Linearity of Certain Periodic Summations September 1, 2020 Take the function g ( x ) = 3 √ x . We all know what this looks like, however, if we translate it one to the left, and one to the right, and add all three together (dividing it by a constant to prevent divergence) we get Figure 1. This isn’t that spectacular, it does follow y = x for a while, but not that closely and it quickly diverges. However, if we generalize this, and we translate g ( x ) 100 times to the left, and 100 times to the right, and we add all of those together (again dividing by a constant), we get Figure 2. Figure 1: Figure 2: This was our inspiration for this proof. 1 Definitions: k ∈ Z + , m ∈ Z + , x ∈ R g : R → R : g ( x ) is k + 1 times differentiable everywhere, continuous, and odd. g n has the same properties of g but is not equal to g This is useful when we have multiple components to an equation that need to act like g lim x →∞ g ( k ) ( x ) = 0 , and lim x →∞ g ( k − 1) ( x ) 6 = 0 Properties of g ( x ) : g ( x ) = g 1 ( nx ) , g ( x ) = g 2 ( x ) + g 3 ( x ) Next we’ll define m ∑ n = − m = ∑ since it clears the clutter in fractions, as seen: f : R → R : f ( x ) = lim m →∞ m ∑ n = − m g ( x − n ) m ∑ n = − m g (1 − n ) = lim m →∞ ∑ g ( x − n ) ∑ g (1 − n ) Note that m ∑ n = − m g (1 − n ) = g ( m ) + g ( m + 1) because g ( x ) is odd, but having it in the summation form reminds us it’s only there to make f (1) = 1, and prevent f from diverging. Theorem: f ( x ) = x 2 Lemma 1: f (0) = 0 and f (1) = 1 g (0) = 0 because g ( x ) is odd and continuous f (0) = lim m →∞ g ( − m ) + · · · + g (0) + · · · + g ( m ) ∑ g (1 − n ) = lim m →∞ g (0) + g (1) − g ( − 1) + · · · + g ( m ) − g ( m ) ∑ g (1 − n ) = lim m →∞ g (0) ∑ g (1 − n ) = 0 because g ( x ) is odd ∴ f (0) = 0 f (1) = lim m →∞ ∑ g (1 − n ) ∑ g (1 − n ) = 1 ∴ f (1) = 1 Lemma 2: ∀ x ∈ Z : f ( x ) = x f ′ ( x + 1) = lim m →∞ ∑ g ′ ( x + 1 + n ) ∑ g (1 − n ) = lim m →∞ g ′ ( x − m + 1) + · · · + g ′ ( x + m + 1) g ( x + m ) + g ( x + m + 1) = lim m →∞ − g ′ ( m − x ) + g ′ ( m + x + 1) g ( m + x ) + g ( m + x + 1) + f ′ ( x ) = 0 − 0 2 c + f ′ ( x ) or = c − c 2 ∞ + f ′ ( x ) or = ∞ − ∞ 2 ∞ + f ′ ( x ) = f ′ ( x ) or H = lim m →∞ g ( k − 1) ( m + x + 1) − g ( k − 1) ( m − x ) g ( k − 2) ( m + x ) + g ( k − 2) ( m + x + 1) + f ′ ( x ) = f ′ ( x ) or = c − c ∞ + f ′ ( x ) = f ′ ( x ) ∴ f ′ ( x + 1) = f ′ ( x ) f ( x + 1) = f ( x ) + c 1 f (1) = f (0) + c 1 1 = 0 + c 1 because Lemma 1 ∴ f ( x + 1) = f ( x ) + 1 ∴ ∀ x ∈ Z : f ( x ) = x because Lemma 1 3 Lemma 3: lim x →∞ g ( x ) = ±∞ = ⇒ f ( x ) = x f ( k +1) ( x ) = lim m →∞ ∑ g ( k +1) ( x − n ) ∑ g (1 − n ) lim m →∞ ∑ g (1 − n ) = lim m →∞ ( g ( m ) + g ( m + 1)) because g ( x ) is odd = ±∞ because lim x →∞ g ( x ) = ±∞ lim m →∞ ∑ g ( k +1) ( x − n ) converges iff lim m →∞ ∫ m − m g ( k +1) ( x − n ) dn converges lim m →∞ ∫ m − m g ( k +1) ( x − n ) dn = lim m →∞ ( − g ( k ) ( x − m ) + g ( k ) ( x + m )) = 0 because lim x →∞ g ( k ) ( x ) = 0 and g is odd ∴ lim m →∞ ∑ g ( k +1) ( x − n ) = c 6 = ±∞ f ( k +1) ( x ) = lim m →∞ c ±∞ = 0 f ( x ) = c a x a + c a − 1 x a − 1 + · · · + c 1 x + c 0 = x because Lemma 2 ∴ lim x →∞ g ( x ) = ±∞ = ⇒ f ( x ) = x 4 Theorem: f ( x ) = x Suppose g ( x ) = g 1 ( x )+ g 2 ( x ) such that lim x →∞ g 1 ( x ) = ∞ and lim x →∞ g 2 ( x ) = −∞ Note that ∀ s ∈ R ∪ ( −∞ , ∞ ) ∃ g | lim x →∞ g ( x ) = s f ( x ) = lim m →∞ ∑ g ( x − n ) ∑ g (1 − n ) = lim m →∞ ∑ g 1 ( x − n ) + ∑ g 2 ( x − n ) ∑ g (1 − n ) = lim m →∞ ∑ g 1 (1 − n ) ∑ g 1 ( x − n ) ∑ g 1 (1 − n ) + ∑ g 2 (1 − n ) ∑ g 2 ( x − n ) ∑ g 2 (1 − n ) ∑ g (1 − n ) = lim m →∞ ∑ g 1 (1 − n ) · x + ∑ g 2 (1 − n ) · x ∑ g (1 − n ) because Lemma 3 = x lim m →∞ ∑ g (1 − n ) ∑ g (1 − n ) because x is independent of m = x 5