Download PDF by Russell L. Herman: A Course in Mathematical Methods for Physicists

By Russell L. Herman

ISBN-10: 1466584688

ISBN-13: 9781466584686

Advent and ReviewWhat Do i must comprehend From Calculus?What i want From My Intro Physics Class?Technology and TablesAppendix: Dimensional AnalysisProblemsFree Fall and Harmonic OscillatorsFree FallFirst Order Differential EquationsThe uncomplicated Harmonic OscillatorSecond Order Linear Differential EquationsLRC CircuitsDamped OscillationsForced SystemsCauchy-Euler EquationsNumerical recommendations of ODEsNumerical Read more...

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

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Additional resources for A Course in Mathematical Methods for Physicists

Sample text

U When you become proficient at integration, you can bypass some of these steps. In the next example we try to demonstrate the thought process involved in using substitution without explicitly using the substitution variable. 6. Evaluate dx As with the previous example, one sees that the derivative of 9 + 4x2 is proportional to x, which is in the numerator of the integrand. Thus, a substitution would give an integrand of the form u−1/2 . So, we expect the answer to be proportional to √ √ u = 9 + 4x2 .

23. Evaluate sec3 θ dθ using hyperbolic function substitution. du First, we consider the transformation sec θ = cosh u with dθ = cosh u . Then, sec3 θ dθ = du . 7, leading to sec3 θ dθ = 2 tan−1 eu + C. introduction and review 23 While correct, this is not the form usually encountered. Instead, we make the slightly different transformation tan θ = sinh u. Since sec2 θ = 1 + tan2 θ, we find sec θ = cosh u. As before, we find dθ = du . cosh u Using this transformation and several identities, the integral becomes sec3 θ dθ cosh2 u du = = = = = = 1 (1 + cosh 2u) du 2 1 1 u + sinh 2u 2 2 1 (u + sinh u cosh u) 2 1 cosh−1 (sec θ ) + tan θ sec θ 2 1 (sec θ tan θ + ln(sec θ + tan θ )) .

I θ − + −... 2! 4! 3! 5! We recognize the expansions in the parentheses as those for the cosine and sine functions. Thus, we end with Euler’s Formula. We further derive relations from this result, that will be important for our next studies. From Euler’s Formula we have that for integer n, einθ = cos(nθ ) + i sin(nθ ). introduction and review We also have einθ = eiθ n 33 = (cos θ + i sin θ )n . Equating these two expressions, we are led to de Moivre’s Formula, named after Abraham de Moivre (1667–1754), (cos θ + i sin θ )n = cos(nθ ) + i sin(nθ ).

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


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