Question

Investigate the sequence \[x_{n+1}=\frac{ 1 }{ x _{n}}\]

Answer 1

so multiply both sides by xn
xn^2 + xn = 1 ,
xn^2 + xn -1 = 0
this is a quadratic in xn

Answer 2

xn = [ sqrt(5 ) +- 1] / 2

Answer 3

xn = [ -1 + sqrt(5)] /2 , [-1 - sqrt(5)] / 2 ,

Answer 4

oh, woops, didnt see the xn+1 , but it comes out the same

Answer 5

hey you changed your question

Answer 6

this new sequence oscillates

Answer 7

lets say the sequence started at some value \(x_0=a\).
now list what the rest of the terms in the sequence would be given this starting point.

Answer 8

the sequence was , i thought
x_n+1 + 1 = 1/x_n

Answer 9

ok sorry i misread, my font no good

Answer 10

the sequence will be a, 1/a, a, 1/a, ...

Answer 11

sorry, I do not understand.

Answer 12

\[x_{1}=a+1???\]

Answer 13

the "n" subscript represent the term in the sequence.
so when n=0, we have \(x_0\) which I said lets say it has some value "a": \(x_0=a\)
so the next term will be \(x_1\) and we know we can work this out using your formula:\[x_{n+1}=\frac{1}{x_n}\]therefore:\[x_1=\frac{1}{x_0}=\frac{1}{a}\]does that make sense?

Answer 14

the formula you have is basically saying:

next term = 1 divided by previous term

Answer 15

@asnaseer I kept getting 1/a

Answer 16

lets see what the next term would be...\[x_2=\frac{1}{x_1}=?\]

Answer 17

x\[x{_{2}}=1/x _{1}=1/x _{0}\]

Answer 18

no

Answer 19

ok

Answer 20

the formula says:\[x_{n+1}=\frac{1}{x_n}\]so, for n=0, we get:\[x_1=\frac{1}{x_0}\]for n=1, we get:\[x_2=\frac{1}{x_1}\]for n=2, we get:\[x_3=\frac{1}{x_2}\]etc...

Answer 21

and we started with: \(x_0=a\).
so: \(\displaystyle x_1=\frac{1}{x_0}=\frac{1}{a}\)
and: \(\displaystyle x_2=\frac{1}{x_1}=\frac{1}{\frac{1}{a}}=1\div\frac{1}{a}=1\times\frac{a}{1}=a\)

Answer 22

yeah I get it.

Answer 23

so the first three terms of the sequence would be:\[a,\frac{1}{a},a\]

Answer 24

hopefully you can now see that it just oscillates between \(a\) and \(\displaystyle\frac{1}{a}\)

Answer 25

alos note that the sequence can NEVER have \(a=0\) otherwise you would have \(\displaystyle\frac{1}{0}=\infty\)

Answer 26

yes I see that.

Answer 27

another interesting observation you could make is what happens if \(a=1\) or \(a=-1\)?

Answer 28

I will try that now.

Answer 29

is that \[x_{1}=a??\]

Answer 30

no, if \(x_0=a\) then \(\displaystyle x_1=\frac{1}{a}\)

Answer 31

so, if a=1, what will \(x_0=?\)

Answer 32

and what will \(x_1=?\)

Answer 33

confused.

Answer 34

remember your formula generates the following sequence if \(x_0=a\):\[a,\frac{1}{a},a,\frac{1}{a},a,\frac{1}{a},...\]do you agree with this?

Answer 35

yes.

Answer 36

so, if say we let \(a=2\), then we get the sequence:\[2,\frac{1}{2},2,\frac{1}{2},2,\frac{1}{2},...\]agreed?

Answer 37

so \[x _{0}=1\]

Answer 38

and \[x_{1}=1/1=1?\] if a=1?

Answer 39

yes, so what will the sequence look like if \(a=1\)?

Answer 40

111111111

Answer 41

remember to put comma's between each term

Answer 42

ok

Answer 43

so if \(a=1\) we get a sequence were every term equals 1.
similarly, if \(a=-1\), we get a sequence where every term is equal to -1.

Answer 44

thanks you. very much,

Answer 45

yw :)

Answer 46

very interesting math.

Answer 47

I'm glad you find it interesting - it makes the subject much easier to learn if you find it interesting. :)