Numerical Methods

Introduction This chapter focuses on using some numerical methods to solve problems We will look at finding the region where a root lies We will learn what iteration is and how it solves equations

Teachings for Exercise 4A

Numerical Methods You need to be able to show approximations for roots graphically and numerically Show that the equation: x3 - 3x2 + 3x – 4 = 0 has a root between 2 and 3  Draw the equation 5 4 3 2 1 -4 -3 -2 -1 1 2 3 4 -1 -2 -3 -4 There is a root between 2 and 3 IMPORTANT POINT:  Either side of a root, the y value changes sign… 4A

Numerical Methods f(x) = ex + 2x - 3 f(0.5) = e0.5 + 2(0.5) - 3 You need to be able to show approximations for roots graphically and numerically Show numerically that the equation: ex + 2x – 3 = 0 has a root between x = 0.5 and x = 0.6… f(x) = ex + 2x - 3 f(0.5) = e0.5 + 2(0.5) - 3 f(0.5) =-0.351… f(0.6) = e0.6 + 2(0.6) - 3 f(0.6) = 0.022 The sign has swapped so the function has passed through the x-axis  Hence, there is a root between x = 0.5 and x = 0.6 4A

Numerical Methods You need to be able to show approximations for roots graphically and numerically In General, if there is an interval where f(x) changes sign, there is a root in that interval. The only exception is if there is a discontinuity in f(x), such as on the graph y = 1/x On this graph, there is a missing value where the sign change would take place y = 1/x 4A

Numerical Methods You need to be able to show approximations for roots graphically and numerically a) On the same set of axes, sketch the graphs of y = lnx and y = 1/x. Hence show why the equation: lnx = 1/x has only one root. b) Show that this root lies in the interval: 1.7 < x < 1.8 y = 1/x y = lnx Only 1 intersection 4A

Numerical Methods lnx = 1/x lnx - 1/x = 0 ln(1.7) - 1/1.7 You need to be able to show approximations for roots graphically and numerically a) On the same set of axes, sketch the graphs of y = lnx and y = 1/x. Hence show why the equation: lnx = 1/x has only one root. b) Show that this root lies in the interval: 1.7 < x < 1.8 lnx = 1/x Subtract 1/x lnx - 1/x = 0 ln(1.7) - 1/1.7 ln(1.8) - 1/1.8 = -0.0576 = 0.0322 The sign has changed from negative to positive, so the root must be in this interval! 4A

Teachings for Exercise 4B

Numerical Methods You are able to work out lots of different types of sum: For example, Multiplying. You could use the grid method Addition or Subtraction. You use the column method You can divide using the bus shelter You can also deal with powers and fractions …But how would you work out square roots? You could use a calculator, but what is the calculator doing? 4B

Numerical Methods Calculating Square Roots Your calculator will use a method called ‘iteration’ This involves putting a ‘guess’ number into a formula, and getting an answer The answer is then put back into the formula, to get another answer This process is repeated many times, and each time, the answer becomes more accurate This formula will calculate a square root xn is the first guess xn + 1 is the answer which will be put in again a is the number being rooted So if we want to calculate √3… 4B

Using iteration to find √3 Numerical Methods Using iteration to find √3 Calculating Square Roots Your calculator will use a method called ‘iteration’ This involves putting a ‘guess’ number into a formula, and getting an answer The answer is then put back into the formula, to get another answer This process is reapeated many times, and each time, the answer becomes more accurate a = 3, we can use 2 as our ‘first guess’ (x0) Type into the calculator (this would once have been done by hand!) a = 3, we now use 1.75 (x1) Type into the calculator (this would once have been done by hand!) 4B

Using iteration to find √3 Numerical Methods Using iteration to find √3 Calculating Square Roots Your calculator will use a method called ‘iteration’ This involves putting a ‘guess’ number into a formula, and getting an answer The answer is then put back into the formula, to get another answer This process is reapeated many times, and each time, the answer becomes more accurate a = 3, use the previous answer again Make use of the Ans button on your calculator. After this you can effectively just keep pressing the = sign! Keep going like this and you get closer to the exact answer to √3! 4B

can be written in the form: Numerical Methods You can use iteration to find approximations for f(x) = 0, to any desired degree of accuracy… a) Show that: can be written in the form: Add 4x, Subtract 1 Divide by x 4B

Numerical Methods You can use iteration to find approximations for f(x) = 0, to any desired degree of accuracy… b) Use the iteration formula: to find, to 2 decimal places, a root of the equation: Use x0 = 3 You do not need to write this out every single time, but a few examples are needed After a few iterations, the first few decimal places stop changing  3.73 to 2dp 4B

But why does this work? What exactly is going on… Numerical Methods You can use iteration to find approximations for f(x) = 0, to any desired degree of accuracy… But why does this work? What exactly is going on… When we find the roots of this equation, we are really finding where 2 equations cross each other… 4B

Numerical Methods You can use iteration to find approximations for f(x) = 0, to any desired degree of accuracy… But why does this work? What exactly is going on… This equation was rearranged to: When solving this, we are looking at the intersections of two different graphs… 4B

Look at the x values that solve each pair of equations… Numerical Methods You can use iteration to find approximations for f(x) = 0, to any desired degree of accuracy… Look at the x values that solve each pair of equations… The point to understand from this is that solving a rearranged equation is the same as solving the original one! 4B

We will now zoom in on part of the graph… Numerical Methods You can use iteration to find approximations for f(x) = 0, to any desired degree of accuracy… Now we know that solving the rearranged equation will still give the correct values, we can see what is happening through iteration… We will now zoom in on part of the graph… 4B

Numerical Methods You can use iteration to find approximations for f(x) = 0, to any desired degree of accuracy…  It is important to realise that different starting numbers can result in convergence to different roots  Different rearrangements for iteration can also yield different results  It is also possible that this will not work and the answers diverge away from any roots. In this case, the correction is to change the starting number ! We assumed x = 3 (Left side) on our first guess  This gave us a value of 3.66 on the right side We then took this for our new x-value (Left side)  Putting into the right side gave us a new value of 3.72 The process causes the x-values to converge at a root… 4B

Numerical Methods 4B

Can be written either as: Numerical Methods You can use iteration to find approximations for f(x) = 0, to any desired degree of accuracy… Show that: Can be written either as: or Add 5x, Add 3 Square root Add 5x Divide by 5 4B

Numerical Methods You can use iteration to find approximations for f(x) = 0, to any desired degree of accuracy… Show that the iteration formulae: Give different roots of the equation: Use x0 = 5 x4 is usually enough unless specified! 4B

Numerical Methods You can use iteration to find approximations for f(x) = 0, to any desired degree of accuracy… Show that the iteration formulae: Give different roots of the equation: Use x0 = 5 Sometimes you will have to go further in order to find answers to a given number of decimal places… 4B

Summary We have looked at various numerical methods that can be used to approximate solutions to equations We have seen how to rearrange formulae for iteration We have seen how iteration works…