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# Change of sign method --- interval bisection method

Extracts from this document...

Introduction

Winnie Zheng

Pure Mathematics 2 Coursework

Change of sign method --- interval bisection method

Introduction:

When I am looking for the roots of the equation, actually what I want is the values of x for which the graph y=f(x) crosses the x-axis. As the graph f(x) crosses the x-axis, there is going to be a sign change of y value on the two sides of the root. Hence provided the function gives a continuous graph, if there is a sign change in a located interval, I will know that the interval contains one root.

For interval bisection method, what I am actually going to do is to divide the interval into two parts, then take the half which contains sign changes of y.

Y=(x-2)(x-4)(x-6)+1 From the graph we can see that there are three roots lying in the interval (1,2); (4,5); (5,6). I am going to focus on the root lying in the interval (1,2) I will show the steps on a spreadsheet:

 a b (a+b)/2 f(x) error 1 1.00000000 2.00000000 1.50000000 -4.62500000 0.50000000 2 1.50000000 2.00000000 1.75000000 -1.39062500 0.25000000 3 1.75000000 2.00000000 1.87500000 -0.09570313 0.12500000 4 1.87500000 2.00000000 1.93750000 0.47631836 0.06250000 5 1.87500000 1.93750000 1.90625000 0.19644165 0.03125000 6 1.87500000 1.90625000 1.89062500 0.05191422 0.01562500 7 1.87500000 1.89062500 1.88281250 -0.02150679 0.00781250 8 1.88281250 1.89062500 1.88671875 0.01530045 0.00390625 9 1.88281250 1.88671875 1.88476563 -0.00307896 0.00195313 10 1.88476563 1.88671875 1.88574219 0.00611680 0.00097656 11 1.88476563 1.88574219 1.88525391 0.00152043 0.00048828 12 1.88476563 1.88525391 1.88500977 -0.00077889 0.00024414 13 1.88500977 1.88525391 1.88513184 0.00037087 0.00012207 14 1.88500977 1.88513184 1.88507080 -0.00020399 0.00006104 15 1.88507080 1.88513184 1.88510132 0.00008345 0.00003052 16 1.88507080 1.88510132 1.88508606 -0.00006027 0.00001526 17 1.88508606 1.88510132 1.88509369 0.00001159 0.00000763 18 1.88508606 1.88509369 1.88508987 -0.00002434 0.00000381

The formulae I used in the spreadsheet are shown below:

 a b (a+b)/2 f(x) error 1 1 2 =1/2*(B2+C2) =(D2-2)*(D2-4)*(D2-6)+1 =1/2*(C2-B2) 2 =IF(E2<0,D2,B2) =IF(E2<0,C2,D2) =1/2*(B3+C3) =(D3-2)*(D3-4)*(D3-6)+1 =1/2*(C3-B3) 3 =IF(E3<0,D3,B3) =IF(E3<0,C3,D3) =1/2*(B4+C4) =(D4-2)*(D4-4)*(D4-6)+1 =1/2*(C4-B4) 4 =IF(E4<0,D4,B4) =IF(E4<0,C4,D4) =1/2*(B5+C5) =(D5-2)*(D5-4)*(D5-6)+1 =1/2*(C5-B5) 5 =IF(E5<0,D5,B5) =IF(E5<0,C5,D5) =1/2*(B6+C6) =(D6-2)*(D6-4)*(D6-6)+1 =1/2*(C6-B6) 6 =IF(E6<0,D6,B6) =IF(E6<0,C6,D6) =1/2*(B7+C7) =(D7-2)*(D7-4)*(D7-6)+1 =1/2*(C7-B7) 7 =IF(E7<0,D7,B7) =IF(E7<0,C7,D7) =1/2*(B8+C8) =(D8-2)*(D8-4)*(D8-6)+1 =1/2*(C8-B8) 8 =IF(E8<0,D8,B8) =IF(E8<0,C8,D8) =1/2*(B9+C9) =(D9-2)*(D9-4)*(D9-6)+1 =1/2*(C9-B9) 9 =IF(E9<0,D9,B9) =IF(E9<0,C9,D9) =1/2*(B10+C10) =(D10-2)*(D10-4)*(D10-6)+1 =1/2*(C10-B10) 10 =IF(E10<0,D10,B10) =IF(E10<0,C10,D10) =1/2*(B11+C11) =(D11-2)*(D11-4)*(D11-6)+1 =1/2*(C11-B11) 11 =IF(E11<0,D11,B11) =IF(E11<0,C11,D11) =1/2*(B12+C12) =(D12-2)*(D12-4)*(D12-6)+1 =1/2*(C12-B12) 12 =IF(E12<0,D12,B12) =IF(E12<0,C12,D12) =1/2*(B13+C13) =(D13-2)*(D13-4)*(D13-6)+1 =1/2*(C13-B13) 13 =IF(E13<0,D13,B13) =IF(E13<0,C13,D13) =1/2*(B14+C14) =(D14-2)*(D14-4)*(D14-6)+1 =1/2*(C14-B14) 14 =IF(E14<0,D14,B14) =IF(E14<0,C14,D14) =1/2*(B15+C15) =(D15-2)*(D15-4)*(D15-6)+1 =1/2*(C15-B15) 15 =IF(E15<0,D15,B15) =IF(E15<0,C15,D15) =1/2*(B16+C16) =(D16-2)*(D16-4)*(D16-6)+1 =1/2*(C16-B16) 16 =IF(E16<0,D16,B16) =IF(E16<0,C16,D16) =1/2*(B17+C17) =(D17-2)*(D17-4)*(D17-6)+1 =1/2*(C17-B17) 17 =IF(E17<0,D17,B17) =IF(E17<0,C17,D17) =1/2*(B18+C18) =(D18-2)*(D18-4)*(D18-6)+1 =1/2*(C18-B18) 18 =IF(E18<0,D18,B18) =IF(E18<0,C18,D18) =1/2*(B19+C19) =(D19-2)*(D19-4)*(D19-6)+1 =1/2*(C19-B19)

