By Russell L. Herman

ISBN-10: 1466584688

ISBN-13: 9781466584686

Creation and ReviewWhat Do i must recognize From Calculus?What i would like From My Intro Physics Class?Technology and TablesAppendix: Dimensional AnalysisProblemsFree Fall and Harmonic OscillatorsFree FallFirst Order Differential EquationsThe easy Harmonic OscillatorSecond Order Linear Differential EquationsLRC CircuitsDamped OscillationsForced SystemsCauchy-Euler EquationsNumerical options of ODEsNumericalRead more...

summary: creation and ReviewWhat Do i must understand From Calculus?What i want From My Intro Physics Class?Technology and TablesAppendix: Dimensional AnalysisProblemsFree Fall and Harmonic OscillatorsFree FallFirst Order Differential EquationsThe basic Harmonic OscillatorSecond Order Linear Differential EquationsLRC CircuitsDamped OscillationsForced SystemsCauchy-Euler EquationsNumerical ideas of ODEsNumerical ApplicationsLinear SystemsProblemsLinear AlgebraFinite Dimensional Vector SpacesLinear TransformationsEigenvalue ProblemsMatrix formula of Planar SystemsApplicationsAppendix: Diagonali

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12 4 So, how does one show that these answers are the same? √ √ Let’s focus on the factor 2 + 3. We seek to write this in the form c + d 3. Equating the two expressions and squaring, we have 2+ √ √ = ( c + d 3)2 3 √ = c2 + 3d2 + 2cd 3. 27) √ In order to solve for c and d, it would seem natural to equate the coefficients of 3 and the remaining terms. 28) 2cd = 1. 29) Solving the second equation for d = 1/2c, and substituting the result into the first equation, we find 4c4 − 8c2 + 3 = 0. This fourth-order equation has four solutions: √ √ 2 6 c=± ,± 2 2 and √ √ 2 6 b=± ,± .

1 3 = 5 7 3, 3 . The resulting series is 1 ∑∞ n =1 n . = This is the harmonic series, and thus it does not converge. 5 Inserting x = 3 , we get the alternating harmonic series. This series does converge. So, we have convergence on [ 53 , 73 ). However, it is only conditionally convergent at the left endpoint, x = 53 . 36. Find an expansion of f ( x ) = x+ 2 about x = 1. Instead of explicitly computing the Taylor series expansion for this function, we can make use of an already known function. We first write f ( x ) as a function of x − 1, as we are expanding about x = 1; that is, we are seeking a series whose terms are powers of x − 1.

We apply integration by parts by letting U = x and dV = sin x dx. This gives dU = dx and V = − cos x. Therefore, we have π 0 = − x cos x x sin x dx = π + sin x π 0 π + π 0 cos x dx 0 = π. 67) The final result is π 0 x2 cos x dx = −2π. There are other ways to compute integrals of this type. First of all, there is the Tabular Method to perform integration by parts. A second method is to use differentiation of parameters under the integral. We will demonstrate this using examples. 10. Compute the integral 0 x2 cos x dx using the Tabular Method.

### A Course in Mathematical Methods for Physicists by Russell L. Herman

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