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Taylor Polynomials at Nonzero Centers

Build a Taylor polynomial at a nonzero center by calculating the function’s derivatives there and matching them with the polynomial.

Advanced 9 min read

What You'll Learn

  • Write the degree-n Taylor polynomial centered at a using derivatives evaluated at a.
  • Explain why matching derivatives determines each coefficient.
  • Build and check Taylor polynomials for exponential functions at nonzero centers.
  • Build and check a Taylor polynomial for a reciprocal function away from its singularity.
  • Recognize why using powers of x instead of powers of x minus a gives the wrong center.

Why the Center Matters

A polynomial can model a function near a chosen input value. When that value is not zero, the polynomial is built from powers of the distance from the center, \(x-a\), rather than powers of \(x\). The derivatives of the function at \(a\) determine the coefficients.

This choice lets the polynomial match the function’s value, slope, and higher derivatives at the selected point. For example, a Taylor polynomial centered at \(a=2\) is designed to match the function at \(x=2\), not at \(x=0\).

Key idea: To construct a Taylor polynomial centered at \(x=a\), evaluate the function’s derivatives at \(a\), then use those values as coefficients of powers of \(x-a\), with each derivative value divided by the corresponding factorial.

Definition and Formula

Suppose a function \(f\) has derivatives through order \(n\) at \(x=a\). 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 \(f^{(0)}(a)\) means \(f(a)\), and \(0!=1\). Written term by term, the same formula is

$$ T_n(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\cdots+\frac{f^{(n)}(a)}{n!}(x-a)^n. $$

The polynomial has degree at most \(n\): its \(n\)th-degree coefficient could be zero. The center \(a\) can be any point where the required derivatives exist. For instance, \(1/x\) has a Taylor polynomial centered at \(a=2\), since its derivatives exist at \(2\), even though the function is not defined at \(x=0\).

Why the Derivative Formula Works

The formula is chosen so that the polynomial has the same value and derivatives as \(f\) at the center, through order \(n\). To see why the coefficients have to take this form, begin with a general polynomial expressed in powers of \(x-a\):

$$ P_n(x)=c_0+c_1(x-a)+c_2(x-a)^2+\cdots+c_n(x-a)^n. $$

When we evaluate \(P_n\) at \(a\), every positive power of \(x-a\) becomes zero. Thus \(P_n(a)=c_0\). When we differentiate once and evaluate at \(a\), the constant term vanishes, terms with powers above one still contain a positive power of \(x-a\), and only the linear term contributes. Therefore \(P_n'(a)=c_1\).

More generally, after differentiating \(j\) times, terms with powers below \(j\) have disappeared, terms with powers above \(j\) still contain a positive power of \(x-a\) at the center, and the \(j\)th-power term contributes \(j!c_j\). Consequently,

$$ P_n^{(j)}(a)=j!c_j. $$

To make \(P_n\) match \(f\) and its derivatives at \(a\), set \(P_n^{(j)}(a)=f^{(j)}(a)\). Solving the displayed equation for \(c_j\) gives \(c_j=f^{(j)}(a)/j!\). This works for every \(j\) from \(0\) through \(n\), proving the formula and showing why the coefficients are determined by the derivatives at the center.

Theorem: For \(0\leq j\leq n\), the Taylor polynomial satisfies \(T_n^{(j)}(a)=f^{(j)}(a)\). It is the unique polynomial of degree at most \(n\) with these matching derivative values at \(a\).

For uniqueness, any polynomial of degree at most \(n\) can be written in powers of \(x-a\). The matching condition forces each coefficient \(c_j\) to equal \(f^{(j)}(a)/j!\), leaving no other possible polynomial.

Worked Examples

Worked Example: An Exponential Function Centered at 1

Find the degree-3 Taylor polynomial for \(f(x)=e^x\) centered at \(a=1\), and check its derivatives at the center.

Every derivative of \(e^x\) is \(e^x\). Thus the values needed are \(f(1)=e\), \(f'(1)=e\), \(f''(1)=e\), and \(f'''(1)=e\). Substitute each into the formula:

$$ \begin{aligned} T_3(x) &=f(1)+f'(1)(x-1)+\frac{f''(1)}{2!}(x-1)^2+\frac{f'''(1)}{3!}(x-1)^3\\ &=e+e(x-1)+\frac{e}{2}(x-1)^2+\frac{e}{6}(x-1)^3. \end{aligned} $$

To check, differentiate the polynomial:

$$ \begin{aligned} T_3'(x)&=e+e(x-1)+\frac{e}{2}(x-1)^2,\\ T_3''(x)&=e+e(x-1),\\ T_3'''(x)&=e. \end{aligned} $$

At \(x=1\), these give \(T_3(1)=e\), \(T_3'(1)=e\), \(T_3''(1)=e\), and \(T_3'''(1)=e\). Each equals the corresponding derivative of \(e^x\) at 1, as required.

Worked Example: A Reciprocal Function Centered at 2

Find the degree-3 Taylor polynomial for \(f(x)=1/x\) centered at \(a=2\), and verify the matches. The function and its derivatives are defined at 2; the center must not be its singularity at \(x=0\).

