edited posts

_posts/2025-12-08-2s-complement.md

@@ -6,7 +6,7 @@ categories: cs, binary, bits, negative numbers, derivation, proof, math, 2's com
 permalink: /2s-complement
 ---
 
-## Preamble
+## Introduction
 
 If you have ever taken CS in high school or college, you have probably been taught about how to represent negative numbers in binary using 2's complement.
 

_posts/2025-12-10-2nd-diff.md

@@ -1,34 +1,31 @@
 ---
 layout: post
-title: 2nd Difference Formula Proof
+title: 2nd Difference Derivation & Proof
 date: 2025-12-10
-categories: math, 2nd difference, second difference, proof, derivation, recursion, closed-form solution
+categories: math, constant 2nd difference, constant second difference, proof, derivation, recurrence, recursion, closed-form solution, nth term
 permalink: /2nd-diff
 ---
 
-## 1. Preamble<a name="preamble"></a>
+## 1. Introduction<a name="intro"></a>
 
-During junior high school, I was introduced to a formula for finding the n<sup>th</sup> term of a certain sequence with a second difference that went something like:
+During a math class in junior high school, I was introduced to a formula for finding the n<sup>th</sup> term of a certain sequence with a constant second difference that went something like:
 
 T<sub>n</sub> = 1/2(n)<sup>2</sup> - 1/2(n) + 1, for a sequence `1, 2, 4, 7 ...` and n >= 1.
 
-At first, the formula seemed quite random (similar to the quadratic formula) and I could not wrap my head around it at the time.
+The teacher never really explained it though, so it was just a mysterious formula to me at the time.
 
-I wondered how adding a second difference could make the trivial arithmetic formula so complex, and I was inspired by a textbook I was reading ("Concrete Mathematics" by Donald Knuth), so I tried to derive the formula myself.
-I also did not like to use things I did not understand, and there is no better way to understand something than to make it yourself from scratch.
-
-I wrote this blog post to share in hopes that someone may find it useful.
+Later in college, I was inspired by a textbook I was reading (Chapter 1 of Concrete Mathematics by Donald Knuth on Recurrence Relations) and remembered the mysterious formula, so I tried to derive it myself to better understand it.
 
 ## 2. What is a 2<sup>nd</sup> Difference?
 
 Say there is a sequence `1, 2, 4, 7 ...`.
 You may have come across quite a few of these types of sequences on IQ tests and whatnot, and it is easy to find that the fifth term is 11 and so on.
 
-However, it is obvious at a glance that the sequence does not follow the familiar arithmetic or geometric formulas, so we cannot yet find an arbitrary n<sup>th</sup> term of the sequence without knowing the previous terms.
-Finding the general formula for this will be the goal of this derivation.
+However, it is obvious at a glance that the sequence does not follow the familiar formulas for finding the n<sup>th</sup> term of arithmetic or geometric sequences, so we cannot yet find an arbitrary n<sup>th</sup> term of the sequence without first knowing the previous terms.
+Finding the general closed-form solution for this will be the goal of this derivation.
 
 The difference or ratio between every term and its subsequent term is not constant, unlike what is needed for a+(n-1)d or ar<sup>n-1</sup> to work properly.
-Instead, the difference (which we will now call the 'first difference') increases by a constant amount as the sequence goes on.
+Instead, the difference (which we will call the 'first difference') increases by a constant amount as the sequence goes on.
 
 We can see this clearly if we construct a new sequence from the first differences:
 
@@ -44,15 +41,15 @@ so the sequence of first differences is:
 
 `1, 2, 3 ...`.
 
-Unlike the original sequence, it is easy to see that this sequence has a constant difference of one, so the n<sup>th</sup> term can be found using the familiar formula (albeit with some notation modifications):
+Unlike the original sequence, it is easy to see that this sequence has a constant difference of one, so the n<sup>th</sup> term can be found using the familiar formula (albeit with some modifications to notation):
 
 (d<sub>1</sub>)<sub>n</sub> = a<sub>2</sub> + (n - 1)d<sub>2</sub>, for n >= 1,
 
-where (d<sub>1</sub>)<sub>n</sub> is the n<sup>th</sup> first difference, a<sub>2</sub> is the first term of the sequence of first differences (a<sub>1</sub> would be the first term of the original sequence), and d<sub>2</sub> is the difference for the sequence of first differences (or the second difference).
+where (d<sub>1</sub>)<sub>n</sub> is the n<sup>th</sup> first difference, a<sub>2</sub> is the first term of the sequence of first differences (a<sub>1</sub> would be the first term of the original sequence), and d<sub>2</sub> is the constant difference between each term in the sequence of first differences (which we will call the 'second difference').
 
