By Russell L. Herman
Advent and ReviewWhat Do i have to recognize 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 options of ODEsNumerical ApplicationsLinear SystemsProblemsLinear AlgebraFinite Dimensional Vector SpacesLinear TransformationsEigenvalue ProblemsMatrix formula of Planar SystemsApplicationsAppendix: Diagonali. Read 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 options 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
This example was fairly simple using the transformation sec θ = cosh u. Another common integral that arises often is integrations of sec3 θ. In a typical calculus class, this integral is evaluated using integration by parts. ) In the next example, we will show how hyperbolic function substitution is simpler. 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.
1 . 1−x But, 1−1 x = (1 − x )−1 . This is a binomial to a power, but the power is not an integer. 109). This example suggests that our sum may no longer be finite. 110) and see if the resulting series makes sense. However, we quickly run into problems with the coefficients in the series. Consider the coefficient for r = 1 in an expansion of (1 + x )−1 . This is given by (−1)! (−1)! −1 = = . 1! 1! 1 But what is (−1)!? By definition, it is (−1)! = (−1)(−2)(−3) · · · . Andreas Freiherr von Ettingshausen (1796–1878) was a German mathematician and physicist who in 1826 intron duced the notation .
1 1 This expansion is easily done by noting that x+ 2 = ( x −1)+3 . Factoring out a 3, we can rewrite this expression as a sum of a geometric series. Namely, we use the expansion for g(z) = = 1 1+z 1 − z + z2 − z3 + . . 100) 31 32 mathematical methods for physicists and then we rewrite f ( x ) as f (x) = = = = 1 x+2 1 ( x − 1) + 3 1 1 3[1 + 3 ( x − 1)] 1 1 . 1 3 1 + 3 ( x − 1) Note that f ( x ) = 13 g( 13 ( x − 1)) for g(z) = f (x) = 1 1 1 − ( x − 1) + 3 3 1 1+ z . 101) So, the expansion becomes 2 − 1 ( x − 1) 3 3 +...