4 ms·

Sorry that it took me two days to get back to this; if you're not reading the comment threads, that's okay. I did indeed forget the one statement which makes t

by drostie 15y ago

Sorry that it took me two days to get back to this; if you're not reading the comment threads, that's okay. I did indeed forget the one statement which makes this rigorous (and nontrivial): technically this definition for [ΔF]ᵢ = Fᵢ − Fᵢ₋₁ is undefined for [ΔF]₀ , because our sequences don't have a -1'th element. For these we take [ΔF]₀ = F₀, which in some sense means that F₋₁ = 0. I don't know why you want to require that "F₀ = F₁ = 0, etc." -- perhaps you were, in the back of your mind, iterating through the negative numbers, thinking that they were positive. ^_^;;

JadeNB 15y ago
Sure, I'm OK with the implicit extension by 0 to the left. No, my problem is with the statement that ΣΔF = F. Indeed, since Δ annihilates constants, it's not possible to recover a function from its Δ-image, by Σ or any other means; just as it's not possible to recover a function from its derivative. For example, ΣΔ1 = 0.
- drostie 15y ago
  Here is the proof that ΣΔF = F. First we define our "sequences", which are just functions from N to R; N includes 0. Next we define operators, which I will say always take precedence over function arguments. We define two operators in particular: Δg(i) = g(0) if i = 0, else g(i) − g(i − 1) Σg(i) = Σ{n: 0 → i} g(n) Note that Δg(0) = g(0) by these definitions, and note that Σg(0) = g(0) as well. Call these the "zeroth element relations." (Also note that I am saying Σg(0) where perhaps [Σg](0) would be more clear.) I will observe a very simple lemma which embeds Σg in a recursive form: Σg(n + 1) = Σg(n) + g(n + 1) for any g, any n in N. On to the proof. ΣΔF(0) = F(0) by composing those two zeroth element relations. Also, if ΣΔF(n) = F(n) for any n, then I claim that ΣΔF(n + 1) = F(n + 1). Proof: ΣΔF(n + 1) = ΣΔF(n) + ΔF(n + 1), by the lemma. ΣΔF(n + 1) = F(n) + ΔF(n + 1), because "if ΣΔF(n) = F(n)". ΣΔF(n + 1) = F(n) + F(n + 1) − F(n), by definition of Δ. ΣΔF(n + 1) = F(n + 1) Since it holds for n=0, and whenever it holds for n, it holds for n + 1, by induction this holds for all n in N and therefore, by induction: ΣΔF = F The reverse argument is also pretty simple. It's also obviously true for n = 0 due to the zeroth element relations. For n > 0 we don't even need induction: ΔΣF(n) = ΣF(n) − ΣF(n − 1), by definition of Δ ΔΣF(n) = ΣF(n − 1) + F(n) − ΣF(n − 1), by lemma ΔΣF(n) = F(n) ΔΣF = F. If you still don't believe me, here it all is in Python: >>> sigma = lambda seq: lambda i: sum(seq(u) for u in range(0, i + 1)) >>> delta = lambda seq: lambda i: seq(0) if i == 0 else seq(i) - seq(i - 1) >>> expand = lambda seq: [seq(i) for i in range(0, 7)] >>> constant = lambda N: lambda i: N >>> expand(constant(3)) [3, 3, 3, 3, 3, 3, 3] >>> expand(sigma(constant(3))) [3, 6, 9, 12, 15, 18, 21] >>> expand(delta(constant(3))) [3, 0, 0, 0, 0, 0, 0] >>> expand(sigma(delta(constant(3)))) [3, 3, 3, 3, 3, 3, 3] >>> expand(delta(sigma(constant(3)))) [3, 3, 3, 3, 3, 3, 3] Enjoy. :-)
  - JadeNB 15y ago
    You're right; sorry. I was stuck (as my comments indicate) on thinking of the classical setting of calculus, where functions can differ by a constant; but, since your functions are implicitly extended by 0, they cannot so differ! That is, unlike the constant function 0 : Z -> R, the 'constant' function 1 : N -> R isn't really constant once we regard it as a function Z -> R. P.S. I am a mathematician; Python convinces me of nothing but special cases. :-)

by drostie 15y ago

JadeNB 15y ago

Sure, I'm OK with the implicit extension by 0 to the left. No, my problem is with the statement that ΣΔF = F. Indeed, since Δ annihilates constants, it's not possible to recover a function from its Δ-image, by Σ or any other means; just as it's not possible to recover a function from its derivative. For example, ΣΔ1 = 0.

drostie 15y ago
Here is the proof that ΣΔF = F. First we define our "sequences", which are just functions from N to R; N includes 0. Next we define operators, which I will say always take precedence over function arguments. We define two operators in particular: Δg(i) = g(0) if i = 0, else g(i) − g(i − 1) Σg(i) = Σ{n: 0 → i} g(n) Note that Δg(0) = g(0) by these definitions, and note that Σg(0) = g(0) as well. Call these the "zeroth element relations." (Also note that I am saying Σg(0) where perhaps [Σg](0) would be more clear.) I will observe a very simple lemma which embeds Σg in a recursive form: Σg(n + 1) = Σg(n) + g(n + 1) for any g, any n in N. On to the proof. ΣΔF(0) = F(0) by composing those two zeroth element relations. Also, if ΣΔF(n) = F(n) for any n, then I claim that ΣΔF(n + 1) = F(n + 1). Proof: ΣΔF(n + 1) = ΣΔF(n) + ΔF(n + 1), by the lemma. ΣΔF(n + 1) = F(n) + ΔF(n + 1), because "if ΣΔF(n) = F(n)". ΣΔF(n + 1) = F(n) + F(n + 1) − F(n), by definition of Δ. ΣΔF(n + 1) = F(n + 1) Since it holds for n=0, and whenever it holds for n, it holds for n + 1, by induction this holds for all n in N and therefore, by induction: ΣΔF = F The reverse argument is also pretty simple. It's also obviously true for n = 0 due to the zeroth element relations. For n > 0 we don't even need induction: ΔΣF(n) = ΣF(n) − ΣF(n − 1), by definition of Δ ΔΣF(n) = ΣF(n − 1) + F(n) − ΣF(n − 1), by lemma ΔΣF(n) = F(n) ΔΣF = F. If you still don't believe me, here it all is in Python: >>> sigma = lambda seq: lambda i: sum(seq(u) for u in range(0, i + 1)) >>> delta = lambda seq: lambda i: seq(0) if i == 0 else seq(i) - seq(i - 1) >>> expand = lambda seq: [seq(i) for i in range(0, 7)] >>> constant = lambda N: lambda i: N >>> expand(constant(3)) [3, 3, 3, 3, 3, 3, 3] >>> expand(sigma(constant(3))) [3, 6, 9, 12, 15, 18, 21] >>> expand(delta(constant(3))) [3, 0, 0, 0, 0, 0, 0] >>> expand(sigma(delta(constant(3)))) [3, 3, 3, 3, 3, 3, 3] >>> expand(delta(sigma(constant(3)))) [3, 3, 3, 3, 3, 3, 3] Enjoy. :-)
- JadeNB 15y ago
  You're right; sorry. I was stuck (as my comments indicate) on thinking of the classical setting of calculus, where functions can differ by a constant; but, since your functions are implicitly extended by 0, they cannot so differ! That is, unlike the constant function 0 : Z -> R, the 'constant' function 1 : N -> R isn't really constant once we regard it as a function Z -> R. P.S. I am a mathematician; Python convinces me of nothing but special cases. :-)