Question

Show that `r= (1+ sqrt5)/2` for the fibonacci sequence?

`r= (1+ sqrt5)/2`

The teacher wants us to prove this using the quadratic formula.

Answer 1

See below

Explanation:

I suppose you want to prove that

`lim_{n\to\infty} \frac{F_{n+1}}{F_n} = \varphi`

where `F_n` is the `n`-th term of the Fibonacci series, and `\varphi` is the number `\frac{1+sqrt{5}}{2}`

This result can be proven as follows: suppose that the ratio of two consective numbers converges to some number `x`, i.e. suppose that

`lim_{n\to\infty} \frac{F_{n+1}}{F_n} = x`

now use the recursive definition of Fibonacci numbers, i.e. `F_{n+1} = F_n+F_{n-1}` to write

`\frac{F_{n+1}}{F_n} = \frac{F_n+F_{n-1}}{F_n} = 1+frac{F_{n-1}}{F_n}`

so,

`lim_{n\to\infty} \frac{F_{n+1}}{F_n} = lim_{n\to\infty}1+frac{F_{n-1}}{F_n}`

since we're supposing that the ratio of two consecutive numbers converges to `x`, the previous expression becomes, when `n \to \infty`,

`x = 1+1/x`

which is the equation solved by `\frac{1\pm\sqrt{5}}{2}`.

Since `\frac{1-\sqrt{5}}{2}<0`, and we're considering ratios between positive numbers, the only possible solution is the one we're looking for.

Answer 2

See explanation...

Explanation:

The Fibonacci sequence is defined recursively by:

`{ (F_0 = 0), (F_1 = 1), (F_(n+2) = F_(n+1) + F_n) :}`

If a geometric sequence with general term `a_n = a r^(n-1)` satifies `a_(n+2) = a_(n+1) + a_n`, then we have:

`ar^(n+1) = ar^n+ar^(n-1)`

and hence:

`r^(n+1) = r^n+r^(n-1)`

for any integer `n`.

In particular, putting `n=1`, we find:

`r^2=r+1`

and hence:

`0 = r^2-r-1`

`color(white)(0) = (r-1/2)^2-5/4`

`color(white)(0) = (r-1/2)^2-(sqrt(5)/2)^2`

`color(white)(0) = (r-1/2-sqrt(5)/2)(r-1/2+sqrt(5)/2)`

That is:

`r = 1/2+-sqrt(5)/2`

Hence we find that the general term of any sequence satisfying `a_(n+2) = a_(n+1)+a_n` is expressible in the form:

`a_n = A(1/2+sqrt(5)/2)^n+B(1/2-sqrt(5)/2)^n`

for some `A, B` to be determined.

In particular, for the Fibonacci sequence, we have `F_0 = 0` and `F_1 = 1`, so we want `A, B` satisfying:

`0 = F_0 = A(1/2+sqrt(5)/2)^0+B(1/2-sqrt(5)/2)^0 = A+B`

`1 = F_1`

`color(white)(1) = A(1/2+sqrt(5)/2)^1+B(1/2-sqrt(5)/2)^1`

`color(white)(1) = 1/2(A+B)+sqrt(5)/2(A-B)`

`color(white)(1) = sqrt(5)A`

So `A=1/sqrt(5)` and `B=-1/sqrt(5)` giving us:

`F_n = ((1/2+sqrt(5)/2)^n - (1/2-sqrt(5)/2)^n)/sqrt(5)`

Note that:

`abs(1/2-sqrt(5)/2) < 1`

while:

`abs(1/2+sqrt(5)/2) > 1`

Hence the ratio between consecutive terms tends to `1/2+sqrt(5)/2`