# | Video | Duration |
---|---|---|
1 | Part 1: Introduction | 11:57 |
2 | Part 2: Stopping criteria and convergence condition | 28:19 |
3 | Part 3: Method applications | 15:15 |
Part 3: Method applications
Hi, now that we have seen what the Newton’s Method is used for, how it works and we have defined the three most important stop criteria and analyzed how to choose the x0 to guarantee the convergence of the method in the search interval. In this video we will present the different types of applications where the Newton’s Method can be useful as well. Specifically, we will explain how to solve square roots and inverses using simple calculus, and transcendental equations, which by definition imply they cannot be solved algebraically.
But before beginning with new content, we are going to solve the exercises we laid out at the end of the previous video
Make the calculus that prove that [0,75,1] is a search interval that guarantees the convergence of the Newton’s Method for the function f(x)= 6x^3-2x^2-2x-1 and choose the possible value of X0.
The following calculus would be necessary to confirm that [0,75, 1] is a search interval that guarantees the convergence of the Newton’s Method:
Let’s remember the formula of Convergence Condition:
|x*-x0|< 2/max |f’’(x)/f’(x)| We already saw that since the functions were increasing monotonically we had to calculate f’’(1)=32 i f’(0,75)=5,125
Therefore, |x*-x0|<2/6,244=0,32 approximately.
In this case, the convergence is fulfilled, because even in the worst of the cases, in which x* would be one of the extremes, for instance, x*=0,75, if we took x0 on the other extreme, in other words x0=1, its difference would be smaller than 0,32, which is what the formula requires.
Since in X0 we could take any x E [0,75,1] value, for instance X0=1.
It is important to point out that in case the functions were not increasing monotonically or decreasing monotonically like in the examples we have seen until now, you would have to find the maximum and the minimum of the functions deriving and equaling to 0 like we usually do.
Find a search interval that guarantees the convergence of the Newton’s Method for the function f(x)=x^3-2x-5 For Bolzano’s Theorem we have already seen a possible search interval that could be [2,3]. Let’s then see if this interval guarantees the convergence of the method for this function:
Since it is also about an increasing monotonic function, we calculate f’(2) and f’’(3) that will be the values that will maximize the denominator of the formula. This way we have that:
|x*-x0|<2/1,8=1,111 (which is bigger than 1, difference between superior extreme and inferior extreme of the interval we had defined).
Therefore, we can confirm that this interval is correct and that, therefore, taking any value of x0 E [2,3] the Newton’s Method would approximate us the zero correctly.
Number the three stop criteria we have explained and explain them with your own words. The stop criteria we have introduced are the following:
That the absolute error (or, as an alternative, a relative error) between two consecutive iterations is smaller than a certain Tolerance T predefined by the user. (Insert formula).
That, for some of the iterations, the absolute value of the image xk is smaller than a certain Tolerance T predefined by the user, understanding that the closer f(xk) is to 0 the closer to the solution we will be.
That we have gotten to k=kmax, where kmax is a maximum number of iterations prefixed by the user.
We want the Newton’s Method to solve the square root of a number (which is not trivial). How would you set out the general equation? And in case this number was pi, would you know how to choose a suitable x0?
Let’s see how we would apply the Newton’s Method to solve the square roots:
If a E [0, +inf) from which we want to calculate root (a) doing only algebraic operations. Obviously, root(a) is the positive zero of the function f(x) =x^2 – a, therefore we can make:
Xn+1= Xn – f(Xn)/f’(Xn) = Xn – (Xn^2-a)/2Xn = (Xn^2 + a)/2Xn
In this case we can start by any X0 > 0 and, we can give root(a) with the accuracy we want only addition, multiplication and division.
In the specific case of wanting to calculate the root of pi by the Newton’s Method, and knowing that root(pi)>1 because pi>1, we can take X0=2. Then:
X0= 2,000000000000000
X1= 1,785398163397448
X2= 1,772500774675609
X3= 1,772453851526627
X4=1,772453850905516
Another possible application of the Newton’s Method is the calculus of inverses so that:
If a E R \{0} from which we want to calculate 1/a without dividing, we can manipulate the equation so that if x=1/a <-> a=1/x <-> a-1/x = 0. Therefore, calculating 1/a is equivalent to finding the zero of the function f(x)=a-1/x.
If we apply the Newton’s Method we will have:
Xn+1= Xn – f(x)/f’(x) = Xn – (a-1/Xn)/(1/Xn^2) = Xn (2-aXn)
And we will be able to calculate the value of 1/a with the accuracy we want by only multiplying and subtracting.
For example, if we want to calculate 1/pi the same way, since pi>1, we know that 0 < 1/pi < 1 and therefore we can take x0=0,5. Then:
X0=0,500000000000000
X1=0,214601836602552
X2=0,284520927641251
X3=0,314723149582307
X4=0,318269470601345
X5=0,318309881052253
X6=0,318309886183791
Finally, another application that you might find interesting is the solution of transcendental equations.
These kinds of equations are normally native of calculating the point of intersection between the graphs of two functions. For instance, knowing where sin(x) cuts log(x) ( sin(x)=log(X) ) is not an equation that can be algebraically solved, and therefore the solutions must be approximated numerically. In general, it is about isolating the two functions on one side of the equation to be able to equal to zero and thus be able to apply the Newton’s Method.
In this specific case, it would be:
Sin(x)=log(x) <-> sin(x)-log(x)=0
And therefore our f(x)= sin(x)-log(x) and the problem will have turned into finding the roots of this function. To choose a suitable X0, we would have to make a little graphic study of the function, and if you do it you will be able to conclude that a possible search interval could be [1,3]. If we take, for instance, X0=2, then:
X0=2,0000000000000000
X1=2,235934063889129
X2=2,219185521531423
X3=2,219107150643726
X4= 2,219107148913746
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