Month: January 2024

Here is the promised geometric visualization of the fact that the $n$th square number is the sum of the first $n$ odd numbers. I hope it may be a bit more enjoyable for readers, who I know are mostly Olly well-wishers rather than math enthusiasts.

I actually developed this visualization between the first and second proofs I shared yesterday. As I knew the proof by induction would probably practically write itself, I got it out of the way first. I then moved on to look at the problem geometrically, hoping to gain insight that I could use to write a more illuminating proof. This geometric approach turned out to be more closely related to the proof by induction than to the second proof I came up with later, though.

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Claim: The $n$th square number is the sum of the first $n$ odd numbers.

First of all, how would we represent a square number visually? One way is as a collection of dots arranged in a square, as shown below.

Second, how can we represent an odd number visually? Well, every odd number is one more than an even number, and any even number, being divisible by 2, can be represented as two rows of the same number of dots. Thus, an odd number can be represented as two rows of the same number of dots, plus one more. (Except for 1, which is simply 1 dot.)

Now imagine rearranging each odd number into an L shape as shown below. These L-shaped figures also have two rows of the same number of dots, plus one more.

Next, notice how the series of L-shaped figures representing the odd numbers nest together to form a square.

Thus, the claim holds at least up to $n=5$. To draw a square representing the next square number, you would add a row and a column to this square, which you can see would be the same as adding another L-shaped figure representing the next largest odd number. So the claim must hold for each higher $n$ as well.

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I also played around with a similar visualization using even numbers. The L-shaped figures representing odd numbers can be adapted to instead represent even numbers by removing the dot in the corner. When you do that, the nested figures look like this:

The center diagonal is missing from each square. Since it contains a number of dots equal to the side length of the square, the number of dots left in the $n$th square is $n^2-n$.

Accounting for the fact that the first even number, 2, corresponds to the second square, this leads to the formula

$$\sum_{i=1}^n2i=(n+1)^2-(n+1)=(n+1)[(n+1)-1]=(n+1)(n)\text{.}$$

This is what you would expect given the well known formula for the sum of the first $n$ natural numbers and can also be proven by induction. I also checked the sum of this formula and the one from yesterday. The sum of the first $n$ odd natural numbers plus the sum of the first $n$ even natural numbers should equal the sum of the first $2n$ natural numbers, which it does.

$$n^2+(n+1)(n)=n[n+(n+1)]=n(2n+1)=\frac{2n(2n+1)}{2}$$

Today was a fruitful day for math. I watched more of the Essence of Calculus series, did a couple of pages of algebra review exercises, then started on the book The Joy of X by Steven Strogatz, which I bought earlier in the week. Near the beginning of that book, the author mentions that the $n$th square number is the sum of the first $n$ odd numbers. I stopped reading at that point and went to see if I could prove the claim. I was able to come up with two proofs, as well as a way of visualizing the problem geometrically that led me to a formula for the sum of the first $n$ even numbers. I also checked that the sum of the two formulas yielded the well known one for the sum of the first $n$ natural numbers, which it did.

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Proposition:
$$\sum_{i=1}^{n}(2i-1)=n^2\text{ for all }n\in\mathbb{N}\text{.}$$

The proof by induction is easy…

Proof 1 (by induction):

When $n=1$, $\sum_{i=1}^{n}(2i-1)=\sum_{i=1}^{1}(2i-1)=2(1)-1=1=1^2$.

Assume that $\sum_{i=1}^{n-1}(2i-1)=(n-1)^2$.

It follows that $\sum_{i=1}^{n}(2i-1)=\sum_{i=1}^{n-1}(2i-1)+(2n-1)=(n-1)^2+2n-1=$ $(n^2-2n+1)+2n-1=n^2$.

Therefore, by induction, the proposition holds for all $n\in\mathbb{N}$.

Proof 2:

For any $n\in\mathbb{N}$, $\sum_{i=1}^{n}(2i-1)=[2n-1]+[2(n-1)-1]+[2(n-2)-1]+…+[2(1)-1]$.

Since there are $n$ terms in the sum, there are $n$ instances of $-1$.

Thus, $\sum_{i=1}^{n}(2i-1)=2[n+(n-1)+(n-2)+…+1]-n$.

Using the well known formula for the sum of the first $n$ natural numbers, it follows that $\sum_{i=1}^{n}(2i-1)=2[\frac{n(n+1)}{2}]-n=n(n+1)-n=n[(n+1)-1]=n^2$.

Thus, the proposition holds for all $n\in\mathbb{N}$.

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Stay tuned tomorrow for a geometric visualization of the problem and the formula for the sum of the first $n$ even numbers.