Why Products Require Care
A Taylor polynomial describes a function near a chosen center using powers of the displacement from that center. When two functions are multiplied, their Taylor polynomials can be multiplied too—but the coefficient of a given power in the product may come from several pairs of terms.
For instance, the coefficient of \(h^2\) in a product can receive contributions from a constant times a quadratic term, two linear terms, and a quadratic term times a constant. The rule that collects all these contributions is called coefficient convolution. It gives a systematic way to find the Taylor polynomial of a product without repeatedly differentiating that product from scratch.
Taylor Polynomials and the Product Rule
Let \(f\) and \(g\) have continuous derivatives through order \(n\) on an open interval containing \(a\). Their degree-\(n\) Taylor polynomials centered at \(a\) are
Here, \(f^{(0)}=f\), and \(0!=1\). The degree-\(n\) Taylor polynomial of \(fg\) is determined by the derivatives of the product through order \(n\). Its coefficient of \((x-a)^k\) is
Each term in the sum pairs a degree-\(j\) contribution from \(f\) with a degree-\((k-j)\) contribution from \(g\). All such pairs have combined degree \(k\).
Terms of degree greater than \(n\) in the polynomial product cannot change any coefficient of degree at most \(n\). They are discarded for this calculation; the theorem does not assert that the full product of the two degree-\(n\) Taylor polynomials equals the Taylor polynomial of \(fg\).
Why the Convolution Rule Holds
The key fact is the Leibniz formula for derivatives of a product. For each integer \(k\ge 0\), if the necessary derivatives exist, then
This identity is a statement about the functions at every \(x\) in an interval, not just about their values at the center. That distinction matters: an equality of values at one point cannot be differentiated to obtain an identity for the next derivative.
For \(k=0\), the sum has one term, so it says \((fg)(x)=f(x)g(x)\), which is true.
Assume the Leibniz formula holds for every \(x\) at order \(k\). Differentiating its right-hand side term by term gives $$ \sum_{j=0}^{k}\binom{k}{j} \left(f^{(j+1)}(x)g^{(k-j)}(x)+f^{(j)}(x)g^{(k-j+1)}(x)\right). $$
In the first part of the sum, set \(r=j+1\); in the second, set \(r=j\). For each interior term, the resulting coefficient is $$ \binom{k}{r-1}+\binom{k}{r}=\binom{k+1}{r}. $$ The end terms have coefficient \(1\). Thus the sum becomes the Leibniz formula at order \(k+1\).
Set \(x=a\), then divide the order-\(k\) formula by \(k!\). Since \(\binom{k}{j}/k!=1/(j!(k-j)!)\), the coefficient of \((x-a)^k\) has the stated convolution form.
This proves the product rule for Taylor coefficients. The assumption of enough derivatives ensures that each differentiation in the induction is valid through the required order.
Worked Product Approximations
Worked Example: \(e^x\cos x\) Through Degree Three
Find the degree-three Maclaurin polynomial for \(h(x)=e^x\cos x\). The relevant polynomials are
Multiply and collect terms through degree three. The constant coefficient is \(1\). The coefficient of \(x\) is \(1\). For \(x^2\), the contributions are \(\frac12\cdot 1\) and \(1\cdot(-\frac12)\), giving \(\frac12-\frac12=0\). For \(x^3\), the contributions are \(\frac16\cdot 1\) and \(1\cdot(-\frac12)\), giving \(\frac16-\frac12=-\frac13\). Therefore,
Check by differentiating \(h\). We have \(h(0)=1\), \(h'(x)=e^x(\cos x-\sin x)\), so \(h'(0)=1\). Also \(h''(x)=-2e^x\sin x\), so \(h''(0)=0\), and \(h'''(x)=-2e^x(\sin x+\cos x)\), so \(h'''(0)=-2\). The Taylor coefficients are \(1\), \(1\), \(0/2!\), and \(-2/3!=-1/3\), as required.
