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# I am going to solve equations by using three different numerical methods in this coursework and compare the methods which are Bisection, Newton-Raphson, and Rearranging method.

Extracts from this document...

Introduction

Introduction-Solving equations by numerical methods

Numerical Methodsare used for solving equations which solutions are not possible to be solved by algebraic or analytical methods. Now I am going to solve equations by using three different numerical methods in this coursework and compare the methods which are Bisection, Newton-Raphson, and Rearranging method. The comparison is about their ease of use and speed of convergence. In this coursework, I will use some software which are Microsoft Excel (used for calculations) and Autograph 3.0 (used for drawing graphs) to help me complete the coursework.

Bisection method

Bisection Method is looking for a sign change of a continuous function. Actually, we couldn’t use some normal methods which are algebraic or analytical methods to solve some particular cases. I have chosen an equation y=2x³-3x²-8x+7 which is a non-trivial equation. By using the autograph, we can see the graph (Below) cross the x axis and there are three roots. Now take the root in the interval [-2,-1] and start by taking the mid-point of the interval,-1.5.

f(-1.5)=5.5,so f(-1.5)>0.Since f(-2)<0,the root is in[-2,-1.5].

Now take the mid-point of this second interval,-1.75.

f(-1.75)=1.09375,so f(-1.75)>0.Since f(-2)<0,the root is in[-2,-1.75].

And if we continue this method, than we can find out the root that I have shown on the graph.

I am going to combine the graph and the figure I have worked out in Excel. As the figures show blow:

 A B f(a)<0 f(b)>0 (a+b)/2 y=2x³-3x²-8x+7 -2 -1 -5 10 -1.5 5.5 -2 -1.5 -5 5.5 -1.75 1.09375 -2 -1.75 -5 1.09375 -1.875 -1.73047 -1.875 -1.75 -1.73046875 1.09375 -1.8125 -0.26416 -1.8125 -1.75 -0.264160156 1.09375 -1.78125 0.428162 -1.8125 -1.78125 -0.264160156 0.428162 -1.796875 0.085365 -1.8125 -1.796875 -0.264160156 0.085365 -1.8046875 -0.08855 -1.804688 -1.796875 -0.088553429 0.085365 -1.80078125 -0.00138 -1.800781 -1.796875 -0.001383424 0.085365 -1.798828125 0.042044 -1.800781 -1.798828125 -0.001383424 0.042044 -1.799804688 0.020343 -1.800781 -1.799804688 -0.001383424 0.020343 -1.800292969 0.009483 -1.800781 -1.800292969 -0.001383424 0.009483 -1.800537109 0.004051 -1.800781 -1.800537109 -0.001383424 0.004051 -1.80065918 0.001334 -1.800781 -1.80065918 -0.001383424 0.001334 -1.800720215 -2.5E-05 -1.80072 -1.80065918 -2.47371E-05 0.001334 -1.800689697 0.000655 -1.80072 -1.800689697 -2.47371E-05 0.000655 -1.800704956 0.000315 -1.80072 -1.800704956 -2.47371E-05 0.000315 -1.800712585 0.000145 -1.80072 -1.800712585 -2.47371E-05 0.000145 -1.8007164 6.02E-05 -1.80072 -1.8007164 -2.47371E-05 6.02E-05 -1.800718307 1.77E-05 -1.80072 -1.800718307 -2.47371E-05 1.77E-05 -1.800719261 -3.5E-06 -1.800719 -1.800718307 -3.5084E-06 1.77E-05 -1.800718784 7.11E-06

In this spreadsheet, a and b are the two values of intervals. (a+b)/2 is the mid-point.

By using the Excel, we can easily find the x value in the many terms satisfy my required degree of accuracy which is answer to 4 decimal places.

Root is -1.8007 to 4d.p

Error bounds is -1.8007±0.00005

Root bounds is -1.80075<x<-1.80065

Check     X           Y

-1.80075     -0.000688(negative)

-1.80065      0.001538(positive)

Therefore, my answer is correct and it lies between this interval.

