Taylor formula remainder: various expressions

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

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

1)

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

(Integralform)

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 fromthe 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 ${\int }_a^x\frac{f^{(n+1)}(t)}{n!}{(x-t)}^n𝑑t$ by parts.

${\displaystyle {\int }_a^x}{\displaystyle \frac{f^{(n+1)}(t)}{n!}}{(x-t)}^n𝑑t=$

$=$${{\displaystyle \frac{f^{(n)}(t)}{n!}}{(x-t)}^n|}_a^x+{\displaystyle {\int }_x^a}{\displaystyle \frac{f^{(n)}(t)}{n!}}n{(x-t)}^{n-1}𝑑t$

$=$$-{\displaystyle \frac{f^{(n)}(a)}{n!}}{(x-a)}^n+{\displaystyle {\int }_x^a}{\displaystyle \frac{f^{(n)}(t)}{(n-1)!}}{(x-t)}^{n-1}𝑑t$

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

$=$$f(x)-T_{n-1,a}-{\displaystyle \frac{f^{(n)}(a)}{n!}}{(x-a)}^n$

$=$$f(x)-T_{n,a}(x)=R_{n,a}(x).$

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

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

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

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

$=$${\displaystyle \frac{f^{(n+1)}(\eta )}{n!}}{(x-\eta )}^{n-p+1}{\displaystyle {\int }_a^x}{(x-t)}^{p-1}𝑑t$

$=$${\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 theintegral)

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:

${\displaystyle {\int }_a^b}{\displaystyle \frac{f^{(n+1)}(x)}{(n+1)!}}{(b-x)}^{n+1}𝑑x=$

$=$${{\displaystyle \frac{f^{(n)}(x)}{(n+1)!}}{(b-x)}^{n+1}|}_a^b+{\displaystyle {\int }_a^b}{\displaystyle \frac{f^{(n)}(x)}{(n+1)!}}(n+1){(b-x)}^n𝑑x$

$=$$-{\displaystyle \frac{f^{(n)}(a)}{(n+1)!}}{(b-a)}^{n+1}+{\displaystyle {\int }_a^b}{\displaystyle \frac{f^{(n)}(x)}{n!}}{(b-x)}^n𝑑x$

$=$$\mathrm{\dots }=-{\displaystyle \sum _{k=0}^n}{\displaystyle \frac{f^{(k)}(a)}{(k+1)!}}{(b-a)}^{k+1}+{\displaystyle {\int }_a^b}f(x)𝑑x$

$=$$-{\displaystyle \sum _{k=0}^n}{\displaystyle \frac{f^{(k)}(a)}{k!}}{\displaystyle {\int }_a^b}{(x-a)}^k𝑑x+{\displaystyle {\int }_a^b}f(x)𝑑x$

$=$${\displaystyle {\int }_a^b}\left(-{\displaystyle \sum _{k=0}^n}{\displaystyle \frac{f^{(k)}(a)}{k!}}{(x-a)}^k+f(x)\right)𝑑x={\displaystyle {\int }_a^b}{R}_{n,a}(x)𝑑x.$

Note:

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

Let

$\varphi (t)={\sum }_{k=0}^n\frac{f^{(k)}(t)}{k!}{(x-t)}^k$

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

Let’s now compute ${\varphi }^{\prime }(t).$

${\varphi }^{\prime }(t)={\displaystyle \sum _{k=0}^n}{\displaystyle \frac{1}{k!}}\left[f^{(k+1)}(t){(x-t)}^k-f^{(k)}(t)k{(x-t)}^{k-1}\right]$

$=$${\displaystyle \sum _{k=0}^n}{\displaystyle \frac{f^{(k+1)}(t)}{k!}}{(x-t)}^k-{\displaystyle \sum _{k=1}^n}{\displaystyle \frac{f^{(k)}(t)}{(k-1)!}}{(x-t)}^{k-1}$

$=$${\displaystyle \sum _{k=0}^n}{\displaystyle \frac{f^{(k+1)}(t)}{k!}}{(x-t)}^k-{\displaystyle \sum _{k=0}^{n-1}}{\displaystyle \frac{f^{(k+1)}(t)}{k!}}{(x-t)}^k$

$=$${\displaystyle \frac{f^{(n+1)}(t)}{n!}}{(x-t)}^n.$

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

${\displaystyle {\int }_a^x}{\displaystyle \frac{f^{(n+1)}(t)}{n!}}{(x-t)}^n𝑑t={{\displaystyle \frac{f^{(n)}(t)}{n!}}{(x-t)}^n|}_a^x+{\displaystyle {\int }_a^x}{\displaystyle \frac{f^{(n)}(t)}{n!}}n{(x-t)}^{n-1}𝑑t$

$=$$-{\displaystyle \frac{f^{(n)}(a)}{n!}}{(x-a)}^n+{\displaystyle {\int }_a^x}{\displaystyle \frac{f^{(n)}(t)}{(n-1)!}}{(x-t)}^{n-1}𝑑t$

$=$$-{\displaystyle \frac{f^{(n)}(a)}{n!}}{(x-a)}^n-{\displaystyle \frac{f^{(n-1)}(a)}{(n-1)!}}{(x-a)}^{n-1}+{\displaystyle {\int }_a^x}{\displaystyle \frac{f^{(n-1)}(t)}{(n-2)!}}{(x-t)}^{n-2}𝑑t$

$=$$\mathrm{\dots }=-{\displaystyle \frac{f^{(n)}(a)}{n!}}{(x-a)}^n-{\displaystyle \frac{f^{(n-1)}(a)}{(n-1)!}}{(x-a)}^{n-1}-\mathrm{\dots }-{\displaystyle \frac{f^{(n-k+1)}(a)}{(n-k+1)!}}{(x-a)}^{n-k+1}+{\displaystyle {\int }_a^x}{\displaystyle \frac{f^{(n-k+1)}(t)}{(n-k)!}}{(x-t)}^{n-k}𝑑t$

$\mathrm{\dots }$

$=$$-{\displaystyle \sum _{k=1}^n}{\displaystyle \frac{f^{(k)}(a)}{k!}}{(x-a)}^k+{\displaystyle {\int }_a^x}{f}^{\prime }(t)𝑑t$

$=$$-{\displaystyle \sum _{k=1}^n}{\displaystyle \frac{f^{(k)}(a)}{k!}}{(x-a)}^k+f(x)-f(a)$

$=$$-{\displaystyle \sum _{k=0}^n}{\displaystyle \frac{f^{(k)}(a)}{k!}}{(x-a)}^k+f(x).$

that is

$f(x)$$=$${\displaystyle \sum _{k=0}^n}{\displaystyle \frac{f^{(k)}(a)}{k!}}{(x-a)}^k+{\displaystyle {\int }_a^x}{\displaystyle \frac{f^{(n+1)}(t)}{n!}}{(x-t)}^n𝑑t$

$=$${\displaystyle \sum _{k=0}^n}{\displaystyle \frac{f^{(k)}(a)}{k!}}{(x-a)}^k+R_{n,a}(x).$