Introduction
A Taylor polynomial records the values of a function and some of its derivatives at a chosen center. When that center is zero, the polynomial is called a Maclaurin polynomial. Its coefficients are determined by derivatives evaluated at zero.
The definition is compact, but using it carefully requires attention to the order of the polynomial, the factorial in each coefficient, and which derivatives are needed. In this tutorial, we build low-degree examples from the derivative formula and verify that their derivatives at zero match those of the original functions.
Definition and Formula
Written term by term, the formula begins
The coefficient of \(x^k\) is \(f^{(k)}(0)/k!\). Each coefficient uses the derivative of the matching order, evaluated at zero. In particular, the constant term is \(f(0)\), and the coefficient of \(x\) is \(f'(0)\).
The formula requires the derivatives appearing in it to exist at zero. It defines a polynomial with specified derivative values at zero; by itself, it does not guarantee a particular error size or that a function equals its Maclaurin polynomial nearby.
Why the Derivative Formula Works
The formula follows from the requirement that the polynomial match the function’s derivatives at zero. To see this, write a polynomial of order at most \(n\) with unknown coefficients \(c_k\):
For a fixed index \(j\), differentiate \(Q\) exactly \(j\) times and evaluate at zero. A term \(c_kx^k\) contributes zero if \(k<j\), because its \(j\)th derivative is zero. If \(k>j\), its \(j\)th derivative still contains a positive power of \(x\), so its value at zero is zero. The only term that contributes is \(c_jx^j\), whose \(j\)th derivative is the constant \(j!c_j\). Therefore \(Q^{(j)}(0)=j!c_j\).
For each \(j\) from \(0\) through \(n\), require \(Q^{(j)}(0)=f^{(j)}(0)\).
Since \(Q^{(j)}(0)=j!c_j\), the matching condition gives \(j!c_j=f^{(j)}(0)\).
Because \(j!\ne 0\), \(c_j=f^{(j)}(0)/j!\). Substituting these coefficients into \(Q\) gives the Maclaurin formula.
This also verifies the formula directly: the \(j\)th derivative of \(P_n\), evaluated at zero, is \(j!\) times its \(x^j\) coefficient. Hence
Thus the polynomial matches the function’s value and derivatives through order \(n\) at zero. The factorial is essential: it compensates for the factor produced when \(x^j\) is differentiated \(j\) times.
Worked Examples
Worked Example: A Degree-Two Polynomial from Derivatives
Find the degree-two Maclaurin polynomial for \(f(x)=x^3+2x^2-x+4\).
We need the function, its first derivative, and its second derivative at zero:
Use the formula through order two:
Check the derivative values. The polynomial satisfies \(P_2(0)=4\), \(P_2'(x)=-1+4x\), so \(P_2'(0)=-1\), and \(P_2''(x)=4\), so \(P_2''(0)=4\). These equal \(f(0)\), \(f'(0)\), and \(f''(0)\), respectively. In this example, \(f(x)-P_2(x)=x^3\); the omitted term does not change any of the matched derivatives through order two at zero.
Worked Example: A Degree-Four Maclaurin Polynomial
Let \(f(x)=\arctan x\). Find its degree-four Maclaurin polynomial.
Differentiate successively:
Differentiate \(f''\) using the product rule:
Differentiating once more gives
Substitute the five derivative values, including \(f(0)\), into the formula:
Check by differentiating the polynomial: \(P_4'(x)=1-x^2\), \(P_4''(x)=-2x\), \(P_4'''(x)=-2\), and \(P_4^{(4)}(x)=0\). At zero, these derivatives, together with \(P_4(0)=0\), are \(0,1,0,-2,0\), matching the corresponding values of \(f\).
Worked Example: A Rational Function Through Order Three
Find the degree-three Maclaurin polynomial for \(g(x)=\dfrac{x^2}{1+x}\), which is defined near zero.
Rewrite the function to make its derivatives straightforward:
Now evaluate the derivatives at zero:
Therefore
Check the match: \(P_3(0)=0\), \(P_3'(x)=2x-3x^2\) gives \(P_3'(0)=0\), \(P_3''(x)=2-6x\) gives \(P_3''(0)=2\), and \(P_3'''(x)=-6\) gives \(P_3'''(0)=-6\). Each value agrees with the corresponding derivative of \(g\) at zero.
Order and Actual Degree
The subscript \(n\) indicates that the formula includes terms from the constant term through \(x^n\). It does not promise that the coefficient of \(x^n\) is nonzero. If that coefficient is zero and the polynomial is nonzero, its actual degree is less than \(n\). A zero coefficient of a lower power does not by itself lower the degree.
For instance, \(x^3+0x^2+2x\) has degree three: its \(x^3\) coefficient is nonzero. In the degree-four example above, the \(x^4\) coefficient is zero and the resulting polynomial is nonzero of degree three. It remains correctly named the degree-four Maclaurin polynomial because it was constructed using derivatives through order four.
Common Mistake: Omitting the Factorial
A frequent error is to use the derivative value itself as the coefficient of \(x^k\). For the function \(g\) in the last example, that mistake would put \(2x^2\) in the polynomial because \(g''(0)=2\). But the coefficient is \(g''(0)/2!=1\), not \(2\). The same issue occurs for the cubic term: the coefficient is \(g'''(0)/3!=-6/6=-1\), not \(-6\).
The incorrect expression \(2x^2-6x^3\) does not pass the derivative check: its second derivative at zero is \(4\), not \(g''(0)=2\), and its third derivative at zero is \(-36\), not \(g'''(0)=-6\). Dividing each derivative value by its matching factorial gives \(x^2-x^3\), whose derivatives at zero do match.
Why Maclaurin Polynomials Matter
Maclaurin polynomials organize information about a function at a particularly simple center, zero. The coefficients give a precise record of the function’s local derivative data: the constant term records its value, the linear coefficient records its first derivative, and higher coefficients encode higher derivatives with the required factorial adjustment.
They are also a useful setting for practicing the general Taylor-polynomial method. Before using a polynomial to estimate function values, one must distinguish construction from accuracy: the derivative formula determines the polynomial, while a separate error analysis is needed to justify how close it is to the function away from zero. The calculations here establish the polynomial and its derivative-matching property, not an error bound.
Check Your Understanding
For each question, use the derivative formula and check the relevant derivative values at zero.
- Write the degree-two Maclaurin formula using \(f(0)\), \(f'(0)\), and \(f''(0)\).
- For \(h(x)=x^2+3x+1\), find its degree-two Maclaurin polynomial.
- Why does the coefficient of \(x^j\) in a Maclaurin polynomial equal \(f^{(j)}(0)/j!\)?
- If a degree-four Maclaurin polynomial has zero coefficient of \(x^4\), what can be concluded about its actual degree, assuming it is nonzero?
- Does a zero coefficient of \(x^2\) alone imply that a nonzero degree-four polynomial has degree less than four? Explain.
