I'm proving a remark in problem I.12.10 from textbook Analysis I by Amann/Escher.
The remark is as follows:
$$(x_{n}-x_{0}) a_{n}+a_{n-1}-b_{n-1}=0$$
Here are relevant definitions from my textbook:
Remark 12.10(b)
My issue:
- The authors define $f\left[x_{0}, \ldots, x_{n}\right]$ by $$f\left[x_{0}, \ldots, x_{n}\right] := \sum_{j=0}^{n} \frac{f\left(x_{j}\right)}{\prod_{k=0 \atop k \neq j}^{n}\left(x_{j}-x_{k}\right)}$$
I guess $$f\left[x_{1}, \ldots, x_{n}\right] = f\left[x_{n}, \ldots, x_{1}\right] = \sum_{j=1}^{n} \frac{f\left(x_{j}\right)}{\prod_{k=1 \atop k \neq j}^{n}\left(x_{j}-x_{k}\right)}$$
I'm not sure whether my guess is correct or not.
- The authors said that $a_{n}=b_{n}$ for $1 \leq n \leq m$. Then $(x_{n}-x_{0}) a_{n}+a_{n-1}-b_{n-1}=0$ becomes $(x_{n}-x_{0}) a_{n}=0$, which is trivially wrong.
Could you please elaborate more on these points?
Here is the answer I sketched in the comments, in more detail.
Background.
Let $\mathbb{N}=\left\{ 0,1,2,\ldots\right\} $.
Fix a field $\mathbb{K}$. Let $V$ be a $\mathbb{K}$-vector space. (In your setting, $V=\mathbb{K}$, but there is nothing gained from this specialization.) If $\lambda\in\mathbb{K}$ is nonzero and if $v\in V$, then $\dfrac{v}{\lambda}$ shall denote the vector $\lambda^{-1}v\in V$.
Fix a map $f:\mathbb{K}\rightarrow V$.
If $n\in\mathbb{N}$, and if $x_{0},x_{1},\ldots,x_{n}$ are any $n+1$ distinct elements of $\mathbb{K}$, then we define a vector $f\left[ x_{0},x_{1} ,\ldots,x_{n}\right] \in V$ by setting \begin{equation} f\left[ x_{0},x_{1},\ldots,x_{n}\right] =\sum_{j=0}^{n}\dfrac{f\left( x_{j}\right) }{\prod\limits_{\substack{k\in\left\{ 0,1,\ldots,n\right\} ;\\k\neq j}}\left( x_{j}-x_{k}\right) }. \label{darij1.eq.def-fbra} \tag{1} \end{equation}
Everything I am doing can be generalized even further by letting $\mathbb{K}$ be any commutative ring (instead of a field). In this case, instead of requiring $x_{0},x_{1},\ldots,x_{n}$ to be distinct, we need to require that their pairwise differences $x_{i}-x_{j}$ for $i\neq j$ are invertible in $\mathbb{K}$. Other than that, the arguments work the same. But I will keep to the case when $\mathbb{K}$ is a field, just to keep my statements shorter.
Let me restate \eqref{darij1.eq.def-fbra} as follows: If $n\in\mathbb{N}$, and if $a_{0},a_{1},\ldots,a_{n}$ are any $n+1$ distinct elements of $\mathbb{K}$, then \begin{equation} f\left[ a_{0},a_{1},\ldots,a_{n}\right] =\sum_{j=0}^{n}\dfrac{f\left( a_{j}\right) }{\prod\limits_{\substack{k\in\left\{ 0,1,\ldots,n\right\} ;\\k\neq j}}\left( a_{j}-a_{k}\right) }. \label{darij1.eq.def-fbra-a} \tag{2} \end{equation} Indeed, this follows by applying \eqref{darij1.eq.def-fbra} to $x_{i}=a_{i}$.
