Why squaring a power series has a pattern
A power series is an infinite sum, but the coefficient of any one power in its square is determined by only finitely many terms. Finding that coefficient is therefore a finite calculation. The calculations also have a useful symmetry: in a product, the pair of terms contributing \(a_k a_{n-k}\) appears in the opposite order as \(a_{n-k}a_k\).
This symmetry makes squaring a particularly clear application of the Cauchy product. It helps organize the arithmetic, provided we count the middle term correctly when the power is even.
The coefficient formula for a square
Let a power series centered at \(c\) be \(f(x)=\sum_{n=0}^{\infty}a_n(x-c)^n\), with radius of convergence \(R\). For \(|x-c|<R\), the series converges absolutely, so its product with itself can be grouped by powers of \(x-c\). Write the resulting coefficients as \(b_n\):
To obtain the coefficient of \((x-c)^n\), choose a term of degree \(k\) from the first factor and a term of degree \(n-k\) from the second. The coefficient is consequently the sum of all such products.
This is a finite sum for each \(n\). To exploit its symmetry, pair the term indexed by \(k\) with the term indexed by \(n-k\). Since multiplication of real or complex numbers is commutative, these two terms have equal values.
If \(n\) is odd, there is no unpaired middle term. If \(n\) is even, the middle term occurs when \(k=n/2\); it must be included once, not twice. Thus the symmetric versions of the formula are:
For \(n=0\), the sum in the even formula has no terms, leaving \(b_0=a_0^2\). This convention makes the formula valid at the constant term as well.
Why the formula is valid inside the radius
The coefficient calculation is not only a formal rearrangement. For a fixed \(x\) with \(|x-c|<R\), absolute convergence gives \(\sum_{n=0}^{\infty}|a_n(x-c)^n|<\infty\). The sum of the absolute values of all terms in the product is therefore finite:
Absolute convergence permits grouping the product terms according to their total degree \(i+j\). For total degree \(n\), the possible pairs are \((0,n),(1,n-1),\ldots,(n,0)\). Their sum is exactly the convolution formula.
Each product term has the form \(a_i a_j(x-c)^{i+j}\).
To get degree \(n\), use precisely the pairs of indices with \(i+j=n\).
The pairs \((k,n-k)\) and \((n-k,k)\) contribute the same product; retain a single middle term when \(n\) is even.
The resulting power series represents \(f(x)^2\) for \(|x-c|<R\). This guarantees convergence of the squared-series representation throughout the original series’ interval of absolute convergence. It does not by itself assert that the new series has exactly the same radius: the radius can be larger if its coefficients have cancellations or other special structure.
Worked Example: Computing several coefficients from given terms
Worked Example: Computing several coefficients from given terms
Suppose the coefficients through degree four are \(a_0=1\), \(a_1=2\), \(a_2=-1\), \(a_3=3\), and \(a_4=0\). Find the coefficients \(b_0,\ldots,b_4\) in the square, for a series centered at \(c\). Coefficients of higher degree in the original series cannot affect these first five results.
For degree zero, \(b_0=a_0a_0=1\).
For degree one, \(b_1=a_0a_1+a_1a_0=(1)(2)+(2)(1)=4\).
For degree two, pair the two outer terms and retain the middle term once: \(b_2=2a_0a_2+a_1^2=2(1)(-1)+(2)^2=-2+4=2\).
For degree three, pair the two contributions: \(b_3=2(a_0a_3+a_1a_2)=2((1)(3)+(2)(-1))=2(3-2)=2\).
For degree four, pair the outer terms and include the middle term: \(b_4=2a_0a_4+2a_1a_3+a_2^2=2(1)(0)+2(2)(3)+(-1)^2=0+12+1=13\).
Check by directly expanding the original series only as far as needed. Writing \(h=x-c\), its initial terms are \(1+2h-h^2+3h^3+0h^4+\cdots\). The contributions through degree four in its square are:
The direct expansion gives the same coefficients \(1,4,2,2,13\) as the convolution calculations.
