Computing a Taylor Polynomial from Derivative Data

Reading Derivative Data as Polynomial Coefficients

A table of function values and derivatives at one point contains exactly the information needed to construct a Taylor polynomial centered at that point. The main task is to place each value in the correct position: the \(k\)-th derivative belongs with the \(k\)-th power of the distance from the center, divided by \(k!\).

This procedure works for any center, not only zero. If the center is \(a\), the powers are \(1\), \(x-a\), \((x-a)^2\), and so on. Keeping the center visible helps prevent a common mistake: using powers of \(x\) when the data are given at a point other than zero.

Key idea: To construct the degree-\(n\) Taylor polynomial at \(x=a\), put \(f^{(k)}(a)\) over \(k!\) and multiply by \((x-a)^k\), for each \(k\) from \(0\) to \(n\).

The Formula and Its Ingredients

Suppose the function \(f\) has derivatives at \(a\) through order \(n\), and those derivative values are known. The Taylor polynomial of degree \(n\) centered at \(a\) is

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

Here \(f^{(0)}(a)\) means \(f(a)\), the function value itself. The factorial is defined by \(0!=1\), and for a positive integer \(k\), \(k!=k(k-1)\cdots 2\cdot1\). Thus the first terms, written out, are

P_n(x) = f(a) +f'(a)(x-a) +\frac{f''(a)}{2!}(x-a)^2 +\frac{f'''(a)}{3!}(x-a)^3 +\cdots +\frac{f^{(n)}(a)}{n!}(x-a)^n.

The term for \(k=0\) is just \(f(a)\), since \((x-a)^0=1\) and \(0!=1\). Each later term pairs a derivative with the same-order power: the second derivative with \((x-a)^2\), the third derivative with \((x-a)^3\), and so forth.

If a table lists derivative values from order zero through order \(n\), it supplies \(n+1\) entries and determines a polynomial of degree at most \(n\). “At most” matters: if the highest-order derivative value is zero, the coefficient of that power is zero, so the resulting polynomial may have lower degree.

Why Each Derivative Is Divided by a Factorial

The factorial is not an extra adjustment to memorize without reason. It is precisely what makes the polynomial’s \(k\)-th derivative at the center equal the supplied \(k\)-th derivative value.

To see this, write a polynomial in powers of \(x-a\) as \(P(x)=\sum_{j=0}^{n}c_j(x-a)^j\). When it is differentiated \(k\) times and evaluated at \(x=a\), a term with \(j<k\) has already differentiated to zero. A term with \(j>k\) still contains a positive power of \(x-a\), so it is zero at \(a\). Only the \(j=k\) term remains, and its \(k\)-th derivative is \(k!c_k\). Therefore

P^{(k)}(a)=k!c_k.

To make that derivative equal \(f^{(k)}(a)\), the coefficient must be \(c_k=f^{(k)}(a)/k!\). Applying this reasoning for each \(k\) gives the Taylor polynomial formula and shows why the derivative values must be divided by their corresponding factorials.

Coefficient check: In a polynomial written in powers of \(x-a\), the coefficient of \((x-a)^k\) must be \(f^{(k)}(a)/k!\) if the polynomial is to have \(k\)-th derivative \(f^{(k)}(a)\) at the center.

A Reliable Procedure for Using a Table

Before calculating, identify the center \(a\) and the highest derivative order listed. Then transcribe the data carefully: the value \(f(a)\) is order zero, \(f'(a)\) is order one, \(f''(a)\) is order two, and so on. Match the order in the table to both the power and the factorial in the formula.

Find the center.
Read the point \(a\) where the function and derivative values are given.

Find the degree.
If the data go through \(f^{(n)}(a)\), use powers through \((x-a)^n\).

Write one term per table entry.
For order \(k\), use \(f^{(k)}(a)(x-a)^k/k!\).

Simplify and check.
Differentiate the polynomial and evaluate its derivatives at \(a\) to confirm the supplied data.

Worked Examples

Worked Example: A Cubic from Data at a Nonzero Center

Suppose the derivative data are given at \(a=2\):

Quantity	Value
\(f(2)\)	\(3\)
\(f'(2)\)	\(-2\)
\(f''(2)\)	\(8\)
\(f'''(2)\)	\(-6\)

The highest derivative listed is the third derivative, so construct a degree-three polynomial. Use \(x-2\) for the powers because the center is \(2\):

\begin{aligned} P_3(x) &=f(2)+f'(2)(x-2) +\frac{f''(2)}{2!}(x-2)^2 +\frac{f'''(2)}{3!}(x-2)^3\\ &=3-2(x-2)+\frac{8}{2}(x-2)^2+\frac{-6}{6}(x-2)^3\\ &=3-2(x-2)+4(x-2)^2-(x-2)^3. \end{aligned}

Check the data by differentiating:

\begin{aligned} P_3(2)&=3,\\ P_3'(x)&=-2+8(x-2)-3(x-2)^2, & P_3'(2)&=-2,\\ P_3''(x)&=8-6(x-2), & P_3''(2)&=8,\\ P_3'''(x)&=-6, & P_3'''(2)&=-6. \end{aligned}

All four values match the table, so the polynomial has the required derivative data at \(x=2\).

