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# Different methods of solving equations compared. From the Excel tables of each method, we know that method 1 (change of sign method) takes 28 steps to find the root, while method 2 (Newton-Raphson method) and method 3 (rearrangement) take 4 and 17 steps r

Extracts from this document...

Introduction

Numerical Methods

Method 1: Change of sign method

Equation 1: → Use the Autograph and it gives us the overall view of the graph of the function. If we zoom in on the x and y axes, we can see that the graph shows that the equation has a root between x=0 and x=1. Bisect the interval between 0 and 1, then bisect the new interval where the sign changes and we can see that the two boundary lines get closer and closer to each other and will meet at a point finally (in theory). However, in practice, we are not able to find out the exact values of the roots but we can still use this method to find the approximate values. Excel can help us with that.

As shown in the table below by inputting initial values of a=0, b=1, we can calculate that f(a)>0, f(b)<0. Then divide the interval in half and get the midpoint c, where c= =0.5, and also the value of f(c). Since f(c)>0, we then can get a subinterval between c and b. The previous steps are then repeated. This method is applied recursively to the subinterval where the sign change occurs.

The table below shows the data with accuracy of 8 decimal places.

 a f(a) b f(b) c f( c) Max. error 0.00000000 2.60000000 1.00000000 -0.80000000 0.50000000 1.00000000 1.70000000 0.50000000 1.00000000 1.00000000 -0.80000000 0.75000000 0.10625000 0.90000000 0.75000000 0.10625000 1.00000000 -0.80000000 0.87500000 -0.34765625 0.45312500 0.75000000 0.10625000 0.87500000 -0.34765625 0.81250000 -0.12060547 0.22695313 0.75000000 0.10625000 0.81250000 -0.12060547 0.78125000 -0.00711670 0.11342773 0.75000000 0.10625000 0.78125000 -0.00711670 0.76562500 0.04958649 0.05668335 0.76562500 0.04958649 0.78125000 -0.00711670 0.77343750 0.02123928 0.02835159 0.77343750 0.02123928 0.78125000 -0.00711670 0.77734375 0.00706232 0.01417799 0.77734375 0.00706232 0.78125000 -0.00711670 0.77929688 -0.00002694 0.00708951 0.77734375 0.00706232 0.77929688 -0.00002694 0.77832031 0.00351775 0.00354463 0.77832031 0.00351775 0.77929688 -0.00002694 0.77880859 0.00174542 0.00177235 0.77880859 0.00174542 0.77929688 -0.00002694 0.77905273 0.00085924 0.00088618 0.77905273 0.00085924 0.77929688 -0.00002694 0.77917480 0.00041615 0.00044309 0.77917480 0.00041615 0.77929688 -0.00002694 0.77923584 0.00019460 0.00022155 0.77923584 0.00019460 0.77929688 -0.00002694 0.77926636 0.00008383 0.00011077 0.77926636 0.00008383 0.77929688 -0.00002694 0.77928162 0.00002844 0.00005539 0.77928162 0.00002844 0.77929688 -0.00002694 0.77928925 0.00000075 0.00002769 0.77928925 0.00000075 0.77929688 -0.00002694 0.77929306 -0.00001310 0.00001385 0.77928925 0.00000075 0.77929306 -0.00001310 0.77929115 -0.00000617 0.00000692 0.77928925 0.00000075 0.77929115 -0.00000617 0.77929020 -0.00000271 0.00000346 0 .77928925 0.00000075 0.77929020 -0.00000271 0.77928972 -0.00000098 0.00000173 0.77928925 0.00000075 0.77928972 -0.00000098 0.77928948 -0.00000012 0.00000087 0.77928925 0.00000075 0.77928948 -0.00000012 0.77928936 0.00000032 0.00000043 0.77928936 0.00000032 0.77928948 -0.00000012 0.77928942 0.00000010 0.00000022 0.77928942 0.00000010 0.77928948 -0.00000012 0.77928945 -0.00000001 0.00000011 0.77928942 0.00000010 0.77928945 -0.00000001 0.77928944 0.00000005 0.00000005 0.77928944 0.00000005 0.77928945 -0.00000001 0.77928945 0.00000002 0.00000003 0.77928945 0.00000002 0.77928945 -0.00000001 0.77928945 0.00000001 0.00000001

Error bound: ±0.000000005 (9dp)

The table below shows part of the formulas used in the Excel.

