# 4.9 Newton’s method  (Page 5/7)

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## Key concepts

• Newton’s method approximates roots of $f\left(x\right)=0$ by starting with an initial approximation ${x}_{0},$ then uses tangent lines to the graph of $f$ to create a sequence of approximations ${x}_{1},{x}_{2},{x}_{3}\text{,…}.$
• Typically, Newton’s method is an efficient method for finding a particular root. In certain cases, Newton’s method fails to work because the list of numbers ${x}_{0},{x}_{1},{x}_{2}\text{,…}$ does not approach a finite value or it approaches a value other than the root sought.
• Any process in which a list of numbers ${x}_{0},{x}_{1},{x}_{2}\text{,…}$ is generated by defining an initial number ${x}_{0}$ and defining the subsequent numbers by the equation ${x}_{n}=F\left({x}_{n-1}\right)$ for some function $F$ is an iterative process. Newton’s method is an example of an iterative process, where the function $F\left(x\right)=x-\left[\frac{f\left(x\right)}{{f}^{\prime }\left(x\right)}\right]$ for a given function $f.$

For the following exercises, write Newton’s formula as ${x}_{n+1}=F\left({x}_{n}\right)$ for solving $f\left(x\right)=0.$

$f\left(x\right)={x}^{2}+1$

$f\left(x\right)={x}^{3}+2x+1$

$F\left({x}_{n}\right)={x}_{n}-\frac{{x}_{n}{}^{3}+2{x}_{n}+1}{3{x}_{n}{}^{2}+2}$

$f\left(x\right)=\text{sin}\phantom{\rule{0.1em}{0ex}}x$

$f\left(x\right)={e}^{x}$

$F\left({x}_{n}\right)={x}_{n}-\frac{{e}^{{x}_{n}}}{{e}^{{x}_{n}}}$

$f\left(x\right)={x}^{3}+3x{e}^{x}$

For the following exercises, solve $f\left(x\right)=0$ using the iteration ${x}_{n+1}={x}_{n}-cf\left({x}_{n}\right),$ which differs slightly from Newton’s method. Find a $c$ that works and a $c$ that fails to converge, with the exception of $c=0.$

$f\left(x\right)={x}^{2}-4,$ with ${x}_{0}=0$

$|c|>0.5$ fails, $|c|\le 0.5$ works

$f\left(x\right)={x}^{2}-4x+3,$ with ${x}_{0}=2$

What is the value of $\text{“}c\text{”}$ for Newton’s method?

$c=\frac{1}{{f}^{\prime }\left({x}_{n}\right)}$

For the following exercises, start at

a. ${x}_{0}=0.6$ and

b. ${x}_{0}=2.$

Compute ${x}_{1}$ and ${x}_{2}$ using the specified iterative method.

${x}_{n+1}={x}_{n}{}^{2}-\frac{1}{2}$

${x}_{n+1}=2{x}_{n}\left(1-{x}_{n}\right)$

a. ${x}_{1}=\frac{12}{25},{x}_{2}=\frac{312}{625};$ b. ${x}_{1}=-4,{x}_{2}=-40$

${x}_{n+1}=\sqrt{{x}_{n}}$

${x}_{n+1}=\frac{1}{\sqrt{{x}_{n}}}$

a. ${x}_{1}=1.291,{x}_{2}=0.8801;$ b. ${x}_{1}=0.7071,{x}_{2}=1.189$

${x}_{n+1}=3{x}_{n}\left(1-{x}_{n}\right)$

${x}_{n+1}={x}_{n}{}^{2}+{x}_{n}-2$

a. ${x}_{1}=-\frac{26}{25},{x}_{2}=-\frac{1224}{625};$ b. ${x}_{1}=4,{x}_{2}=18$

${x}_{n+1}=\frac{1}{2}{x}_{n}-1$

${x}_{n+1}=|{x}_{n}|$

a. ${x}_{1}=\frac{6}{10},{x}_{2}=\frac{6}{10};$ b. ${x}_{1}=2,{x}_{2}=2$

For the following exercises, solve to four decimal places using Newton’s method and a computer or calculator. Choose any initial guess ${x}_{0}$ that is not the exact root.

${x}^{2}-10=0$

${x}^{4}-100=0$

$3.1623\phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.2em}{0ex}}-3.1623$

${x}^{2}-x=0$

${x}^{3}-x=0$

$0,-1\phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.2em}{0ex}}1$

$x+5\phantom{\rule{0.1em}{0ex}}\text{cos}\left(x\right)=0$

$x+\text{tan}\left(x\right)=0,$ choose ${x}_{0}\in \left(-\frac{\pi }{2},\frac{\pi }{2}\right)$

$0$

$\frac{1}{1-x}=2$

$1+x+{x}^{2}+{x}^{3}+{x}^{4}=2$

$0.5188\phantom{\rule{0.2em}{0ex}}\text{or}\phantom{\rule{0.2em}{0ex}}-1.2906$

${x}^{3}+{\left(x+1\right)}^{3}={10}^{3}$

$x={\text{sin}}^{2}\left(x\right)$

$0$

For the following exercises, use Newton’s method to find the fixed points of the function where $f\left(x\right)=x;$ round to three decimals.

