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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
Hass15th EditionThomas' CalculusISBN: 9780137616077Non è quello che usi tu?Cambia libro di testo
Capitolo 7, Problema 7.8.13

13. When is a polynomial f(x) of at most the order of a polynomial g(x) as x→∞? Give reasons for your answer.

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Understand the concept of the order of a polynomial as \( x \to \infty \). The order is determined by the highest power of \( x \) in the polynomial, which dominates the behavior of the polynomial for very large values of \( x \).
Let \( f(x) \) and \( g(x) \) be polynomials with degrees \( n \) and \( m \) respectively, where \( n \) and \( m \) are the highest powers of \( x \) in \( f(x) \) and \( g(x) \).
To say that \( f(x) \) is of at most the order of \( g(x) \) as \( x \to \infty \) means that the growth rate of \( f(x) \) does not exceed that of \( g(x) \). This happens if and only if \( n \leq m \).
This is because for large \( x \), the term with the highest power dominates, so if \( n > m \), then \( f(x) \) grows faster than \( g(x) \), and if \( n \leq m \), then \( f(x) \) grows at most as fast as \( g(x) \).
Therefore, the reason is that the degree of \( f(x) \) must be less than or equal to the degree of \( g(x) \) for \( f(x) \) to be of at most the order of \( g(x) \) as \( x \to \infty \).

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