The geometric series 1/(1−x) = Σxⁿ converges for |x| < 1. What is the interval of convergence for the series of 1/(1−3x) obtained by substituting 3x for x?
A|x| < 1, because substitution does not change the radius of convergence
B|x| < 1/3, because the condition |3x| < 1 becomes |x| < 1/3
C|x| < 3, because the substitution stretches the interval by a factor of 3
DAll real x, because rational functions are defined everywhere
The convergence condition for the geometric series is |substituted variable| < 1. After substituting 3x, the condition becomes |3x| < 1, which simplifies to |x| < 1/3. Option A is the classic error: it copies the original radius without rederiving it after substitution. The radius always scales inversely with any constant multiplier on x — multiplying x by 3 shrinks the interval of convergence by a factor of 3.
Question 2 Multiple Choice
Why does the Maclaurin series for sin(x) contain only odd-power terms (x, x³, x⁵, ...) with no even-power terms?
ABy coincidence in the pattern of derivatives at x = 0
BBecause sin(x) is an odd function, and odd functions have only odd-power Maclaurin series
CBecause even-power terms would cause the series to diverge
DBecause the denominators n! grow too fast for even powers to contribute
An odd function satisfies f(−x) = −f(x). For a Maclaurin series f(x) = Σaₙxⁿ, this forces aₙ = 0 for all even n — otherwise the even terms would not cancel under x → −x. Since sin(−x) = −sin(x), all even coefficients must vanish. The even-power terms are absent by structural necessity, not coincidence. Similarly, cos(x) is even (cos(−x) = cos(x)), so its series has only even powers. This structural insight lets you reconstruct the form of these series quickly even if you forget the exact coefficients.
Question 3 True / False
Substituting x² into the series for e^x gives a new series valid mainly for |x| < 1, since e^x itself converges mainly on a bounded interval.
TTrue
FFalse
Answer: False
The series for e^x converges for all real x (its radius of convergence is infinite). Substituting x² gives e^(x²) = Σ(x²)ⁿ/n! = Σx^(2n)/n!, which also converges for all real x, since x² is finite for any finite x. The misconception arises from confusing e^x with the geometric series 1/(1−x), which does have a finite radius. After substitution into e^x, the only question is whether x² is finite — which it always is.
Question 4 True / False
Differentiating the geometric series 1/(1−x) = Σxⁿ term by term yields the series for 1/(1−x)².
TTrue
FFalse
Answer: True
Differentiating both sides: the left side gives 1/(1−x)². The right side differentiates term by term to Σnx^(n−1). This is valid within the radius of convergence |x| < 1, where power series can be differentiated term by term. This technique — differentiating a known series to obtain a new one — is one of the three core manipulation strategies (with substitution and integration) for building new series without starting from scratch.
Question 5 Short Answer
Why must you rederive the interval of convergence after substituting into a known Taylor series? Give a specific example.
Think about your answer, then reveal below.
Model answer: The convergence condition depends on the magnitude of the variable in the series. After substitution, the variable changes, so the condition must be re-expressed in terms of the new variable. Example: 1/(1−x) = Σxⁿ converges for |x| < 1. Substituting x² gives 1/(1−x²) = Σx^(2n), convergent when |x²| < 1, i.e., |x| < 1 — same interval here. But substituting 2x gives 1/(1−2x) = Σ(2x)ⁿ, convergent when |2x| < 1, i.e., |x| < 1/2. The substitution changed the interval.
This is a direct application of the chain: 'convergence condition for the original series' → 'apply the substitution' → 'solve for x.' Skipping this step is the most common source of errors when deriving new series. The interval can expand (if you substitute something smaller than x) or contract (if you substitute something larger), so the original radius of convergence is never automatically inherited.