 ## 3. Derivation of the Formula
 
-Using the approach we naturally used to find that 11 was the fifth term in the original sequence earlier, we can model the sequence as a recursive problem:
+Taking inspiration from how we intuitively found that 11 was the fifth term in the original sequence earlier, we model the sequence as a recurrence problem:
 
 T<sub>1</sub> = a<sub>1</sub>;
 
@@ -64,13 +61,13 @@ Note that it is (d<sub>1</sub>)<sub>n-1</sub> and not (d<sub>1</sub>)<sub>n</sub
 
 so it is
 
-(d<sub>1</sub>)<sub>n</sub> = T<sub>n+1</sub> - T<sub>n</sub>
+(d<sub>1</sub>)<sub>n</sub> = T<sub>n+1</sub> - T<sub>n</sub>,
 
 and not
 
 (d<sub>1</sub>)<sub>n</sub> = T<sub>n</sub> - T<sub>n-1</sub>.
 
-Moving on, we expand---or "unfold"---the recursive formula to try and spot some patterns we could use to find a closed-form solution:
+Moving on: we expand---or "unfold"---the recursive formula to try and spot some patterns we could use to find a closed-form solution:
 
 T<sub>n</sub> = T<sub>n-1</sub> + (d<sub>1</sub>)<sub>n-1</sub>
 
@@ -141,16 +138,32 @@ T<sub>n</sub> = a<sub>1</sub> + (n - 1)a<sub>2</sub> + 1/2(n - 2)(n - 1)d<sub>2<
 
 = 1/2(n)<sup>2</sup> - 1/2(n) + 1.
 
-Which matches what I wrote on the [preamble](/2nd-diff#preamble).
-You can try out a few examples to check for any counter-examples, but you will not know for certain if it will always be correct unless you prove it, which will be talked about in the next section.
+Which matches what I wrote in the [introduction](/2nd-diff#intro).
+You can try out a few examples to check for any counter-examples, but you will not know for certain if it will always be correct unless you prove it, which is what we will do in the next section.
 
 ## 4. Proof by Mathematical Induction
 
 Continuing from the previous section: the derivation makes sense, and we can try a few examples, but how do we know for sure that there is no counter-example somewhere we did not check?
 We need to be mathematically rigorous to show our formula will always work.
 
-Fortunately, there is a convenient method called "proof by induction" that fits nicely for our sequences situation.
-If we show that our formula is true for a base case and also that it is true for any n if it was true for (n - 1), we can confidently conclude that the formula is true for all n.
+Fortunately, there is a convenient method called "proof by induction" that fits nicely into our sequences situation.
+If we show that our formula is true for a base case and also that it is true for any n if it was true for (n - 1), we can confidently conclude that the formula is true for all positive integer n.
+
+### 4.2 Base Case<a name="base"></a>
+
+Plugging in the formula for T<sub>1</sub>:
+
+T<sub>1</sub> = a<sub>1</sub> + (n - 1)a<sub>2</sub> + 1/2(n - 2)(n - 1)d<sub>2</sub>
+
+= a<sub>1</sub> + (1 - 1)a<sub>2</sub> + 1/2(1 - 2)(1 - 1)d<sub>2</sub>, for n = 1
+
+= a<sub>1</sub> + (0)a<sub>2</sub> + 1/2(-1)(0)d<sub>2</sub>, for n = 1
+
+= a<sub>1</sub> + 0 + 0
+
+= a<sub>1</sub>.
+
+Which matches our definition, hence proving that the formula is true for n = 1.
 
 ### 4.1 Induction Step<a name="step"></a>
 
@@ -195,28 +208,11 @@ T<sub>n</sub> = a<sub>1</sub> + ((n - 1) - 1)a<sub>2</sub> + 1/2((n - 1) - 2)((n
 = a<sub>1</sub> + (n - 1)a<sub>2</sub> + 1/2(n - 2)(n - 1)d<sub>2</sub>
 
 That matches our formula, and hence proves that the formula will hold true for any n if (n - 1) is true.
-However, we are still missing our base case (n = 1), which is covered in the next subsection.
-
-### 4.2 Base Case
-
-Plugging in the formula for T<sub>1</sub>:
-
-T<sub>1</sub> = a<sub>1</sub> + (n - 1)a<sub>2</sub> + 1/2(n - 2)(n - 1)d<sub>2</sub>
-
-= a<sub>1</sub> + (1 - 1)a<sub>2</sub> + 1/2(1 - 2)(1 - 1)d<sub>2</sub>, for n = 1
-
-= a<sub>1</sub> + (0)a<sub>2</sub> + 1/2(-1)(0)d<sub>2</sub>, for n = 1
-
-= a<sub>1</sub> + 0 + 0
-
-= a<sub>1</sub>.
 
-Which matches our definition, and hence proves that the formula is true for n = 1.
-Combining that with the earlier [induction step](/2nd-diff#step), we get that it is also true for n = 2, n - 1 = 1.
-Then that also shows that the formula is true for n = 3, n - 1 = 2.
-This then continues on infinitely, which shows that our formula holds true for any n.
+### 4.3 Conclusion
 
-QED.
+Combining the [base case](/2nd-diff#base) with the [induction step](/2nd-diff#step), we get that it is also true for n = 2 (as it is true for n = 1), which also shows that the formula is true for n = 3 (as it is true for n = 2) and so on.
+This continues infinitely, showing our formula holds true for any positive integer n. QED.
 
 ## 5. Conclusion
 

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