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Ch. 7 - Transcendental Functions
Hass - Thomas' Calculus 15th Edition
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7장, 문제 7.AAE.17

17. Even-odd decompositions
b. If f(x) = f_E(x) + f_O(x) is the sum of an even function f_E(x) and an odd function f_O(x), then show that
f_E(x) = (f(x)+f(-x))/2 and f_O(x) = (f(x)-f(-x))/2

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Recall the definitions of even and odd functions: an even function satisfies \(f_E(x) = f_E(-x)\), and an odd function satisfies \(f_O(x) = -f_O(-x)\).
Given that \(f(x) = f_E(x) + f_O(x)\), write the expression for \(f(-x)\) by substituting \(-x\) into the function: \(f(-x) = f_E(-x) + f_O(-x)\).
Use the properties of even and odd functions to rewrite \(f(-x)\) as \(f(-x) = f_E(x) - f_O(x)\), since \(f_E(-x) = f_E(x)\) and \(f_O(-x) = -f_O(x)\).
Add the two equations \(f(x) = f_E(x) + f_O(x)\) and \(f(-x) = f_E(x) - f_O(x)\) to isolate \(f_E(x)\): \(f(x) + f(-x) = 2 f_E(x)\).
Similarly, subtract \(f(-x)\) from \(f(x)\) to isolate \(f_O(x)\): \(f(x) - f(-x) = 2 f_O(x)\). Then solve for \(f_E(x)\) and \(f_O(x)\) to get \(f_E(x) = \frac{f(x) + f(-x)}{2}\) and \(f_O(x) = \frac{f(x) - f(-x)}{2}\).

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주요 개념

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Even and Odd Functions

An even function satisfies f(x) = f(-x) for all x, meaning its graph is symmetric about the y-axis. An odd function satisfies f(-x) = -f(x), showing symmetry about the origin. Understanding these definitions is essential to decompose any function into even and odd parts.
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Function Decomposition

Any function f(x) can be expressed as the sum of an even function f_E(x) and an odd function f_O(x). This decomposition helps analyze the function's symmetry properties and simplifies integration and other operations by separating symmetric components.
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Algebraic Manipulation for Decomposition

To find f_E(x) and f_O(x), use the definitions of even and odd functions to form equations involving f(x) and f(-x). Adding and subtracting these equations isolates the even and odd parts, leading to the formulas f_E(x) = (f(x)+f(-x))/2 and f_O(x) = (f(x)-f(-x))/2.
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