Could anyone answer me some questions on the method of bisection, please?

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In my math textbook the subjects are presented as a few examples (from 6 to 11 or 12 usually) and the more than $50$ exercises at the end. So one of those examples was about the method of bisection and I didn't understand well enough, I have some doubts. The example reads: Find a solution of the equation $x^5-x=3$ in $(0,2$) to within an accuracy of 0.1 by repeatedly dividing intervals in half and testing each half for a root.



And in the solution it solves the equation for 1 and 2. Did it solve for $1$ and $2$ because they're in $(0,2)$? Could it have picked a number greater than $2$?
I'm working on an exercise and I need to understand how this method works. Any help will be appreciated. Thank you.




calculus




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    In my math textbook the subjects are presented as a few examples (from 6 to 11 or 12 usually) and the more than $50$ exercises at the end. So one of those examples was about the method of bisection and I didn't understand well enough, I have some doubts. The example reads: Find a solution of the equation $x^5-x=3$ in $(0,2$) to within an accuracy of 0.1 by repeatedly dividing intervals in half and testing each half for a root.



    And in the solution it solves the equation for 1 and 2. Did it solve for $1$ and $2$ because they're in $(0,2)$? Could it have picked a number greater than $2$?
    I'm working on an exercise and I need to understand how this method works. Any help will be appreciated. Thank you.




    calculus




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      up vote
      0
      down vote

      favorite









      up vote
      0
      down vote

      favorite











      In my math textbook the subjects are presented as a few examples (from 6 to 11 or 12 usually) and the more than $50$ exercises at the end. So one of those examples was about the method of bisection and I didn't understand well enough, I have some doubts. The example reads: Find a solution of the equation $x^5-x=3$ in $(0,2$) to within an accuracy of 0.1 by repeatedly dividing intervals in half and testing each half for a root.



      And in the solution it solves the equation for 1 and 2. Did it solve for $1$ and $2$ because they're in $(0,2)$? Could it have picked a number greater than $2$?
      I'm working on an exercise and I need to understand how this method works. Any help will be appreciated. Thank you.




      calculus




      share|cite|improve this question













      In my math textbook the subjects are presented as a few examples (from 6 to 11 or 12 usually) and the more than $50$ exercises at the end. So one of those examples was about the method of bisection and I didn't understand well enough, I have some doubts. The example reads: Find a solution of the equation $x^5-x=3$ in $(0,2$) to within an accuracy of 0.1 by repeatedly dividing intervals in half and testing each half for a root.



      And in the solution it solves the equation for 1 and 2. Did it solve for $1$ and $2$ because they're in $(0,2)$? Could it have picked a number greater than $2$?
      I'm working on an exercise and I need to understand how this method works. Any help will be appreciated. Thank you.





      calculus





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      edited Jul 25 at 1:14









      Moo

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      asked Jul 25 at 0:57









      Kevin

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          You need to start with values on both sides, i.e. with $x_1$ and $x_2$ such that $x_1^5-x_1lt3$ and $x_2^5-x_2gt3$, so that you can then narrow in on the root. The values should also be in the interval in which you want to find a root; otherwise the bisection might lead to a root outside the interval (since it can lead to any value that's between your initial values).






          share|cite|improve this answer




























            up vote
            0
            down vote













            We have $x^5-x=3$



            The question has told you there is an $x$ bigger than $0$ and less than $2$ for which $x^5-x=3$. Your task is to find that root to a suitable level of accuracy.



            First plug in $x=0$. We get $0^5-0=0$



            Then plug in $x=2$. We get $2^5-2=30$



            Now plug in $x=1$, the midpoint of these two values. We get $1^5-1=0$.



            Note that the answer we want $(3)$ lies between the results of $x=1$ and $x=2$. So we shrink our interval to $(1,2)$.



            Now we plug in the midpoint of these two values, $x=1.5$. We get $1.5^5-1.5=6.09375$.



            The answer we want $(3)$ is between $0$ and $6.09375$, so we now shrink our interval to between $x=1$ and $x=1.5$



            Then use the midpoint of these $(x=1.25)$ and go from there...






            share|cite|improve this answer





















            • So you can't choose x bigger than 3, right?
              – Kevin
              Jul 25 at 1:16










            • The question specifies you should start with $0$ and $2$, but either way, $2^5-2>3$, and therefore $3^5-3>3$ also, so there is no point.
              – Rhys Hughes
              Jul 25 at 1:18

















            up vote
            0
            down vote













            If you have an equation, and an interval, and both of the following are fulfilled:



            1. Both sides of the equation are defined and continuous for all $x$ on that interval

            2. One side of the equation has greater value on one end of the interval, while the other side of the equation has greater value on the other end of the interval

            Then the equation has (at least) one solution in that interval. And this is the case for your equation and interval. We want to narrow down the space of possible locations of that solution.



