Bisection Method for Approximating Zeros of a Function f We...

Bisection Method for Approximating Zeros of a Function $f$ We begin with two consecutive integers, $a$ and $a+1,$ such that $f(a)$ and $f(a+1)$ are of opposite sign. Evaluate $f$ at the midpoint $m_{1}$ of $a$ and $a+1 .$ If $f\left(m_{1}\right)=0,$ then $m_{1}$ is the zero of $f,$ and we are finished. Otherwise, $f\left(m_{1}\right)$ is of opposite sign to either $f(a)$ or $f(a+1) .$ Suppose that it is $f(a)$ and $f\left(m_{1}\right)$ that are of opposite sign. Now evaluate $f$ at the midpoint $m_{2}$ of $a$ and $m_{1} .$ Repeat this process until the desired degree of accuracy is obtained. Note that each iteration places the zero in an interval whose length is half that of the previous interval. Use the bisection method to approximate the zero of $f(x)=8 x^{4}-2 x^{2}+5 x-1$ in the interval [0,1] correct to three decimal places. [Hint: The process ends when both endpoints agree to the desired number of decimal places.

Transcript

00:01 We are going to use the byte section method to approximate the 0 of the function of 8 x to the 4th minus 2x square plus 5x minus 1 in the interval 0 1.

00:18 And we want the answer correct to 3 decimal places.

00:24 So if you graph this function over the interval 01, we verify that there is only 1 root.

00:29 In fact, because f is a polynomial function, so it is continuous in.

00:38 The real numbers so in particular in the interval 0 1 so we can ensure that there exists to root there if the function change signs at the endpoints so it is continuous in or on okay let me write this better it is continuous on 01 and on the other hand we have that f at 0 is negative 1 which is negative and f at 1 that is evaluating the function at the end points of the interval at 1 is 8 minus 2 plus 5 minus 1 and that is 6 11 10 which is positive so the function changes sign on the end points and is continuous then f has at least the 0 on the interval 1 a list a zero in zero one if we graph the function we can verify there is only one zero graphically we can verify that there is only one zero of f in zero one okay so we know there is only one zero there and we apply the we want to apply by section method so we calculate the midpoint so the intervals ab through the method let's say here are we going to apply the method applying a section in the bisection method you find the sub intervals that is the interval ab is changing is redefined at each iteration so i'm going to write some of them we start with ab, which is the integral 01, and when we do the first iteration, i found that the function changes sign at or on the interval 0 .5.

04:53 That is, at 0, the function is negative, as we saw here.

04:58 And if you evaluate the function at 0 .5, you will see that is positive.

05:03 So a remain the same, but b is redefined at 0 .5.

05:09 That's the first iteration.

05:10 Second, iteration we find that now at 0 .5 at 0 is negative and at 0 .25 which is a midpoint is positive so a stays being equal to 0 and at 0 .25 which is a midpoint between 0 and 0 .5 the function is positive so change of side is between 0 and 0 .25 then that's the second iteration and now third we get a bit point of zero zero point five is 0 .1 to 5 and now at 0 .1 to 5 the sign is negative so we get 0 .1 to 5 and sorry 0 .25 that is when we calculate the midpoint of this interval here which is 0 .125 at that point 0 .125 the function is negative so we get to redefine for the first time the value of a and now b is the same because there the function is positive and if we look at another iteration we get 0 .1875 point of this interval over here and 0 .25.

07:12 That is keep on being positive at 0 .25 and it's negative at the midpoint of this new interval.

07:19 And you can continue this way.

07:24 And then you find the iteration where the endpoints, a and b, have the same three decimals.

07:34 That is what we want...

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