Taylor's theorem

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

The polynomial appearing in Taylor's theorem is the k-th order Taylor polynomial

${\displaystyle P_{k}(x)=f(a)+f'(a)(x-a)+{\frac {f''(a)}{2!}}(x-a)^{2}+\cdots +{\frac {f^{(k)}(a)}{k!}}(x-a)^{k}}$

of the function f at the point a. The Taylor polynomial is the unique "asymptotic best fit" polynomial in the sense that if there exists a function h_k : R → R and a k-th order polynomial p such that

${\displaystyle f(x)=p(x)+h_{k}(x)(x-a)^{k},\quad \lim _{x\to a}h_{k}(x)=0,}$

then p = P_k. Taylor's theorem describes the asymptotic behavior of the remainder term

${\displaystyle R_{k}(x)=f(x)-P_{k}(x),}$

which is the approximation error when approximating f with its Taylor polynomial. Using the little-o notation, the statement in Taylor's theorem reads as

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

Under stronger regularity assumptions on f there are several precise formulas for the remainder term R_k of the Taylor polynomial, the most common ones being the following.

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

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

for some number ξ between a and x. This version covers the Lagrange and Cauchy forms of the remainder as special cases, and is proved below using Cauchy's mean value theorem.

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

Due to absolute continuity of f^(k) on the closed interval between a and x, its derivative f^(k+1) exists as an L¹-function, and the result can be proven by a formal calculation using fundamental theorem of calculus and integration 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 that f is (k + 1)-times continuously differentiable in an interval I containing a. Suppose that there are real constants q and Q such that

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

throughout I. Then the remainder term satisfies the inequality[11]

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

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

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

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

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

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

Approximation of e^x (blue) by its Taylor polynomials P_k of order k = 1,…,7 centered at x = 0 (red).

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:

${\displaystyle e^{0}=1,\qquad {\frac {d}{dx}}e^{x}=e^{x},\qquad e^{x}>0,\qquad x\in \mathbb {R} .}$

(⁎)

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

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

where ξ is some number between 0 and x. Since e^x is increasing by (⁎), we can simply use e^x ≤ 1 for x ∈ [−1, 0] to estimate the remainder on the subinterval [−1, 0]. To obtain an upper bound for the remainder on [0,1], we use the property e^ξ < e^x for 0<ξ<x to estimate

${\displaystyle e^{x}=1+x+{\frac {e^{\xi }}{2}}x^{2}<1+x+{\frac {e^{x}}{2}}x^{2},\qquad 0<x\leq 1}$

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

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

simply by maximizing the numerator and minimizing the denominator. Combining these estimates for e^x we see that

${\displaystyle |R_{k}(x)|\leq {\frac {4|x|^{k+1}}{(k+1)!}}\leq {\frac {4}{(k+1)!}},\qquad -1\leq x\leq 1,}$

so the required precision is certainly reached, when

${\displaystyle {\frac {4}{(k+1)!}}<10^{-5}\quad \Longleftrightarrow \quad 4\cdot 10^{5}<(k+1)!\quad \Longleftrightarrow \quad k\geq 9.}$

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

${\displaystyle e^{x}=1+x+{\frac {x^{2}}{2!}}+\cdots +{\frac {x^{9}}{9!}}+R_{9}(x),\qquad |R_{9}(x)|<10^{-5},\qquad -1\leq x\leq 1.}$

For instance, this approximation provides a decimal expression e ≈ 2.71828, correct up to five decimal places.