            The idea of the bisection method is the following: the middle of your interval is $x=1$, so we test the equation there. Now there are three possible cases:



            1. The sides are equal. In that case we're done, as $x=1$ is a solution.

            2. The side which is larger at $x=1$ is the side which is smaller at $x=2$. In that case, by the same reason as above, there most be a solution somewhere in $(1,2)$. Repeat this process, bisecting at $x=1.5$.

            3. The side which is larger for $x=1$ is the side which is larger for $x=2$. In that case it is also the side which is smaller for $x=0$. By the same reason as above, there must be a solution somewhere in $(0,1)$. Repeat this process, bisecting at $x=0.5$.

            There is no real reason to pick $x=2$ rather than $x=0$, but for the argument to work correctly, you have to pick one of the end points of your interval.






            share|cite|improve this answer





















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              3 Answers
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              up vote
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              You need to start with values on both sides, i.e. with $x_1$ and $x_2$ such that $x_1^5-x_1lt3$ and $x_2^5-x_2gt3$, so that you can then narrow in on the root. The values should also be in the interval in which you want to find a root; otherwise the bisection might lead to a root outside the interval (since it can lead to any value that's between your initial values).






              share|cite|improve this answer

























                up vote
                0
                down vote













                You need to start with values on both sides, i.e. with $x_1$ and $x_2$ such that $x_1^5-x_1lt3$ and $x_2^5-x_2gt3$, so that you can then narrow in on the root. The values should also be in the interval in which you want to find a root; otherwise the bisection might lead to a root outside the interval (since it can lead to any value that's between your initial values).






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  You need to start with values on both sides, i.e. with $x_1$ and $x_2$ such that $x_1^5-x_1lt3$ and $x_2^5-x_2gt3$, so that you can then narrow in on the root. The values should also be in the interval in which you want to find a root; otherwise the bisection might lead to a root outside the interval (since it can lead to any value that's between your initial values).






                  share|cite|improve this answer













                  You need to start with values on both sides, i.e. with $x_1$ and $x_2$ such that $x_1^5-x_1lt3$ and $x_2^5-x_2gt3$, so that you can then narrow in on the root. The values should also be in the interval in which you want to find a root; otherwise the bisection might lead to a root outside the interval (since it can lead to any value that's between your initial values).







                  share|cite|improve this answer













                  share|cite|improve this answer



                  share|cite|improve this answer











                  answered Jul 25 at 1:08









                  joriki

                  164k10180328




                  164k10180328




















                      up vote
                      0
                      down vote













                      We have $x^5-x=3$



                      The question has told you there is an $x$ bigger than $0$ and less than $2$ for which $x^5-x=3$. Your task is to find that root to a suitable level of accuracy.



                      First plug in $x=0$. We get $0^5-0=0$



                      Then plug in $x=2$. We get $2^5-2=30$



                      Now plug in $x=1$, the midpoint of these two values. We get $1^5-1=0$.



                      Note that the answer we want $(3)$ lies between the results of $x=1$ and $x=2$. So we shrink our interval to $(1,2)$.



                      Now we plug in the midpoint of these two values, $x=1.5$. We get $1.5^5-1.5=6.09375$.



                      The answer we want $(3)$ is between $0$ and $6.09375$, so we now shrink our interval to between $x=1$ and $x=1.5$



                      Then use the midpoint of these $(x=1.25)$ and go from there...






                      share|cite|improve this answer





















                      • So you can't choose x bigger than 3, right?
                        – Kevin
                        Jul 25 at 1:16










                      • The question specifies you should start with $0$ and $2$, but either way, $2^5-2>3$, and therefore $3^5-3>3$ also, so there is no point.
                        – Rhys Hughes
                        Jul 25 at 1:18














                      up vote
                      0
                      down vote













                      We have $x^5-x=3$



                      The question has told you there is an $x$ bigger than $0$ and less than $2$ for which $x^5-x=3$. Your task is to find that root to a suitable level of accuracy.



                      First plug in $x=0$. We get $0^5-0=0$



                      Then plug in $x=2$. We get $2^5-2=30$



                      Now plug in $x=1$, the midpoint of these two values. We get $1^5-1=0$.



                      Note that the answer we want $(3)$ lies between the results of $x=1$ and $x=2$. So we shrink our interval to $(1,2)$.



