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Ch. 11 - Power Series
Briggs - Calculus: Early Transcendentals 3rd Edition
Briggs3rd EditionCalculus: Early TranscendentalsISBN: 9780136847243Non è quello che usi tu?Cambia libro di testo
Capitolo 11, Problema 11.2.54

Differentiating and integrating power series Find the power series representation for g centered at 0 by differentiating or integrating the power series for f (perhaps more than once). Give the interval of convergence for the resulting series.


g(x) = x/(1 + x²)² using f(x) = 1/(1 + x²)

Guida verificata passo dopo passo
1
Start with the given function \( f(x) = \frac{1}{1 + x^2} \). Recognize that this is a geometric series of the form \( \frac{1}{1 - (-x^2)} \), which can be expanded as a power series centered at 0: \( f(x) = \sum_{n=0}^\infty (-1)^n x^{2n} \).
To find \( g(x) = \frac{x}{(1 + x^2)^2} \), notice that it can be expressed in terms of the derivative of \( f(x) \). Differentiate \( f(x) \) with respect to \( x \) to relate it to \( g(x) \). Use the chain rule: \( f'(x) = \frac{d}{dx} \left( \frac{1}{1 + x^2} \right) = -\frac{2x}{(1 + x^2)^2} \).
Rearrange the derivative expression to isolate \( g(x) \): \( g(x) = \frac{x}{(1 + x^2)^2} = -\frac{1}{2} f'(x) \). This means \( g(x) \) can be represented as \( -\frac{1}{2} \) times the derivative of the power series for \( f(x) \).
Differentiate the power series term-by-term: \( f(x) = \sum_{n=0}^\infty (-1)^n x^{2n} \) implies \( f'(x) = \sum_{n=1}^\infty (-1)^n 2n x^{2n-1} \). Then multiply by \( -\frac{1}{2} \) to get the power series for \( g(x) \): \( g(x) = -\frac{1}{2} f'(x) = \sum_{n=1}^\infty (-1)^{n+1} n x^{2n-1} \).
Determine the interval of convergence. Since the original series for \( f(x) \) converges for \( |x| < 1 \), and differentiation does not change the radius of convergence, the power series for \( g(x) \) also converges for \( |x| < 1 \).

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