 𝑥 𝑗 = ∑
𝑥
(𝑗 + 1)!
𝑗!
Other than the change of index from i to j, this is the formula we are looking for,
and the proof is complete.
1.2 Corollary
For any n>0, the nth derivative of f(x) evaluated at x = 0 is:
𝑓 (𝑛) (0) = (𝑛 − 1)!
1.2.1 Proof
From the previous section, we know that:
∞
𝑓 (𝑛) (𝑥 ) = ∑
𝑖=0
(𝑛 + 𝑖 − 1)! 𝑖
𝑥
𝑖!
For any exponent i>0, xi = 0, so:
𝑓 (𝑛) (0) =
2
(𝑛 + 0 − 1)! 0
𝑥 = (𝑛 − 1)!
0!
Defining F(x)
Let:
𝐹(𝑥) = 𝑓 2 (𝑥) = (−ln(1 − 𝑥))2
2.1 Theorem
For n ≥ 1, the nth derivative of F(x) is given by:
𝐹 (𝑛) (𝑥) =
𝑎𝑛 𝑙𝑛(1 − 𝑥) + 𝑏𝑛
(1 − 𝑥)𝑛
Where a1 = -2 and b1 = 0, and for n > 1, an = (n – 1)an-1 and bn = (n – 1)bn-1 –
an-1.
2.1.1 Proof
𝐹(𝑥) = 𝑓 2 (𝑥) = (−ln(1 − 𝑥))2 = (ln(1 − 𝑥))2
𝐹 (1) (𝑥) = 2(ln(1 − 𝑥)) ∙
1
−2ln(1 − 𝑥)
∙ (−1) =
1−𝑥
1−𝑥
So, when n = 1, F(n)(x) is defined as described above, with a1 = -2 and b1 = 0. If
we assume that the formula is true for a given n, and show that implies that the
formula holds for n+1, then the proof is complete by induction.
If:
𝐹 (𝑛) (𝑥) =
𝑎𝑛 𝑙𝑛(1 − 𝑥) + 𝑏𝑛
(1 − 𝑥)𝑛
Then, by the quotient rule, its derivative is given by:
−𝑎𝑛
(1 − 𝑥)𝑛 ∙
− (𝑎𝑛 𝑙𝑛(1 − 𝑥) + 𝑏𝑛 ) ∙ 𝑛(1 − 𝑥)𝑛−1 ∙ (−1)
1
−
𝑥
(𝑛+1) ( )
𝐹
𝑥 =
(1 − 𝑥)2𝑛
𝐹
(𝑛+1) (𝑥)
(1 − 𝑥)𝑛−1 (−𝑎𝑛 ) + (1 − 𝑥)𝑛−1 (𝑛𝑎𝑛 𝑙𝑛(1 − 𝑥) + 𝑛𝑏𝑛 )
=
(1 − 𝑥)2𝑛
𝐹 (𝑛+1) (𝑥) =
(−𝑎𝑛 ) + (𝑛𝑎𝑛 𝑙𝑛(1 − 𝑥) + 𝑛𝑏𝑛 ) 𝑛𝑎𝑛 𝑙𝑛(1 − 𝑥) + 𝑛𝑏𝑛 − 𝑎𝑛
=
(1 − 𝑥)2𝑛−𝑛+1
(1 − 𝑥)𝑛+1
Which is what we wanted to show.
2.2 Corollary
In the formula above, an = -2(n – 1)!.
2.2.1 Proof
We know that a1 = -2, and that for n > 1, an = (n – 1)an-1. So, for n = 2, a2 = (2 –
1)a1 = -2. Note that -2(2 – 1)! = -2 as well. Similarly, for n = 3, a3 = (3 – 1)a2 =
2(-2) = -4, and so is -2(3 – 1)! = -2(2). Again, by induction, assume that an = 2(n – 1)!, and that an+1 = ((n + 1) – 1)an = nan = (n)(-2(n – 1)!) = -2(n)(n-1)!
= -2n! = -2((n + 1) – 1)!, and the proof is complete.
2.3 Corollary
In the formula above, for n > 1,
𝑛−1
𝑏𝑛 = ∑
𝑖=1
(𝑛 − 1)!