Below shows the formulae for using in the Excel:

 A B f(a)<0 f(b)>0 (a+b)/2 y=2x³-3x²-8x+7 -2 -1 =2*B4^3-3*B4^2-8*B4+7 =2*C4^3-3*C4^2-8*C4+7 =(B4+C4)/2 =2*F4^3-3*F4^2-8*F4+7 =IF(G4>0,B4,F4) =IF(G4>0,F4,C4) =2*B4^3-3*B4^2-8*B4+8 =2*C4^3-3*C4^2-8*C4+8 =(B4+C4)/3 =2*F4^3-3*F4^2-8*F4+8 =IF(G4>0,B4,F5) =IF(G4>0,F4,C5) =2*B4^3-3*B4^2-8*B4+9 =2*C4^3-3*C4^2-8*C4+9 =(B4+C4)/4 =2*F4^3-3*F4^2-8*F4+9 =IF(G4>0,B4,F6) =IF(G4>0,F4,C6) =2*B4^3-3*B4^2-8*B4+10 =2*C4^3-3*C4^2-8*C4+10 =(B4+C4)/5 =2*F4^3-3*F4^2-8*F4+10 =IF(G4>0,B4,F7) =IF(G4>0,F4,C7) =2*B4^3-3*B4^2-8*B4+11 =2*C4^3-3*C4^2-8*C4+11 =(B4+C4)/6 =2*F4^3-3*F4^2-8*F4+11 =IF(G4>0,B4,F8) =IF(G4>0,F4,C8) =2*B4^3-3*B4^2-8*B4+12 =2*C4^3-3*C4^2-8*C4+12 =(B4+C4)/7 =2*F4^3-3*F4^2-8*F4+12 =IF(G4>0,B4,F9) =IF(G4>0,F4,C9) =2*B4^3-3*B4^2-8*B4+13 =2*C4^3-3*C4^2-8*C4+13 =(B4+C4)/8 =2*F4^3-3*F4^2-8*F4+13 =IF(G4>0,B4,F10) =IF(G4>0,F4,C10) =2*B4^3-3*B4^2-8*B4+14 =2*C4^3-3*C4^2-8*C4+14 =(B4+C4)/9 =2*F4^3-3*F4^2-8*F4+14 =IF(G4>0,B4,F11) =IF(G4>0,F4,C11) =2*B4^3-3*B4^2-8*B4+15 =2*C4^3-3*C4^2-8*C4+15 =(B4+C4)/10 =2*F4^3-3*F4^2-8*F4+15 =IF(G4>0,B4,F12) =IF(G4>0,F4,C12) =2*B4^3-3*B4^2-8*B4+16 =2*C4^3-3*C4^2-8*C4+16 =(B4+C4)/11 =2*F4^3-3*F4^2-8*F4+16 =IF(G4>0,B4,F13) =IF(G4>0,F4,C13) =2*B4^3-3*B4^2-8*B4+17 =2*C4^3-3*C4^2-8*C4+17 =(B4+C4)/12 =2*F4^3-3*F4^2-8*F4+17 =IF(G4>0,B4,F14) =IF(G4>0,F4,C14) =2*B4^3-3*B4^2-8*B4+18 =2*C4^3-3*C4^2-8*C4+18 =(B4+C4)/13 =2*F4^3-3*F4^2-8*F4+18 =IF(G4>0,B4,F15) =IF(G4>0,F4,C15) =2*B4^3-3*B4^2-8*B4+19 =2*C4^3-3*C4^2-8*C4+19 =(B4+C4)/14 =2*F4^3-3*F4^2-8*F4+19 =IF(G4>0,B4,F16) =IF(G4>0,F4,C16) =2*B4^3-3*B4^2-8*B4+20 =2*C4^3-3*C4^2-8*C4+20 =(B4+C4)/15 =2*F4^3-3*F4^2-8*F4+20 =IF(G4>0,B4,F17) =IF(G4>0,F4,C17) =2*B4^3-3*B4^2-8*B4+21 =2*C4^3-3*C4^2-8*C4+21 =(B4+C4)/16 =2*F4^3-3*F4^2-8*F4+21 =IF(G4>0,B4,F18) =IF(G4>0,F4,C18) =2*B4^3-3*B4^2-8*B4+22 =2*C4^3-3*C4^2-8*C4+22 =(B4+C4)/17 =2*F4^3-3*F4^2-8*F4+22 =IF(G4>0,B4,F19) =IF(G4>0,F4,C19) =2*B4^3-3*B4^2-8*B4+23 =2*C4^3-3*C4^2-8*C4+23 =(B4+C4)/18 =2*F4^3-3*F4^2-8*F4+23 =IF(G4>0,B4,F20) =IF(G4>0,F4,C20) =2*B4^3-3*B4^2-8*B4+24 =2*C4^3-3*C4^2-8*C4+24 =(B4+C4)/19 =2*F4^3-3*F4^2-8*F4+24 =IF(G4>0,B4,F21) =IF(G4>0,F4,C21) =2*B4^3-3*B4^2-8*B4+25 =2*C4^3-3*C4^2-8*C4+25 =(B4+C4)/20 =2*F4^3-3*F4^2-8*F4+25 =IF(G4>0,B4,F22) =IF(G4>0,F4,C22) =2*B4^3-3*B4^2-8*B4+26 =2*C4^3-3*C4^2-8*C4+26 =(B4+C4)/21 =2*F4^3-3*F4^2-8*F4+26 =IF(G4>0,B4,F23) =IF(G4>0,F4,C23) =2*B4^3-3*B4^2-8*B4+27 =2*C4^3-3*C4^2-8*C4+27 =(B4+C4)/22 =2*F4^3-3*F4^2-8*F4+27