Worked Example: A Quartic with a Zero First Derivative

At \(a=1\), suppose the table gives \(f(1)=2\), \(f'(1)=0\), \(f''(1)=-4\), \(f'''(1)=12\), and \(f^{(4)}(1)=24\). The data go through order four, so the polynomial is

\begin{aligned} P_4(x) &=2+0(x-1)+\frac{-4}{2!}(x-1)^2 +\frac{12}{3!}(x-1)^3 +\frac{24}{4!}(x-1)^4\\ &=2-2(x-1)^2+2(x-1)^3+(x-1)^4. \end{aligned}

The zero first derivative gives a zero linear term; it does not change the powers or the factorials for the later terms. To check, differentiate and evaluate at \(x=1\):

\begin{aligned} P_4'(x)&=-4(x-1)+6(x-1)^2+4(x-1)^3, & P_4'(1)&=0,\\ P_4''(x)&=-4+12(x-1)+12(x-1)^2, & P_4''(1)&=-4,\\ P_4'''(x)&=12+24(x-1), & P_4'''(1)&=12,\\ P_4^{(4)}(x)&=24, & P_4^{(4)}(1)&=24. \end{aligned}

Also, substituting \(x=1\) into \(P_4\) gives \(P_4(1)=2\). Thus the polynomial reproduces every entry in the table.

Worked Example: A Quadratic Centered at a Negative Number

Suppose \(a=-1\), \(f(-1)=0\), \(f'(-1)=5\), and \(f''(-1)=-2\). Since the data stop at the second derivative, use a degree-two polynomial. The powers are based on \(x-(-1)=x+1\):

\begin{aligned} P_2(x) &=f(-1)+f'(-1)(x+1)+\frac{f''(-1)}{2!}(x+1)^2\\ &=0+5(x+1)+\frac{-2}{2}(x+1)^2\\ &=5(x+1)-(x+1)^2. \end{aligned}

Differentiate to check the data:

\begin{aligned} P_2(-1)&=5(0)-0^2=0,\\ P_2'(x)&=5-2(x+1), & P_2'(-1)&=5,\\ P_2''(x)&=-2, & P_2''(-1)&=-2. \end{aligned}

The value and both derivatives agree with the supplied data. The center can be negative; the same formula applies as long as the powers use the distance from that center.

Common Pitfall: Using the Derivative as the Coefficient

A frequent error is to put the derivative value directly in front of its power, forgetting to divide by the factorial. For example, if \(f'''(a)=12\), writing \(12(x-a)^3\) as the cubic term is incorrect. Differentiating that term three times gives \(12\cdot3!=72\), so its third derivative at \(a\) is \(72\), not \(12\).

The correct coefficient is \(12/3!=12/6=2\). The cubic term is therefore \(2(x-a)^3\), whose third derivative is \(2\cdot3!=12\), as required. A quick way to catch this mistake is to differentiate each term the number of times indicated by its order and check the value at the center.

Another related error is using \(x^k\) instead of \((x-a)^k\). That is correct only when the center is zero. For data at \(a=2\), for instance, the linear term is \(f'(2)(x-2)\), not \(f'(2)x\). At the center, \(x-2=0\), which makes it straightforward to check the intended derivative data.

What the Construction Does—and Does Not—Require

For this calculation, the essential input is a center and a consistent list of derivative values at that point. The formula constructs a polynomial whose value and derivatives through the listed order match those data at the center. If the data stop at order \(n\), no values of derivatives above order \(n\) are needed to build \(P_n\).

The construction is a precise algebraic task: place each supplied derivative over its factorial, attach the matching power of \(x-a\), and add the terms. It does not, by itself, establish how closely the polynomial matches the function at other points. Here the goal is to compute the polynomial from the derivative data and verify that its derivatives at the center are correct.

Takeaway: Read the derivative order carefully. Order \(k\) contributes \(\frac{f^{(k)}(a)}{k!}(x-a)^k\). After forming the polynomial, differentiation at \(x=a\) provides a direct check against the table.

Check Your Understanding

Use the derivative order, center, and factorial carefully in each question.

If \(f(3)=4\), \(f'(3)=-1\), and \(f''(3)=6\), write the degree-two Taylor polynomial centered at \(3\).
For a Taylor polynomial centered at \(a=-2\), what power of \(x\) appears in the term corresponding to \(f'''(-2)\), and what factorial divides its coefficient?
If \(f^{(4)}(a)=48\), what is the coefficient of \((x-a)^4\) in the degree-four Taylor polynomial?
Explain why a zero value of \(f'(a)\) makes the linear term zero but does not change the quadratic term’s factorial.
Differentiate the polynomial \(P_2(x)=2+3(x-1)-2(x-1)^2\) and check \(P_2(1)\), \(P_2'(1)\), and \(P_2''(1)\).