4

There is an obvious interpretation of $\sum_{i=1}^n i$ as just a function $\mathbb N \to \mathbb N$:

$$n \mapsto \underbrace{1 + 2 + 3 + \ldots}_{n}$$

However since the sum in hand equals

$$\frac{n(n+1)}{2}$$

which is polynomial, one can treat it as a much more general function than just $\mathbb N \to \mathbb N$. Thus such looking weird things as

$$\sum_{i=1}^{\bigl(\begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \bigr)} i = \frac{1}{2} \begin{pmatrix} a^2+a+bc & (a+d+1)b \\ (a+d+1)c & d^2 +d + bc \end{pmatrix}$$

make sense if properly interpreted. Actually I didn't calculate it by hand, I just entered the sum above as it is (not the polynomial, the real $\sum$) in my TI-89.

I know of one valid interpretation of the sum in this case:

$$\sum_{i=1}^n i = I(n+1) - I(1)$$

where $I = \frac{1}{2}(i^2-i)$ is the antidifference of $i$. That is it is defined as an inverse operation.

Now I ask if it is possible to interpret

$$\sum_{i=1}^n i^s$$

as a real sum (with some sort of addition $+$), but to allow $n$ to be something more than a number? So I could sum from $1$ to $\frac{d}{dx}$.

Yrogirg
  • 3,609
  • I suppose that if you're dealing with some system of well-ordered operators/matrices/objects $i$ such that $i^s$ is well-defined for all such $i$ and with some sort of sensible addition...you certainly could. I'm curious what the context is that would make one want to (for instance) sum such things from $1$ to $\frac{d}{dx}$. – Cameron Buie Jun 18 '12 at 17:58
  • +1, Subing a matrice in the upper index! I have been keeping an eye on how far one can push the identities by applying it to objects that they were not considered for in the first place. I guess you know about sin( A ) , cos( A ) etc. where A is a matrix then. – jimjim Jun 18 '12 at 22:04

2 Answers2

4

I would strongly disagree that the sum you display 'makes sense'. The values $\displaystyle{\sum_{i=1}^n i}$ and $n(n+1)/2$ are the same as polynomials in $n$, but the sum operator has a syntax and a semantics associated with it that your formula for $\displaystyle{\sum_{i=1}^{\bigl(\begin{smallmatrix} a & b \\ c & d \end{smallmatrix} \bigr)} i}$ completely ignores. In particular, it's easy to talk about summing over numbers, and even relatively straightforward to talk about summing over matrices - but what entities are you summing over, even abstractly, that assign meaning to a sum with lower limit a number and upper limit a matrix?

There are cases where this sort of symbolic manipulation can turn out to be remarkably powerful (in particular, you might want to have a look at the Umbral calculus, which formalizes some of the notions that you're talking about), but a lot of care has to be taken when applying them, because unexpected surprises can sometimes lurk outside the original domain. For instance, it's reasonable to say "The value of $\int_a^b 1/x\ dx$ is $\log(b)-\log(a)$" when $a$ and $b$ are positive reals, but you have to be careful about saying "The value of $\int_{-i}^i 1/z\ dz$ is $\log(i)-\log(-i)$", because neither of the expressions in the second statement has a unique meaning. In spite of how it can sometimes appear, mathematics is much more than just manipulating symbols, and an understanding of the semantics you're working with can save you from many potential pitfalls.

Steven Stadnicki
  • 51,746
2

It is not clear to me what values you want $s$ to take so I'll assume it's a non-negative integer. In that case, Faulhaber's formula gives $$\sum_{i=1}^n i^s = \frac{1}{s+1} \sum_{j=0}^s (-1)^j {s+1 \choose j} B_j n^{s+1-j}$$

where $B_j$ are the Bernoulli numbers. So $n$ can take values in any ring in which $s+1$ and the denominators of the relevant Bernoulli numbers are invertible, at least.

Qiaochu Yuan
  • 419,620