i.
If the sequence {pn} approximates a solution p of (1), then
|pn - p| =< b-a / 2^n
where a and be are endpoints of the interval
use the estimate to find the number of iterations N required so that the solution to (a) using the bisection method is correct to the actual solution to within 10^-5
this refers to the Analysis (Wikipedia) of the bisection method.
In simpler terms, the distance from our approximation (pn) to the actual answer (p) must be equal to or less than the distance between the two endpoints divided by the number of times we have halved the interval
(ex. if our initial interval is [0,1] by the 10th iteration, our estimated value MUST be within (1/2^10) or 0.0009765 of the exact answer)
so the question is asking what is the smallest value N with which 0.5-0 /2^n =< 10^-5
to find the answer through excel I made
A1 = 0.5
and then made the progressive rule for A2
=A1/2
and then made all of the B column equal to 10^-5
(notice however that the value of row 1 is equal to n=0, so row An is actually the value of our n-1 iteration)
after dragging the excel worksheet down:
| 1 | 0.500000 | 0.000010 |
| 2 | 0.250000 | 0.000010 |
| 3 | 0.125000 | 0.000010 |
| .. | .... | ... |
| 15 | 0.000031 | 0.000010 |
| 16 | 0.000015 | 0.000010 |
| 17 | 0.000008 | 0.000010 |
| 18 | 0.000004 | 0.000010 |
| 19 | 0.000002 | 0.000010 |
| 20 | 0.000001 | 0.000010 |
at row 17 column A becomes less than column B: therefore our answer is N=16
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ii.
Given that an approximate value of the root is p=0.36623819, use the values for pn that you obtained in excel for part(a) to find |pn-p|
I am not entirely sure exactly what is being asked; does he want us to use the 16th iteration we arrived at for i or can/should we be more precise?
in that case we end up with |0.36622826 - 0.36623819| = 0.00000993
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iii. How does that compare to 10^-5
0.00000993 < 0.00001
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