Incalculus,Taylor's theorem gives an approximation of a$k$-timesdifferentiable function around a given point by apolynomial of degree$k$, called the$k$-th-orderTaylor polynomial. For asmooth function, the Taylor polynomial is the truncation at the order$k$ of theTaylor series of the function. The first-order Taylor polynomial is thelinear approximation of the function, and the second-order Taylor polynomial is often referred to as thequadratic approximation.^[1] There are several versions of Taylor's theorem, some giving explicit estimates of the approximation error of the function by its Taylor polynomial.

Taylor's theorem is named after the mathematicianBrook Taylor, who stated a version of it in 1715,^[2] although an earlier version of the result was already mentioned in1671 byJames Gregory.^[3]

Taylor's theorem is taught in introductory-level calculus courses and is one of the central elementary tools inmathematical analysis. It gives simple arithmetic formulas to accurately compute values of manytranscendental functions such as theexponential function andtrigonometric functions.It is the starting point of the study ofanalytic functions, and is fundamental in various areas of mathematics, as well as innumerical analysis andmathematical physics. Taylor's theorem also generalizes tomultivariate andvector valued functions. It provided the mathematical basis for some landmark early computing machines:Charles Babbage'sDifference Engine calculated sines, cosines, logarithms, and other transcendental functions by numerically integrating the first 7 terms of their Taylor series.

If a real-valuedfunction$f(x)$  isdifferentiable at the point$x=a$ , then it has alinear approximation near this point. This means that there exists a functionh₁(x) such that

$${\displaystyle f(x)=f(a)+f^{\prime }(a)(x-a)+h_1(x)(x-a),\underset{x\to a}{lim}{h}_1(x)=0.}$$ 

Here

$${\displaystyle P_1(x)=f(a)+f^{\prime }(a)(x-a)}$$ 

is the linear approximation of$f(x)$  forx near the pointa, whose graph$y=P_1(x)$  is thetangent line to the graph$y=f(x)$  atx =a. The error in the approximation is:$${\displaystyle R_1(x)=f(x)-P_1(x)=h_1(x)(x-a).}$$ 

Asx tends to a, this error goes to zero much faster than${\displaystyle (x-a)}$ , making${\displaystyle f(x)\approx P_1(x)}$  a useful approximation.

For a better approximation to$f(x)$ , we can fit aquadratic polynomial instead of a linear function:

$${\displaystyle P_2(x)=f(a)+f^{\prime }(a)(x-a)+\frac{f^{″}(a)}{2}(x-a{)}^2.}$$ 

Instead of just matching one derivative of$f(x)$  at$x=a$ , this polynomial has the same first and second derivatives, as is evident upon differentiation.

Taylor's theorem ensures that thequadratic approximation is, in a sufficiently small neighborhood of$x=a$ , more accurate than the linear approximation. Specifically,

$${\displaystyle f(x)=P_2(x)+h_2(x)(x-a{)}^2,\underset{x\to a}{lim}{h}_2(x)=0.}$$ 

Here the error in the approximation is

$${\displaystyle R_2(x)=f(x)-P_2(x)=h_2(x)(x-a{)}^2,}$$ 

which, given the limiting behavior of${\displaystyle h_2}$ , goes to zero faster than${\displaystyle (x-a{)}^2}$  asx tends to a.

Similarly, we might get still better approximations tof if we usepolynomials of higher degree, since then we can match even more derivatives withf at the selected base point.

In general, the error in approximating a function by a polynomial of degreek will go to zero much faster than${\displaystyle (x-a{)}^k}$  asx tends to a. However, there are functions, even infinitely differentiable ones, for which increasing the degree of the approximating polynomial does not increase the accuracy of approximation: we say such a function fails to beanalytic atx = a: it is not (locally) determined by its derivatives at this point.

