What does the term $c_x$ mean in the theorem of Taylor's remainder?

Question

Theorem (Taylor's remainder) Assume that $f(x)$ has $n+1$ continuous derivatives on an interval $\alpha\leq x\leq\beta$, and let the point $a$ belong to that interval. For the Taylor polynomial $p_n(x)$, let $R_n(x) \equiv f(x)-p_n(x)$ denote the remainder or error in approximating $f(x)$ by $p_n(x)$. Then $$R_n(x) = \frac{(x-a)^{n+1}}{(x+1)!}f^{(n+1)}(c_x),\text{ }\alpha\leq x\leq\beta$$ with $c_x$ an unknown point between $a$ and $x$.

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Taylor polynomial: $$p_n(x) = \sum_{j=0}^n\frac{(x-a)^j}{j!}f^{(j)}(a)$$

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1. Why is there an unknown point $c_x$ in the theorem above and what does $c_x$ mean?
2. Why do we know that $c_x$ is a point between $a$ and $x$?

Answer 1

Maybe the best way to see where $c_x$ "comes from", is to do the proof:

Take $x_0=0$ for convenience, suppose $f:\mathbb R\to \mathbb R$ has (at least) $n+1$ derivatives at $0$ and let $p(x)=\sum^n_{k=0}\frac{f^{(k)}(0)}{k!}x^k$ be the Taylor (well, MacLaurin) polynomial for $f$ of dgree $n$.

The idea is that $p$ should be a good approximation to $f$ in some interval $I$ containing $0$. That is, $f(x)-p(x)$ should be small in this interval. We want to see how small it is.

So, fix $b\in I$. We will approximate $f(b)$ using $p$. The trick is to choose $K$ so that

$f(b)-p(b)-Kb^{n+1}=0$, and set $T(x)=f(x)-p(x)-Kx^{n+1}$.

The first thing to notice is that $T^{(k)}(0)=0$ for all $0\le k\le n$. That is, the first $n$ derivatives of $T$ at $0$ are all equal to $0$.

Then, since by construction, $T(b)=f(b)-p(b)-Kb^{n+1}=0$, we have $T(b)=T(0)=0$ and Rolle's theorem applies to give a $0<c_1<b$ such that $T'(c_1)=0$. But we also have $T'(0)=0$ so Rolle applies again, and we get a $0<c_2<c_1<b$ such that $T''(c_2)=0.$

By now we see what is happening: we are getting a sequence of numbers $0<c_k<c_{k-1}<b$ with the property that $T^{(k)}(c_k)=0.$ This process continues until at the $n^{th}$ step, we get a $0<c_{n+1}<\cdots<b$ such that $T^{n+1}(c_{n+1})=0.$ Then, since $T^{n+1}(x)=f^{n+1}(x)-0-K(n+1)!$, we get $0=f^{n+1}(c_{n+1})-0-K(n+1)!$, and therefore

$K=\frac{f^{n+1}(c_{n+1})}{(n+1)!}$.

The upshot of all this is that we have now, since $T(b)=0,$

$f(b)=p(b)+\frac{f^{n+1}(c_n)}{(n+1)!}b^n=\sum^n_{k=0}\frac{f^{(k)}(0)}{k!}b^k+\frac{f^{n+1}(c_{n+1})}{(n+1)!}b^{n+1}.$

So, by getting $K$ in terms of the $(n+1)^{th}$ derivative of $f$ at a point in $I$, we have found a bound on the error we make by replacing $f$ by $p$. And we also know where $c_{n+1}$ comes from: it is the last point we get from $n$ repetitions of Rolle's theorem.

Answer 2

It means that there is some point between $a$ and $x$ for which the equation is true. This is exactly like the unknown point in the mean value theorem. In fact, the mean value theorem is just the $n=0$ case of Taylor's theorem.

The application of this is that if we know bounds on $f^{(n+1)}$ between $a$ and $x$ then we can use these to get bound on the value of $f(x).$ That is if we know $m\leq|f^{(n+1)}(x)|\leq M,$ then we can bound the error in the Taylor polynomial approximation.