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I thought about a generalization for the formula $$\frac{x^n - y^n}{x - y} = x^{n - 1} + yx^{n - 2} + \ldots + y^{n - 1}$$ It can be written as $$\frac{x^n - y^n}{x - y} = x^{n - 1} + yx^{n - 2} + \ldots + y^{n - 1} = \sum_{i + j = n - 1}x^iy^j$$

So we would like to generalize: $$\sum_{i_1 + i_2 + i_3 + ... +i_k = n - 1}{x_1}^{i_1}{x_2}^{i_2}{x_3}^{i_3} \cdots {x_k}^{i_k}$$ For example" $$\sum_{i + j + k = n - 1}{x}^{i}{y}^{j}{z}^{k} = \sum_{k=0}^{n - 1}{z}^{k} \sum_{i + j = n - k - 1}{x}^{i}{y}^{j} = \sum_{k=0}^{n - 1}{z}^{k} \frac{x^{n - k} - y^{n - k}}{x -y} = \frac{1}{x-y} \left(\frac{x^{n + 1} - z^{n + 1}}{x - z} - \frac{y^{n + 1} - z^{n + 1}}{y - z}\right)$$

It seems that the generalized expression is the divided difference of $x^{n + k - 2}$ in the points $x_1, x_2, \ldots, x_k$.

Does anyone have an idea how to prove it?

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  • $\begingroup$ Be careful. $\sum_{k=0}^{n-1} z^k x^{n-k} = x \frac{z^n-x^n}{z-x}$ $\endgroup$
    – vanna
    Aug 31, 2012 at 12:07
  • $\begingroup$ I think it's unlikely (but don't let this stop you!). The 2-variable formula you quote is just the homogenization of a 1-variable formula for the sum of a geometric sequence. $\endgroup$
    – lhf
    Aug 31, 2012 at 12:16
  • $\begingroup$ vanna - I was right: The sum I wrote is the same sum as k=0 to n, since for k = n we get z^n(1 - 1)/(x - y) = 0 $\endgroup$
    – darkl
    Aug 31, 2012 at 12:23
  • $\begingroup$ @DarkL : right. I thought you separated the sums without changing summation to $n$. $\endgroup$
    – vanna
    Aug 31, 2012 at 12:39

2 Answers 2

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I think that this formula is what you are looking for. If $\mathbf x = (x_1,\dotsc,x_r)$, then

$$ \sum_{|I|=n} \mathbf{x}^I = \sum_i\frac{x_i^{n}}{\prod_{j\neq i}(1-\frac{x_j}{x_i})}. $$

With 1 variable, it gives $$ x^n = x^n, $$ for two, $$ \sum_{i+j = n} x^i y^j = \frac{x^{n+1}}{x-y} + \frac{y^{n+1}}{y-x}, $$ and for three, if gives $$ \sum_{i+j+k=n} x^i y^j z^k= \frac{x^{n+2}}{(x-y)(x-z)} + \frac{y^{n+2}}{(y-x)(y-z)} + \frac{z^{n+2}}{(z-x)(z-y)}. $$

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  • $\begingroup$ Thanks, but can you show how this confirms that the sum is the divided difference of x^(n + k - 1) in the requested points? $\endgroup$
    – darkl
    Aug 31, 2012 at 13:59
  • $\begingroup$ @DarkL — Can you explicit what you mean ? $\endgroup$
    – Lierre
    Aug 31, 2012 at 14:12
  • $\begingroup$ $$\sum_{i_1 + i_2 + i_3 + ... +i_k = n}{x_1}^{i_1}{x_2}^{i_2}{x_3}^{i_3} \cdots {x_k}^{i_k} = f[x_1, x_2, ..., x_k]$$ where $$f(x) = x^{n + k - 1}$$ and f[,,..,] is the divided difference: see en.wikipedia.org/wiki/Divided_differences $\endgroup$
    – darkl
    Aug 31, 2012 at 17:27
  • $\begingroup$ @DarkL — Thanks, I didn't know this notion. But the answer is yes because of the formula (on the wikipage) $f[x_0,\dots,x_n] = \sum_{j=0}^{n} \frac{f(x_j)}{\prod_{k\in\{0,\dots,n\}\setminus\{j\}} (x_j-x_k)}$ $\endgroup$
    – Lierre
    Aug 31, 2012 at 17:50
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Found a simple proof: in induction:

Assume

$$\sum_{i_1 + i_2 + i_3 + ... +i_k = n}{x_1}^{i_1}{x_2}^{i_2}{x_3}^{i_3} \cdots {x_k}^{i_k} = f[x_1,x_2,..,x_k]$$

For $$f(x) = x^{n + k - 1}$$

For every n.

We'll show $$\sum_{i_1 + i_2 + i_3 + ... +i_k + i_{k + 1} = n}{x_1}^{i_1}{x_2}^{i_2}{x_3}^{i_3} \cdots {x_k}^{i_k}{x_{k+1}}^{i_{k+1}} = g[x_1,x_2,..,x_k,x_{k+1}]$$

where $$g(x) = x^{n + k}$$

Proof: by induction on k: $$\sum_{i_1 + i_2 + i_3 + ... +i_k = n + 1}{x_1}^{i_1}{x_2}^{i_2}{x_3}^{i_3} \cdots {x_k}^{i_k} = g[x_1,x_2,..,x_k]$$ $$\sum_{i_1 + i_2 + i_3 + ... +i_k = n + 1}{x_{k+1}}^{i_1}{x_2}^{i_2}{x_3}^{i_3} \cdots {x_k}^{i_k} = g[x_{k+1},x_2,..,x_k] = g[x_2,..,x_k,x_{k+1}]$$

Then by the definition of the divided difference: $$ g[x_1,..,x_k,x_{k+1}] = \frac{g[x_1,..,x_k] - g[x_2,..,x_k,x_{k+1}]}{x_1 - x_{k+1}}$$

But then

$$g[x_1,..,x_k,x_{k+1}] =$$ $$ \frac{1}{x_1 - x_{k + 1}}\cdot\sum_{i_1 + i_2 + i_3 + ... +i_k = n + 1}\left({x_1}^{i_1}{x_2}^{i_2}{x_3}^{i_3} \cdots {x_k}^{i_k} - {x_{k+1}}^{i_1}{x_2}^{i_2}{x_3}^{i_3} \cdots {x_k}^{i_k}\right) = \sum_{i_1 + i_2 + i_3 + ... +i_k = n + 1}\frac{{x_1}^{i_1} - {x_{k + 1}}^{i_1}}{x_1 - x_{k + 1}} \left({x_2}^{i_2}{x_3}^{i_3} \cdots {x_k}^{i_k} \right)$$ $$ = \sum_{i_1 + i_2 + i_3 + ... +i_k = n + 1}\sum_{j_1 + j_2 = i_1 - 1} {x_1}^{j_1}{x_2}^{i_2}{x_3}^{i_3} \cdots {x_k}^{i_k}{x_{k+1}}^{j_2} $$

$$= \sum_{i_1 + i_2 + i_3 + ... +i_k + i_{k + 1} = n}{x_1}^{i_1}{x_2}^{i_2}{x_3}^{i_3} \cdots {x_k}^{i_k}{x_{k+1}}^{i_{k+1}}$$

That's nice (and directly from the definition), but if someone has a more geometrical explanation for this I'll be glad to hear.

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