Taylor Remainder Terms

What a Taylor Remainder Measures

In “Proof of Taylor’s Theorem,” the error after approximating $f(x)$ by its degree-$n$ Taylor polynomial was expressed using the $(n+1)$st derivative at an intermediate point. Here we examine the remainder itself as an object that can be tracked across different polynomial orders. This makes an important distinction visible: the next Taylor term is a particular coefficient evaluated at the center, while the remainder is the full error after truncation.

Fix a center $a$, and write $T_{n,f,a}$ for the degree-$n$ Taylor polynomial of $f$ at $a$, as in “Higher-Order Taylor Polynomials.” When the required derivatives exist, define the signed remainder by subtracting that polynomial from the function. The word “signed” matters: a remainder can be positive or negative, and its absolute value is the size of the error.

Definition (Taylor Remainder): Suppose $f$ has derivatives through order $n$ on an open interval containing $a$. For $x$ in that interval, the remainder after the degree-$n$ Taylor polynomial at $a$ is $$ R_{n,f,a}(x)=f(x)-T_{n,f,a}(x). $$ In particular, $R_{n,f,a}(a)=0$. The absolute approximation error is $\lvert R_{n,f,a}(x)\rvert$.

The Taylor polynomial consists of the terms $$ \frac{f^{(j)}(a)}{j!}(x-a)^j,\qquad j=0,1,\ldots,n. $$ The term of degree $j$ is determined by the $j$th derivative at the center, but the remainder $R_{n,f,a}(x)$ also reflects all the behavior of $f$ between the center and the target. In particular, the remainder is not generally equal to the first omitted Taylor term.

How Remainders Change with the Order

Increasing the degree by one adds exactly one term to the Taylor polynomial. Subtracting the two polynomial approximations from the same function gives an exact relation between their remainders. This identity needs no estimate and no intermediate point.

Theorem (Consecutive Taylor Remainders): Suppose $f$ has derivatives through order $n+1$ on an open interval containing $a$, where $n\geq0$. For every $x$ in that interval, $$ R_{n,f,a}(x) = \frac{f^{(n+1)}(a)}{(n+1)!}(x-a)^{n+1} + R_{n+1,f,a}(x). $$

Proof. By the definition of the Taylor polynomials, $$ T_{n+1,f,a}(x) = T_{n,f,a}(x) + \frac{f^{(n+1)}(a)}{(n+1)!}(x-a)^{n+1}. $$ Using $R_{n,f,a}(x)=f(x)-T_{n,f,a}(x)$ and $R_{n+1,f,a}(x)=f(x)-T_{n+1,f,a}(x)$, we obtain $$ R_{n,f,a}(x)-R_{n+1,f,a}(x) = T_{n+1,f,a}(x)-T_{n,f,a}(x) = \frac{f^{(n+1)}(a)}{(n+1)!}(x-a)^{n+1}. $$ Rearranging proves the identity. At $x=a$, both remainders and the added term are zero, so the identity holds there as well. $\square$

The displayed term is the increment in the approximation when the degree increases from $n$ to $n+1$. Equivalently, it is the amount by which the signed remainder changes. The identity does not say that the old remainder equals that term: the new remainder remains on the right-hand side.

Worked Example: Tracking Exponential Remainders

Let $f(x)=e^x$, with center $a=0$, and evaluate at $x=1$. Since every derivative of $e^x$ equals $e^x$, every derivative at zero equals $1$. Thus $$ T_{0,f,0}(1)=1,\qquad T_{1,f,0}(1)=1+1=2,\qquad T_{2,f,0}(1)=1+1+\frac{1}{2}=\frac{5}{2}. $$ Consequently, $$ R_{0,f,0}(1)=e-1,\qquad R_{1,f,0}(1)=e-2,\qquad R_{2,f,0}(1)=e-\frac{5}{2}. $$ For the transition from degree zero to degree one, the added term is $f'(0)(1-0)=1$. Directly, $$ R_{0,f,0}(1)-R_{1,f,0}(1)=(e-1)-(e-2)=1. $$ For the next transition, the added term is $f''(0)(1-0)^2/2!=1/2$, and $$ R_{1,f,0}(1)-R_{2,f,0}(1)=(e-2)-\left(e-\frac{5}{2}\right)=\frac{1}{2}. $$ Both calculations verify the consecutive-remainder identity. They also show why the remainder after degree one is not the degree-two term: $R_{1,f,0}(1)=e-2$, whereas that term is $1/2$.

Differentiating a Remainder

Remainders at neighboring degrees have another useful relationship: differentiation lowers the Taylor order by one. The result follows from differentiating the function and the Taylor polynomial separately. It lets us connect the change of an error as $x$ varies to an error for the derivative function.

