By Edwin Zondervan

"This booklet emphasizes the deriviation and use of a number of numerical equipment for fixing chemical engineering difficulties. The algorithms are used to resolve linear equations, nonlinear equations, usual differential equations and partial differential equations. additionally it is chapters on linear- and nonlinear regression and ond optimizaiton. MATLAB is followed because the programming setting during the book.�Read more...

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**Extra info for A numerical primer for the chemical engineer**

**Sample text**

The general recipe for LU factorization is as follows: 1. Write down a permutation matrix. 2. Write down the matrix to decompose. 32 A Numerical Primer for the Chemical Engineer 3. Promote the largest value in the column diagonal. 4. Eliminate all elements below the diagonal. 5. Move on to the next column and move the largest elements to the diagonal. 6. Eliminate the elements below the diagonal. 7. Repeat steps 5 and 6. 8. Write down L, U , and P . Let’s do an example: 1. Write down a permutation matrix (initially the identity matrix: 1 0 0 P = 0 1 0 .

Subsequently, solve the system with A−1 b. b Given is the system Ax = b, with 1 1 A= 2 1 1 0 0 1 . 1 Prove that A is singular. Find a b for which this system does not have a solution, and find a b for which b has an infinite number of solutions. 24 A Numerical Primer for the Chemical Engineer Exercise 2 Calculate the eigenvalues of 1 A= 0 0 the following matrices: 1 0 1 2 3 −1 0 ; B = 2 3 1 . 0 2 3 2 1 Exercise 3 Given this matrix: −2 2 −1 3 −1 , A= 7 −4 −4 −2 prove that the characteristic equation is given by: −λ3 − λ2 + 30λ + 72 = 0.

By Gaussian elimination and U , so that A11 A12 A21 A22 A31 A32 we could factor the matrix A into two matrices, L A13 1 A23 = ⋆ A33 ⋆ 0 0 ⋆ 1 0 0 ⋆ 1 0 ⋆ ⋆ ⋆ ⋆ . 14) Then we could solve each right-hand side using only forward and back substitution. 15) we could rewrite A in terms of L and U : LU x = b. 16 and solve by forward substitution as: Ly = b. 17) Elimination methods 31 And subsequently we solve by back substitution: U x = y. 18) So, how do we decompose A as given before?