 By Arieh Iserles

The once a year e-book Acta Numerica has proven itself because the top discussion board for the presentation of definitive reports of present numerical research issues. The invited papers, by means of leaders of their respective fields, let researchers and graduate scholars to speedy take hold of fresh traits and advancements during this box. Highlights of this year's quantity are articles on area decomposition, mesh adaption, pseudospectral equipment, and neural networks.

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Acta Numerica 1994: Volume 3

The once a year ebook Acta Numerica has tested itself because the best discussion board for the presentation of definitive reports of present numerical research themes. The invited papers, via leaders of their respective fields, enable researchers and graduate scholars to speedy take hold of contemporary traits and advancements during this box.

Extra resources for Acta Numerica 1994: Volume 3

Sample text

Ii) The condition 0 < [x - al < ~ (excluding the possibility x = a) is used rather than the condition [x - al < ~ as in the definition of continuity since J may not be defined at a itself. PROBLEMS In Problems 1 through 8 the functions are continuous at the value a given. In each case find a value ~ corresponding to the given value of e so that the definition of continuity is satisfied. Draw a graph. 1. 01 2. 01 3. 01 4. 1 5. 01 6. f(x) = x 3 - 7. f(x) = x 3 + 3x, a = 8. 2. Limits 35 In Problems 9 through 17 the functions are defined in an interval about the given value of a but not at a.

EXAMPLE 2. Solve for x: 3 x -<5 (x :1= 0). Solution. Since we don't know in advance whether x is positive or negative, we cannot multiply by x unless we impose additional conditions. We therefore separate the problem into two cases: (i) x is positive, and (ii) x is negative. The desired solution set can be written as the union of the sets S, and S2 defined by s, = {x:~ < 5 and x> o}, S2 = {x:~ < 5 and x< o}. Now S, X E <::> 3 < 5x <::> X <::> x>l <::> 3 > 5x <::> X X < O. Similarly, X E S2 x>0 and >!

We assume that L > M and reach a contradiction. Let us define e = (L - M)/2; then from the definition of limit there are positive numbers (jl and (j2 such that and If(x) - LI < e for all x satisfying 0 < Ix - al < (jl Ig(x) - MI < s for all x satisfying 0 < I x - al < (j2. We choose a positive number (j which is smaller than (jl and (j2 and furthermore so small that f(x) ~ g(x) for 0 < Ix - al < (j. In this interval, we have M - e < g(x) < M Since M +e and + s = L - s, it follows that g(x) < M + s = L - e < f(x) < L + e.