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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 11.3.25b

Taylor series and interval of convergence


b. Write the power series using summation notation.


f(x) = ln (x − 2), a = 3

Guida verificata passo dopo passo
1
Identify the function and the center of the Taylor series expansion. Here, the function is \(f(x) = \ln(x - 2)\) and the center is \(a = 3\).
Rewrite the function in terms of \((x - a)\) to express it as a power series centered at \(x = 3\). Set \(u = x - 3\), so that \(x - 2 = (x - 3) + 1 = u + 1\).
Recall the Taylor series expansion for \(\ln(1 + u)\) around \(u = 0\), which is \(\ln(1 + u) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{u^n}{n}\) for \(|u| < 1\).
Substitute back \(u = x - 3\) into the series to write \(f(x)\) as a power series centered at \(x = 3\): \(f(x) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(x - 3)^n}{n}\).
Express the final answer in summation notation: \(f(x) = \sum_{n=1}^{\infty} (-1)^{n+1} \frac{(x - 3)^n}{n}\), which represents the power series expansion of \(\ln(x - 2)\) about \(x = 3\).

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