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Let <math>x_1 := 8</math> and <math>x_{n+1} := \frac{1}{2}x_n + 2</math> for <math>n \in N</math> | Let <math>x_1 := 8</math> and <math>x_{n+1} := \frac{1}{2}x_n + 2</math> for <math>n \in N</math> | ||
| − | a) Use induction to show that(<math>x_n</math>) is bounded below by 4. | + | a) Use induction to show that (<math>x_n</math>) is bounded below by 4. |
-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</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> -Half of <math>x_n</math> is greater than 2, so adding 2 to both sides, the left hand side is greater than 4. | + | <math>x_n > 4 \Rightarrow \frac{1}{2}x_n > 2</math> |
| − | <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>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> | ||
| + | -Half of <math>x_n</math> is greater than 2, so adding 2 to both sides, the left hand side is greater than 4. | ||
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| + | <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 | ||
| + | |||
| + | -More specifically, 4 is the infimum of the sequence | ||
| + | |||
| + | b) Show that sequence (<math>x_n</math>) is monotone | ||
| + | -The sequence is bounded below by 4, from above | ||
| + | |||
| + | <math>\forall y > 4 \in \Re, (y - \frac{1}{2}y) > 2</math> | ||
| + | -For all real y greater than 4, the distance, or length, between y and half of y is greater than 2 units | ||
| + | |||
| + | <math>\forall y > 4 \in \Re, y > \frac{1}{2}y + 2</math> | ||
| + | -Then for all real y greater than 4, adding 2 units to half of y is always less than the original y | ||
| + | |||
| + | <math>x_n > \frac{1}{2}x_n + 2 \Rightarrow x_n > x_{n+1}</math> | ||
| + | -Therefore, <math>x_n</math> always being greater than 4, <math>x_n</math> is always greater than <math>x_{n+1}</math> | ||
| + | -The sequence is therefore decreasing, and by '''3.3.1 Definition''', the sequence is monotone | ||
| + | |||
| + | c) What conclusions can you draw from parts a) and b) about the convergence of (<math>x_n</math>)? | ||
| + | -The sequence (<math>x_n</math>) is monotone and bounded on (4,8], so it is convergent by '''3.3.2 Monotone Convergence Theorem''' | ||
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| + | d) If the sequence (<math>x_n</math>) is convergent, what is the limit of the sequence? | ||
| + | <math>\lim_{x\to\infty}(x_n) = 4</math> | ||
| + | -By part '''(b)''' of '''3.3.2 Monotone Convergence Theorem''', for (<math>x_n</math>) being a bounded decreasing sequence, the limit | ||
| + | of the sequence is equal to its infimum | ||
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| + | |||
| + | '''PROBLEM 3''' | ||
Latest revision as of 12:01, 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 $) is bounded below by 4 -More specifically, 4 is the infimum of the sequence
b) Show that sequence ($ x_n $) is monotone
-The sequence is bounded below by 4, from above
$ \forall y > 4 \in \Re, (y - \frac{1}{2}y) > 2 $
-For all real y greater than 4, the distance, or length, between y and half of y is greater than 2 units
$ \forall y > 4 \in \Re, y > \frac{1}{2}y + 2 $
-Then for all real y greater than 4, adding 2 units to half of y is always less than the original y
$ x_n > \frac{1}{2}x_n + 2 \Rightarrow x_n > x_{n+1} $ -Therefore, $ x_n $ always being greater than 4, $ x_n $ is always greater than $ x_{n+1} $ -The sequence is therefore decreasing, and by 3.3.1 Definition, the sequence is monotone
c) What conclusions can you draw from parts a) and b) about the convergence of ($ x_n $)?
-The sequence ($ x_n $) is monotone and bounded on (4,8], so it is convergent by 3.3.2 Monotone Convergence Theorem
d) If the sequence ($ x_n $) is convergent, what is the limit of the sequence?
$ \lim_{x\to\infty}(x_n) = 4 $ -By part (b) of 3.3.2 Monotone Convergence Theorem, for ($ x_n $) being a bounded decreasing sequence, the limit of the sequence is equal to its infimum
PROBLEM 3
