Question

how to estimate the number of iterations needed in bisection method in order to have the answer correct to given number of decimal places?

Answer 1

hi do you have books on this subject

Answer 2

u mean ebuks?

Answer 3

no, school books

Answer 4

yep. but it dosent xplain much abt da above ques. bt i gotta do it 4 ny assignment

Answer 5

are you in algebra or in numerical analysis?

Answer 6

numerical analysis

Answer 7

ah ok,,,,,

Answer 8

can u hrlp me????

Answer 9

[Bisection Method

Bisection Method Algorithm:

Input \epsilon (i.e level of accuracy)
Set a, b, (The interval)
Compute midpoint=\frac{a+b}{2}
If f(midpoint)=0, midpoint=root
If f(midpoint)>0 (i.e function is positive at the midpoint), replace midpoint with the endpoint where function was positive
If f(midpoint)<0 (i.e negative), replace midpoint with endpoint where function was negative
If (b-a)>\epsilon (if root is not found and level of desired accuracy is not reached, repeat 3-6)

Answer 10

The equation


t3 + 4 t2 - 1 = 0

has a positive root r in [0,1]. f(0)<0 and f(1)>0.
Since f(0.5) = 0.125 the root is in [0, 0.5].
Since f(0.25) = -0.73 the root is in [0.25, 0.5].
Since f(0.375) = -0.38 the root is in [0.375, 0.5].
Since f(0.4375) = -0.15 the root is in [0.4375, 0.5].
Since f(0.46875) = -0.018 the root is in [0.46875, 0.5].
Since f(0.484375) = 0.05 the root is in [0.46875, 0.484375].
... and so we approach the root 0.472834

Answer 11

how can i knw da number of iterations needed b4 solving it?

Answer 12

take a good guess first

Answer 13

if f(c) is positive on your first guess, then on your 2nd guess changes sign then the root is in between

Answer 14

do you have a sample problem there?

Answer 15

yep wait i'l post it here

Answer 16

use your calculator,,ok post them

Answer 17

It is required to find \[\sqrt[3]{2}\] to five decimal places, by solving the eqn \[x^3-2=0\] by means of the method of bisection. Estimate the number of iterations that will be required, and show this is so by performing the iterations.

Answer 18

f(2)=6
f(1)==1
here you know its between 1 and 2, so (1+2)/2=1.5 now try 1.5

Answer 19

f(1.5)=1.375 so (1+1.5)/2=1.25

Answer 20

yeah i get it. but it says 'estimate the number of iterations needed' ??????

Answer 21

yes count your number of iteration till you get closer to the root

Answer 22

:( didnt get it?

Answer 23

The method selects the subinterval that is a bracket as a new interval to be used in the next step. In this way the interval that contains a zero of f is reduced in width by 50% at each step. The process is continued until the interval is sufficiently small.

Answer 24

The method is guaranteed to converge to a root of f if f is a continuous function on the interval [a, b] and f(a) and f(b) have opposite signs. The absolute error is halved at each step so the method converges linearly, which is comparatively slow.

Specifically, if p1 = (a+b)/2 is the midpoint of the initial interval, and pn is the midpoint of the interval in the nth step, then the difference between pn and a solution p is bounded by

Answer 25

absolute value of[Pn-P]< or =abs[b-a]/2^n

Answer 26

That formula can be used to determine in advance the number of iterations that the bisection method would need to converge to a root to within a certain tolerance.

Answer 27

thnks a lot..

Answer 28

is it clearer now?

Answer 29

yep.. i was wandering how to find da number of iteratins b4 solvng.. my txt buk dosent mention abt it

Answer 30

Initialization: The bisection method is initialized by specifying the function f(x), the interval [a,b], and the tolerance > 0.

We also check whether f(a) = 0 or f(b) = 0, and if so return the value of a or b and exit.

Loop: Let m = (a + b)/2 be the midpoint of the interval [a,b]. Compute the signs of f(a), f(m), and f(b).

If any are zero, return the corresponding point and exit.

Assuming none are zero, if f(a) and f(m) have opposite sides, replace b by m, else replace a by m.

If the length of the [a,b] is less than , return the value of a and exit.

Analysis: When we enter the loop f(a) and f(b) have opposite sign. It follows that either f(m) and f(a) have opposite sign or f(m) and f(b) have opposite sign. Thus the initial conditions are still satisfied each time we enter the loop.

The length of the initial interval is (b - a). After one time through the loop the length is (b - a)/2, after two times it is (b - a)/4, and after n passes through the loop, the length of the remaining interval is (b - a)/2n. No matter how small , eventually (b - a)/2^n <E . In fact we can solve this inequality for n:

(b - a)/2^n <E
2^n >(b - a)/E
n ln 2 >ln(b - a) - ln(E)
n>[ln(b - a) - ln(E)]/ln 2.

Thus the algorithm terminates after at most M passes through the loop where M is the first integer larger than [ln(b - a) - ln(E)]/ln 2.

Answer 31

Starting with the interval [1,2], find sqrt(2) to within two decimal places (to within an error of .01).

Answer 32

try that for yourself for practice

Answer 33

solve f(x) = x^2 -2.

Answer 34

the final result is the approximation 1.41406 for the sqrt(2).

Answer 35

wait il do it n post da answer here

Answer 36

make a table that have like this:

a b b-a m=(a+b)/2 f(a) f(b) f(m)

Answer 37

answer it like this:
a b b-a m=(a+b)/2 f(a) f(b) f(m)
1) 1 2 1 1.5 -1 2 25
2)
3)
5)
6)
7)
8)1.41406 1.42187 .0078125 -.0004343 .0217285

Answer 38

bro i gt 2 go nw.. can gve me ur email so i cn mail it 2 u? or shall i post it here when im back?

Answer 39

ok

Answer 40

or you can post em if you want

Answer 41

ok..thnkx

Answer 42

yw good luck now