(−𝑎𝑖 )
𝑖!
2.3.1 Proof
We know that b1 = 0, and also that bn = (n – 1)bn-1 – an-1. Thus, b2 = (2 – 1)(0)
– (a1) = -a1, which means that:
2−1
(2 − 1)!
(−𝑎𝑖 )
𝑖!
𝑏2 = ∑
𝑖=1
Again, we’ll proceed with a proof by induction. Suppose that the formula holds
for n:
𝑛−1
𝑏𝑛 = ∑
𝑖=1
(𝑛 − 1)!
(−𝑎𝑖 )
𝑖!
From the theorem, we know that bn+1 = (n + 1 – 1)bn+1-1 – an+1-1 = nbn – an.
Substituting for bn we get:
𝑛−1
𝑏𝑛+1 = 𝑛 ∑
𝑖=1
𝑏𝑛+1
𝑏𝑛+1
𝑛−1
𝑛
𝑖=1
𝑖=𝑛
𝑛(𝑛 − 1)!
𝑛(𝑛 − 1)!
(−𝑎𝑖 ) + ∑
(−𝑎𝑖 )
=∑
𝑖!
𝑖!
𝑛−1
𝑛
𝑖=1
𝑖=𝑛
𝑛!
𝑛!
= ∑ (−𝑎𝑖 ) + ∑ (−𝑎𝑖 )
𝑖!
𝑖!
𝑛
𝑏𝑛+1
(𝑛 − 1)!
(−𝑎𝑖 ) + (−𝑎𝑛 )
𝑖!
𝑛!
= ∑ (−𝑎𝑖 ) =
𝑖!
𝑖=1
(𝑛+1)−1
∑
𝑖=1
((𝑛 + 1) − 1)!
(−𝑎𝑖 )
𝑖!
Which is what we wanted to show, so the proof is complete.
2.4 Corollary
We can also write bn as:
𝑛−1
𝑏𝑛 = 2(𝑛 − 1)! ∑
𝑖=1
1
𝑖
2.4.1 Proof
From above,
𝑛−1
𝑏𝑛 = ∑
𝑖=1
(𝑛 − 1)!
(−𝑎𝑖 )
𝑖!
But ai = -2(i – 1)!, so -ai = 2(i – 1)!, which means that:
𝑛−1
𝑏𝑛 = ∑
𝑖=1
𝑛−1
𝑏𝑛 = ∑
𝑖=1
(𝑛 − 1)!
∙ 2 ∙ (𝑖 − 1)!
𝑖!
2(𝑛 − 1)!
𝑖
But 2(n – 1)! is constant for all values of i, and can be factored out of the
summation, so:
𝑛−1
𝑏𝑛 = 2(𝑛 − 1)! ∑
𝑖=1
1
𝑖
2.5 Corollary
For any value of n ≥ 1, F(n)(0) = bn and we can write:
𝑛−1
𝐹 (𝑛) (0) = 2(𝑛 − 1)! ∑
𝑖=1
1
𝑖
We already showed that:
𝐹 (𝑛) (𝑥) =
𝑎𝑛 𝑙𝑛(1 − 𝑥) + 𝑏𝑛
(1 − 𝑥)𝑛
Which means that:
𝑛−1
𝑎𝑛 𝑙𝑛(1 − 0) + 𝑏𝑛
1
𝐹 (𝑛) (0) =
=
𝑏
=
2(𝑛
−
1)!
∑
𝑛
(1 − 0)𝑛
𝑖
𝑖=1
3 Define Some Useful Functions
3.1 P(x)
Define the function P(x) as:
1
𝑃(𝑥) = ∑ 𝑥 𝑖
𝑖
𝑖
Where i is an odd and prime number. Some useful properties of the function
P(x) are as follows.
3.1.1 Property of P(x)
If n is not an odd prime number (i.e., if n is either an even number or an odd
number that isn’t prime), then P(n)(0) = 0.
3.1.2 Property of P(x)
If n is an odd prime number, then P(n)(0) = (n – 1)!
