Choosing a Center and Degree for a Local Model

Introduction

A Taylor polynomial is a local model: it uses a function’s value and derivatives at one center to approximate the function nearby. Two choices shape that model. The center determines where the polynomial is anchored, and the degree determines how many derivative-based terms it includes.

These choices should match the task. If values are needed near \(x=4\), a model centered at \(4\) may be more useful than one centered at \(0\). If a linear model misses noticeable curvature, a higher degree may improve the approximation. Neither choice is automatic: the function must have the required derivatives at the center, and a higher degree is not, by itself, a guarantee of a particular accuracy.

What the Center and Degree Specify

Let \(f\) have derivatives through order \(n\) at a point \(a\), where \(n\) is a nonnegative integer. Its degree-\(n\) Taylor polynomial centered at \(a\) is

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

Here \(f^{(0)}(a)\) means \(f(a)\), and \(0!=1\). The center is \(a\); the degree is \(n\). The term with \(k=0\) gives the function’s value at the center. The term with \(k=1\) describes the slope there. The term with \(k=2\) incorporates curvature, and higher-degree terms incorporate further local derivative information.

Definition: A Taylor polynomial of degree \(n\) at \(a\) is the polynomial above, formed from the function’s derivatives at \(a\) through order \(n\). Its local model is intended for inputs \(x\) near \(a\), where the function and the polynomial can be meaningfully compared.

The powers \((x-a)^k\) help explain the role of the center. At \(x=a\), every term with \(k\geq 1\) vanishes. Near \(a\), those powers are small, especially for higher \(k\). If the target input is far from the center, the polynomial is being used farther from the point where it matches the function’s derivative information.

Why the Polynomial Matches at Its Center

The Taylor polynomial is constructed so its derivatives at the center agree with those of \(f\) through the chosen degree. This makes the center a precise anchor, rather than just a convenient place to write down a formula.

Result: If \(f\) has derivatives through order \(n\) at \(a\), then \(T_n^{(j)}(a;a)=f^{(j)}(a)\) for each integer \(j\) with \(0\leq j\leq n\).

To see why, consider a single term of the polynomial, \(c_k(x-a)^k\), where \(c_k=f^{(k)}(a)/k!\). After taking \(j\) derivatives, that term is zero if \(j>k\). If \(j<k\), it still contains a positive power of \(x-a\), so it is zero at \(x=a\). When \(j=k\), its \(j\)th derivative is the constant \(k!c_k=f^{(k)}(a)\). Thus, at \(x=a\), only the term indexed by \(k=j\) contributes to the \(j\)th derivative. This proves the stated match for every \(j\) from \(0\) to \(n\).

How to Choose a Center

Begin with the input values where an approximation will be used. A useful center is generally near those values, because the polynomial is designed to describe behavior locally around that point. If several target values lie in a small interval, a center near the middle of that interval may be a practical choice; if there is one main target, a nearby center is often natural.

Also check that the function has the needed derivatives at the proposed center. A simple center such as \(0\) is not automatically suitable. The center should be selected from the function’s domain and with the required differentiability in mind. When several centers are possible, it can help to prefer one where the derivative values are manageable or already known.

Finally, compare the target’s distance from the center with the size of the terms in the polynomial. The displacement \(h=x-a\) is the input measured from the center. When \(h\) is small, the higher powers \(h^2,h^3,\ldots\) are often progressively smaller. That observation helps explain why a local polynomial can work well nearby, but it does not by itself prove how accurate the approximation is.

How to Choose a Degree

The degree controls which local features are included. A degree-zero polynomial uses only the function’s value at the center. A degree-one polynomial adds the slope and gives a linear model. A degree-two polynomial also matches curvature. Still higher degrees match more derivatives. Choose a degree based on the detail the task needs, rather than treating the largest available degree as automatically necessary.

A practical approach is to calculate successive approximations at the target input. If adding a term changes the answer substantially, that term matters for the current model. If it changes the answer only slightly, the lower-degree result may already be adequate for a rough estimate. This comparison is a guide, not a rigorous error guarantee: a small latest term alone does not establish a bound on the difference between the polynomial and the function.

Locate the target input.
Identify the value, or range of values, where the approximation will be used.

Select a suitable center.
Choose a nearby point where the function has the derivatives required for the model.

Set a starting degree.
Use a low degree for a simple local description; include higher terms when slope alone or curvature does not capture enough detail.

Evaluate and compare.
Compute the polynomial at the target, then check how the estimate changes when another term is included.

Worked Examples

Worked Example: Choosing a Degree for Sine Near Zero

Approximate \(\sin(0.2)\) using Taylor polynomials centered at \(0\), and compare degrees one and three. Since the target \(0.2\) is near zero and the derivatives of sine there are simple, zero is a convenient center.

The degree-one polynomial uses \(\sin(0)=0\) and \(\sin'(0)=\cos(0)=1\):

$$ T_1(x;0)=0+1x=x. $$

Thus \(T_1(0.2;0)=0.2\). For degree three, also use \(\sin''(0)=0\) and \(\sin'''(0)=-1\):

T_3(x;0)=0+x+0\frac{x^2}{2!}-\frac{x^3}{3!} =x-\frac{x^3}{6}.

Since \(0.2^3=0.008\) and \(0.008/6=0.001333333\ldots\),

T_3(0.2;0)=0.2-0.001333333\ldots=0.198666666\ldots.