 A f(a) b f(b) c f( c) 0 =0.2*(A2-4)*(A2+2)*(2*A2-1)+1 1 =0.2*(C2-4)*(C2+2)*(2*C2-1)+1 =(A2+C2)/2 =0.2*(E2-4)*(E2+2)*(2*E2-1)+1 =IF(F2>0,E2,A2) =0.2*(A3-4)*(A3+2)*(2*A3-1)+1 =IF(F2<0,E2,C2) =0.2*(C3-4)*(C3+2)*(2*C3-1)+1 =(A3+C3)/2 =0.2*(E3-4)*(E3+2)*(2*E3-1)+1 =IF(F3>0,E3,A3) =0.2*(A4-4)*(A4+2)*(2*A4-1)+1 =IF(F3<0,E3,C3) =0.2*(C4-4)*(C4+2)*(2*C4-1)+1 =(A4+C4)/2 =0.2*(E4-4)*(E4+2)*(2*E4-1)+1

Middle

0.68896484

-0.00016641

0.00025309

0.68896484

-0.00016641

0.68945313

0.00008753

0.68920898

-0.00003965

0.00012697

0.68920898

-0.00003965

0.68945313

0.00008753

0.68933105

0.00002389

0.00006359

0.68920898

-0.00003965

0.68933105

0.00002389

0.68927002

-0.00000790

0.00003177

0.68927002

-0.00000790

0.68933105

0.00002389

0.68930054

0.00000799

0.00001589

0.68927002

-0.00000790

0.68930054

0.00000799

0.68928528

0.00000005

0.00000794

0.68927002

-0.00000790

0.68928528

0.00000005

0.68927765

-0.00000393

0.00000397

0.68927765

-0.00000393

0.68928528

0.00000005

0.68928146

-0.00000194

0.00000199

0.68928146

-0.00000194

0.68928528

0.00000005

0.68928337

-0.00000095

0.00000099

0.68928337

-0.00000095

0.68928528

0.00000005

0.68928432

-0.00000045

0.00000050

0.68928432

-0.00000045

0.68928528

0.00000005

0.68928480

-0.00000020

0.00000025

0.68928480

-0.00000020

0.68928528

0.00000005

0.68928504

-0.00000008

0.00000012

0.68928504

-0.00000008

0.68928528

0.00000005

0.68928516

-0.00000002

0.00000006

0.68928516

-0.00000002

0.68928528

0.00000005

0.68928522

0.00000002

0.00000003

0.68928516

-0.00000002

0.68928522

0.00000002

0.68928519

0.00000000

0.00000002

0.68928519

0.00000000

0.68928522

0.00000002

0.68928520

0.00000001

0.00000001

0.68928519

0.00000000

0.68928520

0.00000001

0.68928520

0.00000000

0.00000000

0.68928519

0.00000000

0.68928520

0.00000000

0.68928519

0.00000000

0.00000000

0.68928519

0.00000000

0.68928519

0.00000000

0.68928519

0.00000000

0.00000000

Since this method cannot find out all the roots, we say that it fails in this case. For the example above, it is because the three roots lie too close together. We usually ignore the other two roots when we find out one in the interval since we didn’t expect them all in such a small interval.

Method 2: Newton-Raphson method

Equation 2: → Here shows the overall view of the graph of the function. Zoom in on the axes we can clearly see that using the Newton-Raphson method gives us one root efficiently. How does the Newton-Raphson method actually work? The graph above shows a part of a function (the blue curve). Suppose we have an estimated value of a root, xn. Draw a tangent at where x=xn, which is shown in red, we can get another estimated root xn+1 which is a better approximation.

Since , we can deduce that .

With the help of Excel, we can get the approximate value of the root shown above within just a few steps.

The table below shows the data with accuracy of 8 decimal places.

 x f(x) f'(x) 0.00000000 -1.00000000 6.00000000 0.16666667 -0.13425926 4.41666667 0.19706499 -0.00413017 4.14585394 0.19806121 -0.00000437 4.13707266 0.19806226 0.00000000 4.13706334

And here’s the formulas used in the table:

 x f(x) f'(x) 0 =A2*(A2-2)*(A2-3)-1 =3*A2^2-10*A2+6 =A2-(B2/C2) =A3*(A3-2)*(A3-3)-1 =3*A3^2-10*A3+6 =A3-(B3/C3) =A4*(A4-2)*(A4-3)-1 =3*A4^2-10*A4+6 =A4-(B4/C4) =A5*(A5-2)*(A5-3)-1 =3*A5^2-10*A5+6 =A5-(B5/C5) =A6*(A6-2)*(A6-3)-1 =3*A6^2-10*A6+6

