Composing Series with Substitutions

Why Substitution Changes the Validity Condition

A known power series can become a useful expansion for a more complicated function by replacing its input with another expression. The important point is that the original series converges according to its input, not directly according to \(x\). After substitution, we must determine which \(x\)-values make that input fall within the original convergence interval.

For example, if a series in \(u\) converges when \(-1 \leq u \leq 1\), then substituting \(u=2x-1\) requires solving \(-1 \leq 2x-1 \leq 1\). It does not automatically give the same interval for \(x\). The substituted expression may shrink, shift, or otherwise change the set of valid inputs.

Core rule: If a known series represents \(F(u)\) for \(u\) in its convergence interval \(I\), then substituting \(u=g(x)\) gives a series for \(F(g(x))\) at every \(x\) for which \(g(x)\in I\). Boundary values must be included only when the original series converges there.

The Substitution Principle

Suppose a known power series has the form

F(u)=\sum_{n=0}^{\infty}c_nu^n

and represents \(F(u)\) for inputs in a convergence interval \(I\). Replacing \(u\) by \(g(x)\) gives

F(g(x))=\sum_{n=0}^{\infty}c_n\bigl(g(x)\bigr)^n, \qquad \text{provided } g(x)\in I.

This is a direct substitution: each occurrence of \(u\) in the known series is replaced by the same expression \(g(x)\). The resulting terms may involve powers of a polynomial or another function of \(x\). If those powers are expanded or rearranged to make a conventional power series in \(x\), that is an additional algebraic step; the substitution itself does not change the original convergence requirement.

If the known series has radius of convergence \(R\), then it converges for \(|u|<R\). Consequently, \(|g(x)|<R\) is a sufficient condition for the substituted series to converge. The original series may also converge at one or both endpoints \(u=\pm R\); those endpoints require a separate check using the known interval of convergence.

Why the Substitution Works

The justification comes from the meaning of a series representation. For each allowed input \(u\), the partial sums approach the value \(F(u)\). Once \(u\) is set equal to \(g(x)\), those same partial sums approach \(F(g(x))\) whenever \(g(x)\) is an allowed input. The convergence condition is therefore transferred through the inner function.

The geometric series gives a direct derivation. For \(u\neq 1\), its finite partial sum satisfies

1+u+u^2+\cdots+u^N = \frac{1-u^{N+1}}{1-u}.

When \(|u|<1\), the powers \(u^{N+1}\) approach zero as \(N\) increases. Taking the limit of the partial sums gives

\sum_{n=0}^{\infty}u^n=\frac{1}{1-u}, \qquad |u|<1.

Now let \(u=g(x)\). For every \(x\) such that \(|g(x)|<1\), the same partial-sum identity applies, and \(g(x)^{N+1}\) approaches zero. Thus

\sum_{n=0}^{\infty}\bigl(g(x)\bigr)^n = \frac{1}{1-g(x)}, \qquad |g(x)|<1.

The restriction is not an optional detail: it is what makes the limiting step valid. At an endpoint where \(|g(x)|=1\), this derivation does not establish convergence, so the endpoint must be tested separately.

Worked Example: A Geometric Series with a Quadratic Input

Worked Example: Expand a Rational Function by Substitution

Find a geometric-series expansion for \(\frac{1}{1+x^2}\), and determine its interval of convergence.

Start with the known geometric series

\frac{1}{1-u}=\sum_{n=0}^{\infty}u^n, \qquad |u|<1.

Set \(u=-x^2\). The left side becomes

\frac{1}{1-(-x^2)} = \frac{1}{1+x^2}.

The right side becomes

\sum_{n=0}^{\infty}(-x^2)^n = \sum_{n=0}^{\infty}(-1)^n x^{2n}.

The original condition is \(|u|<1\), so here it becomes

|-x^2|<1 \quad\Longleftrightarrow\quad x^2<1 \quad\Longleftrightarrow\quad -1<x<1.

At \(x=1\), the terms are \((-1)^n\), which do not approach zero, so the series diverges. At \(x=-1\), \(x^{2n}=1\), so the terms are again \((-1)^n\) and do not approach zero. The series therefore diverges at both endpoints.

The result is

\frac{1}{1+x^2} = 1-x^2+x^4-x^6+\cdots = \sum_{n=0}^{\infty}(-1)^n x^{2n}, \qquad -1<x<1.

As a check, the geometric partial-sum formula with \(u=-x^2\) gives

\sum_{n=0}^{N}(-x^2)^n = \frac{1-(-x^2)^{N+1}}{1+x^2}.

For \(|x|<1\), the term \((-x^2)^{N+1}\) approaches zero, so the partial sums approach \(\frac{1}{1+x^2}\), as required.

Worked Example: A Logarithmic Series and Included Endpoints

Worked Example: Expand \(\ln(1+x^2)\)

Use the known series

\ln(1+u) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}u^n}{n}, \qquad -1<u\leq 1.

Substitute \(u=x^2\). The function becomes \(\ln(1+x^2)\), and the series becomes

\ln(1+x^2) = \sum_{n=1}^{\infty} \frac{(-1)^{n+1}x^{2n}}{n}.

