H3 #3 c

ci. the estimate for a fixed point {pn} with an initial approximation p0 is |pn-p| =< k^n / 1-k * |p1-p0| where k is the constant above (strictly less than 1)

I don't understand the question exactly since he doesn't give a value for k, but says it is "strictly less than 1"
also it seems as though his given equation is off, it looks as if the left side of the inequality is implying the distance from pn to the actual number p, however Wikipedia gives the full equation:



ii. Given that an approximate value of the fixed point of g is 1.94712296, use the values for pn obtained in Excel for the fixed point iteration of g to find |pn-p|

this is the same issue as 2bii in that he doesn't specify a maximum number of iterations, if you go down far enough, the |pn-p| will always be essentially zero...
in one case, using p0=1 to the 20th iteration: (1.94712296 - 1.94712296) = 0
it might be better to not go so far: the 15th iteration: (1.94712261- 1.94712296) = 0.0000004


3ciii. How does this error compare to 10^-5

0.0000004 < 0.00001