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# ?

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Answer 1 · 2016-07-08T04:44:55

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

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

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 · 2016-07-08T10:25:53

See below

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