Introduction
A power series is written using two kinds of information: powers of a variable measured from a center, and coefficients attached to those powers. Reading the notation accurately means identifying both parts. This is straightforward when every series is written in exactly the same format, but a shifted starting index or a pattern of skipped powers can hide the coefficient sequence.
The central form to recognize is \(\sum_{n=0}^{\infty} c_n(x-a)^n\). The number \(a\) is the center, and \(c_n\) is the coefficient of the \(n\)th power of \(x-a\). The summation index is a label; it need not itself be the exponent appearing in every version of the series.
Recognizing the Center
Look at the variable expression being raised to successive powers. In \((x-a)^n\), the center is \(a\), the number subtracted from \(x\). The sign inside the parentheses matters: \(x+4\) is \(x-(-4)\), so its center is \(-4\), not \(4\).
This identification is about the base of the powers, not about the coefficients. A large coefficient, a factorial, or a sign pattern does not change the center. For instance, the expression \(7(-1)^n(x-3)^n\) has center \(3\); the coefficient of the \(n\)th power is \(7(-1)^n\).
When the base is written in another but equivalent way, rewrite it mentally in the form \(x-a\). For example, \(2-x=-(x-2)\). Thus powers of \(2-x\) are powers about the center \(2\), with the extra signs incorporated into their coefficients.
Recognizing the Coefficients
Once the powers are identified, attach to each power the factor multiplying it. In the standard form, \(c_0\) multiplies \((x-a)^0\), \(c_1\) multiplies \((x-a)^1\), and so on. Since \((x-a)^0=1\), the first term is the constant coefficient \(c_0\).
Writing out the first few terms is a useful way to check the reading. It makes the correspondence between the exponent and its coefficient visible. This is a notation check, not a test for convergence: no conclusion about where a series converges follows merely from identifying its center and coefficients.
Worked Example: Reading a Standard Power Series
Identify the center and coefficient sequence of
First express the base in the form \(x-a\): \(x+2=x-(-2)\). Therefore the center is \(a=-2\).
The factor multiplying \((x+2)^n\) is \((2n+1)/4^n\), so the coefficient with index \(n\) is
Compute the first three coefficients:
The series therefore begins
Check: The coefficient of \((x+2)^2\) in the displayed series is \(5/16\), which agrees with the formula \(c_2=(2\cdot2+1)/4^2=5/16\). The base \(x+2=x-(-2)\) also confirms the center \(-2\).
When the Index and Exponent Differ
A series may be indexed by \(n\) while the power is \(n-1\), \(n+2\), or another expression. In that case, do not assume that the index is the power number. Instead, match each term to its actual power of \(x-a\).
For a common pattern, suppose the index begins at \(r\) and the exponent is \(n-r\). Set \(k=n-r\). Then \(n=k+r\), and as \(n\) runs through \(r,r+1,r+2,\ldots\), \(k\) runs through \(0,1,2,\ldots\). Substituting this relation into both the coefficient and the power gives the standard coefficient sequence.
Identify \(a\) from the factor \((x-a)\).
Set the displayed exponent equal to the standard power index \(k\).
Express its original index in terms of \(k\), then check the first terms.
Worked Example: Reindexing a Shifted Power
Identify the center and coefficient sequence of
The base is \(x-5\), so the center is \(5\). The exponent is \(n-1\), so set \(k=n-1\). This gives \(n=k+1\). The original starting value \(n=1\) corresponds to \(k=0\).
Replace \(n\) by \(k+1\) in the coefficient:
Consequently, the coefficient of \((x-5)^k\) is
To check, take the first two original terms. At \(n=1\), the term is \(-1\cdot(x-5)^0=-1\). At \(n=2\), it is \(\frac12(x-5)^1\). The reindexed coefficients give \(c_0=(-1)^1/1=-1\) and \(c_1=(-1)^2/2=1/2\), matching both terms.
Check: For the third term, the original index is \(n=3\), giving \((-1)^3/3=-1/3\) and exponent \(3-1=2\). The reindexed formula gives \(c_2=(-1)^3/3=-1/3\), as required.
