# Taylor formula remainder: various expressions

Let $f:\mathbb{R}\rightarrow\mathbb{R}$ be an $n+1$ times differentiable function, and let $T_{n,a}(x)$ its $n^{th\text{ }}$-degree Taylor polynomial;

Then the following expressions for the remainder $R_{n,a}(x)=f(x)-T_{n,a}(x)$ hold:

1)

 $R_{n,a}(x)=\frac{1}{n!}\int_{a}^{x}f^{(n+1)}(t)(x-t)^{n}dt$

(Integral form)

2)

 $R_{n,a}(x)=\frac{f^{(n+1)}(\eta)}{n!p}(x-\eta)^{n-p+1}(x-a)^{p}$

for a $\eta(x)\in(a,x)$ and $\forall p>0$ (Schlömilch form)

3)

 $R_{n,a}(x)=\frac{f^{(n+1)}(\xi)}{n!}(x-\xi)^{n}(x-a)$

for a $\xi(x)\in(a,x)$  (Cauchy form)

4)

 $R_{n,a}(x)=\frac{f^{(n+1)}(\vartheta)}{(n+1)!}(x-a)^{n+1}$

for a $\vartheta(x)\in(a,x)$  (Lagrange form)

Moreover the following result holds for the integral of the remainder from the center point  $a$ to an arbitrary point $b$:

5)

 $\int_{a}^{b}R_{n,a}(x)dx=\int_{a}^{b}\frac{f^{(n+1)}(x)}{(n+1)!}(b-x)^{n+1}dx$

Proof:

1) Let’s proceed by induction.

$n=0.$ $\int_{a}^{x}f^{{}^{\prime}}(t)dt=f(x)-f(a)=R_{0,a}(x)$, since $T_{0,a}(x)=$ $f(a)$.

Let’s take it for true that $R_{n-1,a}(x)=\frac{1}{(n-1)!}\int_{a}^{x}f^{(n)}(t)(x-t)^{n-1}dt$, and let’s compute $\int_{a}^{x}\frac{f^{(n+1)}(t)}{n!}(x-t)^{n}dt$ by parts.

 $\displaystyle\int_{a}^{x}\frac{f^{(n+1)}(t)}{n!}(x-t)^{n}dt=$ $\displaystyle=$ $\displaystyle\frac{f^{(n)}(t)}{n!}(x-t)^{n}|_{a}^{x}+\int_{x}^{a}\frac{f^{(n)}% (t)}{n!}n(x-t)^{n-1}dt$ $\displaystyle=$ $\displaystyle-\frac{f^{(n)}(a)}{n!}(x-a)^{n}+\int_{x}^{a}\frac{f^{(n)}(t)}{(n-% 1)!}(x-t)^{n-1}dt$ $\displaystyle=$ $\displaystyle R_{n-1,a}(x)-\frac{f^{(n)}(a)}{n!}(x-a)^{n}$ $\displaystyle=$ $\displaystyle f(x)-T_{n-1,a}-\frac{f^{(n)}(a)}{n!}(x-a)^{n}$ $\displaystyle=$ $\displaystyle f(x)-T_{n,a}(x)=R_{n,a}(x).$

2) Let’s write the remainder in the integral form this way:

 $R_{n,a}(x)=\frac{1}{n!}\int_{a}^{x}f^{(n+1)}(t)(x-t)^{n}dt=\frac{1}{n!}\int_{a% }^{x}f^{(n+1)}(t)(x-t)^{n-p+1}(x-t)^{p-1}dt$

Now, since $(x-t)^{p-1}$ doesn’t change sign between $a$ and $x$, we can apply the integral Mean Value theorem (http://planetmath.org/IntegralMeanValueTheorem). So a point $\eta(x)\in(a,x)$ exists such that

 $\displaystyle R_{n,a}(x)=\frac{1}{n!}\int_{a}^{x}f^{(n+1)}(t)(x-t)^{n-p+1}(x-t% )^{p-1}dt$ $\displaystyle=$ $\displaystyle\frac{f^{(n+1)}(\eta)}{n!}(x-\eta)^{n-p+1}\int_{a}^{x}(x-t)^{p-1}dt$ $\displaystyle=$ $\displaystyle\frac{f^{(n+1)}(\eta)}{n!p}(x-\eta)^{n-p+1}(x-a)^{p}$

(Note that the condition $p>0$ is needed to ensure convergence of the integral)

3) and 4) are obtained from Schlömilch form by plugging in $p=1$ and $p=n+1$ respectively.