Taylor's theorem is of asymptotic nature: it only tells us that the error$R_k$  in anapproximation by a$k$ -th order Taylor polynomialP_k tends to zero faster than any nonzero$k$ -th degreepolynomial as$x\to a$ . It does not tell us how large the error is in any concreteneighborhood of the center of expansion, but for this purpose there are explicit formulas for the remainder term (given below) which are valid under some additional regularity assumptions onf. These enhanced versions of Taylor's theorem typically lead touniform estimates for the approximation error in a small neighborhood of the center of expansion, but the estimates do not necessarily hold for neighborhoods which are too large, even if the functionf isanalytic. In that situation one may have to select several Taylor polynomials with different centers of expansion to have reliable Taylor-approximations of the original function (see animation on the right.)

There are several ways we might use the remainder term:

1. Estimate the error for a polynomialP_k(x) of degreek estimating$f(x)$  on a given interval (a –r,a +r). (Given the interval and degree, we find the error.)
2. Find the smallest degreek for which the polynomialP_k(x) approximates$f(x)$  to within a given error tolerance on a given interval (a −r,a +r) . (Given the interval and error tolerance, we find the degree.)
3. Find the largest interval (a −r,a +r) on whichP_k(x) approximates$f(x)$  to within a given error tolerance. (Given the degree and error tolerance, we find the interval.)

The precise statement of the most basic version of Taylor's theorem is as follows:

The polynomial appearing in Taylor's theorem is the$\mathit{k}$ -th order Taylor polynomial

$${\displaystyle P_k(x)=f(a)+f^{\prime }(a)(x-a)+\frac{f^{″}(a)}{2!}(x-a{)}^2+\cdots +\frac{f^{(k)}(a)}{k!}(x-a{)}^k}$$ 

of the functionf at the pointa. The Taylor polynomial is the unique "asymptotic best fit" polynomial in the sense that if there exists a functionh_k :R →R and a$k$ -th order polynomialp such that

$${\displaystyle f(x)=p(x)+h_k(x)(x-a{)}^k,\underset{x\to a}{lim}{h}_k(x)=0,}$$ 

thenp = P_k. Taylor's theorem describes the asymptotic behavior of theremainder term

$${\displaystyle R_k(x)=f(x)-P_k(x),}$$ 

which is theapproximation error when approximatingf with its Taylor polynomial. Using thelittle-o notation, the statement in Taylor's theorem reads as

$${\displaystyle R_k(x)=o(|x-a{|}^k),x\to a.}$$ 

Under stronger regularity assumptions onf there are several precise formulas for the remainder termR_k of the Taylor polynomial, the most common ones being the following.

These refinements of Taylor's theorem are usually proved using themean value theorem, whence the name. Additionally, notice that this is precisely themean value theorem when$k=0$ . Also other similar expressions can be found. For example, ifG(t) is continuous on the closed interval and differentiable with a non-vanishing derivative on the open interval between$a$  and$x$ , then

$${\displaystyle R_k(x)=\frac{f^{(k+1)}(\xi )}{k!}(x-\xi {)}^k\frac{G(x)-G(a)}{G^{\prime }(\xi )}}$$ 

for some number$\xi$  between$a$  and$x$ . This version covers the Lagrange and Cauchy forms of the remainder as special cases, and is proved below usingCauchy's mean value theorem. The Lagrange form is obtained by taking${\displaystyle G(t)=(x-t{)}^{k+1}}$  and the Cauchy form is obtained by taking${\displaystyle G(t)=t-a}$ .

The statement for the integral form of the remainder is more advanced than the previous ones, and requires understanding ofLebesgue integration theory for the full generality. However, it holds also in the sense ofRiemann integral provided the (k + 1)th derivative off is continuous on the closed interval [a,x].

Due to theabsolute continuity off^(k) on theclosed interval between$a$  and$x$ , its derivativef^(k+1) exists as anL¹-function, and the result can beproven by a formal calculation using thefundamental theorem of calculus andintegration by parts.

