Prove $f[x_0,...,x_{n-1}] = \sum_{i=0}^{n-1}x_i$ if $f(x) = x^n$ for $n\in\mathbb{N}$

Question

Definition The divided difference of function $f$ at points $x_0, x_1, ..., x_k$ is defined recursively as $$ f[x]=f(x), \qquad f[x_0,...,x_k] =\frac{f[x_1,...,x_k]-f[x_0,...,x_{k-1}]}{x_k-x_0}, \quad k \ge 1 $$

Prove that if $f(x) = x^n$ for $n\in\mathbb{N}$, then $f[x_0,...,x_{n-1}] = \sum_{i=0}^{n-1}x_i$.

I initially thought about induction but the problem is that change in $n$ causes change in $f$ and I was unable to derive a useful formula for dependency between $a^{n-1}-b^{n-1}$ and $a^n-b^n$.

Example If $f(x) = x^3$ then $$f[a,b,c]=\frac{f[b,c]-f[a,b]}{c-a}=\frac{\frac{f[c]-f[b]}{c-b}-\frac{f[b]-f[a]}{b-a}}{c-a}=\frac{\frac{c^3-b^3}{c-b}-\frac{a^3-b^3}{a-b}}{c-a}=a+b+c$$

Answer 1

Here's a sketch of one solution.First see that $f[x_0,\cdots,x_{n-1}]$ is a polynomial of degree $1$. Use the fact that they are symmetric to see that we must have

$$f[x_0,\cdots,x_{n-1}]=k\sum_{i=0}^{n-1} x_i$$

for some $k$. Now use their mean value theorem to conclude.

Answer 2

We consider Newton's form of the interpolating polynomial. Let $f(x) = x^n$ and let $\{x_j\}_{j=0}^n$ denote $n+1$ distinct nodes. Then $$x^n = f[x_0] + f[x_0,x_1](x-x_0) + \dots + f[x_0,\dotsc,x_{n-1}] \prod_{i=0}^{n-2}(x-x_i) + f[x_0,\dotsc,x_n] \prod_{i=0}^{n-1}(x-x_i).$$ We will exploit the fact that the coefficient of $x^{n-1}$ is $0$. We begin by studying the structure of the different terms. The last term is \begin{align} & f[x_0,\dotsc,x_n] \prod_{i=0}^{n-1}(x-x_i) \\ & = f[x_0,\dotsc,x_n] x^n - f[x_0,\dotsc,x_n] (x_0 + x_1 + \dots + x_{n-1}) x^{n-1} + O(x^{n-2}), \end{align} where $O(x^{n-2})$ means "terms with $x$-degree $\leq n-2$". It is possible to conclude that $$f[x_0,\dotsc,x_n] = 1,$$ because all other terms have order at most $n-1$ and the coefficient of $x^n$ is $1$ (alternatively, see this post). As for the second to last term we have $$ f[x_0,\dotsc,x_{n-1}] \prod_{i=0}^{n-2}(x-x_i) = f[x_0,\dotsc,x_{n-1}] x^{n-1} + O(x^{n-2}).$$ Since all other terms have degree at most $n-2$ we conclude that $$ f[x_0,\dotsc,x_{n-1}] = x_0 + x_1 + \dots + x_{n-1}.$$ It is worth stressing our use of fact that two polynomials are identical if and only their coefficients are identical. This completes the analysis.

Answer 3

The Liebniz rule states that $$ (fg)[x_0,\dots,x_n] = f[x_0]g[x_0,\dots,x_n] + f[x_0,x_1]g[x_1,\dots,x_n] + \dots + f[x_0,\dots,x_n]g[x_n]. $$

Proceed by induction on $n$. For convenience, let $f_n(x) := x^n$. Notice that the base case is trivial.

