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Unformatted text preview: Now since 1/c  1 we know that 1/c  n Ý as n Ý and we conclude that c  n
Therefore c n 0 as n Ý and we conclude that
n
lim 1 c 1 .
nÝ 1
1c
c 0 as n Ý. 4. Suppose that x n is a given sequence of real numbers, that x 1 0 and that the equation
8x 31 6x n 1
n
holds for every positive integer n.
a. Use Scientific Notebook to work out the first twenty terms in the sequence x n .
To use Scientific Notebook for this purpose, place your cursor in the equation
fx
and click on the button 3 6x 1
8 in your computing toolbar to supply the definition of the function f to Scientific Notebook. Then open the Compute menu, scroll down to Calculus and move
across to Iterate. Fill the dialogue box giving the name of the function as f, the number of
iterations as 20 and the starting value as 0. a. Prove that x n 1 for every positive integer n.
We use mathematical induction. For each positive integer n we take p n to be the assertion that
x n 1. Since x 1 0, the assertion p 1 is true. Now suppose that n is any positive integer for
which the statement p n is true. Then
8x 31 6x n 1 6 1 1
n
and so
7 1
8
from which it follows that the assertion p n1 is also true. We deduce from mathematical
induction that the assertion p n is true for every positive integer n.
x n 1 3 b. Prove that the sequence x n is strictly increasing.
Once again we use mathematical induction. For each positive integer n we take p n to be the
assertion that x n x n1 . Since x 1 0 1 x 2 , the assertion p 1 is true. Now suppose that n is
2
any positive integer for which the assertion p n is true. Since
x n 2 3 6x n1 1
8 146 3 6x n 1 x
n 1
8 we see that the assertion p n1 must also be true. We deduce from mathematical induction that
the assertion p n is true for every positive integer n.
c. Deduce that the sequence x n is convergent and discuss its limit. Assuming an unofficial knowledge of
the trigonometric functions, prove that the limit of the sequence x n is cos .
9 Solution to part d: We write the limit of this sequence as x. from the fact that
8x 31 6x n 1
n
for each n we obtain
8x 3 6x 1 0.
This equation has one positive solution and two negative solutions and from the fact that
8 cos 3 6 cos 1 2 4 cos 3 3 cos
1
9
9
9
9
2 cos 3
1 2 cos 1 0
3
9
we see that the positive root is cos .
9
5. In this exercise we study the sequence x n defined by the equation
n
xn 1 1
n
for every integer n 1. You will probably want to make use of the binomial theorem when you do this
exercise.
a. Ask Scientific Notebook to make a 2D plot of the graph of the function f defined by the equation
f n 1 1
n
for 1 n n 100. b. Prove that x n 3 for every n. Solution: For each n 2 we see that
n xn 1 1
n n n
j j0 1
n 1
n 1 j n
j2 n 11 1 1 j2
n 2
j2 1 n j1
j! n j nn 11 2 1
n j 1 1
j! 1 1
j! n n 1 2
j! 1 3
2j 1 j2 c. Prove that the sequence x n is increasing. Solution: For each n 2 we see that
n 1 1 1...
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This note was uploaded on 11/26/2012 for the course MATH 2313 taught by Professor Staff during the Fall '08 term at Texas El Paso.
 Fall '08
 STAFF
 Math, Calculus

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