Numerical Analysis Bisection Method Problem

**wvlilgurl** · January 18th 2009, 12:18 PM

Use the bisection methos to find solutions accurate to within 10^-2 for x^3-7x^2+14x-6=0 for each interval [0,1],[1,3.2],[3.2,4].

I don't know how to find the number of iterations necessary for each interval. Thanks!

**Constatine11** · January 18th 2009, 12:28 PM

Originally Posted by wvlilgurl

Use the bisection methos to find solutions accurate to within 10^-2 for x^3-7x^2+14x-6=0 for each interval [0,1],[1,3.2],[3.2,4].

I don't know how to find the number of iterations necessary for each interval. Thanks!

The interval length is halved at every iteration so after n steps the inteval length is $a\times 2^{-n}$ where $a$ is the initial interval length.

So if we want the $n$ -th interval to be less than or equal $10^{-2}$ we need:

$10^{-2}\ge a\times 2^{-n}$

.

**wvlilgurl** · January 19th 2009, 11:12 AM

for the interval [0,1] they get 7
for the interval [1,3.2} they get 8

I got the first one at 7 but I cannot get 8 for the second one. DO you knwo how they got 8?

**Constatine11** · January 19th 2009, 11:32 AM

Originally Posted by wvlilgurl

for the interval [0,1] they get 7
for the interval [1,3.2} they get 8

I got the first one at 7 but I cannot get 8 for the second one. DO you knwo how they got 8?

In case 2 we have $a=2.2$ , so we seek an integer n such that:

$10^{-2} \ge 2.2\times 2^{-n}$

which we can use trial and error to find that the smallest integer $n$ such that this holds is $8$

.

**wvlilgurl** · January 19th 2009, 11:39 AM

still dont get it. I get 6

**Constatine11** · January 19th 2009, 11:46 AM

Originally Posted by Constatine11

In case 2 we have $a=2.2$ , so we seek an integer n such that:

$10^{-2} \ge 2.2\times 2^{-n}$

which we can use trial and error to find that the smallest integer $n$ such that this holds is $8$

.

Put $n=7,\ 2^{-7}\approx 0.0078$ so $2.2 \times 2^{-7}\approx 0.017>0.01$

Put $n=8,\ 2^{-8}\approx 0.0039$ so $2.2 \times 2^{-8}\approx0.0086 <0.01$
.

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