Why Repeated Factors Change the Coefficients
A geometric series represents one reciprocal linear factor. When a denominator contains a repeated factor, multiplying copies of that geometric series makes the coefficients grow: the same power can be formed in several different ways. Counting those ways gives a general formula for the expansion.
For example, multiplying two copies of \(1+u+u^2+\cdots\), the coefficient of \(u^2\) is not \(1\). It is \(3\), because \(u^2\) can be formed as \(u^2\cdot 1\), \(u\cdot u\), or \(1\cdot u^2\). The more general count is a binomial coefficient.
Deriving the Repeated-Factor Formula
For \(|u|<1\), the geometric series gives \((1-u)^{-1}=\sum_{k=0}^{\infty}u^k\). Therefore, for a positive integer \(m\),
Each term in the product is determined by choosing one exponent from each of its \(m\) factors. To produce \(u^n\), the chosen exponents must be nonnegative integers \(k_1,\ldots,k_m\) satisfying \(k_1+\cdots+k_m=n\). Thus the coefficient of \(u^n\) is the number of such choices.
That number is \(\binom{n+m-1}{m-1}\). To see why, represent the \(n\) units of exponent as \(n\) identical marks in a row. Insert \(m-1\) separators to divide them into \(m\) groups. A group may be empty, corresponding to an exponent \(k_i=0\). There are \(n+m-1\) positions in the combined arrangement of marks and separators, and choosing the \(m-1\) separator positions determines the groups. Hence there are \(\binom{n+m-1}{m-1}\) choices.
For \(|u|<1\), each geometric series converges absolutely. The product of a finite number of absolutely convergent series can be multiplied using its Cauchy product, so this count gives the coefficients of the product.
The convergence condition is inherited from the geometric factors: every factor converges when \(|u|<1\). In an expression where \(u\) depends on another variable, impose that condition on the substituted value of \(u\).
Worked Example: Expanding a Cubed Geometric Factor
Worked Example: Expanding \((1-u)^{-3}\)
Use the formula with \(m=3\). The coefficient of \(u^n\) is \(\binom{n+2}{2}\), which equals \(\frac{(n+1)(n+2)}{2}\). Therefore, for \(|u|<1\),
The first coefficients are obtained by substituting \(n=0,1,2,3\):
Thus the series begins \(1+3u+6u^2+10u^3+\cdots\). To check it, let \(c_n=\frac{(n+1)(n+2)}{2}\). The coefficients of \(u^0,u^1,u^2\) in \((1-u)^3\sum_{n=0}^{\infty}c_nu^n\) are respectively \(1\), \(3-3(1)=0\), and \(6-3(3)+3(1)=0\). For \(n\geq 3\), the coefficient is \(c_n-3c_{n-1}+3c_{n-2}-c_{n-3}=0\): the third finite difference of the quadratic polynomial \(c_n\) is zero. The product is therefore \(1\), as required.
Scaling and Shifting the Factor
A repeated linear factor can often be put into the form \(1-u\) by factoring out a constant. For example, write \(a+bx=a\left(1+\frac{b}{a}x\right)\), provided \(a\neq 0\). Then use the theorem with \(u=-\frac{b}{a}x\), and require \(\left|\frac{b}{a}x\right|<1\). This gives the expansion around \(x=0\) on the interval where the substituted geometric series converges.
Worked Example: A Scaled Repeated Factor
Worked Example: Expanding \((2+x)^{-2}\)
Factor out \(2\) from \(2+x\), then square the factor:
Use the repeated-factor formula with \(m=2\) and \(u=-x/2\). Its convergence condition \(|u|<1\) becomes \(|x|<2\). The coefficient is \(\binom{n+1}{1}=n+1\), so
The first terms are \(\frac14-\frac14x+\frac{3}{16}x^2-\frac18x^3+\cdots\). For a check, set \(b_n=(-1)^n(n+1)/2^n\), so the series for \((1+x/2)^{-2}\) has coefficients \(b_n\). Multiplication by \((1+x/2)^2\) gives constant coefficient \(b_0=1\), coefficient of \(x\) equal to \(b_1+b_0= -1+1=0\), and for \(n\geq2\) the coefficient is \(b_n+b_{n-1}+\frac14b_{n-2}=0\). Indeed, substituting the values gives \(\frac{(-1)^n}{2^n}\big((n+1)-2n+(n-1)\big)=0\). The product is \(1\), confirming the expansion. The outside factor \(1/4\) then gives \((2+x)^{-2}\).
