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Proof of Barnett’s Identity Jacob Barnett Department of Mathematical Physics Perimeter Institute for Theoretical Physics May 2014 Abstract In order for vortex equations on moduli spaces of Zn orbifolds with N = 1 spacetime supersymmetry of hyper-Kahler Hirzebruch surfaces fibered over a lens space to be gauge mediated, one requires, ∞ X

k=

k=0

eiπ 11.999...

This identity is proved via the thrice iterated application of integration operators, and by evaluating non-trivial zeros of real valued second degree polynomials of a single variable.

1

Proof

To begin with, we first want to prove that the sequence of partial sums (sn ) P n x2n+1 of the sum n∈N |(−1) (2n+1)! | in R satisfies the property that ∃ some x ∈ R such that ∀ε > 0, ∃N ∈ N : |sn − x| < ε ∀n ≥ N. We have 2n+1 X |x|2n+1 X (−1)n x = (2n + 1)! (2n + 1)! n∈N

n∈N

But, X |x|2n+1 (2n + 1)!

n∈N

is just the terms of:

X |x|n n!

n∈N

for odd n.

1

Thus,

∞ n X |x|2n+1 X |x| < (2n + 1)! n = 0 n!

n∈N

However,

X |x|n = exp |x| n!

n∈N

which is from the Taylor series expansion of e|x| , and the result follows from application of the Squeeze theorem. P n x2n The exact same argument can be used to show that n∈N (−1) (2n)! also has the same properety that the sequence of partial sums (sn ) satisfy the condition that in R, ∃ some x ∈ R such that ∀ε > 0, ∃N ∈ N : |sn − x| < ε ∀n ≥ N. Now, when we have, X

(−1)

n

2n+1 X θ2n n θ (−1) +i (2n)! (2n + 1)! n∈N

n∈N

we get an intriguing result. Since both sums are absolutely convergent they can be summed, 2n 2n+1 X n θ n θ = (−1) + i (−1) (2n)! (2n + 1)! n∈N

2n+1

2n

=

X n∈N

(iθ) (iθ) + (2n)! (2n + 1)!

!

X (iθ)n n!

n∈N

= eiθ Let θ = π. Then since P

n∈N

(−1)

n

P

n∈N

(−1)

θ2n = −1 and (2n)!

n

θ2n+1 = 0 we have , (2n + 1)! 2n 2n+1 X X n π n π (−1) +i (−1) +1=0 (2n)! (2n + 1)! n∈N

n∈N

2

For the second part of the proof, consider the set of rational numbers r such that r < 0, or r < 0.9, or r < 0.99, or r is less than some other number of the 1 n ) . This set is the number 0.999... ∈ R. Every element of 0.999... form 1 − ( 10 is less than 1, so it is an element of the real number 1. Conversely, an element of 1 is a rational number ab < 1, which implies a 1 < 1 − ( )b b 10 Since 0.999... and 1 contain the same rational numbers, they are the same set. Finally, given what was deduced above and since, eiπ =

X n∈N

(−1)

n

2n+1 X π 2n n π +i (−1) (2n)! (2n + 1)! n∈N

Barnett’s identity can thus be seen as an immediate collorary of Brady Haran’s astounding result [1]. Thus, ∞ X k=0

k=

eiπ 11.999...

3

References [1] B. Haran, “An Astounding Result”, Journal of Advanced Research in Pure Mathematics. Series B, 2013.

4

View more...
k=

k=0

eiπ 11.999...

This identity is proved via the thrice iterated application of integration operators, and by evaluating non-trivial zeros of real valued second degree polynomials of a single variable.

1

Proof

To begin with, we first want to prove that the sequence of partial sums (sn ) P n x2n+1 of the sum n∈N |(−1) (2n+1)! | in R satisfies the property that ∃ some x ∈ R such that ∀ε > 0, ∃N ∈ N : |sn − x| < ε ∀n ≥ N. We have 2n+1 X |x|2n+1 X (−1)n x = (2n + 1)! (2n + 1)! n∈N

n∈N

But, X |x|2n+1 (2n + 1)!

n∈N

is just the terms of:

X |x|n n!

n∈N

for odd n.

1

Thus,

∞ n X |x|2n+1 X |x| < (2n + 1)! n = 0 n!

n∈N

However,

X |x|n = exp |x| n!

n∈N

which is from the Taylor series expansion of e|x| , and the result follows from application of the Squeeze theorem. P n x2n The exact same argument can be used to show that n∈N (−1) (2n)! also has the same properety that the sequence of partial sums (sn ) satisfy the condition that in R, ∃ some x ∈ R such that ∀ε > 0, ∃N ∈ N : |sn − x| < ε ∀n ≥ N. Now, when we have, X

(−1)

n

2n+1 X θ2n n θ (−1) +i (2n)! (2n + 1)! n∈N

n∈N

we get an intriguing result. Since both sums are absolutely convergent they can be summed, 2n 2n+1 X n θ n θ = (−1) + i (−1) (2n)! (2n + 1)! n∈N

2n+1

2n

=

X n∈N

(iθ) (iθ) + (2n)! (2n + 1)!

!

X (iθ)n n!

n∈N

= eiθ Let θ = π. Then since P

n∈N

(−1)

n

P

n∈N

(−1)

θ2n = −1 and (2n)!

n

θ2n+1 = 0 we have , (2n + 1)! 2n 2n+1 X X n π n π (−1) +i (−1) +1=0 (2n)! (2n + 1)! n∈N

n∈N

2

For the second part of the proof, consider the set of rational numbers r such that r < 0, or r < 0.9, or r < 0.99, or r is less than some other number of the 1 n ) . This set is the number 0.999... ∈ R. Every element of 0.999... form 1 − ( 10 is less than 1, so it is an element of the real number 1. Conversely, an element of 1 is a rational number ab < 1, which implies a 1 < 1 − ( )b b 10 Since 0.999... and 1 contain the same rational numbers, they are the same set. Finally, given what was deduced above and since, eiπ =

X n∈N

(−1)

n

2n+1 X π 2n n π +i (−1) (2n)! (2n + 1)! n∈N

Barnett’s identity can thus be seen as an immediate collorary of Brady Haran’s astounding result [1]. Thus, ∞ X k=0

k=

eiπ 11.999...

3

References [1] B. Haran, “An Astounding Result”, Journal of Advanced Research in Pure Mathematics. Series B, 2013.

4

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