Similarly, starting with another two points, we can get the approximate values of the other two roots. The tables below show the data with accuracy of 8 decimal places as well.

 x f(x) f'(x) x f(x) f'(x) 2.00000000 -1.00000000 -2.00000000 3.00000000 -1.00000000 3.00000000 1.50000000 0.12500000 -2.25000000 3.33333333 0.48148148 6.00000000 1.55555556 -0.00137174 -2.29629630 3.25308642 0.03168108 5.21684957 1.55495818 -0.00000012 -2.29589698 3.24701358 0.00017529 5.15915579 1.55495813 0.00000000 -2.29589694 3.24697960 0.00000001 5.15883361 3.24697960 0.00000000 5.15883360

Conclusion

-0.00848389

0.39843750

0.01212597

0.39843750

0.01212597

0.40625000

-0.00848389

0.40234375

0.00181472

0.40234375

0.00181472

0.40625000

-0.00848389

0.40429688

-0.00333621

0.40234375

0.00181472

0.40429688

-0.00333621

0.40332031

-0.00076114

0.40234375

0.00181472

0.40332031

-0.00076114

0.40283203

0.00052669

0.40283203

0.00052669

0.40332031

-0.00076114

0.40307617

-0.00011725

0.40283203

0.00052669

0.40307617

-0.00011725

0.40295410

0.00020471

0.40295410

0.00020471

0.40307617

-0.00011725

0.40301514

0.00004373

0.40301514

0.00004373

0.40307617

-0.00011725

0.40304565

-0.00003676

0.40301514

0.00004373

0.40304565

-0.00003676

0.40303040

0.00000348

0.40303040

0.00000348

0.40304565

-0.00003676

0.40303802

-0.00001664

0.40303040

0.00000348

0.40303802

-0.00001664

0.40303421

-0.00000658

0.40303040

0.00000348

0.40303421

-0.00000658

0.40303230

-0.00000155

0.40303040

0.00000348

0.40303230

-0.00000155

0.40303135

0.00000097

0.40303135

0.00000097

0.40303230

-0.00000155

0.40303183

-0.00000029

0.40303135

0.00000097

0.40303183

-0.00000029

0.40303159

0.00000034

0.40303159

0.00000034

0.40303183

-0.00000029

0.40303171

0.00000003

0.40303171

0.00000003

0.40303183

-0.00000029

0.40303177

-0.00000013

0.40303171

0.00000003

0.40303177

-0.00000013

0.40303174

-0.00000005

0.40303171

0.00000003

0.40303174

-0.00000005

0.40303172

-0.00000001

0.40303171

0.00000003

0.40303172

-0.00000001

0.40303171

0.00000001

0.40303171

0.00000001

0.40303172

-0.00000001

0.40303172

0.00000000

0.40303171

0.00000001

0.40303172

0.00000000

0.40303172

0.00000000

0.40303172

0.00000000

0.40303172

0.00000000

0.40303172

0.00000000

Use Autograph and Excel to find the required root, and we can see that it gives us the same value of 0.40303172 (correct to 8dp).

Newton-Raphson method

Equation 3: We are required to find the same root which lies between x=0 and x=1. x f(x) f'(x) 0.00000000 1.00000000 -2.00000000 0.50000000 -0.25000000 -2.50000000 0.40000000 0.00800000 -2.64000000 0.40303030 0.00000373 -2.63752066 0.40303172 0.00000000 -2.63751948

It also finds the same value of the root which is corrected to 8dp successfully.

From the Excel tables of each method, we know that method 1 (change of sign method) takes 28 steps to find the root, while method 2 (Newton-Raphson method) and method 3 (rearrangement) take 4 and 17 steps respectively. In terms of speed of convergence, we can say that the Newton-Raphson method is the most efficient one.

However, if we compare them in terms of ease of use with available hardware and software, the change of sign method is the easiest one to use, since it involves least calculation. In change of sign method, we just need the original equation, however, in Newton-Raphson method, we need to calculate its derivative and in rearrangement method we need to rearrange the equation to get g(x). It can be illustrated more clearly in the following table.

 steps formulas involved Change of sign method 28 f(x) Newton-Raphson method 4 f(x), f’(x) Rearrangement 17 f(x), g(x)

This student written piece of work is one of many that can be found in our AS and A Level Core & Pure Mathematics section.

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