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

Numerical method, find the newton raphson iterative formula for a reciproca...

find the newton raphson iterative formula for a reciprocal of a number N and hence find the value of 1/23

Sequences - calculus, Sequences Let us start off this section along wi...

Sequences Let us start off this section along with a discussion of just what a sequence is. A sequence is nothing much more than a list of numbers written in a particular orde

Arithmetic sequence, find a30 given that the first few terms of an arithmet...

find a30 given that the first few terms of an arithmetic sequence are given by 6,12,18...

Graphical understanding of derivatives, Graphical Understanding of Derivati...

Graphical Understanding of Derivatives: A ladder 26 feet long is leaning against a wall. The ladder begins to move such that the bottom end moves away from the wall at a const

Natural numbers, To begin with we have counting numbers. These ...

To begin with we have counting numbers. These numbers are also known as natural numbers and are denoted by a symbol 'N'. These numbers are obtai

Determine all possible solutions to ivp, Determine all possible solutions t...

Determine all possible solutions to the subsequent IVP. y' = y ? y(0) = 0 Solution : First, see that this differential equation does NOT satisfy the conditions of the th

Naming fractions greater than 1, the 10 miles assigned to the chess club st...

the 10 miles assigned to the chess club start at the 10 mile point and go to the 20 mile point when the chess club members have cleaned 5/8 of their 10 mile section between which m

Geometry, Given: ??????? is supp. to ??????? ???? ????? bisects ??????? ?...

Given: ??????? is supp. to ??????? ???? ????? bisects ??????? ???? ????? bisects ??????? Prove: ??????? is a rt. ?

L''hospital''s rule, L'Hospital's Rule Assume that we have one of the g...

L'Hospital's Rule Assume that we have one of the given cases, where a is any real number, infinity or negative infinity.  In these cases we have, Therefore, L'H

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