3.2 Z(x) = P2(x)
Define the function Z(x) = P2(x) = (P(x))2. Please note that the only reason for
introducing another function is so that we can easily write out “the nth
derivative of P2(x)” as Z(n)(x). In general, we can just refer to P2(x), which we
can write as:
1
1
1
𝑃2 (𝑥) = (∑ 𝑥 𝑖 ) ∙ (∑ 𝑥 𝑖 ) = ∑ ∑ 𝑥 𝑖+𝑗
𝑖
𝑖
𝑖𝑗
𝑖
𝑖
𝑖
𝑗
Where i and j are odd prime numbers. Suppose we wrote the series expansion
of P2(x) as:
𝑃2 (𝑥) = ∑ 𝑝𝑛 𝑥 𝑛
𝑛
We can see from the previous formula that the exponents n must all be of the
form i + j, where i and j are odd prime numbers. That is, n is an even number
that is the sum of two odd prime numbers.
If the Goldbach Conjecture is true, then every even n > 4 is the sum of two odd
prime numbers. This further implies that pn > 0 for every even n > 4 and 0
otherwise. Finally, this also implies, using Z(x) = P2(x), that Z(n)(0) > 0 for every
even n > 4, and 0 otherwise.
However, if the Goldbach Conjecture is false, then there must exist some even n
> 4 that is not the sum of two odd prime numbers. That is, there is some even
n > 4 where pn = 0 and Z(n)(0) = 0.
See the definition of H(x) in the next section.
3.3 H(x)
Define the function H(x) as:
1
𝐻(𝑥) = ∑ 𝑥 𝑖
𝑖
𝑖
Where i > 4 is an even number which is not the sum of two odd prime numbers.
Using the function Z(x) = P2(x) from above, i > 4 is an even number such that
Z(i)(0) = 0.
3.3.1 Property of H(x)
If n > 4 is an even number that is the sum of two odd primes, or n > 4 is an odd
number, or n ≤ 4, then H(n)(0) = 0.
3.3.2 Property of H(x)
If n > 4 is an even number that is not the sum of two odd primes, then H(n)(0)
= (n – 1)!
3.3.3 Property of H(x)
If the Goldbach Conjecture is true, then H(x) = 0.
3.3.4 Property of H(x)
If the Goldbach Conjecture is false, then there must be some smallest even
number, M, such that M is not the sum of two odd primes. In that case, H(M)(0)
= (M – 1)!, and if n < M then H(n)(0) = 0.
3.4 G(x)
Define the function G(x) as:
1
𝐺(𝑥) = ∑ 𝑥 𝑖
𝑖
𝑖
Where i > 1 is an odd non-prime number. Some useful properties of the
function G(x) are as follows.
3.4.1 Property of G(x)
If n is not an odd non-prime number (i.e., if n is either an even number or an
odd number that isn’t prime or n=1), then G(n)(0) = 0.
3.4.2 Property of G(x)
If n is an odd non-prime number, then G(n)(0) = (n – 1)!
3.5 Q(x)
Define the function G(x) as:
1
𝑄(𝑥) = ∑ 𝑥 𝑖
𝑖
𝑖
Where i = 2, i = 4, or i > 4 is an even number which is the sum of two odd
primes. Some useful properties of the function Q(x) are as follows; they are
closely related to the properties of H(x).
3.5.1 Property of Q(x)
If n > 4 is an even number that is not the sum of two odd primes, or n > 4 is an
odd number, or n < 4, then Q(n)(0) = 0.
3.5.2 Property of Q(x)
If n = 2, or n = 4, or n > 4 is an even number that is the sum of two odd primes,
then Q(n)(0) = (n – 1)!
3.5.3 Property of Q(x)
If the Goldbach Conjecture is true, then H(x) = 0, and all even exponents n ≥ 2
are accounted for in the expansion of Q(x). That is, Q(n)(0) > 0 for all even n ≥
2.
3.5.4 Property of Q(x)
If the Goldbach Conjecture is false, then there must be some smallest even
number, M, such that M is not the sum of two odd primes. In that case, Q(M)(0)
= 0, and if n is even and 2 ≤ n < M then Q(n)(0) = (n – 1)!.
