Power Series Mastery

From Reciprocal Series to Composition

The Local Power-Series Reciprocal theorem shows how to build a new series by expanding a geometric series and controlling the resulting coefficients. A broader version of the same question is this: when can one power series be substituted into another, and how can we justify the coefficients produced by that substitution?

The key is to make the inner series small in absolute value. The outer series then converges absolutely at the values taken by the inner series. Absolute convergence also controls the many products created by substitution, so they can be collected by powers of the original variable. This gives a local composition theorem and a practical formula for its coefficients.

Let the outer series be centered at $b$, and write the inner series as its value $b$ at $a$ plus a nonconstant part:

F(y)=\sum_{n=0}^{\infty}c_n(y-b)^n, \qquad G(x)=b+H(x), \qquad H(x)=\sum_{m=1}^{\infty}d_m(x-a)^m.

The composition $F(G(x))$, when defined, is obtained by substituting $H(x)$ for $y-b$. In other words, it is the sum of $\sum_{n=0}^{\infty}c_nH(x)^n$. The constant term $b$ in $G$ is essential: it ensures the substituted series is centered at the center of the outer series.

The Local Composition Theorem

Theorem (Local Composition of Power Series): Suppose $F(y)=\sum_{n=0}^{\infty}c_n(y-b)^n$ has radius of convergence $R>0$, and $G(x)=b+\sum_{m=1}^{\infty}d_m(x-a)^m$ has positive radius of convergence. There is an $r>0$ such that $F(G(x))$ is represented for $|x-a|\leq r$ by a power series $$ F(G(x))=\sum_{k=0}^{\infty}\gamma_k(x-a)^k $$ that converges absolutely there. Its coefficients are $$ \gamma_0=c_0,\qquad \gamma_k=\sum_{n=1}^{k}c_n \sum_{\substack{m_1,\ldots,m_n\geq1\\m_1+\cdots+m_n=k}} d_{m_1}\cdots d_{m_n} \quad (k\geq1). $$

Proof. Choose $t$ with $0<t<R$. By Convergence Inside and Outside the Radius, $\sum_{n=0}^{\infty}|c_n|t^n$ converges. Apply the Small-Radius Tail Bound from the previous tutorial to the nonconstant part of $G$. It gives an $r>0$, smaller than the radius of convergence of that series, such that

h(r):=\sum_{m=1}^{\infty}|d_m|r^m<t.

For $|x-a|\leq r$, the series for $H(x)$ converges absolutely, and

|H(x)|\leq\sum_{m=1}^{\infty}|d_m||x-a|^m \leq h(r)<t<R.

Thus the outer series converges absolutely at $G(x)=b+H(x)$. To control the coefficient construction, consider all terms generated by expanding $c_nH(x)^n$. The sum of their absolute values, weighted by their powers of $r$, is at most

\sum_{n=0}^{\infty}|c_n|h(r)^n \leq\sum_{n=0}^{\infty}|c_n|t^n <\infty.

For each fixed $n$, the Cauchy Product Theorem gives the coefficients of $H(x)^n$. More explicitly, when $n\geq1$, the terms contributing to degree $k$ are precisely the products $d_{m_1}\cdots d_{m_n}$ for which $m_1+\cdots+m_n=k$. Each $m_i\geq1$, so there are no such products when $k<n$. For a fixed $k$, only $n\leq k$ can contribute, and there are only finitely many choices of $m_1,\ldots,m_n$. Therefore the coefficient at degree $k$ is exactly the stated finite formula for $\gamma_k$, with constant coefficient $\gamma_0=c_0$.

The finite coefficient calculation identifies the terms at each degree. The absolute bound above justifies collecting all terms by degree: the total absolute sum is finite, so this regrouping does not change the sum. Consequently, for $|x-a|\leq r$,

F(G(x)) =\sum_{n=0}^{\infty}c_nH(x)^n =\sum_{k=0}^{\infty}\gamma_k(x-a)^k.

