Approximating solutions to equations newtons method, Mathematics

Assignment Help:

Approximating solutions to equations : In this section we will look at a method for approximating solutions to equations. We all know that equations have to be solved on occasion and actually we've solved out quite a few equations by ourselves to this point.  In all the instances we've looked at to this instance we were capable to in fact find the solutions, however it's not always probable to do that exactly and/or do the work by hand.

That is where this application comes into play.  Therefore, let's see what this application is all about.

1141_Newton’s Method.png

Let's assume that we desire to approximate the solution to f (x) = 0 and let's also assume that we have somehow found an initial approximation to this solution say, x0. This initial approximation is perhaps not all that good and therefore we'd like to discover a better approximation. It is easy enough to do.  Firstly we will get the tangent line to f ( x )at x0.

y = f ( x0 ) + f ′ ( x0 ) ( x - x0 )

Now, take a look at the graph below.

The blue line (if you're reading this in color anyway...) is the tangent line at x0. We can illustrate that this line will cross the x-axis much closer to the actual solution to the equation than x0 does.  Let's call this point where the tangent at x0 crosses the x-axis x1 and we'll utilizes this point as our new approximation to the solution.

Therefore, how do we determine this point? Well we know it's coordinates, ( x1 ,0) , and we know that it's on the tangent line therefore plug this point into the tangent line & solve out for x1 as follows,

0 = f ( x0 ) + f ′ ( x0 ) ( x1 - x0 )

x - x0 = -  f (x0 ) /f ′ ( x0 )

x1 = x0  - (f ( x0 ) /f ′ ( x0 ))

Therefore, we can determine the new approximation provided the derivative isn't zero at the original approximation.

Now we repeat the whole procedure to determine an even better approximation. We build up the tangent line to f ( x ) at x1 and utilizes its root, that we'll call x2, as a new approximation to the actual solution.  If we do it we will arrive at the given formula.

                  x2= x1 - (f ( x1 ) /f ′ ( x1 ))

This point is also illustrated on the graph above and we can illustrated from this graph that if we continue following this procedure will get a sequence of numbers which are getting very close the real solution. This procedure is called Newton's Method.


Related Discussions:- Approximating solutions to equations newtons method

Laws of logarithms, express each logariths in terms of log3 P and log3 Q. 1...

express each logariths in terms of log3 P and log3 Q. 1. log3 P^2 Q^3

Write triangles named by the lengths of their sides, Write Triangles Named ...

Write Triangles Named by the Lengths of Their Sides? An equilateral triangle is a triangle with three congruent sides. All three sides of this triangle are the same lengt

Project, report on shares and dividend using newspaper

report on shares and dividend using newspaper

Fermats last theorem, Explain Fermats Last Theorem? How to solve problems u...

Explain Fermats Last Theorem? How to solve problems under Fermats Last Theorem?

2 step equations, What is a two step equation that equals 8 ?

What is a two step equation that equals 8 ?

Find integer if sum of two consecutive odd integers is -112, The sum of two...

The sum of two consecutive odd integers is -112. What is the larger integer? Two consecutive odd integers are numbers in order such as 3 and 5 or -31 and -29, that are each 2 n

AREA, How do you find the distributive property any faster?

How do you find the distributive property any faster?

Factoring out the greatest common factor, Factoring out the greatest common...

Factoring out the greatest common factor of following polynomials.                    8x 4 - 4 x 3 + 10 x 2  Solution Primary we will notice that we can factor out a

Linear equation in two variables., draw the graph of following pair of line...

draw the graph of following pair of linear equation:-2y=4x-6

Write Your Message!

Captcha
Free Assignment Quote

Assured A++ Grade

Get guaranteed satisfaction & time on delivery in every assignment order you paid with us! We ensure premium quality solution document along with free turntin report!

All rights reserved! Copyrights ©2019-2020 ExpertsMind IT Educational Pvt Ltd