$\text{sin}\phantom{\rule{0.1em}{0ex}}x$

$\text{tan}\left(x\right)$ on $x=\left(\frac{\pi }{2},\frac{3\pi }{2}\right)$

$4.493$

${e}^{x}-2$

$\text{ln}\left(x\right)+2$

$0.159,3.146$

Newton’s method can be used to find maxima and minima of functions in addition to the roots. In this case apply Newton’s method to the derivative function ${f}^{\prime }\left(x\right)$ to find its roots, instead of the original function. For the following exercises, consider the formulation of the method.

To find candidates for maxima and minima, we need to find the critical points ${f}^{\prime }\left(x\right)=0.$ Show that to solve for the critical points of a function $f\left(x\right),$ Newton’s method is given by ${x}_{n+1}={x}_{n}-\frac{{f}^{\prime }\left({x}_{n}\right)}{f\text{″}\left({x}_{n}\right)}.$

What additional restrictions are necessary on the function $f?$

We need $f$ to be twice continuously differentiable.

For the following exercises, use Newton’s method to find the location of the local minima and/or maxima of the following functions; round to three decimals.

Minimum of $f\left(x\right)={x}^{2}+2x+4$

Minimum of $f\left(x\right)=3{x}^{3}+2{x}^{2}-16$

$x=0$

Minimum of $f\left(x\right)={x}^{2}{e}^{x}$

Maximum of $f\left(x\right)=x+\frac{1}{x}$

$x=-1$

Maximum of $f\left(x\right)={x}^{3}+10{x}^{2}+15x-2$

Maximum of $f\left(x\right)=\frac{\sqrt{x}-\sqrt[3]{x}}{x}$

$x=5.619$

Minimum of $f\left(x\right)={x}^{2}\text{sin}\phantom{\rule{0.1em}{0ex}}x,$ closest non-zero minimum to $x=0$

Minimum of $f\left(x\right)={x}^{4}+{x}^{3}+3{x}^{2}+12x+6$

$x=-1.326$

For the following exercises, use the specified method to solve the equation. If it does not work, explain why it does not work.

Newton’s method, ${x}^{2}+2=0$

Newton’s method, $0={e}^{x}$

There is no solution to the equation.

Newton’s method, $0=1+{x}^{2}$ starting at ${x}_{0}=0$

Solving ${x}_{n+1}=\text{−}{x}_{n}{}^{3}$ starting at ${x}_{0}=-1$

It enters a cycle.

For the following exercises, use the secant method , an alternative iterative method to Newton’s method. The formula is given by

${x}_{n}={x}_{n-1}-f\left({x}_{n-1}\right)\frac{{x}_{n-1}-{x}_{n-2}}{f\left({x}_{n-1}\right)-f\left({x}_{n-2}\right)}.$

Find a root to $0={x}^{2}-x-3$ accurate to three decimal places.

Find a root to $0=\text{sin}\phantom{\rule{0.1em}{0ex}}x+3x$ accurate to four decimal places.

$0$

Find a root to $0={e}^{x}-2$ accurate to four decimal places.

Find a root to $\text{ln}\left(x+2\right)=\frac{1}{2}$ accurate to four decimal places.

$-0.3513$

Why would you use the secant method over Newton’s method? What are the necessary restrictions on $f?$

For the following exercises, use both Newton’s method and the secant method to calculate a root for the following equations. Use a calculator or computer to calculate how many iterations of each are needed to reach within three decimal places of the exact answer. For the secant method, use the first guess from Newton’s method.

$f\left(x\right)={x}^{2}+2x+1,{x}_{0}=1$

Newton: $11$ iterations, secant: $16$ iterations

$f\left(x\right)={x}^{2},{x}_{0}=1$

$f\left(x\right)=\text{sin}\phantom{\rule{0.1em}{0ex}}x,{x}_{0}=1$

Newton: three iterations, secant: six iterations

$f\left(x\right)={e}^{x}-1,{x}_{0}=2$

$f\left(x\right)={x}^{3}+2x+4,{x}_{0}=0$

Newton: five iterations, secant: eight iterations

In the following exercises, consider Kepler’s equation regarding planetary orbits, $M=E-\epsilon \phantom{\rule{0.1em}{0ex}}\text{sin}\left(E\right),$ where $M$ is the mean anomaly, $E$ is eccentric anomaly, and $\epsilon$ measures eccentricity.

Use Newton’s method to solve for the eccentric anomaly $E$ when the mean anomaly $M=\frac{\pi }{3}$ and the eccentricity of the orbit $\epsilon =0.25;$ round to three decimals.

