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Newton's Method and Hoptial's Rule

  1. Jul 24, 2011 #1
    Hi all I have these problem on my test review and even after guessing one right I have no idea how one would arrive at that answer so your help is greatly appreciated.
    1. The problem statement, all variables and given/known data

    1.find the limit as x approaches 0 in the equation[(3/x^4)-(4/x^2)]

    2.Use Newton's method to find the positive value of which satisfies . Compute enough approximations so that your answer is within .05 of the exact answer.




    3. The attempt at a solution
    For the limit problem I simply tried taking the derivative of the two equations and subtracting them which gave me 0/0. I know the answer to the problem is infinity but I have no idea how one would arrive at that solution?

    On the second problem I really dont have a clue. I dont know how to apply that to the normal newtons method formula to a problem worded like this any help is appreciated.
     
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  3. Jul 24, 2011 #2

    HallsofIvy

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    [tex]\frac{1}{x^4}- \frac{4}{x^2}= \frac{1- 4x^2}{x^4}[/tex].
    Apply L'Hopital's rule to that.

    "Use Newton's method to find the positive value of which satisfies ."
    I have no clue what that means. There seem to be words missing after "positive value of" and "satisfies".
     
  4. Jul 24, 2011 #3
    sorry it should read.
    Use Newton's method to find the positive value of x which satisfies x=1.5cos(x) . Compute enough approximations so that your answer is within .05 of the exact answer. (You may use any starting point you deem appropriate.)
     
  5. Jul 24, 2011 #4
    The only thing is my problem is 3/(x^4) instead of 1/(x^4) I tried to apply hopitals rule to your example and I came out with just zero after taking the derivative of the result and then plugging in zero back into the derivative.
     
  6. Jul 24, 2011 #5

    SammyS

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    @ redbird555,

    Does L'Hopital's rule apply to [itex]\displaystyle \lim_{x\to 0}\,\frac{3- 4x^2}{x^4}\ ?[/itex] Why not?

    You should be able to do this without L'Hopital .
     
  7. Jul 24, 2011 #6
    It would and your right it wouldnt even really need to apply hereI could just plug in zero. However if I do that I get a divided by 0 answer and I know the answer to the problem is a negative infinity so I'm not sure whats up.
     
  8. Jul 24, 2011 #7

    SammyS

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    Looks like +∞ to me.
     
  9. Jul 24, 2011 #8
    ugghhh sorry I had a brain fart for a second it is positive infinity. thanks for all the help guys does anyone know how to complete the 2nd one I'm completely lost?? I relisted the problem with the correct info.
     
  10. Jul 24, 2011 #9

    SammyS

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    Do you know Newton's method ?

    To get a starting value, graph:
    y=x

    and

    y=cos(x)​

    On the same set of axes & see approx where they intersect.
     
  11. Jul 24, 2011 #10
    yes i understand how newtons method works but I think the y=1.5cosx is confusing me.
    I should be able to take any value for x and plug it in so lets say I take 3. Now that per newtons method 3-[1.5cos(3)/(-3sin(3)/2)] this gives me -4.01525 and that is not correct which is why I'm stumped
     
    Last edited: Jul 24, 2011
  12. Jul 24, 2011 #11

    SammyS

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    x = 3 is a poor choice.

    See the graph I suggested. attachment.php?attachmentid=37499&stc=1&d=1311559888.gif

    What are you using for your function?

    What's the derivative?
     

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  13. Jul 24, 2011 #12
    ok lol im either very confused or there was some mis-communication. I'm using 1.5cos(x) as the function. so by that graph 1 would look like a logical choice therefor newtons method reads x1-[f(x)/f'(x)] so I would get 1-[1.5cos(1)/(3sin(1)/2)] correct? If so that leaves me with a final value of 1.64209 which is not correct either. I've done quite a few newtons method problems on this hw and was able to do them fine thats why I keep thinking its some organization mistake in the equation.
     
  14. Jul 24, 2011 #13

    SammyS

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    From one of your earlier posts:
    I thought you were solving x = 1.5 cos(x) .
     
  15. Jul 24, 2011 #14
    Well its asking for an x value that would satisfy that equation and as far as I could gather thats what I was trying to solve for in my calculations? I just need to figure it out before 1130 seeing as thats when my hw is due and this is the last problem I have
     
  16. Jul 24, 2011 #15

    SammyS

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    Newton's method finds a root of a function. You need a function, f(x), such that f(x) = 0 for the value of x that solves your equation. such a function is:

    f(x) = x - 1.5cos(x)

    Find f ' (x) for that function.

    Then [itex]\displaystyle x_1 = x_0-\frac{f(x_0)}{f'(x_0)}\,.[/itex]

    In general: [itex]\displaystyle x_{n+1} = x_n-\frac{f(x_n)}{f'(x_n)}\,.[/itex]
     
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