Find the golden ratio in surd form

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    I have this question:
    The golden ratio appears in many natural situations. It can be found by looking at the ratio of success in terms of the Fibonacci sequence:
    1,1,2,3,5,8,13,21,34
    2/1, 3/2, 5/3, 8/5
    2, 1.5, 1.67, 1.6... the golden ratio is about equal to 1.6.
    Find it in surd form.

    So far I've figured that 2 should be a, then 3 should be b, 5 should be a+b, and so on. But I need help on how I can make this into a quadratic maybe and express it as a surd. Thanks...
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    (Original post by Yewdraconis)
    I have this question:
    The golden ratio appears in many natural situations. It can be found by looking at the ratio of success in terms of the Fibonacci sequence:
    1,1,2,3,5,8,13,21,34
    2/1, 3/2, 5/3, 8/5
    2, 1.5, 1.67, 1.6... the golden ratio is about equal to 1.6.
    Find it in surd form.

    So far I've figured that 2 should be a, then 3 should be b, 5 should be a+b, and so on. But I need help on how I can make this into a quadratic maybe and express it as a surd. Thanks...
    I don't think you will get anywhere with that.

    I've seen something like this at Cambrige over a year ago, but they started later down with my method, so I think my method is correct however I don't have much experience with these recurrence things and infinities, so I'll give a shot at explaining this. Anyway:

    Second sequence. Observe:

    \displaystyle 1, \frac{2}{1}, \frac{3}{2}, \frac{5}{3}, \frac{8}{5}, ...

    \displaystyle \Rightarrow 1, 1+\frac{1}{1}, 1+\frac{1}{2}, 1+ \frac{2}{3}, 1+\frac{3}{5}, ...

    So:

    x_1=1

    x_2=\frac{2}{1}=1+\frac{1}{1}

    x_3=\frac{3}{2}=1+\frac{1}{2}

    x_4=\frac{5}{3}=1+\frac{2}{3}

    x_5=\frac{8}{5}=1+\frac{3}{5}

    We can observe that n^{th} term, with x_1=1, is given by:

    \displaystyle x_n=1+\frac{1}{x_{n-1}}=1+ \frac{1}{1+\frac{1}{x_{n-2}}}=1+\frac{1}{1+\frac{1}{1+...  }} for n > 1, n \in \mathbb{Z}

    Now we get the golden ratio as we go further and further down the sequence, so n \rightarrow \infty

    Meaning \displaystyle \phi = \lim_{n \rightarrow \infty} (x_n)

    As this sequence converges onto a particular value (which is our golden ratio!), it is safe to assume that after an infinite amount of steps, we get:

    \displaystyle X=1+\frac{1}{X}

    ...as the difference between x_{n} and x_{n-1} becomes 0 at infinity, hence they are the same, referring to this value as X.

    At this point just solve the equation and you get \phi = X (take the positive value, as our sequence is strictly positive).

    Feel free to ask if you don't understand something.
    Anyone can also feel free to correct me if I made a mistake, not too familiar with recurrence relations and limits combined into one as I said.
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    (Original post by Yewdraconis)
    I have this question:
    The golden ratio appears in many natural situations. It can be found by looking at the ratio of success in terms of the Fibonacci sequence:
    1,1,2,3,5,8,13,21,34
    2/1, 3/2, 5/3, 8/5
    2, 1.5, 1.67, 1.6... the golden ratio is about equal to 1.6.
    Find it in surd form.

    So far I've figured that 2 should be a, then 3 should be b, 5 should be a+b, and so on. But I need help on how I can make this into a quadratic maybe and express it as a surd. Thanks...

    let the nth term be b and the n+1th term be a

    n+2th term=a+b

    the limit of the sequence is when the ratio of the terms n to n+1 and n+1 to n+2 is the same

    a/b=(a+b)/a
    a/b=1+b/a
    let a/b=x
    x=1+1/x
    solve for x and the solutions will be the forward ratio and the backward ratio
    RDKgames's way is a lot better and works for far more types of sequences but I think this one is a bit easier to read
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    (Original post by RDKGames)
    I don't think you will get anywhere with that.

    I've seen something like this at Cambrige over a year ago, but they started later down with my method, so I think my method is correct however I don't have much experience with these recurrence things and infinities, so I'll give a shot at explaining this. Anyway:

    Second sequence. Observe:

    \displaystyle 1, \frac{2}{1}, \frac{3}{2}, \frac{5}{3}, \frac{8}{5}, ...

    \displaystyle \Rightarrow 1, 1+\frac{1}{1}, 1+\frac{1}{2}, 1+ \frac{2}{3}, 1+\frac{3}{5}, ...

    So:

    x_1=1

    x_2=\frac{2}{1}=1+\frac{1}{1}

    x_3=\frac{3}{2}=1+\frac{1}{2}

    x_4=\frac{5}{3}=1+\frac{2}{3}

    x_5=\frac{8}{5}=1+\frac{3}{5}

    We can observe that n^{th} term, with x_1=1, is given by:

    \displaystyle x_n=1+\frac{1}{x_{n-1}}=1+ \frac{1}{1+\frac{1}{x_{n-2}}}=1+\frac{1}{1+\frac{1}{1+...  }} for n > 1, n \in \mathbb{Z}

    Now we get the golden ratio as we go further and further down the sequence, so n \rightarrow \infty

    Meaning \displaystyle \phi = \lim_{n \rightarrow \infty} (x_n)

    As this sequence converges onto a particular value (which is our golden ratio!), it is safe to assume that after an infinite amount of steps, we get:

    \displaystyle X=1+\frac{1}{X}

    ...as the difference between x_{n} and x_{n-1} becomes 0 at infinity, hence they are the same, referring to this value as X.

    At this point just solve the equation and you get \phi = X (take the positive value, as our sequence is strictly positive).

    Feel free to ask if you don't understand something.
    Anyone can also feel free to correct me if I made a mistake, not too familiar with recurrence relations and limits combined into one as I said.
    Thanks this makes some sense, I think I can work the rest out. Thanks
 
 
 
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