It is often useful in practice to be able to estimate the remainder term appearing in the Taylor approximation, rather than having an exact formula for it. Suppose thatf is(k + 1)-times continuously differentiable in an intervalI containinga. Suppose that there are real constantsq andQ such that

$${\displaystyle q\le f^{(k+1)}(x)\le Q}$$ 

throughoutI. Then the remainder term satisfies the inequality^[11]

$${\displaystyle q\frac{(x-a{)}^{k+1}}{(k+1)!}\le R_k(x)\le Q\frac{(x-a{)}^{k+1}}{(k+1)!},}$$ 

ifx >a, and a similar estimate ifx <a. This is a simple consequence of the Lagrange form of the remainder. In particular, if

$${\displaystyle |f^{(k+1)}(x)|\le M}$$ 

on an intervalI = (a −r,a +r) with some${\displaystyle r>0}$  , then

$${\displaystyle |R_k(x)|\le M\frac{|x-a{|}^{k+1}}{(k+1)!}\le M\frac{r^{k+1}}{(k+1)!}}$$ 

for allx∈(a −r,a +r). The second inequality is called auniform estimate, because it holds uniformly for allx on the interval(a −r,a +r).

Suppose that we wish to find the approximate value of the function$f(x)=e^x$  on the interval$[-1,1]$  while ensuring that the error in the approximation is no more than 10⁻⁵. In this example we pretend that we only know the following properties of the exponential function:

From these properties it follows that$f^{(k)}(x)=e^x$  for all$k$ , and in particular,$f^{(k)}(0)=1$ . Hence the$k$ -th order Taylor polynomial of$f$  at$0$  and its remainder term in the Lagrange form are given by

$${\displaystyle P_k(x)=1+x+\frac{x^2}{2!}+\cdots +\frac{x^k}{k!},R_k(x)=\frac{e^{\xi }}{(k+1)!}{x}^{k+1},}$$ 

where$\xi$  is some number between 0 andx. Sincee^x is increasing by (★), we can simply use$e^x\le 1$  for$x\in [-1,0]$  to estimate the remainder on the subinterval${\displaystyle [-1,0]}$ . To obtain an upper bound for the remainder on${\displaystyle [0,1]}$ , we use the property  for  to estimate

 

using the second order Taylor expansion. Then we solve fore^x to deduce that

$${\displaystyle e^x\le \frac{1+x}{1-\frac{x^2}{2}}=2\frac{1+x}{2-x^2}\le 4,0\le x\le 1}$$ 

simply by maximizing thenumerator and minimizing thedenominator. Combining these estimates fore^x we see that

$${\displaystyle |R_k(x)|\le \frac{4|x{|}^{k+1}}{(k+1)!}\le \frac{4}{(k+1)!},-1\le x\le 1,}$$ 

so the required precision is certainly reached, when

 

(Seefactorial or compute by hand the values$9!=362880$  and$10!=3628800$ .) As a conclusion, Taylor's theorem leads to the approximation

 

For instance, this approximation provides adecimal expression${\displaystyle e\approx 2.71828}$ , correct up to five decimal places.

LetI ⊂R be anopen interval. By definition, a functionf :I →R isreal analytic if it is locally defined by a convergentpower series. This means that for everya ∈ I there exists somer > 0 and a sequence of coefficientsc_k ∈ R such that(a −r,a +r) ⊂I and

 

In general, theradius of convergence of a power series can be computed from theCauchy–Hadamard formula

$${\displaystyle \frac{1}{R}=\underset{k\to \mathrm{\infty }}{lim sup}|c_k{|}^{\frac{1}{k}}.}$$ 

This result is based on comparison with ageometric series, and the same method shows that if the power series based ona converges for someb ∈R, it must convergeuniformly on theclosed interval$[a-r_b,a+r_b]$ , where$r_b=\left|b-a\right|$ . Here only the convergence of the power series is considered, and it might well be that(a −R,a +R) extends beyond the domainI off.

The Taylor polynomials of the real analytic functionf ata are simply the finite truncations

$${\displaystyle P_k(x)=\sum _{j=0}^k{c}_j(x-a{)}^j,c_j=\frac{f^{(j)}(a)}{j!}}$$ 

of its locally defining power series, and the corresponding remainder terms are locally given by the analytic functions

 

Here the functions

 

are also analytic, since their defining power series have the same radius of convergence as the original series. Assuming that[a −r,a +r] ⊂I andr < R, all these series converge uniformly on(a −r,a +r). Naturally, in the case of analytic functions one can estimate the remainder term$R_k(x)$  by the tail of the sequence of the derivativesf′(a) at the center of the expansion, but usingcomplex analysis also another possibility arises, which is describedbelow.

