Why Scaling Changes a Taylor Series
A Taylor series can be useful beyond the function for which it was first derived. If a known expansion represents \(f(u)\), replacing \(u\) by \(cx\) gives a series for \(f(cx)\). This simple substitution changes both the coefficients and the interval on which the series converges.
The radius of convergence changes in a predictable way when \(c\ne 0\). Endpoint behavior, however, cannot be read from the radius alone. The original endpoint results must be mapped through the substitution, with special care when \(c\) is negative.
The Scaling Rule
Suppose a power series centered at zero is known to represent \(f(u)\) on its interval of convergence:
For a constant \(c\), substitution gives
For \(c\ne 0\), the new coefficient of \(x^n\) is \(a_nc^n\). More importantly, the series on the right is the original series evaluated at \(u=cx\). This observation determines its convergence set, including its endpoints.
Assume the original series has a finite positive radius \(R\). Its new radius after the substitution \(u=cx\), for \(c\ne 0\), is \(R/|c|\). The absolute value matters because a radius is nonnegative; a negative \(c\) reverses the order of the mapped endpoints, not the radius itself.
Why the Radius Scales by \(1/|c|\)
The radius formula follows from the root test, or equivalently the Cauchy–Hadamard formula. Let \(L=\limsup_{n\to\infty}|a_n|^{1/n}=1/R\). The coefficients of the scaled series are \(b_n=a_nc^n\), so
The Cauchy–Hadamard formula says that the radius for coefficients \(b_n\) is the reciprocal of this limit superior. Therefore the new radius is \(R/|c|\). Equivalently, the root test gives convergence when \(|c||x|/R<1\), or \(|x|<R/|c|\), and divergence when \(|x|>R/|c|\). At equality, the root test does not decide convergence: the original endpoint series must be checked.
If \(c=0\), the substitution gives \(f(0)\), a constant series, and the resulting series converges for every \(x\). The radius formula \(R/|c|\) is not used in that case.
Worked Example: Scaling a Logarithm by a Negative Constant
Worked Example: Find the series and interval for \(\ln(1-2x)\)
The known logarithm series is
Set \(u=-2x\). Substitution gives
For the interval, apply the original condition \(-1<u\leq 1\) to \(u=-2x\):
The inequality directions reverse when dividing by \(-2\). The interval is therefore \([-1/2,1/2)\). Check its endpoints in the series. At \(x=-1/2\), \(u=1\), so the series is \(\sum_{n=1}^{\infty}(-1)^{n+1}/n=\ln 2\), which converges. At \(x=1/2\), \(u=-1\), so the series is \(-\sum_{n=1}^{\infty}1/n\), which diverges. The function itself is undefined there because \(\ln(1-2(1/2))=\ln 0\).
As a check, the first terms are \(-2x-2x^2-\frac83x^3-\cdots\), matching the substituted expression. At the included endpoint \(x=-1/2\), the series becomes \(1-\frac12+\frac13-\frac14+\cdots=\ln 2\), as required by \(\ln(1-2(-1/2))=\ln 2\).
Endpoint Mapping and Sign of the Scale
Suppose the original series converges on an interval whose endpoints are \(-R\) and \(R\), with either endpoint possibly included. If \(c>0\), the endpoint \(x=R/c\) maps to \(u=R\), and \(x=-R/c\) maps to \(u=-R\). If \(c<0\), these correspondences reverse: \(x=R/|c|\) maps to \(u=-R\), while \(x=-R/|c|\) maps to \(u=R\).
This reversal is why simply writing “the new interval is \((-R/|c|,R/|c|)\)” is incomplete. That only describes the interior. The exact interval depends on which original endpoints converge, and on which endpoint each new value maps to.
