Power Series and Taylor Expansions Free Online FlashCards

What is a power series centered at a?

A power series centered at a is an infinite series of the form Σ from n=0 to ∞ cₙ(x−a)ⁿ, where the cₙ are constants.

What makes a power series Maclaurin?

A Maclaurin power series is a power series centered at 0: Σ from n=0 to ∞ cₙxⁿ.

State the geometric-series identity and its convergence condition.

The geometric series is Σ from n=0 to ∞ rⁿ = 1/(1−r), and it converges when |r|<1.

How do radius and interval of convergence differ?

The radius of convergence R describes the distance from the center where convergence holds inside |x−a|<R. The interval of convergence is the complete set of convergent x-values, including any endpoints that pass testing.

How should power-series endpoints be handled?

Test x=a−R and x=a+R separately. The ratio test usually determines only the open interval and is inconclusive at the endpoints.

Find the convergence interval for Σ (x−2)ⁿ/[(n+1)3ⁿ].

For Σ (x−2)ⁿ/[(n+1)3ⁿ], the radius is R=3 and the interval of convergence is [−1,5). The left endpoint converges by the Alternating Series Test; the right diverges as a harmonic series.

What happens to convergence under term-by-term calculus?

Inside the original open interval of convergence, a power series can be differentiated or integrated term by term. Both resulting series retain the same radius, though endpoint behavior may change.

What is the Taylor-series formula centered at a?

The Taylor series of f centered at a is Σ from n=0 to ∞ [f⁽ⁿ⁾(a)/n!](x−a)ⁿ.

Do derivatives of all orders guarantee a Taylor representation?

No. Having derivatives of every order does not guarantee that f equals its Taylor series; the Taylor remainder must approach zero on the interval considered.

Give the Maclaurin series and domain for eˣ.

eˣ = Σ from n=0 to ∞ xⁿ/n!, and this Maclaurin series converges for every real x.

How can 1/(2+x) be represented as a power series?

1/(2+x) = Σ from n=0 to ∞ (−1)ⁿxⁿ/2ⁿ⁺¹ for |x|<2. Rewrite it as (1/2)·1/(1+x/2) and use the geometric series.

State the Lagrange error bound for a Taylor polynomial.

If |f⁽ⁿ⁺¹⁾(t)|≤M between a and x, then the Lagrange bound is |Rₙ(x)|≤M|x−a|ⁿ⁺¹/(n+1)!.