Middle

-3.41573E-05

1.269745

1.269775

-3.41573E-05

4.17701E-05

1.26976

3.80883E-06

1.269745

1.26976

-3.41573E-05

3.80883E-06

1.269753

-1.51736E-05

1.269753

1.26976

-1.51736E-05

3.80883E-06

1.269756

-5.68226E-06

1.269756

1.26976

-5.68226E-06

3.80883E-06

1.269758

-9.36675E-07

1.269758

1.26976

-9.36675E-07

3.80883E-06

1.269759

1.43609E-06

This is a bisection failure. As the graph shows, we can't solve the other roots. We can only find one root, so the failure exists.

Newton-Raphson Method:

First of all, I am trying to estimate the root, and I start to work with tangents to find a more accurate estimation. We have to continue this method until the root is found to more than 5 decimal places. To be safe, it is necessary to evaluate f(x) at both these points and show that one value is positive and the other is negative (sign change).

The equation f(x) =0 is solved using the iteration:

I have chosen an equation y=4x³-x²-12x+2 which is a non-trivial equation and I am going to use Newton-Raphson method to solve it. By using the Autograph, see blow:

We can see the roots are in the intervals [-2,-1], [0, 1] and [1, 2].

I am going to show one root graphically which lies between [-2,-1].  Afterwards, I will combine my equation with

Now I am using the Excel to help me solve it accurately.

 Xr f(Xr) f'(Xr) X(r+1) -5 -463 298 -3.446308725 -3.446308725 -132.249178 137.4171434 -2.483916588 -2.483916588 -35.6643303 67.00593259 -1.95166028 -1.95166028 -8.12437758 37.61105474 -1.735649917 -1.735649917 -1.09912868 27.62106745 -1.695856789 -1.695856789 -0.03431212 25.90287655 -1.694532143 -1.69453214 -3.745E-05 25.8463345 -1.69453069 -1.694530694 -4.48E-11 25.84627268 -1.694530694 -1.694530694 3.55271E-15 25.84627268 -1.694530694 -1.694530694 -7.1054E-15 25.84627268 -1.694530694 -1.694530694 3.55271E-15 25.84627268 -1.694530694

In this spreadsheet, Xr = estimated value Xr.f'(Xr) =the value after differentiated f (Xr)

By using the Excel, we can easily find the x value in the many terms satisfy my required degree of accuracy which is answer to 7 decimal places.

Root is -1.694532 to 6d.p.

