Why Algebraic Series Manipulations Help
A function may be built from simpler functions whose Taylor series are already known. Rather than calculate every derivative of the new function, you can often add, subtract, or multiply the known series and keep only the powers needed. This can save work, especially when a product would require repeated applications of the product rule.
The key is to track powers carefully. To find a Taylor polynomial through degree \(n\), terms of degree greater than \(n\) cannot affect its coefficients. But when multiplying, a term of degree \(n\) can arise from several different pairs of lower-degree terms, so those contributions must all be included.
Adding and Subtracting Known Series
Suppose two functions have power series about the same center \(a\), and both series converge on some interval around \(a\). Write \(h=x-a\). If \(f(x)=\sum_{k=0}^{\infty}c_kh^k\) and \(g(x)=\sum_{k=0}^{\infty}d_kh^k\), then on the interval where both series converge, their sum and difference are represented by adding or subtracting matching coefficients.
Therefore, to obtain the degree-\(n\) Taylor polynomial of \(f\pm g\), combine the coefficients only for \(k=0,1,\ldots,n\). A missing power has coefficient zero; do not shift the coefficients to fill the gap.
Worked Example: Adding Exponential and Sine Series
Find the Maclaurin polynomial of degree \(4\) for \(f(x)=e^x+\sin x\).
Use the known series, retaining terms through degree \(4\):
Write the sine series with its zero coefficients visible through degree \(4\):
Add the coefficients of each power:
Thus the degree-\(4\) Maclaurin polynomial is
The cubic terms cancel, but that does not change the requested degree: the polynomial is still specified through degree \(4\). As a check, the constant and linear coefficients match \(f(0)=1\) and \(f'(0)=e^0+\cos 0=2\); the quadratic coefficient is \(f''(0)/2!=1/2\), and the cubic coefficient is \(f'''(0)/3!=0\). Also \(f^{(4)}(0)/4!=1/24\), so the computed coefficients agree with the derivative definition.
Multiplying Series: The Coefficient Rule
For a product, the coefficient of \(h^k\) is formed from every pair of terms whose exponents add to \(k\). If \(f(x)=\sum_{i=0}^{\infty}c_i h^i\) and \(g(x)=\sum_{j=0}^{\infty}d_j h^j\), then the product coefficient at degree \(k\) is
This is sometimes called the convolution rule for coefficients. To find a product through degree \(n\), calculate this sum for \(k=0,1,\ldots,n\). Terms with exponent greater than \(n\) may be discarded after all contributions to degrees at most \(n\) have been included.
Why Truncation Gives the Correct Coefficients
The coefficient rule follows by distributing the factors. A term \(c_i h^i\) multiplied by a term \(d_j h^j\) gives \(c_i d_j h^{i+j}\). It contributes to degree \(k\) exactly when \(i+j=k\), which means \(j=k-i\). The possible indices are \(i=0,1,\ldots,k\), giving the stated sum.
There is also a useful way to justify discarding higher powers. Let \(P_n\) and \(Q_n\) be the respective degree-\(n\) truncations of two series. Near the center, write their functions as \(f=P_n+R_f\) and \(g=Q_n+R_g\), where the remainders have no terms below degree \(n+1\). In power-series language, \(R_f\) and \(R_g\) are \(O(h^{n+1})\): their size is bounded by a constant times \(|h|^{n+1}\) sufficiently near \(h=0\). Then
The polynomials \(P_n\) and \(Q_n\) remain bounded near \(h=0\). Therefore \(P_nR_g\) and \(Q_nR_f\) are \(O(h^{n+1})\), while \(R_fR_g\) is \(O(h^{2n+2})\), which is also \(O(h^{n+1})\). Thus every term involving a remainder begins at degree \(n+1\) or higher. The coefficients through degree \(n\) in \(fg\) are exactly those found by multiplying \(P_n\) and \(Q_n\) and discarding powers above \(n\).
Worked Example: Multiplying Exponential and Cosine Series
Find the degree-\(4\) Maclaurin polynomial for \(f(x)=e^x\cos x\).
