Taylor Error Estimates For Nonzero Centers

Why the Center Matters

A Taylor polynomial is built from the value of a function and its derivatives at a chosen center. The center need not be zero. When the target value $x$ is near a nonzero center $a$, the powers in the approximation are powers of the displacement $x-a$, not powers of $x$. To certify accuracy, we must also bound the relevant derivative on the interval from $a$ to $x$.

This distinction matters in applications: the function may be well behaved near the chosen center even when a bound based on a larger interval would be unnecessarily weak. A valid estimate depends on the interval actually connecting the center to the target, as well as on the derivative bound over that interval.

Definition: If $f$ has $n$ derivatives at $a$, its degree-$n$ Taylor polynomial centered at $a$ is $$ T_n(x)=\sum_{k=0}^{n}\frac{f^{(k)}(a)}{k!}(x-a)^k. $$ Here $f^{(0)}(a)=f(a)$ and $0!=1$. The center is nonzero when $a\ne 0$.

The formula uses derivatives evaluated at the center $a$, but the polynomial is evaluated at the target $x$. If $x=a$, every term with $k\ge 1$ vanishes and $T_n(a)=f(a)$. For $x\ne a$, the size of the displacement $|x-a|$ affects the remainder.

The Lagrange Bound Between the Center and Target

Suppose $f$ has $n+1$ continuous derivatives on a closed interval containing both $a$ and $x$. Taylor’s theorem with the Lagrange form of the remainder states that for some $\xi$ between $a$ and $x$,

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

In particular, if $\left|f^{(n+1)}(t)\right|\le M_{n+1}$ everywhere on the interval between $a$ and $x$, then taking absolute values gives the error estimate

\left|f(x)-T_n(x)\right| \le \frac{M_{n+1}|x-a|^{n+1}}{(n+1)!}.

The interval here is the closed interval with endpoints $a$ and $x$, regardless of which endpoint is larger. The derivative bound must hold throughout that interval. A bound valid only at the center, or only at the target, is not enough.

Theorem: If $f^{(n+1)}$ is continuous on the interval between $a$ and $x$, and $\left|f^{(n+1)}(t)\right|\le M_{n+1}$ there, then the degree-$n$ Taylor approximation at $a$ has error at most $$ \frac{M_{n+1}|x-a|^{n+1}}{(n+1)!}. $$ For $x=a$, the error is zero.

Why the Bound Works

The remainder formula identifies the error as a derivative value at some intermediate point, multiplied by the displacement to the power $n+1$, and divided by $(n+1)!$. The derivative bound controls the intermediate derivative, while the absolute value removes any dependence on the sign of the remainder.

Start with the remainder.
For some $\xi$ between $a$ and $x$, the error is $f^{(n+1)}(\xi)(x-a)^{n+1}/(n+1)!$.

Use the derivative bound.
Since $\xi$ lies in the interval where the bound holds, $\left|f^{(n+1)}(\xi)\right|\le M_{n+1}$.

Take absolute values.
It follows that $\left|f(x)-T_n(x)\right|\le M_{n+1}|x-a|^{n+1}/(n+1)!$, as claimed.

This reasoning explains the two main ingredients in a certified estimate: a valid bound for the next derivative and the distance from the center to the target. The derivative order is $n+1$, one higher than the degree of the Taylor polynomial.

Worked Example: Certifying the First Degree That Meets a Tolerance

Worked Example: Finding the First Certified Degree

Let $f(t)=0.005(t-2)+0.02(t-2)^2+0.01(t-2)^3$. Approximate $f(2.5)$ with Taylor polynomials centered at $a=2$, and use Lagrange bounds to find the first degree certified to have error at most $0.00001$.

Set $u=t-2$. The target displacement is $h=2.5-2=0.5$, and the interval is $2\le t\le 2.5$, or $0\le u\le 0.5$. The derivatives are

f'(t)=0.005+0.04u+0.03u^2,\qquad f''(t)=0.04+0.06u,\qquad f'''(t)=0.06,\qquad f^{(4)}(t)=0.

On the interval, $f'(t)$ increases from $0.005$ to $0.0325$, so $M_1=0.05$ is a valid bound. Also $f''(t)\le 0.04+0.06(0.5)=0.07$, so take $M_2=0.07$; take $M_3=0.06$ and $M_4=0$.

For degree zero, $T_0(2.5)=f(2)=0$. Its bound is

\left|f(2.5)-T_0(2.5)\right| \le \frac{0.05(0.5)}{1!}=0.025=\frac{0.1}{2^2}>0.00001.

For degree one, $f'(2)=0.005$, so $T_1(2.5)=0+0.005(0.5)=0.0025$. Its bound is

\left|f(2.5)-T_1(2.5)\right| \le \frac{0.07(0.5)^2}{2!} =\frac{0.0175}{2}=0.00875>0.00001.

For degree two, $f''(2)=0.04$, and therefore

T_2(2.5)=0+0.005(0.5)+\frac{0.04}{2!}(0.5)^2 =0.0025+0.005=0.0075.

The degree-two error bound is

\left|f(2.5)-T_2(2.5)\right| \le \frac{0.06(0.5)^3}{3!} =\frac{0.0075}{6}=0.00125>0.00001.

For degree three, the polynomial includes every term of $f$, so $T_3(2.5)=0.0025+0.005+0.00125=0.00875$. Since $f^{(4)}(t)=0$, its error bound is $0$, which meets the tolerance. Direct substitution gives $f(2.5)=0.005(0.5)+0.02(0.25)+0.01(0.125)=0.00875$, confirming the approximation exactly. The degree-zero case has been checked as well as degrees one and two; thus degree three is the first certified degree.

