Taylor's Theorem

From Derivative Matching to an Error Formula

The previous tutorial defined the Taylor polynomial $T_{n,f,a}$, which matches a function’s derivatives through order $n$ at the center $a$. That matching explains how to construct the polynomial, but it does not by itself say how accurately the polynomial approximates the function at another point $x$. Taylor’s Theorem supplies that missing information: it expresses the difference $f(x)-T_{n,f,a}(x)$ using the next derivative at an intermediate point.

The particular version used here is often called Taylor’s Theorem with the Lagrange form of the remainder. It is useful both as an exact identity and as a route to an error bound. Its hypotheses must hold throughout the interval connecting $a$ and $x$, not only at the center. The proof is the subject of the next tutorial; here we state the theorem and develop consequences that make it practical.

Taylor’s Theorem with Lagrange Remainder

Theorem (Taylor’s Theorem, Lagrange Form): Let $n$ be a nonnegative integer, let $I$ be an open interval, and let $f:I\to\mathbb{R}$ have derivatives through order $n+1$ on $I$. If $a,x\in I$ and $x\ne a$, then there is a point $c$ strictly between $a$ and $x$ such that $$ f(x)=T_{n,f,a}(x)+\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}. $$ Equivalently, the remainder $R_{n,f,a}(x)=f(x)-T_{n,f,a}(x)$ satisfies $$ R_{n,f,a}(x)=\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}. $$ At $x=a$, the remainder is zero by the definition of the Taylor polynomial.

The point $c$ depends on $f$, $a$, $x$, and $n$; the theorem does not generally identify it explicitly. It lies between the two points, regardless of their order. Thus if $x<a$, then $x<c<a$. The signed power $(x-a)^{n+1}$ is retained in the formula. When taking an absolute value to estimate the error, its contribution becomes $|x-a|^{n+1}$.

The theorem requires derivatives through order $n+1$ on the interval, which is one order more than the derivatives used to define $T_{n,f,a}$. It is precisely this next derivative that controls the error. For $n=0$, the theorem says $f(x)=f(a)+f'(c)(x-a)$, the familiar form of the Mean Value Theorem. For larger $n$, the earlier derivatives are incorporated into the polynomial and the remaining difference is controlled by the next one.

A General Error Bound

The exact remainder contains the unknown intermediate point $c$. Often the derivative at that point is not known, but its magnitude can be bounded throughout the interval. This gives an estimate that does not require finding $c$.

Theorem (Taylor Remainder Bound): Assume the hypotheses of Taylor’s Theorem. Suppose $M\geq 0$ satisfies $|f^{(n+1)}(t)|\leq M$ for every $t$ between $a$ and $x$. Then $$ |f(x)-T_{n,f,a}(x)|\leq \frac{M}{(n+1)!}|x-a|^{n+1}. $$

Proof. If $x=a$, then $f(x)-T_{n,f,a}(x)=0$, so the inequality holds. Suppose $x\ne a$. Taylor’s Theorem gives a point $c$ between $a$ and $x$ such that

f(x)-T_{n,f,a}(x) =\frac{f^{(n+1)}(c)}{(n+1)!}(x-a)^{n+1}.

Since $c$ is between the two endpoints, the assumed derivative bound applies at $c$. Taking absolute values therefore yields

|f(x)-T_{n,f,a}(x)| =\frac{|f^{(n+1)}(c)|}{(n+1)!}|x-a|^{n+1} \leq\frac{M}{(n+1)!}|x-a|^{n+1}.

This proves the bound. $\square$

The estimate separates two sources of error: the size of the next derivative and the distance from the center raised to the power $n+1$. A larger order can improve the power of the distance, but the derivative bound and factorial must also be taken into account. A bound is useful only when its hypothesis holds on the whole segment between $a$ and $x$.

Using the Remainder in Examples

Worked Example: Approximating the Sine Function Near Zero

Take $f(x)=\sin x$, center $a=0$, and order $n=3$. The derivatives at zero are $f(0)=0$, $f'(0)=1$, $f''(0)=0$, and $f'''(0)=-1$. Hence

T_{3,f,0}(x)=x-\frac{x^3}{3!}=x-\frac{x^3}{6}.

The fourth derivative is $f^{(4)}(t)=\sin t$, so $|f^{(4)}(t)|\leq 1$ for all real $t$. The Taylor remainder bound with $M=1$ gives

\left|\sin x-\left(x-\frac{x^3}{6}\right)\right| \leq\frac{|x|^4}{4!}=\frac{|x|^4}{24}.

For $x=0.2=1/5$, the polynomial gives $1/5-(1/5)^3/6=1/5-1/750=149/750$. The error is at most

\frac{(1/5)^4}{24}=\frac{1}{625\cdot24}=\frac{1}{15000}.

This is an error guarantee, not a claim that the actual error equals $1/15000$. Taylor’s Theorem guarantees that it is no greater than that amount.

Worked Example: A Quadratic Approximation to the Exponential

Let $f(x)=e^x$, with center $a=0$ and order $n=2$. Since $f(0)=f'(0)=f''(0)=1$, its Taylor polynomial is

T_{2,f,0}(x)=1+x+\frac{x^2}{2}.

The third derivative is $f^{(3)}(t)=e^t$. To estimate the error at $x=0.1$, note that the intermediate point $c$ lies in $(0,0.1)$. Since $e^t$ is increasing, $e^c\leq e^{0.1}$, so the remainder formula gives

\left|e^{0.1}-\left(1+0.1+\frac{(0.1)^2}{2}\right)\right| \leq \frac{e^{0.1}}{3!}(0.1)^3 =\frac{e^{0.1}}{6000}.