Let I ⊂ R be an open interval. By definition, a function f : I → R is real analytic if it is locally defined by a convergent power series. This means that for every a ∈ I there exists some r > 0 and a sequence of coefficients c_k ∈ R such that (a − r, a + r) ⊂ I and

${\displaystyle f(x)=\sum _{k=0}^{\infty }c_{k}(x-a)^{k}=c_{0}+c_{1}(x-a)+c_{2}(x-a)^{2}+\cdots ,\qquad |x-a|<r.}$

In general, the radius of convergence of a power series can be computed from the Cauchy–Hadamard formula

${\displaystyle {\frac {1}{R}}=\limsup _{k\to \infty }|c_{k}|^{\frac {1}{k}}.}$

This result is based on comparison with a geometric series, and the same method shows that if the power series based on a converges for some b ∈ R, it must converge uniformly on the closed interval [a − r_b, a + r_b], where r_b = |b − a|. Here only the convergence of the power series is considered, and it might well be that (a − R,a + R) extends beyond the domain I of f.

The Taylor polynomials of the real analytic function f at a are simply the finite truncations

${\displaystyle P_{k}(x)=\sum _{j=0}^{k}c_{j}(x-a)^{j},\qquad c_{j}={\frac {f^{(j)}(a)}{j!}}}$

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

${\displaystyle R_{k}(x)=\sum _{j=k+1}^{\infty }c_{j}(x-a)^{j}=(x-a)^{k}h_{k}(x),\qquad |x-a|<r.}$

Here the functions

${\displaystyle {\begin{aligned}&h_{k}:(a-r,a+r)\to \mathbb {R} \\&h_{k}(x)=(x-a)\sum _{j=0}^{\infty }c_{k+1+j}\left(x-a\right)^{j}\end{aligned}}}$

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 and r < 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 derivatives f′(a) at the center of the expansion, but using complex analysis also another possibility arises, which is described below.

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

One might think of the Taylor series

${\displaystyle f(x)\approx \sum _{k=0}^{\infty }c_{k}(x-a)^{k}=c_{0}+c_{1}(x-a)+c_{2}(x-a)^{2}+\cdots }$

of an infinitely many times differentiable function f : R → R as its "infinite order Taylor polynomial" at a. Now the estimates for the remainder imply that if, for any r, the derivatives of f are known to be bounded over (a − r, a + r), then for any order k and for any r > 0 there exists a constant M_k,r > 0 such that

${\displaystyle |R_{k}(x)|\leq M_{k,r}{\frac {|x-a|^{k+1}}{(k+1)!}}}$

(⁎⁎)

for every x ∈ (a − r,a + r). Sometimes the constants M_k,r can be chosen in such way that M_k,r is bounded above, for fixed r and all k. Then the Taylor series of f converges uniformly to some analytic function

${\displaystyle {\begin{aligned}&T_{f}:(a-r,a+r)\to \mathbb {R} \\&T_{f}(x)=\sum _{k=0}^{\infty }{\frac {f^{(k)}(a)}{k!}}\left(x-a\right)^{k}\end{aligned}}}$

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

The limit function T_f is by definition always analytic, but it is not necessarily equal to the original function f, even if f is infinitely differentiable. In this case, we say f is a non-analytic smooth function, for example a flat function:

${\displaystyle {\begin{aligned}&f:\mathbb {R} \to \mathbb {R} \\&f(x)={\begin{cases}e^{-{\frac {1}{x^{2}}}}&x>0\\0&x\leq 0.\end{cases}}\end{aligned}}}$

Using the chain rule repeatedly by mathematical induction, one shows that for any order k,

${\displaystyle f^{(k)}(x)={\begin{cases}{\frac {p_{k}(x)}{x^{3k}}}\cdot e^{-{\frac {1}{x^{2}}}}&x>0\\0&x\leq 0\end{cases}}}$

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

• The Taylor series of f converges uniformly to the zero function T_f(x) = 0, which is analytic with all coefficients equal to zero.
• The function f is unequal to this Taylor series, and hence non-analytic.
• For any order k ∈ N and radius r > 0 there exists M_k,r > 0 satisfying the remainder bound (⁎⁎) above.

However, as k increases for fixed r, the value of M_k,r grows more quickly than r^k, and the error does not go to zero.