The same bound proves absolute convergence of the resulting series on this interval. This establishes both the representation and its convergence. $\square$

The coefficient formula is finite at every degree, but the finiteness alone is not a convergence proof. The absolute bound is what makes it legitimate to sum all the expanded powers of $H$ and then collect terms by degree.

Worked Applications

Worked Example: The Exponential of the Sine Series

Let $E(u)=\sum_{n=0}^{\infty}u^n/n!$ be the exponential series, and let $S(x)$ be the sine series. We want a local power series for $E(S(x))$ near $0$. The sine series has no constant term, and at $r=1/2$ its absolute sum is bounded by

\sum_{j=0}^{\infty}\frac{(1/2)^{2j+1}}{(2j+1)!} \leq\sum_{n=1}^{\infty}\frac{(1/2)^n}{n!} <\sum_{n=1}^{\infty}(1/2)^n=1.

The last inequality is strict because $1/n!<1$ for $n\geq2$. The outer exponential series has infinite radius of convergence, so the composition theorem guarantees an absolutely convergent power series on $|x|\leq1/2$.

To find the first terms, use $S(x)=x-x^3/6+\cdots$ and $E(u)=1+u+u^2/2+u^3/6+\cdots$. The linear coefficient is $1$. At degree $2$, only $S(x)^2/2$ contributes, with coefficient $1/2$. At degree $3$, the contributions are $-1/6$ from $S(x)$ and $1/6$ from $S(x)^3/6$; they cancel. Thus

E(S(x))=1+x+\frac{x^2}{2}+0x^3+\cdots \qquad (|x|\leq1/2).

The cancellation at degree $3$ illustrates why the coefficient formula is more reliable than guessing a pattern from the outer coefficients alone.

Worked Example: A Binomial Series with a Quadratic Inner Series

Consider the outer binomial series $(1+u)^{1/2}$, which has radius of convergence $1$, and substitute $u=x+x^2$. At $r=1/3$, the absolute sum of the inner series is

\frac13+\left(\frac13\right)^2=\frac49<1.

Therefore the composition theorem applies on $|x|\leq1/3$. The first outer terms are $1+u/2-u^2/8+u^3/16+\cdots$. Substituting $u=x+x^2$, the coefficient of $x$ is $1/2$. The coefficient of $x^2$ is $1/2-1/8=3/8$, since the $u/2$ term contributes $x^2/2$ and $-u^2/8$ contributes $-x^2/8$. At degree $3$, the contributions are $-2/8$ from $-u^2/8$ and $1/16$ from $u^3/16$. Hence

(1+x+x^2)^{1/2} =1+\frac{x}{2}+\frac{3x^2}{8}-\frac{3x^3}{16}+\cdots \qquad (|x|\leq1/3).

For example, the degree $3$ coefficient checks directly as $-2/8+1/16=-4/16+1/16=-3/16$. The convergence conclusion comes from the smallness of the inner series, not merely from the formal substitution.

Worked Example: The Logarithm of One Plus the Sine Series

The logarithm series centered at $1$ is $\ln(1+u)=\sum_{n=1}^{\infty}(-1)^{n+1}u^n/n$, with radius $1$. As in the first example, the absolute sum of the sine series at $r=1/2$ is less than $1$. Thus substituting $u=S(x)$ produces a convergent power series on $|x|\leq1/2$.

For the first terms, $S(x)=x-x^3/6+\cdots$. The term $S(x)$ contributes $x-x^3/6$. The term $-S(x)^2/2$ contributes $-x^2/2$ and has no degree $3$ term, because $S(x)^2=x^2+$ terms of degree at least $4$. The term $S(x)^3/3$ contributes $x^3/3$. Combining these contributions gives

\ln(1+S(x)) =x-\frac{x^2}{2}+\left(-\frac16+\frac13\right)x^3+\cdots =x-\frac{x^2}{2}+\frac{x^3}{6}+\cdots \qquad (|x|\leq1/2).

This calculation also shows why the center of the outer series matters. The logarithm series used here is centered at $1$, and the inner function $1+S(x)$ takes the value $1$ at $x=0$, exactly matching that center.

