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                                                     hand side is greater than 4.
 
                                                     hand side is greater than 4.
 
     <math>\frac{1}{2}x_n + 2 = x_{n+1}</math>, therefore <math>x_{n+1}</math> > 4, so the series (<math>x_n</math>) is bounded below by 4
 
     <math>\frac{1}{2}x_n + 2 = x_{n+1}</math>, therefore <math>x_{n+1}</math> > 4, so the series (<math>x_n</math>) is bounded below by 4
 +
 +
b) Show that sequence (<math>x_n</math>) is monotone
 +
    -The sequence is bounded below by 4, from above
 +
   
 +
    <math>\forall x > 4 \in \Re, (x - \frac{1}{2}x) > 2</math> 
 +
    -For all real x greater than 4, the distance, or length, between x and half of x is greater than 2 units
 +
 +
    <math>\forall x > 4 \in \Re, x > \frac{1}{2}x + 2</math>
 +
    -Then for all real x greater than 4, adding 2 units to half of x is always less than the original x

Revision as of 11:29, 18 April 2010

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PROBLEM 1

1.a. A sequence ($ x_n $) is said to be a Cauchy sequence if

     -Choice 2 by 3.5.1 Definition

1.b. The statement of the Bolzano-Weierstrass theorem is:

     -Choice 3 by 3.4.8 Theorem

1.c. Let $ f: A \mapsto \Re $. Suppose that $ (a,\infty) \subset A $ for some $ a \in \Re $. We say the limit of f as $ x \rightarrow \infty $ and write $ \lim_{x\to\infty}f = L $

     -Choice 5 by 4.3.10 Definition

1.d. Let $ A \subset \Re $, let $ f: A \mapsto \Re $, and let $ c \in A $. We say that f is continuous at c if

     -Choice 4 by 5.5.1 Definition


PROBLEM 2

Let $ x_1 := 8 $ and $ x_{n+1} := \frac{1}{2}x_n + 2 $ for $ n \in N $

a) Use induction to show that($ x_n $) is bounded below by 4.

    -Base case $ x_1 = 8 $, so $ x_1 > 4 $
    -Assume $ x_n > 4 $; is $ x_{n+1} > 4 $?
    $ x_n > 4  \Rightarrow  \frac{1}{2}x_n > 2 $  $ x_n $ is greater than 4, so half of $ x_n $ is greater than 2
    $ \frac{1}{2}x_n > 2    \Rightarrow    \frac{1}{2}x_n + 2> 2 + 2    \Rightarrow   \frac{1}{2}x_n + 2 > 4 $  Half of $ x_n $ is greater than 2, so adding 2 to both sides, the left 
                                                    hand side is greater than 4.
    $ \frac{1}{2}x_n + 2 = x_{n+1} $, therefore $ x_{n+1} $ > 4, so the series ($ x_n $) is bounded below by 4

b) Show that sequence ($ x_n $) is monotone

    -The sequence is bounded below by 4, from above
    
    $ \forall x > 4 \in \Re, (x - \frac{1}{2}x) > 2 $  
    -For all real x greater than 4, the distance, or length, between x and half of x is greater than 2 units
    $ \forall x > 4 \in \Re, x > \frac{1}{2}x + 2 $
    -Then for all real x greater than 4, adding 2 units to half of x is always less than the original x

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