Why Add Fourth- and Higher-Degree Terms?
A linear model can describe a constant slope, and quadratic and cubic models can describe some curved patterns. But some relationships change direction or curvature in ways those lower-degree forms cannot represent adequately. Adding fourth- or higher-degree powers gives a polynomial fit more flexibility to follow such patterns within the predictor range being studied.
That flexibility comes with an interpretive challenge. In a polynomial model, the contribution of a power depends on the predictor value, and several powers work together to determine the fitted curve. A coefficient such as the coefficient of \(x^4\) is not, by itself, a complete description of the curve’s shape or of an overall “fourth-degree effect.”
This model describes a mean response as a polynomial function of the predictor; it does not say that individual observations lie exactly on the curve. In the usual regression formulation, \(Y_i=\beta_0+\beta_1x_i+\cdots+\beta_dx_i^d+\varepsilon_i\), where the errors have conditional mean zero, are independent across observations, and have a common variance. These conditions support the usual least-squares interpretation and inference. The examples below use fitted equations to interpret predictions; they do not require a claim that the relationship is causal.
The coefficient \(b_k\) multiplies \(x^k\), so its units are response units divided by predictor units to the \(k\)th power. But the term’s contribution at a particular predictor value is \(b_kx^k\), in response units. The coefficient alone does not give that contribution without specifying \(x\). Further, it is the sum of all terms that determines the fitted response.
Interpreting a Fourth-Degree Term in Context
Consider a fitted model that includes powers through \(x^4\). Holding its other coefficients fixed, the fourth-degree term adds \(b_4x^4\) to the fitted value at \(x\). This is a useful accounting of the fitted equation at that point. It is not automatically the same as the change in predictions from fitting a new model without the fourth-degree term: refitting a model can change the other estimated coefficients too.
Worked Example: A Quartic Contribution at Different Predictor Values
Suppose a hypothetical fitted model predicts a response in points from dose, with \(x\) measured in 10-mg increments relative to a reference dose. Its fitted equation is $$ \hat{y}(x)=70+3x-0.8x^2+0.2x^3+0.6x^4. $$ At \(x=1\), the fourth-degree term contributes \(0.6(1)^4=0.6\) points. The fitted value is $$ \hat{y}(1)=70+3-0.8+0.2+0.6=73.0\text{ points}. $$ If we hold the other displayed coefficients fixed and omit just the quartic term, the corresponding value is \(70+3-0.8+0.2=72.4\) points. Thus, at \(x=1\), that term adds \(0.6\) points to this equation’s fitted value.
At \(x=2\), the quartic contribution is \(0.6(2)^4=0.6(16)=9.6\) points, and the full fitted value is $$ \hat{y}(2)=70+6-3.2+1.6+9.6=84.0\text{ points}. $$ The contrast with the lower-degree terms held fixed is \(9.6\) points at this predictor value. It would be incorrect to report \(0.6\) points as the quartic contribution everywhere: \(b_4x^4\) changes with \(x\).
The example’s point-by-point accounting does not establish that the quartic term is important for the data overall, nor does it isolate a causal effect. It explains what this term contributes within the particular fitted equation at selected values. Assessing an added term statistically requires an appropriate model comparison or inference; it cannot be settled from the coefficient’s magnitude alone.
Interpret the Higher-Degree Terms Together
For a model of degree \(d\), write the fitted value as the sum of a lower-degree part and a higher-degree part: $$ \hat{y}(x)=\underbrace{b_0+b_1x+b_2x^2+b_3x^3}_{\text{terms through degree 3}}+ \underbrace{b_4x^4+\cdots+b_dx^d}_{\text{terms of degree 4 and higher}}. $$ At a specified \(x\), the second bracket gives the combined contribution of all the fourth- and higher-degree terms. A positive or negative coefficient within that bracket does not determine the sign of the whole sum: the powers can reinforce or offset one another.
Worked Example: Combining Fourth- and Fifth-Degree Terms
Suppose a hypothetical model for a response measured in points uses a predictor \(x\) in 10-mg increments from a reference dose: $$ \hat{y}(x)=50+4x-x^2+0.5x^3+2x^4-x^5. $$ The combined contribution from degrees four and five is \(g(x)=2x^4-x^5\), in points.
At \(x=-1\), \(g(-1)=2(1)-(-1)=3\) points. The lower-degree part is \(50+4(-1)-(-1)^2+0.5(-1)^3=44.5\) points, so the full fitted value is \(44.5+3=47.5\) points.
