| Line 26: | Line 26: | ||
-Base case <math>x_1 = 8</math>, so <math>x_1 > 4</math> | -Base case <math>x_1 = 8</math>, so <math>x_1 > 4</math> | ||
-Assume <math>x_n > 4</math>; is <math>x_{n+1} > 4</math>? | -Assume <math>x_n > 4</math>; is <math>x_{n+1} > 4</math>? | ||
| − | <math>x_n > 4 \Rightarrow \frac{1}{2}x_n > 2</math> | + | <math>x_n > 4 \Rightarrow \frac{1}{2}x_n > 2</math> <math>x_n</math> is greater than 4, so half of <math>x_n</math> is greater than 2 |
| − | <math>\frac{1}{2}x_n > 2 \Rightarrow \frac{1}{2}x_n + 2> 2 + 2 \Rightarrow \frac{1}{2}x_n + 2 > 4</math> | + | <math>\frac{1}{2}x_n > 2 \Rightarrow \frac{1}{2}x_n + 2> 2 + 2 \Rightarrow \frac{1}{2}x_n + 2 > 4</math> Half of <math>x_n</math> is greater than 2, so adding 2 to both sides, the left |
| − | <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 | + | 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 | ||
Revision as of 11:21, 18 April 2010
Use this page to submit solutions for the test 2. Extra credit will be given to those who will submit solutions, notice and help correct mistakes in the solutions submitted by other students, develop alternative solutions, participate actively in the discussion.
Please, if you made any contributions to the content of this page, also notify the professor by sending a short e-mail.
go back to the Discussion Page
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