Your first question is asking why every positive integer $n$ and every $n$ distinct elements $x_{1} ,x_{2},\ldots,x_{n}$ of $\mathbb{K}$ satisfy \begin{equation} f\left[ x_{1},x_{2},\ldots,x_{n}\right] =\sum_{j=1}^{n}\dfrac{f\left( x_{j}\right) }{\prod\limits_{\substack{k\in\left\{ 1,2,\ldots,n\right\} ;\\k\neq j}}\left( x_{j}-x_{k}\right) }. \label{darij1.eq.2} \tag{3} \end{equation}
Proof of \eqref{darij1.eq.2}: Let $n$ be a positive integer, and let $x_{1} ,x_{2},\ldots,x_{n}$ be $n$ distinct elements of $\mathbb{K}$. Then, \eqref{darij1.eq.def-fbra-a} (applied to $n-1$ and $x_{i+1}$ instead of $n$ and $a_{i}$) yields \begin{align*} f\left[ x_{1},x_{2},\ldots,x_{n}\right] & =\sum_{j=0}^{n-1}\dfrac{f\left( x_{j+1}\right) }{\prod\limits_{\substack{k\in\left\{ 0,1,\ldots,n-1\right\} ;\\k\neq j}}\left( x_{j+1}-x_{k+1}\right) }\\ & =\sum_{j=0}^{n-1}\dfrac{f\left( x_{j+1}\right) }{\prod \limits_{\substack{k\in\left\{ 0,1,\ldots,n-1\right\} ;\\k+1\neq j+1}}\left( x_{j+1}-x_{k+1}\right) }\\ & \qquad\left( \text{since the statement "}k\neq j\text{" is equivalent to "}k+1\neq j+1\text{"}\right) \\ & =\sum_{j=0}^{n-1}\dfrac{f\left( x_{j+1}\right) }{\prod \limits_{\substack{k\in\left\{ 1,2,\ldots,n\right\} ;\\k\neq j+1}}\left( x_{j+1}-x_{k}\right) }\\ & \qquad\left( \text{here, we have substituted }k\text{ for }k+1\text{ in the product}\right) \\ & =\sum_{j=1}^{n}\dfrac{f\left( x_{j}\right) }{\prod\limits_{\substack{k\in \left\{ 1,2,\ldots,n\right\} ;\\k\neq j}}\left( x_{j}-x_{k}\right) }\\ & \qquad\left( \text{here, we have substituted }j\text{ for }j+1\text{ in the sum}\right) . \end{align*} This proves \eqref{darij1.eq.2}. $\blacksquare$
Your second question is about proving the following claim:
I don't understand the hint that Amann and Escher give for this proposition, but fortunately you can just as well prove it by straightforward manipulation of sums:
Proof of Proposition 1. We have $0\neq n$ (since $n$ is positive) and thus $x_{0}\neq x_{n}$ (since $x_{0},x_{1},\ldots,x_{n}$ are distinct). Thus, $x_{0}-x_{n}\neq0$.
Applying \eqref{darij1.eq.def-fbra} to $n-1$ instead of $n$, we obtain \begin{equation} f\left[ x_{0},x_{1},\ldots,x_{n-1}\right] =\sum_{j=0}^{n-1}\dfrac{f\left( x_{j}\right) }{\prod\limits_{\substack{k\in\left\{ 0,1,\ldots,n-1\right\} ;\\k\neq j}}\left( x_{j}-x_{k}\right) }. \label{darij1.pf.prop1.1} \tag{4} \end{equation}