5) Let’s start from the right-end side:

 $\displaystyle\int_{a}^{b}\frac{f^{(n+1)}(x)}{(n+1)!}(b-x)^{n+1}dx=$ $\displaystyle=$ $\displaystyle\frac{f^{(n)}(x)}{(n+1)!}(b-x)^{n+1}|_{a}^{b}+\int_{a}^{b}\frac{f% ^{(n)}(x)}{(n+1)!}(n+1)(b-x)^{n}dx$ $\displaystyle=$ $\displaystyle-\frac{f^{(n)}(a)}{(n+1)!}(b-a)^{n+1}+\int_{a}^{b}\frac{f^{(n)}(x% )}{n!}(b-x)^{n}dx$ $\displaystyle=$ $\displaystyle...=-\sum_{k=0}^{n}\frac{f^{(k)}(a)}{(k+1)!}(b-a)^{k+1}+\int_{a}^% {b}f(x)dx$ $\displaystyle=$ $\displaystyle-\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}\int_{a}^{b}(x-a)^{k}dx+\int_% {a}^{b}f(x)dx$ $\displaystyle=$ $\displaystyle\int_{a}^{b}\left(-\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^{k}+f% (x)\right)dx=\int_{a}^{b}R_{n,a}(x)dx.$

Note:

1) The proof of the integral form could also be stated as follow:

Let

$\phi(t)=\sum_{k=0}^{n}\frac{f^{(k)}(t)}{k!}(x-t)^{k}$

Then $\phi(x)=f(x)$ and $\phi(a)=T_{n,a}(x)$, so that $R_{n,a}(x)=\phi(x)-\phi(a)=\int_{a}^{x}\phi^{\prime}(t)dt.$

Let’s now compute $\phi^{\prime}(t).$

 $\displaystyle\phi^{\prime}(t)=\sum_{k=0}^{n}\frac{1}{k!}\left[f^{(k+1)}(t)(x-t% )^{k}-f^{(k)}(t)k(x-t)^{k-1}\right]$ $\displaystyle=$ $\displaystyle\sum_{k=0}^{n}\frac{f^{(k+1)}(t)}{k!}(x-t)^{k}-\sum_{k=1}^{n}% \frac{f^{(k)}(t)}{(k-1)!}(x-t)^{k-1}$ $\displaystyle=$ $\displaystyle\sum_{k=0}^{n}\frac{f^{(k+1)}(t)}{k!}(x-t)^{k}-\sum_{k=0}^{n-1}% \frac{f^{(k+1)}(t)}{k!}(x-t)^{k}$ $\displaystyle=$ $\displaystyle\frac{f^{(n+1)}(t)}{n!}(x-t)^{n}.$

2) From the integral form of the remainder it is possible to obtain the entire Taylor formula; indeed, repeatly integrating by parts, one gets:

 $\displaystyle\int_{a}^{x}\frac{f^{(n+1)}(t)}{n!}(x-t)^{n}dt=\frac{f^{(n)}(t)}{% n!}(x-t)^{n}|_{a}^{x}+\int_{a}^{x}\frac{f^{(n)}(t)}{n!}n(x-t)^{n-1}dt$ $\displaystyle=$ $\displaystyle-\frac{f^{(n)}(a)}{n!}(x-a)^{n}+\int_{a}^{x}\frac{f^{(n)}(t)}{(n-% 1)!}(x-t)^{n-1}dt$ $\displaystyle=$ $\displaystyle-\frac{f^{(n)}(a)}{n!}(x-a)^{n}-\frac{f^{(n-1)}(a)}{(n-1)!}(x-a)^% {n-1}+\int_{a}^{x}\frac{f^{(n-1)}(t)}{(n-2)!}(x-t)^{n-2}dt$ $\displaystyle=$ $\displaystyle...=-\frac{f^{(n)}(a)}{n!}(x-a)^{n}-\frac{f^{(n-1)}(a)}{(n-1)!}(x% -a)^{n-1}-...-\frac{f^{(n-k+1)}(a)}{(n-k+1)!}(x-a)^{n-k+1}+\int_{a}^{x}\frac{f% ^{(n-k+1)}(t)}{(n-k)!}(x-t)^{n-k}dt$ $\displaystyle...$ $\displaystyle=$ $\displaystyle-\sum_{k=1}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^{k}+\int_{a}^{x}f^{% \prime}(t)dt$ $\displaystyle=$ $\displaystyle-\sum_{k=1}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^{k}+f(x)-f(a)$ $\displaystyle=$ $\displaystyle-\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^{k}+f(x).$

that is

 $\displaystyle f(x)$ $\displaystyle=$ $\displaystyle\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^{k}+\int_{a}^{x}\frac{f^% {(n+1)}(t)}{n!}(x-t)^{n}dt$ $\displaystyle=$ $\displaystyle\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^{k}+R_{n,a}(x).$
Title Taylor formula remainder: various expressions TaylorFormulaRemainderVariousExpressions 2013-03-22 15:53:57 2013-03-22 15:53:57 gufotta (12050) gufotta (12050) 19 gufotta (12050) Result msc 41A58