4 Revisiting f(x) = -ln(1 – x)
Previously, we defined the function f(x) as:
∞
1
𝑓(𝑥) = − ln(1 − 𝑥) = ∑ 𝑥 𝑖
𝑖
𝑖=1
In the last section, we defined four useful functions that were similar to f(x),
but used only a subset of the index i:
• P(x): i is an odd prime number
• H(x): i > 4 is an even number which is not the sum of two odd prime
numbers
• G(x): i > 1 is an odd non-prime number
• Q(x): i = 2, i = 4, or i > 4 is an even number which is the sum of two odd
primes
There is no overlap in these definitions. That is, any given value of i only
appears in one of the functions. Further, every i, other than i = 1, appears in one
of the four functions. That is, any i > 1 is either odd (and appears in P(x) if it is
prime and G(x) if it is not prime) or even (and appears in H(x) if i > 4, but it is
not the sum of two odd prime numbers, and Q(x) otherwise).
This means that we can also write the function f(x) as:
∞
1
𝑓(𝑥) = ∑ 𝑥 𝑖 = 𝑥 + 𝑃(𝑥) + 𝐻(𝑥) + 𝐺(𝑥) + 𝑄(𝑥)
𝑖
𝑖=1
5 Revisiting F(x) = f2(x)
We previously defined the function F(x) as:
𝐹(𝑥) = 𝑓 2 (𝑥) = (−ln(1 − 𝑥))2
Which we can now re-write as:
𝐹(𝑥) = (𝑥 + 𝑃(𝑥) + 𝐻(𝑥) + 𝐺(𝑥) + 𝑄(𝑥))2
In order to prove the Goldbach Conjecture, we only need to consider the even
terms of F(x). Each of the six components of f(x) is either and odd function or
an even function:
• Odd: x, P(x), G(x)
• Even: H(x), Q(x)
When expanding F(x), even terms will only arise from multiplying Odd times
Odd, or Even times Even. That is, we can expand F(x) to be:
𝐹(𝑥) = 𝑥 2 + 2𝑥𝑃(𝑥) + 2𝑥𝐺(𝑥) + 𝑃2 (𝑥) + 𝐻2 (𝑥) + 2𝑃(𝑥)𝐺(𝑥) + 𝐺 2 (𝑥)
+ 2𝐻(𝑥)𝑄(𝑥) + 𝑄2 (𝑥) + {𝑜𝑑𝑑 𝑡𝑒𝑟𝑚𝑠}
6 Proof of The Goldbach Conjecture, by Contradiction
Suppose that the Goldbach Conjecture is false. Let M > 4 be the smallest even
number that is not the sum of two odd primes. In the previous section, we
showed that F(x) can be stated in terms of the functions P(x), G(x), H(x) and
Q(x). However, we also previously showed that:
𝑀−1
𝐹 (𝑀) (0) = 2(𝑀 − 1)! ∑
𝑖=1
1
𝑖
If we can show that our newly defined form of F(x):
𝐹(𝑥) = 𝑥 2 + 2𝑥𝑃(𝑥) + 2𝑥𝐺(𝑥) + 𝑃2 (𝑥) + 𝐻2 (𝑥) + 2𝑃(𝑥)𝐺(𝑥) + 𝐺 2 (𝑥)
+ 2𝐻(𝑥)𝑄(𝑥) + 𝑄2 (𝑥) + {𝑜𝑑𝑑 𝑡𝑒𝑟𝑚𝑠}
Leads to the contradiction that:
𝑀−1
𝐹 (𝑀) (0) ≠ 2(𝑀 − 1)! ∑
𝑖=1
1
𝑖
Then the Goldbach Conjecture will be proved.
We will calculate F(M)(0) by looking at the each of the following components of
F(x) separately, and in the following order:
•
•
•
•
•
•
F0(x) = {odd terms}
F1(x) = 2xP(x)
F2(x) = 2xG(x)
F3(x) = P2(x)
F4(x) = H2(x)
2P(x)G(x) + G2(x), which we’ll actually consider as the sum of:
o F5(x) = P(x)G(x)
o F6(x) = G(x)[P(x) + G(x)]
• F7(x) = 2H(x)Q(x)
• F8(x) = Q2(x)
6.1 General Leibniz Rule
If u(x) and v(x) are both n-times differentiable functions, then the product
u(x)v(x) is also n-times differentiable, and the nth derivative is given by:
𝑛
(𝑢(𝑥)𝑣(𝑥))(𝑛) = ∑
𝑖=0
𝑛!