Now, fix $j\in\left\{ 0,1,\ldots,n-1\right\} $. Then, $j\neq n$, so that $x_{j}\neq x_{n}$ (since $x_{0},x_{1},\ldots,x_{n}$ are distinct), and thus $x_{j}-x_{n}\neq0$. Hence, we can divide by $x_{j}-x_{n}$. Moreover, $n$ is a $k\in\left\{ 0,1,\ldots,n\right\} $ satisfying $k\neq j$ (since $n\neq j$); therefore, the product $\prod\limits_{\substack{k\in\left\{ 0,1,\ldots ,n\right\} ;\\k\neq j}}\left( x_{j}-x_{k}\right) $ contains a factor for $k=n$. Splitting off this factor, we obtain \begin{equation} \prod\limits_{\substack{k\in\left\{ 0,1,\ldots,n\right\} ;\\k\neq j}}\left( x_{j}-x_{k}\right) =\left( x_{j}-x_{n}\right) \cdot\prod \limits_{\substack{k\in\left\{ 0,1,\ldots,n-1\right\} ;\\k\neq j}}\left( x_{j}-x_{k}\right) . \end{equation} Thus, \begin{align} \left( x_{j}-x_{n}\right) \cdot\dfrac{f\left( x_{j}\right) } {\prod\limits_{\substack{k\in\left\{ 0,1,\ldots,n\right\} ;\\k\neq j}}\left( x_{j}-x_{k}\right) } & =\left( x_{j}-x_{n}\right) \cdot\dfrac{f\left( x_{j}\right) }{\left( x_{j}-x_{n}\right) \cdot\prod\limits_{\substack{k\in \left\{ 0,1,\ldots,n-1\right\} ;\\k\neq j}}\left( x_{j}-x_{k}\right) }\nonumber\\ & =\dfrac{f\left( x_{j}\right) }{\prod\limits_{\substack{k\in\left\{ 0,1,\ldots,n-1\right\} ;\\k\neq j}}\left( x_{j}-x_{k}\right) } . \label{darij1.pf.prop1.2} \tag{5} \end{align}
Forget that we fixed $j$. We thus have proven \eqref{darij1.pf.prop1.2} for each $j\in\left\{ 0,1,\ldots,n-1\right\} $. Now, \eqref{darij1.pf.prop1.1} becomes \begin{align*} f\left[ x_{0},x_{1},\ldots,x_{n-1}\right] & =\sum_{j=0}^{n-1} \underbrace{\dfrac{f\left( x_{j}\right) }{\prod\limits_{\substack{k\in \left\{ 0,1,\ldots,n-1\right\} ;\\k\neq j}}\left( x_{j}-x_{k}\right) } }_{\substack{=\left( x_{j}-x_{n}\right) \cdot\dfrac{f\left( x_{j}\right) }{\prod\limits_{\substack{k\in\left\{ 0,1,\ldots,n\right\} ;\\k\neq j}}\left( x_{j}-x_{k}\right) }\\\text{(by \eqref{darij1.pf.prop1.2})}}}\\ & =\sum_{j=0}^{n-1}\left( x_{j}-x_{n}\right) \cdot\dfrac{f\left( x_{j}\right) }{\prod\limits_{\substack{k\in\left\{ 0,1,\ldots,n\right\} ;\\k\neq j}}\left( x_{j}-x_{k}\right) }. \end{align*} Comparing this with \begin{align*} & \sum_{j=0}^{n}\left( x_{j}-x_{n}\right) \cdot\dfrac{f\left( x_{j}\right) }{\prod\limits_{\substack{k\in\left\{ 0,1,\ldots,n\right\} ;\\k\neq j}}\left( x_{j}-x_{k}\right) }\\ & =\sum_{j=0}^{n-1}\left( x_{j}-x_{n}\right) \cdot\dfrac{f\left( x_{j}\right) }{\prod\limits_{\substack{k\in\left\{ 0,1,\ldots,n\right\} ;\\k\neq j}}\left( x_{j}-x_{k}\right) }+\underbrace{\left( x_{n} -x_{n}\right) }_{=0}\cdot\dfrac{f\left( x_{n}\right) }{\prod \limits_{\substack{k\in\left\{ 0,1,\ldots,n\right\} ;\\k\neq n}}\left( x_{n}-x_{k}\right) }\\ & \qquad\left( \text{here, we have split off the addend for }j=n\text{ from the sum}\right) \\ & =\sum_{j=0}^{n-1}\left( x_{j}-x_{n}\right) \cdot\dfrac{f\left( x_{j}\right) }{\prod\limits_{\substack{k\in\left\{ 0,1,\ldots,n\right\} ;\\k\neq j}}\left( x_{j}-x_{k}\right) }, \end{align*} we obtain \begin{equation} f\left[ x_{0},x_{1},\ldots,x_{n-1}\right] =\sum_{j=0}^{n}\left( x_{j} -x_{n}\right) \cdot\dfrac{f\left( x_{j}\right) }{\prod \limits_{\substack{k\in\left\{ 0,1,\ldots,n\right\} ;\\k\neq j}}\left( x_{j}-x_{k}\right) }. \label{darij1.pf.prop1.x0xn-1} \tag{6} \end{equation}
