Prove that approximate error is always grater than true error in bisection method
Answers
Answered by
0
have a brief question related to an example in my textbook. In my book, the following theorem on Bisection Method is presented:
If [a0,b0],[a1,b1],...,[an,bn]...[a0,b0],[a1,b1],...,[an,bn]... denote the intervals in the bisection method, then the limits limn→∞anlimn→∞an, and limn→∞bnlimn→∞bn exist, are equal, and represent a zero of ff. If r=limn→∞cnr=limn→∞cn and cn=12(an+bn)cn=12(an+bn), then
|r−cn|≤2−(n+1)(b0−a0)|r−cn|≤2−(n+1)(b0−a0)
Next, there is the following example:
Suppose that the bisection method is started with the interval [50,63][50,63]. How many steps should be taken to compute a root with relative accuracy of one part in 10−1210−12?
OK, so if I were going to solve this, I would have used the theorem above and thought that we must have:
2−(n+1)(63−50)≤10−122−(n+1)(63−50)≤10−12
and then solve this for nn. However, the book example says:
The stated requirement on relative accuracy means that
|r−cn|/|r|≤10−12|r−cn|/|r|≤10−12
We know that r≥50r≥50, and thus it suffices to secure the inequality
|r−cn|/50≤10−12|r−cn|/50≤10−12
By means of the theorem above, we infer that the following condition is sufficent:
2−(n+1)⋅(13/50)≤10−122−(n+1)⋅(13/50)≤10−12
Solving this for nn, we conclude that n≥37n≥37.
OK, so what I don't understand here is why the example begins by writing |r−cn|/|r|≤10−12|r−cn|/|r|≤10−12instead of just |r−cn|≤10−12|r−cn|≤10−12. What is the motivation for including the |r||r| in the denominator on the left side of the inequality?
If someone could explain this to me, I would be very grateful!
If [a0,b0],[a1,b1],...,[an,bn]...[a0,b0],[a1,b1],...,[an,bn]... denote the intervals in the bisection method, then the limits limn→∞anlimn→∞an, and limn→∞bnlimn→∞bn exist, are equal, and represent a zero of ff. If r=limn→∞cnr=limn→∞cn and cn=12(an+bn)cn=12(an+bn), then
|r−cn|≤2−(n+1)(b0−a0)|r−cn|≤2−(n+1)(b0−a0)
Next, there is the following example:
Suppose that the bisection method is started with the interval [50,63][50,63]. How many steps should be taken to compute a root with relative accuracy of one part in 10−1210−12?
OK, so if I were going to solve this, I would have used the theorem above and thought that we must have:
2−(n+1)(63−50)≤10−122−(n+1)(63−50)≤10−12
and then solve this for nn. However, the book example says:
The stated requirement on relative accuracy means that
|r−cn|/|r|≤10−12|r−cn|/|r|≤10−12
We know that r≥50r≥50, and thus it suffices to secure the inequality
|r−cn|/50≤10−12|r−cn|/50≤10−12
By means of the theorem above, we infer that the following condition is sufficent:
2−(n+1)⋅(13/50)≤10−122−(n+1)⋅(13/50)≤10−12
Solving this for nn, we conclude that n≥37n≥37.
OK, so what I don't understand here is why the example begins by writing |r−cn|/|r|≤10−12|r−cn|/|r|≤10−12instead of just |r−cn|≤10−12|r−cn|≤10−12. What is the motivation for including the |r||r| in the denominator on the left side of the inequality?
If someone could explain this to me, I would be very grateful!
Answered by
0
see above u will get it ✔️✔️✔️✔️
Attachments:
Similar questions