Error bounds is -1.69453±0.000005.

Root bounds is -1.694535<x<-1.694525.

Check     X           Y

-1.694535-0.00011128 (negative)

-1.6945250.000147178 (positive)

Therefore, my answer is correct and it is in the bounds.

Below shows the formulae for using in the Excel:

 Xr f(Xr) f'(Xr) X(r+1) -5 =4*B4^3-B4^2-12*B4+2 =12*B4^2-2*B4-12 =B4-C4/D4 =E4 =4*B4^3-B4^2-12*B4+3 =12*B4^2-2*B4-13 =B4-C4/D5 =E5 =4*B4^3-B4^2-12*B4+4 =12*B4^2-2*B4-14 =B4-C4/D6 =E6 =4*B4^3-B4^2-12*B4+5 =12*B4^2-2*B4-15 =B4-C4/D7 =E7 =4*B4^3-B4^2-12*B4+6 =12*B4^2-2*B4-16 =B4-C4/D8 =E8 =4*B4^3-B4^2-12*B4+7 =12*B4^2-2*B4-17 =B4-C4/D9 =E9 =4*B4^3-B4^2-12*B4+8 =12*B4^2-2*B4-18 =B4-C4/D10 =E10 =4*B4^3-B4^2-12*B4+9 =12*B4^2-2*B4-19 =B4-C4/D11 =E11 =4*B4^3-B4^2-12*B4+10 =12*B4^2-2*B4-20 =B4-C4/D12 =E12 =4*B4^3-B4^2-12*B4+11 =12*B4^2-2*B4-21 =B4-C4/D13 =E13 =4*B4^3-B4^2-12*B4+12 =12*B4^2-2*B4-22 =B4-C4/D14

By using the Excel, I can find the other two roots of the equation y=4x³-x²-12x+2.

 Xr f(Xr) f'(Xr) X(r+1) 0 2 -12 0.166666667 0.166666667 -0.00925926 -12 0.165895062 0.165895062 5.9354E-07 -12.001536 0.165895111 0.165895111 2.44249E-15 -12.001536 0.165895111 0.165895111 0 -12.001536 0.165895111 0.165895111 0 -12.001536 0.165895111 0.165895111 0 -12.001536 0.165895111

This root is in interval [0, 1]

Root is 0.1658951 to 7d.p.

Error bounds are 0.1658951±0.00000005.

Root bounds is 0.16589505<x<0.16589515.

Check     X           Y

0.165895057.34295E-07(positive)

0.16589515-4.6586E-07 (negative)

 Xr f(Xr) f'(Xr) X(r+1) 4 194 172 2.872093023 2.872093023 54.05260858 81.24283396 2.206771469 2.206771469 13.63540073 42.02454085 1.882308656 1.882308656 2.545935114 26.75241321 1.787142097 1.787142097 0.192065226 22.75223831 1.7787005 1.7787005 0.001454566 22.40790463 1.778635587 1.778635587 8.5724E-08 22.4052634 1.778635583 1.778635583 0 22.40526329 1.778635583 1.778635583 0 22.40526329 1.778635583 1.778635583 0 22.40526329 1.778635583

This root is in interval [1, 2]

Root is 1.7786356 to 7d.p

Error bounds are 1.7786356±0.00000005.

Root bounds is 1.77863555<x<1.77863565.

Check     X           Y

1.778635651.4965E-06 (positive)

1.77863555-7.4402E-07 (negative)

Therefore, the roots of this equation are -1.694532 to 6d.p, 0.1658951 to 7d.p and 1.7786356 to 7d.p

Newton-Raphson failure:

I am going to show the failure of Newton-Raphson method. I have chosen a non-trivial equation

y=12x³-7x²-4x-0.2. And the graph shows:

