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Power Series Structure · Tutorial 718 of 998

How Variable Scaling Changes the Radius

Learn how scaling the variable shifts a power series’ center and multiplies or divides its radius, then apply the rules to convergence intervals.

Beginner 10 min read

What You'll Learn

  • Derive the new center and radius after replacing the variable by a multiple of itself.
  • Derive the corresponding rules when the variable is divided by a nonzero constant.
  • Track how positive and negative scaling affect interval endpoints.
  • Find the convergence intervals of scaled geometric-type series.
  • Check endpoint inclusion directly for the geometric examples.

Introduction

A power series is organized around a center: its terms involve powers of the distance from that center. Scaling the variable changes that distance, so it can change both the series’ center and its radius of convergence.

The effect is systematic. Replacing a variable by \(kx\) compresses or stretches distances in the opposite way from replacing it by \(x/k\). Keeping track of the center as well as the radius makes the rule easier to remember and prevents a common scaling error.

The Scaling Rule

Suppose a power series in the variable \(u\) is centered at \(c\):

$$ \sum_{n=0}^{\infty} a_n(u-c)^n. $$

Let its radius of convergence be \(R\), where \(0<R<\infty\), and let \(k\ne 0\) be a real constant. A point is inside the radius when its distance from \(c\) is less than \(R\). The substitution determines how that distance is measured in terms of \(x\).

Theorem: Replacing \(u\) by \(kx\) gives a series centered at \(c/k\), with radius \(R/|k|\). Replacing \(u\) by \(x/k\) gives a series centered at \(kc\), with radius \(|k|R\). These statements concern the radius; endpoint inclusion is carried over by the substitution and must be determined separately when it is not already known.

The absolute value matters because the radius is a nonnegative distance. If \(k\) is negative, the substitution also reverses the order of points on the real line, but it does not make the radius negative.

Why the Rule Works

For the substitution \(u=kx\), rewrite each term so that the power is measured from the new center:

$$ a_n(kx-c)^n = a_n\left(k\left(x-\frac{c}{k}\right)\right)^n = a_n k^n\left(x-\frac{c}{k}\right)^n. $$

This shows that the new center is \(c/k\). To determine where the substituted series converges, compare the distances from the old and new centers. The original series converges absolutely when \(|u-c|<R\). With \(u=kx\), this condition becomes \(|kx-c|<R\), or equivalently \(|k|\,|x-c/k|<R\). Thus the distance from the new center must be less than \(R/|k|\).

$$ |kx-c|<R \quad\Longleftrightarrow\quad \left|x-\frac{c}{k}\right|<\frac{R}{|k|}. $$

For \(u=x/k\), the same reasoning gives \(|x/k-c|<R\). Since \(x/k-c=(x-kc)/k\), the condition becomes \(|x-kc|<|k|R\). This proves the second rule as well.

$$ \left|\frac{x}{k}-c\right|<R \quad\Longleftrightarrow\quad |x-kc|<|k|R. $$

These equivalences also explain what happens at the endpoints: the substitution maps each candidate endpoint in \(x\) to the corresponding endpoint in \(u\). If the original series converges at that point, the substituted series does too; if it diverges there, the substituted series diverges there. A negative \(k\) swaps which endpoint corresponds to the left and right sides.

Worked Example: Replacing \(u\) by a Multiple

Worked Example: Replacing \(u\) by a Multiple

Find the convergence interval of \(\displaystyle\sum_{n=0}^{\infty}(-3x-1)^n\).

This is a geometric series with ratio \(q=-3x-1\). A geometric series \(\sum_{n=0}^{\infty}q^n\) converges when \(|q|<1\). Therefore:

$$ |-3x-1|<1 \quad\Longleftrightarrow\quad |3x+1|<1. $$

Solving the inequality gives:

$$ -1<3x+1<1 \quad\Longleftrightarrow\quad -2<3x<0 \quad\Longleftrightarrow\quad -\frac{2}{3}<x<0. $$

The ratio is \(1\) at \(x=-2/3\) and \(-1\) at \(x=0\). At either point, the terms \(q^n\) do not tend to zero, so the series diverges. The convergence interval is \((-2/3,0)\).

Check the scaling rule by viewing this as \(\sum_{n=0}^{\infty}(u-1)^n\) with \(u=-3x\). The original geometric series has center \(1\) and radius \(1\). Here \(k=-3\), so the new center is \(1/(-3)=-1/3\), and the new radius is \(1/3\). The interval \((-1/3-1/3,-1/3+1/3)\) is exactly \((-2/3,0)\).

Worked Example: Replacing \(u\) by a Quotient

Worked Example: Replacing \(u\) by a Quotient

Find the convergence interval of \(\displaystyle\sum_{n=0}^{\infty}\left(\frac{x/2-1}{3}\right)^n\).

Rewrite the ratio to expose the center:

$$ \frac{x/2-1}{3} = \frac{x-2}{6}. $$

The series is geometric with ratio \((x-2)/6\). It converges when the absolute value of that ratio is less than \(1\):

$$ \left|\frac{x-2}{6}\right|<1 \quad\Longleftrightarrow\quad |x-2|<6 \quad\Longleftrightarrow\quad -4<x<8. $$

At \(x=-4\), the ratio is \(-1\), and at \(x=8\), it is \(1\). In both cases the terms fail to tend to zero, so neither endpoint is included. The interval is \((-4,8)\).