                      Now we plug in the midpoint of these two values, $x=1.5$. We get $1.5^5-1.5=6.09375$.



                      The answer we want $(3)$ is between $0$ and $6.09375$, so we now shrink our interval to between $x=1$ and $x=1.5$



                      Then use the midpoint of these $(x=1.25)$ and go from there...






                      share|cite|improve this answer





















                      • So you can't choose x bigger than 3, right?
                        – Kevin
                        Jul 25 at 1:16










                      • The question specifies you should start with $0$ and $2$, but either way, $2^5-2>3$, and therefore $3^5-3>3$ also, so there is no point.
                        – Rhys Hughes
                        Jul 25 at 1:18












                      up vote
                      0
                      down vote










                      up vote
                      0
                      down vote









                      We have $x^5-x=3$



                      The question has told you there is an $x$ bigger than $0$ and less than $2$ for which $x^5-x=3$. Your task is to find that root to a suitable level of accuracy.



                      First plug in $x=0$. We get $0^5-0=0$



                      Then plug in $x=2$. We get $2^5-2=30$



                      Now plug in $x=1$, the midpoint of these two values. We get $1^5-1=0$.



                      Note that the answer we want $(3)$ lies between the results of $x=1$ and $x=2$. So we shrink our interval to $(1,2)$.



                      Now we plug in the midpoint of these two values, $x=1.5$. We get $1.5^5-1.5=6.09375$.



                      The answer we want $(3)$ is between $0$ and $6.09375$, so we now shrink our interval to between $x=1$ and $x=1.5$



                      Then use the midpoint of these $(x=1.25)$ and go from there...






                      share|cite|improve this answer













                      We have $x^5-x=3$



                      The question has told you there is an $x$ bigger than $0$ and less than $2$ for which $x^5-x=3$. Your task is to find that root to a suitable level of accuracy.



                      First plug in $x=0$. We get $0^5-0=0$



                      Then plug in $x=2$. We get $2^5-2=30$



                      Now plug in $x=1$, the midpoint of these two values. We get $1^5-1=0$.



                      Note that the answer we want $(3)$ lies between the results of $x=1$ and $x=2$. So we shrink our interval to $(1,2)$.



                      Now we plug in the midpoint of these two values, $x=1.5$. We get $1.5^5-1.5=6.09375$.



                      The answer we want $(3)$ is between $0$ and $6.09375$, so we now shrink our interval to between $x=1$ and $x=1.5$



                      Then use the midpoint of these $(x=1.25)$ and go from there...







                      share|cite|improve this answer













                      share|cite|improve this answer



                      share|cite|improve this answer











                      answered Jul 25 at 1:11









                      Rhys Hughes

                      3,8681227




                      3,8681227











                      • So you can't choose x bigger than 3, right?
                        – Kevin
                        Jul 25 at 1:16










                      • The question specifies you should start with $0$ and $2$, but either way, $2^5-2>3$, and therefore $3^5-3>3$ also, so there is no point.
                        – Rhys Hughes
                        Jul 25 at 1:18
















                      • So you can't choose x bigger than 3, right?
                        – Kevin
                        Jul 25 at 1:16










                      • The question specifies you should start with $0$ and $2$, but either way, $2^5-2>3$, and therefore $3^5-3>3$ also, so there is no point.
                        – Rhys Hughes
                        Jul 25 at 1:18















                      So you can't choose x bigger than 3, right?
                      – Kevin
                      Jul 25 at 1:16




                      So you can't choose x bigger than 3, right?
                      – Kevin
                      Jul 25 at 1:16












                      The question specifies you should start with $0$ and $2$, but either way, $2^5-2>3$, and therefore $3^5-3>3$ also, so there is no point.
                      – Rhys Hughes
                      Jul 25 at 1:18




                      The question specifies you should start with $0$ and $2$, but either way, $2^5-2>3$, and therefore $3^5-3>3$ also, so there is no point.
                      – Rhys Hughes
                      Jul 25 at 1:18










                      up vote
                      0
                      down vote













                      If you have an equation, and an interval, and both of the following are fulfilled:



                      1. Both sides of the equation are defined and continuous for all $x$ on that interval

                      2. One side of the equation has greater value on one end of the interval, while the other side of the equation has greater value on the other end of the interval

                      Then the equation has (at least) one solution in that interval. And this is the case for your equation and interval. We want to narrow down the space of possible locations of that solution.