Applying the Formula to Rational Functions
A rational function with repeated linear factors can have contributions from several terms, each with its own expansion. If partial fractions express the function as a sum of terms with denominators \((1-u)^m\), expand each term using the theorem and add the resulting series. The convergence condition must hold for every term in the sum. In particular, the interval or disk where all the expansions converge is the intersection of their individual conditions.
The geometric-product formula directly handles the repeated factors. Partial fractions can be used, when needed, to separate different factors; the calculation below illustrates how to verify the decomposition and combine the resulting coefficients.
Worked Example: Combining a Repeated Pole and a Simple Pole
Worked Example: Expanding \(\frac{1}{(1-x)^2(1+x)}\)
First decompose the rational function:
Multiplying both sides by \((1-x)^2(1+x)\) gives \(1=A(1-x)(1+x)+B(1+x)+C(1-x)^2\). Setting \(x=1\) gives \(1=2B\), so \(B=\frac12\). Setting \(x=-1\) gives \(1=4C\), so \(C=\frac14\). Comparing constant terms gives \(A+B+C=1\), hence \(A=1-\frac12-\frac14=\frac14\). Substitution verifies the decomposition: the right-hand side after multiplication by the denominator is \(\frac14(1-x^2)+\frac12(1+x)+\frac14(1-2x+x^2)=1\).
For \(|x|<1\), expand each term. The first uses the geometric series, the second uses the repeated-factor formula with \(m=2\), and the third uses the geometric series with \(u=-x\):
For \(n=0,1,2,3\), the coefficients are \(1,1,2,2\), respectively, so the series begins \(1+x+2x^2+2x^3+\cdots\). To check it, write \(a_n=(2n+3+(-1)^n)/4\). The denominator is \((1-x)^2(1+x)=1-x-x^2+x^3\), so multiplying the series by the denominator gives coefficient \(a_0=1\) for the constant term. The next coefficients are \(a_1-a_0=1-1=0\), \(a_2-a_1-a_0=2-1-1=0\), and \(a_3-a_2-a_1+a_0=2-2-1+1=0\). For \(n\geq3\), \(a_n-a_{n-1}=a_{n-2}-a_{n-3}\): the successive differences alternate between \(0\) and \(1\). Thus every coefficient beyond the constant is zero, confirming that the product is \(1\).
A Common Mistake: Forgetting That Products Add Contributions
A frequent error is to assume that squaring a geometric series leaves all its coefficients equal to \(1\). That would give the incorrect calculation
The coefficient of \(u\) in the product is \(1\cdot u+u\cdot1=2u\), not \(u\). More generally, for \(m=2\), there are \(n+1\) pairs of nonnegative exponents adding to \(n\). The correct expansion is
The coefficient count, rather than the appearance of the original geometric series, determines the answer.
Why the Formula Is Useful
Repeated linear factors occur naturally in rational functions, and their coefficients are not constant as they are for a single geometric factor. The binomial-coefficient formula gives those coefficients directly, without multiplying the series term by term indefinitely. It also makes the convergence requirement visible: after rewriting the denominator in the form \((1-u)^m\), check \(|u|<1\) before using the expansion.
Check Your Understanding
Use the repeated geometric factor formula and state the condition under which each expansion converges.
- What is the coefficient of \(u^5\) in \((1-u)^{-4}\)?
- Write the first four terms of \((1+3x)^{-2}\).
- Why does the coefficient of \(u^n\) in the product of three geometric series count triples of nonnegative integers summing to \(n\)?
- What condition on \(x\) allows the geometric-product expansion of \((4-x)^{-2}\) about \(x=0\)?
- For \(|x|<1\), what is the coefficient of \(x^n\) in \(\frac{1}{(1-x)^2(1+x)}\), as expanded in the worked example?