Assume that the result holds for some positive integer $k$. It can be shown that \begin{equation} f_1[x_0,\dots,x_j] = \begin{cases} 1, & j=1; \\ 0, & j>1. \end{cases} \tag{1} \label{ddi} \end{equation}

Thus, \begin{align} f_{k+1}[x_0,\dots,x_k] &= \left(f_1 f_k\right)[x_0,\dots,x_k] \\ &= f_1[x_0]f_k[x_0,\dots,x_k] + f_1[x_0, x_1]f_k[x_1,\dots,x_k] + \cdots + f_1[x_0,\dots,x_k]f_k[x_k] \\ &= x_0 f_k[x_0,\dots,x_k] + f_1[x_0,x_1]f_k[x_1,\dots,x_k] \tag{by \eqref{ddi}} \\ &= x_0 f_k[x_0,\dots,x_k] + \sum_{j=1}^{k} x_j. \tag{IH} \end{align}

By the mean value theorem for divided differences, $\exists \xi \in \left[ \min_j\{x_j\},\max_j\{x_j\} \right]$ such that $$ f_k[x_0,\dots,x_k] = \frac{f_k^{(k)}(\xi)}{k!}. $$ But $f_k^{(k)}(x) = k!$, so $f_k[x_0,\dots,x_k]=1$. Thus, $$ f_{k+1}[x_0,\dots,x_k] = \sum_{j=0}^k x_j $$ and the result follows by induction.

Answer 4

All of this can be achieved by weak induction.

I'll use $f[x_0,x_1,...,x_k]_n$ to denote $f[x_0,x_1,...,x_k]$ where $f(r)=r^n$.

First, we'll show that $f[s,t]_n=\sum_{a+b=n-1}(s^at^b)$ for all $n$ (where $a$ and $b$ are nonnegative) by simple induction.
(a proof can be found here so you can skip the next paragraph if you want)

Clearly $f[s,t]_1=\sum_{a+b=0}(s^at^b)$ as $$f[s,t]_1=\frac{f[t]_1-f[s]_1}{t-s}=\frac{t-s}{t-s}=1=s^0t^0=\sum_{a+b=0}(s^at^b)$$

Now, suppose for some $p$, $f[s,t]_p=\sum_{a+b=p-1}(s^at^b)$.
Then, we can relate $f[s,t]_{p+1}$ and $f[s,t]_p$ as follows: $$f[s,t]_{p+1}=\frac{f[t]_{p+1}-f[s]_{p+1}}{t-s}=\frac{t^{p+1}-s^{p+1}}{t-s}=\frac{t^pt-s^ps}{t-s}=\frac{t^p(t-s)+s(t^p-s^p)}{t-s}\\=t^p+s\frac{t^p-s^p}{t-s}=t^p+sf[s,t]_p$$

So (by the induction hypothesis): $$f[s,t]_{p+1}=t^p+s\bigg(\sum_{a+b=p-1}(s^at^b)\bigg)=t^p+s\bigg(\sum_{a+b=p-1,\\0\leq a,b \leq p-1}(s^at^b)\bigg)=t^p+\sum_{a+b=p-1,\\0\leq a,b \leq p-1}(s^{a+1}t^b)$$

Using a change of variable from $a+1$ to $c$: $$t^p+\sum_{c+b=p,\\1\leq c\leq p\\0\leq b \leq p-1}(s^ct^b)=s^0t^p+\sum_{c+b=p,\\1\leq c\leq p\\0\leq b \leq p-1}(s^ct^b)=\sum_{c+b=p,\\0\leq c,b\leq p}(s^ct^b)=\sum_{c+b=(p+1)-1}(s^ct^b)\\=\sum_{a+b=(p+1)-1}(s^at^b)$$

That proves the general formula for $f[s,t]_n$.