Middle

0.000000

-11.937282

-0.916521

0.000000

-11.937282

The formula I used in the spreadsheet are shown below:

 x f(x) f'(x) -1 =3*(B3-1)*(B3+1)*(B3+3)+1 =9*B3^2+18*B3-3 =B3-C3/D3 =3*(B4-1)*(B4+1)*(B4+3)+1 =9*B4^2+18*B4-3 =B4-C4/D4 =3*(B5-1)*(B5+1)*(B5+3)+1 =9*B5^2+18*B5-3 =B5-C5/D5 =3*(B6-1)*(B6+1)*(B6+3)+1 =9*B6^2+18*B6-3

From the results I get from the spreadsheet, I can see the root lying between the interval (-1,0) is -0.916521.

2) For the root lying between the interval (0,1):

 x f(x) f'(x) 1.000000 1.000000 24.000000 0.958333 0.031033 22.515625 0.956955 0.000033 22.467057 0.956954 0.000000 22.467005

The formula I used in the spread sheet are shown below:

 x f(x) f'(x) 1 =3*(B3-1)*(B3+1)*(B3+3)+1 =9*B3^2+18*B3-3 =B3-C3/D3 =3*(B4-1)*(B4+1)*(B4+3)+1 =9*B4^2+18*B4-3 =B4-C4/D4 =3*(B5-1)*(B5+1)*(B5+3)+1 =9*B5^2+18*B5-3 =B5-C5/D5 =3*(B6-1)*(B6+1)*(B6+3)+1 =9*B6^2+18*B6-3

From the results I get from the spreadsheet, I can see the root lying between the interval (0,1) is 0.956954.

Problems with Newton-Raphson method:

There are some cases in which the Newton-Raphson method does not work:

1. The function is discontinuous, for example:

y=-1/x +2 If the graph is discontinuous, and unfortunately I’ve chosen a starting point near the edge of one part of the graph, using the Newton-Raphson method, the next guess will go to the other side of the graph, getting further and further away from the root we are looking for.

2) Poor choice of starting value. If my initial value is quite far way from a root, the iteration may be divergent, i.e. moving away from the root. For example:

y= x(x-1)(x-2)+1 Using Newton-Raphson method with the first guess x=0.5, from the graph below we can see that actually our second guess is getting further away from the chosen root, so this method failed. 1. if my initial value is near a turning point of y=f(x), the iteration will diverge as well. For example:

y=(x-2)(x-1)x+1 For this equation, there is a station point of maximum at around x=0.5; if my first guess is 0.4 then the gradient of the tangent at that point is actually nearly zero, i.e. horizontal to the x-axis, hence out next guess will be very far from out first guess and the root we are looking for. If out first guess is actually at the turning point, the tangent will be horizontal to the x-axis, it will never cut the x-axis by any chance. In both of the cases, the Newton-Raphson method fails.