A Short Justification for Reindexing
Reindexing here does not alter the series terms; it gives each term a new label that agrees with its power. More precisely, for each original index \(n\ge r\), there is exactly one \(k=n-r\ge0\), and conversely each \(k\ge0\) gives exactly one \(n=k+r\). Also, the corresponding power satisfies \((x-a)^{n-r}=(x-a)^k\). The coefficient is therefore the original coefficient evaluated at \(n=k+r\).
This equality expresses the same terms in the same order, with the index shifted. The resulting coefficient formula follows directly from the one-to-one correspondence between \(n\) and \(k\); it is not a new convergence claim.
When Some Powers Are Missing
Not every power needs to appear explicitly. A series containing only even powers, for example, can still be described using a coefficient for every nonnegative integer power: assign coefficient zero to each missing power. This gives a complete sequence \(c_0,c_1,c_2,\ldots\), rather than a sequence whose positions are ambiguous.
Worked Example: Reading a Series with Missing Powers
Identify the center and coefficient sequence of
The base is \(x-1\), so the center is \(1\). The powers are \(2n+1\), which are the positive odd integers. To describe the coefficient of each power \(k\), distinguish odd and even values of \(k\).
If \(k=2n+1\) is odd, then \(n=(k-1)/2\), and the coefficient is \(3^n=3^{(k-1)/2}\). If \(k\) is even, that power does not occur, so its coefficient is zero. Thus
Calculate the first six coefficients: \(c_0=0\), \(c_1=3^0=1\), \(c_2=0\), \(c_3=3^1=3\), \(c_4=0\), and \(c_5=3^2=9\). The first terms are therefore
After omitting the zero terms, this reads \((x-1)+3(x-1)^3+9(x-1)^5+\cdots\), which matches the original sum at \(n=0,1,2\). Check: The original term at \(n=2\) is \(3^2(x-1)^5=9(x-1)^5\), agreeing with \(c_5=9\).
Signs and Equivalent Bases
A minus sign in the base can be part of the coefficient pattern once the base is rewritten. For example, \((2-x)^n\) can be written as \((-1)^n(x-2)^n\). Thus, in a series whose terms are \(d_n(2-x)^n\), the center is \(2\), and the coefficient of \((x-2)^n\) is \(d_n(-1)^n\). The sign factor is not a change in the center.
Likewise, the center is not the value that makes an entire coefficient equal to zero, nor a number appearing only in a coefficient formula. It is read from the variable-dependent base of the powers. Keeping the base and coefficient roles separate prevents many sign errors.
| Displayed power | Center | What to notice |
|---|---|---|
| \((x-4)^n\) | \(4\) | The base already has the form \(x-a\). |
| \((x+4)^n\) | \(-4\) | Rewrite \(x+4\) as \(x-(-4)\). |
| \((2-x)^n\) | \(2\) | Rewrite as \((-1)^n(x-2)^n\). |
Common Pitfall: Confusing a Sign or Index with the Center
A frequent mistake is to read the number inside parentheses without accounting for its sign. For instance, calling the center of \((x+4)^n\) equal to \(4\) is incorrect. The definition requires the base to be \(x-a\), and \(x+4=x-(-4)\), so the center is \(-4\).
Another error is to call the coefficient of \((x-a)^{n-1}\) the coefficient with index \(n\). In standard notation, the index names the power. If the exponent is \(n-1\), then the coefficient belongs to power \(k=n-1\). The correction is to set \(k=n-1\), solve \(n=k+1\), and substitute that expression in the coefficient. This gives the coefficient of each power in its proper position.
Check Your Understanding
For each question, identify the center, the relevant coefficient rule, or both. Check your results by comparing the first terms.
- For \(\sum_{n=0}^{\infty}(n+2)(x-3)^n\), what are the center and the coefficient \(c_n\)?
- For \(\sum_{n=1}^{\infty}n(x+1)^{n-1}\), reindex the series and state \(c_k\) and the center.
- For \(\sum_{n=0}^{\infty}2^n(x-4)^{2n}\), what are \(c_0,c_1,c_2,c_3,c_4\)?
- What is the center of a series with powers \((5-x)^n\), and what sign factor appears when those powers are written using \(x-5\)?
- In a series with terms \(A_n(x-a)^{n-2}\) beginning at \(n=2\), express the coefficient \(c_k\) in terms of \(A\).