Taylor's theorem generalizes to functions f : C → C which are complex differentiable in an open subset U ⊂ C of the complex plane. However, its usefulness is dwarfed by other general theorems in complex analysis. Namely, stronger versions of related results can be deduced for complex differentiable functions f : U → C using Cauchy's integral formula as follows.

Let r > 0 such that the closed disk B(z, r) ∪ S(z, r) is contained in U. Then Cauchy's integral formula with a positive parametrization γ(t) = z + re^it of the circle S(z, r) with t ∈ [0, 2π] gives

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

Here all the integrands are continuous on the circle S(z, r), which justifies differentiation under the integral sign. In particular, if f is once complex differentiable on the open set U, then it is actually infinitely many times complex differentiable on U. One also obtains the Cauchy's estimates[12]

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

for any z ∈ U and r > 0 such that B(z, r) ∪ S(c, r) ⊂ U. These estimates imply that the complex Taylor series

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

of f converges uniformly on any open disk B(c, r) ⊂ U with S(c, r) ⊂ U into some function T_f. Furthermore, using the contour integral formulas for the derivatives f^(k)(c),

${\displaystyle {\begin{aligned}T_{f}(z)&=\sum _{k=0}^{\infty }{\frac {(z-c)^{k}}{2\pi i}}\int _{\gamma }{\frac {f(w)}{(w-c)^{k+1}}}\,dw\\&={\frac {1}{2\pi i}}\int _{\gamma }{\frac {f(w)}{w-c}}\sum _{k=0}^{\infty }\left({\frac {z-c}{w-c}}\right)^{k}\,dw\\&={\frac {1}{2\pi i}}\int _{\gamma }{\frac {f(w)}{w-c}}\left({\frac {1}{1-{\frac {z-c}{w-c}}}}\right)\,dw\\&={\frac {1}{2\pi i}}\int _{\gamma }{\frac {f(w)}{w-z}}\,dw=f(z),\end{aligned}}}$

so any complex differentiable function f in an open set U ⊂ C is in fact complex analytic. All that is said for real analytic functions here holds also for complex analytic functions with the open interval I replaced by an open subset U ∈ C and a-centered intervals (a − r, a + r) replaced by c-centered disks B(c, r). In particular, the Taylor expansion holds in the form

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

where the remainder term R_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 oriented Jordan curve γ which parametrizes the boundary ∂W ⊂ U of a region W ⊂ U, one obtains expressions for the derivatives f^(j)(c) as above, and modifying slightly the computation for T_f(z) = f(z), one arrives at the exact formula

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

The important feature here is that the quality of the approximation by a Taylor polynomial on the region W ⊂ U is dominated by the values of the function f itself on the boundary ∂W ⊂ U. Similarly, applying Cauchy's estimates to the series expression for the remainder, one obtains the uniform estimates

${\displaystyle |R_{k}(z)|\leq \sum _{j=k+1}^{\infty }{\frac {M_{r}|z-c|^{j}}{r^{j}}}={\frac {M_{r}}{r^{k+1}}}{\frac {|z-c|^{k+1}}{1-{\frac {|z-c|}{r}}}}\leq {\frac {M_{r}\beta ^{k+1}}{1-\beta }},\qquad {\frac {|z-c|}{r}}\leq \beta <1.}$

Complex plot of f(z) = 1/(1 + z²). Modulus is shown by elevation and argument by coloring: cyan=0, blue = π/3, violet = 2π/3, red = π, yellow=4π/3, green=5π/3.