The Taylor series off will converge in some interval in which all its derivatives are bounded and do not grow too fast ask goes to infinity. (However, even if the Taylor series converges, it might not converge tof, as explained below;f is then said to be non-analytic.)

One might think of the Taylor series

$${\displaystyle f(x)\approx \sum _{k=0}^{\mathrm{\infty }}{c}_k(x-a{)}^k=c_0+c_1(x-a)+c_2(x-a{)}^2+\cdots }$$ 

of an infinitely many times differentiable functionf :R →R as its "infinite order Taylor polynomial" ata. Now theestimates for the remainder imply that if, for anyr, the derivatives off are known to be bounded over (a − r,a + r), then for any orderk and for anyr > 0 there exists a constantM_k,r > 0 such that

for everyx ∈ (a − r,a + r). Sometimes the constantsM_k,r can be chosen in such way thatM_k,r is bounded above, for fixedr and allk. Then the Taylor series offconverges uniformly to some analytic function

 

(One also gets convergence even ifM_k,r is not bounded above as long as it grows slowly enough.)

The limit functionT_f is by definition always analytic, but it is not necessarily equal to the original functionf, even iff is infinitely differentiable. In this case, we sayf is anon-analytic smooth function, for example aflat function:

 

Using thechain rule repeatedly bymathematical induction, one shows that for any order k,

 

for some polynomialp_k of degree 2(k − 1). The function${\displaystyle e^{-\frac{1}{x^2}}}$  tends to zero faster than any polynomial as$x\to 0$ , sof is infinitely many times differentiable andf^(k)(0) = 0 for every positive integerk. The above results all hold in this case:

• The Taylor series off converges uniformly to the zero functionT_f(x) = 0, which is analytic with all coefficients equal to zero.
• The functionf is unequal to this Taylor series, and hence non-analytic.
• For any orderk ∈ N and radiusr > 0 there existsM_k,r > 0 satisfying the remainder bound (★★) above.

However, ask increases for fixedr, the value ofM_k,r grows more quickly thanr^k, and the error does not go to zero.

Taylor's theorem generalizes to functionsf :C →C which arecomplex differentiable in an open subsetU ⊂ C of thecomplex plane. However, its usefulness is dwarfed by other general theorems incomplex analysis. Namely, stronger versions of related results can be deduced forcomplex differentiable functionsf : U → C usingCauchy's integral formula as follows.

Letr > 0 such that theclosed diskB(z, r) ∪ S(z, r) is contained inU. Then Cauchy's integral formula with a positive parametrizationγ(t) =z +re^it of the circleS(z,r) with${\displaystyle t\in [0,2\pi ]}$  gives

$${\displaystyle f(z)=\frac{1}{2\pi i}{\int }_{\gamma }\frac{f(w)}{w-z}dw,f^{\prime }(z)=\frac{1}{2\pi i}{\int }_{\gamma }\frac{f(w)}{(w-z{)}^2}dw,\dots ,f^{(k)}(z)=\frac{k!}{2\pi i}{\int }_{\gamma }\frac{f(w)}{(w-z{)}^{k+1}}dw.}$$ 

Here all the integrands are continuous on thecircleS(z, r), which justifies differentiation under the integral sign. In particular, iff is oncecomplex differentiable on the open setU, then it is actually infinitely many timescomplex differentiable onU. One also obtainsCauchy's estimate^[12]

$${\displaystyle |f^{(k)}(z)|\le \frac{k!}{2\pi }{\int }_{\gamma }\frac{M_r}{|w-z{|}^{k+1}}dw=\frac{k!M_r}{r^k},M_r=\underset{|w-c|=r}{max}|f(w)|}$$ 

for anyz ∈ U andr > 0 such thatB(z, r) ∪ S(c, r) ⊂ U. The estimate implies that thecomplexTaylor series

$${\displaystyle T_f(z)=\sum _{k=0}^{\mathrm{\infty }}\frac{f^{(k)}(c)}{k!}(z-c{)}^k}$$ 

off converges uniformly on anyopen disk$B(c,r)\subset U$  with$S(c,r)\subset U$  into some functionT_f. Furthermore, using thecontour integral formulas for the derivativesf^(k)(c),

