Build a Power Series From a Maclaurin Expansion
A worked explanation of build a power series from a maclaurin expansion, including the key limitation and a checkable example.
Build a Power Series From a Maclaurin Expansion
You know the Maclaurin series for 1/(1−x) is Σ xⁿ for |x|<1. That single known expansion is the starting tool for building a power series from a maclaurin expansion into forms that look nothing like the original. Shifting, scaling, differentiation, and integration each let you turn that base into a new representation without re-deriving coefficients from scratch. The operations are valid term-by-term inside the same radius of convergence, but endpoints require a separate test.
Four Operations That Transform a Known Series
Each operation takes the original Maclaurin series and produces a different function's power series. The radius of convergence stays the same, but endpoint convergence can change. You must re-test endpoints after applying any operation.
Shifting
Replace x in the original series with (x−c). For example, the Maclaurin series for eˣ is Σ xⁿ/n!. Shift it to get eˣ⁻² = Σ (x−2)ⁿ/n!. The new series is centered at 2, not 0, and converges for |x−2| < ∞.
Scaling
Replace x with kx. From sin x = Σ (−1)ⁿ x²ⁿ⁺¹/(2n+1)!, scaling gives sin(3x) = Σ (−1)ⁿ (3x)²ⁿ⁺¹/(2n+1)!. The radius shrinks by factor |k|: the original radius ∞ becomes ∞ still, but a finite radius R becomes R/|k|.
Differentiation
Differentiate the series term-by-term inside its interval. From 1/(1−x) = Σ xⁿ, differentiate to get 1/(1−x)² = Σ n xⁿ⁻¹ for |x|<1. The radius stays 1. The derivative series may converge at an endpoint where the original did not, or vice versa.
Integration
Integrate term-by-term inside the interval. Integrate 1/(1−x) to get −ln(1−x) = Σ xⁿ⁺¹/(n+1) for |x|<1. At x=0.5, the partial sum of the first three terms is 0.5 + 0.125 + 0.0417 = 0.6667, while −ln(0.5) = 0.6931, an error of about 0.0264.
A Worked Example
Start with the known Maclaurin series for 1/(1−x) = Σ xⁿ for |x|<1. To build a representation for ln(1+x), integrate the series for 1/(1+x) = Σ (−1)ⁿ xⁿ for |x|<1. Integration gives ln(1+x) = Σ (−1)ⁿ xⁿ⁺¹/(n+1) = x − x²/2 + x³/3 − x⁴/4 + ... for |x|<1. At x=0.5, the sum of the first four terms is 0.5 − 0.125 + 0.0417 − 0.0156 = 0.4011, while ln(1.5) = 0.4055, an error of 0.0044. The radius is still 1, but at x=1 the alternating harmonic series converges conditionally, whereas the original series for 1/(1+x) converges absolutely at x=1.
Calculation Steps
To verify the calculation, use the Maclaurin polynomial calculator to evaluate the partial sum at x=0.5. The series for −ln(1−x) = Σ xⁿ⁺¹/(n+1) gives S₃ = 0.5 + 0.125 + 0.0417 = 0.6667. The exact value of −ln(0.5) is 0.6931, so the remainder is 0.0264. The Lagrange error bound using the fourth derivative of −ln(1−x) at x=0.5 gives a maximum error of 0.0313, which bounds the actual error. The series converges at x=0.5 because the radius is 1.
Why the Method Works
The question 'How can shifting, scaling, differentiation, or integration turn a known Maclaurin series into a new representation?' is answered by the operations above. Each operation preserves the radius of convergence but may change endpoint behavior. The key is that term-by-term operations are valid inside the interval without re-deriving coefficients. The failure case occurs when you apply the operation outside the radius: the series diverges and the partial sum is meaningless.
| Operation | Resulting Series | Radius of Convergence | Endpoint Change |
|---|---|---|---|
| Shifting x→(x−c) | Σ (x−c)ⁿ | 1 (if original radius 1) | Center shifts; test new endpoints separately |
| Scaling x→kx | Σ kⁿ xⁿ | 1/|k| | Radius shrinks by factor |k| |
| Differentiation | Σ n xⁿ⁻¹ | 1 | May converge at endpoint where original diverged |
| Integration | Σ xⁿ⁺¹/(n+1) | 1 | May converge at endpoint where original diverged |
What Most Often Goes Wrong
The single thing that most often goes wrong is forgetting to test endpoints after applying an operation. A student shifts, scales, differentiates, or integrates a known series, assumes the radius stays the same, and then evaluates at an endpoint where the new series converges but the original did not (or vice versa). The series is not wrong, but the assumption that the interval is unchanged is. Always test x = a ± R separately after any transformation.
Common Questions
What is the first step to shift a Maclaurin series?
Replace x in the original series with (x−c), where c is the new center. For example, start with eˣ = Σ xⁿ/n! and shift to eˣ⁻² = Σ (x−2)ⁿ/n!.
Does scaling change the radius of convergence?
Yes. If the original radius is R, scaling by factor k gives a new radius R/|k|. For example, sin(3x) has radius ∞ from sin x, but a finite R would shrink.
Can I differentiate a power series at an endpoint?
Only if the differentiated series converges at that endpoint. The radius stays the same, but endpoint convergence must be re-tested separately.
What is the error when integrating a series to approximate ln(1.5)?
For ln(1+x) = x − x²/2 + x³/3 − x⁴/4 + ... at x=0.5, the first four terms give 0.4011, actual ln(1.5)=0.4055, error about 0.0044.
Why must I test endpoints after integration?
Integration may produce convergence at an endpoint where the original diverged. For example, integrating 1/(1+x) gives ln(1+x), which converges at x=1, but the original diverges there.
What is the most common mistake with term-by-term operations?
Applying the operation outside the radius of convergence. The series diverges there, and the partial sum is not a valid approximation of the function.
How do I check if a shifted series converges at x=0?
Substitute x=0 into the shifted series. It becomes Σ c_n (−c)ⁿ, which is a constant series. Test its convergence using the ratio test or root test.