From a Derivative Table to a Polynomial
A Taylor polynomial can be built from a list of derivative values at one point. This is useful when a problem provides the function’s values in a table rather than as a formula that is easy to differentiate. The process is systematic: identify the center, use each derivative value in the matching term, and evaluate with the input’s displacement from that center.
The center matters. A table of derivatives at \(x=a\) gives the Taylor polynomial centered at \(a\), not automatically at zero. And the derivatives are not themselves the coefficients: the \(k\)th derivative value must be divided by \(k!\).
In expanded form, the polynomial begins $$ T_n(x)=f(a)+f'(a)(x-a)+\frac{f''(a)}{2!}(x-a)^2+\frac{f'''(a)}{3!}(x-a)^3+\cdots+\frac{f^{(n)}(a)}{n!}(x-a)^n. $$ Each row of a derivative table supplies one of the values in this expression.
How to Read the Table
Before calculating, check which input the table uses. If the table lists values at \(x=a\), that is the center. Then check the requested degree. A degree-\(n\) polynomial needs \(f(a)\) and the derivatives through \(f^{(n)}(a)\). Values for higher derivatives are not needed for that polynomial.
| Table entry | Term in the Taylor polynomial |
|---|---|
| \(f(a)\) | \(f(a)\) |
| \(f'(a)\) | \(f'(a)(x-a)\) |
| \(f''(a)\) | \(\frac{f''(a)}{2!}(x-a)^2\) |
| \(f'''(a)\) | \(\frac{f'''(a)}{3!}(x-a)^3\) |
| \(f^{(k)}(a)\) | \(\frac{f^{(k)}(a)}{k!}(x-a)^k\) |
A reliable construction process is to write the center and degree first, then fill in one term at a time. Once the polynomial is assembled, substitute the requested input into \(x-a\), not into \(x\) alone.
Read the \(x\)-value at which the table gives \(f(a)\) and its derivatives.
For degree \(n\), use the table entries from \(f(a)\) through \(f^{(n)}(a)\).
For each \(k\), multiply \((x-a)^k\) by \(f^{(k)}(a)/k!\).
Substitute the input into the completed polynomial, then verify its derivatives at the center agree with the table entries used.
Why the Factorials Appear
The factorial is needed because differentiating a power \( (x-a)^k \) exactly \(k\) times produces the constant \(k!\). More precisely, if a polynomial is written as \(P(x)=\sum_{k=0}^{n}c_k(x-a)^k\), then its \(j\)th derivative at \(a\) is \(P^{(j)}(a)=j!c_j\). Terms with exponent below \(j\) have become zero after \(j\) derivatives; terms with exponent above \(j\) still contain a positive power of \(x-a\), so they vanish when \(x=a\). Only the \(j\)th term contributes at the center.
Therefore, to make \(P^{(j)}(a)=f^{(j)}(a)\), its coefficient must be \(c_j=f^{(j)}(a)/j!\). This gives the Taylor-polynomial formula directly and explains why each derivative-table value is scaled by its matching factorial.
Worked Example: Construct a Cubic at a Nonzero Center
Suppose the derivative table gives the following values at \(x=2\), and construct the degree-three Taylor polynomial. Then evaluate it at \(x=2.1\).
| Value | At \(x=2\) |
|---|---|
| \(f(2)\) | \(3\) |
| \(f'(2)\) | \(-2\) |
| \(f''(2)\) | \(6\) |
| \(f'''(2)\) | \(-24\) |
The center is \(a=2\), and the degree is \(3\). Thus $$ T_3(x)=f(2)+f'(2)(x-2)+\frac{f''(2)}{2!}(x-2)^2+\frac{f'''(2)}{3!}(x-2)^3. $$ Substitute each table value: $$ T_3(x)=3-2(x-2)+\frac{6}{2}(x-2)^2+\frac{-24}{6}(x-2)^3 =3-2(x-2)+3(x-2)^2-4(x-2)^3. $$ For \(x=2.1\), the displacement is \(x-2=2.1-2=0.1\). Therefore $$ T_3(2.1)=3-2(0.1)+3(0.1)^2-4(0.1)^3 =3-0.2+0.03-0.004=2.826. $$ To check the construction, write \(h=x-2\), so \(T_3=3-2h+3h^2-4h^3\). Its derivatives with respect to \(x\) are $$ T_3'(x)=-2+6h-12h^2,\qquad T_3''(x)=6-24h,\qquad T_3'''(x)=-24. $$ At \(x=2\), \(h=0\), so \(T_3(2)=3\), \(T_3'(2)=-2\), \(T_3''(2)=6\), and \(T_3'''(2)=-24\), as required.
Evaluation Uses the Displacement
For a center \(a\) and an input \(x\), the quantity used in each power is the displacement \(x-a\). If the input is close to the center, this displacement is small. The polynomial is evaluated by substituting that displacement into every term, preserving its sign. The calculation itself does not require a formula for \(f\); the table provides the information needed to form the polynomial.
