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Finding indexes.

by The_Iceflash
Tags: indexes
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The_Iceflash
#1
Feb12-10, 04:07 PM
P: 50
I for the most part have this completed but I have a small question and thus checking if I did this correctly.

1. The problem statement, all variables and given/known data
Given the Sequence = [tex]\frac{n}{n+2}[/tex] [tex]\approx_{\epsilon}[/tex] 1 , for n >> 1

Show what index if [tex]\epsilon[/tex] = .001
" " if [tex]\epsilon[/tex] = .000002
" " for any [tex]\epsilon[/tex] > 0



2. Relevant equations
N/A


3. The attempt at a solution

for [tex]\epsilon[/tex] = .001 I did:

[tex]\left|\frac{n}{n+2}-1\right|< .001[/tex]

[tex]\left|\frac{-2}{n+2}\right| < .001[/tex]

[tex]\frac{2}{n+2} < .001[/tex]

[tex]\frac{2}{n+2} < \frac{1}{1000}[/tex]

[tex]\frac{n+2}{2} > 1000[/tex]

[tex]n+2 > 2000[/tex]

[tex]n > 1998[/tex]

The only issue I have with this one and the next one is that they aren't decreasing and I'm not sure if it needs to be or not:

[tex]\frac{1999}{1999+2} = .9990004998[/tex]

[tex]\frac{2000}{2000+2} = .9990009990[/tex]

for [tex]\epsilon[/tex] = .000002 I did:

[tex]\left|\frac{n}{n+2}-1\right|< .000002[/tex]

[tex]\left|\frac{-2}{n+2}\right| < .000002[/tex]

[tex]\frac{2}{n+2} < .000002[/tex]

[tex]\frac{2}{n+2} < \frac{2}{1000000}[/tex]

[tex]\frac{n+2}{2} > \frac{1000000}{2}[/tex]

[tex]n+2 > 1000000[/tex]

[tex]n > 999998[tex]

for any [tex]\epsilon[/tex] > 0

[tex]\left|\frac{n}{n+2}-1\right|< \epsilon[/tex]

[tex]\left|\frac{-2}{n+2}\right| < \epsilon[/tex]

[tex]\frac{2}{n+2} < \epsilon[/tex]

[tex]\frac{n+2}{2} > \frac{1}{\epsilon}[/tex]

[tex]n+2 > \frac{2}{\epsilon}[/tex]

[tex]n > \frac{2}{\epsilon}-2[/tex]
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Mark44
#2
Feb12-10, 04:31 PM
Mentor
P: 21,397
Quote Quote by The_Iceflash
The only issue I have with this one and the next one is that they aren't decreasing and I'm not sure if it needs to be or not:
Who are "they" in "they aren't decreasing"? The terms in the sequence n/(n + 2) are increasing, but you're looking at the difference n/(n + 2) - 1. This sequence is decreasing.
The_Iceflash
#3
Feb12-10, 04:42 PM
P: 50
Quote Quote by Mark44 View Post
Who are "they" in "they aren't decreasing"? The terms in the sequence n/(n + 2) are increasing, but you're looking at the difference n/(n + 2) - 1. This sequence is decreasing.
The index:

I got n > 1998 as the index so,

[tex]\frac{1999}{1999+2} = .9990004998[/tex]

[tex]\frac{2000}{2000+2} = .9990009990[/tex]

Mark44
#4
Feb12-10, 07:52 PM
Mentor
P: 21,397
Finding indexes.

This is to be expected in a sequence that is increasing, as {n/(n + 2)} is. A larger index gives you a larger value.
HallsofIvy
#5
Feb13-10, 06:09 AM
Math
Emeritus
Sci Advisor
Thanks
PF Gold
P: 39,682
As Mark44 said, that is to be expected. [itex]\epsilon[/itex] being smaller means you must get closer to the limit value which, in turn, means you must go further out in the sequence. As [itex]\epsilon[/itex] gets smaller, you should expect the index, N, to get larger, not smaller.

You may be thinking of function limits, [itex]\lim_{x\to a}f(x)[/itex] where to get closer to the limit, you must get closer to a: smaller [itex]\epsilon[/itex] means smaller [itex]\delta[/itex].

But with [itex]\lim_{n\to \infty} a_n[/itex] or even [itex]\lim_{x\to\infty} f(x)[/itex], your "a" is [itex]\infty[/itex] so you must get "closer to infinity" which means larger.


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