Choosing Expansion Centers and Scaling Inputs

Why the center and scale matter

A Taylor polynomial is built around a particular input, called its expansion center. It usually gives its most useful approximation near that center, because the terms involving powers of the displacement become smaller as the displacement shrinks.

In computation, the inputs of interest may lie in a range that is not centered at zero, or may be clustered around a convenient value. Choosing a center near that range, then expressing inputs as small displacements from it, can make an approximation accurate where it is needed. A rescaled variable can also put the range into a standard interval, which makes the size of the displacement easier to track.

Key idea: Write an input as \(x=c+s z\), where \(c\) is a center and \(s z\) is the displacement from it. Choose \(c\) and \(s\) so the values of \(s z\) remain small over the inputs being approximated.

Centers, shifts, and scales

Let \(f\) be a function with derivatives through order \(n\) near a center \(c\). Its degree-\(n\) Taylor polynomial about \(c\) is

T_n(x;c)=\sum_{k=0}^{n}\frac{f^{(k)}(c)}{k!}(x-c)^k.

The quantity \(x-c\) is the displacement from the center. A shift replaces the original input by a displacement from \(c\). A scale expresses that displacement in units of some chosen length \(s\), by writing \(x=c+s z\). For an interval \([a,b]\), a natural choice is \(c=(a+b)/2\) and \(s=(b-a)/2\), so that \(z\) ranges from \(-1\) to \(1\).

Define the transformed function \(g(z)=f(c+s z)\). The Taylor polynomial for \(g\) about \(z=0\) is a polynomial in the scaled coordinate. The following identity shows exactly how it relates to the Taylor polynomial for \(f\).

Theorem: If \(s\ne 0\), then the degree-\(n\) Taylor polynomial of \(g(z)=f(c+s z)\) about \(z=0\) is \(T_n(c+s z;c)\). In particular, its coefficient of \(z^k\) is \(\frac{f^{(k)}(c)}{k!}s^k\).

Why the transformation works

Each differentiation with respect to \(z\) contributes a factor of \(s\), by the chain rule. Repeated differentiation gives \(g^{(k)}(z)=s^k f^{(k)}(c+s z)\), so \(g^{(k)}(0)=s^k f^{(k)}(c)\). Substituting these derivatives into the Taylor polynomial for \(g\) gives

\sum_{k=0}^{n}\frac{g^{(k)}(0)}{k!}z^k = \sum_{k=0}^{n}\frac{f^{(k)}(c)}{k!}(s z)^k = T_n(c+s z;c).

This is an identity, not an approximation: the transformed polynomial is the same Taylor approximation expressed in a different coordinate. The change of variable can make the range and displacement easier to manage, but it does not automatically improve the error. The actual error still depends on the function’s derivatives and the original displacement \(x-c=s z\).

For error estimates, the Lagrange remainder gives a useful condition. If \(f\) has a continuous derivative of order \(n+1\) throughout the segment between \(c\) and \(x\), and \(|f^{(n+1)}(u)|\le M\) on that segment, then

|f(x)-T_n(x;c)| \le \frac{M|x-c|^{n+1}}{(n+1)!}.

Thus a good center can help by reducing \(|x-c|\), while a good scale helps express and compare those displacements. The derivative bound must still hold on the relevant interval.

Worked examples

Worked Example: Estimating a square root near a convenient center

Estimate \(4\sqrt{15/16}\) using a degree-three Taylor polynomial. Write the expression as \(F(u)=4\sqrt{1+u}\) at \(u=-1/16\). The center \(u=0\) is convenient because the derivatives there are simple, and the target displacement is only \(1/16\).

The first four Taylor terms are obtained from \(F(0)=4\), \(F'(0)=2\), \(F''(0)=-1\), and \(F'''(0)=3/2\):

T_3(u;0) = 4+2u-\frac{u^2}{2}+\frac{u^3}{4}.

At \(u=-1/16\), the individual terms are \(4\), \(-1/8\), \(-1/512\), and \(-1/16384\). Therefore,

T_3\left(-\frac{1}{16};0\right) = 4-\frac18-\frac{1}{512}-\frac{1}{16384} = \frac{63455}{16384} = 3.87298583984375.

To bound the error, differentiate four times:

F^{(4)}(u)=-\frac{15}{4}(1+u)^{-7/2}.

On \([-1/16,0]\), the magnitude is largest at \(u=-1/16\). The exact maximum is \(\frac{15}{4}(15/16)^{-7/2}=4(16/15)^{5/2}\), approximately \(4.70036\). The Lagrange remainder bound is therefore

\left|F\left(-\frac{1}{16}\right)-T_3\left(-\frac{1}{16};0\right)\right| \le \frac{4(16/15)^{5/2}(1/16)^4}{24} \approx 0.0000029884.

As a check, the exact target is \(4\sqrt{15/16}=\sqrt{15}\approx3.8729833462\). Its difference from the polynomial value is approximately \(0.0000024936\), which is below the stated bound.