First write both known series through degree \(4\):
Now collect the products that contribute at each degree. The constant coefficient is \(1\cdot1=1\). The coefficient of \(x\) is \(1\cdot0+1\cdot1=1\). For degree \(2\), the contributions are \(1\cdot(-1/2)\), \(1\cdot0\), and \((1/2)\cdot1\), so the coefficient is \(-1/2+0+1/2=0\).
For degree \(3\), the contributions are \(1\cdot0\), \(1\cdot(-1/2)\), \((1/2)\cdot0\), and \((1/6)\cdot1\). Their sum is \(0-1/2+0+1/6=-1/3\).
For degree \(4\), the contributions are \(1\cdot(1/24)\), \(1\cdot0\), \((1/2)\cdot(-1/2)\), \((1/6)\cdot0\), and \((1/24)\cdot1\). Their sum is \(1/24+0-1/4+0+1/24=-1/6\). Therefore,
Check the result using derivatives at \(0\). For \(f=e^x\cos x\), differentiation gives \(f'=e^x(\cos x-\sin x)\), \(f''=-2e^x\sin x\), \(f'''=-2e^x(\sin x+\cos x)\), and \(f^{(4)}=-4e^x\cos x\). At \(0\), the values are \(f(0)=1\), \(f'(0)=1\), \(f''(0)=0\), \(f'''(0)=-2\), and \(f^{(4)}(0)=-4\). Dividing by the corresponding factorials gives coefficients \(1,1,0,-2/6,-4/24\), which are exactly \(1,1,0,-1/3,-1/6\).
When Many Product Terms Cancel
Alternating coefficients can make a product simpler than either factor. The coefficient rule is still necessary: cancellations should emerge from the sums rather than from guessing a pattern. Geometric series provide a compact example.
Worked Example: Multiplying Two Geometric Series
Use known geometric series to find the Maclaurin polynomial through degree \(5\) for \(f(x)=\frac{1}{1-x}\frac{1}{1+x}\).
For \(|x|<1\), the geometric-series representations are
The degree-\(k\) coefficient is the sum of the products of terms whose degrees add to \(k\). Listing the contributions gives:
So the degree-\(5\) polynomial is
Check by simplifying the function itself: \(\frac{1}{1-x}\frac{1}{1+x}=\frac{1}{(1-x)(1+x)}=\frac{1}{1-x^2}\). Applying the geometric series with \(x^2\) in place of \(x\) gives \(1+x^2+x^4+x^6+\cdots\), whose terms through degree \(5\) are exactly \(1+x^2+x^4\). The simplification confirms the computed coefficients.
A Common Truncation Mistake
A frequent error is to multiply only the terms of the same degree, or to discard terms before checking whether their products contribute to a retained power. For example, consider finding the product through degree \(2\) of \(1+x+x^2+\cdots\) and \(1-x+x^2-\cdots\). Multiplying only matching-degree terms might incorrectly suggest the \(x^2\) coefficient is \(1+1=2\). But the degree-\(2\) coefficient comes from all pairs with degrees adding to \(2\):
The coefficient is \(1\), not \(2\). The middle product matters because its degrees also add to \(2\).
A Reliable Procedure
Rewrite each known series in powers of \(h=x-a\), if the center is \(a\).
For a sum or difference through degree \(n\), retain terms through \(h^n\). For a product, retain all terms that can combine to produce powers through \(h^n\).
Add or subtract matching coefficients; for products, sum all pairwise products whose exponents add to the target degree.
Compare coefficients with derivative values when practical, or simplify the original function and verify that its series has the same retained terms.
Check Your Understanding
Use known series and retain only the requested powers. Pay attention to every contribution to each product coefficient.
- Use the Maclaurin series for \(e^x\) and \(\cos x\) to find the degree-\(3\) polynomial for \(e^x+\cos x\).
- Find the degree-\(3\) Maclaurin polynomial for \(e^x\sin x\) by multiplying the known series.
- For a product of two series, which pairs of exponents contribute to the coefficient of \(h^k\)?
- Why can terms of degree greater than \(n\) be discarded when finding a product through degree \(n\)?
- Find the degree-\(4\) Maclaurin polynomial for \(\frac{1}{1-x}-\frac{1}{1+x}\), using geometric series.