Worked Example: A Rational Function at a Nonzero Center

Worked Example: Approximating a Reciprocal Near Two

Use a Taylor polynomial centered at $a=2$ to approximate $f(2.5)$, where $f(t)=1/t$. Find the first degree certified to have error at most $0.01$.

The interval is $[2,2.5]$, and $h=0.5$. The derivatives are $f'(t)=-t^{-2}$, $f''(t)=2t^{-3}$, and $f'''(t)=-6t^{-4}$. On this interval, their absolute values are largest at $t=2$. Thus we may use $M_1=1/4$, $M_2=1/4$, and $M_3=6/16=3/8$.

For degree zero, $T_0(2.5)=f(2)=1/2$, and the bound is $(1/4)(0.5)=0.125$, which exceeds $0.01$. For degree one, $f'(2)=-1/4$, so $T_1(2.5)=1/2-(1/4)(0.5)=3/8$. Its bound is

\frac{M_2(0.5)^2}{2!} =\frac{(1/4)(1/4)}{2} =\frac{1}{32}=0.03125>0.01.

For degree two, $f''(2)=1/4$, so

T_2(2.5)=\frac12-\frac14(0.5)+\frac{1/4}{2!}(0.5)^2 =\frac12-\frac18+\frac1{32} =\frac{13}{32}.

The degree-two error bound is

\frac{M_3(0.5)^3}{3!} =\frac{(3/8)(1/8)}{6} =\frac{1}{128}=0.0078125<0.01.

Therefore degree two is the first degree certified to meet the tolerance: degrees zero and one fail their bounds, and degree two passes. Checking against the exact value, $f(2.5)=1/(5/2)=2/5=0.4$, while $13/32=0.40625$. The actual absolute error is $13/32-2/5=1/160=0.00625$, which is below both the certified bound $1/128$ and the requested tolerance.

Worked Example: Approximating to the Left of the Center

Worked Example: A Square Root Centered at Four

Approximate $\sqrt{3}$ using the degree-two Taylor polynomial for $f(t)=\sqrt{t}$ centered at $a=4$, and certify the error. The target is to the left of the center, so the interval to use is $[3,4]$, and $h=3-4=-1$.

The derivatives at $4$ are $f(4)=2$, $f'(4)=1/4$, and $f''(4)=-1/32$. Hence

T_2(3)=2+\frac14(-1)+\frac{-1/32}{2!}(-1)^2 =2-\frac14-\frac1{64} =\frac{111}{64}=1.734375.

For $t>0$, $f'''(t)=3/(8t^{5/2})$, which decreases as $t$ increases. On $[3,4]$, it is therefore bounded by $M_3=3/(8\cdot 3^{5/2})=1/(24\sqrt{3})$. The Lagrange estimate is

\left|\sqrt{3}-\frac{111}{64}\right| \le \frac{M_3|{-1}|^3}{3!} =\frac{1}{144\sqrt{3}} \approx 0.004009.

For a direct check, $\sqrt{3}\approx 1.732051$, so the absolute error is approximately $1.734375-1.732051=0.002324$, below the bound. The displayed decimals are rounded to six places; the exact Taylor approximation is $111/64$, and the inequality itself uses the exact bound $1/(144\sqrt{3})$.

Choosing a Degree Reliably

For a prescribed tolerance $\varepsilon>0$, a degree $n$ is certified if a valid derivative bound gives

\frac{M_{n+1}|x-a|^{n+1}}{(n+1)!}\le \varepsilon.

The bound may depend on $n$, because each degree uses a different derivative. Therefore, do not assume that a familiar pattern in the bounds applies unless it has been justified. Check the derivative bound and the inequality for each candidate degree.

To identify the first certified degree, start with $n=0$ and proceed upward. A degree that fails the bound is not certified by that particular estimate; this does not prove that the actual error exceeds the tolerance. Conversely, a bound below the tolerance guarantees the actual error is below it. This distinction between an actual error and a certified upper bound is essential.

Common Pitfall: Bounding the Derivative Only at the Center

A tempting but invalid shortcut is to evaluate $f^{(n+1)}$ only at $a$ and use that value as $M_{n+1}$. The Lagrange remainder involves a derivative at an intermediate point $\xi$, not necessarily at the center. Unless you have shown that the derivative’s absolute value is bounded by its center value throughout the interval, that substitution is unjustified.

For example, for $f(t)=1/t$ centered at $2$, $f'''(2)=-3/8$, so its absolute value is $3/8$. In this particular example, that value does bound $\left|f'''(t)\right|$ on $[2,2.5]$, because $6/t^4$ decreases there. But the validity comes from checking the whole interval, not merely evaluating at $2$. For another function, the derivative could become larger between the endpoints.

Takeaway: A Taylor error estimate at a nonzero center uses the displacement $|x-a|$ and a bound for the next derivative over the entire interval joining $a$ and $x$. To find the first certified degree, include degree zero and test each lower degree before claiming that a later one is first.

Check Your Understanding

For each question, identify the center, target, relevant interval, and derivative order before calculating.

State the Lagrange error bound for a degree-$n$ Taylor polynomial centered at $a$, including the condition required of the derivative.
If $a=5$ and $x=4.6$, what is the interval on which the derivative bound must hold, and what is the displacement magnitude?
For a degree-four Taylor approximation, which derivative must be bounded?
Why does evaluating the next derivative only at the center generally fail to provide a valid remainder bound?
When seeking the first degree certified to meet a tolerance, why should degree zero be checked?