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a(n) = (a(n-1)**2-2)/(2*a(n-1)-3)

This recursive equation seems to be increasing linearly with any a(1) that is entered first. There is no oscillation, there is no limit for the situations where a1=1, 2, or -1. I am not sure at the moment how to prove that for all values of a1 that it will be the same for all of them. It seems that the sequence will have the same slope no matter what the initial a1 is, but depending on what a1 is, will depend on what b would be for y=mx+b. At the beginning of the sequence for each of the a1 I tested, there appears to be a little discontinuity, but it quickly goes to a line. I even tested what would happen for a1 = 500 and a1 = -500 and they still seem to have the same slope as the other ones. 

It turns out that my code is completely wrong as I did the problem by hand and obtained completely different results. I have learned that I should be using a while loop for my code instead of a for loop. I am sure that I could use a for loop, but clearly I was not using it correctly since what I described above does not match what I should have gotten. What I have discovered is that this sequence reminds me of iterative methods used to find the square roots of a function. I tried many different values of a(n-1) and for every single value so far, the sequence converges either to 2 or 1. I have plugged in negative integers and very large numbers to test if for whatever real number I plug in for a(n-1) that the sequence will converge to either 1 or 2. It seems that for any number less than 1.5, the sequence will converge to 1. I tested a(n-1)= 1.5 and I got an error in my code. This is because plugging 1.5 in for a(n-1) into the recursive sequence gives 0.25/0 and 0 cannot be in the denominator. Any real number for a(n-1) greater than 1.5 will make the recursive sequence converge to 2. 

Here are the arrays formed by a(n-1)=1.5001, 1.6, 1.55, and 1.49999 respectively. 

Here is the code I used to obtain the correct results.

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Fibonacci numbers’ ratio

I am setting out to find a recursive formula for the ratio of 2 successive terms of the fibonacci sequence. 

The fibonacci sequence recursive formula is f(n)=f(n-1) + f(n-2), f(1)=1, f(1)=1

Some thoughts on the fibonacci numbers: 

1, 1, 2, 3, 5, 8, 13, 21, 34, 55, 89

 

1/1=1

2/1=2

3/2=1.5

5/3=1.66666

8/5=1.6

13/8=1.625

21/13=1.6153846154

34/21=… and so on

 

Here is some code I typed up in python that graphs the ratio of two successive terms of the fibonacci sequence. 

This is a line graph with all of the points visible.

This is a line graph showing most of the points, but only from y=1.618 to y=1.6181. This pattern seems to be oscillating and converges to the golden ratio.

This formula will be represented by r(n) with r(1)=1, r(2)=2, r(3)=1.5, r(4)=1.666667, r(5)=1.6, r(6)=1.625 and so on. 

r(n) = f(n+1)/f(n), so writing this in terms of r(n-1) seems to be a little tricky. 

Maybe something like r(n) = r(n-1) + (-1)^n*r(n-1)

After doing some research, there is a recursive sequence for rn that is only in terms of r(n-1). The sequence is rn = 1 + 1/r(n-1). This sequence oscillates and is bounded by 1 ≤ rn ≤ 2. Even when the difference between r(n) and r(n+1) is very small as n goes to infinity, the sequence will still oscillate. I found out while trying to figure out a formula for r(n) that to go from r(n) to r(n+1) is +1/f(n)f(n+1) then -1/f(n+1)f(n+2) then +1/f(n+2)f(n+3) and so on (fn is the fibonacci series). This series does converge to a limit and that limit is the golden ratio or (1+sqrt(5))/2. After a quick search on what a Cauchy sequence is, I believe r(n) is one. From what I observe and read, a Cauchy sequence is a sequence that oscillates and the outputs of the sequence become closer and closer as n becomes larger and larger. 

 

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Square root of 5 rational?

I am setting out to find out why the square root of 5 is irrational. It may be obvious that there is no whole number or integer that when squared is equal to 5, but why the square root of 5 is irrational, is a different problem.

I found that there are many ways of proving that the square root of 2 is irrational including proofs by contradiction and by constructive proofs, so it may be easier to figure out how the square root of 5 is irrational.

Things that will be easy are that there are lots of resources to use in order to formulate my own proofs and gather ideas to figure out how to solve the problem.

The thing that will be hardest for me to solve this problem is that I am bad at proofs in general. I have a hard time understanding the logic behind why proofs are written the way they are and why certain strategies are used.

Here is my thought process on how I went about it. A rational number is one that can be written as m/n, m, n in the set of integers. So I let the square root of 5 equal m/n which means 5=m²/n². And in order to get an odd quotient, m² and n² must be odd. Thus m and n are odd because the square root of an odd square integer is odd since an odd integer times an odd integer is odd. So I tried substituting m for 2k+1 and n for 2j+1 and that did not work. Then I tried substituting m for 5x and n for 5y. That did not work either. I tried seeing if there was some kind of pattern with square numbers, but I decided that it would take too long to write and think about all of the square numbers and their relation to one another. Finally I tried substituting things for other known variables and assumed that m and n are coprimes because or else there would be a seemingly infinite amount of possible rational numbers that would equal the square root of 5. 

Proof 1: 

This is a proof by contradiction that will prove the square root of 5 is irrational. 

Assume that the square root of 5 is rational. 

Let sqrt(5)=m/n, m and n in the set of positive integers and m and n are coprime which means that m and n have no other common factors besides 1. Then 5=m²/n², 5n²=m². This means that 5 is a factor of m. Therefore 5c=m for some integer c → m²=25c². Then substitute 25c² for m² in 5n²=m², 5n²=25c² → n²=5c² therefore 5 is a factor of n as well as m. This is a contradiction because we stated that m and n only have a common factor of 1, but that is not true. 

Conclusion: It was a lot easier to form a proof after looking at my notes. Discrete math was quite helpful while writing this and understanding others’ works.