Why the Center and the Order Matter
A Taylor polynomial is built from information about a function at one chosen input, called the center. Its terms use powers of the displacement from that center. The chosen center and the chosen degree therefore determine how the polynomial is written and which derivative values are included.
The notation keeps two roles separate: \(a\) is the fixed center, while \(x\) is the variable. As \(x\) varies, the displacement \(x-a\) varies; the center \(a\) does not. This distinction matters even when a calculation is short, because the powers in the polynomial are powers of \(x-a\), not generally powers of \(x\).
The formula includes one term for each integer \(k\) from \(0\) through \(n\), inclusive. In particular, the coefficient of \((x-a)^k\) in \(T_n\) is \(f^{(k)}(a)/k!\) for each \(k\) with \(0\le k\le n\). The constant term is \(f(a)\), because \((x-a)^0=1\). When \(n\ge1\), the coefficient of \(x-a\) is \(f'(a)\); when \(n\ge2\), the coefficient of \((x-a)^2\) is \(f''(a)/2!\).
These statements concern only terms that the chosen polynomial includes. A degree-zero Taylor polynomial, for example, has no \(x-a\) term. Defining \(T_n\) requires derivative values only through order \(n\); no derivative of order greater than \(n\) occurs in the definition.
What the Symbols Tell You
The symbol \(T_n\) records the order \(n\) chosen for the polynomial. The input \(x\) is the variable at which we evaluate that polynomial. The fixed input \(a\) identifies the center, and the values \(f(a),f'(a),\ldots,f^{(n)}(a)\) supply the data for its coefficients.
The use of \(x-a\) makes the center visible in every nonconstant term. At \(x=a\), the displacement is zero, so all positive powers of \(x-a\) vanish. That is why substituting \(x=a\) into the defining formula gives \(T_n(a)=f(a)\). For \(x\) away from \(a\), the powers \((x-a)^k\) describe the terms according to their order \(k\).
The factorial \(k!\) is the product of the positive integers from \(1\) through \(k\), with \(0!=1\). Thus \(1!=1\), \(2!=2\), and \(3!=6\). The denominator is part of the coefficient: it is not an instruction to divide the whole polynomial or the power \((x-a)^k\) by \(k!\) again.
Worked Examples
Worked Example: Writing a Third-Order Taylor Polynomial from Derivative Data
Suppose \(f\) has the following values at the center \(a=2\):
For \(n=3\), the definition includes the terms \(k=0,1,2,3\). Substitute the center and derivative values into the formula:
Here \(a=2\) remains fixed, while \(x\) is free to vary. As a check, substituting \(x=2\) gives \(T_3(2)=1=f(2)\). Also, differentiating the displayed polynomial gives \(T_3'(2)=4\), \(T_3''(2)=-6\), and \(T_3'''(2)=12\), as shown in the verification below.
Worked Example: A First-Order Polynomial Uses Only Two Derivative Values
Suppose \(g(5)=-2\) and \(g'(5)=3\), and consider the degree-one Taylor polynomial at \(a=5\). The definition uses \(k=0\) and \(k=1\) only:
No second derivative is needed to define this \(T_1\). At the center, \(T_1(5)=-2=g(5)\). Differentiating gives \(T_1'(x)=3\), so \(T_1'(5)=3=g'(5)\). There is no quadratic term in this polynomial, and its definition makes no claim about \(g''(5)\).
Worked Example: The Degree-Zero Case
Let \(h(x)=|x|\), choose \(a=0\), and take \(n=0\). The function is defined at \(0\), with \(h(0)=0\). The degree-zero formula contains only \(k=0\):
This definition needs the value of \(h\) at the center but no derivative there. Indeed, \(h(x)=|x|\) is not differentiable at \(0\), yet its degree-zero Taylor polynomial is defined. The direct check is \(T_0(0)=0=h(0)\). This example also shows why one should not expect a term involving \(x-a\) in \(T_0\): the sum stops at \(k=0\).
Why the Factorial Appears
The factorial normalizes a power of the displacement. To see the role of that normalization, consider the single term \(c(x-a)^k/k!\), where \(c\) is a constant and \(k\) is a nonnegative integer. Differentiate this term \(k\) times. Each differentiation lowers the power by one and contributes the power that was present before the differentiation.
For \(k\ge1\), differentiating \((x-a)^k\) once gives \(k(x-a)^{k-1}\). Repeating this process \(k\) times contributes the factors \(k,k-1,\ldots,1\), whose product is \(k!\).
The \(k\)th derivative of \((x-a)^k\) is the constant \(k!\). Therefore the \(k\)th derivative of \(c(x-a)^k/k!\) is \(c\), including when evaluated at \(x=a\).
Dividing by \(k!\) makes the \(k\)th derivative at the center equal to the coefficient \(c\), rather than \(k!c\). For \(k=0\), the term is the constant \(c\), and the same conclusion holds because \(0!=1\).
This calculation explains the scale of the factorial denominator in the definition. It does not say that every Taylor polynomial is a good approximation for every \(x\), nor does the definition itself guarantee an infinite series representation. Questions about approximation error and convergence require additional results.
Reading the Index Range Correctly
The upper index \(n\) specifies exactly how many orders of derivative data are included: the orders \(0\) through \(n\). This is important when interpreting the coefficients. For every included \(k\), the coefficient of \((x-a)^k\) is \(f^{(k)}(a)/k!\). If \(n\ge1\), that includes the linear coefficient \(f'(a)\); if \(n\ge2\), it includes the quadratic coefficient \(f''(a)/2!\).
If the highest included derivative value is zero, the corresponding term vanishes. For instance, if \(f^{(n)}(a)=0\), then the coefficient of \((x-a)^n\) is zero, so the polynomial may have degree less than \(n\). The name “degree-\(n\) Taylor polynomial” identifies the order used in its construction; it does not require the resulting polynomial to have degree exactly \(n\).
The definition also applies when \(n=0\). In that case, \(T_0(x)=f(a)\), a constant polynomial. The value of the function at the center is enough; no positive-order derivative is included. More generally, a degree-\(n\) formula cannot call for derivative data beyond order \(n\), because its sum has no terms with \(k>n\).
Common Mistake: Using the Wrong Power or Omitting the Factorial
Suppose the center is \(a=2\), the degree is \(2\), and the given data are \(f(2)=1\), \(f'(2)=4\), and \(f''(2)=6\). A common error is to write \(1+4x+6x^2\). This uses powers of \(x\) instead of powers of the displacement \(x-2\), and it omits the factorial denominator on the quadratic coefficient.
Apply the definition term by term instead:
Check the corrected expression at the center: \(T_2(2)=1\). Its first derivative is \(T_2'(x)=4+6(x-2)\), so \(T_2'(2)=4\). Its second derivative is \(T_2''(x)=6\), so \(T_2''(2)=6\). By contrast, the incorrect expression \(1+4x+6x^2\) has value \(33\) at \(x=2\), not \(1\), and does not use the stated center correctly.
Check Your Understanding
Use the definition and pay close attention to the order and center in each question.
- Write the general formula for the degree-\(n\) Taylor polynomial of \(f\) at \(a\), and state which derivative orders it uses.
- For a degree-two Taylor polynomial, what is the coefficient of \((x-a)^2\)? Which derivative value supplies it?
- Suppose only \(f(a)\) is known and \(f\) has no derivative at \(a\). Which Taylor polynomial can still be defined from that information?
- Explain why a degree-\(n\) Taylor polynomial can have actual degree smaller than \(n\).
- If \(a=3\), should the terms be written using powers of \(x\) or powers of \(x-3\)? Explain the roles of \(a\) and \(x\).
