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Theorem (Taylor's remainder) Assume that $f(x)$ has $n+1$ continuous derivatives on an interval $alphaleq xleqbeta$, 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
$$R_n(x) = frac(x-a)^n+1(x+1)!f^(n+1)(c_x),text alphaleq xleqbeta$$
with $c_x$ an unknown point between $a$ and $x$.

Taylor polynomial:
$$p_n(x) = sum_j=0^nfrac(x-a)^jj!f^(j)(a)$$

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 $x$?

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

Take $x_0=0$ for convenience, suppose $f:mathbb Rto mathbb R$ has (at least) $n+1$ derivatives at $0$ and let $p(x)=sum^n_k=0fracf^(k)(0)k!x^k$ 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 $bin I$. 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 $0le kle n$. 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

$K=fracf^n+1(c_n+1)(n+1)!$.

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

$f(b)=p(b)+fracf^n+1(c_n)(n+1)!b^n=sum^n_k=0fracf^(k)(0)k!b^k+fracf^n+1(c_n+1)(n+1)!b^n+1.$

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.

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 $mleq|f^(n+1)(x)|leq M,$ then we can bound the error in the Taylor polynomial approximation.