Download Mathematics Formulary by J. Wevers PDF
By J. Wevers
Read Online or Download Mathematics Formulary PDF
Best industrial & technical books
Multivariate Datenanalyse GERMAN
In vielen Fachgebieten, wie z. B. der Lebensmittelchemie, der pharmazeutischen oder biotechnologischen Industrie fallen immer mehr Daten an, die ausgewertet werden m? ssen. Klassische Verfahren gelangen hierbei schnell an ihre Grenzen. Die multivariate Datenanalyse besch? ftigt sich mit Verfahren, mit denen guy aus einer F?
Content material: Enzymes for fuels and chemical feedstocks / ok. Grohmann and Michael E. Himmel -- Enzymes in pulp and paper processing / L. Viikari, A. Kantelinen, M. Rättö, and J. Sundquist -- Enzymes for anaerobic municipal strong waste disposal / Christopher J. Rivard, William S. Adney, and Michael E. Himmel -- Thermostable saccharidases : new resources, makes use of, and biodesigns / J.
- Methods and Applications of Cycloaddition Reactions in Organic Syntheses
- Value creation in the pharmaceutical industry : the critical path to innovation
- Block Copolymers: Synthetic Strategies, Physical Properties, and Applications
- Poor Man's James Bond
Extra info for Mathematics Formulary
Sample text
A harmonic function with a normal derivative of 0 on the boundary of an area is constant within that area. The Dirichlet problem is: ∇2 u(x ) = −f (x ) , x ∈ R , u(x ) = g(x ) for all x ∈ S. It has a unique solution. The Neumann problem is: ∇2 u(x ) = −f (x ) , x ∈ R , ∂u(x ) = h(x ) for all x ∈ S. ∂n 32 Mathematics Formulary by ir. A. Wevers The solution is unique except for a constant. The solution exists if: f (x )d3 V = − R h(x )d2 A S A fundamental solution of the Laplace equation satisfies: ∇2 u(x ) = −δ(x ) This has in 2 dimensions in polar coordinates the following solution: u(r) = ln(r) 2π This has in 3 dimensions in spherical coordinates the following solution: u(r) = 1 4πr The equation ∇2 v = −δ(x − ξ ) has the solution v(x ) = 1 4π|x − ξ | After substituting this in Green’s 2nd theorem and applying the sieve property of the δ function one can derive Green’s 3rd theorem: u(ξ ) = − ∇2 u 3 1 d V + r 4π 1 4π R 1 ∂u ∂ −u r ∂n ∂n 1 r d2 A S The Green function G(x, ξ ) is defined by: ∇2 G = −δ(x − ξ ), and on boundary S holds G(x, ξ ) = 0.
2. Introduce: x˙ = u and y˙ = v, this leads to x ¨ = u˙ and y¨ = v. ˙ This transforms a n-dimensional set of second order equations into a 2n-dimensional set of first order equations. 44 Mathematics Formulary by ir. A. 1 Quadratic forms in IR2 The general equation of a quadratic form is: xT Ax + 2xT P + S = 0. Here, A is a symmetric matrix. , λn ) holds: uT Λu + 2uT P + S = 0, so all cross terms are 0. u = (u, v, w) should be chosen so that det(S) = +1, to maintain the same orientation as the system (x, y, z).
We define the notations Cρ+ = {z||z| = ρ, (z) ≥ 0} and Cρ− = {z||z| = ρ, (z) ≤ 0} and M + (ρ, f ) = max+ |f (z)|, M − (ρ, f ) = max− |f (z)|. We assume that z∈Cρ z∈Cρ f (z) is analytical for (z) > 0 with a possible exception of a finite number of singular points which do not lie on the real axis, lim ρM + (ρ, f ) = 0 and that the integral exists, than ρ→∞ ∞ f (x)dx = 2πi Resf (z) in (z) > 0 −∞ Replace M + by M − in the conditions above and it follows that: ∞ f (x)dx = −2πi Resf (z) in (z) < 0 −∞ Jordan’s lemma: let f be continuous for |z| ≥ R, (z) ≥ 0 and lim M + (ρ, f ) = 0.