Physics 1 Problem Solutions 17

Physics 1 Problem Solutions 17 - Chapt. 1 Quantities,...

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Unformatted text preview: Chapt. 1 Quantities, Units, Conversions, Measurement, Significant Figures, and Uncertainties 13 sufficiently small (x − x0 ). The smaller (x − x0 ) is the less important the higher order terms. In physics (and other fields no doubt), one is often interested in the behavior near some important point x0 , and so truncates the Taylor’s series to find an simple approximate expression for the neighborhood of x0 . For example f (x) = f (x0 ) , f (x0 ) + (x − x0 )f ′ (x0 ) , 0th order or constant approximation; 1st order or linear approximation; 2 f (x0 ) + (x − x0 )f ′ (x0 ) + (x − x0 ) f ′′ (x0 ) , 2 2nd order or quadratic approximation. The 0th order expansion, approximates the function as a constant which is the function value at x0 ; the 1st order expansion approximates the function by a line that is tangent to the function at x0 ; the 2nd order expansion approximates the function by a quadratic that is tangent to the function x0 . a) Draw a general function on the Cartesian plane and at a general point x0 schematically show how the 0th, 1st, and 2nd order approximations to the function behave. b) Taylor expand f (x) = 1 1+x to 2nd order about x0 = 0. (Here x − x0 is just x, of course.) c) Taylor expand √ 1+x f (x) = √ 1 1+x f (x) = to 2nd order about x = 0. d) Taylor expand to 2nd order about x = 0. e) Taylor expand f (x) = sin(x) to 3rd order about x = 0. Note that x must be in radians for the Taylor’s expansion to work for the trigonometric functions. This is because the derivatives of these functions are derived using radians. f) Taylor expand f (x) = cos(x) to 2nd order about x = 0. ...
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