# 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 $\text{}20.$

can you give me a problem for function. a trigonometric one
state and prove L hospital rule
I want to know about hospital rule
Faysal
If you tell me how can I Know about engineering math 1( sugh as any lecture or tutorial)
Faysal
I don't know either i am also new,first year college ,taking computer engineer,and.trying to advance learning
Amor
if you want some help on l hospital rule ask me
it's spelled hopital
Connor
hi
BERNANDINO
you are correct Connor Angeli, the L'Hospital was the old one but the modern way to say is L 'Hôpital.
Leo
I had no clue this was an online app
Connor
Total online shopping during the Christmas holidays has increased dramatically during the past 5 years. In 2012 (t=0), total online holiday sales were $42.3 billion, whereas in 2013 they were$48.1 billion. Find a linear function S that estimates the total online holiday sales in the year t . Interpret the slope of the graph of S . Use part a. to predict the year when online shopping during Christmas will reach \$60 billion?
what is the derivative of x= Arc sin (x)^1/2
y^2 = arcsin(x)
Pitior
x = sin (y^2)
Pitior
differentiate implicitly
Pitior
then solve for dy/dx
Pitior
thank you it was very helpful
morfling
questions solve y=sin x
Solve it for what?
Tim
you have to apply the function arcsin in both sides and you get arcsin y = acrsin (sin x) the the function arcsin and function sin cancel each other so the ecuation becomes arcsin y = x you can also write x= arcsin y
Ioana
what is the question ? what is the answer?
Suman
there is an equation that should be solve for x
Ioana
ok solve it
Suman
are you saying y is of sin(x) y=sin(x)/sin of both sides to solve for x... therefore y/sin =x
Tyron
or solve for sin(x) via the unit circle
Tyron
what is unit circle
Suman
a circle whose radius is 1.
Darnell
the unit circle is covered in pre cal...and or trigonometry. it is the multipcation table of upper level mathematics.
Tyron
what is function?
A set of points in which every x value (domain) corresponds to exactly one y value (range)
Tim
what is lim (x,y)~(0,0) (x/y)
limited of x,y at 0,0 is nt defined
Alswell
But using L'Hopitals rule is x=1 is defined
Alswell
Could U explain better boss?
emmanuel
value of (x/y) as (x,y) tends to (0,0) also whats the value of (x+y)/(x^2+y^2) as (x,y) tends to (0,0)
NIKI
can we apply l hospitals rule for function of two variables
NIKI
why n does not equal -1
Andrew
I agree with Andrew
Bg
f (x) = a is a function. It's a constant function.
proof the formula integration of udv=uv-integration of vdu.?
Find derivative (2x^3+6xy-4y^2)^2
no x=2 is not a function, as there is nothing that's changing.
are you sure sir? please make it sure and reply please. thanks a lot sir I'm grateful.
The
i mean can we replace the roles of x and y and call x=2 as function
The
if x =y and x = 800 what is y
y=800
800
Bg
how do u factor the numerator?
Nonsense, you factor numbers
Antonio
You can factorize the numerator of an expression. What's the problem there? here's an example. f(x)=((x^2)-(y^2))/2 Then numerator is x squared minus y squared. It's factorized as (x+y)(x-y). so the overall function becomes : ((x+y)(x-y))/2
The
The problem is the question, is not a problem where it is, but what it is
Antonio
I think you should first know the basics man: PS
Vishal
Yes, what factorization is
Antonio
Antonio bro is x=2 a function?
The
Yes, and no.... Its a function if for every x, y=2.... If not is a single value constant
Antonio
you could define it as a constant function if you wanted where a function of "y" defines x f(y) = 2 no real use to doing that though
zach
Why y, if domain its usually defined as x, bro, so you creates confusion
Antonio
Its f(x) =y=2 for every x
Antonio
Yes but he said could you put x = 2 as a function you put y = 2 as a function
zach
F(y) in this case is not a function since for every value of y you have not a single point but many ones, so there is not f(y)
Antonio
x = 2 defined as a function of f(y) = 2 says for every y x will equal 2 this silly creates a vertical line and is equivalent to saying x = 2 just in a function notation as the user above asked. you put f(x) = 2 this means for every x y is 2 this creates a horizontal line and is not equivalent
zach
The said x=2 and that 2 is y
Antonio
that 2 is not y, y is a variable 2 is a constant
zach
So 2 is defined as f(x) =2
Antonio
No y its constant =2
Antonio
what variable does that function define
zach
the function f(x) =2 takes every input of x within it's domain and gives 2 if for instance f:x -> y then for every x, y =2 giving a horizontal line this is NOT equivalent to the expression x = 2
zach
Yes true, y=2 its a constant, so a line parallel to y axix as function of y
Antonio
Sorry x=2
Antonio
And you are right, but os not a function of x, its a function of y
Antonio
As function of x is meaningless, is not a finction
Antonio
yeah you mean what I said in my first post, smh
zach
I mean (0xY) +x = 2 so y can be as you want, the result its 2 every time
Antonio
OK you can call this "function" on a set {2}, but its a single value function, a constant
Antonio
well as long as you got there eventually
zach
2x^3+6xy-4y^2)^2 solve this
femi
moe
volume between cone z=√(x^2+y^2) and plane z=2
Fatima
It's an integral easy
Antonio
V=1/3 h π (R^2+r2+ r*R(
Antonio