Worked Example: \((1+x)\ln(1+x)\) Through Degree Three
Find the degree-three Maclaurin polynomial for \(q(x)=(1+x)\ln(1+x)\). The logarithm is defined near \(x=0\), and its degree-three polynomial is
The other factor is the exact polynomial \(1+x\). Multiplication gives
Discard the degree-four term. Since \(1-\frac12=\frac12\) and \(-\frac12+\frac13=-\frac16\),
For a check, differentiate the product: \(q'(x)=\ln(1+x)+1\), \(q''(x)=1/(1+x)\), and \(q'''(x)=-1/(1+x)^2\). At zero, \(q(0)=0\), \(q'(0)=1\), \(q''(0)=1\), and \(q'''(0)=-1\). These give coefficients \(0\), \(1\), \(1/2\), and \(-1/6\), matching the result.
Worked Example: \(\sin x\cos x\) Near \(x=\pi/4\)
Find the degree-two Taylor polynomial of \(\sin x\cos x\) centered at \(a=\pi/4\). Write \(h=x-\pi/4\), and set \(r=\sqrt{2}/2\). Using the sine and cosine values and derivatives at the center,
The constant coefficient of their product is \(r^2=1/2\). The coefficient of \(h\) is \(r(-r)+r(r)=0\). The coefficient of \(h^2\) has three contributions:
Thus the degree-two polynomial is
Check directly: \(\sin x\cos x=\frac12\sin(2x)\). At \(x=\pi/4\), the function has value \(1/2\), its first derivative is \(\cos(2x)=0\), and its second derivative is \(-2\sin(2x)=-2\). The quadratic Taylor coefficient is \(-2/2!=-1\), confirming the calculation.
A Common Coefficient Error
A frequent mistake is to omit the contribution from a pair of different-degree terms, or to omit the binomial coefficient when using the derivative form of the product rule. For example, take \(f(x)=g(x)=e^x\) at zero. Since \(f(0)=g(0)=f'(0)=g'(0)=f''(0)=g''(0)=1\), an incorrect second-derivative calculation might write
The product rule applied twice has a factor of \(2\) on the middle term:
Therefore the coefficient of \(x^2\) is \(4/2!=2\), not \(3/2\). Convolution shows the same result: the degree-two coefficient is \(\frac12\cdot1+1\cdot1+1\cdot\frac12=2\). In the polynomial multiplication, the two cross-products \(x\cdot x\) account for the middle contribution.
Using the Rule Reliably
For a requested degree \(n\), it is useful to organize the calculation by total degree. To find a coefficient of degree \(k\), combine only pairs of degrees that add to \(k\). This keeps the calculation manageable and prevents irrelevant higher-degree terms from obscuring the answer.
| Target degree | Contributions from the two factors |
|---|---|
| 0 | degree 0 and degree 0 |
| 1 | degrees 0 and 1; degrees 1 and 0 |
| 2 | degrees 0 and 2; degrees 1 and 1; degrees 2 and 0 |
| 3 | degrees 0 and 3; degrees 1 and 2; degrees 2 and 1; degrees 3 and 0 |
The center must be consistent throughout. At a general center \(a\), the powers are powers of \(x-a\), and the coefficients come from derivatives evaluated at \(a\). At the Maclaurin center \(a=0\), those powers are simply powers of \(x\). The convolution principle is identical in either case.
This method is especially useful when the factors have familiar Taylor polynomials or when their derivatives are already known. It avoids recalculating every derivative of the product and makes clear exactly which terms determine the requested degree. Its scope here is products: combining Taylor polynomials for sums and scalar multiples uses the corresponding coefficient-by-coefficient operations.
Check Your Understanding
For each question, identify the coefficients or terms needed before carrying out the calculation.
- State the convolution formula for the coefficient of \((x-a)^k\) in the Taylor polynomial of \(fg\).
- When multiplying two degree-four Taylor polynomials, which pairs of degrees contribute to the coefficient of \((x-a)^3\)?
- Find the degree-two Maclaurin polynomial for \(e^x\sin x\) using product coefficients.
- Why must the Leibniz identity be stated for a variable \(x\) before differentiating it in an induction proof?
- For \(f(x)=g(x)=e^x\), what is the degree-two coefficient in the Taylor polynomial of \(fg\), and how can convolution verify it?
