TheCalcHater
  • TheCalcHater
Anyone good at calc and newton's methods?
Mathematics
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
TheCalcHater
  • TheCalcHater
randomzombie23
  • randomzombie23
i am
randomzombie23
  • randomzombie23
wats the qestion and i will answer it for you

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TheCalcHater
  • TheCalcHater
It is posted in a comment above. Here it is again.
randomzombie23
  • randomzombie23
ok
randomzombie23
  • randomzombie23
hold on i found it
randomzombie23
  • randomzombie23
23
TheCalcHater
  • TheCalcHater
can you explain?
TheCalcHater
  • TheCalcHater
jigglypuff314
  • jigglypuff314
for part b of that wouldn't you just plug in 4(1 - (1/20)) + 5 = ? and compare it to 9
TheCalcHater
  • TheCalcHater
That would be 8.8 so I would just say that it is .2 off?
jigglypuff314
  • jigglypuff314
indeed :)
TheCalcHater
  • TheCalcHater
ok can you check what i have for part a @dan815 and @zepdrix helped me with a.
1 Attachment
jigglypuff314
  • jigglypuff314
yes, that looks right ^_^
TheCalcHater
  • TheCalcHater
ok thanks. Jess do you have any clue about the newtons method one?
TheCalcHater
  • TheCalcHater
http://tutorial.math.lamar.edu/Classes/CalcI/NewtonsMethod.aspx I found this I don't know if it helps...
jigglypuff314
  • jigglypuff314
I honestly don't know, sorry but I remember @Loser66 knowing how to do those perhaps you can ask them when they come online :)
TheCalcHater
  • TheCalcHater
yeah hopefully... I know @sleepyjess said she would take a look, and she is currently viewing so.... (crosses fingers)
jigglypuff314
  • jigglypuff314
I don't believe sleepyjess would be able to help you as she has not taken calculus yet :P
TheCalcHater
  • TheCalcHater
yeah she just sent me a message and she thought i was asking about newton's laws XD
jigglypuff314
  • jigglypuff314
I wonder... @mathmale could you try this calculus question please?
TheCalcHater
  • TheCalcHater
@ParthKohli @whpalmer4 @mathmate @Astrophysics
TheCalcHater
  • TheCalcHater
@satellite73
mathmale
  • mathmale
I'm helping another student at the moment, but will help you now and later, as much as I can.
TheCalcHater
  • TheCalcHater
ok thank you. When ever you get the chance I will appreciate it :)
mathmale
  • mathmale
Tell me, please, about your previous experience with Newton's Method.
TheCalcHater
  • TheCalcHater
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.
mathmale
  • mathmale
What's Newton's Method for?
mathmale
  • mathmale
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.
mathmale
  • mathmale
The formula used in Newton's Method is\[x _{2}=x _{1}-\frac{ f(x _{1}) }{ f '(x _{1}) }\]
TheCalcHater
  • TheCalcHater
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.
mathmale
  • mathmale
x sub 1 would be your guess or starting value in approximating the root. f(x) represents the function, and f '(x) its derivative.
mathmale
  • mathmale
could you write this from memory? If not, please at least copy this formula down onto paper for reference: "Formula for Newton's Method".
TheCalcHater
  • TheCalcHater
ok i copied it down.
mathmale
  • mathmale
Type out the given function, f(x). Find and type out the derivative of the given function, f '(x).
TheCalcHater
  • TheCalcHater
f(x)=x^3+x+1 f'(x)=3x^2+1
mathmale
  • mathmale
good. Now, you are to choose a starting value for x from the interval [-1,0]. Do that now, please. \[starting .value =x _{1}= ?\]
TheCalcHater
  • TheCalcHater
1
mathmale
  • mathmale
Is 1 within the given interval? ;(
TheCalcHater
  • TheCalcHater
oh.... so -1 would work?
mathmale
  • mathmale
Yes.
mathmale
  • mathmale
Now follow the formula. Write \[x _{2}=( ? ) - \frac{ f (?) }{ f '(?) }\]
mathmale
  • mathmale
where \[x _{1}=-1\]
TheCalcHater
  • TheCalcHater
ok give me a moment.
mathmale
  • mathmale
Please type in or draw your work, so I can see how you've gotten your results.
TheCalcHater
  • TheCalcHater
like this?
1 Attachment
mathmale
  • mathmale
Yes, except that you need to substitute -1 for the 'x' in the formulas.
TheCalcHater
  • TheCalcHater
i got -.75 or -3/4
mathmale
  • mathmale
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 \[x _{2}is known . and .is -3/4\]
mathmale
  • mathmale
doing this will produce a new approx. root, which is the goal of this problem.
TheCalcHater
  • TheCalcHater
so like this..... I got -.7325581
1 Attachment
TheCalcHater
  • TheCalcHater
so it wants me to approximate that to the 4th decimal place?
mathmale
  • mathmale
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.
TheCalcHater
  • TheCalcHater
oh.... Can you type out your steps so i can see where i went wrong? (I assume that is what you are doing now)
mathmale
  • mathmale
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.
TheCalcHater
  • TheCalcHater
so plug the -06860 into the f(x)/f'(x) equation we have been using?
mathmale
  • mathmale
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.
TheCalcHater
  • TheCalcHater
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?
mathmale
  • mathmale
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.
mathmale
  • mathmale
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.
mathmale
  • mathmale
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.
TheCalcHater
  • TheCalcHater
Ok one last question before we leave... the -.73 whatever would be our answer?
mathmale
  • mathmale
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.
mathmale
  • mathmale
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.
mathmale
  • mathmale
bye for now!
mathmate
  • mathmate
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.

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