Question

How do you prove that the 8-sd approximation to the value of the infinite continued fraction `0.0123456789/(1+0.0123456789/(1+0.0123456789/(1+...))) =0.012196914` ?

Answer 1

Express as a solution of a quadratic, applying the quadratic formula to find value:

`(-1+sqrt(1.0493827156))/2~~0.012196914`

Explanation:

Let `k = 0.0123456789`.

We want to find the value of `x` given by:

`x = k/(1+k/(1+k/(1+...)))`

Then:

`k/x = k -: (k/(1+k/(1+k/(1+...)))) = 1+k/(1+k/(1+k/(1+...))) = 1+x`

Multiplying both ends by `x` and transposing, we get:

`x+x^2 = k`

Subtract `k` from both sides and rearrange slightly to get:

`x^2+x-k = 0`

Using the quadratic formula:

`x = (-1+-sqrt(1+4k))/2`

Since `k > 0`, we require `x > 0` too. So the only suitable root of this quadratic is:

`x = (-1+sqrt(1+4k))/2`

`= (-1+sqrt(1+(4*0.0123456789)))/2`

`=(-1+sqrt(1.0493827156))/2`

`~~0.01219691418437889686`

`~~0.012196914` to `8` s.d.

Answer 2

See below

Explanation:

It is an infinite expansion
`y = x/(1+x/(1+x/(1+x/(cdots))))` so

`y = x/(1+y)->x=y(1+y)`

but `y = 0.012196914`

then

`x = 0.012345678711123395`

but

`x_0 = 0.0123456789`

then `abs(x_0-x) = 1.88877*10^-10` within the required precision