 Xr f(Xr) f'(Xr) X(r+1) 0.5 -2.45 -2 -0.725 -0.725 -5.5523125 25.0725 -0.503549706 -0.503549706 -1.49291202 12.17793891 -0.380958188 -0.380958188 -0.35552885 6.5580637 -0.326745715 -0.326745715 -0.05896778 4.417899459 -0.313398246 -0.313398246 -0.00331416 3.923440018 -0.312553538 -0.312553538 -1.3038E-05 3.892579256 -0.312550189 -0.31255019 -2.048E-10 3.89245699 -0.31255019 -0.312550189 0 3.892456989 -0.312550189 -0.312550189 0 3.892456989 -0.312550189 -0.312550189 0 3.892456989 -0.312550189 -0.312550189 0 3.892456989 -0.312550189

Root is -0.3125502 to 7d.p

Error bounds is -0.3125502±0.00000005

Root bounds is -0.31255025<x<-0.31255015

Check     X           Y

-0.31255025-2.3749E-07(negative)

-0.312550151.51756E-07(positive)

As the graph shows, it occurs failure. 0.5 is chosen for Xr and it is near the turning point .By using the Newton Raphson Method, I can't get the point I want, but the further one. Therefore, it is failure.

Rearranging equation method:

This method is rearranging the equation f(x) =0 into form x=g(x).Thus y=x and y=g(x) can cross together, and then we can get a single value which can be estimated for the root.

I have chosen an equation y=2x³-3x²-5x-2 which is a non-trivial equation. By using the Autograph, the graph has been shown blow:

The equation y=2x³-3x²-5x-2 can be rearranged to different forms. But it could exist the failure, therefore I have to try. When 0=2x³-3x²-5x-2 rearranged to 5x=2x³-3x²-2, and we can gain y=x and g(x) = (2x³-3x²-2)/5.

And using the Autograph to draw the y=x and y=g(x). See blow:

I am going to combine the graph and the figure I have worked out in Excel. As the figures show blow:

 x x=(2x³+3x²–2)/5 n g'(x) 0 -0.4 0 0 -0.4 -0.3296 1 -0.288 -0.3296 -0.349140895 2 -0.26516 -0.3491409 -0.343884402 3 -0.27269 -0.3438844 -0.345312735 4 -0.27075 -0.34531273 -0.344925628 5 -0.27129 -0.34492563 -0.345030617 6 -0.27114 -0.34503062 -0.345002148 7 -0.27118 -0.34500215 -0.345009868 8 -0.27117 -0.34500987 -0.345007774 9 -0.27117 -0.34500777 -0.345008342 10 -0.27117 -0.34500834 -0.345008188 11 -0.27117 -0.34500819 -0.34500823 12 -0.27117 -0.34500823 -0.345008218 13 -0.27117 -0.34500822 -0.345008221 14 -0.27117 -0.34500822 -0.345008221 15 -0.27117

Conclusion

-0.34501

4E-06

-0.34500885

-0.345001221

4E-06

-4.4E-05

-0.34501

-2E-05

Root is -0.34501 to 5d.p

Error bounds is -0.34501±0.000005

Root bounds is -0.345015<x<-0.345005

Check     X           Y

-0.345015     4.3088E-05 (positive)

-0.345005   -2.04707E-05 (negative)

Newton-Raphson methods

 Xr f(Xr) f'(Xr) X(r+1) -1 4 -5 -0.2 -0.2 -0.896 -5.96 -0.35034 -0.35034 0.033886 -6.3656 -0.34501 -0.34501 2.57E-05 -6.35587 -0.34501 -0.34501 1.53E-11 -6.35587 -0.34501 -0.34501 0 -6.35587 -0.34501 -0.34501 0 -6.35587 -0.34501

Root is -0.34501 to 5d.p

Error bounds is-0.34501±0.000005

Root bounds is -0.345015<x<-0.345005

Check     X           Y

-0.3450154.3088E-05 (positive)

-0.345005-2E-05 (negative)

Rearranging equation methods

 x x=(2x³+3x²–2)/5 n g'(x) 0 -0.4 0 0 -0.4 -0.3296 1 -0.288 -0.3296 -0.349140895 2 -0.26516 -0.3491409 -0.343884402 3 -0.27269 -0.3438844 -0.345312735 4 -0.27075 -0.34531273 -0.344925628 5 -0.27129 -0.34492563 -0.345030617 6 -0.27114 -0.34503062 -0.345002148 7 -0.27118 -0.34500215 -0.345009868 8 -0.27117 -0.34500987 -0.345007774 9 -0.27117 -0.34500777 -0.345008342 10 -0.27117 -0.34500834 -0.345008188 11 -0.27117 -0.34500819 -0.34500823 12 -0.27117 -0.34500823 -0.345008218 13 -0.27117 -0.34500822 -0.345008221 14 -0.27117 -0.34500822 -0.345008221 15 -0.27117

Root is -0.3450082 to 7d.p.