Use Newton’s method to solve for the eccentric anomaly $E$ when the mean anomaly $M=\frac{3\pi }{2}$ and the eccentricity of the orbit $\epsilon =0.8;$ round to three decimals.

$E=4.071$

The following two exercises consider a bank investment. The initial investment is $\text{}10,000.$ After $25$ years, the investment has tripled to $\text{}30,000.$

Use Newton’s method to determine the interest rate if the interest was compounded annually.

Use Newton’s method to determine the interest rate if the interest was compounded continuously.

$4.394\text{%}$

The cost for printing a book can be given by the equation $C\left(x\right)=1000+12x+\left(\frac{1}{2}\right){x}^{2\text{/}3}.$ Use Newton’s method to find the break-even point if the printer sells each book for $\text{}20.$

what is the power rule
how do i deal with infinity in limits?
f(x)=7x-x g(x)=5-x
Awon
5x-5
Verna
what is domain
difference btwn domain co- domain and range
Cabdalla
x
Verna
The set of inputs of a function. x goes in the function, y comes out.
Verna
where u from verna
Arfan
If you differentiate then answer is not x
Raymond
domain is the set of values of independent variable and the range is the corresponding set of values of dependent variable
Champro
what is functions
give different types of functions.
Paul
how would u find slope of tangent line to its inverse function, if the equation is x^5+3x^3-4x-8 at the point(-8,1)
pls solve it i Want to see the answer
Sodiq
ok
Friendz
differentiate each term
Friendz
why do we need to study functions?
to understand how to model one variable as a direct relationship to another variable
Andrew
integrate the root of 1+x²
use the substitution t=1+x. dt=dx √(1+x)dx = √tdt = t^1/2 dt integral is then = t^(1/2 + 1) / (1/2 + 1) + C = (2/3) t^(3/2) + C substitute back t=1+x = (2/3) (1+x)^(3/2) + C
navin
find the nth differential coefficient of cosx.cos2x.cos3x
determine the inverse(one-to-one function) of f(x)=x(cube)+4 and draw the graph if the function and its inverse
f(x) = x^3 + 4, to find inverse switch x and you and isolate y: x = y^3 + 4 x -4 = y^3 (x-4)^1/3 = y = f^-1(x)
Andrew
in the example exercise how does it go from -4 +- squareroot(8)/-4 to -4 +- 2squareroot(2)/-4 what is the process of pulling out the factor like that?
Andrew
√(8) =√(4x2) =√4 x √2 2 √2 hope this helps. from the surds theory a^c x b^c = (ab)^c
Barnabas
564356
Myong
can you determine whether f(x)=x(cube) +4 is a one to one function
Crystal
one to one means that every input has a single output, and not multiple outputs. whenever the highest power of a given polynomial is odd then that function is said to be odd. a big help to help you understand this concept would be to graph the function and see visually what's going on.
Andrew
one to one means that every input has a single output, and not multiple outputs. whenever the highest power of a given polynomial is odd then that function is said to be odd. a big help to help you understand this concept would be to graph the function and see visually what's going on.
Andrew
can you show the steps from going from 3/(x-2)= y to x= 3/y +2 I'm confused as to how y ends up as the divisor
step 1: take reciprocal of both sides (x-2)/3 = 1/y step 2: multiply both sides by 3 x-2 = 3/y step 3: add 2 to both sides x = 3/y + 2 ps nice farcry 3 background!
Andrew
first you cross multiply and get y(x-2)=3 then apply distribution and the left side of the equation such as yx-2y=3 then you add 2y in both sides of the equation and get yx=3+2y and last divide both sides of the equation by y and you get x=3/y+2
Ioana
Multiply both sides by (x-2) to get 3=y(x-2) Then you can divide both sides by y (it's just a multiplied term now) to get 3/y = (x-2). Since the parentheses aren't doing anything for the right side, you can drop them, and add the 2 to both sides to get 3/y + 2 = x
Melin
thank you ladies and gentlemen I appreciate the help!
Robert
keep practicing and asking questions, practice makes perfect! and be aware that are often different paths to the same answer, so the more you familiarize yourself with these multiple different approaches, the less confused you'll be.
Andrew
please how do I learn integration
they are simply "anti-derivatives". so you should first learn how to take derivatives of any given function before going into taking integrals of any given function.
Andrew
best way to learn is always to look into a few basic examples of different kinds of functions, and then if you have any further questions, be sure to state specifically which step in the solution you are not understanding.
Andrew
example 1) say f'(x) = x, f(x) = ? well there is a rule called the 'power rule' which states that if f'(x) = x^n, then f(x) = x^(n+1)/(n+1) so in this case, f(x) = x^2/2
Andrew
great noticeable direction
Isaac
limit x tend to infinite xcos(π/2x)*sin(π/4x)
can you give me a problem for function. a trigonometric one