𝑢(𝑖) (𝑥)𝑣 (𝑛−𝑖) (𝑥)
𝑖! (𝑛 − 𝑖)!
Most of the functions Fi(x) defined in the previous section are products of two
other functions, so we can use the General Leibniz rule to calculate their
derivatives. We’ll only be considering Mth derivatives valued at x = 0, so we
will use this version of the General Leibniz Rule:
𝑀
(𝑢(0)𝑣(0))(𝑀)
=∑
𝑖=0
𝑀!
𝑢(𝑖) (0)𝑣 (𝑀−𝑖) (0)
𝑖! (𝑀 − 𝑖)!
6.2 The Mth Derivatives of Fi(0)
6.2.1 F0(0) = {odd terms}
M is an even number, and F0(x) is by definition and odd function. This means
that:
(𝑀)
𝐹0
(0) = 0
6.2.2 F1(x) = 2xP(x)
Let u(x) = 2x. This means that:
𝑢(1) (𝑥) = 2
For all values of x, and that:
𝑢(𝑖) (𝑥) = 0
For all values of x, and for all i > 1. Further, note that u(0) = 2(0) = 0, which
means that:
(𝑀)
𝐹1
(0) =
𝑀!
𝑢(1) (0)𝑃(𝑀−1) (0)
(𝑀 − 1)!
Note that the value of P(M-1)(0) depends on whether or not (M – 1) is prime. P(M1)(0) = (M – 2)! if (M – 1) is prime, and P(M-1)(0) = 0 if (M – 1) is not prime. Thus
we have:
(𝑀)
𝐹1
(0) = 2𝑀(𝑀 − 2)!
If (M – 1) is prime and:
(𝑀)
𝐹1
(0) = 0
If (M – 1) is not prime.
6.2.3 F2(x) = 2xG(x)
Similar to the previous section, let u(x) = 2x. This means that:
𝑢(1) (𝑥) = 2
For all values of x, and that:
𝑢(𝑖) (𝑥) = 0
For all values of x, and for all i > 1. Further, note that u(0) = 2(0) = 0, which
means that:
(𝑀)
𝐹2
(0) =
𝑀!
𝑢(1) (0)𝐺 (𝑀−1) (0)
(𝑀 − 1)!
Note that the value of G(M-1)(0) depends on whether or not (M – 1) is prime. G(M1)(0) = (M – 2)! if (M – 1) is not prime, and G(M-1)(0) = 0 if (M – 1) is prime. Thus
we have:
(𝑀)
𝐹2
(0) = 2𝑀(𝑀 − 2)!
If (M – 1) is not prime and:
(𝑀)
𝐹2
(0) = 0
If (M – 1) is prime.
6.2.4 Important Note on F1(x) + F2(x) = 2xP(x) + 2xG(x)
Note that from the previous two sections, we can observe that regardless of
whether (M – 1) is prime,
(𝑀)
𝐹1
(0) + 𝐹2(𝑀) (0) = 2𝑀(𝑀 − 2)!
6.2.5 F3(x) = P2(x)
Using the General Leibniz Rule, and writing F3(x) = P(x)P(x), we have:
𝑀
(𝑀)
𝐹3 (0)
=∑
𝑖=0
𝑀!
𝑃(𝑖) (0)𝑃(𝑀−𝑖) (0)
𝑖! (𝑀 − 𝑖)!
P(i)(0) = 0 when i is not an odd prime. However, we have assumed that M is not
the sum of two odd primes. This, if i is prime, then (M – i) is not prime. So for
all i, either P(i)(0) = 0 or P(M-i)(0) = 0 (or both), and so the product P(i)(0)P(Mi)(0) = 0, and:
(𝑀)
𝐹3
(0) = 0
6.2.6 F4(x) = H2(x)
Using the General Leibniz Rule, and writing F4(x) = H(x)H(x), we have:
𝑀
(𝑀)
𝐹4 (0)
=∑
𝑖=0
𝑀!
