Why Multiplication Shifts the Coefficients
A power series is organized by powers of its centered variable. If its center is \(c\), write the series as \(\sum_{n=0}^{\infty} a_n(x-c)^n\). Multiplying by \(x-c\) raises each power by one. The coefficients themselves do not change; their positions do.
This simple shift is useful because many products involving a power series reduce to it. For example, the original variable satisfies \(x=c+(x-c)\), so multiplying by \(x\) combines a scaled copy of the original series with a shifted copy.
Throughout, the index begins at \(n=0\). Any identity between a series and its product is understood at values of \(x\) where the original series converges. The coefficient rules below follow by multiplying its terms by a finite polynomial and collecting equal powers.
The Centered-Variable Shift Rule
Let \(F(x)=\sum_{n=0}^{\infty}a_n(x-c)^n\), and set \(u=x-c\) temporarily to make the powers easier to see. Multiplication by \(u\) gives
The first term is \(a_0u\), not a constant term. To express the result as a series in the usual powers \(u^n\), reindex by setting \(k=n+1\). Then \(n=k-1\), and the index \(n=0,1,2,\ldots\) corresponds to \(k=1,2,3,\ldots\).
The coefficient at degree zero is zero because multiplication by \(x-c\) creates no constant term. For each degree \(n\geq1\), the new coefficient is the old coefficient from degree \(n-1\). This is the precise meaning of “shift up one degree.”
Worked Example: Multiplying by the Centered Variable
Consider \(F(x)=\sum_{n=0}^{\infty}(n+1)(x-3)^n\). Find the coefficients of \((x-3)F(x)\).
Here the center is \(c=3\), and \(a_n=n+1\). The shift rule gives \(b_0=0\), and for \(n\geq1\), \(b_n=a_{n-1}\). Thus \(b_n=(n-1)+1=n\).
Check by multiplying the first terms directly. The original begins \(F(x)=1+2(x-3)+3(x-3)^2+4(x-3)^3+\cdots\). Therefore
which agrees with the coefficient rule: the constant coefficient is zero, and the coefficients of degrees one through four are \(1,2,3,4\).
Multiplication by the Original Variable
The shift rule applies directly to \(x-c\). To multiply by \(x\), express \(x\) as the sum of the center and the centered variable:
Distributing \(x\) across \(F(x)\) gives one copy of the original series scaled by \(c\), plus one copy shifted by a degree. In coefficient form, the degree-zero coefficient is \(ca_0\). At each degree \(n\geq1\), the coefficient combines \(ca_n\) from the scaled original series and \(a_{n-1}\) from the shift.
Worked Example: Multiplying by \(x\) at a Nonzero Center
Let \(F(x)=1+2(x-2)-(x-2)^2+4(x-2)^3\). Find the coefficients of \(xF(x)\) as a polynomial in \(x-2\).
The center is \(c=2\). Write \(u=x-2\), so \(x=2+u\). First multiply by the constant part \(2\):
Next multiply by \(u\), which raises each degree by one:
Add the two polynomials and collect like powers:
Substitute \(u=x-2\) to write the result in the centered variable:
Check by distributing \(2+u\) directly: \((2+u)(1+2u-u^2+4u^3)=2+5u+0u^2+7u^3+4u^4\). In particular, the degree-two coefficient cancels because the contributions \(-2u^2\) and \(2u^2\) sum to zero.
Linear and Polynomial Factors
A general linear factor can be written as \(\alpha(x-c)+\beta\), where \(\alpha\) and \(\beta\) are constants. Multiplication by this factor combines a scaled shift with a scaled copy of the original series. If the product is \(\sum_{n=0}^{\infty}b_n(x-c)^n\), then
The same idea extends to a polynomial factor. Let \(P(u)=\sum_{j=0}^{m}p_ju^j\), where \(u=x-c\). Multiplying \(F(x)=\sum_{n=0}^{\infty}a_nu^n\) by \(P(u)\), each term \(p_ju^j\) shifts the series by \(j\) degrees and scales it by \(p_j\). Therefore the coefficient \(b_n\) of \(u^n\) is
This is a finite sum for each \(n\), since the polynomial has only \(m+1\) terms. It is the same coefficient-collection principle used for products of power series, here simplified because one factor has only finitely many nonzero coefficients.
