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# Mathematical equations can be solved in many ways; however some equations cannot be solved algebraically. I am going to show the three methods of solving these types of equations numerically

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Introduction

| Page                Milan Vashi

Maths Core 3: Coursework

## Introduction

Mathematical equations can be solved in many ways; however some equations cannot be solved algebraically. I am going to show the three methods of solving these types of equations numerically, also I will show each method working, and each method failing.

The three methods are:

• Change Of Sign Method
• Rearranging  f(x)= 0 Into The Form x=g(x)
• Newton-Raphson Method

## Change of Sign Method

### Change of Sign Method Working

I am going to solve the equation f(x) = 0, where f(x) =  . The graph of y=f(x) is shown here.     In order to calculate the value of the root in [0.3, 0.4] to 3 decimal places I must check whether the value is closer to 0.309 than 0.310 meaning that the value 0.3095 needs to be calculated. If this value is negative then the root is 0.309 to 3 decimal places however if the value is positive then the root would have to be given as 0.310.       The value of f(x) when x=0.3095 is 0.00614708 meaning that I can deduce that the value of the root in [0, 1] to 3 decimal places is 0.310.

The error bounds are 0.3095 and 0.310.

## Change of Sign Method Failing

Consider the equation f(x) =0, where f(x) =  Middle

The sequence formed is converging; however it converges to the incorrect root meaning that it is considered as a failure.  , therefore  . As we know that the root <-3, g’(x)>1 meaning that it fails as the range which enables the method to succeed -1<g’(x) <1 is not fulfilled by this equation.

### Rearranging f(x) =0 into x=g(x) working

I am going to solve the equation f(x) =0 where f(x) =  . The graph of y=f(x) is shown below. •   •  •  •  By starting with the value of -3 we are now able to start finding the root in [-3,-2]. The iterations are shown below:

•  =-3.301927  The initial estimate for the root was -3. To find U2, this value must be substituted into the equation of the curve, g(x). This produces the value -3.301927 which can be seen on the y-axis. To find U3, this new value must be substituted into the equation of the curve. So we go across to the line y=x, which reflects this value onto the x-axis, and then down to the curve. This produces the value -3.382999. I then repeated the process, going across to the line and down to the curve each time.

Conclusion

·         Gives an estimate of the point faster than the other two methods.

·         It allows us to find the intervals of the values very easily.

·         It has the greatest possibility of failing.

·         Very hard to make the method fail.

·         Slow to generate the final value.

·         Algebra is required to complete this method.

·         Calculus is required to complete this method.

Excel

·         Easy to make tables.

·         Easy to make tables.

·         Easy to make tables.

·         Easy to fill down the rows with the same formula.

·         Easy to fill down the rows with the same formula.

·         Easy to fill down the rows with the same formula.

·         The formulas that need to be inputted are quite fiddly and mistakes are easily made when inputting data.

·         Rather fiddly to do as 2 different equations are required for the formula for one cell.

Autograph

·         Allows us to produce graphs of given equations and also allows multiple zooms to be taken of the said graphs.

·         Allows us to produce graphs of given equations and also allows multiple zooms to be taken of the said graphs.

·         Allows us to produce graphs of given equations and also allows multiple zooms to be taken of the said graphs.

·         There is a pre-programmed function which allows us to complete the iteration required with relative ease.

·         There is a pre-programmed function which allows us to complete the iteration required with relative ease.

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