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whuber
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###Edit###

Michael E2, in an extension to his answer, reminds us of the value of simplifying multiple sums. His generalization further clarifies the nature of these operations. Because the variable name "$x$" does not explicitly appear in the $s_i$ notation, the variable name is immaterial. What matters are the index names, only insofar as they are used to connect an $x$ with its enclosing summation: as I remarked at the outset, as bound variables they must disappear at the end.

Thus, for example, we could write sum[i^3, i] for $\sum_{i=1}^n x_i^3$: the first argument to sum is a polynomial (or rational function, even) and the second one is a symbol to indicate what is varying over the summation.

A comparable extension is trivial to make: just include the second argument explicitly in the definition of sum. So that we can reproduce the previous solution (where x was the only variable), we can have the second argument default to x if it's missing. Here is the entire (generalized) solution:

ClearAll[sum, s, n];
sum[a_] := sum[a, x];
sum[x_, x_] := Subscript[s, 1]; sum[x_^p_, x_] := Subscript[s, p];
sum[Times[a_, y__], x_] /; FreeQ[a, x] := a sum[Times[y], x];
sum[a_ + b_, x_] := sum[a, x] + sum[b, x];
sum[a_, x_] /; FreeQ[a, x] := n a; 
expand[a_] /; ! FreeQ[a, sum] := Map[Expand, a, Infinity]; expand[a_] := a;

For example, compute $\sum_{k=1}^n\sum_{j=1}^n\sum_{i=1}^n (x_i+x_j+x_k - 3\bar{x})^2$, where $\bar{x} = \sum_{i=1}^n x_i / n$ is the average:

sum[sum[sum[(i + j + k - 3 sum[i, i]/n)^3, i], j], k]

$6 s_1^3-9 n s_1 s_2+3 n^2 s_3$

(Notice that it was no problem for i to appear in two summations.) You do have to be a little careful not to go overboard and stretch the use of sum too far:

sum[x^2, x^2] // expand

$s_1$

This surprising result is correct: the symbol x^2 in the second argument is being used to represent the variable summed over; in terms of this pattern, the first argument x^2 is the first power, whence the result is indeed $s_1$.

One can rather easily go further with generalizations, depending on the need. The next big step would be to implement multivariate sums, such as $\sum_{i=1}^n\sum_{j=1}^n x_i y_j^2$, perhaps to be written $s_1(x)s_2(y)$ or--assuming a specific ordering $(x,y)$ of the variables--even as $s_{1,2}$. As we appear to have gone beyond the scope of the present question, I leave these generalizations to interested readers. I hope that the examples given here have shown how stripping the notation down to its essentials can lead to simple and clear solutions.

###Edit###

Michael E2, in an extension to his answer, reminds us of the value of simplifying multiple sums. His generalization further clarifies the nature of these operations. Because the variable name "$x$" does not explicitly appear in the $s_i$ notation, the variable name is immaterial. What matters are the index names, only insofar as they are used to connect an $x$ with its enclosing summation: as I remarked at the outset, as bound variables they must disappear at the end.

Thus, for example, we could write sum[i^3, i] for $\sum_{i=1}^n x_i^3$: the first argument to sum is a polynomial (or rational function, even) and the second one is a symbol to indicate what is varying over the summation.

A comparable extension is trivial to make: just include the second argument explicitly in the definition of sum. So that we can reproduce the previous solution (where x was the only variable), we can have the second argument default to x if it's missing. Here is the entire (generalized) solution:

ClearAll[sum, s, n];
sum[a_] := sum[a, x];
sum[x_, x_] := Subscript[s, 1]; sum[x_^p_, x_] := Subscript[s, p];
sum[Times[a_, y__], x_] /; FreeQ[a, x] := a sum[Times[y], x];
sum[a_ + b_, x_] := sum[a, x] + sum[b, x];
sum[a_, x_] /; FreeQ[a, x] := n a; 
expand[a_] /; ! FreeQ[a, sum] := Map[Expand, a, Infinity]; expand[a_] := a;

