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In a solution to a certain Rolle's theorem problem, I have $x_k=a+\frac{k}{n}(b-a)$ and $$ 0=f(b)-f(a)=\sum_{k=1}^nf(x_k)-f(x_{k-1})=\frac{b-a}n\sum_{k=1}^n\frac{f(x_k)-f(x_{k-1})}{x_k-x_{k-1}} $$

How exactly does this work? I mean, let's say $$\sum_{k=1}^nf(x_k)-f(x_{k-1})$$ has $$f(x_2)-f(x_{2-1})$$ as one of it's terms. How will $x_k=a+\frac{k}{n}(b-a)$ be used in this? Also, how we define $n$ there? Does it mean that $k$ is divided by $n$? Is $n$ supposed to be some large finite number?

Excuse me if the question sounds silly.

pikarin-g
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    The question does not make much sense. The second line is not valid in general. What exactly is the question? Are you supposed to find $f$ so that line 2 is valid? – Kavi Rama Murthy Aug 09 '19 at 23:22
  • it is solution of certain rolle's theorem problem, it supposed have multiple $c$ with $c1<c2..c even number$ and the total sum of all $c$ is equal to zero – pikarin-g Aug 09 '19 at 23:25
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    Please try to make the titles of your questions more informative. For example, Why does $a<b$ imply $a+c<b+c$? is much more useful for other users than A question about inequality. From How can I ask a good question?: Make your title as descriptive as possible. In many cases one can actually phrase the title as the question, at least in such a way so as to be comprehensible to an expert reader. You can find more tips for choosing a good title here. – Shaun Aug 09 '19 at 23:27
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    It is not clear what you want. Do you want to find a function that satisfies the first equation? – NoChance Aug 09 '19 at 23:54
  • Question is incomprehensible. Please think it through, decide precisely what it is you want to know, and express your needs carefully. – Gerry Myerson Aug 10 '19 at 00:05
  • @pikarin-g I edited your question text to add the context you provided in your first comment, i.e., it comes from the solution to a certain Rolle's theorem problem, and did some rewording. I tried to keep my changes to a fairly minimum amount to help ensure I didn't change your meaning, but also hopefully enough so the question is fairly clear to anybody reading it. Please read what it says now and, if I accidentally misrepresented anything you're asking about, please accept my apology and make any appropriate changes to correct it. Thanks. – John Omielan Aug 10 '19 at 00:21
  • @pikarin-g One thing I didn't change, though, is your question title. Please change that yourself to make it more descriptive of what you're asking in the question. Thanks. – John Omielan Aug 10 '19 at 00:22

2 Answers2

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You're given that

$$x_k=a+\frac{k}{n}(b-a) \tag{1}\label{eq1}$$

Also, it's stated that

$$0=f(b)-f(a)=\sum_{k=1}^nf(x_k)-f(x_{k-1})=\frac{b-a}n\sum_{k=1}^n\frac{f(x_k)-f(x_{k-1})}{x_k-x_{k-1}} \tag{2}\label{eq2}$$

It seems you're asking about how & why \eqref{eq2} works. Well, you're dealing with a function $f(x)$, with $x \in [a,b]$ and $f(a) = f(b)$. The interval $[a,b]$ is divided into $n$ equal parts, for some $n \in \mathbb{N}$, with the $k$'th point ($0 \le k \le n$) being given by \eqref{eq1}. Note that

$$\begin{equation}\begin{aligned} \sum_{k=1}^n \left(f(x_k)-f(x_{k-1})\right) & = (f(x_1) - f(x_0)) + (f(x_2) - f(x_1)) + \ldots + (f(x_n) - f(x_{n-1})) \\ & = f(x_n) - f(x_0) \\ & = f(b) - f(a) \end{aligned}\end{equation}\tag{3}\label{eq3}$$

Note this is a Telescoping series, so all of the terms except for $f(x_0)$ and $f(x_n)$ cancel each other. Next, from \eqref{eq1}, you have that

$$\begin{equation}\begin{aligned} x_k - x_{k-1} & = \left(a + \frac{k}{n}(b-a)\right) - \left(a + \frac{k - 1}{n}(b-a)\right) \\ & = \frac{b - a}{n} \end{aligned}\end{equation}\tag{4}\label{eq4}$$

Thus, the last part of \eqref{eq2}, i.e.,

$$\frac{b-a}n\sum_{k=1}^n\frac{f(x_k)-f(x_{k-1})}{x_k-x_{k-1}} \tag{5}\label{eq5}$$

is basically multiplying the middle part by the RHS of \eqref{eq4}, then dividing by the LHS of \eqref{eq4}, and moving that last factor into the summation (which you can do as it's just a constant).

John Omielan
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  • @pikarin-g You're welcome. As the question comments say, it's somewhat hard to understand what you're asking based on your wording. I'm glad I was able to determine you're mainly asking to understand how that second line of equations works and I was able to be of some help. – John Omielan Aug 10 '19 at 00:08
  • l'm truly sorry , up until know i'm only dealing with simple application of rolle theorem . so frankly i'm not sure how using right words for this question – pikarin-g Aug 10 '19 at 00:13
  • @pikarin-g No worries. As I commented above, I've edited your question text to try to make it more easily understood. One thing I also did, and you should do in the future, is start with where the question comes from, i.e., it being regarding a solution to a certain Rolle's theorem problem. Even just that bit of extra information would help provide the context to make it easier for people to understand what you're asking about, even if your wording is possibly somewhat confusing. – John Omielan Aug 10 '19 at 00:25
  • hello jon, once again thanks of your concern regarding my question. I'm truly appreciate it. one that left me unsure, just like I'd wrote in comment that this function should satisfy 2 conditions.First, is that the sum of all derivative of $c$ is equal with zero and you already explained this with telescoping theorem.Second $c1<c2<..<cn$ . How do i prove this later part ? bacause just like you said that the interval is splits $evenly$... – pikarin-g Aug 10 '19 at 08:02
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    @pikarin-g I only just noticed you wrote the comment because I chose to check on this question again. If you wish for me to be notified, please use @ followed by my name. As for your question, by $c1 \lt c2 \lt \ldots \lt cn$, I assume the $c_i$ are the $x_k$ in your original equation. Well, you can see from my eq. (4) the difference between each one is $\frac{b-a}{n}$. Since $b \gt a$ and $n \gt 0$, this shows $x_k \gt x_{k-1}$, i.e., $x_{k-1} \lt x_k$. Also, as I stated & you said as well, the distance between each $x_{k-1}$ and $x_k$ is a constant, i.e., the interval is split evenly. – John Omielan Aug 10 '19 at 08:49
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For those inexperienced in $\Sigma$ notation, try writing a particular example. Say $n=4$. Then $x_0 = a$ and $x_4 = b$.
So $$ \sum_{k=1}^n \big(f(x_k)-f(x_{k-1}\big) \\= \big(f(x_1)-f(x_0)\big)+\big(f(x_2)-f(x_1)\big)+\big(f(x_3)-f(x_2)\big)+\big(f(x_4)-f(x_3)\big) \\ = f(x_4)-f(x_0) = f(b)-f(a). $$

GEdgar
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