I. Basic math.
 II. Pricing and Hedging.
 III. Explicit techniques.
 IV. Data Analysis.
 V. Implementation tools.
 VI. Basic Math II.
 VII. Implementation tools II.
 1 Calculational Linear Algebra.
 2 Wavelet Analysis.
 3 Finite element method.
 4 Construction of approximation spaces.
 A. Finite element.
 B. Averaged Taylor polynomial.
 C. Stable space splittings.
 D. Frames.
 E. Tensor product splitting.
 F. Sparse tensor product. Cure for curse of dimensionality.
 5 Time discretization.
 6 Variational inequalities.
 VIII. Bibliography
 Notation. Index. Contents.

## Finite element.

efinition

(Finite element) 1. "Finite element" is a triple where the "element domain" is a closed bounded subset of , the "space of shape functions" is a finite dimensional space of functions defined on and the "nodal variables" are a basis of the dual space .

2. The basis of the space with the property

 (Dual basis)
is called the "nodal basis" of .

3. The set "determines" the space if for any

Example

(Constant finite element) , is the 1-dimensional space of constant functions,

Example

(1-dimensional Lagrange element) , is the 2-dimensional space of linear functions,

Example

(2-dimensional Lagrange element) is a triangle in with vertices , is the 3-dimensional space of linear functions,

Proposition

(Identification of nodal variables 1) Let be an -dimensional vector space and is a subset of the dual space . Then the following statements are equivalent:

1. is a basis for .

2. For any ,

Proof

. Assume the contrary: holds but Then we introduce a s.t. Such is linearly independent from . We obtained a contradiction.

. Assume the contrary: is not a basis. Since has elements it has to be linearly dependent: there is a rearrangement of indexes and numbers s.t. Therefore we reduced the number of indexes in the condition (2): Let be a basis of . For any there are numbers such that Thus reads as follows: there are numbers , s.t. for any set of numbers This means that the matrix would be invertible if complemented to an matrix with an arbitrary row . We obtained a contradiction.

Proposition

(Identification of nodal variables 2) Let , be a polynomial of degree that vanishes on the hyperplane . Then where the is a polynomial of degree .

Proof

We perform an affine change of coordinates from to so that the plane is given by . Let We have thus the has the form for some polynomial of degree .

Example

(Linear tensor product element) Let be a rectangle with vertices , be a set of linear polynomials of coordinates and is given by

We establish that forms a nodal basis by noting that on any edge a function from is 1-degree polynomial of one variable. Hence, if then such function vanishes on every edge. But then, by the proposition ( Identification of nodal variables 2 ), where are 1-degree polynomials and , identify two adjacent edges. Thus by examining on one vertex not on and . Hence forms a nodal basis by the proposition ( Identification of nodal variables 1 ).

Definition

(Interpolant) Let be a finite element and be a corresponding nodal basis. We define the "interpolant" operator: for any function such that the operation is well defined for .

Proposition

(Properties of interpolant) The operator of the definition ( Interpolant ) has the following properties:

1. is a linear operator.

2. , .

3. . In particular, .

Proof

Direct verification.

 Notation. Index. Contents.