Here the polynomial value is $1+0.1+0.005=1.105$. The estimate is valid because the derivative bound was checked on the entire interval from $0$ to $0.1$, which contains $c$.

Worked Example: Bounding the Error for a Logarithm

Consider $f(x)=\ln(1+x)$, centered at zero, and use the Taylor polynomial of order two. On the interval $(-1,\infty)$,

f(0)=0,\qquad f'(x)=\frac{1}{1+x},\qquad f''(x)=-\frac{1}{(1+x)^2},

so $f'(0)=1$, $f''(0)=-1$, and

T_{2,f,0}(x)=x-\frac{x^2}{2}.

The third derivative is $f^{(3)}(t)=2/(1+t)^3$. At $x=1/2$, the intermediate point $c$ lies in $(0,1/2)$, so $1+c\geq1$ and $0<f^{(3)}(c)\leq2$. Thus

\left|\ln(3/2)-\left(\frac12-\frac{(1/2)^2}{2}\right)\right| \leq\frac{2}{3!}\left(\frac12\right)^3 =\frac{2}{6\cdot8} =\frac{1}{24}.

The polynomial value is $1/2-1/8=3/8$. The exact remainder is positive in this case, since $f^{(3)}(c)>0$ and $x-a>0$, but the displayed estimate only requires its magnitude.

What Continuity Adds Near the Center

A uniform bound gives a quantitative estimate, while continuity of the next derivative gives more precise information about the leading behavior of the remainder near the center. The following consequence says that, after subtracting the contribution from the next derivative at $a$, what remains is smaller than $|x-a|^{n+1}$ in the limiting sense.

Theorem (Local Refinement of the Taylor Remainder): Suppose $f$ has derivatives through order $n+1$ on an open interval containing $a$, and $f^{(n+1)}$ is continuous at $a$. Then, as $x\to a$, $$ f(x)-T_{n,f,a}(x) =\frac{f^{(n+1)}(a)}{(n+1)!}(x-a)^{n+1} +o\bigl(|x-a|^{n+1}\bigr). $$

Proof. For $x\ne a$, Taylor’s Theorem supplies a point $c_x$ strictly between $a$ and $x$ such that

f(x)-T_{n,f,a}(x) =\frac{f^{(n+1)}(c_x)}{(n+1)!}(x-a)^{n+1}.

Because $c_x$ lies between $a$ and $x$, we have $|c_x-a|\leq|x-a|$, and therefore $c_x\to a$ as $x\to a$. Subtract the proposed leading term and divide by $|x-a|^{n+1}$. The absolute value of the resulting difference is

\frac{|f^{(n+1)}(c_x)-f^{(n+1)}(a)|}{(n+1)!} \frac{|(x-a)^{n+1}|}{|x-a|^{n+1}} = \frac{|f^{(n+1)}(c_x)-f^{(n+1)}(a)|}{(n+1)!}.

The ratio of powers is one for $x\ne a$. By continuity of $f^{(n+1)}$ at $a$ and the fact that $c_x\to a$, the final expression tends to zero. This is exactly the asserted $o(|x-a|^{n+1})$ relation. $\square$

This refinement distinguishes a bound from a leading-term description. The bound says the remainder is no larger than a specified multiple of $|x-a|^{n+1}$. The refinement says that, after the next-derivative term at the center is removed, the remaining difference divided by $|x-a|^{n+1}$ tends to zero. Continuity at $a$ is essential for this conclusion as stated.

Applying the Theorem Carefully

A reliable application can be organized around the following checks:

Choose the center and order.
Write down $a$, $n$, and the Taylor polynomial $T_{n,f,a}$ using derivatives at $a$.

Check the interval.
Verify that the needed derivatives exist throughout an interval containing both $a$ and the target point $x$.

Use the next derivative.
For an exact remainder formula, retain $f^{(n+1)}(c)$ at an intermediate point; for a bound, find a valid bound on that derivative over the whole segment.

Keep the power and factorial together.
The remainder has denominator $(n+1)!$ and factor $(x-a)^{n+1}$, not $(x-a)^n$.

A common mistake is to use a derivative bound that holds only at the center. Taylor’s Theorem evaluates the derivative at an intermediate point, so a bound at $a$ alone does not control the remainder. Another is to treat the theorem’s formula as though it gave a known value of $c$. In most applications, $c$ remains unknown; a bound on the derivative over the segment is what turns the identity into a usable estimate.

Taylor’s Theorem therefore complements the derivative-matching characterization from “Higher-Order Taylor Polynomials.” Matching determines the polynomial, while the Lagrange remainder relates it to the function away from the center. The next tutorial proves the theorem itself.

Check Your Understanding

Use Taylor’s Theorem and its consequences to answer the following questions.

In the Lagrange form of the remainder, where must the point $c$ lie relative to $a$ and $x$?
What bound follows if $|f^{(n+1)}(t)|\leq M$ throughout the segment from $a$ to $x$?
Why is a derivative bound only at the center generally insufficient for that error estimate?
For $f(x)=\sin x$, what error bound does the order-three Taylor polynomial at zero give at $x=0.1$?
What additional hypothesis gives the local refinement involving $f^{(n+1)}(a)$?

What You'll Learn