Worked Example: Scaling the Arctangent Series
Worked Example: Find the series and interval for \(\arctan(3x)\)
The arctangent Maclaurin series is
Substitute \(u=3x\):
The convergence condition maps to \(-1\leq 3x\leq 1\). Dividing by the positive number \(3\) gives \(-1/3\leq x\leq 1/3\). Both endpoints are included in the original arctangent series. Explicitly, at \(x=1/3\) the series becomes \(1-\frac13+\frac15-\frac17+\cdots=\arctan(1)=\pi/4\). At \(x=-1/3\) it becomes \(-1+\frac13-\frac15+\frac17-\cdots=\arctan(-1)=-\pi/4\). Each endpoint series converges by the alternating-series test.
As a direct check, substituting \(x=1/3\) into the scaled function gives \(\arctan(3(1/3))=\arctan(1)=\pi/4\); the series gives the same value. At \(x=-1/3\), both give \(-\pi/4\). The radius has changed from \(1\) to \(1/3\), and the endpoint checks confirm that the interval is closed.
Worked Example: Scaling the Arcsine Series
Worked Example: Find the series and interval for \(\arcsin(2x)\)
The arcsine Maclaurin series, including its endpoint convergence and endpoint values, is
Substitute \(u=2x\). The first coefficients are obtained as follows: for \(n=0\), the coefficient is \(1\); for \(n=1\), it is \(2!/(4\cdot 1\cdot 3)=1/6\); for \(n=2\), it is \(4!/(16\cdot 4\cdot 5)=3/40\). Thus
The original interval condition is \(-1\leq 2x\leq 1\), so \(-1/2\leq x\leq 1/2\). Both endpoints are included. At \(x=1/2\), the series is the original arcsine series at \(u=1\), and its sum is \(\arcsin(1)=\pi/2\). At \(x=-1/2\), it is the original series at \(u=-1\), and its sum is \(\arcsin(-1)=-\pi/2\).
For a check, the scaled function at the endpoints gives \(\arcsin(2(1/2))=\pi/2\) and \(\arcsin(2(-1/2))=-\pi/2\), agreeing with the mapped series. The radius is \(1/2\), with both endpoints included.
A Common Mistake: Keeping the Same Endpoint Labels
A frequent error is to scale the radius correctly but then assume the endpoint convergence is unchanged at the same signed endpoint. With a negative scale, the endpoint labels swap. For example, the logarithm series converges at \(u=1\) and diverges at \(u=-1\). Under \(u=-2x\), the endpoint \(x=-1/2\) maps to \(u=1\), while \(x=1/2\) maps to \(u=-1\).
Another mistake is to conclude endpoint convergence from the radius. The root test establishes behavior strictly inside and strictly outside the radius; it makes no decision at equality. Always return to the original series, substitute the mapped endpoint, and use its known endpoint behavior.
Why Scaling Is Useful
Scaling lets a known expansion answer questions about a function whose input changes at a different rate. It also makes interval restrictions visible: if a known series is valid only for inputs in a particular interval, the scaled input must remain in that interval. This is especially useful for constructing expansions for logarithms and inverse trigonometric functions without repeating their original derivations.
Record its coefficients, radius, and the convergence status of both endpoints.
Replace the original variable by \(cx\), retaining the factor \(c^n\) in each term.
Use the condition on \(cx\); if \(c<0\), reverse the inequality directions.
Map each new endpoint to the original endpoint and use the original series' convergence result.
The reliable principle is simple: scaling transforms the input, so the original convergence set transforms with it. The radius changes by the factor \(1/|c|\), while endpoint inclusion follows the exact mapping \(u=cx\).
Check Your Understanding
Use the original series and the scaling rule to answer each question.
- If a power series has radius \(5\), what is the radius after replacing its variable by \(4x\)?
- For a series that converges on \((-1,1]\), what is the corresponding \(x\)-interval after substituting \(u=-3x\)?
- Why can the root test not determine convergence at the new radius endpoints?
- Substitute \(u=2x\) into the arctangent Maclaurin series and state its convergence interval.
- For the logarithm series \(\ln(1+u)\), which new endpoint maps to the convergent endpoint \(u=1\) after \(u=-2x\)?
