What is the point of convergence for a recursive sequence in a plane?

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In summary, the sequence of points \{P_i\}_{i=0}^\infty on a plane is defined recursively by a_1 = 2 and a_{n+1} = 2 - \frac{a_n}{4}. The x-coordinate of the point P = \lim_{i→\infty} P_i is found to be 1.6, and the sequence is bounded by 1 ≤ a_n ≤ 2. However, the sequence is not strictly increasing or decreasing, but it does approach a certain point.
  • #1
toothpaste666
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Homework Statement



Let [itex]\{P_i\}_{i=0}^\infty [/itex] be a sequence of points on a plane. Suppose [itex]P_i[/itex]s are placed as on the picture below, so that [itex] |P_0 P_1|=2, |P_1 P_2|=1, |P_2 P_3|=.5, |P_3P_4|=.25[/itex], ... Find the coordinate of the point [itex]P = \lim_{i→\infty} P_i [/itex]

attachment.php?attachmentid=68646&d=1397459162.png


Homework Equations


The Attempt at a Solution



here are the points [itex] P_0: (0,0) P_1: (2,0) P_2: (2, 1) P_3: (1.5, 1) P_4: (1.5, .75) P_5: (1.625, .75) [/itex]

lets examine the x values first:
2, 1.5, 1.625

this is a sequence defined recursively by:

[itex] a_1 = 2 [/itex]

[itex] a_{n+1} = 2 - \frac{a_n}{4} [/itex]

[itex] L = \lim_{a_n\rightarrow\infty} a_n = \lim_{a_n\rightarrow\infty} a_{n+1}
= \lim_{an\rightarrow\infty} 2-\frac{a_n}{4} [/itex]

which means that

[itex] L = 2-\frac{L}{4} [/itex]

[itex] 4L = 8 - L [/itex]

[itex] 5L = 8 [/itex]

[itex] L = \frac{8}{5} [/itex]

[itex] L = 1.6 [/itex]

so 1.6 would be the x coordinate of the point.

I then would follow a similar process to find the y coordinate, but before I do that I just want to make sure what I have so far is correct.
 

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  • #2
also forgive me for not proving the sequence is bounded and decreasing (I left it out to save time and narrow the focus) but i have done so by mathematical induction
 
  • #3
Yes it is correct but also prove it is bounded and decreasing.
 
  • #4
Ok so it turns out one of my induction proofs was wrong. I can prove it is bounded but this sequence is not strictly increasing or decreasing. it hops back and forth between increasing and decreasing but it IS approaching a certain point. Is there any way to prove that it has a limit?
 
  • #5
the proof that it is bounded:

[itex] 1 ≤ a_n ≤ 2 [/itex]

[itex] \frac{1}{4} ≤ \frac{a_n}{4} ≤ \frac{2}{4} [/itex]

[itex] -\frac{1}{4} ≥ -\frac{a_n}{4} ≥ -\frac{2}{4} [/itex]

[itex] 2-\frac{1}{4} ≥ 2-\frac{a_n}{4} ≥ 2-\frac{2}{4} [/itex]

[itex] 1.75 ≥ a_{n+1} ≥ 1.5 [/itex]

[itex] 1 ≤ 1.5 ≤ a_{n+1} ≤ 1.75 ≤ 2 [/itex]

[itex] 1 ≤ a_{n+1} ≤ 2 [/itex]
 

1. What is a recursive sequence?

A recursive sequence is a sequence of numbers where each term is defined by a function that uses one or more previous terms in the sequence. This means that the value of each term depends on the values of the previous terms.

2. How is a recursive sequence different from a regular arithmetic sequence?

A regular arithmetic sequence has a constant difference between each term, while a recursive sequence has a variable difference that depends on the function used to define each term.

3. How can I find the nth term in a recursive sequence?

To find the nth term in a recursive sequence, you can use the recursive formula provided. This formula will allow you to calculate the value of each term in the sequence, starting with the first term and working your way up to the nth term.

4. Can a recursive sequence have a negative term?

Yes, a recursive sequence can have negative terms. This will depend on the function used to define each term, as well as the values of the previous terms in the sequence.

5. Are there any real-life applications of recursive sequences?

Yes, recursive sequences have many real-life applications, such as in population growth, financial investments, and computer algorithms. They are also used in various fields of science, including physics, biology, and chemistry.

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