# Barnetts Identity Pdf1

September 22, 2017 | Author: Roger Penrose | Category: Element (Mathematics), Infinity, Pi, Sequence, Analysis

#### Description

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.

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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.

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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

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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 . Thus, ∞ X k=0

k=

eiπ 11.999...

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

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