 

so anycomplex differentiable functionf in an open setU ⊂ C is in factcomplex analytic. All that is said for real analytic functionshere holds also for complex analytic functions with the open intervalI replaced by an open subsetU ∈ C anda-centered intervals (a − r, a + r) replaced byc-centered disksB(c, r). In particular, the Taylor expansion holds in the form

$${\displaystyle f(z)=P_k(z)+R_k(z),P_k(z)=\sum _{j=0}^k\frac{f^{(j)}(c)}{j!}(z-c{)}^j,}$$ 

where the remainder termR_k is complex analytic. Methods of complex analysis provide some powerful results regarding Taylor expansions. For example, using Cauchy's integral formula for any positively orientedJordan curve$\gamma$  which parametrizes the boundary$\mathrm{\partial }W\subset U$  of a region$W\subset U$ , one obtains expressions for the derivativesf^(j)(c) as above, and modifying slightly the computation forT_f(z) =f(z), one arrives at the exact formula

$${\displaystyle R_k(z)=\sum _{j=k+1}^{\mathrm{\infty }}\frac{(z-c{)}^j}{2\pi i}{\int }_{\gamma }\frac{f(w)}{(w-c{)}^{j+1}}dw=\frac{(z-c{)}^{k+1}}{2\pi i}{\int }_{\gamma }\frac{f(w)dw}{(w-c{)}^{k+1}(w-z)},z\in W.}$$ 

The important feature here is that the quality of the approximation by a Taylor polynomial on the region$W\subset U$  is dominated by the values of the functionf itself on the boundary$\mathrm{\partial }W\subset U$ . Similarly, applying Cauchy's estimates to the series expression for the remainder, one obtains the uniform estimates

 

The function

 

isreal analytic, that is, locally determined by its Taylor series. This function was plottedabove to illustrate the fact that some elementary functions cannot be approximated by Taylor polynomials in neighborhoods of the center of expansion which are too large. This kind of behavior is easily understood in the framework of complex analysis. Namely, the functionf extends into ameromorphic function

 

on the compactified complex plane. It has simple poles at$z=i$  and$z=-i$ , and it is analytic elsewhere. Now its Taylor series centered atz₀ converges on any discB(z₀,r) withr < |z − z₀|, where the same Taylor series converges atz ∈ C. Therefore, Taylor series off centered at 0 converges onB(0, 1) and it does not converge for anyz ∈C with |z| > 1 due to the poles ati and −i. For the same reason the Taylor series off centered at 1 converges on$B(1,\sqrt{2})$  and does not converge for anyz ∈ C with$\left|z-1\right|>\sqrt{2}$ .

A functionf:Rⁿ →R isdifferentiable ata ∈Rⁿif and only if there exists alinear functionalL :Rⁿ →R and a functionh :Rⁿ →R such that

$${\displaystyle f(\mathit{x})=f(\mathit{a})+L(\mathit{x}-\mathit{a})+h(\mathit{x})\Vert \mathit{x}-\mathit{a}\Vert ,\underset{\mathit{x}\to \mathit{a}}{lim}h(\mathit{x})=0.}$$ 

If this is the case, then$L=df(\mathit{a})$  is the (uniquely defined)differential off at the pointa. Furthermore, then thepartial derivatives off exist ata and the differential off ata is given by

$${\displaystyle df(\mathit{a})(\mathit{v})=\frac{\mathrm{\partial }f}{\mathrm{\partial }{x}_1}(\mathit{a})v_1+\cdots +\frac{\mathrm{\partial }f}{\mathrm{\partial }{x}_n}(\mathit{a})v_n.}$$ 