How the First Nonzero Terms Behave

Composition does more than produce coefficients: it also explains how quickly a composition can vanish at its center. The first nonzero term of the outer series determines a power of the inner series. If the inner series itself begins only at a higher degree, that degree is multiplied by the outer order.

Proposition (Leading Term of a Composition): Suppose $$ F(b+u)-F(b)=c_pu^p+\text{terms of degree greater than }p $$ with $p\geq1$ and $c_p\neq0$, and suppose $$ G(x)-b=d_q(x-a)^q+\text{terms of degree greater than }q $$ with $q\geq1$ and $d_q\neq0$. Then the first nonzero term of $F(G(x))-F(b)$ is $$ c_pd_q^p(x-a)^{pq}. $$

Proof. In the composition formula, the outer terms of index less than $p$ vanish after subtracting $F(b)$. The term of outer index $p$ is $c_p(G(x)-b)^p$. Since the first nonzero term of $G(x)-b$ is $d_q(x-a)^q$, the first nonzero term of its $p$th power is $d_q^p(x-a)^{pq}$. Its coefficient in the composition is therefore $c_pd_q^p$, which is nonzero. Every term of outer index $n>p$ has degree at least $nq>pq$, so none can contribute at degree $pq$ or below. Thus the displayed term is the first nonzero one. $\square$

Worked Example: A Composition That Vanishes to Fourth Order

Take $F(u)=C(u)$, the cosine series, and $G(x)=x^2+x^3$, centered at $0$. The first nonconstant term of $C(u)-C(0)$ is $-u^2/2$, so $p=2$ and $c_p=-1/2$. The first term of $G(x)$ is $x^2$, so $q=2$ and $d_q=1$. The proposition predicts the first nonzero term of $C(G(x))-1$ to be

-\frac12\cdot 1^2\,x^{2\cdot2}=-\frac{x^4}{2}.

Indeed, $C(u)=1-u^2/2+u^4/24-\cdots$, and

-\frac12(x^2+x^3)^2 =-\frac{x^4}{2}-x^5-\frac{x^6}{2}.

All subsequent outer terms have degree at least $8$ after substitution, so none changes the degree $4$ term. The composition theorem supplies convergence on a sufficiently small interval, and the leading-term proposition explains why the first change from $1$ occurs at degree $4$.

Why Convergence Control Matters

A formal substitution can give a plausible list of coefficients without proving that the resulting series converges or represents the intended function. In the composition theorem, two separate facts are needed. First, the inner series must stay within the radius of convergence of the outer series. Second, the products generated by expanding the powers of the inner series must have a finite total absolute sum, so that collecting them by degree is valid.

The proof secures both facts by choosing $r$ so that $\sum_{m\geq1}|d_m|r^m<t<R$. This bound controls the inner series even when its coefficients have mixed signs, and it bounds all the expanded outer terms by $\sum_{n\geq0}|c_n|t^n$. A radius that works for convergence need not be the largest possible radius; the theorem asserts a local representation, not a complete classification of where the composition converges.

Takeaway: To compose power series locally, make the absolute sum of the inner nonconstant series smaller than a radius where the outer series converges absolutely. This yields a convergent composition series, finite coefficient formulas, and a direct way to find its leading nonzero term.

Check Your Understanding

Use the composition theorem and its coefficient formula to answer the following questions.

Why is it useful to write the inner series as $G(x)=b+H(x)$, where $b$ is the center of the outer series?
What estimate ensures that $F(G(x))$ lies within the convergence region of the outer series?
Why are only finitely many outer powers able to contribute to the coefficient of $(x-a)^k$?
If the first nonzero terms of $F(b+u)-F(b)$ and $G(x)-b$ are $c_3u^3$ and $d_2(x-a)^2$, respectively, what is the first nonzero term of $F(G(x))-F(b)$?
For the expansion of $(1+x+x^2)^{1/2}$, identify the two contributions to the coefficient of $x^3$.

What You'll Learn