Thinking Mathematically: Chapter 7 Solutions

Number Spirals

[!info] Number Spirals Numbers are written consecutively in a spiral as follows: Extend the spiral and write down any questions that occur to you.

I had several questions:

• Can I find a direct formula for the numbers on the top-left diagonal (1, 7, 21, 43, …)?
• Given a row and column can I calculate the number?
• How can I determine the position of square numbers?

I decided to focus on the first question.

The first path I took got very messy, but it provoked the insight  I needed for a much simpler, more beautiful solution.

Entry: The First Approach (The Winding Path)

My first instinct was to specialize  and get a feel for the problem’s structure. I noticed the spiral was built on expanding squares of side lengths 1x1, 3x3, 5x5, and so on.

I tried to find a formula by counting the “edges” I had to traverse to get from one diagonal number to the next. For example, to get from 1 to 7:

1. Traverse 3 sides of the inner 1x1 square’s “shell” (the path 2-3-4). (3 edges)
2. “Escape” to the next square (the path 4 to 5). (1 edge)
3. Traverse a partial side of the 3x3 square to get to 7 (the path 5-6-7). (2 edges)
4. Total difference: 3+1+2=6. And sure enough, 1+6=7.

This seemed to work! I tried it for the jump from 7 to 21 (on the 3x3 square) and got a difference of 14, which was also correct (7+14=21).

This led me down a long algebraic path. I introduced  notation:

• k = The “step number” (k=1 for the 1x1 square, k=2 for the 3x3, etc.)
• Dk​ = The k-th diagonal number (D1​=1,D2​=7,…)
• Nstart​ = The side length of the square we start on.

After a lot of algebra, I found that the difference between terms was Δ=8k−10. This gave me a recursive formula: Dk​=Dk−1​+8k−10

By using summation formulas, I eventually found a “closed” formula: $D_k​=4k^2−6k+3$

Attack: The Second Approach (The Geometric “Why”)

I had an answer, but I was STUCK! . The formula worked, but it felt messy and not harmonious. It didn’t give me an intuitive sense of  the “why.” Why was it quadratic?

AHA! Instead of tracing the winding path, what if I just looked at the completed squares?

Suddenly, the structure was perfectly clear!

1. The Square Size: The side length is just the sequence of odd numbers: 1, 3, 5, 7, … The formula for the k-th odd number is m=2k−1.
2. The Largest Number (Cmax​): The largest number in each square is just its total number of cells, which is the area. So, $C_{max} ​= N^2 = (2k−1)^2$.
3. The Offset: The offset (the number of steps back from the largest number to my diagonal number) was just the sequence of even numbers: 0, 2, 4, 6, … The formula for this is $O_k​= 2(k−1) = 2k−2$.

This gave me a new conjecture: $D_k​=(Largest Number)−(Offset) D_k​=(2k−1)^2−(2k−2)$

Now, the final test. I simplified this new formula: $D_k​=(4k^2−4k+1)−(2k−2)​ = 4k^2−4k+1−2k+2 ​=4k^2−6k+3$

It was the exact same formula!

Review: What I Learned

This was a perfect lesson in mathematical thinking.

• The “Why” is Key: The algebraic solution from the first approach was correct, but it wasn’t satisfying. It didn’t explain why the formula was quadratic.
• Don’t Fear the Mess: My first “messy” attempt wasn’t a failure. It was the Entry  phase. Getting stuck and feeling that the solution wasn’t “right” is what forced me to change my perspective and find the more elegant Attack.
• Trust Your Intuition: I found two ways to understand this intuitively:
  1. The “Neighbor” Idea: An even number is 2k. Its neighbor, one step before it, is always 2k−1. This works perfectly for k=1, which gives 2(1)−1=1.
  2. The “Leftover” Idea: An odd number is a set of pairs plus one “leftover” (2k+1). This is also a valid formula for odd numbers, but it’s “out of sync” with our k=1 starting point (it gives 3). Both are correct, but 2k−1 was the right one for this problem.