For a scaling check, start from \(\sum_{n=0}^{\infty}((u-1)/3)^n\), which has center \(1\) and radius \(3\), and substitute \(u=x/2\). This is the rule \(u=x/k\) with \(k=2\). The new center is \(2(1)=2\), and the new radius is \(2(3)=6\), matching the interval just found.

Scaling and Interval Endpoints

A radius describes the distance from the center to the boundary of the region of absolute convergence. It gives an open interval first; whether either endpoint is included depends on the behavior of the series at that point. Scaling does not change that endpoint behavior by itself: it transports each endpoint through the substitution.

For example, if a series in \(u\) converges for \(u\) in an interval with endpoints \(A\) and \(B\), then substituting \(u=kx\) means asking which \(x\) satisfy \(kx\) in that interval. When \(k>0\), the endpoint order is preserved. When \(k<0\), the order is reversed. This set-based viewpoint works whether the endpoints are included or excluded.

Key takeaway: Under \(u=kx\), divide the old radius by \(|k|\). Under \(u=x/k\), multiply it by \(|k|\). In either case, transform the center using the same substitution, and map endpoint inclusion through that substitution.

Worked Example: Negative Scaling and Reversed Endpoints

Worked Example: Negative Scaling and Reversed Endpoints

Find the convergence interval of \(\displaystyle\sum_{n=0}^{\infty}\left(\frac{-3x-1}{2}\right)^n\).

The ratio is \(q=(-3x-1)/2\). Apply the geometric convergence condition:

$$ \left|\frac{-3x-1}{2}\right|<1 \quad\Longleftrightarrow\quad |-3x-1|<2 \quad\Longleftrightarrow\quad -2<-3x-1<2. $$

Add \(1\) to all parts, then divide by \(-3\), reversing both inequality signs:

$$ -1<-3x<3 \quad\Longleftrightarrow\quad -1<x<\frac{1}{3}. $$

The ratio is \(1\) at \(x=-1\) and \(-1\) at \(x=1/3\), so at both endpoints the terms do not tend to zero. The interval is \((-1,1/3)\).

To verify the center and radius another way, start with \(\sum_{n=0}^{\infty}((u-1)/2)^n\), which has center \(1\) and radius \(2\), and substitute \(u=-3x\). The new center is \(-1/3\), and the new radius is \(2/3\). Thus the open interval is \((-1/3-2/3,-1/3+2/3)=(-1,1/3)\). The negative scaling reverses the endpoint correspondence: \(x=-1\) maps to \(u=3\), while \(x=1/3\) maps to \(u=-1\).

A Common Scaling Mistake

Suppose a series centered at \(c\) has radius \(R\), and the variable is replaced by \(kx\). A common error is to multiply the radius by \(|k|\), perhaps because each term contains a factor of \(k\). That gives the wrong direction: the substitution \(u=kx\) makes a change in \(x\) correspond to a change \(|k|\) times as large in \(u\).

Pitfall: If the original radius is \(2\) and \(u=4x\), it is incorrect to claim the new radius is \(8\). The condition is \(|4x-c|<2\), so \(|x-c/4|<1/2\). The new center is \(c/4\), and the new radius is \(2/4=1/2\).

By contrast, with \(u=x/4\), the condition \(|x/4-c|<2\) is equivalent to \(|x-4c|<8\), so the new radius is \(8\). The substitution itself tells you which rule to use: multiplying the variable by \(4\) divides the radius by \(4\); dividing the variable by \(4\) multiplies it by \(4\).

What If the Scaling Constant Is Zero?

The formulas above require \(k\ne0\), since they involve division by \(k\) and a reversible change of variable. If \(u=0x\), the substituted expression no longer varies with \(x\). For instance, \(\sum a_n(0x-c)^n\) is the same numerical series for every \(x\), provided that series is defined and converges. It is then a constant function of \(x\), not a series whose radius follows from the nonzero scaling rule.

Why Scaling Is Useful

Scaling lets you recognize a familiar series inside a new expression. Rather than restarting the radius calculation from scratch, identify the original center and radius, determine the substitution, and transform both. For geometric-type series, the ratio also gives a direct check: impose \(|q(x)|<1\), solve the inequality carefully, and check any boundary points by examining whether the terms tend to zero.

The two checks reinforce one another. The center-and-radius rule is efficient for general power series, while the ratio inequality is often especially transparent for a geometric series. If their results disagree, revisit the substitution, absolute value, or direction of an inequality after division by a negative number.

Check Your Understanding

For each question, identify the substitution and track the center and radius before deciding whether any endpoints are included.

  1. A power series centered at \(u=2\) has radius \(5\). After substituting \(u=3x\), what are its new center and radius?
  2. A series centered at \(u=-1\) has radius \(4\). After substituting \(u=x/2\), what are its new center and radius?
  3. Find the open interval where \(\displaystyle\sum_{n=0}^{\infty}(2x-4)^n\) converges. What happens at the two endpoints?
  4. For \(\displaystyle\sum_{n=0}^{\infty}\left(\frac{x}{5}+1\right)^n\), find the convergence interval and check both endpoints.
  5. Explain why the radius is divided by \(|k|\) for the substitution \(u=kx\), rather than multiplied by \(|k|\).

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