Finding Indexes for \frac{n}{n+2} \approx 1 with varying epsilon values

  • Thread starter The_Iceflash
  • Start date
In summary: The index:I got n > 1998 as the index so,\frac{1999}{1999+2} = .9990004998\frac{2000}{2000+2} = .9990009990This is to be expected in a sequence that is increasing, as {n/(n + 2)} is. A larger index gives you a larger value.As Mark44 said, that is to be expected. \epsilon being smaller means you must get closer to the limit value which, in turn, means you must go further out in the sequence. As \epsilon gets smaller, you should expect the index, N, to get larger, not smaller.Who
  • #1
The_Iceflash
50
0
I for the most part have this completed but I have a small question and thus checking if I did this correctly.

Homework Statement


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



Homework Equations


N/A


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]
 
Last edited:
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  • #2
The_Iceflash said:
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.
 
  • #3
Mark44 said:
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]
 
  • #4
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
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.
 

What is the purpose of finding indexes for \frac{n}{n+2} \approx 1 with varying epsilon values?

The purpose of finding indexes for \frac{n}{n+2} \approx 1 with varying epsilon values is to determine the values of n where the expression \frac{n}{n+2} is approximately equal to 1 within a certain margin of error, represented by the epsilon value. This can help in understanding the behavior of the expression and its relationship with n.

How do you calculate the indexes for \frac{n}{n+2} \approx 1 with varying epsilon values?

To calculate the indexes for \frac{n}{n+2} \approx 1 with varying epsilon values, you can start by setting up the equation \frac{n}{n+2} = 1 + \epsilon, where \epsilon represents the margin of error. Then, you can solve for n by setting \epsilon to different values and finding the corresponding values of n that satisfy the equation. This can be done using algebraic manipulation or through trial and error.

What is the significance of using varying epsilon values in finding indexes for \frac{n}{n+2} \approx 1?

The use of varying epsilon values allows for a more comprehensive understanding of how the expression \frac{n}{n+2} behaves as n changes. It also helps in determining the range of values for n where the expression is approximately equal to 1 with different levels of accuracy. This can provide insights into the limitations and precision of the expression.

What are some potential applications of finding indexes for \frac{n}{n+2} \approx 1 with varying epsilon values?

Some potential applications of finding indexes for \frac{n}{n+2} \approx 1 with varying epsilon values include analyzing the convergence of series, approximating the value of a limit, and understanding the accuracy of mathematical models. It can also be useful in fields such as finance, engineering, and physics where precise calculations are necessary.

Is there a specific range of epsilon values that should be used in finding indexes for \frac{n}{n+2} \approx 1?

There is no specific range of epsilon values that should be used as it depends on the specific context and purpose of the calculation. In some cases, a smaller epsilon value may be preferred for higher accuracy, while in others, a larger epsilon value may be sufficient. It is important to consider the trade-off between accuracy and practicality when choosing epsilon values.

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