At \(x=1\), \(g(1)=2(1)-1=1\) point. The lower-degree part is \(50+4-1+0.5=53.5\) points, giving a full fitted value of \(53.5+1=54.5\) points.
At \(x=2\), the two higher-degree terms cancel: \(g(2)=2(16)-32=0\) points. The lower-degree part is \(50+8-4+4=58\) points, so the full fitted value is \(58+0=58\) points. Neither coefficient alone describes the combined contribution at all three values. In particular, the positive coefficient of \(x^4\) does not mean the fourth- and fifth-degree terms always add a positive amount.
How Higher Powers Change Local Shape
Derivatives provide a concise way to see how the powers work together. For the fitted polynomial, the slope at \(x\) is $$ \hat{y}'(x)=b_1+2b_2x+3b_3x^2+\cdots+db_dx^{d-1}. $$ The curvature, measured by the second derivative, is $$ \hat{y}''(x)=2b_2+6b_3x+12b_4x^2+\cdots+d(d-1)b_dx^{d-2}. $$ Thus, a fourth-degree coefficient contributes \(4b_4x^3\) to the slope and \(12b_4x^2\) to the curvature; a fifth-degree term also contributes to both, with its own dependence on \(x\). The local slope and curvature are sums, not readings of one coefficient in isolation.
Worked Example: Reading Local Curvature from the Whole Model
Suppose a hypothetical fitted response, measured in units, is $$ \hat{y}(x)=20+2x-1.5x^2+0.4x^3+0.05x^4, $$ where \(x\) is measured in 10-mg increments from a reference dose. Differentiating gives $$ \hat{y}'(x)=2-3x+1.2x^2+0.2x^3,\qquad \hat{y}''(x)=-3+2.4x+0.6x^2. $$ At \(x=0\), the curvature is \(\hat{y}''(0)=-3\) response units per squared predictor unit. At \(x=2\), it is $$ \hat{y}''(2)=-3+2.4(2)+0.6(2^2)=-3+4.8+2.4=4.2 $$ response units per squared predictor unit. The positive quartic coefficient does not imply positive curvature at every predictor value: at \(x=0\), the combined curvature is negative. At \(x=2\), the sum of the terms gives positive curvature. These are local descriptions of this fitted curve, not claims about the response outside the range of observed predictor values.
A Short Justification for Added Shape Flexibility
Each extra power adds a term to the slope and curvature, so the model can represent patterns unavailable to a lower-degree polynomial. A degree-\(d\) polynomial has a derivative of degree at most \(d-1\). Unless that derivative is identically zero, it can have at most \(d-1\) distinct real roots: a nonzero polynomial of degree \(m\) cannot have more than \(m\) distinct roots, because each distinct root supplies a linear factor. Since a stationary point must satisfy \(\hat{y}'(x)=0\), a degree-\(d\) polynomial can have at most \(d-1\) stationary points. Some of those points may not be local maxima or minima, and many polynomials have fewer. The bound explains why higher degree permits more possible changes in local shape; it does not guarantee that a fitted curve will use all that flexibility.
Common Mistake
A common error is to say, “The positive coefficient of \(x^4\) means the fitted response rises more and more quickly.” That reasoning treats one coefficient as if it controlled the whole curve. In fact, the \(x^4\) term’s slope contribution is \(4b_4x^3\), which depends on \(x\), and the total slope also includes contributions from every lower- and higher-degree term. The correction is to calculate or describe the complete fitted value, slope, or curvature at the predictor values relevant to the question. Likewise, a term’s contribution at one \(x\) should not be presented as its contribution at every \(x\), or as evidence of causation.
Check Your Understanding
Use the definitions and calculations above to answer these questions.
- In a model containing \(b_4x^4\), what is the contribution of that term at a specified value \(x\), and why does it generally vary with \(x\)?
- A fitted model contains \(2x^4-x^5\). Calculate the combined contribution of these terms at \(x=1\) and \(x=2\). What does the comparison show?
- Why is the coefficient \(b_4\) not, by itself, a description of the fitted curve’s slope or curvature?
- What is the largest possible number of distinct stationary points of a nonconstant polynomial of degree six? Explain using the degree of its derivative.
- Why does calculating the contribution of a term while holding other coefficients fixed differ from comparing two separately fitted models?