Rearranging f(x)=0 in the form x=g(x)

Introduction:

This method is actually finding a single value or point of an estimate value of the root rather than identifying the interval in which the root lies.

The method involves rearranging the function f(x)=0 in the form x=g(x). Drawing the graphs y=x and y=g(x) on the same graph: The root of the function f(x)=0 is indeed the intersection of the two lines.

The iteration I am going to use in this method is actually based on:

xn+1=g(xn)

Presenting the iteration graphically, we will have either a ‘staircase’ or a ‘cobweb’ diagram.

For the equation:  f(x)=y=x³+6x²+9x+1 rearrange f(x)=0, I get x=-(x³+6x²+1)/9

drawing y=x and g(x)=y=-(x³+6x²+1)/9 on the same graph, I get the graph below: By using the rearranging f(x)=0 into x=g(x) method, it provides the iterative formula xn+1=(-xn³-6xn²-1)/9.  It is shown in the purple line in the graph below. I chose x=-4 as my starting point, find the corresponding value of g(-4). Next, we take this value g(-4) as the next guess, this means that x2=g(-4) and move the point horizontally towards the right to the line y=x and find the value of g(x2), this process continues until both x and g(x) are the same value.

Showing the method on a spreadsheet:

 n x g(x) 1 -4.0000000000 -3.6666666667 2 -3.6666666667 -3.5967078189 3 -3.5967078189 -3.5655250877 4 -3.5655250877 -3.5499338288 5 -3.5499338288 -3.5417564282 6 -3.5417564282 -3.5373669161 7 -3.5373669161 -3.5349823162 8 -3.5349823162 -3.5336786021 9 -3.5336786021 -3.5329633716 10 -3.5329633716 -3.5325702506 11 -3.5325702506 -3.5323539521 12 -3.5323539521 -3.5322348755 13 -3.5322348755 -3.5321693010 14 -3.5321693010 -3.5321331835 15 -3.5321331835 -3.5321132887 16 -3.5321132887 -3.5321023293 17 -3.5321023293 -3.5320962919 18 -3.5320962919 -3.5320929660 18 -3.5320929660 -3.5320911338

The formula I used in the spreadsheet are as follows:

 n x g(x) 1 -4 =-(B2^3+6*B2^2+1)/9 2 =C2 =-(B3^3+6*B3^2+1)/9 3 =C3 =-(B4^3+6*B4^2+1)/9 4 =C4 =-(B5^3+6*B5^2+1)/9 5 =C5 =-(B6^3+6*B6^2+1)/9 6 =C6 =-(B7^3+6*B7^2+1)/9 7 =C7 =-(B8^3+6*B8^2+1)/9 8 =C8 =-(B9^3+6*B9^2+1)/9 9 =C9 =-(B10^3+6*B10^2+1)/9 10 =C10 =-(B11^3+6*B11^2+1)/9 11 =C11 =-(B12^3+6*B12^2+1)/9 12 =C12 =-(B13^3+6*B13^2+1)/9 13 =C13 =-(B14^3+6*B14^2+1)/9 14 =C14 =-(B15^3+6*B15^2+1)/9 15 =C15 =-(B16^3+6*B16^2+1)/9 16 =C16 =-(B17^3+6*B17^2+1)/9 17 =C17 =-(B18^3+6*B18^2+1)/9 18 =C18 =-(B19^3+6*B19^2+1)/9 18 =C19 =-(B20^3+6*B20^2+1)/9

Conclusion

=B5^3-12*B5^2+44*B5-47

=3*B5^2-24*B5+44

5

=B5-C5/D5

=B6^3-12*B6^2+44*B6-47

=3*B6^2-24*B6+44

6

=B6-C6/D6

=B7^3-12*B7^2+44*B7-47

=3*B7^2-24*B7+44

From the table I get the root of f(x)=0 is x=1.88509246

1. using the rearranging f(x)=0 in the form x=g(x) method;

for this method, I am going to rearrange f(x) = x3-12x2+44x-47=0 in the form x=(-x3+12x2+47)/44, i.e. g(x)= (-x3+12x2+47)/44

draw y=x and y=g(x)= (-x3+12x2+47)/44 on the same graph, I get the following: using the iteration xn+1=g(xn), doing on the spreadsheet with a starting point x=3 I got the following table:

