Taylor series and interval of convergence

b. Write the power series using summation notation.

f(x) = (1 + x²)⁻¹, a = 0

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Recognize that the function given is \( f(x) = (1 + x^2)^{-1} \), which resembles the form of a geometric series \( \frac{1}{1 - r} = \sum_{n=0}^{\infty} r^n \) when \( |r| < 1 \).

Rewrite the function to match the geometric series form by identifying \( r = -x^2 \), so \( f(x) = \frac{1}{1 - (-x^2)} \).

Express the power series as a summation using the geometric series formula: \[ f(x) = \sum_{n=0}^{\infty} (-x^2)^n \].

Simplify the term inside the summation to get \( (-1)^n x^{2n} \), so the power series becomes \[ f(x) = \sum_{n=0}^{\infty} (-1)^n x^{2n} \].

Note that the interval of convergence is determined by \( |r| < 1 \), which means \( |-x^2| = |x|^2 < 1 \), so the interval of convergence is \( |x| < 1 \).

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주요 개념

질문에 올바르게 답하기 위해 반드시 이해해야 하는 핵심 개념들은 다음과 같습니다.

Taylor Series Expansion

A Taylor series represents a function as an infinite sum of terms calculated from the derivatives of the function at a single point. For a function f(x) centered at a = 0, it is expressed as f(x) = Σ (f⁽ⁿ⁾(0)/n!) xⁿ, where f⁽ⁿ⁾(0) is the nth derivative evaluated at 0. This allows approximation of functions using polynomials.

Power Series and Summation Notation

A power series is an infinite series of the form Σ cₙ (x - a)ⁿ, where cₙ are coefficients and a is the center. Summation notation compactly expresses this infinite sum, making it easier to manipulate and analyze. Writing a function as a power series involves finding the coefficients cₙ that match the function's behavior.

Interval of Convergence

The interval of convergence is the set of x-values for which a power series converges to the function. It is determined by testing the radius of convergence, often using the ratio or root test. Understanding this interval is crucial to knowing where the power series accurately represents the function.

Taylor series and interval of convergenceb. Write the power series using summation notation.f(x) = (1 + x²)⁻¹, a = 0