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

Mr, i needed help with algebra

i needed help with algebra

Parity to De-Skew, Consider the following proposal to deskew a skewed bitst...

Consider the following proposal to deskew a skewed bitstream from a TRNG. Consider the bitstream to be a sequence of groups ot n bits for some n > 2. Take the first n bits, and o

Pair of linear equations in two variables, a lending library has a fixed ch...

a lending library has a fixed charge for the first three days and an additional charge for each day thereafter. sam paid Rs 27 for a bookkept for 7 days while jaan paid Rs 21 for t

Graphs of sin x and cos x, Q. Graphs of Sin x and Cos x ? Ans. The...

Q. Graphs of Sin x and Cos x ? Ans. The sine and cosine functions are related to the path that an object might take around a circle. Suppose a dolphin was swimming over

What was the dow at the end of the day after the 2% drop, The Dow Jones Ind...

The Dow Jones Industrial Average fell 2% presently. The Dow began the day at 8,800. What was the Dow at the end of the day after the 2% drop? The Dow lost 2%, so it is worth 9

Customary units of length, Eileen needs 9 feet of fabric to make a skirt. I...

Eileen needs 9 feet of fabric to make a skirt. If Eileen has 18 feet of fabric how many skirts can she make?

Demerits and merits -the arithmetic mean or a.m, Demerits and merits of the...

Demerits and merits of the measures of central tendency The arithmetic mean or a.m Merits i.  It employs all the observations given ii. This is a very useful

Triangle inequalities, poa is a straight line in circle,wher o is center of...

poa is a straight line in circle,wher o is center of circle,b is any pointjoined with p.prove that pa>pb

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