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

By Russell L. Herman

ISBN-10: 1466584688

ISBN-13: 9781466584686

Creation and ReviewWhat Do i must be aware of 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 suggestions of ODEsNumerical Read more...

summary: creation and ReviewWhat Do i have to be aware of From Calculus?What i would like 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 recommendations of ODEsNumerical ApplicationsLinear SystemsProblemsLinear AlgebraFinite Dimensional Vector SpacesLinear TransformationsEigenvalue ProblemsMatrix formula of Planar SystemsApplicationsAppendix: Diagonali

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56) and having to figure out what was differentiated in order to get the given function. As you may recall from the Fundamental Theorem of Calculus, the integral is the inverse operation to differentiation: df dx = f ( x ) + C. 57) It is not always easy to evaluate a given integral. In fact, some integrals are not even doable! However, you learned in calculus that there are some methods that could yield an answer. com, or a fancy calculator, you should know a few basic integrals and know how to use tables for some of the more complicated ones.

For our example, du = 2x dx. Looking at the integral, part of the integrand can be written as x dx = 12 u du. Then, the integral becomes √ x x2 +1 dx = 1 2 du √ . u The substitution has converted our integral into an integral over u. Also, this integral is doable! It is one of the integrals we should know. Namely, we can write it as 1 du 1 √ = u−1/2 du. 2 2 u This is now easily finished after integrating and using the substitution variable to give x 1 u1/2 √ dx = + C = x2 + 1 + C. 2 12 x2 + 1 Note that we have added the required integration constant, and that the derivative of the result easily gives the original integrand (after employing the Chain Rule).

However, if we used the partial fraction expansion 1 1 1 = − , n ( n + 1) n n+1 then we find that the kth partial sum can be written as k sk = 1 n ( n + 1) n =1 = ∑ ∑ k n =1 = 1 1 − n n+1 1 1 − 1 2 + 1 1 − 2 3 +···+ 1 1 − k k+1 . 93) We see that there are many cancellations of neighboring terms, leading to the series collapsing (like a retractable telescope) to something manageable: sk = 1 − Taking the limit as k → ∞, we find ∑∞ n =1 1 . k+1 1 n ( n +1) = 1. A rearrangement of terms in an infinite series is allowed when the series is absolutely convergent.

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


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