𝐻(𝑖) (0)𝐻(𝑀−𝑖) (0)
𝑖! (𝑀 − 𝑖)!
For any value of i < M, H(i)(0) = 0. When i = M, H(M-i)(0) = 0, which means that:
(𝑀)
𝐹4
(0) = 0
6.2.7 F5(x) = P(x)G(x)
Using the General Leibniz Rule, we have:
𝑀
(𝑀)
𝐹5 (0)
=∑
𝑖=0
𝑀!
𝑃(𝑖) (0)𝐺 (𝑀−𝑖) (0)
𝑖! (𝑀 − 𝑖)!
If i is an odd prime number, then P(i)(0) = (i – 1)!, and P(i)(0) = 0 otherwise.
Because have assumed that M is not the sum of two odd prime numbers, we
know that when i is an odd prime, (M – i) is odd but not prime. That is, when i
is an odd prime, G(M-i)(0) = (M – i – 1)!. Thus we can write:
(𝑀)
𝐹5
(0) = ∑
𝑀!
𝑀!
(𝑖 − 1)! (𝑀 − 𝑖 − 1)! = ∑
𝑖! (𝑀 − 𝑖)!
𝑖(𝑀 − 𝑖)
Where the summation is over all odd primes i < M.
6.2.8 F6(x) = G(x)[P(x) + G(x)]
Using the General Leibniz Rule, we have:
𝑀
(𝑀)
𝐹6 (0)
=∑
𝑖=0
𝑀!
𝐺 (𝑖) (0)[𝑃(𝑀−𝑖) (0) + 𝐺 (𝑀−𝑖) (0)]
𝑖! (𝑀 − 𝑖)!
If i is an odd number and not prime, then G(i)(0) = (i – 1)!, and G(i)(0) = 0
otherwise. But if i is an odd number and not prime, then (M – i) is also an odd
number and can either be prime or not. If (M – i) is prime, then:
𝑃(𝑀−𝑖) (0) = (𝑀 − 𝑖 − 1)!
and
𝐺 (𝑀−𝑖) (0) = 0
If (M – i) is not prime, then:
𝑃(𝑀−𝑖) (0) = 0
and
𝐺 (𝑀−𝑖) (0) = (𝑀 − 𝑖 − 1)!
Whether (M – i) is prime or not we know that:
𝑃(𝑀−𝑖) (0) + 𝐺 (𝑀−𝑖) (0) = (𝑀 − 𝑖 − 1)!
Which means that:
(𝑀)
𝐹6
(0) = ∑
𝑀!
𝑀!
(𝑖 − 1)! (𝑀 − 𝑖 − 1)! = ∑
𝑖! (𝑀 − 𝑖)!
𝑖(𝑀 − 𝑖)
Where the summation is over all odd non-primes 1 < i < M.
6.2.9 F7(x) = 2H(x)Q(x)
Using the General Leibniz Rule, we have:
𝑀
(𝑀)
𝐹7 (0)
= 2∑
𝑖=0
𝑀!
𝐻(𝑖) (0)𝑄(𝑀−𝑖) (0)
𝑖! (𝑀 − 𝑖)!
However, if i < M, then H(i)(0) = 0. Thus, we can re-write the above as:
(𝑀)
𝐹7
(0) = 2𝐻(𝑀) (0)𝑄(0) (0)
However, Q(0)(0) = Q(0) = 0, which means that:
(𝑀)
𝐹7
(0) = 0
6.2.10
F8(x) = Q2(x)
Using the General Leibniz Rule, and writing F8(x) = Q(x)Q(x), we have:
𝑀
(𝑀)
𝐹8 (0)
=∑
𝑖=0
𝑀!
𝑄(𝑖) (0)𝑄(𝑀−𝑖) (0)
𝑖! (𝑀 − 𝑖)!
