Question

Anyone good at calc and newton's methods?

Answer 1

i am

Answer 2

wats the qestion and i will answer it for you

Answer 3

It is posted in a comment above. Here it is again.

Answer 4

ok

Answer 5

hold on i found it

Answer 6

23

Answer 7

can you explain?

Answer 8

for part b of that wouldn't you just plug in 4(1 - (1/20)) + 5 = ?
and compare it to 9

Answer 9

That would be 8.8 so I would just say that it is .2 off?

Answer 10

indeed :)

Answer 11

ok can you check what i have for part a @dan815 and @zepdrix helped me with a.

Answer 12

yes, that looks right ^_^

Answer 13

ok thanks. Jess do you have any clue about the newtons method one?

Answer 14

http://tutorial.math.lamar.edu/Classes/CalcI/NewtonsMethod.aspx I found this I don't know if it helps...

Answer 15

I honestly don't know, sorry
but I remember @Loser66 knowing how to do those
perhaps you can ask them when they come online :)

Answer 16

yeah hopefully... I know @sleepyjess said she would take a look, and she is currently viewing so.... (crosses fingers)

Answer 17

I don't believe sleepyjess would be able to help you as she has not taken calculus yet :P

Answer 18

yeah she just sent me a message and she thought i was asking about newton's laws XD

Answer 19

I wonder... @mathmale could you try this calculus question please?

Answer 20

@ParthKohli @whpalmer4 @mathmate @Astrophysics

Answer 21

@satellite73

Answer 22

I'm helping another student at the moment, but will help you now and later, as much as I can.

Answer 23

ok thank you. When ever you get the chance I will appreciate it :)

Answer 24

Tell me, please, about your previous experience with Newton's Method.

Answer 25

i really don't have any... This is the first time i am learning it and it is why i am looking for help this is one of the practice questions.

Answer 26

What's Newton's Method for?

Answer 27

I see you're working on another problem, which I certainly understand because it'd been 21 minutes since I last responded.

Newton's method is for the purpose of approximating roots. A root is an x value for which (a given function of x = 0). This x is also the x-coordinate of the point where the graph of the function crosses the x-axis.

Answer 28

The formula used in Newton's Method is\[x _{2}=x _{1}-\frac{ f(x _{1}) }{ f '(x _{1}) }\]

Answer 29

I am done helping that student and can work on this one now :). I know that when i asked this question yesterday someone said that we pick a value and we picked 10 and they asked this "so we have f(x) we take a guess call it g1 (g1,f(g1)) is a point on the curve what is the tangent line there?" and i got really confused.

I would like you to list the steps and explain each one and then ill ask questions.

Answer 30

x sub 1 would be your guess or starting value in approximating the root. f(x) represents the function, and f '(x) its derivative.

Answer 31

could you write this from memory? If not, please at least copy this formula down onto paper for reference: "Formula for Newton's Method".

Answer 32

ok i copied it down.

Answer 33

Type out the given function, f(x).
Find and type out the derivative of the given function, f '(x).

Answer 34

f(x)=x^3+x+1
f'(x)=3x^2+1

Answer 35

good. Now, you are to choose a starting value for x from the interval [-1,0]. Do that now, please. \[starting .value =x _{1}= ?\]

Answer 36

1

Answer 37

Is 1 within the given interval? ;(

Answer 38

oh.... so -1 would work?

Answer 39

Yes.

Answer 40

Now follow the formula. Write \[x _{2}=( ? ) - \frac{ f (?) }{ f '(?) }\]

Answer 41

where \[x _{1}=-1\]

Answer 42

ok give me a moment.

Answer 43

Please type in or draw your work, so I can see how you've gotten your results.

Answer 44

like this?

Answer 45

Yes, except that you need to substitute -1 for the 'x' in the formulas.

Answer 46

i got -.75 or -3/4

Answer 47

I 'm not actually doing the problem myself, but that -.75 or -3/4 looks very reasonable. Note that it's close to your starting value, -1. Good!

Now you have to repeat the same procedure, EXCEPT that now \[x _{2}is known . and .is -3/4\]

Answer 48

doing this will produce a new approx. root, which is the goal of this problem.

Answer 49

so like this..... I got -.7325581

Answer 50

so it wants me to approximate that to the 4th decimal place?

Answer 51

Yes, give your answer to 4 decimal places.
Your result is in the ballpark, but in doing the problem myself, I 've obtained a slightly larger answer.

Answer 52

oh.... Can you type out your steps so i can see where i went wrong? (I assume that is what you are doing now)

Answer 53

You are given the function f(x); you can check your approx roots by subst. them into this function. Try subst. your x=-.7326 first. Your function value should be close to zero.
Next, try subst. x=-06860. Your function value should be even closer to 0.

Answer 54

so plug the -06860 into the f(x)/f'(x) equation we have been using?

Answer 55

No. Your purpose here is to approx. a root of the given function (not of its derivative). Therefore, to test the accuracy of -0.6860 as an approx root, evaluate the original function at x. If the result is close to 0, then the chances are excellent that that's a good and correct approx. root.

Answer 56

ok im a bit confused..... You had me plugging the values into the equation you gave above it was like the xsub2 (?)-f(?)/f'(?) so you are asking me to plug in -.6860 into what?

Answer 57

It all goes back to what you are doing here.
Your job is to find an approx root for the given function f(x).

The formula \[x _{1}=x-\frac{ f(x) }{ f'(x) }\]
is for that purpose.;

And y ou have used it correctly.

Now I'm asking you to VERIFY that you do have a good approximation to a zero of that function.

Answer 58

this step is optional. I've asked you to subs. the approx root into f(x) to see whether the result is 0 or near 0. If this is too confusing, drop it.

Answer 59

You've got a good beginning understanding of this process.

Why not turn away from your computer and see whether you can obtain the first 2 or 3 approximations to a root of f(x) between -1 and 0 on your own, as a self-check?

glad to work with you. However, I' need to get off the 'Net really soon.
I'd be happy to continue later if you like and if we can find a time to meet online.

Answer 60

Ok one last question before we leave... the -.73 whatever would be our answer?

Answer 61

the very first time you look for an approx., you got a result of -.75 (or -.73, or whatever). Go thru the approx. process once more, to obtain a "second approximation." that'd be your final answer.

Answer 62

I'd be glad to go over other examples of applications of Newton's Method with you, later on, if you want add'l practice.

Answer 63

bye for now!

Answer 64

Newton's method is a second order method to find the roots of non-linear equations.

The single-variable version shown above will require an initial approximation (x0) to calculate a better approximation (x1).The better approximation (x1) will be used to calculate a still better approximation (x2), and so on.

It is a second order (tangent) method because the number of accurate digits doubles after each iteration provide a close initial approximation is used.

Unfortunately the method does not converge for all initial values. A close initial approximation x0 is usually obtained by graphing or prior information. Convergence is very slow with multiple roots.

Denoting the successive approximations by x0 (initial), x1, x2,..
we use
\(x_{n+1}=x_n-f(x_n)/f'(x_n)\)
repeatedly until successive approximations match, which gives an idea of the accuracy of the answer.
\(\color{red}{Example:}\)
Find roots of \(f(x)=x^\color{red}{5}+x+1\)
f'(x)=5x^4+1
let x0=-1.
x1=-1-f(-1)/f'(-1)=-1-(-1)/(6)=-5/6=-0.833333
x2=-5/6-(-0.235211/3.411265)=-0.764382
x3=-0.755025
x4=-0.754878
x5=-0.754878
So we conclude that the solution is -0.754878 (to 6 dec. figures), and notice the doubling of accuracy between x3 and x4.