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$1^2 = 1$, $2^2 = 4$, $3^2 = 9$, $4^2 = 16$, $5^2 = 25$, etc...

Looking at the difference between those square values, we get: 3, 5, 7, 9, etc...

The difference from one (integer) square to the next increases by 2 without fault (let's assume).

Why is that? Why is there that pattern of increases by 2? What is it due to? What is the source of it?

I can "see" squaring visually as the construction of an actual square and I have drawn subdivided squares within squares to see the pattern unfold, but I just don't understand how to explain that increase by 2; I can't trace it, essentially.

jeremy radcliff
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    Hint. Can you calculate $(x+1)^2) - x^2$? – Ethan Bolker May 07 '15 at 16:36
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    $(n+1)^2-n^2=2n+1$ – hamid kamali May 07 '15 at 16:36
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    It's very much of the same origin as the second derivative of $f(x)=x^2$. – Jyrki Lahtonen May 07 '15 at 16:55
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    @JyrkiLahtonen, sorry I don't see it, do you mind explaining the connection? – jeremy radcliff May 07 '15 at 16:59
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    Not him, but a derivative is infinitesimal difference whereas what you are looking at finite differences. – David Etler May 07 '15 at 22:12
  • @DavidEtler, yes but I'm not sure where he's getting the "same origin" part from. I can see that the slope at each successive integer value of x is double that of the previous one, but I'm not sure I understand the connection in this context. – jeremy radcliff May 08 '15 at 16:09
  • @jeremyradcliff We can show for $f(x) = x^2$, $D f(x) = \lim_{h\to0}\frac{(x+h)^2 - x^2}{h} = \lim_{h\to0}\frac{x^2 + 2xh + h^2 - x^2}{h} = 2x$ and $D^2f(x) = \lim_{h\to0}\frac{2(x+h) - (2x)}{h}= \color{red}{\lim_{h\to0} \frac{2h}{h} = 2}$. The key importance is in the second derivative: though it is a limit of values for $h$, the expression inside the limit is a constant. A finite difference ($a_{n+1} - a_n$) is essentially a difference quotient with $h=1$. Here, the expression is $2$ regardless of $h$. – David Etler May 08 '15 at 22:06

5 Answers5

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For a visual intuition.... The inner blue region increments the gray square by one, but to make the next increment you need the extra yellow unit squares to get to the following square.

enter image description here

Joffan
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We have the series $$ x_n = n^2 $$ and the differences $$ \Delta x_n = x_{n+1} - x_n = (n+1)^2 - n^2 = n^2 + 2n + 1 - n^2 = 2n + 1 $$ and the difference of differences $$ \Delta x_{n+1} - \Delta x_n = (2(n+1)+1) - (2n+1) = 2n + 3 - 2n - 1 = 2 $$

mvw
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The difference between two adjacent squares, $(a+1)^2$ and $a^2$:

$$\begin{align} (a+1)^2 - a^2 &= a^2 + 2a + 1 - a^2 \\ &= 2a + 1 \end{align}$$

So for each increase in $a$ by $1$, the delta between squares gets $2$ bigger.

Joffan
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Hint: $(a+1)^2-a^2=a^2+2a+1-a^2=2a+1$

callculus42
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Let $a$ be some natural number, and then let's think about $a+1$.
Since $a^2 = a \cdot a$, we have $(a+1)^2 = a^2 + 2a + 1$.

Then $(a+1)^2 - a^2 = 2a + 1$.

And then note for the numbers $1,2,3,\ldots$ applying $2a+1$ gives $2\cdot 1 + 1, 2 \cdot 2 + 1, 3 \cdot 2 + 1,\ldots$ Which is: $$3,5,7,9,\ldots$$

Newb
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