Now suppose you have shown that $f[x_0,x_1,...,x_{k-1}]=\sum_{i_0+i_1+...+i_{k-1}=n-k+1}(x_0^{i_0}x_1^{i_1}...x_{k-1}^{i_{k-1}})$ for the given $n$ for some $k$.
(again, $i_0,i_1,...,i_{k-1}$ are all nonnegative)(the case for $k=2$ was just proved above)

Then: $$f[x_0,x_1,...,x_k]=\frac{f[x_1,x_2,...,x_k]-f[x_0,x_1,...,x_{k-1}]}{x_k-x_0}\\=\frac{\sum_{i_1+i_2+...+i_k=n-k+1}(x_1^{i_1}x_2^{i_2}...x_k^{i_k})-\sum_{i_1+i_2+...+i_k=n-k+1}(x_0^{i_1}x_1^{i_2}...x_{k-1}^{i_k})}{x_k-x_0}\\=\frac{\sum_{i_1+i_2+...+i_k=n-k+1}(x_1^{i_1}x_2^{i_2}...x_{k-1}^{i_{k-1}}x_k^{i_k})-\sum_{i_1+i_2+...+i_k=n-k+1}(x_1^{i_1}x_2^{i_2}...x_{k-1}^{i_{k-1}}x_0^{i_k})}{x_k-x_0}$$

In the above step we have simply relabeled the indices of the powers in the second term to our advantage- all that matters is that whenever the powers appearing on the terms $x_1,x_2,...,x_{k-1}$ are the same in both sums, the powers appearing on $x_k$ and $x_0$ will necessarily be the same (because the sum of all powers $=n-k+1$)

$$\\=\frac{\sum_{i_1+i_2+...+i_k=n-k+1}(x_1^{i_1}x_2^{i_2}...x_{k-1}^{i_{k-1}}(x_k^{i_k}-x_0^{i_k}))}{x_k-x_0}\\=\sum_{i_1+i_2+...+i_k=n-k+1}(x_1^{i_1}x_2^{i_2}...x_{k-1}^{i_{k-1}}\bigg(\frac{x_k^{i_k}-x_0^{i_k}}{x_k-x_0}\bigg))\\=\sum_{i_1+i_2+...+i_k=n-k+1}(x_1^{i_1}x_2^{i_2}...x_{k-1}^{i_{k-1}}\;f[x_0,x_k]_{i_k})\\=\sum_{i_1+i_2+...+i_k=n-k+1}(x_1^{i_1}x_2^{i_2}...x_{k-1}^{i_{k-1}}\bigg(\sum_{j_1+j_2=i_k-1}x_0^{j_1}x_k^{j_2}\bigg))$$

Noting that $i_1+i_2+...i_{k-1}+(j_1+j_2)=i_1+i_2+...+i_{k-1}+(i_k-1)=(n-k+1)-1=n-k\\=n-(k+1)+1$ and relabeling indices again, we get:

$$\sum_{i_0+i_1+...+i_{k-1}+i_k=n-(k+1)+1}(x_0^{i_0}x_1^{i_1}...x_k^{i_k})$$

So that does the general formula for $f[x_0,x_1,...,x_k]_n$ for us (for all $n$ and $k$).

Now, your question is about what happens when $k=n-1$. Well, in that case: $$f[x_0,x_1,...,x_{n-1}]_n=\sum_{i_0+i_1+...+i_{n-1}=n-n+1}(x_0^{i_0}x_1^{i_1}...x_{n-1}^{i_{n-1}})=\sum_{i_0+i_1+...+i_{n-1}=1}(x_0^{i_0}x_1^{i_1}...x_{n-1}^{i_{n-1}})$$

Since all of $i_0,i_1,...,i_{n-1}$ are to be nonnegative integers, the only possibilities for their values are for one of the $i_k$'s to be 1 and the rest to be zero.

So we get $$f[x_0,x_1,...,x_{n-1}]_n\\=x_0^1x_1^0x_2^0...x_{n-1}^0+x_0^0x_1^1x_0^0...x_{n-1}^0+x_0^0x_1^0x_2^1...x_{n-1}^0+\;.\;.\;.\; +x_0^0x_1^0x_2^0...x_{n-1}^1\\=x_0+x_1+...+x_{n-1}=\sum_{i=0}^{n-1}x_i$$

(please comment or edit for any corrections)