# Numerical Methods Essay Sample

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Try it free!The place in which the graph of a line crosses the x axis is known as the root of the equation. It is not always possible to find the solution of an equation by algebraic or analytical methods such as factorising. This applies to equations such as y=3×3-11x+7. To solve equations such as these, numerical methods such as change of sign, x=g(x) and Newton-Raphson can be used to give estimates

give estimates of the roots.

Change of Sign Method

The Change of sign method is a method used to look for when a sequence of numbers in the boundary of a root change from negative the positive or vice versa. This change means that the root of the equation is somewhere between the interval where there is a change of sign.

This is the graph of the equation y=3×3-11x+7

There are 3 roots to the equation y=3×3-11x+7, this is illustrated by the three intersections with the x axis. There appears to be a root between 0 and 1.

By taking increments of 0.1 between 0 and 1 it will be possible to use decimal search to look for a change in sign. This will make it possible to find an approximation for to the root between 0 and 1.

This table shows the results of the numbers in increments of 0.1 between 0 and 1. Upon looking at the results we can see that there is a change of sign between 0.7 and 0.8, this shown by the graph below.

This method is repeated again with intervals of 0.01 between 0.7 and 0.8

The table shows that there is a change if sign between 0.75 and 1.76. This means that the root of the equation lies between 0.75 and 0.76. This is shown by the graph below.

Once again this method is repeated with intervals of 0.001 between 0.75 and 0.76.

The able shows that there is change of sign from 0.752 to 0.753. This means that the root of the equation must lie between 0.752 and 0.753. Our estimate of the root is with a maximum error of ï¿½ 0.0005. This is illustrated by the graph below.

This graph shows the equation of the line y=xï¿½+10xï¿½+4.8x+0.576

The roots appear to be between -1 and 0. By taking increments of 0.1 between -1 and 0 it will be possible to use decimal search to attempt to look for a change in sign.

As you can see from the table there is no change of sign. As the graph of y=xï¿½+10xï¿½+4.8x+0.576 is a repeated root the change of sign method, and thus Change of Sign, cannot be used to find the root. Repeated roots such as y=xï¿½+10xï¿½+4.8x+0.576 cause the change of sign method to fail as there will be no change of sign.

Newton-Raphson Method

The Newton Raphson method is a fixed point iteration method. It uses an estimate of the root as a starting point.

The graph shows the equation of the line y=3x-9x+4

In order to find each root I can use the Newton Raphson method. The first root I shall attempt to find is the root at point C. The initial estimate (starting point) is x=2.

The tangent to the point x=2 crosses the x axis at 1.63. This gives a second estimate of 1.63.

It is possible to iterate again to find a more accurate estimate of the root at point C.

The tangent to the point x=1.63 crosses the x axis at the point x=1.472. This gives a third estimate of 1.474. The tangent to the point 1.474 crosses the x axis at the point 1.442.

The Newton Raphson formula is:

In order to use this formula to investigate the roots of the equation y=3xï¿½-9x+4 in must first differentiate the formula. y=3xï¿½-9x+4, when differentiated, is =9×2-9. This means that my formula for to find the roots would be:

As stated earlier, there appears to be a root between 1 and 2 at the point, C. This means that x=2 would be a suitable starting point.

The table below shows the result to the Newton Raphson formula when using 2 as the starting point.

I shall now use the Change of Sign method to establish the error bounds.

The root at the point B appears to be between 0 and 1. As a result of this x=0 would be a suitable starting point.

The table below shows the result to the Newton Raphson Formula when using 0 as the starting point.

I shall now use the Change of Sign method to establish the error bounds.

The root at the point, C, appears to be between -2 and -1. As a result of this x=-2 would be a suitable starting point.

The table below shows the result to the Newton Raphson Formula when using -2 as the starting point.

I shall now use the Change of Sign method to establish the error bounds.

The Newton Raphson Method does not work for all equations. There are some equations in which the Newton Raphson method fails.

This is the graph of the equation y-log(x+3)-x

At the point x=-3 the gradient is infinity. As the Newton Raphson method relies on the gradient of the points it is not possible to use the Newton Raphson method the find the roots of the equation y=log(x+3)-x.

I shall now attempt to use the Newton Raphson formula to find the roots of the equation. When differentiated, y=log(x+3)-x is

This does not work as an equation so it is impossible to find the root of this equation using the Newton Raphson method.

X=g(x) Method

The x=g(x) method is another fixed point iteration method. This means that an estimate of the root is need as a starting point. The x=g(x) method uses the gradient of the equation to find estimates of the roots. Each iteration from the starting point gives a better estimation of the root. To use the x=g(x) method the equation is first written in the form x=g(x) and the intersection between x=g(x) and y=x gives the point to the root or f(x).

This is the graph of the equation y=x5+2xï¿½-10x+6

In order to find the root between 0 and 1 using the x=g(x) method it is necessary to find x=g(x). For this equation it is.

This is the graph of the equations y=x5+2xï¿½-10x+6, and y=x.

I have zoomed in on the intersection between and y=x. This makes the relationship between the intersection between and y=x and the root of y=x5+2xï¿½-10x+6 more apparent. The x co-ordinates for the intersection between and y=x give a root of the equation y=x5+2xï¿½-10x+6.

This is the first iteration to find the root of the equation. The point x=1 is the starting point and it gives the estimate x=0.9.

After 15 iterations, they eventually converge on the point x=0.675839. This means that the final estimate of the root for the curve, y=x5+2xï¿½-10x+6, is 0.67584.