Introduce themulti-index notation

$${\displaystyle |\alpha |={\alpha }_1+\cdots +{\alpha }_n,\alpha !={\alpha }_1!\cdots {\alpha }_n!,{\mathit{x}}^{\alpha }=x_1^{{\alpha }_1}\cdots x_n^{{\alpha }_n}}$$ 

forα ∈Nⁿ andx ∈Rⁿ. If all the$k$ -th orderpartial derivatives off :Rⁿ →R are continuous ata ∈Rⁿ, then byClairaut's theorem, one can change the order of mixed derivatives ata, so the short-hand notation

$${\displaystyle D^{\alpha }f=\frac{{\mathrm{\partial }}^{|\alpha |}f}{\mathrm{\partial }{\mathit{x}}^{\alpha }}=\frac{{\mathrm{\partial }}^{{\alpha }_1+\dots +{\alpha }_n}f}{\mathrm{\partial }{x}_1^{{\alpha }_1}\cdots \mathrm{\partial }{x}_n^{{\alpha }_n}}}$$ 

for the higher orderpartial derivatives is justified in this situation. The same is true if all the (k − 1)-th order partial derivatives off exist in some neighborhood ofa and are differentiable ata.^[13] Then we say thatf isktimes differentiable at the point a.

Using notations of the preceding section, one has the following theorem.

If the functionf :Rⁿ →R isk + 1 timescontinuously differentiable in aclosed ball${\displaystyle B=\{\mathbf{y}\in {\mathbb{R}}^n:\Vert \mathbf{a}-\mathbf{y}\Vert \le r\}}$  for some${\displaystyle r>0}$ , then one can derive an exact formula for the remainder in terms of(k+1)-th orderpartial derivatives off in this neighborhood.^[15] Namely,

 

In this case, due to thecontinuity of (k+1)-th orderpartial derivatives in thecompact setB, one immediately obtains the uniform estimates

$${\displaystyle \left|R_{\beta }(\mathit{x})\right|\le \frac{1}{\beta !}\underset{|\alpha |=|\beta |}{max}\underset{\mathit{y}\in B}{max}|D^{\alpha }f(\mathit{y})|,\mathit{x}\in B.}$$ 

For example, the third-order Taylor polynomial of a smooth function${\displaystyle f:{\mathbb{R}}^2\to \mathbb{R}}$  is, denoting${\displaystyle \mathit{x}-\mathit{a}=\mathit{v}}$ ,

 

Let^[16]

 

where, as in the statement of Taylor's theorem,

$${\displaystyle P(x)=f(a)+f^{\prime }(a)(x-a)+\frac{f^{″}(a)}{2!}(x-a{)}^2+\cdots +\frac{f^{(k)}(a)}{k!}(x-a{)}^k.}$$ 

It is sufficient to show that

$${\displaystyle \underset{x\to a}{lim}{h}_k(x)=0.}$$ 

The proof here is based on repeated application ofL'Hôpital's rule. Note that, for each$j=0,1,...,k-1$ ,${\displaystyle f^{(j)}(a)=P^{(j)}(a)}$ . Hence each of the first$k-1$  derivatives of the numerator in${\displaystyle h_k(x)}$  vanishes at${\displaystyle x=a}$ , and the same is true of the denominator. Also, since the condition that the function$f$  be$k$  times differentiable at a point requires differentiability up to order$k-1$  in a neighborhood of said point (this is true, because differentiability requires a function to be defined in a whole neighborhood of a point), the numerator and its$k-2$  derivatives are differentiable in a neighborhood of$a$ . Clearly, the denominator also satisfies said condition, and additionally, doesn't vanish unless$x=a$ , therefore all conditions necessary for L'Hôpital's rule are fulfilled, and its use is justified. So

 

where the second-to-last equality follows by the definition of the derivative at$x=a$ .

Let${\displaystyle f(x)}$  be any real-valued continuous function to be approximated by the Taylor polynomial.