Error bounds is -0.3450082±0.00000005.

Root bounds is -0.34500825<x<-0.34500815.

Check     X                   Y

-0.34500825-0.345008213

-0.34500815-0.34500824

X                 g'(x)

-0.34500825    -0.271173069

-0.34500815     -0.271173032

After I used each method to solve the same equation, I found that difference in the speed of convergence.

In bisection method, I have to use 18 iterations.

In Newton-Raphson method, I have to use 4 iterations.

In rearranging equation method, I have to use 15 iterations.

Now we can see the Newton-Raphson method is the fastest and the bisection method is the slowest.

All of the methods require using software, which is autograph and algebraic, spreadsheet to find out the root. And see blow there are two boxes:

 Method Algebraic Spreadsheet Ease of use Bisection method No algebraic work 6 formulas Easy Newton-Raphson method Work out the gradient 4 formulas Fair Rearranging equation method Rearrangeto 4 formulas Fair
 Method Advantages Disadvantages Bisection method Safe and accurate Slow Has error bounds Can’t find complex root Newton-Raphson method Require one guess only Require differentiation Quick Require calculator Rearranging equation method Easy to understand Require rearranging equation

In the whole project, using the Excel and Autograph is much more efficient and easier.

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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This result agrees with the Newton Raphson method which gave the root to 5 decimal places as 1.44528 which, when rounded to 3 decimal places, gives 1.445 Using the fixed point iteration by rearrangement method, the function can be rearranged to give xn+1 = V((2xn3+7xn+11)/13)

1. ## Change of Sign Method.

The following table was constructed using a Microsoft Excel spreadsheet with the formulae displayed, again applying the iterative formula: xr+1 = xr - f(xr) The values obtained are as follows: It is evident that the 'new x value' tends towards 0.

2. ## Solving equations by numerical methods - The Interval Bisection method

0.33203125 0.000195401 0.00390625 0.328125 0.33203125 0.330078125 -7.2169E-05 0.001953125 0.330078125 0.33203125 0.331054688 6.23853E-05 0.000976563 0.330078125 0.331054688 0.330566406 -4.69913E-06 0.000488281 The method is applied successfully to find one of the roots of the equation x�-1.8x�+x-0.17=0. But the graph drawn to a suitable scale immediately shows that there is more than one root on the same interval.

1. ## Newton Raphson Method for Solving 6x3+7x2-9x-7=0

The f(-2) = -9. This will be the last root for this equation. The error bounds and the final answer will be stated in this root. The first approximation is -2. The tangent at this point on the curve passes the x ? axis at -1.7429.

2. ## C3 COURSEWORK - comparing methods of solving functions

1/3 X3 X2 X1 Newton Raphson method does not work everytime. Newton Raphson method will fail when I consider operating the process on the function f(x) = x1/3, start with the initial value x=-3. As you can see, from the start point of -3, the values of Xn become further and further away from the root between -3.52695 and -3.52685.

1. ## Numerical integration coursework

Again, if the Simpson?s rule with n strips has a strip width of h, then the Simpson?s rule with 2n strips has a strip width of h/2. Hence: That shows that doubling n will reduce the error by a factor of 16, therefore the error multiplier is 0.0625.

2. ## Solving Equations Using Numerical Methods

This allowed me to work out that the three real roots of the function were between the integers of -1 and 7. These are:-1 < ? > 0 0 < ? > 1 6 < ? > 7 The first root I have chosen to use is Root ?. • Over 160,000 pieces
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