Worked Example: Squaring the geometric power series
Worked Example: Squaring the geometric power series
Consider \(f(x)=\sum_{n=0}^{\infty}x^n\), which converges for \(|x|<1\). Here \(c=0\) and \(a_n=1\) for every \(n\geq0\). For a fixed degree \(n\), the convolution has \(n+1\) terms, each equal to \(1\). Hence \(b_n=n+1\), and the squared series is \(\sum_{n=0}^{\infty}(n+1)x^n\) on this interval.
The symmetry formula shows the same count. If \(n=2m+1\) is odd, there are \(m+1\) paired contributions, each contributing \(2\), so \(b_n=2(m+1)=2m+2=n+1\). If \(n=2m\) is even, there are \(m\) pairs and one middle term, so \(b_n=2m+1=n+1\).
Check the first coefficients by direct expansion: \((1+x+x^2+x^3+\cdots)^2\) has constant term \(1\), coefficient \(1+1=2\) for \(x\), coefficient \(1+1+1=3\) for \(x^2\), and coefficient \(1+1+1+1=4\) for \(x^3\). These agree with \(n+1\) for \(n=0,1,2,3\). In particular, within \(|x|<1\), the squared series represents the square of the original geometric series.
Worked Example: A series with only even powers
Worked Example: A series with only even powers
Let \(g(x)=\sum_{m=0}^{\infty}x^{2m}=1+x^2+x^4+x^6+\cdots\), which converges for \(|x|<1\). Its coefficient sequence in ordinary powers of \(x\) is \(a_{2m}=1\) and \(a_{2m+1}=0\).
For degree \(2n\), the nonzero terms in the convolution come from even indices \(2j\) and \(2(n-j)\), with \(j=0,\ldots,n\). There are \(n+1\) such terms, each equal to \(1\), so \(b_{2n}=n+1\). For odd degree \(2n+1\), each pair of indices has one odd index; its coefficient is zero. Therefore \(b_{2n+1}=0\).
Consequently, the square is \(\sum_{n=0}^{\infty}(n+1)x^{2n}\) for \(|x|<1\). Check the initial terms directly: \((1+x^2+x^4+\cdots)^2=1+2x^2+3x^4+4x^6+\cdots\). The coefficients of \(x,x^3,x^5,\ldots\) are zero, and the first even-degree coefficients agree with \(n+1\).
A common mistake: doubling the middle term
Symmetry allows paired terms to be doubled, but it does not allow every term to be doubled. When \(n\) is even, the index \(k=n/2\) pairs with itself. Doubling this term counts the same product twice.
The correct calculation counts the outer pair twice in total, but counts the middle product once:
Direct multiplication checks the correction: the degree-two terms are \(1\cdot 3x^2\), \(2x\cdot 2x\), and \(3x^2\cdot1\), whose coefficients sum to \(3+4+3=10\).
What to keep in view
To square a power series, first identify its center and coefficient sequence. Then compute each desired coefficient from the finite convolution; use symmetry to pair equal products, taking care to include an even-degree middle term only once. A direct expansion of the first few terms is an effective check on indexing and arithmetic.
The convergence condition matters too. The identity between the square of the function and the convolution series is guaranteed inside the original radius, where the power series converges absolutely. Endpoint behavior or a possibly larger radius of the squared series requires separate analysis; it should not be inferred merely from the coefficient formula.
Check Your Understanding
Use the convolution formula and, where requested, check your result by direct expansion.
- If \(a_0=2\), \(a_1=-1\), and \(a_2=3\), calculate \(b_0\), \(b_1\), and \(b_2\).
- For an arbitrary coefficient sequence, write \(b_5\) using symmetry and explain why there is no middle term to add separately.
- For an arbitrary coefficient sequence, write \(b_6\) using symmetry and identify the unpaired middle term.
- If \(a_{2m}=1\) and \(a_{2m+1}=0\) for all \(m\geq0\), which coefficients of the square must vanish? Explain using the convolution indices.
- Why does absolute convergence inside the original radius justify grouping the product terms by total degree?