Worked Example: Multiplying by a Quadratic Factor
Let \(F(x)=\sum_{n=0}^{\infty}\frac{1}{n+1}(x-1)^n\). Find the first five coefficients of \(\big((x-1)^2+2(x-1)-1\big)F(x)\).
Set \(u=x-1\). The polynomial factor is \(u^2+2u-1\), so \(p_0=-1\), \(p_1=2\), and \(p_2=1\). The coefficients of \(F\) are \(a_n=\frac{1}{n+1}\). The product coefficients are \(b_n=a_{n-2}+2a_{n-1}-a_n\), omitting any term whose index is negative.
For \(n=0\), only the term \(-a_0\) occurs:
For \(n=1\), the terms are \(2a_0-a_1\):
For \(n=2\), all three terms occur:
For \(n=3\):
For \(n=4\):
Thus the product begins \(-1+\frac32u+\frac53u^2+\frac{11}{12}u^3+\frac{19}{30}u^4+\cdots\). Check the coefficient formula: at degree four, for example, the contributions are \(a_2\) from \(u^2F\), \(2a_3\) from \(2uF\), and \(-a_4\) from \(-F\); their sum is \(19/30\), as computed.
A Reliable Coefficient-Collection Method
When the factor is a polynomial, a small coefficient table can make the shifts visible. For a quadratic \(p_0+p_1u+p_2u^2\), the contributions to degree \(n\) are \(p_0a_n\), \(p_1a_{n-1}\), and \(p_2a_{n-2}\). A term is absent whenever its coefficient index would be negative.
Write \(u=x-c\), keeping the center \(c\) fixed throughout.
For example, \(x=c+u\), and a quadratic factor can be written \(p_0+p_1u+p_2u^2\).
The term \(p_ju^j\) contributes \(p_ja_{n-j}\) to degree \(n\).
Compute the constant and first few coefficients directly to catch missing or misindexed terms.
Common Mistake: Shifting in the Wrong Direction
A frequent error is to claim that multiplication by \(x-c\) makes the new degree-\(n\) coefficient equal to \(a_{n+1}\). That reverses the shift. The coefficient \(a_n\) originally multiplies \((x-c)^n\); after multiplication it multiplies \((x-c)^{n+1}\). Thus it becomes the coefficient at degree \(n+1\), not degree \(n\).
For instance, if \(F(x)=a_0+a_1(x-c)+a_2(x-c)^2+\cdots\), then
The constant coefficient is \(0\), and the degree-one coefficient is \(a_0\). The incorrect reversed rule would put \(a_1\) at degree one and could even assign a nonzero constant coefficient. Checking these first two terms exposes the error immediately.
Why This Rule Is Useful
Multiplication by a centered variable is one of the simplest operations on a power series, but it gives a practical way to assemble more complicated products. Linear factors require only the original and shifted coefficient sequences. A polynomial of degree \(m\) requires at most \(m+1\) contributions to each product coefficient. This lets you calculate a product without expanding every term afresh.
The center matters. The factor \(x-c\) creates the clean one-place shift, whereas \(x\) also includes the unshifted contribution \(cF(x)\). Rewriting factors in powers of \(x-c\) before calculating keeps this distinction explicit and reduces indexing mistakes.
These coefficient identities describe the product wherever the original series converges. They do not require multiplying two infinite coefficient sequences: the extra factor here is a finite polynomial, so each resulting coefficient is obtained from finitely many terms.
Check Your Understanding
Use the coefficient rules to answer each question. In each case, write the product in powers of the centered variable.
- If \(F(x)=\sum_{n=0}^{\infty}a_n(x-4)^n\), what are the constant and degree-\(n\) coefficients of \((x-4)F(x)\)?
- If \(F(x)=\sum_{n=0}^{\infty}a_n(x+1)^n\), express the coefficients of \(xF(x)\) using the \(a_n\).
- For \(F(x)=1+3(x-2)+2(x-2)^2\), calculate the coefficients of \((x-2)F(x)\).
- For a factor \(p_0+p_1(x-c)\), which terms contribute to the degree-\(n\) coefficient of its product with \(\sum_{n=0}^{\infty}a_n(x-c)^n\)?
- Why must a coefficient with a negative index be omitted when using the polynomial coefficient formula?