                      The idea of the bisection method is the following: the middle of your interval is $x=1$, so we test the equation there. Now there are three possible cases:



                      1. The sides are equal. In that case we're done, as $x=1$ is a solution.

                      2. The side which is larger at $x=1$ is the side which is smaller at $x=2$. In that case, by the same reason as above, there most be a solution somewhere in $(1,2)$. Repeat this process, bisecting at $x=1.5$.

                      3. The side which is larger for $x=1$ is the side which is larger for $x=2$. In that case it is also the side which is smaller for $x=0$. By the same reason as above, there must be a solution somewhere in $(0,1)$. Repeat this process, bisecting at $x=0.5$.

                      There is no real reason to pick $x=2$ rather than $x=0$, but for the argument to work correctly, you have to pick one of the end points of your interval.






                      share|cite|improve this answer

























                        up vote
                        0
                        down vote













                        If you have an equation, and an interval, and both of the following are fulfilled:



                        1. Both sides of the equation are defined and continuous for all $x$ on that interval

                        2. One side of the equation has greater value on one end of the interval, while the other side of the equation has greater value on the other end of the interval

                        Then the equation has (at least) one solution in that interval. And this is the case for your equation and interval. We want to narrow down the space of possible locations of that solution.



                        The idea of the bisection method is the following: the middle of your interval is $x=1$, so we test the equation there. Now there are three possible cases:



                        1. The sides are equal. In that case we're done, as $x=1$ is a solution.

                        2. The side which is larger at $x=1$ is the side which is smaller at $x=2$. In that case, by the same reason as above, there most be a solution somewhere in $(1,2)$. Repeat this process, bisecting at $x=1.5$.

                        3. The side which is larger for $x=1$ is the side which is larger for $x=2$. In that case it is also the side which is smaller for $x=0$. By the same reason as above, there must be a solution somewhere in $(0,1)$. Repeat this process, bisecting at $x=0.5$.

                        There is no real reason to pick $x=2$ rather than $x=0$, but for the argument to work correctly, you have to pick one of the end points of your interval.






                        share|cite|improve this answer























                          up vote
                          0
                          down vote










                          up vote
                          0
                          down vote









                          If you have an equation, and an interval, and both of the following are fulfilled:



                          1. Both sides of the equation are defined and continuous for all $x$ on that interval

                          2. One side of the equation has greater value on one end of the interval, while the other side of the equation has greater value on the other end of the interval

                          Then the equation has (at least) one solution in that interval. And this is the case for your equation and interval. We want to narrow down the space of possible locations of that solution.



                          The idea of the bisection method is the following: the middle of your interval is $x=1$, so we test the equation there. Now there are three possible cases:



                          1. The sides are equal. In that case we're done, as $x=1$ is a solution.

                          2. The side which is larger at $x=1$ is the side which is smaller at $x=2$. In that case, by the same reason as above, there most be a solution somewhere in $(1,2)$. Repeat this process, bisecting at $x=1.5$.

                          3. The side which is larger for $x=1$ is the side which is larger for $x=2$. In that case it is also the side which is smaller for $x=0$. By the same reason as above, there must be a solution somewhere in $(0,1)$. Repeat this process, bisecting at $x=0.5$.

                          There is no real reason to pick $x=2$ rather than $x=0$, but for the argument to work correctly, you have to pick one of the end points of your interval.






                          share|cite|improve this answer













                          If you have an equation, and an interval, and both of the following are fulfilled:



                          1. Both sides of the equation are defined and continuous for all $x$ on that interval

                          2. One side of the equation has greater value on one end of the interval, while the other side of the equation has greater value on the other end of the interval

                          Then the equation has (at least) one solution in that interval. And this is the case for your equation and interval. We want to narrow down the space of possible locations of that solution.



                          The idea of the bisection method is the following: the middle of your interval is $x=1$, so we test the equation there. Now there are three possible cases:



                          1. The sides are equal. In that case we're done, as $x=1$ is a solution.

                          2. The side which is larger at $x=1$ is the side which is smaller at $x=2$. In that case, by the same reason as above, there most be a solution somewhere in $(1,2)$. Repeat this process, bisecting at $x=1.5$.

                          3. The side which is larger for $x=1$ is the side which is larger for $x=2$. In that case it is also the side which is smaller for $x=0$. By the same reason as above, there must be a solution somewhere in $(0,1)$. Repeat this process, bisecting at $x=0.5$.

                          There is no real reason to pick $x=2$ rather than $x=0$, but for the argument to work correctly, you have to pick one of the end points of your interval.







                          share|cite|improve this answer













                          share|cite|improve this answer



                          share|cite|improve this answer











                          answered Jul 25 at 1:13









                          Arthur

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