For example, compute $\sum_{k=1}^n\sum_{j=1}^n\sum_{i=1}^n (x_i+x_j+x_k - 3\bar{x})^2$, where $\bar{x} = \sum_{i=1}^n x_i / n$ is the average:

sum[sum[sum[(i + j + k - 3 sum[i, i]/n)^3, i], j], k]

$6 s_1^3-9 n s_1 s_2+3 n^2 s_3$

(Notice that it was no problem for i to appear in two summations.) You do have to be a little careful not to go overboard and stretch the use of sum too far:

sum[x^2, x^2] // expand

$s_1$

This surprising result is correct: the symbol x^2 in the second argument is being used to represent the variable summed over; in terms of this pattern, the first argument x^2 is the first power, whence the result is indeed $s_1$.

One can rather easily go further with generalizations, depending on the need. The next big step would be to implement multivariate sums, such as $\sum_{i=1}^n\sum_{j=1}^n x_i y_j^2$, perhaps to be written $s_1(x)s_2(y)$ or--assuming a specific ordering $(x,y)$ of the variables--even as $s_{1,2}$. As we appear to have gone beyond the scope of the present question, I leave these generalizations to interested readers. I hope that the examples given here have shown how stripping the notation down to its essentials can lead to simple and clear solutions.

Source Link
whuber
whuber
  • 20.6k
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  • 61
  • 113

As bound variables, $r$ and $i$ must play no role in the expansion and so they shouldn't even appear in our solution. It should be equally evident that $n$ is just along for the ride as a placeholder for the upper limit; we could call it anything, and therefore we may call it nothing and ignore it except in expressions where it will necessarily appear in the output. Notice, too, that $x$ disappears under the summation: it really just serves as a placeholder for arguments of other expressions being summed.

Consequently, expressions like "$\sum_{i=1}^n f(x_i)$" should be understood as operations on $f$ itself. What properties should they have? Only the obvious ones implied by the ring operations $+$ and $\times$ in the algebra of polynomials in the $x_i$. This gives just four simple rules, after we establish the unique roles of the symbols "$x$", "$s$", and "$n$":

ClearAll[sum, x, s, n];

  1. Converting sums of powers of $x$ into the power sum variables $s_i$:

     sum[x] := Subscript[s, 1]; sum[x^p_] := Subscript[s, p];

  2. Additivity:

     sum[a_ + b_] := sum[a] + sum[b];

  3. Linearity (with respect to "scalars" which do not have any $x_i$ in them):

     sum[Times[a_, y__]] /; FreeQ[a, x] := a sum[Times[y]];

  4. The effect of summing constant values (this is the only place $n$ need appear):

     sum[a_] /; FreeQ[a, x] := n a;

That should do it. But to perform the algebra, we need to expand algebraic combinations of everything possible into powers of $x$ (and otherwise leave everything else alone):

expand[a_] /; ! FreeQ[a, sum] := Map[Expand[#, x] &, a, Infinity];
expand[a_] := a;

###Examples###

The question:

sum[x ((sum[x] - x)^2 - sum[x^2])] // expand

$s_1^3-3 s_1 s_2+s_3$

Variance:

sum[(x - sum[x]/n)^2] // expand

$-\frac{s_1^2}{n}+s_2$

Skewness:

sum[(x - sum[x]/n)^3] / sum[(x - sum[x]/n)^2] ^(3/2) // expand

$\frac{\frac{2 s_1^3}{n^2}-\frac{3 s_1 s_2}{n}+s_3}{\left(-\frac{s_1^2}{n}+s_2\right){}^{3/2}}$

etc.

We might worry about misapplications, such as to non-polynomial functions. Not to fear:

sum[Exp[(x + 1)^2]] // expand

$\text{sum}\left[e^{1+2 x+x^2}\right]$

The expansion proceeds insofar as it can, but sum does not know how to go any further, and so stops. sum is also ignorant of other non-polynomial objects, but when they can be converted to polynomials, it works:

sum[Series[Exp[x], {x, 0, 4}]  // Normal] // expand

$n+s_1+\frac{s_2}{2}+\frac{s_3}{6}+\frac{s_4}{24}$

(This is a series expansion through order four of $\sum_{i=1}^n e^{x_i}$.)