Worked Example: Shifting a reciprocal to a distant input range

Consider \(f(x)=1/(1+x)\) for \(4\le x\le5\). Expanding at zero would place these inputs far from the center and would not give a useful geometric-series expansion there. Instead, shift to \(c=4\) and set \(z=x-4\), so \(0\le z\le1\). Then

f(4+z)=\frac{1}{5+z} =\frac15\frac{1}{1+z/5}.

Using the geometric expansion through degree three gives the polynomial

P_3(z)=\frac15\left(1-\frac z5+\frac{z^2}{25}-\frac{z^3}{125}\right).

At \(x=4.5\), the shifted input is \(z=0.5\). The terms inside the parentheses are \(1\), \(-0.1\), \(0.01\), and \(-0.001\), so

P_3(0.5)=\frac15(0.909)=0.1818.

For a direct error check, the finite geometric identity gives \(1/(1+q)=1-q+q^2-q^3+q^4/(1+q)\). With \(q=z/5\), the exact remainder is \(z^4/[625(5+z)]\). At \(z=0.5\), it is \(0.0625/[625(5.5)]=0.0000181818\ldots\). Indeed, \(f(4.5)=1/5.5=0.18181818\ldots\), and the difference from \(0.1818\) is \(0.00001818\ldots\).

Here the shift is essential: it puts the whole target range within one unit of the expansion center. The calculation also shows why an expansion should be checked on the intended range, rather than assumed to work just because its formula is familiar.

Worked Example: Scaling a short interval to a standard coordinate

Approximate \(\ln x\) on \(0.9\le x\le1.1\). Choose center \(c=1\) and scale \(s=0.1\), setting \(x=1+0.1z\), so \(-1\le z\le1\). The transformed function is \(g(z)=\ln(1+0.1z)\). Its degree-three Taylor polynomial at zero is

P_3(z)=0.1z-\frac{(0.1z)^2}{2}+\frac{(0.1z)^3}{3}.

At the right endpoint \(x=1.1\), \(z=1\), and the polynomial evaluates to

P_3(1)=0.1-0.005+\frac{0.001}{3} =0.0953333333\ldots.

To verify the error across this scaled interval, use \(f(u)=\ln(1+u)\) in the original displacement \(u=x-1\). Its fourth derivative is \(-6(1+u)^{-4}\). On \([-0.1,0.1]\), its magnitude is at most \(6/(0.9)^4\approx9.14465\). Since \(|u|\le0.1\), the degree-three remainder is bounded by

\frac{(6/(0.9)^4)(0.1)^4}{24} \approx0.00003811.

At the tested endpoint, \(\ln(1.1)\approx0.0953101798\); the actual error is about \(0.00002315\), within the bound. Scaling makes the full input range the simple interval \([-1,1]\), while the corresponding displacement in the original variable remains at most \(0.1\).

Choosing a useful center in practice

For a fixed interval, the midpoint is often a sensible starting point because it minimizes the largest distance to the endpoints. But it is not automatically the best center for every function or every task. A nearby point may have simpler derivatives, a known function value, or a larger region on which the needed derivatives are well behaved. The function’s domain also matters: an interval that crosses a singularity or leaves the domain cannot be handled by a derivative bound that assumes smoothness throughout.

A practical choice balances several considerations:

Distance: Keep the largest displacement from the center small over the input range.
Derivatives: Prefer a center and interval where the required derivatives are manageable and bounded.
Known values: A center at which the function and its derivatives are easy to evaluate may reduce computational work.
Coordinates: A scale can map the range to a convenient variable interval, such as \([-1,1]\), without changing the underlying approximation.

Scaling is not a way to evade an error estimate. If \(x=c+s z\), then the displacement is \(s z\), and the derivative of order \(k\) of the transformed function contains the factor \(s^k\). These factors must be included when working directly with \(g(z)\); otherwise, the polynomial coefficients or error estimate will be wrong.

A common scaling mistake

Suppose \(x=1+0.1z\), and we use the quadratic Taylor polynomial for \(\ln(1+u)\), namely \(u-u^2/2\). A common mistake is to substitute \(z\) directly and write \(z-z^2/2\). That ignores the scale \(0.1\), making the answer ten times too large at leading order.

The correction is to substitute the actual displacement \(u=0.1z\) into every term:

u-\frac{u^2}{2} = 0.1z-\frac{(0.1z)^2}{2} = 0.1z-0.005z^2.

At \(z=1\), the incorrect expression gives \(0.5\), while the correctly transformed polynomial gives \(0.095\). This is also a useful check on the theorem: each degree-\(k\) coefficient must acquire the factor \(s^k\).

Check Your Understanding

Use the relationship between the original displacement and the scaled coordinate to answer these questions.

For \(x\in[2,6]\), what midpoint center and half-width scale map the interval to \(z\in[-1,1]\)?
If \(g(z)=f(c+s z)\), what is \(g^{(k)}(0)\) in terms of \(f^{(k)}(c)\) and \(s\)?
Why does changing coordinates not, by itself, guarantee a smaller Taylor error?
For \(x=3+0.2z\), what is the displacement from the center when \(z=-1\)?
What smoothness and derivative-bound conditions are needed to use the Lagrange remainder bound on an interval?