12
$\begingroup$

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?

  • 0
    @DarkL : right. I thought you separated the sums without changing summation to $n$.2012-08-31

2 Answers 2

7

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)}. $

  • 0
    @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)}$2012-08-31
2

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.