We have assumed that for all even numbers 0 < i < M, Q(i)(0) = (i – 1)!. Note
that for all such i, (M – i) is also an even number such that 0 < (M – i) < M,
which means that Q(M-i)(0) = (M – i – 1)!. Finally, because M is an even number
that is not the sum of two odd primes, Q(M)(0) = 0. This means that:
(𝑀)
𝐹8
(0) = ∑
𝑀!
𝑀!
(𝑖 − 1)! (𝑀 − 𝑖 − 1)! = ∑
𝑖! (𝑀 − 𝑖)!
𝑖(𝑀 − 𝑖)
Where the summation is over all even numbers 0 < i < M.
6.2.11
Important Note on F5(x) + F6(x) + F8(x)
For each of the functions F5(x), F6(x), and F8(x), the Mth derivatives valued at x
= 0 are of the same general form:
(𝑀)
𝐹𝑗
(0) = ∑
𝑀!
𝑖(𝑀 − 𝑖)
Where the summations are taken over different values of i for j = 5, 6, and 8:
• j = 5: odd primes i < M
• j = 6: odd non-primes 1 < i < M
• j = 8: even numbers 0 < i < M
Every value of i such that 1 < i < M is either an odd prime, an odd non-prime,
or and even number. Further, there is no overlap in the values of i included in
the three functions. That is:
𝑀−1
(𝑀)
𝐹5 (0)
+
(𝑀)
(𝑀)
𝐹6 (0) + 𝐹8 (0)
=∑
𝑖=2
𝑀!
𝑖(𝑀 − 𝑖)
6.3 Putting it all back together again
If we write the function F(x) as:
8
𝐹(𝑥) = ∑ 𝐹𝑖 (𝑥)
𝑖=0
Then we can also write:
8
𝐹
(𝑀) (𝑥)
(𝑀)
(𝑥)
(𝑀)
(0)
= ∑ 𝐹𝑖
𝑖=0
As well as:
8
𝐹
(𝑀) (0)
= ∑ 𝐹𝑖
𝑖=0
In the previous sections, we showed that:
(𝑀)
𝐹0
(0) = 0
(𝑀)
(0) + 𝐹2(𝑀) (0) = 2𝑀(𝑀 − 2)!
(𝑀)
(0) = 0
(𝑀)
(0) = 0
𝐹1
𝐹3
𝐹4
𝑀−1
(𝑀)
𝐹5 (0)
+
(𝑀)
(𝑀)
𝐹6 (0) + 𝐹8 (0)
=∑
𝑖=2
(𝑀)
𝐹7
𝑀!
𝑖(𝑀 − 𝑖)
(0) = 0
Which means that:
𝑀−1
𝐹 (𝑀) (0) = 2𝑀(𝑀 − 2)! + ∑
𝑖=2
𝑀!
𝑖(𝑀 − 𝑖)
We previously showed that:
𝑀−1
𝐹 (𝑀) (0) = 2(𝑀 − 1)! ∑
𝑖=1
1
𝑖
But in the next section, we show that:
𝑀−1
𝑀−1
𝑖=1
𝑖=2
1
𝑀!
2(𝑀 − 1)! ∑ = 𝑀(𝑀 − 2)! + ∑
𝑖
𝑖(𝑀 − 𝑖)
And this,
𝑀−1
𝑀−1
𝑖=1
𝑖=2
1
𝑀!
2(𝑀 − 1)! ∑ ≠ 2𝑀(𝑀 − 2)! + ∑
𝑖
𝑖(𝑀 − 𝑖)
And we have a contradiction, which means that the proof is complte.
7 Fun with Summations
In this section, we’ll show that:
𝑛−1
𝑛−1
𝑖=1
𝑖=2
1
𝑛!