On the other hand, fix $j\in\left\{ 1,2,\ldots,n\right\} $. Then, $j\neq0$, so that $x_{j}\neq x_{0}$ (since $x_{0},x_{1},\ldots,x_{n}$ are distinct), and thus $x_{j}-x_{0}\neq0$. Hence, we can divide by $x_{j}-x_{0}$. Moreover, $0$ is a $k\in\left\{ 0,1,\ldots,n\right\} $ satisfying $k\neq j$ (since $0\neq j$); therefore, the product $\prod\limits_{\substack{k\in\left\{ 0,1,\ldots,n\right\} ;\\k\neq j}}\left( x_{j}-x_{k}\right) $ contains a factor for $k=0$. Splitting off this factor, we obtain \begin{equation} \prod\limits_{\substack{k\in\left\{ 0,1,\ldots,n\right\} ;\\k\neq j}}\left( x_{j}-x_{k}\right) =\left( x_{j}-x_{0}\right) \cdot\prod \limits_{\substack{k\in\left\{ 1,2,\ldots,n\right\} ;\\k\neq j}}\left( x_{j}-x_{k}\right) . \end{equation} Thus, \begin{align} \left( x_{j}-x_{0}\right) \cdot\dfrac{f\left( x_{j}\right) } {\prod\limits_{\substack{k\in\left\{ 0,1,\ldots,n\right\} ;\\k\neq j}}\left( x_{j}-x_{k}\right) } & =\left( x_{j}-x_{0}\right) \cdot\dfrac{f\left( x_{j}\right) }{\left( x_{j}-x_{0}\right) \cdot\prod\limits_{\substack{k\in \left\{ 1,2,\ldots,n\right\} ;\\k\neq j}}\left( x_{j}-x_{k}\right) }\nonumber\\ & =\dfrac{f\left( x_{j}\right) }{\prod\limits_{\substack{k\in\left\{ 1,2,\ldots,n\right\} ;\\k\neq j}}\left( x_{j}-x_{k}\right) } . \label{darij1.pf.prop1.5} \tag{7} \end{align}
Forget that we fixed $j$. We thus have proven \eqref{darij1.pf.prop1.5} for each $j\in\left\{ 1,2,\ldots,n\right\} $. Now, \eqref{darij1.eq.2} becomes \begin{align} f\left[ x_{1},x_{2},\ldots,x_{n}\right] & =\sum_{j=1}^{n} \underbrace{\dfrac{f\left( x_{j}\right) }{\prod\limits_{\substack{k\in \left\{ 1,2,\ldots,n\right\} ;\\k\neq j}}\left( x_{j}-x_{k}\right) } }_{\substack{=\left( x_{j}-x_{0}\right) \cdot\dfrac{f\left( x_{j}\right) }{\prod\limits_{\substack{k\in\left\{ 0,1,\ldots,n\right\} ;\\k\neq j}}\left( x_{j}-x_{k}\right) }\\\text{(by \eqref{darij1.pf.prop1.5})}}}\\ & =\sum_{j=1}^{n}\left( x_{j}-x_{0}\right) \cdot\dfrac{f\left( x_{j}\right) }{\prod\limits_{\substack{k\in\left\{ 0,1,\ldots,n\right\} ;\\k\neq j}}\left( x_{j}-x_{k}\right) }. \end{align} Comparing this with \begin{align*} & \sum_{j=0}^{n}\left( x_{j}-x_{0}\right) \cdot\dfrac{f\left( x_{j}\right) }{\prod\limits_{\substack{k\in\left\{ 