Theorem (Derivative of a Taylor Remainder): Let $n\geq1$, and suppose $f$ has derivatives through order $n$ on an open interval containing $a$. Then $R_{n,f,a}$ is differentiable there, and $$ \frac{d}{dx}R_{n,f,a}(x)=R_{n-1,f',a}(x). $$

Proof. By definition, $R_{n,f,a}(x)=f(x)-T_{n,f,a}(x)$, so its derivative is $f'(x)-T_{n,f,a}'(x)$. The Taylor polynomial is $$ T_{n,f,a}(x)=\sum_{j=0}^{n}\frac{f^{(j)}(a)}{j!}(x-a)^j. $$ Differentiating term by term removes the constant term and gives $$ T_{n,f,a}'(x) = \sum_{j=1}^{n}\frac{f^{(j)}(a)}{(j-1)!}(x-a)^{j-1}. $$ Set $k=j-1$. Then $k$ ranges from $0$ to $n-1$, and $f^{(j)}(a)=f'^{(k)}(a)$. Therefore $$ T_{n,f,a}'(x) = \sum_{k=0}^{n-1}\frac{f'^{(k)}(a)}{k!}(x-a)^k = T_{n-1,f',a}(x). $$ It follows that $$ R_{n,f,a}'(x) = f'(x)-T_{n-1,f',a}(x) = R_{n-1,f',a}(x), $$ as claimed. $\square$

Worked Example: A Sine Remainder and Its Derivative

Take $f(x)=\sin x$, $a=0$, and $n=2$. At zero, $f(0)=0$, $f'(0)=1$, and $f''(0)=0$. Hence $T_{2,f,0}(x)=x$, and $$ R_{2,f,0}(x)=\sin x-x. $$ The derivative is $R_{2,f,0}'(x)=\cos x-1$. For the derivative function $f'(x)=\cos x$, the degree-one Taylor polynomial at zero is $T_{1,f',0}(x)=1$, because $f'(0)=1$ and $f''(0)=0$. Thus $$ R_{1,f',0}(x)=\cos x-1=R_{2,f,0}'(x). $$ For example, at $x=\pi/3$, the remainder is $$ R_{2,f,0}\left(\frac{\pi}{3}\right)=\frac{\sqrt{3}}{2}-\frac{\pi}{3}, $$ and its derivative there is $$ R_{2,f,0}'\left(\frac{\pi}{3}\right)=\frac{1}{2}-1=-\frac{1}{2}. $$ The derivative remainder gives the same value, since $\cos(\pi/3)-1=1/2-1=-1/2$.

The Next Taylor Term Is Not the Whole Error

The consecutive-remainder identity separates the old remainder into the next Taylor term and the remainder after adding that term. This is exact, but it does not make the old remainder equal to the added term. To estimate the remaining part, one may use Taylor’s Theorem or its Taylor Remainder Bound, established earlier. Under the hypotheses of Taylor’s Theorem, for $x\ne a$ there is a point $c$ strictly between $a$ and $x$ such that $$ R_{n,f,a}(x) = \frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}. $$ The derivative in this representation is evaluated at $c$, not necessarily at $a$. The next Taylor term, by contrast, uses $f^{(n+1)}(a)$. Those values need not be equal.

For the logarithm, the difference is visible even at a simple target. Let $f(x)=\ln(1+x)$, $a=0$, and $x=1/2$. The derivatives at zero through order three give $$ T_{2,f,0}(x)=x-\frac{x^2}{2}, \qquad T_{3,f,0}(x)=x-\frac{x^2}{2}+\frac{x^3}{3}. $$ At $x=1/2$, these are $$ T_{2,f,0}\left(\frac12\right)=\frac12-\frac18=\frac38, \qquad T_{3,f,0}\left(\frac12\right)=\frac38+\frac{1}{24}=\frac{5}{12}. $$ Therefore $$ R_{2,f,0}\left(\frac12\right)=\ln\left(\frac32\right)-\frac38, \qquad R_{3,f,0}\left(\frac12\right)=\ln\left(\frac32\right)-\frac{5}{12}. $$ Their difference is $$ R_{2,f,0}\left(\frac12\right)-R_{3,f,0}\left(\frac12\right) = \left(-\frac38\right)-\left(-\frac{5}{12}\right) = -\frac{9}{24}+\frac{10}{24} = \frac{1}{24}, $$ which is exactly the next Taylor term, $(1/2)^3/3=1/24$. The original degree-two remainder still includes $R_{3,f,0}(1/2)$; it is not just $1/24$.

This distinction matters when interpreting an approximation. A small next Taylor term alone does not, by the consecutive-remainder identity, prove that the current error is small: the remainder after adding that term also contributes. To obtain a guaranteed error bound, use an applicable remainder estimate, such as the Taylor Remainder Bound. Conversely, the exact identity is useful even when no numerical bound is available, because it shows precisely how truncation errors at successive orders differ.

Using the Remainder Relationships

The two relationships proved here serve different purposes. The consecutive-order identity is algebraic: it compares approximations of different degrees at the same target point. The derivative identity is functional: it describes how the error at one degree changes as the target point moves. Together they help organize calculations without assuming that a remainder has a particular sign or that successive errors decrease in magnitude.

A frequent pitfall is to treat the Lagrange form as though its intermediate point were the center. It is generally an unknown point between $a$ and $x$, and it can depend on $x$ and on the order $n$. Another is to infer an error estimate solely from the size of the last displayed term. Such an inference needs additional justification; Taylor’s Theorem with a derivative bound supplies one when its hypotheses hold. The signed remainder definition itself requires no such bound and makes sense whenever the function and polynomial are defined.

Check Your Understanding

Use the definition of the remainder and the two proved relationships to answer these questions.

Write the consecutive Taylor remainder identity for the change from degree $n$ to degree $n+1$.
Why does that identity not say that $R_{n,f,a}(x)$ equals the next Taylor term?
For $n\geq1$, which Taylor remainder equals the derivative of $R_{n,f,a}$?
If $f^{(n+1)}(a)=0$, what does the consecutive-remainder identity imply about $R_{n,f,a}(x)$ and $R_{n+1,f,a}(x)$?
In the Lagrange remainder formula, why should its intermediate point not be replaced automatically by $a$?

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