2(𝑛 − 1)! ∑ = 𝑛(𝑛 − 2)! + ∑
𝑖
𝑖(𝑛 − 𝑖)
To use this in the previous section, substitute M for n. We will start by showing
that:
𝑛
𝑛
𝑖=1
𝑖=1
1
𝑛+1
∑ =∑
𝑖
2𝑖(𝑛 + 1 − 𝑖)
Which we’ll do by showing that:
𝑛
𝑛
𝑖=1
𝑖=1
𝑛+1
1
−∑ =0
∑
2𝑖(𝑛 + 1 − 𝑖)
𝑖
So, let:
𝑛
𝑛
𝑖=1
𝑖=1
𝑛+1
1
𝑆𝑛 = ∑
−∑
2𝑖(𝑛 + 1 − 𝑖)
𝑖
𝑛
𝑆𝑛 = ∑ (
𝑖=1
𝑛
𝑆𝑛 = ∑ (
𝑖=1
𝑛
𝑆𝑛 = ∑
𝑖=1
𝑛
𝑆𝑛 = ∑
𝑖=1
𝑛+1
1
− )
2𝑖(𝑛 + 1 − 𝑖) 𝑖
𝑛+1
2(𝑛 + 1 − 𝑖)
−
)
2𝑖(𝑛 + 1 − 𝑖) 2𝑖(𝑛 + 1 − 𝑖)
𝑛 + 1 − 2𝑛 − 2 + 2𝑖
2𝑖(𝑛 + 1 − 𝑖)
2𝑖 − 𝑛 − 1
2𝑖(𝑛 + 1 − 𝑖)
For convenience, let:
𝑠𝑛,𝑖 =
2𝑖 − 𝑛 − 1
2𝑖(𝑛 + 1 − 𝑖)
So we can also write:
𝑛
𝑆𝑛 = ∑ 𝑠𝑛,𝑖
𝑖=1
However, we can easily show that:
𝑠𝑛,𝑛+1−𝑖 = −𝑠𝑛,𝑖
𝑠𝑛,𝑛+1−𝑖 =
2(𝑛 + 1 − 𝑖) − 𝑛 − 1
2(𝑛 + 1 − 𝑖)(𝑛 + 1 − (𝑛 + 1 − 𝑖))
𝑠𝑛,𝑛+1−𝑖 =
𝑛 + 1 − 2𝑖
−(2𝑖 − 𝑛 − 1)
=
2(𝑛 + 1 − 𝑖)(𝑖) 2𝑖(𝑛 + 1 − 𝑖)
𝑠𝑛,𝑛+1−𝑗 = −𝑠𝑛,𝑗
This means that, when n is even, we can pair up the terms in Sn such that the
sum of each pair is 0, which means that:
𝑛
𝑆𝑛 = ∑ 𝑠𝑛,𝑖 = 0
𝑖=1
When n is odd, we can pair up all the terms except for one, which is when i = n
+ 1 – i, or i = (n+1)/2. However, in that case:
𝑠𝑛,(𝑛+1)/2
𝑛+1
2(
)−𝑛−1
𝑛+1−𝑛−1
2
=
=
=0
𝑛+1
𝑛+1
𝑛+1
(𝑛 + 1) (
2(
+1−(
2 )
2 ) (𝑛
2 ))
So, whether n is even or odd, we have:
𝑛
𝑆𝑛 = ∑ 𝑠𝑛,𝑖 = 0
𝑖=1
This means that:
𝑛
𝑛
𝑖=1
𝑖=1
1
𝑛+1
∑ =∑
𝑖
2𝑖(𝑛 + 1 − 𝑖)
Which also means that:
𝑛−1
𝑛−1
𝑖=1
𝑖=1
1
𝑛
∑ =∑
𝑖
2𝑖(𝑛 − 𝑖)
Multiplying both sides by 2(n – 1)!:
𝑛−1
𝑛−1
𝑖=1
𝑖=1
1
𝑛!
2(𝑛 − 1)! ∑ = ∑
𝑖
𝑖(𝑛 − 𝑖)
Pulling out the value i = 1 from the Summation on the right side, we get:
𝑛−1
𝑛−1
𝑖=1
𝑖=2
1
𝑛!
𝑛!
2(𝑛 − 1)! ∑ =
+∑
𝑖 𝑛−1
𝑖(𝑛 − 𝑖)
𝑛−1
𝑛−1
𝑖=1
𝑖=2
1
𝑛!
2(𝑛 − 1)! ∑ = 𝑛(𝑛 − 2)! + ∑
𝑖
𝑖(𝑛 − 𝑖)