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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 11.1.65c

Explain why or why not Determine whether the following statements are true and give an explanation or counterexample.


c. Only even powers of x appear in the nth−order Taylor polynomial for f(x)=√(1+x²) centered at 0.

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Recall that the nth-order Taylor polynomial of a function \(f(x)\) centered at 0 (Maclaurin polynomial) is given by: \[T_n(x) = \sum_{k=0}^n \frac{f^{(k)}(0)}{k!} x^k,\] where \(f^{(k)}(0)\) is the \(k\)th derivative of \(f\) evaluated at 0.
Consider the function \(f(x) = \sqrt{1 + x^2}\). Notice that \(f(x)\) is an even function because \(f(-x) = \sqrt{1 + (-x)^2} = \sqrt{1 + x^2} = f(x)\).
Since \(f(x)\) is even, its Taylor series expansion around 0 will contain only even powers of \(x\). This is a general property: the Taylor series of an even function centered at 0 contains only even powers, and the Taylor series of an odd function contains only odd powers.
To confirm this, you can compute the first few derivatives of \(f(x)\) at 0 and observe that all derivatives of odd order vanish at 0, i.e., \(f^{(1)}(0) = 0\), \(f^{(3)}(0) = 0\), etc., which means the coefficients of odd powers are zero.
Therefore, the nth-order Taylor polynomial for \(f(x) = \sqrt{1 + x^2}\) centered at 0 contains only even powers of \(x\), making the statement true.

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