Step 1: Let$F$  and$G$  be functions. Set$F$  and$G$  to be

$${\displaystyle \begin{array}{r}F(x)=f(x)-\sum _{k=0}^{n-1}\frac{f^{(k)}(a)}{k!}(x-a{)}^k\end{array}}$$ 

$${\displaystyle \begin{array}{r}G(x)=(x-a{)}^n\end{array}}$$ 

Step 2: Properties of$F$  and$G$ :

 

Similarly,

$${\displaystyle \begin{array}{r}{F}^{\prime }(a)=f^{\prime }(a)-f^{\prime }(a)-\frac{f^{″}(a)}{(2-1)!}(a-a{)}^{(2-1)}-...-\frac{f^{(n-1)}(a)}{(n-2)!}(a-a{)}^{n-2}=0\end{array}}$$ 

 

Step 3: Use Cauchy Mean Value Theorem

Let${\displaystyle f_1}$  and${\displaystyle g_1}$  be continuous functions on${\displaystyle [a,b]}$ . Since  so we can work with the interval${\displaystyle [a,x]}$ . Let${\displaystyle f_1}$  and${\displaystyle g_1}$  be differentiable on${\displaystyle (a,x)}$ . Assume${\displaystyle g_1^{\prime }(x)\ne 0}$  for all${\displaystyle x\in (a,b)}$ .Then there exists${\displaystyle c_1\in (a,x)}$  such that

$${\displaystyle \begin{array}{r}\frac{f_1(x)-f_1(a)}{g_1(x)-g_1(a)}=\frac{f_1^{\prime }(c_1)}{g_1^{\prime }(c_1)}\end{array}}$$ 

Note:${\displaystyle G^{\prime }(x)\ne 0}$  in${\displaystyle (a,b)}$  and${\displaystyle F(a),G(a)=0}$  so

$${\displaystyle \begin{array}{r}\frac{F(x)}{G(x)}=\frac{F(x)-F(a)}{G(x)-G(a)}=\frac{F^{\prime }(c_1)}{G^{\prime }(c_1)}\end{array}}$$ 

for some${\displaystyle c_1\in (a,x)}$ .

This can also be performed for${\displaystyle (a,c_1)}$ :

$${\displaystyle \begin{array}{r}\frac{F^{\prime }(c_1)}{G^{\prime }(c_1)}=\frac{F^{\prime }(c_1)-F^{\prime }(a)}{G^{\prime }(c_1)-G^{\prime }(a)}=\frac{F^{″}(c_2)}{G^{″}(c_2)}\end{array}}$$ 

for some${\displaystyle c_2\in (a,c_1)}$ .This can be continued to${\displaystyle c_n}$ .

This gives a partition in${\displaystyle (a,b)}$ :

 

with

$${\displaystyle \frac{F(x)}{G(x)}=\frac{F^{\prime }(c_1)}{G^{\prime }(c_1)}=\cdots =\frac{F^{(n)}(c_n)}{G^{(n)}(c_n)}.}$$ 

Set${\displaystyle c=c_n}$ :

$${\displaystyle \frac{F(x)}{G(x)}=\frac{F^{(n)}(c)}{G^{(n)}(c)}}$$ 

Step 4: Substitute back

$${\displaystyle \begin{array}{r}\frac{F(x)}{G(x)}=\frac{f(x)-\sum _{k=0}^{n-1}\frac{f^{(k)}(a)}{k!}(x-a{)}^k}{(x-a{)}^n}=\frac{F^{(n)}(c)}{G^{(n)}(c)}\end{array}}$$ 

By the Power Rule, repeated derivatives of${\displaystyle (x-a{)}^n}$ ,${\displaystyle G^{(n)}(c)=n(n-1)...1}$ , so:

$${\displaystyle \frac{F^{(n)}(c)}{G^{(n)}(c)}=\frac{f^{(n)}(c)}{n(n-1)\cdots 1}=\frac{f^{(n)}(c)}{n!}.}$$ 

This leads to:

$${\displaystyle \begin{array}{r}f(x)-\sum _{k=0}^{n-1}\frac{f^{(k)}(a)}{k!}(x-a{)}^k=\frac{f^{(n)}(c)}{n!}(x-a{)}^n\end{array}.}$$