The function

${\displaystyle {\begin{aligned}&f:\mathbb {R} \to \mathbb {R} \\&f(x)={\frac {1}{1+x^{2}}}\end{aligned}}}$

is real analytic, that is, locally determined by its Taylor series. This function was plotted above 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 function f extends into a meromorphic function

${\displaystyle {\begin{aligned}&f:\mathbb {C} \cup \{\infty \}\to \mathbb {C} \cup \{\infty \}\\&f(z)={\frac {1}{1+z^{2}}}\end{aligned}}}$

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 at z₀ converges on any disc B(z₀, r) with r < |z − z₀|, where the same Taylor series converges at z ∈ C. Therefore, Taylor series of f centered at 0 converges on B(0, 1) and it does not converge for any z ∈ C with |z| > 1 due to the poles at i and −i. For the same reason the Taylor series of f centered at 1 converges on B(1, √2) and does not converge for any z ∈ C with |z − 1| > √2.

A function f: Rⁿ → R is differentiable at a ∈ Rⁿ if and only if there exists a linear functional L : Rⁿ → R and a function h : Rⁿ → R such that

${\displaystyle f({\boldsymbol {x}})=f({\boldsymbol {a}})+L({\boldsymbol {x}}-{\boldsymbol {a}})+h({\boldsymbol {x}})\lVert {\boldsymbol {x}}-{\boldsymbol {a}}\rVert ,\qquad \lim _{{\boldsymbol {x}}\to {\boldsymbol {a}}}h({\boldsymbol {x}})=0.}$

If this is the case, then L = df(a) is the (uniquely defined) differential of f at the point a. Furthermore, then the partial derivatives of f exist at a and the differential of f at a is given by

${\displaystyle df({\boldsymbol {a}})({\boldsymbol {v}})={\frac {\partial f}{\partial x_{1}}}({\boldsymbol {a}})v_{1}+\cdots +{\frac {\partial f}{\partial x_{n}}}({\boldsymbol {a}})v_{n}.}$

Introduce the multi-index notation

${\displaystyle |\alpha |=\alpha _{1}+\cdots +\alpha _{n},\quad \alpha$ !=\alpha _{1}!\cdots \alpha _{n}!,\quad {\boldsymbol {x}}^{\alpha }=x_{1}^{\alpha _{1}}\cdots x_{n}^{\alpha _{n}}}

for α ∈ Nⁿ and x ∈ Rⁿ. If all the k-th order partial derivatives of f : Rⁿ → R are continuous at a ∈ Rⁿ, then by Clairaut's theorem, one can change the order of mixed derivatives at a, so the notation

${\displaystyle D^{\alpha }f={\frac {\partial ^{|\alpha |}f}{\partial x_{1}^{\alpha _{1}}\cdots \partial x_{n}^{\alpha _{n}}}},\qquad |\alpha |\leq k}$

for the higher order partial derivatives is justified in this situation. The same is true if all the (k − 1)-th order partial derivatives of f exist in some neighborhood of a and are differentiable at a.[13] Then we say that f is k times differentiable at the point a.

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

If the function f : Rⁿ → R is k + 1 times continuously differentiable in a closed ball ${\displaystyle B=\{\mathbf {y} \in \mathbb {R} ^{n}:\left\|\mathbf {a} -\mathbf {y} \right\|\leq r\}}$ for some ${\displaystyle r>0}$, then one can derive an exact formula for the remainder in terms of (k+1)-th order partial derivatives of f in this neighborhood.[15] Namely,

${\displaystyle {\begin{aligned}&f({\boldsymbol {x}})=\sum _{|\alpha |\leq k}{\frac {D^{\alpha }f({\boldsymbol {a}})}{\alpha$ !}}({\boldsymbol {x}}-{\boldsymbol {a}})^{\alpha }+\sum _{|\beta |=k+1}R_{\beta }({\boldsymbol {x}})({\boldsymbol {x}}-{\boldsymbol {a}})^{\beta },\\&R_{\beta }({\boldsymbol {x}})={\frac {|\beta |}{\beta !}}\int _{0}^{1}(1-t)^{|\beta |-1}D^{\beta }f{\big (}{\boldsymbol {a}}+t({\boldsymbol {x}}-{\boldsymbol {a}}){\big )}\,dt.\end{aligned}}}