I shall now attempt to use the Change of Sign method to establish error bounds.

However, the x=g(x) method does not work for all equations. Some equations cause the iterations to diverge instead of converge.

This is the graph of the equation of the curve, y=5×5+6×3+9×2-15x-2

I shall use the x=g(x) method to attempt to find root between 0.5 and 1.

In order to find this root, it is necessary to find x=g(x). For this equation x=g(x) is. To find this root I shall use 1 as a starting point.

When using the x=g(x) method, it is not possible to find the root of the equation y=2×5+6×3=15x+2. This is because the x=g(x) method uses the gradient of the points to find the equation of the root. As the gradient of the curve at the points used to find the root is greater than 1 the iterations diverge as opposed to converge. This means that the x=g(x) method fails.

Comparison of Methods

This is the graph of the equation y=3×3-11x+7

There are 3 roots to the equation y=3×3-11x+7, this is illustrated by the three intersections with the x-axis. There appears to be a root between 0 and 1.

Upon using the change of sign method with decimal search the root was found to be 0.75265 with a maximum error of ï¿½ 0.000005.

I shall use the Newton Raphson method to solve the same equation. In order to use the Newton Raphson Method I must first differentiate the equation. y=3×3-11x+7, when differentiated, is. This means that the formula to solve this equation is:

. I will now use this formula to try to find the root between 0 and 1. As with the Change of Sign method, I will use x=0 as my starting point.

Therefore, the root of the equation, y=3×3-11x+7, is 0.7526 with an error bound of ï¿½0.00005.

I shall now use the x=g(x) method to solve the equation y=3×3-11x+7. As with the Change of Sign and Newton Raphson methods I shall use the point x=0 as my starting point. In order to use the x=g(x) method I must first find x=g(x). For the equation, y=3×3-11x+7, x=g(x) is.

Speed of Convergence

In order to find an estimate of the root using the change of sign method it took 35 calculations, this is far more calculations than the other numerical methods. However, as the formula used in the calculations and the intervals between the points could easily be replicated using the software (Microsoft Excel), this did not take a large amount of time. If this method was to be carried out manually, it would take an extremely long amount time. The x=g(x) method took fewer calculations to find he root but not as few as the Newton Raphson method. To find an estimate of the root to 5 decimal places it took 15 iterations using the x=g(x) method. As with the Change of Sign method, the software made it easy to replicate the formula used. Unlike the Change of Sign method, before using the x-g(x) method it is necessary to calculate g(x) first. This adds additional time to the procedure making the g(x) method slightly slower. If this method was to be carried out manually it would probably take much more time than when using software.

The fastest procedure in terms of speed of convergence was the Newton Raphson Method. To find an estimate of the root to 5 decimal places using the Newton Raphson method it took only 5 iterations. As with the other methods, this method was made easier using software. If this method was to be carried out manually it would probably take more time than the other two methods in terms of using the calculations as the formulas for the Newton Raphson method is the most complex. However overall, as it would take the smallest amount of time to converge, the Newton Raphson method would still be the fastest. To conclude the slowest numerical method appears to be the change of sign method as it takes the most calculations to find a root and it is the most tedious. The second slowest numerical method appears to be the x=g(x) method. His is because as with like the Change of Sign method it takes many calculations, in the case of x=g(x), iterations, to find the roots. The fast numerical method appears to be the Newton Raphson method; this is because it takes the fewest number of iterations to find a root.

The Ease of Use of the Software

To carry out my calculations for the Change of Sign, x=g(x) and Newton Raphson methods, I used the spreadsheet programme Microsoft Excel. In order to use the decimal search method to look for a change of sign it was very simple to input my formulae and intervals between my points into Excel. Another feature of Microsoft Excel that made carrying out these calculations easier was the Auto fill function. This is what made it possible the replicate the formulas with great ease. As a result of these various features, using the Change of Sign method was very simple. Using the Newton Raphson Method was also made simpler using Microsoft Excel. As with change of sign, I used Microsoft Excel to replicate the formulae I used. This was also made simpler using the Auto fill function. As the formulae for the Newton Raphson method was more complex than the formula for the Change of Sign method, initially typing the formula was slightly more tedious. In addition to being more tedious, it was more difficult to correct an error when typing the formula for the Newton Raphson method as it was harder to find the error. This is also because the formula was more complex.

Like the previous methods, I used Microsoft Excel when using the x=g(x) method. This method, like the others, was simple to use with Microsoft Excel. It was simple to type in the formula I used as well as the replicate the same formula. It was not as tedious as when using the Newton Raphson typing in this formula. To conclude, I believe the method that was easiest to use with the software was the x=g(x). This is because, it required the least work and using the method was not as tedious. It was easy to notice any typing errors in my formula as the formula was not greatly complex. Although an element of manual work was required to use the x=g(x) method (re-arranging the equation) it was partially eliminated by the simplicity of typing in the x=g(x) equation and replicating it.

The second easiest method to use with the software was the Decimal search method. Although this required slightly more work than the other two methods, the software made this additional work less time-consuming. As the formulae and intervals between the numbers could easily be replicated the decimal search method was fairly simple to use in conjunction with the software I used. The method that was the least easy to use with the software at hand was the Newton Raphson method. This is mainly because of the complexity of the formula. As a result of this complexity it was more difficult to notice any typing error made when inputting the formulae used. This method was also not as simple because the software does not calculate the whole formula. An element of manual work is still required to differentiate the formula and substitute the different areas of the equation into the Newton Raphson Formula.

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