Worked Example: Use a Fourth-Degree Polynomial
At \(x=-1\), suppose a table gives \(f(-1)=2\), \(f'(-1)=0\), \(f''(-1)=-8\), \(f'''(-1)=12\), and \(f^{(4)}(-1)=48\). Construct the degree-four Taylor polynomial and evaluate it at \(x=-0.8\).
Here \(a=-1\), so the powers are powers of \(x-(-1)=x+1\). Use the formula through the fourth derivative: $$ T_4(x)=2+0(x+1)+\frac{-8}{2!}(x+1)^2+\frac{12}{3!}(x+1)^3+\frac{48}{4!}(x+1)^4. $$ Since \(2!=2\), \(3!=6\), and \(4!=24\), $$ T_4(x)=2-4(x+1)^2+2(x+1)^3+2(x+1)^4. $$ At \(x=-0.8\), the displacement is \(-0.8-(-1)=0.2\). Substitution gives $$ T_4(-0.8)=2-4(0.2)^2+2(0.2)^3+2(0.2)^4 =2-0.16+0.016+0.0032=1.8592. $$ For a check, set \(h=x+1\). Then \(T_4=2-4h^2+2h^3+2h^4\), and $$ T_4'(x)=-8h+6h^2+8h^3,\quad T_4''(x)=-8+12h+24h^2,\quad T_4'''(x)=12+48h,\quad T_4^{(4)}(x)=48. $$ At the center \(x=-1\), \(h=0\). These give \(T_4(-1)=2\), \(T_4'(-1)=0\), \(T_4''(-1)=-8\), \(T_4'''(-1)=12\), and \(T_4^{(4)}(-1)=48\), matching every entry used.
A Table with a Zero Derivative
A derivative-table entry may be zero. Its corresponding term is then zero and can be omitted. This does not mean that the polynomial’s degree must be lower: a later derivative may still produce a nonzero term of higher degree.
Worked Example: Keep the Degree When a Term Vanishes
At \(x=1\), suppose \(f(1)=1\), \(f'(1)=2\), \(f''(1)=0\), \(f'''(1)=-6\), and \(f^{(4)}(1)=24\). Construct the degree-four Taylor polynomial and evaluate it at \(x=0.9\).
With \(a=1\), the formula gives $$ T_4(x)=1+2(x-1)+\frac{0}{2!}(x-1)^2+\frac{-6}{3!}(x-1)^3+\frac{24}{4!}(x-1)^4. $$ Using \(3!=6\) and \(4!=24\), and dropping the zero term, $$ T_4(x)=1+2(x-1)-(x-1)^3+(x-1)^4. $$ At \(x=0.9\), \(x-1=-0.1\), so $$ T_4(0.9)=1+2(-0.1)-(-0.1)^3+(-0.1)^4 =1-0.2+0.001+0.0001=0.8011. $$ To check, let \(h=x-1\). Then \(T_4=1+2h-h^3+h^4\), so $$ T_4'(x)=2-3h^2+4h^3,\quad T_4''(x)=-6h+12h^2,\quad T_4'''(x)=-6+24h,\quad T_4^{(4)}(x)=24. $$ At \(x=1\), \(h=0\), and the value and derivatives are \(1,2,0,-6,24\), exactly as in the table.
A Common Pitfall: Substituting the Input Instead of the Displacement
Suppose a polynomial is centered at \(2\) and the requested input is \(2.1\). The substitution for its powers is \(x-2=0.1\), not \(2.1\). Using \(2.1\) directly would evaluate the polynomial at the wrong displacement and generally produce a different result. In the cubic example above, the correct substitution is $$ T_3(2.1)=3-2(0.1)+3(0.1)^2-4(0.1)^3=2.826, $$ whereas substituting \(2.1\) for every occurrence of \(x-2\) would calculate \(3-2(2.1)+3(2.1)^2-4(2.1)^3\), which is not \(T_3(2.1)\). Keep the expression \(x-a\) intact until the input and center have both been identified.
What the Polynomial Represents
The derivative table determines a unique Taylor polynomial of the requested degree by fixing each coefficient. The result is a polynomial that matches the function’s value and the listed derivatives at the center. At a nearby input, its value is the Taylor-polynomial estimate obtained from those data. The calculation produces the polynomial value; by itself, the table does not provide the exact value of the function at that nearby input.
Check Your Understanding
Use the derivative-table formula carefully. For each question, identify the center and use the stated degree.
- At \(x=3\), a table gives \(f(3)=4\), \(f'(3)=5\), and \(f''(3)=8\). Write the degree-two Taylor polynomial centered at \(3\).
- For the polynomial in Question 1, what displacement should be used to evaluate it at \(x=3.2\)?
- At \(x=0\), suppose \(f(0)=2\), \(f'(0)=-3\), and \(f''(0)=12\). Write the degree-two Taylor polynomial and evaluate it at \(x=0.1\).
- Why does the coefficient of \((x-a)^4\) use \(f^{(4)}(a)/4!\), rather than just \(f^{(4)}(a)\)?
- If \(f''(a)=0\) but \(f'''(a)\ne 0\), what happens to the quadratic term, and can the cubic term still appear?