In this case, due to the continuity of (k+1)-th order partial derivatives in the compact set B, one immediately obtains the uniform estimates

${\displaystyle \left|R_{\beta }({\boldsymbol {x}})\right|\leq {\frac {1}{\beta$ !}}\max _{|\alpha |=|\beta |}\max _{{\boldsymbol {y}}\in B}|D^{\alpha }f({\boldsymbol {y}})|,\qquad {\boldsymbol {x}}\in B.}

For example, the third-order Taylor polynomial of a smooth function f: R² → R is, denoting x − a = v,

${\displaystyle {\begin{aligned}P_{3}({\boldsymbol {x}})=f({\boldsymbol {a}})+{}&{\frac {\partial f}{\partial x_{1}}}({\boldsymbol {a}})v_{1}+{\frac {\partial f}{\partial x_{2}}}({\boldsymbol {a}})v_{2}+{\frac {\partial ^{2}f}{\partial x_{1}^{2}}}({\boldsymbol {a}}){\frac {v_{1}^{2}}{2!}}+{\frac {\partial ^{2}f}{\partial x_{1}\partial x_{2}}}({\boldsymbol {a}})v_{1}v_{2}+{\frac {\partial ^{2}f}{\partial x_{2}^{2}}}({\boldsymbol {a}}){\frac {v_{2}^{2}}{2!}}\\&+{\frac {\partial ^{3}f}{\partial x_{1}^{3}}}({\boldsymbol {a}}){\frac {v_{1}^{3}}{3!}}+{\frac {\partial ^{3}f}{\partial x_{1}^{2}\partial x_{2}}}({\boldsymbol {a}}){\frac {v_{1}^{2}v_{2}}{2!}}+{\frac {\partial ^{3}f}{\partial x_{1}\partial x_{2}^{2}}}({\boldsymbol {a}}){\frac {v_{1}v_{2}^{2}}{2!}}+{\frac {\partial ^{3}f}{\partial x_{2}^{3}}}({\boldsymbol {a}}){\frac {v_{2}^{3}}{3!}}\end{aligned}}}$

Let[16]

${\displaystyle h_{k}(x)={\begin{cases}{\frac {f(x)-P(x)}{(x-a)^{k}}}&x\not =a\\0&x=a\end{cases}}}$

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

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

It is sufficient to show that

${\displaystyle \lim _{x\to a}h_{k}(x)=0.}$

The proof here is based on repeated application of L'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'Hopital's rule are fulfilled, and its use is justified. So

${\displaystyle {\begin{aligned}\lim _{x\to a}{\frac {f(x)-P(x)}{(x-a)^{k}}}&=\lim _{x\to a}{\frac {{\frac {d}{dx}}(f(x)-P(x))}{{\frac {d}{dx}}(x-a)^{k}}}=\cdots =\lim _{x\to a}{\frac {{\frac {d^{k-1}}{dx^{k-1}}}(f(x)-P(x))}{{\frac {d^{k-1}}{dx^{k-1}}}(x-a)^{k}}}\\&={\frac {1}{k!}}\lim _{x\to a}{\frac {f^{(k-1)}(x)-P^{(k-1)}(x)}{x-a}}\\&={\frac {1}{k!}}(f^{(k)}(a)-f^{(k)}(a))=0\end{aligned}}}$

where the 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

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

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

Step 2: Properties of F and G:

${\displaystyle {\begin{aligned}F(a)&=f(a)-f(a)-f'(a)(a-a)-...-{\frac {f^{(n-1)}(a)(a-a)^{n-1}}{(n-1)!}}=0\\G(a)&=(a-a)^{n}=0\end{aligned}}}$

Similarly,

${\displaystyle {\begin{aligned}F'(a)=f'(a)-f'(a)-{\frac {2f''(a)(a-a)}{1!}}-...-{\frac {f^{(n-2)}(a)(n-1)(a-a)^{n-2}}{(n-1)!}}=0\end{aligned}}}$

${\displaystyle {\begin{aligned}G'(a)=n(a-a)^{n-1}=0\end{aligned}}}$

. . .