Differentiate successively:

$$ \begin{aligned} f(x)&=x^{-1}, & f'(x)&=-x^{-2},\\ f''(x)&=2x^{-3}, & f'''(x)&=-6x^{-4}. \end{aligned} $$

Now evaluate each derivative at 2:

$$ \begin{aligned} f(2)&=\frac12, & f'(2)&=-\frac14,\\ f''(2)&=\frac{2}{8}=\frac14, & f'''(2)&=-\frac{6}{16}=-\frac38. \end{aligned} $$

Use these values, dividing the second- and third-derivative terms by \(2!=2\) and \(3!=6\):

$$ \begin{aligned} T_3(x) &=\frac12-\frac14(x-2)+\frac{1/4}{2}(x-2)^2+\frac{-3/8}{6}(x-2)^3\\ &=\frac12-\frac14(x-2)+\frac18(x-2)^2-\frac1{16}(x-2)^3. \end{aligned} $$

Differentiate to check:

$$ \begin{aligned} T_3'(x)&=-\frac14+\frac14(x-2)-\frac3{16}(x-2)^2,\\ T_3''(x)&=\frac14-\frac38(x-2),\\ T_3'''(x)&=-\frac38. \end{aligned} $$

At \(x=2\), the polynomial and its derivatives are \(1/2\), \(-1/4\), \(1/4\), and \(-3/8\), respectively. These match \(f(2)\), \(f'(2)\), \(f''(2)\), and \(f'''(2)\).

Worked Example: An Exponential with a Chain-Rule Factor

Find the degree-2 Taylor polynomial for \(g(x)=e^{2x}\) centered at \(a=1\). This example shows why derivatives must be calculated before they are evaluated.

Using the chain rule, \(g'(x)=2e^{2x}\), and differentiating again gives \(g''(x)=4e^{2x}\). At the center, the values are \(g(1)=e^2\), \(g'(1)=2e^2\), and \(g''(1)=4e^2\). Therefore,

$$ \begin{aligned} T_2(x) &=g(1)+g'(1)(x-1)+\frac{g''(1)}{2!}(x-1)^2\\ &=e^2+2e^2(x-1)+\frac{4e^2}{2}(x-1)^2\\ &=e^2\left(1+2(x-1)+2(x-1)^2\right). \end{aligned} $$

Differentiate the polynomial: \(T_2'(x)=2e^2+4e^2(x-1)\) and \(T_2''(x)=4e^2\). At \(x=1\), the polynomial and these derivatives are \(e^2\), \(2e^2\), and \(4e^2\), matching \(g(1)\), \(g'(1)\), and \(g''(1)\).

A Common Centering Mistake

A frequent error is to calculate derivative values at a nonzero center but then write powers of \(x\) instead of powers of \(x-a\). For the degree-3 polynomial for \(e^x\) centered at 1, the incorrect expression would be

$$ e+ex+\frac{e}{2}x^2+\frac{e}{6}x^3. $$

This does not even match the function’s value at the center: at \(x=1\), the expression is \(e(1+1+1/2+1/6)=8e/3\), not \(e\). The correction is to use \(x-1\) in every nonconstant term. Substituting \(x=1\) then makes all those terms zero, leaving the correct value \(e\).

Check before finishing: At the center \(x=a\), every positive power of \(x-a\) must vanish. Therefore the constant value of the Taylor polynomial there must be \(f(a)\).

What the Polynomial Guarantees

The derivative-matching property is exact: the Taylor polynomial and the function have the same value and first \(n\) derivatives at the center. That is what the construction guarantees. It does not, by itself, say how accurate the polynomial is at a particular point away from the center or whether an associated infinite Taylor series converges to the function. Those are separate questions.

For AP problems, a reliable construction is to identify the center and degree, compute the required derivatives, evaluate each derivative at the center, and then substitute into the formula using powers of \(x-a\). Checking the polynomial’s derivatives at \(a\) can catch errors in the center, factorials, or derivative values.

1
Identify the center and degree.
Record \(a\) and \(n\); the polynomial will use powers through \((x-a)^n\).
2
Calculate and evaluate derivatives.
Find \(f^{(k)}(a)\) for each integer \(k\) from \(0\) through \(n\).
3
Build the polynomial.
Use \(f^{(k)}(a)/k!\) as the coefficient of \((x-a)^k\).
4
Check at the center.
Differentiate the polynomial as needed and verify that its values at \(a\) match the function’s derivatives there.

Check Your Understanding

For each question, use the derivative formula and keep the center visible in every power.

  1. Write the general formula for the degree-2 Taylor polynomial of \(f\) centered at \(a\).
  2. Find the degree-2 Taylor polynomial for \(e^x\) centered at \(a=2\).
  3. Find the degree-2 Taylor polynomial for \(1/x\) centered at \(a=1\).
  4. Why does the coefficient of \((x-a)^j\) have to be \(f^{(j)}(a)/j!\)?
  5. What is wrong with using powers of \(x\) in a Taylor polynomial centered at a nonzero \(a\)?

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