0,1,\ldots,n\right\} ;\\k\neq j}}\left( x_{j}-x_{k}\right) }\\ & =\sum_{j=1}^{n}\left( x_{j}-x_{0}\right) \cdot\dfrac{f\left( x_{j}\right) }{\prod\limits_{\substack{k\in\left\{ 0,1,\ldots,n\right\} ;\\k\neq j}}\left( x_{j}-x_{k}\right) }+\underbrace{\left( x_{0} -x_{0}\right) }_{=0}\cdot\dfrac{f\left( x_{0}\right) }{\prod \limits_{\substack{k\in\left\{ 0,1,\ldots,n\right\} ;\\k\neq0}}\left( x_{0}-x_{k}\right) }\\ & \qquad\left( \text{here, we have split off the addend for }j=0\text{ from the sum}\right) \\ & =\sum_{j=1}^{n}\left( x_{j}-x_{0}\right) \cdot\dfrac{f\left( x_{j}\right) }{\prod\limits_{\substack{k\in\left\{ 0,1,\ldots,n\right\} ;\\k\neq j}}\left( x_{j}-x_{k}\right) }, \end{align*} we obtain \begin{equation} f\left[ x_{1},x_{2},\ldots,x_{n}\right] =\sum_{j=0}^{n}\left( x_{j} -x_{0}\right) \cdot\dfrac{f\left( x_{j}\right) }{\prod \limits_{\substack{k\in\left\{ 0,1,\ldots,n\right\} ;\\k\neq j}}\left( x_{j}-x_{k}\right) }. \end{equation} Subtracting this equality from \eqref{darij1.pf.prop1.x0xn-1}, we obtain \begin{align*} & f\left[ x_{0},x_{1},\ldots,x_{n-1}\right] -f\left[ x_{1},x_{2} ,\ldots,x_{n}\right] \\ & =\sum_{j=0}^{n}\left( x_{j}-x_{n}\right) \cdot\dfrac{f\left( x_{j}\right) }{\prod\limits_{\substack{k\in\left\{ 0,1,\ldots,n\right\} ;\\k\neq j}}\left( x_{j}-x_{k}\right) }-\sum_{j=0}^{n}\left( x_{j} -x_{0}\right) \cdot\dfrac{f\left( x_{j}\right) }{\prod \limits_{\substack{k\in\left\{ 0,1,\ldots,n\right\} ;\\k\neq j}}\left( x_{j}-x_{k}\right) }\\ & =\sum_{j=0}^{n}\underbrace{\left( \left( x_{j}-x_{n}\right) -\left( x_{j}-x_{0}\right) \right) }_{=x_{0}-x_{n}}\cdot\dfrac{f\left( x_{j}\right) }{\prod\limits_{\substack{k\in\left\{ 0,1,\ldots,n\right\} ;\\k\neq j}}\left( x_{j}-x_{k}\right) }\\ & =\sum_{j=0}^{n}\left( x_{0}-x_{n}\right) \cdot\dfrac{f\left( x_{j}\right) }{\prod\limits_{\substack{k\in\left\{ 0,1,\ldots,n\right\} ;\\k\neq j}}\left( x_{j}-x_{k}\right) }\\ & =\left( x_{0}-x_{n}\right) \cdot\underbrace{\sum_{j=0}^{n}\dfrac{f\left( x_{j}\right) }{\prod\limits_{\substack{k\in\left\{ 0,1,\ldots,n\right\} ;\\k\neq j}}\left( x_{j}-x_{k}\right) }}_{\substack{=f\left[ x_{0} ,x_{1},\ldots,x_{n}\right] \\\text{(by \eqref{darij1.eq.def-fbra})}}}\\ & =\left( x_{0}-x_{n}\right) \cdot f\left[ x_{0},x_{1},\ldots,x_{n}\right] . \end{align*} We can divide both sides of this equality by $x_{0}-x_{n}$ (since $x_{0} -x_{n}\neq0$), and thus obtain \begin{equation} \dfrac{f\left[ x_{0},x_{1},\ldots,x_{n-1}\right] -f\left[ x_{1} ,x_{2},\ldots,x_{n}\right] }{x_{0}-x_{n}}=f\left[ x_{0},x_{1},\ldots ,x_{n}\right] . \end{equation} This proves Proposition 1. $\blacksquare$