The original convergence interval requires \(-1<x^2\leq 1\). Since \(x^2\geq 0\) for every real \(x\), the lower inequality is always satisfied. The upper inequality gives

x^2\leq 1 \quad\Longleftrightarrow\quad -1\leq x\leq 1.

The endpoint \(u=1\) is included in the known interval. Specifically, the series at \(x=1\) is \(\sum_{n=1}^{\infty}(-1)^{n+1}/n\), which converges to \(\ln 2\). At \(x=-1\), the powers \(x^{2n}\) are also \(1\), so the same convergent series results. At each endpoint, the function value is \(\ln(1+1)=\ln 2\).

At \(x=0\), every term is zero and the sum is \(\ln(1)=0\), which also checks the formula at a simple interior input. Therefore the expansion is valid on the full interval

-1\leq x\leq 1.

Worked Example: A Shifted Input and Its Convergence Interval

Worked Example: Expand \(\arctan(2x-1)\)

The inverse-tangent series is

\arctan(u) = \sum_{n=0}^{\infty} \frac{(-1)^n u^{2n+1}}{2n+1}, \qquad -1\leq u\leq 1.

Substitute \(u=2x-1\). This gives

\arctan(2x-1) = \sum_{n=0}^{\infty} \frac{(-1)^n(2x-1)^{2n+1}}{2n+1}.

To find the valid \(x\)-values, require the substituted input to lie in the original interval:

-1\leq 2x-1\leq 1.

Add \(1\) to all three parts and then divide by \(2\):

0\leq 2x\leq 2 \quad\Longleftrightarrow\quad 0\leq x\leq 1.

The endpoints are allowed because the inverse-tangent series converges at \(u=-1\) and \(u=1\). At \(x=0\), the input is \(-1\), and the series sums to \(-\frac{\pi}{4}=\arctan(-1)\). At \(x=1\), the input is \(1\), and the series sums to \(\frac{\pi}{4}=\arctan(1)\). Thus the expansion represents the function on \(0\leq x\leq 1\).

In particular, the center \(x=\frac12\) makes the inner input zero. The condition \(|2x-1|<1\) describes the strict interior \(0<x<1\); the separately known endpoint convergence extends the interval to include \(0\) and \(1\).

Finding the Valid Neighborhood

When the original convergence interval is given, the central task is to take its preimage under the inner function: solve for the \(x\)-values whose substituted inputs lie in that interval. In simple examples this is an inequality. For a more complicated \(g(x)\), there may be several separate intervals of valid inputs, or the condition may need to be described without simplifying it further.

Write the known representation.
Record the function, series variable, and exact convergence interval for that variable.

Substitute consistently.
Replace each occurrence of the series variable with the same expression \(g(x)\), including in the function on the left.

Translate the convergence condition.
Require \(g(x)\) to belong to the original interval, then solve for \(x\).

Check boundary inputs.
If solving the strict radius condition produces endpoints, use the known endpoint behavior of the original series to decide whether to include them.

A Common Mistake: Treating the Radius as the Whole Interval

A frequent error is to convert the strict condition \(|u|<R\) into a complete answer and ignore endpoint information. For the logarithmic example, using only the radius condition gives

|x^2|<1 \quad\Longrightarrow\quad -1<x<1.

That is the guaranteed interior, but it is not the full interval of convergence for \(\ln(1+x^2)\). The original logarithmic series also converges at \(u=1\), and \(x^2=1\) when \(x=-1\) or \(x=1\). The corrected interval is \([-1,1]\).

The reverse mistake is also possible: including endpoints merely because the inequality has a boundary value. For the geometric example, substituting \(x=1\) produces terms \((-1)^n\), which do not approach zero. Since a convergent series must have terms that approach zero, this endpoint cannot be included. Endpoint conclusions depend on the original series, not on whether the substituted expression looks harmless.

Why Composition Is Useful

Substitution lets a familiar expansion describe functions whose inputs are more complicated than the series variable itself. It can produce approximations for rational, logarithmic, and inverse-tangent expressions, among many others. It also makes the domain of an expansion explicit: the series is useful only where its inner input satisfies the original convergence condition.

Keep the two tasks distinct. First, substitute to obtain a candidate representation. Then determine where that representation converges by tracing the input back to the original interval, checking endpoints separately. This avoids both an unjustifiably large interval and an unnecessarily small one.

Check Your Understanding

For each question, state the substituted convergence condition and pay particular attention to any endpoints.

Substitute \(u=3x\) into \(\frac{1}{1-u}=\sum_{n=0}^{\infty}u^n\). For which \(x\)-values does the resulting series converge?
Using the logarithmic series from this tutorial, find the interval on which its substitution \(u=1-2x\) converges.
Substitute \(u=x^2-1\) into the inverse-tangent series. Determine the \(x\)-interval from the original convergence interval.
Why does \(|g(x)|<R\) guarantee convergence for a series with radius \(R\), while a point satisfying \(|g(x)|=R\) needs separate analysis?
For \(\ln(1+x^2)\), what is the value of the series at each endpoint, and why are those endpoints included?