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

How to join as maths expert, Sir, I am a Maths teacher from kolkata,India....

Sir, I am a Maths teacher from kolkata,India.i want to join your website as Maths'' expert.Please guide me as to how to join your website and earn some money. I will be really grat

Calculus, I need help fast with my calculus work

I need help fast with my calculus work

Absolute mean deviation-measures of central tendency, Illustration 1 I...

Illustration 1 In a described exam the scores for 10 students were given as: Student Mark (x) |x-x¯| A 60

Algebra, Multiple response question.Zack puts a mug of water ni his microwa...

Multiple response question.Zack puts a mug of water ni his microwave oven. He knows that the final temperature of the water will be a function of the number of seconds he heats the

Coprime positive integer, 6 male students and 3 female students sit around ...

6 male students and 3 female students sit around a round table. The probability that no 2 female students sit beside each other can be expressed as a/b, where a and b are coprime p

Compute standard and variance deviation, A firm is manufacturing 45,000 uni...

A firm is manufacturing 45,000 units of nuts. The probability of having a defective nut is 0.15 Compute the given i. The expected no. of defective nuts ii. The standard an

How tall was peter when he turned 15, Peter was 60 inches tall on his thirt...

Peter was 60 inches tall on his thirteenth birthday. By the time he turned 15, his height had increased 15%. How tall was Peter when he turned 15? Find 15% of 60 inches and add

Find the value of x of eagle , A fox and an eagle lived at the top of a cli...

A fox and an eagle lived at the top of a cliff of height 6m, whose base was at a distance of 10m from a point A on the ground. The fox descends the cliff and went straight to the p

Chapter problem temperature around the globe.., predict whether there is a ...

predict whether there is a relationship between the mean January temperatures of a city in North America and the city''s position west of the prime meridian.

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