Register to reply 
Finding indexes.by The_Iceflash
Tags: indexes 
Share this thread: 
#1
Feb1210, 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] 


#2
Feb1210, 04:31 PM

Mentor
P: 21,397




#3
Feb1210, 04:42 PM

P: 50

I got n > 1998 as the index so, [tex]\frac{1999}{1999+2} = .9990004998[/tex] [tex]\frac{2000}{2000+2} = .9990009990[/tex] 


#4
Feb1210, 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.



#5
Feb1310, 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. 


Register to reply 
Related Discussions  
Anyone know of experiments to calculate the refractive indexes of liquids?  Introductory Physics Homework  4  
Grammar (ref index/indexes/indices) and appendix quesrtion  Academic Guidance  4  
Citation indexes for mathematical physics  Math & Science Software  2  
Changing the summation indexes in double sums.  Calculus & Beyond Homework  1  
Citation indexes for mathematical physics  Math & Science Software  2 