${\displaystyle {\begin{aligned}G^{(n-1)}(a)=F^{(n-1)}(a)=0\end{aligned}}}$

Step 3: Use Cauchy Mean Value Theorem

Let ${\displaystyle f_{1}}$ and ${\displaystyle g_{1}}$ be continuous functions on ${\displaystyle [a,b]}$. Since ${\displaystyle a<x<b}$ 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 ${\displaystyle g_{1}'(x)\neq 0}$ for all ${\displaystyle x\in (a,b)}$. Then there exists ${\displaystyle c_{1}\in (a,x)}$ such that

${\displaystyle {\begin{aligned}{\frac {f_{1}(x)-f_{1}(a)}{g_{1}(x)-g_{1}(a)}}={\frac {f_{1}'(c_{1})}{g_{1}'(c_{1})}}\end{aligned}}}$

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

${\displaystyle {\begin{aligned}{\frac {F(x)}{G(x)}}={\frac {F(x)-F(a)}{G(x)-G(a)}}={\frac {F'(c_{1})}{G'(c_{1})}}\end{aligned}}}$

for some ${\displaystyle c_{1}\in (a,x)}$.

This can also be performed for ${\displaystyle (a,c_{1})}$:

${\displaystyle {\begin{aligned}{\frac {F'(c_{1})}{G'(c_{1})}}={\frac {F'(c_{1})-F'(a)}{G'(c_{1})-G'(a)}}={\frac {F''(c_{2})}{G''(c_{2})}}\end{aligned}}}$

for some ${\displaystyle c_{2}\in (a,c_{1})}$. This can be continued to ${\displaystyle c_{n}}$.

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

${\displaystyle {\begin{aligned}a<c_{n}<c_{n-1}<...<c_{1}<x\end{aligned}}}$

with

${\displaystyle {\begin{aligned}{\frac {F(x)}{G(x)}}={\frac {F'(c_{1})}{G'(c_{1})}}=...={\frac {F^{(n)}(c_{n})}{G^{(n)}(c_{n})}}\end{aligned}}}$

Set ${\displaystyle c=c_{n}}$:

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

Step 4: Substitute back

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

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

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

This leads to:

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

By rearranging, we get:

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

or because ${\displaystyle c_{n}=a}$ eventually:

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

Let G be any real-valued function, continuous on the closed interval between a and x and differentiable with a non-vanishing derivative on the open interval between a and x, and define

${\displaystyle F(t)=f(t)+f'(t)(x-t)+{\frac {f''(t)}{2!}}(x-t)^{2}+\cdots +{\frac {f^{(k)}(t)}{k!}}(x-t)^{k}.}$

For ${\displaystyle t\in [a,x]}$. Then, by Cauchy's mean value theorem,

${\displaystyle {\frac {F'(\xi )}{G'(\xi )}}={\frac {F(x)-F(a)}{G(x)-G(a)}}}$

(⁎⁎⁎)

for some ξ on the open interval between a and x. Note that here the numerator F(x) − F(a) = R_k(x) is exactly the remainder of the Taylor polynomial for f(x). Compute

${\displaystyle {\begin{aligned}F'(t)={}&f'(t)+{\big (}f''(t)(x-t)-f'(t){\big )}+\left({\frac {f^{(3)}(t)}{2!}}(x-t)^{2}-{\frac {f^{(2)}(t)}{1!}}(x-t)\right)+\cdots \\&\cdots +\left({\frac {f^{(k+1)}(t)}{k!}}(x-t)^{k}-{\frac {f^{(k)}(t)}{(k-1)!}}(x-t)^{k-1}\right)={\frac {f^{(k+1)}(t)}{k!}}(x-t)^{k},\end{aligned}}}$

plug it into (⁎⁎⁎) and rearrange terms to find that

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