 n x g(x) 1 3 2.909090909 2 2.909090909 2.816696264 3 2.816696264 2.724052084 4 2.724052084 2.632540869 5 2.632540869 2.543614417 6 2.543614417 2.458694812 7 2.458694812 2.379066113 8 2.379066113 2.305774054 9 2.305774054 2.23955144 10 2.23955144 2.180782149 11 2.180782149 2.129507776 12 2.129507776 2.085471395 13 2.085471395 2.048185863 14 2.048185863 2.017011601 15 2.017011601 1.991230661 16 1.991230661 1.970108315 17 1.970108315 1.952938344 18 1.952938344 1.939072123 19 1.939072123 1.927933996 20 1.927933996 1.919026346 21 1.919026346 1.911927722 22 1.911927722 1.906286852 23 1.906286852 1.901814605 24 1.901814605 1.89827533 25 1.89827533 1.895478454 26 1.895478454 1.89327079 27 1.89327079 1.891529794 28 1.891529794 1.890157808 29 1.890157808 1.889077232 30 1.889077232 1.888226552 31 1.888226552 1.887557093 32 1.887557093 1.887030397 33 1.887030397 1.886616109 34 1.886616109 1.886290296 35 1.886290296 1.886034098 36 1.886034098 1.885832661 37 1.885832661 1.885674295 38 1.885674295 1.885549798 39 1.885549798 1.885451931 40 1.885451931 1.885375002 41 1.885375002 1.885314533 42 1.885314533 1.885267004 43 1.885267004 1.885229646 44 1.885229646 1.885200282

The formulae I used in the spreadsheet are as follows:

 n x g(x) 1 3 =(-B2^3+12*B2^2+47)/44 2 =C2 =(-B3^3+12*B3^2+47)/44 3 =C3 =(-B4^3+12*B4^2+47)/44 4 =C4 =(-B5^3+12*B5^2+47)/44 5 =C5 =(-B6^3+12*B6^2+47)/44 6 =C6 =(-B7^3+12*B7^2+47)/44 7 =C7 =(-B8^3+12*B8^2+47)/44 8 =C8 =(-B9^3+12*B9^2+47)/44 9 =C9 =(-B10^3+12*B10^2+47)/44 10 =C10 =(-B11^3+12*B11^2+47)/44 11 =C11 =(-B12^3+12*B12^2+47)/44 12 =C12 =(-B13^3+12*B13^2+47)/44 13 =C13 =(-B14^3+12*B14^2+47)/44 14 =C14 =(-B15^3+12*B15^2+47)/44 15 =C15 =(-B16^3+12*B16^2+47)/44 16 =C16 =(-B17^3+12*B17^2+47)/44 17 =C17 =(-B18^3+12*B18^2+47)/44 18 =C18 =(-B19^3+12*B19^2+47)/44 19 =C19 =(-B20^3+12*B20^2+47)/44 20 =C20 =(-B21^3+12*B21^2+47)/44 21 =C21 =(-B22^3+12*B22^2+47)/44 22 =C22 =(-B23^3+12*B23^2+47)/44 23 =C23 =(-B24^3+12*B24^2+47)/44 24 =C24 =(-B25^3+12*B25^2+47)/44 25 =C25 =(-B26^3+12*B26^2+47)/44 26 =C26 =(-B27^3+12*B27^2+47)/44 27 =C27 =(-B28^3+12*B28^2+47)/44 28 =C28 =(-B29^3+12*B29^2+47)/44 29 =C29 =(-B30^3+12*B30^2+47)/44 30 =C30 =(-B31^3+12*B31^2+47)/44 31 =C31 =(-B32^3+12*B32^2+47)/44 32 =C32 =(-B33^3+12*B33^2+47)/44 33 =C33 =(-B34^3+12*B34^2+47)/44 34 =C34 =(-B35^3+12*B35^2+47)/44 35 =C35 =(-B36^3+12*B36^2+47)/44 36 =C36 =(-B37^3+12*B37^2+47)/44 37 =C37 =(-B38^3+12*B38^2+47)/44 38 =C38 =(-B39^3+12*B39^2+47)/44 39 =C39 =(-B40^3+12*B40^2+47)/44 40 =C40 =(-B41^3+12*B41^2+47)/44 41 =C41 =(-B42^3+12*B42^2+47)/44 42 =C42 =(-B43^3+12*B43^2+47)/44 43 =C43 =(-B44^3+12*B44^2+47)/44 44 =C44 =(-B45^3+12*B45^2+47)/44

According to the spreadsheet above, I got the root of f(x) =0 x=1.885229646

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