For a check, the polynomials have the required derivative values at zero: \(T_1(0;0)=0\) and \(T_1'(0;0)=1\); also \(T_3''(0;0)=0\) and \(T_3'''(0;0)=-1\). Numerically, \(\sin(0.2)\approx0.1986693308\). The degree-one estimate differs by about \(0.0013306692\), while the degree-three estimate differs by about \(0.0000026641\). In this case the added cubic term substantially improves the estimate.

Worked Example: Moving the Center Near the Inputs

Approximate \(\sqrt{4.1}\) using a Taylor polynomial centered at \(4\). This center is near the target, and the function \(f(x)=\sqrt{x}\) has the needed derivatives around \(4\).

At \(x=4\), \(f(4)=2\), \(f'(x)=1/(2\sqrt{x})\), so \(f'(4)=1/4\). Also \(f''(x)=-1/(4x^{3/2})\), so \(f''(4)=-1/32\). Therefore the degree-one and degree-two polynomials are

T_1(x;4)=2+\frac{x-4}{4}, \qquad T_2(x;4)=2+\frac{x-4}{4}-\frac{(x-4)^2}{64}.

At the target, \(h=4.1-4=0.1\). The degree-one estimate is

T_1(4.1;4)=2+\frac{0.1}{4}=2+0.025=2.025.

For degree two, \(h^2=0.01\) and \(0.01/64=0.00015625\), so

$$ T_2(4.1;4)=2+0.025-0.00015625=2.02484375. $$

Check the construction by differentiating: \(T_2(4;4)=2\), \(T_2'(x;4)=1/4-(x-4)/32\), so \(T_2'(4;4)=1/4\), and \(T_2''(x;4)=-1/32\). These match \(f(4)\), \(f'(4)\), and \(f''(4)\). Since \(\sqrt{4.1}\approx2.0248456731\), the degree-one estimate differs by about \(0.0001543269\), and the degree-two estimate differs by about \(0.0000019231\). The comparison suggests that curvature is useful at this input.

Worked Example: Comparing Degrees for a Reciprocal Function

Approximate \(1/1.2\) with Taylor polynomials for \(f(x)=1/x\) centered at \(1\). The center is close to the input, and \(f\) has derivatives of every order at \(1\).

The derivative formula is \(f^{(k)}(x)=(-1)^k k!x^{-k-1}\), so \(f^{(k)}(1)/k!=(-1)^k\). With \(h=x-1\), this gives

T_1(x;1)=1-h,\qquad T_2(x;1)=1-h+h^2,\qquad T_3(x;1)=1-h+h^2-h^3.

At \(x=1.2\), \(h=0.2\), \(h^2=0.04\), and \(h^3=0.008\). Thus the three estimates are

$$ T_1(1.2;1)=1-0.2=0.8, $$

$$ T_2(1.2;1)=1-0.2+0.04=0.84, $$

$$ T_3(1.2;1)=1-0.2+0.04-0.008=0.832. $$

Each polynomial has the stated derivative data at the center: its \(k\)th derivative at \(1\) is \((-1)^k k!\) for every included order \(k\). The exact value is \(1/1.2=10/12=5/6=0.833333\ldots\). The absolute differences are approximately \(0.0333333\) for degree one, \(0.0066667\) for degree two, and \(0.0013333\) for degree three. Here each added term moves the estimate closer to the exact value, but this example does not establish that every added term will do so for every function and input.

A Common Mistake: Choosing a Center Without Checking Derivatives

A tempting but invalid choice is to center a Taylor polynomial for \(\sqrt{x}\) at \(0\), simply because zero is a familiar center. The function is defined at zero, but its derivative there is not finite:

f'(0)=\lim_{h\to0^+}\frac{\sqrt{h}-\sqrt{0}}{h} =\lim_{h\to0^+}\frac{1}{\sqrt{h}},

and this limit is unbounded. Therefore the ordinary degree-one Taylor polynomial at zero cannot be formed using a finite value of \(f'(0)\). Writing a linear Taylor term there would be an error. For an input such as \(4.1\), centering at \(4\) is a valid alternative: the function has the required derivatives there, and the target is nearby.

A second caution concerns degree. A higher-degree polynomial contains more derivative information, but it should not be called “accurate enough” merely because it is higher degree. Compare computed estimates when that is useful, and distinguish that evidence from a proven error bound. The center, the input’s distance from it, and the function’s behavior all matter.

Why These Choices Matter

Choosing a center and degree is part of building a useful local description, not merely a preliminary calculation. The center says where the model is anchored. The degree says which derivative features are represented. Together they determine whether the polynomial reflects the function’s behavior near the inputs that matter.

For a rough estimate close to a center, a low-degree model may be sufficient. If curvature matters, a linear model may not be enough. If the target is far from a proposed center, consider whether a different center would better fit the intended use. In every case, check differentiability at the center and treat comparisons between polynomial degrees as practical evidence unless a separate error analysis justifies a precise accuracy claim.

Check Your Understanding

Use the definitions and examples above to answer these questions.

What information does the center of a Taylor polynomial specify, and what information does its degree specify?
Why is a center near the target input often useful for a local approximation?
For a degree-two Taylor polynomial, which function data must match at the center?
Why can comparing successive polynomial estimates be helpful, and what does that comparison fail to prove on its own?
Why can \(\sqrt{x}\) not have an ordinary degree-one Taylor polynomial centered at zero?