4

While wasting time scrolling on instagram recently I scrolled past the identity

$$\sum_{i=1}^{n} i^3 = \left[\sum_{i=1}^{n}i \right]^2.$$

It caught my attention because I don't recall ever stumbling across this identity in undergrad or grad school. A proof of this identity falls out naturally by induction, invoking the known identity for the sum of first $n$ natural numbers, but what I would like to ask MSE is - is there an algebraic or geometric way to provide intuition for the conjecture?

Allow me to explain what I mean; the identity $\sum_{i=1}^{n}i = n(n+1)/2$ can also be easily proven by induction, but before formally proving the identity one might arrive at the conjecture by geometric intuition (drawing dots in a triangle to represent the summation, adding in the reflection of the triangle to make a rectangle, identifying the pattern $n(n=1)/2$. One might also arrive at the conjecture by algebraic intuition from writing out the sum, suppose for $n = 8$, $$1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 = (8+1) + (7+2) + (6+3) + (5+4)$$ which is $n/2$ groups of $n+1$.

So although I can easily prove the original identity, does anyone have a clever way to intuit the conjecture, had I not seen the identity a priori?

Prince M
  • 4,113
  • 2
    Squared triangular number on Wikipedia has an image. – peterwhy Jul 27 '23 at 00:42
  • 3
    It is perhaps also worth noting that there is a continuous version: If we integrate x^3 from 0 to c, we get c^4 / 4. And if we integrate x from 0 to c and then square the resullt, we also get c^4 / 4. – Dan Asimov Jul 27 '23 at 01:13
  • 1
    Integrating x from 0 to c gives the area of the triangle iT n R^2 with vertices at (0,0), (0,c), amd (c,c). Integrating x^3 from 0 to c gives the 4-dimensional volume of the solid in 4-space that is the volume of the cartesian product T x T, and so it must be the square of the area of T. – Dan Asimov Jul 27 '23 at 01:33
  • The comment section on this question is far superior to the answers. Thanks! – Prince M Jul 27 '23 at 19:40

1 Answers1

0

First, induction is a bad way to do this kind of question. It depends on knowing a formula for sums of powers in advance that is pulled out of a hat. Here's a summary of the first lecture of a mini-course on the calculus of finite differences. For a function $f$ with domain the integers, we define the forward difference by $$\Delta f(x)=f(x+1)-f(x)$$. Then we see that $$\sum_{x=a}^b\Delta f(x)=f(b+1)-f(a).$$ So if we wish to sum a function $g(x)$, if we can find a function $f(x)$ such that $g(x)=\Delta f(x)$, we have the answer. Second, we need factorial powers. We define $$x^{(0)}=1, x^{(1)}=x, x^{(n)}=x(x-1)…(x-n+1).$$ We can, by recursion on the degree, express $x^n$ as a polynomial in factorial powers. We can, again by recursion on the degree, express a factorial power as a polynomial in ordinary powers. Third, we have the formula $$x^{(n)}=\Delta \frac{x^{(n+1)}}{n+1}. $$. Putting it all together, $\sum_{i=1}^ni$ is a polynomial in $n$ of degree 2 with leading coefficient ½ and $\sum_{i=1}^ni^3$ is a polynomial in $n$ of degree 4 with leading coefficient ¼.

P. Lawrence
  • 5,768